RMSEO NIN1lf romoN
ATextbook of
les
SJ. tts
Dr. :R.K. Bans t
C py
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First Edition : Sept. 1983
Ninth Edition : 2005
Reprint , 2006, 2007, 2008, 2009
Revised Nrnth Edition : 2010
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CONTENTS
Chapter
Pages
Chapter L PrOllerties of Fluids
1-34
1.1. Introduction
1.2. Properties of Fluids
1.2. 1.
Density or Mass Density
1.2.2.
Spedfic Weight or Weight Density
1.2.3.
1.2.4.
Spocific Volume
Spedfic Gravity
Solved Problems 1.1- 1.2
2
3
Kinematic Viscosity
Newton's Law of Viscosity
1.3.3.
1.3.4.
Variation of Viscosity with Temperature
1.3.5.
Types of Fluids
Solved Problems 1.3-1.19
Thermodynamic Properties
1.4. 1.
Dimension of R
1.4.2.
Isothermal Process
1.4.3.
Adiabatic Process
Universal Gas Constant
1.4 .4.
Solved Problems 1.20-1.22
Compressibility and Bulk Modulus
Solved Problems 1.23-1.24
Surface Tension and Capillarity
1.6. 1.
Surface Tension on Liquid Droplet
1.6.2.
Surface Tension on a H"llow Bubble
1.6.3.
Surface Tension on a Liquid Jet
Solved Problems 1.25- 1.27
Capillarity
1.6.4.
Solved Problems 1.2&--1.32
Vapour Pre~sure and Cavitation
Highlights
Exercise
1.3 .2.
1.5.
1.6.
1.7.
1
1
2
2
l.3. Viscosity
Units of Viscosity
1.3. 1.
104.
1
Chapter 2. Pressure and Its Measurement
3
5
5
6
6
6
17
18
18
18
19
19
21
22
23
23
"2424
25
26
29
30
30
3....8
2 . 1. Fluid P ressure at a Point
2.2. Pascal"s Law
2.3. Pressure Vllristion in a Fluid at Rest
Solved Problems 2.1-2.7
35
35
36
37
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2.4. Absolute. Gauge, Atmospheric and Vacuum
Pressures
Solved Problem 2.8
2.5. Measurement of Pressure
2.5.1.
Manomet<lrs
2.5.2.
Me<:hanical Gauges
2.6. Simple Manometers
2.6.1.
Piezomete r
2.6.2.
U-tube Manometer
Solved Problems 2.9-2. 13
Single Column Manometer
2.6.3.
Solved Problem 2.14
2.7. Differential Manometers
2.7. 1.
U-tube Differential Manometer
Solved Problems 2.15- 2.17
2.7.2.
Inverted U-tube Differential Manometer
Solved Problems 2.18-2.21
2 .8. Pressure at a Point in Compressible Fluid
2.8. 1.
Isothermal Process
2.S.2.
Adiabatic Process
Tl!mperat url! a t any Point in
2.S.3.
Compressible Fluid
Temperature Lapse-Rate (L)
2.S.4.
Solvt.od Problems 2.22-2.26
Highlights
Exercise
Chapter 3. H ydrostatic Forces on Surfaces
3.t. Introduction
3.2. Total Pressure and Cl!n tre of Pressure
3.3. Ve rtical Plane Surface Sub-merged in Liquid
Solved Problems 3.1-3.12
3.4. Horizontal Plane Surface Sub· merged in Liquid
Solved Problem 3.13
3.5. Inclined P lane Surface Sub·merged in Liquid
Solved Problems 3.14(a}-3.21
3.6. Curved Surface Sub-merged in Liquid
Solved P roblems 3.22-3.31
3.7. Total Pressure and Centre of Pressure on
Lock Gates
Solved Problems 3.32-3.33
3.8. Pressure Distribution in II Liquid Subjected to
Constant HorizontaVVertical Acceleration
3.S.1.
Liquid Containers Subject to Constant
Horizontal Acceleration
Solved Problems 3.34-3.36
Liquid Containers Subjected to Constant
3.8.2.
Vertical Acceleration
Solved Problems 3.37-3.38
Highlights
Exercise
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41
42
42
42
43
43
43
43
44
48
50
50
50
51
53
53
56
56
56
58
59
60
64
65
69-130
69
69
69
72
85
86
66
66
97
99
107
109
112
112
115
120
122
124
125
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Ch a pte r 4 . BUOYlincy and Floatlltion
131 - 162
4. 1. Introduction
4.2. Buoyancy
4. 3. Centre of Buoyancy
Solved Problems 4.1-4.6
4 .4 . Meta-centre
4.5. Meta-cent ric Height
4.6. Analytical Method for Meta-Centre Height
Solved Problems 4.7-4,11
4.7. Conditions of Equilibrium of a Floating and
Sub-merged Bodies
4.7. 1.
Stability of a Sub-merged Body
4.7.2.
Stability of a Floating Body
Solved Problems 4.12-4.18
4.8. Experimental Method of Determination of
Me ta-centric Height
Solved Problems 4.19-4.20
4.9. Oscillation (Rolling) of a Floating Body
Solved Problems 4.21-4.22
Highlights
Exercise
Chapte r 5 . Kinematic .. of Flow and Ideal Flow
A. KINEMATI CS OF FLOW
5. 1. Introduction
5.2. Methods of Describing Fluid Motion
5.3. Types of Fluid Flow
5.3.1.
Steady and Unsteady Flows
5.3.2.
Uniform and Non·uniform Flows
5.3.3.
Laminar and Turbulent Flows
5.3.4.
Com pressible and Incompressi ble Flows
5.3.5.
Rotational and lrrotational Flows
5.3.6.
One, two and Three-Dimensional Flows
5.4. Rate of Flow or Discharge (Q )
5.5. Continuity Equation
Solved Problems 5.1-5.5
5.6. Continuity Equation in Three-Dimensions
5.6.1.
Continuity Equation in Cylindrical
Polar Co-ordinates
Solved Problems 5.5A
5.7. Velocity and Acceleration
5.7. 1.
Local Acceleration and Convective
Acceleration
Solved Problems 5.6-5.9
5.S. Velocity Potential Function and Stream Function
5.8. 1.
Velocity Potential Function
5.8.2.
Stream Function
5.8.3.
Equipotential Line
5.8.4.
Line of Constant Stream Function
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131
131
131
131
136
136
137
138
143
143
143
144
1"
155
156
158
159
160
163- 258
163
163
163
163
164
1"
164
165
165
165
165
166
170
171
173
174
175
175
181
181
182
183
183
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Fl ow Net
Relation between Stream Function and
Velocity Potential Function
Solved Problems 5.10---5.17
5.9. Types of Motion
Linear Translation
5.9.1.
5.9.2.
Linear Deformation
5.9.3.
Angular Deformation
or Shear Deformation
5.9.4.
Rotation
5.9.5.
Vorticity
Solved Problems 5. 1S----5. 19
5. 10. Vortex Flow
5.10.1. Forced Vortex Flow
5.10.2. Fre<! Vo rtex Flow
5.10.3. Equation of Motion for Vortex Flow
5.10.4. Equation of Forced Vortex Flow
Solved Problems 5.20---5.25
5.10.5. Closed Cylindrical Vessels
Solved Problems 5.26-5.31
5.10.6. Equation of Free Vortex Flow
Solved Problem 5.32
5.8.5.
5.8.6.
184
18.
184
191
191
191
192
192
192
192
193
193
194
195
196
197
202
202
209
210
(B) lOU!. . ' LOW (POTENTIAL FLOW)
5. 1 1. Introduction
5. 12. Important Cases of Potential Flow
5 .13. U n iform F low
5.13.1. Uniform Flow Parallel tox·Axis
5.13.2. Uniform Potential Flow P afaliel toy-Axis
5. 14. Sou rce Flow
5.15. Sink Flow
Solved Problems 5.33- 5.35
5. 16. Free-Vortex Flow
5. 17. Supcr·!mposed Flow
5. 17.1. Source and Sink Pair
Solved Problems 5.36-5.37
5.17.2. Doublet
Solved Problem 5.38
5.17.3. A Plane Source in a Uniform Flow
(Flow Past a Half-Body)
Solved Problems 5.39-5.41
5.17.4. A Source and Sink Pair in a Uniform F low
(Flow Past a Rankin e Oval Body)
Solved Problem 5.42
5.17.5. A Doublet in a Uniform Flow
(Flow Past a Circular Cylinder)
Solved Problems 5.43- 5.44
Highlights
f.xercise
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210
211
211
211
213
21.
216
216
219
221
221
225
228
231
233
237
241
2..
246
250
252
254
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Chapter 6. Dynam ics
or Fluid
259-316
Flow
6.1. Int rodu ction
6.2. Equ a tion s of Motion
6.3. Euler's Eq uation of Motion
6.4. Bernoulli's Equation from Euler's Equation
6.;'\. As s um ptions
Solved Problems 6. 1- 6.6
6.6. Bernoulli's Equation fo r Real Fluid
Solved Problems 6.7- 6.9
6.7. P ractica l Applications of Bernoulli's Equation
6.7.1.
Vc nlurimctcr
Soh,('od Problems 6.10---6.21
Orifice Meter or Orifice Plate
Solved Problem s 6.22 - 6.23
6.7.3.
Pilot-tube
Solved Problems 6,24---6,28
6.S. The Momentum Equation
Solved Problems 6.29- 6.35
6.9. Moment of Momen tum Equation
Solved Problems 6.36-6.37
6 .10. Free Liquid J ets
Solved Problems 6.38-641
Highlights
6.7,2.
E"en;;ise
7.5.
7.6.
7.7.
7.8.
283
285
266
288
289
298
298
301
303
307
309
317-3;'14
Chapter 7. Orifices and Mouthpi eces
7.1.
7.2.
7.3.
704.
259
259
260
261
261
261
265
266
268
268
270
261
Int roduction
Classifications ofO rifiees
Flow Through an Orifice
Hydraulic Co-efficients
7.4. 1.
Co·efficient of Velocity (Col
7.4.2.
Co·efficient of Contraction (Cel
7.4.3.
Co-efficient of Di~charge (C d)
Solved Problems 7.1 - 7 .2
Experimental Dete rmination of Hydraulic
Co·efficients
7.5.1.
Determination of Co-efficient or Discharge ( C d) '
7.5.2.
Determination of Co-efficient or Velocity (C.)
7.5.3.
Determination of Co-efficient of
Contraction (C , )
Solved Problems 7.3-7.10
Flow Through Large Orifices
7.6.1 .
Discharge Through Large
Hectangula r Orifice
Solved Problems 7. 11- 7.13
Discharge Through Fully Sub·merged Orifice
Solved Problems 7.1 4-7. 15
Discharge Through Partially Sub-merged Orifice
Solved Problem 7. ]6
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317
317
317
31'
318
319
319
319
320
320
321
321
32 1
327
328
328
330
331
331
332
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7.9 . Time of Emptying a Tllnk Through an Orifice
at its Bottom
Solved Problems 7.17- 7.18
7. 10 . Tim e of Emptying a Hemispherica l Tank
&JIved Probl<)ffis 7.19---7.21
7.11. Time of Emptying a Circular Horizontal Tank
Solved Problems 7.22- 7.23
7.12. Classification of Mouthpieces
7.13. Flow Through an External Cylindrical Mouthpiece
Solved Problems 7.24-7.25
7.14. Flow Th rough a Convergent-Divergent Mouthpiece
Solved Problems 7.26-7.28
7.15. Flow Through Internsl or Re-entrant on
332
333
335
336
338
339
341
341
342
344
345
347
349
&>rda's l\fouthpiere
Solved Problem 7.29
Highlights
Exercise
350
352
Ch a p ler 8. Notches an d We ir s
355-386
8.1. l ntroduction
8.2. Cl a ssification of Notches and Weirs
8.3 . Discharge Over a Rectangular Notch or Weir
Solved Problems 8.1-8.3
8.4. Discharge Over II T riangula r Notch or Weir
Solved problems 8.4-8.6
8.5. Advantages of Triangular Notch or
Weir over Rectangular Notch or Weir
8.6. Discharge Over a Trapezoidal Notch or Weir
Solved Problem 8.7
8.7. Discharge Over a Stepped Notch
Solved Problem 8.8
8.8. Effed on Discharge Over a Notch or Weir
Due to Error in the Measurement of Head
8.8.1.
For Rectangular Weir Or Notch
8.8.2.
For Triangular Weir or Notch
Solved Problems 8.9-8.11
8.9. (a) Time Required to Empty a Rese rvoir or a
Tank with II Reetan!,"Ular Weir or Notch
(b) Time Required to Empty a Reservoir or a
Tank with a Triangular Weir or Notch
Solved Problems 8.12-8.1 4
8.10. Velocity of Approach
Solved Problems 8.15-8.19
8.1 1. Empirical Formulae for Discharge Over
Rectangular Weir
Solved Problems 8.20---8.22
8.12. Cipolletti Weir or Notch
Solved Problems 8.23-8.24
8.13. Discharge Over a Broad-C rested Weir
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355
355
356
356
358
359
361
361
362
362
363
364
364
364
365
366
367
368
370
370
374
374
376
377
378
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8. 14. Discharge Over a Narrow·Crested Weir
8.15. Discharge Over an Ogee Weir
8. 16 . Discha rge Over Sub-merged or Drowned Wei r
Solved Problems 8.25-827
Highlights
Ex ercise
Chapter 9. Viscous Flow
379
379
379
380
381
383
387-432
9.1. Introduction
9.2. Flow of Viscous Fluid Through Circular Pipe
Solved Problems 9.1- 9.6
9.3. Flow of Viscous Fluid between Two Parallel Plates
Solved Problems 9.7-9.12
9.4 . Kinetic Energy Corn)dion and Mom!)ntum
Correction Factors
Solved Problem 9.13
9.5. Power Absorbed in Viscous Flow
9.5.1.
Viscous Resis tance of Journal Bearings
Solved Problems 9.14-9.18
Viscous Resistance of Foot-step Bearing
9.5.2.
Solved Problems 9.19--9.20
Viscous Resistance of Collar Bearing
9.5.3.
Solved Problems 9.2 1- 9.22
9.6. Loss of Head Due to Friction in Vi scous Flow
Solved Problems 9.23---9.24
9.7. Movement of Piston in Dash·pot
Solved Problem 9.25
9.8. M(lthods of Dt)t..,rmination of Co-t)fficit)nt of Viscosity
9.8 , l.
Capillary Tube Method
Falling Sph..,re Resistance Method
9.8.2.
Rotating Cylinder Method
9.8.3.
Oriflce TyPtJ Vi scometer
9.8.4.
Solved Problems 9.26---9.32
Highlights
Exercise
Chapter 10. Turbulent Flow
387
387
391
397
400
404
404
407
407
408
411
412
412
413
414
415
417
418
419
419
420
421
422
423
427
429
433-464
10.1. Introduction
10.2. Reynold s Experiment
10.3. Frictional Loss in Pipe Flow
10.3.l. Express ion for Loss of Head Due
to Friction in Pipes
10.3.2. Expression for Co-effici..,nt of Friction
in Terms of Shear Stress
10.4. Sh..,ar Stress in Turbulent Flow
10.4 .1. Reynold s Expression for Tu rbulent
Shear Stress
10.4 .2. Prandtl Mixing Length Theory for
Turhulen t Shear Stress
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433
433
434
434
436
437
437
438
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10.5 . Velocity Distribution in Turbulent Flow in Pipes
10.5.1. Hydrodynamically Smooth and Rough
Boundaries
10.5.2. Velocity Diijtribution for Turbulent Flow
in Smooth P ipes
10.5.3. Velocity Dis tribution for Turbulent Flow
in Rough Pipes
Solved Problems 10.1- 10.4
10.5.4. Velocity Distribution for Turbulent Flow
in Te rms of Average Velocity
Solved Problems 10.5-10.6
10.5.5. Velocity Dis(.ributi()n for Turbul ent Flow
in Smooth Pipes by Power Law
10.6 . Res istance of Smooth and Rough Pipes
Solved Problems 10.7- 10.13
Highligh ts
Exerc ise
Chapter 11. Flow Through Pipell
438
440
441
442
442
446
448
450
450
453
461
462
465-558
11.1. Introduction
11 .2. Loss of Energy in Pipes
11 .3. Loss of Energy (or head ) Due to Friction
Solved Problems 11.1- 11.7
11 .4. Minor Energy (H ead) Losses
11.4.1 . Loss of Head Due to Sudden Enlargement
11.4.2. Loss of Head Due to Sudden Contraction
Solved Problems 11.8--11.14
11.4.3. Loss of Head at the Entrance of a Pipe
11.4.4. Loss of Head at the Exit of Pipe
11.4.5. Loss of Head Due to an Obstruction
in a Pipe
11.4.6. Loss of Head Du e to Bend in P ipe
11.4.7. Loss of Head in Various Pipe Fittings
Solved Problems 11. 15-11.2 1
11 .5. Hydraulic Gradient and Total Energy Line
11.5.1. Hydraulic Gradi en t Lin e
11.5.2. Total Energy Line
Solv<..>d Problems 11.22-11.26
11 .6. Flow Through Syphon
Solved Problems 11.27- 11.29
11 .7. Flow Through Pipes in Series or Flow Through
Compound P ipes
Solved Problems 11.30- 11.30A
11 .8. Equivalent Pipe
Solved Problem 11.31
11 .9 . Flow Through Parallel Pipes
Solved Problems 11.32- 11.41
11 . 10. Flow Through Branched Pipes
Solved Problems 11.42-11.44
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465
465
465
467
471
471
473
474
482
482
482
483
483
483
491
491
491
491
498
498
502
503
507
508
508
509
524
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11 . 11 . Power Transmission Through Pipes
11.11. 1. Condition for Maximum
Transmission of Power
11.11.2. Muximum Efficiency of Transmission
of Power
Solved Problems 11.45-11.47
11.12. Flow Through Nozzles
11.12.1. Power Transmitted Through Nozzle
11.12.2. Condition for Maximum Power
Transmitted Through Nozzle
11.12.3. Diameter of Nozzle for Maximum
Transmission of Pow..,r Through Npzzle
Solved Problems 11.48---11.51
1l.13. Water Hammer in Pipes
11.13.1. Gradual Closure of Valve
11.13.2. Sudden Closure or Valve and Pipe is Rigid
11.13.3. Sudden Closure of Valve and P ipe is Elastic
11.13.4. Time Taken by Pressure Wave to Travel
from the Valve to the Tank and from
Tank to the Valve
Solved Problems 11.52-11.55
11.14. Pipe Network
11.14.1. Hardy Cross Method
Solved Problem 11 .56
Highlights
Exercise
Chapte r 12. Dime n s ional and Mo d e l Ana lysis
12.1. Introduction
12.2. Secondary or Derived Quantities
Solved Problem 12.1
12.3. Dimensio nal Homogeneity
12. 4 . Methods of Dimensional Analys is
12.4.1. Rayleigh's Method
Solved Problems 12.2-12.7
12.4.2. Buckingham's 1!-Theorem
12.4.3. Method of Selecting Repeating Variables
12.4.4. Procedure for oolving Problems by
Buckingham's 1!-Theorem
Solved Problems 12.8- 12.14
12.5. Model Analysis
12 .6. Similitude-Types of Similarities
12.7. Types of Fort.'Cs Acting in Moving ~'luid
12.8. Dimensionless Numbers
12.8.1. Reynold's Number (R .)
12.8.2. Froude's Number (F, )
12.8.3. Euler's Number (E ")
12.8.4. Weber's Number (IV, )
12.8.5. JI.!ach's Number ( M )
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530
531
531
531
535
537
537
538
539
541
542
542
543
545
545
547
548
549
552
554
559-610
559
559
560
561
561
561
562
565
566
566
568
578
579
580
581
581
582
582
582
582
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12.9. Model Laws or Similarity Laws
12.9.1. Reynold's Model Law
Solved Problems 12.15-12.18
12.9.2. "roude Model Law
Solved Problems 12.19-12.27
12.9.3. Eu ler's Model Law
12.9.4. Weber Model Law
12.9.5. Mach Model Law
Solved Problem 12.28
12.10. Model Testing of Partially Sub-me rged Bodies
Solved Problems 12.29-12.32
12.11 . Classification of Model s
12.11.1 . Undistorl.ed Model s
12.11.2. Distorted Mod els
12.11.3. Scale Ratios for Distorted Models
Solved Problem 12.33
Highlights
Exercise
Chapter 13. Boundary Layer Flow
13.1. Introduction
13.2. Definitions
13.2.1. La minar Boundary Layer
13.2.2. Turbulent Boundary Layer
13.2.3. Laminar Sub· layer
13.2.4 . Boundary Laye r Thicknes~ (0)
13.2.5. Displacement Thickness (0· )
13.2.6. Mome ntu m T hickness (9)
13.2.7. Energy Thickness (lin)
Solved P roblems 13. 1- 13.2
lS.3. Drag Force on a Flat Plate Due to Boundary Layer
13.3.1. Local Co·efficient of Drag [C v -1
13.3.2. Average Co·efficient of Drag [C v 1
13.3.3. Boundary Cond itions for the
Veloci ty Profiles
Solved Problems 13.3- 13.12
13.4 . Turbulent Boundary Layer on II Flat Plate
Solved P roblem 13.13
lS.5. Analysis of Turbulent Boundary Layer
lS.6. Total Drag on a Flat Plate Due to Laminar and
Turbulent Boundary Layer
Solved Problems 13. 14-13.17
lS.7. Separation of Boundary Layer
13.7.1. Effect of Pressu re Gradient on
Boundary Layer Separation
13.7.2. Location of Sepa ration Point
Solved Problem 13.18
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583
583
58.
587
590
595
596
596
597
598
600
60.
6"
605
605
606
606
607
611-656
611
612
612
613
613
613
613
615
615
616
619
622
622
622
622
638
638
641
641
642
646
648
649
650
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IL
(xix )
13.7.3.
Methods of Preventing the Separation
of Boundary Laye r
Highlights
Exercise
Chapter 14. Forces on Sub-merged Bodies
14.1. Introduction
14.2. Force Exerted by II Flowing Fluid on
II Stationary Body
14.2.1. Drag
14.2.2. Lift
14.3. Expression for Drag and Li ft
14.3.1. Dimensional Analysis of Drag and Lift
Solved Problems 14. 1- 14.15
14. 3.2. Pressure Drag and Friction Drag
14.3.3. Stream-lined Body
14.3.4. Bluff Body
14.4. Drag on II Sphere
Solved Problem 14.16
14.5 . Terminal Velocity of II Body
Solved Problems 14. 17- 14.20
14.6. Drag on II Cylinder
14.7. Development of Lift on II Circular Cylinder
14.7.1. Flow of Ideal Fluid Over Stationary
Cyl inder
14.7.2. Flow Pattern Around the Cylinder
when II Constant Circulation r is
Imparted to the Cylinder
14.7.3.
Expression for Lift Force Acting on
Rotating Cylinder
14.7.4.
Drag Fonce Acting on a Rotating Cylinde r
14.7.5.
ElI"pression fo r Lift Co·efficient for
Rotating Cylinder
14.7.6.
Location of Stagnation Points for a
Rotating Cyli nde r in a Uniform Flow_field
14.7.7. Magnus E ffect
Solved Problems 14.21- 14.23
14.8. Development of Lift on an Airfoil
14.8.1. Steady-state of a Flying Object
Solved Problems 14.24-14.25
Highlights
Exencise
Chapter 15. Compressible Flow
15.1. Introduction
15.2 . Thermodynamic Relation s
15.2.1. Equation of Sta te
15.2.2.
ElI"pansion and Compression of Perfect Gas
I I
651
651
653
657-692
657
657
658
658
658
659
660
670
671
671
671
672
673
673
677
677
678
678
680
682
682
683
683
683
686
687
687
689
690
693- 736
693
693
893
69.
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IL
(u,
15.3. Basic Equations of Compressible Flow
15.3.1. Continuity Equation
15.3.2. Bernoulli's Equation
Solved Problems 15.1- 15.3
15.3.3. Momentum Equations
15.4. Velocity of Sound or Pressure Wave in II Fluid
15.4.1. Expression for Velocity of Sound
Wave in II Fluid
15.4.2. Veloeity of Sound in Terms of
Bulk Modulus
15.4.3. Velocity of Sound for Isothe rmal Process
15.4.4. Vdoci ty of Sound fo r Adiabatic Process
15.5. Mach Number
Solved Problems 15.4-15.7
15.6 . Propagation of Pressure Waves (or Disturbances)
in II Compressible Fluid
15.6.1. Mach Angle
15.6.2. Zone of Action
15.6.3. Zone or Silence
Solved Problems 15.8- 15.10
15.7. Stagnation Properties
15.7.1. Expression for Stagnation Pressure (p.)
15.7.2. Expression for Stagnation Density (p, )
15.7.3. Expression for Stagnati.m Temperature ( T ,)
Solved Problems 15.11- 15.12
15.8. Area Velocity Relation ship for Compressible Flow
15.9. Flow of Compressible Fluid Through Orifices and
Noz~les Fitted to a Large Tank
15.9.1.
695
695
695
697
702
702
702
704
705
705
705
706
708
709
710
710
710
711
711
715
715
716
718
719
Value of II or !!l. for Maximum Value
p,
721
15.9.3.
of Mass Rate of Flow
Value of V 2 for Maximum Rate of Flow
of Fluid
Maximum Rate of Flow of Fl uid Through
No~zle
722
15.9.4.
Variation of Mass Rate of Flow of Compressible
15.9.2.
Fluid with Pressure ratio
(~ )
Velocity at Outlet of Nozzle for Maximum
Rate of Flow is Equal w Sonic Velocity
Solved Problems 15. 13- 15. 15
15. 10. Ma ss Rate of Flow of Compressible Fluid Through
Venturimeter
Solved Problem 15. 16
IS.H . Piwt·Static Tube in a Compressible Flow
Solved P roblem 15.17
Highlights
Exercise
721
723
15.9.5.
I I
723
724
727
728
730
731
731
734
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(xxil
Chapter 16. Flow in Open
ChRnnel~
16.1. Introduction
16.2. Classification Qf flow in Channels
16.2.1. Steady Flow and Unsteady Flow
16.2.2. Unifonn Flow and Non-uniform Flow
\6.2.3. Laminar Flow and Turbulent Flow
\6.2.4. Sub-critical, Critica l and Super-Critical
~'low
16.3. Disc harge Through Open Channel by Chezy's
Formula
Solved Problems 16.1- 16.7
\6.4. Empirical Formulae for the Value ofChczy's
Constan t
Solved Problems 16.8-16. 12
16.5. Most Economi cal Section of Channels
16.5.1. Most EronomiClii Roctangular Channel
Solved Problems 16.13- 16.15
16.5.2. Most Economical Trapezoidal Channel
Solved Problems 16.16- 16.22
16.5.3. B(lst Side Slope for Most Economical
Trapezoida l Se<;tion
Solved Problems 16.23- \6.24
16.5.4. Flow Through Circular Channel
Solved Problems 16.25-16.29
16.5.5. Most Economical Circular Section
Solved Problems 16.30- 16.32
16.6. Non-Uniform Flow through Open Channels
16.7. Specific Energy and Specific Energy Curve
16. 7.1. Cri t ical Depth (he )
16.7.2. Critical Velocity (Ve )
16.7.3. Minimum Specific E nergy in Terms of
Critical Depth
Solved Problems 16.33-16.35
16.7.4. Critical Flow
16.7.5. Streaming ,"'low or Sub-c ri tical Flow or
Tranquil Flow
16.7.6. Super-Critical Flow or Shooting Flow or
Torrentia l Flow
16.7.7. Alternate Depths
16.7.8. Condition for Maximum Discharge for a
Given Value of Specific Energy
Solved Problems 16.36-16.37
16.8. Hydraulic Jump or Standing Wave
16.8.1. Exprf'ssion for Depth of Hydraulic Jump
16.8.2. Expression for Loss of Energy Due to
Hydraulic Jump
16.8.3. Exprf'ssion fo r Depth of Hydraulic Jump
in Te rms or Upstream Froude Number
I I
737-802
737
737
737
737
738
738
739
740
744
745
749
7"
750
752
754
762
763
766
766
771
775
777
777
779
779
780
780
781
782
782
782
782
782
783
784
786
787
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(xxii)
16.8.4.
Length of Hydraulic Jump
Solved Problems 16.38- 16.42
16.9. Gradually Varied Flow (G.V.F.)
16.9. L Equation of Gradually Varied Flow
Solved Problems 16.43- 16.44
16.9.2. Back Water Curve and Affux
16.9.3. Expression for the Length of Back
Waler Curve
Solved Problem 16.45
Highlights
Exerc ise
Chapter 17. Impact of Jets a nd Jet Pro)Ju\s ion
17.1. Introduction
17.2. Force Exerted by the J et on II Stationary
Vertical Plate
17.2.1. Force Exerted by II Jet on Stationary
Inclined Flat Plate
17.2.2. Force Exerted by II Jet on Stationary
Curved Plate
Solved Problems 17.1- 17.6
17.3. Force Exerted by II Jet on II Hinged Plate
Solved Problems 17.7- 17.10 (a)
17.4. Force Exerted by II Jet on Moving Plates
17.4.1. Force on Flat Vertical Plate Moving
in the Direction of J et
17.4.2. Force on the Inclined Plate Moving in
the Direction of the Jet
Solved Problems 17.ll- 17.13
17.4.3. Force on the Curved Plate when the
Plate is Moving in the Direction of Jet
Solved Problems 17. 14-17.17
17.4.4. Force Exerted by a Jet of Water on an
Unsymmetrical Moving Curved Plate when
Jet Strikes Tangentially at one of the Tips
Solved Problems 17.18- 17.23
17.4.5. Force Exerted by a Jet of Water on a
Series ..,fVanes
17.4.6. Force Exerted on a Series of
Rad ial Curved Vanes
Solv(.od Problems 17.24-17.26
17.5 . Jet Propulsion
17.5.1. Jet Propulsion ofa Tank with an Orifice
Solved Problems 17.27- 17.28
17.5.2. Jet Propulsion of Ships
Solved P roblems 17.29-17.33
Highlights
Exercise
I I
787
787
790
790
792
793
794
795
796
799
B03-&;2
803
803
804
805
807
809
810
814
815
815
816
818
819
823
826
833
834
837
840
841
843
843
844
849
850
Ii
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IL
(r:ciii)
C hapter 18 . Hy d r a u li c Mlic hin es--Turbines
18.1.
18 .2.
18.3.
18.4.
18.5.
18.6 .
18.7.
18.8.
18.9 .
l S. 10.
18. 0 .
18. 12.
IS. 13.
I I
Introduction
Turbines
General Layout of a Hydroe lectric Power Plant
Definitions of Heads and Efficiencies of a Turbine
Classification of Hydraulic T urbines
Pelton Wheel (or Turbine)
18.6.1. Velocity Triangles and Work Don", for
Pelton Wheel
18.6.2. Points to be Remembered for Pel ton Wheel
Solved Problems 18. 1- 18.10
16.6.3. O<)sigrl ofPeilon Wh eel
Solved Problems 18.1 1- 18. 13
Radial Flow Reaction Turbines
18.7.1. Ma in Parts of a Radial Flow
Reaction Turbine
IS.7.2. Inward Radial Flow Turbine
IS.7.3. Degree of Reaction
IS.7.4. Definitions
Solved Problems 18. 14- 18.20
18.7.5. Outward Radia l Flow Reaction Turbine
Solved Problems 18.21- \8.22
Francis Tu rbine
18.8.1. Important Relations for Francis Turbines
Solved Problems 18.23- 18.26
Axial Fl()w Readion Tu rbine
18.9.1. Some Important Point for Propeller
(Kaplan Turbine)
Solved Problems IS.27- 1S.33
Draft-Tube
IS.1O.1. Types of Draft Tubes
18. 10.2. Draft-Tube Theory
IS.10.3. Efficiency ()f Draft-Tube
&lIved Problems IS.33 (0 l-- I S.35
Specific Speed
IS. 11. 1, Derivati()n of the Specific Speed
IS.II.2. Significance of Specific Speed
Solved Problems IS.36- 1S.41
Unit Quantities
18.12.1. Uni t Speed
18.12.2. Unit Discharge
IS.12.3. Unit Powe r
18.12.4. Use of Unit Quantities (N . , Q., Po)
Sol\·(,..1 Problems IS.41 (0 )- 18.45
Characteristic Curves of Hydraulic Turb ines
18.13. 1. Main Characwristic Curves or
Constant Head Curves
18.13.2. Ope rating Characteris tic Curves or
Constant Speed Curves
853-944
853
853
853
853
856
857
859
861
862
873
87.
877
877
878
880
884
88'
892
893
895
896
896
903
905
905
915
915
916
916
917
920
920
921
921
927
927
927
928
928
929
933
933
93.
Ii
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IL
(xxiv)
18.13.3. Constant Efficiency Curves or Musche!
Curves or Iso-Efficiency Curves
18.14. Governing ofl'urbines
Highlights
Exercise
Chapler 19. Centrifugal Pumps
19.1. Introduction
19.2. Main Parts ofa Centrifugal P ump
19.3. Work Done by the Centrifugal Pump
(or by Impnler) on Waler
19.4. Definitions of Heads and Efficiencies of 1\
Centrifugal Pump
Solved Problems \9. 1- 19.12
19.5. Minimum Speed for Starting a Centrifugal Pump
Solved Problems 19. 13- 19. 15
19.6. Multistage Centrifugal Pumps
19.6.1. Multistage Centrifugal Pumps
for High Heads
19.6.2. r.Iultistage Centrifugal Pumps for
High Discharge
Solved Problems 19.16---19.17
19.7. Specific Speed of a Centrifugal Pump (N, )
19.7.1. fo:xprossion for Specific Speed for a Pump
19.8. Model Testing of Centrifugal Pumps
Solved Problems 19. 18- 19.22
19.9. Priming of a Centrifugal Pump
19.10. Characteri~tic Curve~ of Centrifugal Pumps
19.10.1. Main Characteristic Curves
19.10.2. Operating Characteristic Curves
19.10.3. Constant Efficiency Curves
19. 11 . Cavitation
19.11 . 1. Precaution Against Cavitati(1n
\9.11.2. Effects of Cavitation
19.11.3. Hydraulic Machines Subjected to Cavitation
19.11.4. Cavitation in Turbines
19.11.5. Cavitation in Centrifugal Pumps
Solv<.od Problem 19.23
19.12. Maximum Suction Lift (o r Suction Height)
19.13. Net Positive Suction Head (NPS H)
19.14. Cavitation in Centrifugal Pump
Solved Problem 19.24
Highlights
Exerc ise
Chapter 20. Recil,rocating Pumps
20.1. Introduction
20.2. Main Parts of a Reciprocating Pump
20.3. Working of a Reciprocating Pump
I I
935
936
937
939
945-992
945
945
947
948
951
965
966
966
966
969
969
971
971
972
973
976
978
978
979
979
960
960
961
981
981
961
982
983
985
965
966
987
989
993-1040
993
993
994
Ii
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IL
20.3.1.
20.3.2.
20.3.3.
20.4.
20.5.
20.6.
20.7.
20 .8.
20.9 .
20.10.
Discharge Through a Reciprocating Pump
Work Done by Redprocating Pump
Discharge, Work Done and Power
Required to Drive a Doubl e-acting Pum p
Slip of Reciprocating Pump
20.4.1. Negative Slip of the Reciprocating Pump
Classification of Reciprocating Pumps
Solved Problems 20.1- 20.2
Variation of Velocity and Acceleration
in the Suction and De livery Pipes Due to
Acceleration of the Piswn
Effect of Variation of Vel ocity on " riction
in the Suction lind Delivery Pipes
Solved Problem 20.3
Indicator Diagram
20.8.1. ldcal l ndica.lor Diagram
20.8.2. EtTect of Acceleration in Suction and
Delivery Pipes on Indicator Diagram
Solved Problems 20.4-20.9
20.8.3. Effect of I'riction in Suction and Delivery
Pipes on Indi cator Diagram
20.8.4. ElTect of Acceleration and Friction in
Suction and Delivery P ipes on Indicator
Diagram
Solved Problems 20. 10-20.12
20.8.5. Maximum Speed ofa Reciproca ting Pump
Solved Problem 20.13
Air Vessels
Solved Problems 20. 14-20. 18
Comparison between Centrifugal Pumps and
Reciprocating Pumps
Highlight.'<
Exercise
Chapter 21. F l uid System
21.1. Introduction
21.2. The Hydraulic Press
21.2.1.
Mechanical Advantage
21.2.2.
Leverage of the Hydraulic Press
21.2.3. Actual Heavy Hydraulic Press
Solv(.od Problems 21.1 - 21.5
21.3 . The Hydraulic Accumulator
21.3.1. Capacity of Hydraulic Accumulator
Solved Problems 21.6-2 1. 11
21.3.2. Differential Hydraulic Accumulator
21.4. The Hydraulic intensifier
Solved Problems 21. 12-21. 13
2 1.5. The Hydrauli c Ram
Solved Problems 2 1.14-21.15
I I
99<
995
995
996
997
997
997
998
1001
1001
1003
1003
1004
1004
1012
1013
lOIS
1019
1020
1021
1030
1037
1037
1038
104 1- 1070
1041
1041
1042
1042
1042
1043
1045
1046
1047
1051
1051
1053
1053
1055
Ii
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(xxui)
21 .6. The Hydraulic Lift
21.6.1. Direct Acting Hydraulic Lift
21.6.2. Suspended Hydraulic Lift
Solved Problems 21.16- 21.17
21.7. The Hydraulic Crane
Solved Proble ms 21.18-21.20
21.8. The Fluid or Hydraulic Coupling
21.9. The Hyd raulic Torque Converter
21.10. The Air Lift Pump
21.11. The Gear-Wheel Pump
Highlights
Exercise
Objective Type Questions
Appendix
Subject Indel<
I I
1056
1057
1057
1058
1061)
1061)
1063
10'"
1065
IOEm
1067
1068
1071- 1094
1095-1000
1097_1102
Ii
to 1. 1
INTRODUCTION
Fluid mechanics is that branch of sc ience which deals with the behav iour o f the fluids (liquids or
gases) at rest as well as in motion. Thus this branch of scienl'C deals WiTh the stat ic. kinematics and
dy namic aspecTs of fluids. The study of fluids at rest is called fluid statics. The study of fluids in
Illotion. where pressure forces are not considered. is c all ed tluid kinematics and if the pressure forees
are also considered for the flu ids in motion. that branch of science is callt>d fluid dynamics.
to 1.2
PROPERTIES OF FLUIDS
1.2. 1 Density or Mass Density. DensiTy or mass densi ty of a fluid is defincd as the ratio of the
mass of a fluid TO its volume. Th us mass per unit vo lum e of a Iluid is called density. It is denoted by the
symbol p (rho). The unit of mass density in SI urlit is kg per cubic metre. i.p .• kg/m 3. The density of
liquids ma y be cOrlsidercd as COrlSlant while that of gases changes with th e variation o f pressure and
temperature.
Mathematically. Illass de nsity is wr ill en as
p=
The value of density of water is I gm/crn l or I()()(} kg/m 3.
1.2.2 Specific Weight or Weight Dens ity. Specific weig ht or weight density of a fluid is the
ratio between the weight of a fluid to its volume. Thus weight per unit volume of a fluid is called
weight density and it is denoted by the symbo l w.
Thus mathematically.
IV = Weight o f fluid '" ,(,Mc"",co,fcnc'o;cd~);X'i-AC'C"'C""'O"C;OC"7'd"c,c,cQ"g~,c'c'c",-Y
Volumc of fluid
Vo lumc of fluid
Mass of fluid
)
=p
Volu me o f flu id
=p><g
..•
...( 1.1 )
IV= pg
1
I I
Ii
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IL
12 Fluid MC(;hanics
The va lue of specific weight or weig h! density (w) for water is 9.81 x WOO Newlon/II]"' in Sl units.
1.2.3 Specific Volume. Specific volu ille of a fluid is defi ned as the volume of a fluid occupied
by a unit mass or volum e per unit mass of a fluid is callcd specific vo lu11Ie. Mathemat ically. il i~
expressed as
Specific volume
=
Vo lume of fl uid
~M,!","",,,o"r~n~""d;r; '" p
M ass of fl uid
Volume of flu id
Thus specific vol um e is the rec iprocal of mass density. It is c.'pressed as m' fk g. It is commonly
applied 10 gases.
1.2.4 Specific C;uvity. Specific gravity is defined as the rali o of th e weight densit y (or de nsity)
of a fluid to the weight de nsit y (o r den sity) uf a standard fluid. For liquids. the standard fluid is Taken
water and for gases, th e standard fluid is taken air. Specific gravity is nlso called relntivc dens it y. h is
dimension less quantit y and is de noted by tlie symbol S.
Mathem aticall y, S(for liquids) = -;,w",'cig~hc'cd;'e"c'7h~YC(CdC'""e·'ChlYC)Oo"r"tioq~""id"
We igh t de nsi ty (de ns ity) of water
:;;:'"igehc'CdC'C·"C'Ch"YC(~dO'C"C'Ch"YC)OOcf~gc~
o
S(for gases) = -;_W
We ight de nsity (de nsi ty) of air
Thu s wrig ht density of n liqui d = S x
=Sx
The densi ty of a liquid
=Sx
=Sx
Wei ght dt' nsit y of water
lOOOx9.81 N/m J
Density of water
1000 kg/ml .
.. .(\.]A )
If tlie specific gra vity of a fluid is known. then the density of the fluid will be equal to specific
grav ity of fluid multiplied by tli e density of water. For exa mple. tlie specific gravity o f mercury is 13.6.
hence density of mercury = 13.6 x 1000 = 1]600 kg/ml.
Problem 1.1
Co/m/ill" tl/<' sfH'ciflc weigili. delisiTy om/ spnific gral'ily of 0111' lifr" of (/ liquid
... IIid, II'pigll,' 7 N.
Solution. Given:
I
Volume = I litre = - - m
llJOO
Weight = 7 N
(i) Spedfk weight (w)
c-_
(1~) Illl
w7000
I
;
or I li tre
= 7000 N/llr'. AilS.
1
,
= - = - - kg /m = 71.\.5 k g/Ill ' . Ans .
g
(iii) Specific grav ity
.
IIlTe = 1000 m
7CN
= Weig ht = T.c.
Volum e
(ii) Density (p)
,
9.81
'" Densi ty of liqu id '" 7 135
Density of water
1000
I .:
De ns ity of water = 1000 kglm11
'" 0 .7 1.\5. AilS.
I I
Ii
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IL
Properties of Fluids
Problem 1.2
31
Ca/culale tile dellsity. specific weigh! and weight of one lilre of Pe/rot of sped/it
gra..!ty '" 0.7
Solution. Gi ven:
D"'I~'ily
I~
10
rn 3 =0.00 1 rn 3
5 = 0.7
Sp. gravil y
(il
x lOOOem 3 =
Volume: [Iitre =
(p)
Using cq uJlion (I. IA),
= S x 1000 kg/ml '" 0.7 x 1000 == 700 kg/hi ), Am.
Dellsif), (p)
(ii) Specific ...eight (w)
Using equation (1.1),
w
= p x g = 700 x 9.81 Nlrnl = 6867 Ntm J , Ans.
(iii) IVeigl1l ( IV)
We know [hal sp"cific weig h!
,,,,,;,,:hC
'
= .-w
Volume
IV
\I'
... = - - or6867 = - 0.00 1
O.OO[
IV ", 6867 x 0.00 1 = 6.867 N. An s.
..
1,3
VISCOSITY
Vis<:osi ly is d efi ned as the property of a fluid whi<:h offers rcsiswnce tu the movement of o ne layer
of fluid over imolhcr adj ace nt layer of ihe fluid. When twO laye rs of a fluid. a dislanl'C 'dy' aparL move
one over the other at different velocities, say u a nd u + du as show n in Fig. 1.1. Ihe viSC(}~ity togclher
wi th relative velocity cause~ a shear stre~s acting between Ihe nuid laye rs.
Th e lOp layer eauses a shear Slress on the
adjace nt lower layer while the lowe r layer causes
a shear stress on th~ adjacent top layer. This shear
stress is propo rtional to the rate of c han ge of velocit y with respec t to y. It is denoted by symbol
VELOC ITY PROFILE
t (Tau).
"
Mathe maticall y.
,~
d,
dy
-
t~)J
Fig. 1.1 Velocity variation near a mlid boundary.
"
... (1.2)
-
dy
where )J (calk'd mu) is the (.unstant of proportionality and is known as the (.u-cfficient of dynamic viS«)sity
d,
oronl y viM-usity. - represents the rate of she<lr sirain or f:ltc of sheardcfomtatioo or velocity gradient.
dy
From equa tion ( 1.2), we hn ve
].I
1
= -,(
d,
.. .( 1.3 )
If)"
Th us viscosity is a lso defin ed as th e she ar stress required to produce unit rate of shear strain.
1.3.1 Unit$ of Vi$co$ity. Th e uni ts of viscosity is obtained by putt ing the dimcnsions o f the
quantities in cq uation (1.3)
I I
Ii
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IL
14
Fluid MC(;hanics
r orce! Area
Shear st ress
". ~~~~~.~~~~Change o f velocity
1
(Lengeh )
Change o f distance
Time
x Leng th
'" r"'Ofcc/(l.c ngth)' = Force x T ime
1
TI me
In MKS system. force is represented by k.gf and length by InClre (m). in eGS system. force is
represented by dyne and length by em and in SI system force is represented by NewlOn (N) and length
by metre (m).
M KS unit of viscosity
= kgf -scc
m1
= dyne-sec
eGS unit o f viscosity
cm
l
In the aho"c expression Nlm l is also known as Pascal which is represe nted by Pa. Hence N/rn2 = Va
= Pasca l
SI unit of viscosity = Ns/m 2 = Pa s.
Newton sec
SI unit of viscosity
Ns
The uni t of viscusity in eGS is ~ISQ called Poise which is eq u al 10
-sec
"'dyne
' 'em!
C;:=-
The nume rical conversion oflhc unit ofviscosi l y from MKS unit to eGS unit is given beluw:
9.8 1 N-sec
BUl one Newton = one kg (mass) x one
=
(
I':
I kgf=9.8 1 New ton)
-m, ) (acceleration)
~'
(1 000 g m) x ( 100 c m)
scc
l
= 1000 x 100
gm-cm
sec
l
dyne = gm x -c m-, )
= 1000 x 100 dy ne
='
one kgf -sec =9.81 x 100000 dyne-sec
nI l
e l11
2
= 9.81 x 100000
====--,
dyne-sec
100 x IOOxem 2
• 98 . 1 dyne-sec
,d"y,"",~,:c'=-c =
,
= 98. 1 poisc
c m'
em '
Th us for solving numerical prob lems. if viscosity is given in poise. it must be divided by 9&.1
ils equivalent numerical value in MKS.
1.:
1
8m
ml
one Ns
~
I I
=+,"
one kg f -scc • 9.8 1 Ns
Ill "
POiSC)
10
get
= 98.1 poise
9&.1
• - - poise = 10 poisc
9.81
One poise =
1 N,
10 ml
Ii
~ I
IL
51
Properties of Fluids
.
Alt rrnllil' Mdhod . One po lS<: '"
dyncxs
em
gill
But dyn e
l
'"
( l gm X I Cm )
s
,
X- ,
el11
[ em
X -
,-
"
I
I gm
'" ;;::;; '"
O ne poise
l
lOiXI
k
g
I
' -- m
100
=
I
IOpoisc.
"'
1000
1"'otr . (i) In $[ un;l< second i. represented by's" and nOl by 'sec',
(iil If \'iscosity i. gi,'cn in poi,." it mus, be divided by 10 to gel its equivalent numerical ..alue in SJ unil",
Somclimcs a unit of v;"''OSily as centipoise is used where
. ,
I
I ccntlpo,,,, '" The visco,ily of water at
20~C
,
1
P
POIse or ICP. 100
100
is D.OJ poise or 1.0 centipoise.
rtf' '" Centipoise. P '" Poise 1
1.3 .2 I(in e m~ti c Viscos ity . h is deFined as the ratio between Ihe dynamic viscosity and density
of fluid. 11 is de uoted by lhe Greek symbol (v) c a lled 'no . Thus, ma thematica ll y.
\I '"
Vi scos ily
De nsi ty
jJ
...( 1.4)
'" -
p
The units of kinematic viscosity is oblai llcd as
UnitsufjJ
,. URilS of P
Force x Time
--'=~~ =
(Le ngth) ' x Mass J
Force x T ime
M ass
Length
(Length )
Mass x
=
=
Length
, x T ime
(Time)"
1
--7-':';"'--;--
M,,,
( Length
r
Force"" M ass x Ace.
.
= M ass )(
)
Leng th
l
Time
(Le ngth )!
S="Tim e
In MKS and SI. the unit of kinematic viscosity is metre'!scc or m'/sec while in CGS unilS it is
written as (;111'/s. In CGS units. killcm~lic viscosity is also known as stoke.
Thus, olle sloke
=clI1 '/5=
Cen tistoke means
=
1
100
c~or
4
O1'/S= 1O- 01'/s
sloke.
1.3 . 3 Newton ' s Law of Visco s ity. It SImes that1h" she ar sIres.-; (l) on a fluid clemcnt layer is
dircClly proportional to the ral c of shear strain. The 1:01151:1111 of proportionality is called the cocfficielll of viscosi ty. Mathematically, ;1 is c~prcssed as givell by equation (1.2) or as
till
t = !-I - .
(/ )'
I I
Ii
~ I
IL
16 Fluid MC(;hanics
Fluids which obey the above relation arc known as NewtonL'l1i fluids and the fluids which do not
obey tlie above relation arc callcd No n-Newto nia n fluids.
1. 3.4
Variation of Viscosity with Temperature .
Temperature affects the viscosity. The
viscosity of liquids dcneasc s with Ihe increase of tcmpcrmurc while Ihe viscosity of gases inncascs
with the increase of tempermure. This is duc to reason that Ihe viscous forces in a fluid arc duc to
cohesive forces and molecular momentum lransfcr. In liquids. the cOhesive forces predominates
the Illolccular 1110111cn lurn lraJlskr, duc 10 closely packed Illolcculcs and with Ih e increase in
temperature. Ihe cohesive forces decreases wilh the resuh o f decreasing viscosity. Bul in case of
gas.cs the cohesive forces arc small and molecular momentum transfer predominaks. With the
increase in tempera ture, molecular fllom<"nturn transfer increases and h<"nce vis.cosily increases. The
relation between viscosity and temperatu re for liquid.~ and g ases are:
(I) Fo r liquid s, (.I
whe re
(.I
= (.10 (
,l
... ( l.olA)
I
I +O:I+llr
= Viscosity of liquid at
rc, in poiSt:
1-10 = Viscosity of liquid at O°C . in poise
0:..
P " Constants for the
liquid
ror water.
(.10 = 1.79 x 10 J pois.c. (I = 0.03368 and P= 0.000221.
Equation (1.4A) shows that wilh the increas.c of tcmperature. the viscosity dcr rcases.
(ii) For a gas ,
J.I" (.10 + o:r _ PI 2
9
where for air J.lo " 0.000017. 0:= 0.000000056. P'" 0 .1189 X 10- .
... ( I.4B )
Equation (I.4B) shows that with the increase o f temperature. the viscosity increases.
1.3 . S Types of Fluids. The fluids may be classified into the following five types:
I. Ideal nuid.
1. Real fluid.
3. Newtonian fluid.
4. Non -New tonian fluid. and
5. Ideal plastic fluid.
I.
Id"al Fluid. A fluid, whic h is ;ncorll pr<"ssiblc and is
having no I'is.cosily. is known as an ideal fluid. Id~al fluid is
only an imaginary nuid as all the nuids. which exis!. ha ve
some viscosity.
2.
R"nl Fluid. A fluid. whic h possesses viscos il y. is
known as real nuid. All lhe nuids. in aClual practice, are real
fluids.
Ne wtonlun )<' Iuld . A real fluid. in which lhe shear
3.
stress is direc tly proportional to the rate of shear strain (or
velocity gradient). is know n as a Newtonian nuid.
4.
Non -Newtonian Fluid . A real fluid. in whic h the
shear stress is not proportional tu the r:lte of shear strain (or
velocity gradient), known as a Non-Newtonian nuid.
1
IDEAL FLUtD
-
VELOCITY GRADIENT (~~)
Fig . 1.2
Typt$ of flm·d$ .
5.
Ideal Plastic Fluid . A fluid. in which she ar st ress is more than the yield value and shear
stress is propor1ional to the rate of shear strain (or velocity gmdient), is known as ideal plastic fluid.
Problem 1.3
,
If IIII' re/ociry dislribuliOlr O)'e ~ (I pialI' is girol by u = .:. y
J
relocily ill mefre per seCOIrd aI
y " 0 alld y = 0./5
I I
111.
(I
-l
in "'hieil U is lire
disfmrce )" melre ab()re lilt! plme. de/amilre Ille
.~lIe(lr
J·lres.!
{If
Take dYllamic I'iscosify offtllid (IS 8.63 poises.
Ii
~ I
IL
2
Solution. Giwn :
till
,
II= -Y-Y
dy
3
(~:.)~ y_O 01(;:;),.0'"
("")
-
,/ Y
-
"'
",,_0."
("")
Properties of Fluids
,
71
'" ..:. - 2y
3
2
2
- - 2(0)= - =0.667
3
3
2
= - -2x.15=.667-.30=0.367
3
d Y ,_ 0' 5
Value of).l = 8.63 roisc = 8.63 51 units = 0.863 N s/m 2
10
Now shear stress is given by equat ion (1.2) as t =).l
(/11
,/ )'
(I) Shear stress Jt y = 0 is given by
(~)
'" 0.863 x 0.667 '" 0.5756 N/m~. Am.
(Iy y. Q
t o = I-l
(ii) Shear strcs.~ at y = 0.15 III is given by
(l:)y
. O_l~
'"
(;;1y') '" 0.863
Il
x 0.367 '" 03 167 N/m l, AilS.
y_ O l ~
Problem 1.4 A plm/' 0.025 """ (/i.<I"''' from 1I fi-.nl philP, /IIm'n {I/ (,() ("111/,' tII,,1 rl'ff"ir.... " fora "f
2 N pprllll;1 (lrra i.r., 2 Nlm z 10 ",(lima;" 'iii.• sJ'l'rd. f)"'('nl1i,,,, /1", flllid viuos;r), brl»,"'-" 111(' plalt'3.
Solution. Given:
Distance bel ween pl ates.
Velocity of upper
= .025 x 10-] m
1/ = 60 em/s = 0.6 m/s
pl:u~.
N
F= 2.0 2 ,
m
('orce on upper plate.
This is lhe value of shear Slr~ss i.t', ••
lei the fiuid vi.'>Cosily between the plales
Using lhe equalion (1.2). we
whe re
f
dy = .025 nlln
ha,,~ t
==
~
T "'i"''''''''''''
FIXED PLATE
Fig. 1.3
is~.
-d" .
<ly
Ii" = Change of velocily = II - 0 = II '" 0.60 IIl/s
dy '" Change of distaJicc '" .025 x IO- J III
t '" Force per unil area'" 2.0
N
IIll
-;:;~0~.6~0
'"
c
2.0 x .025 X lO-l == 8.33 x 10 .s Ns
111 2
0.60
.025 x 10
5
'" 8.33 x 10- X 10 poise '" tl..~ .~ x 1O -~ IlOi~(' . Ans.
Problem 1.5 A jlal pimp of <ifNI 1.5 x J(/' 111m2 is p"l/n! wilb a sllt'rtf of 0.4 mls ",Ialil'p 10
wJOllwr pl<llrloHllrd (1/ 1I dis/(//ICI' of 0.15 111m from il. Fillillile fora mill po ....'" rf</lIirrd 10 IIwill/aill
Illis spud, if Ilw jlllitl srpllrmillg
is /wl'illg l'iscosiIYa5 J poise.
2.0==~
' '('11/
I I
Ii
~ I
IL
Is Fluid MC(;hanics
Solution. Given:
Area uf the plate.
Speed of plate relative 10 anOlhcr plale. "" '" 0.4 tn/s
Distance ~Iwccn [he plales.liy = 0.15 mill = 0.15 x 10- J III
Viscosity
)..1= I
Using eq uation (1.2) we have '!: " )..l
(r):.
(ii)
Shear force,
Powcr~
F
=
I
N,
poise: - ,.
10 Ill '
d"
I
dy
10
! X area
.4='
x -:~~O~
) = _66.66 2N
.l5xlO
III
= 266.66 x 1.5 = 400 N. AilS.
required to move the pialI' at thc speed 0.4 mlsec
== Fx II =400x0.4 = 160 W . AilS.
Problem 1.6 De/ermilll' Ilu' il//(,lls;l), of shnlr of {III oil/r{/\'illg I'iscosily '" I poiSt'. 'I'llI' oil is used
for illbricwillg Ih(' c/I'lIrlW('I' bl'l"'''1'11 {/ shaft of(liallieler /0 rill {l1II/ its jOllnwl bl'arillg. '/'1,1' c/nlr<lHCl'
is 1.5 111111 amI IIII' slwft rotall'S (1/ 150 r.{!.I11.
I Ns
I poise = - ,
Solullon. Giv"l1 :
Dia. of shaft.
10
D= IOcm=O.1
Ill '
III
Distance be tween shaft and journal bearing.
dy = 1.5 nun = J.S x to
1 III
Speed of shaft.
N = 150 r.p.m.
Tangell1ial speed of shaft is given by
II
=
It DN
== It x 0. 1 x 150 == 0.785 mls
60
60
Using equation ( 1.2).
where (III '" change of ve locity betwee n s haft and bearing'"
I
0.785
10
l.5 xlO
==-x
II -
0 '"
II
..
!
J =.2 ..BNfm. AIIs.
Problem 1 .7 ,a/rllia/(' II", (Iy"m"ic I·iscmily of w, oil ....hid, is uSNI for lubricatioll bNW"CII a
1,Iatp of sic,' O.S III x O.S III amI (III i"diflNI philiP wilh allgl/' of ifldi""titm .woflJ sho",,, ill
Fig. lA. The w..ighl of the S'lllfl'.. plat .. is 300 N (Illd il slilln (IOwlIl{,r illclill~d plmlf Wilh" IIl1iform
I'elocil), ofO.J /Ills. 'fI", thicJ.:III'ss of oil fillll is J.5 111111.
J'IU"rt'
Solution. Given:
Area of plate,
Angle of plane.
Weight of plme.
Velocity of p late.
A = 0.8 x 0.8 = 0.64 111 2
e = 30"
IV= 300 N
II == 0.3 111/s
Fig. 1.4
*
I I
Power=l'x"Nmls=FxulV ( '.· NmJs=Wal1 j
Ii
~ I
IL
Properties of Fluids
91
I" lly" 1.5 111 III " 1.5 x 10 3 III
leI the ViSl:os ity of nuid between plate and inclined plane i s~.
Component of weight IV, along the plane" IV cos 60~ '" 300 oos 60° '" 150 N
Thus the she ar force. F • on the bunum su rface of the plate'" 150 N
Thick.ness of oil film.
F
<= - - " 150 Nfm~
and shear stress.
Area
0.64
Now us in g eq uation ( 1.2). we have
rill
T =).I -
dy
where tlu = change of ve locily = 11 - 0 = " = 0.3 mfs
dy'" [ = 1.5 X IO-J 1ll
150
0.3
O.64 =).I I.5xIO
J
3
150 15 10"= "''''';';'::-;;c;-'" 1.17 N
0.64 x 0.3
x
Problem 1.8
wi,II oil
x
2
s/m '" 1.17 x 10 " 11.7 Jlois('. Ans.
Two IlOri~ollllll l'/mf'S (Iff' plll("'f(1 1.25 em (l1'MI. I/W 51'11("(' bnwulI IIII'm bf'illg jillnf
/ 4 l>oi5"5. CII/ru/ml' 110(' S/WlI' 5/ rl'55 ill oil if UPP'" phil" ; .5 mm'~d wilh fI \-<'10";1)'
('I \'i5,.OS;'Y
0[2.5 ",Is.
Solullon. Giv<,)11 :
Dista nce between plates.
Viscosity.
dy = 1.25 em = 0.0 125 III
Il" 14 poise ==
.!.i
N sJIll 2
10
Ve loc ity of up pe r plate.
1/ "
2.5 Ill/seC.
Shear stress is given by equ ati on (1.2) as, t " Il
where til!" Chan ge of ve locit y be twee n p lmes"
I! -
(I I!
dy
0" "" 2.5 Ill/seC.
dy=0.0 125 m.
14
T"
10 x
25
z
.0125 ,, 2MO N/m . Ans.
Problem 1.9 '11", sP""" bn",rr/l IWo s'I"''''' JIm IJamU,,1 plain is fillt'd \\,;111 (Iii. Eacll shit' of II,,,
plait' ;s 6V UII. 1'11" Illidllt'ss of IIIr oil film is 12.5 111m. n,,, IIPP"f pllllr, ",!ticll I/W ..<'S <1/ 2.5 mt'lfr pr f
,'re r('</"irt's (/ fOfa of 98. J N 10 II/lliW(lill lilt' SfH'(,<!. Dnrflllill" "
(i ) Ihl' dyl/lIIllil' "ismsil), of If,I' oil ill poiSi'. Will
(ii) 1/11' /':illl'lI/(l/il' "is('Osil), of IIII' oil ill slOkr s if 111,. sprl'ific gral'il)' of lIlt' oil is 0.95.
Solution. Given:
Eac h s id e of a square pla te
:. Area.
Th ic kn ess of oil film.
Ve loc ity of upper plate.
~ I
,,60 elll '" 0.60 1ll
1
A " 0.6 x 0.6 " 0.36 m
d y " 12.5 mm '" [2.5 x 10- 3 1ll
11" 2.5 mIse"
I~
~ I
IL
110
Fluid Me<hani cs
Ctmllgc of veloc it y between plm es. <III '" 2.5 mlsec
Force required on upper pl~lc. F ~ 9 8. 1 N
Force F
98.1 N
,. - - = - =
,
Area
A 0.36 m-
She ar st ress.
(i) LeI )l '" Dynamic viscosity o f oil
II"
Usin g eq uatio n (1.2).
t =).l -,,,
)l '"
0'
98.1
-0.3-6
25
"')l)(
98.1 )( 12.5)( 10-
1
'"
CI2~'~X";;'O'"
1.3635
N~
0.36
2.5
rn '
'" 1.3(35)( 10", I.t6.\5 puist'o Ans.
Cln~lS
'"
to poise )
( ii) Sp. gr. of oil. S '" 0.95
leI v'" kine m 3lic viscosity o f o il
Usi ng equation ( 1.IA).
x 1000 '" 0.95 x 1000 '" 950 kg/Ill}
p'" S
Mass de ns it y of oi l.
1.3635 (
Using the relatio rl . v
=
1:. we ge l v =
N~l
III -
P
=.00 1435 m 2/sec = .()() 1435 x 104 C111 2/S
950
(": cm 2/s '" sIDke)
= 14 ..'5 s lok~s. Ans.
Problem 1.10
"PO;/ll in oil
j"
j
Filll/,11l' killfl/Uuir visrosily of 1111 oil/w\'illg (/l'1!5ily 98/ kgllll , TIll' S/W(If
0.2452 NI",l ' Illd ",,'/oril), grad;f'JIl lU I/U" p"ill/ is 0.2 pn .w("(md.
5Iri'.' S (1/
Solutio n. Gi ven:
p" 98 1 kg/lll '
1: = 0.245 2 N/m!
Ma ss densit y.
Shear stress.
Velocity grad ien t.
Usi ng Ihe equation ( 1.2),
~
"y
=0.2 s
(I "
t =j.l -
,/ y
orO.2452"j.l x O.2
j.l " 0.2452 " 1.226 Nslm!
0.200
Kincmatic viS\:osil y v is given by
v=
1:."
p
1.226 =. 125x 10- 2 m !/scc
98 1
" 0. 125 x 10- 2 X 104 I:m 2 /s" 0. 125 x 10 2 cm'/s
= 12.5 en/Is" 12.5 stokl'. Ans.
Problem 1. 11
,·iu(}.lity 0.0.'15
(: cm1/s" stoke)
Dnnmill" Iii" S{J"c ijic gr(l\'ily of" JI"hllllll'ing .. i.l("(uil), 0.05 /1Oi.I" mill kin"",atic
stok~ •.
Solution. Givcn :
W
V'Is.cOSI' 1y. j.l =.
005'
POISC = 0.05 N sJm 2
~ I
I~
~ I
IL
Prope rties of Fluids
11 1
v:: 0.035 sto kes
:: 0.035 cm1ls
:: 0.035 X 10- 4 m 1/s
Kinematic viscosity.
I.l
_4
Usi ng thc relation v:: - .wcgclO.035xIO
P
0.05
I
10
p
= -- x -
0.05
X ~~'--C4" = 14 28.5 kg/m3
10
0.0]5 x [0
Problem 1.12
Density of liquid
14285
:: - - :: 1.4285 ::: 1.4.\. AilS.
Density of wate r
1000
::
Sp. gL of liljuid
/Jell'rmilll' IIIl' I'iscosily of (/ hlfllid havillg til/i'lI/lUil' \-;Seosil)' 6 s/okn 1111<1 specific
gm.-it)' 1.9.
Solution. Given:
Kinematic viscosi ty
Sp. gL of liquid
LeI the v iscosity of liq uid
v:: 6 stokes = 6 cl11'/s:: 6 X 1O- 4 11//S
:: 1.9
.p
D<:nsity of lh ~ liquid
Density of water
Now sp. gr. of a liquid
1.9 :: c"'
='c"="c'Yso~rcn,q=':::
;d
1000
kg
= 1000 x 1.9 = 1900 - ,
Ik nsi ly of liquid
m
v::
.. Usi ng th e relation
1:., we get
p
4
6x 10- = - " -
1900
J.I=6x 10- 4 X 1900= 1.14 Ns/m 1
:: 1.14 x 10:: 11.40 Il oise. Ans.
Problem 1.13
Tilt' l'e/oeiIY distriblliiou for jlow OI'e r " JIM p/lIlr is gil'l'lI by " = -1- )'
II ;5 IIII' \'e/oeil)' ;111111'1'1' IN" secQlul
aJ)'
a (/;5/(IIICI' y 1111'1'" abore IIII' 11/(1/1'. DrlulII;nr IIII' silell' SI'I'SS
(1/
= 0. 15 Ill . Takl' (/),II(11l1ic \'is("o,lily ofjluid ( 1.1 8.6 poiSl'.
Solution. G iven:
11=
(I u
-
If)'
At .1'=0. 15.
Viscosity.
I I
-l ill which
(Iu
-
If .I'
1-1
~
1' _ 1,2
4 '
.
3
= - - 2)'
4
3
= - - 2xO. 15=0.75 - 0.30=0.45
4
= 85 poisc = 8.5 N ~
[0 m-
N~ l
( .: 10 poisc = 1 m
Ii
~ I
IL
112
Fluid Me<hani cs
till
r = '" -
Using eq uatio n ( 1.2).
85
-
~
10
If )'
N
N
x 0.45 - : = 0.,\1'1 25 - : . AilS.
m
In
Problem 1.14 11/1' (/pwlllic viscosity of WI oil. lISt'll for II/brim/ioll bnwrl'u (I SIUlfi 01111 5/1'1'1'(' is
61"';u, TI", .,'mp i., of di"'"N(,T 0.4 III "",I roW/P.' 1lI 190 T.p.m. Cuk ulmr 1111' I}{)""P T /0.'/ ill IIii' brllrill8
fo r ti slnl'(' /""8,11 oJ90m",. TIIr lilickllrH of/lie oil film is 1.5 "''''.
Solution. Gi ve n :
1.5 mm
Viscos ity
j.I=6poi se
6 Ns
~
Dia. o f sha ft.
Speed o f shaft.
f)
10 m
= 0.4
l
m-
y o_~ m
,C': ';::J lKAFT
III
N= 190r.p. m
Sleeve k ng1h.
Thic kn ess of oil fi lm,
L = 90 mm = 90 x 10- )
Tangc llti al vel oc ity of shaft .
/I
Using the re lation
t=l-I dy
where
d
Ns
= 0.6 - ,
SLEEVE
III
I'" 1.5 111m '" 15 x 10- .1 m
Fig. 1.5
'(tDN
IT x 0.4 x 190
= ~=
60
= 3.98 IIl!S
tlu
'/11 = Change of veloc ity = II - 0 = /I '" 3.98 m!s
Ify = Change of d istance = I = 1.5)( 10- 3 1ll
--:-3~.9~8;,. '" 1592N/m'
.=lOx -;l.5x lO J
This is shear
st res.~ On
shaft
Shc ar for(;c o n thc s haft. F '" Shcar stress x Area
'" 1592x1tDxL", 1592x1tx.4x90x lO -3 = 180.0SN
D
0
0.4
T= Forcc x - '" 18 .05 x '" 36.0 1 Nm
Torquc on th e sha ft ,
2
2
_ 21tNT _2 1tX I90 X36.oJ - 7164SW
-
60
-
60
-
.
.
. "'n~ .
Problem 1. 15 If lilt' r('/ocily profil~ of {/ jlllid Ol't'f {/ (,((l/r is p<lf<lbofic wiliJ IlIr I·ent'.\' 20 01/ frolll
IIII' p(<I/(', ...lIerr I(,r " ..(ocily is 120 i'1II/sri'. CO/CII(O/(' Iflr \'('/ocily gradinlls 011.1 ~lIro r SlrfSS('S (1/ II
dislOlI(,(, ofO. 10 olld 20 i'1II frolll Iflr I,ttllf. if IIIl' viscosil)' of liJr jluid is 8.5"oiSf.
Solution. Givcn :
Dista nce of \'c n ex from pl ate = 20 cm
Ve loci ty at vertex.
II = 120 (; m/sc(;
OJ _
211'N
2 11' NT
T)( - - Wan _ - - - Wall
60
I I
u • , 2(l em/Sf!(:
11 '" 85 poise = 8.5 N ~ '" 0.85.
lO m '
Viscosity .
• Power in S.1. unit_ T·
y
Fig. 1.6
'"
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Prope rties of Fluids
The ve locity profile is g ive n parabolic and eq uation o f ve locity profile is
II = <Ii + by + c
where lI, b nlld (" arc constal11s. T heir va lues are dctc nnincd from boundary cond ition s as:
«(I) aly=O.,,=O
(h) 31),=20;;10,11= 110cm/.'i<'c
131
... (1)
,/u
(c) illy=20cm.-=O.
"y
Substitutin g boundary conditi on (ll) in equation (,) . we gd
c = 0.
Boundary cond iti on (b) o n sub stituti o n in (i) gives
2
J 20 = a(201 + b(20) = 400" + 20b
Boundary co ndition (c) 0 11 substi tution in equ ati on (il gi ves
,II. =l,,)' +b
-
dy
... (ii)
...(iii)
-
0= 2 XII X 20
+ b = 4{M + b
Solving L'ijua ti ons (ii) and (iii) for a and b
From eq uation (iii),
b = - 40<1
Substitutin g this va lu e in equati on (ii) , we get
120 = 400<1 + 20 x (- 40a) = 400(/ - 800<1 : - 40011
120
3
tI= - - = - - = - 0.3
- 400
10
b= - 40)« - 0.3)", [2.0
Subslilulin g the va lu es of a. band (' in equat io n (i),
0.3/ + 12y.
II" -
Vrlodty Gradient
-(/ " = - 0.3 )( 2y + 12=-0.6y + 12
Ify
"' Y
(""]
(""]
= O. Veloc ity grad ient. (/ y
at y '" 10c m.
at y : 20 c m.
= - 0.6 )( 0
= - 0.6)( 10+1 2= - 6+ 12: 6/s. AD S.
-
(f y
( ~".]
)
+ 12 = Ills. Ans.
1*°
1 - 10
= - 0.6 )( 20 +1 2: - 12+12= 0. Ans.
y _ 20
Shear Str('sses
Shear stress is g iven by.
I I
t : fl
till
(/ )'
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IL
114
Fluid Me<hani cs
(i) Shear stress at y
(ii) Shear s[rc ....~ at y
- [d"d )
t=J.l[:;"]
= O.
t - lJ
= 10.
y
y
(iii) Shear stress at y '" 20.
t=)J
_.o5x l '0_._l,~Nfrn.
-0"
0'
'
~.o
=O.85x6.0=5.1 N/m
2
•
1 - 10
=O.85xO= O. Ans.
[d"]
Ii)' ,-.
-
Problem 1 .16 A New/ollillll Jill;" is fill ..,} ill IIII' c/CII'<IIICf' bl'lw/>/'/1 {( shaft IIml" collCl'IIlrir sIN'I'e.
Till' s/,,{'\,I' IIIll1illS {/ sP('('(/ of 50 ollis, W/1l'1I II lorn" of 40 N is IIPIJ/inf 10 the S/I'I'I'I' parallel 10 IIII' slmfl.
/){'/('flllillf IIIl' SpUI/ if {/ jora of 200 N is (If'l'lin/.
Solu t ion. Gi ven: Speed of sleeve, fit = 50 em/s
when force.
FI =: 40 N.
Let speed of sleeve is "2 when force. F2 = 200 N.
Using relation
t =
I.l
till
dy
rorce
F
where t = Shear stress'" - - •
Are a
A
<ill '" Change of ve loc ity =
Ily'" Clearance'" )'
/I -
0="
-F =1.1 -"
A
Y
A jJ u
I.
F=--."
A. j.l JIlO yare constanT}
Y
F
f~
...l= ....:.
",
Substituting val ues. we gel 40 ==
200
50
50x 200
40
'" 50 x 5 == 250
cml.~.
Ans.
Problem 1.17 A 15 nil diol/w/f r "erlirol c)'lil/der rotales COI/C<'III rical/)' illsitit' ((lIatller cylilllirr oJ
!lilli/WIer 15.10 nil. 80111 cylindt'rs ort' 25 rill lIigli. Til<' SP((C<' bnw<'<'n III<' c)'lilUlers is jillnl willi 0
liquid whost' I'ismsil), is unknowll. If " torqut' of J2.0 Nm is r..quired 10 rOUlt .. II, .. illll'" cylillll'" lit
100 r,p.III .• dr/erll/illt' Iht' viscosity of lilt' fluid.
Solution. Given:
Diameler of cyli nder
Dia. o f oUler cylinder
Lenglh of cy li nders,
Torque.
I I
== 15cm==O.15m
== 15.10 cm == 0. 151 111
L == 25 cm == 0.25
r == 12.0 N1II
111
Ii
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IL
Prope rties of Fluids
151
N", 100 T.p.m.
Speed,
Let th e viscusit y
""
.
.
.
nDN
IlxQ.15x100
Tan genllal ve loc ity of cyl inder." '" - - '"
'" 0.7854 mfs
60
A = nD x L = It x 0.15 x 0.25 = .1178 rn 2
Surface area of cy linder.
till
Now usin g relation
where ,/" =
1/ -
60
l: :p. -
dy
0 = " = .7854 rnls
15,,',,--,,0,,.' "'
eo: III = .0005
dy = -"0.,,
2
m
I-l x .1854
.0005
Sheaf force.
F = Shear stress x Area =
Torque.
T= Fx -
).1x .7854
x . 11 78
.0005
lJ
2
12.0",
I-l x .7854
X. 11 78 x
.0005
12.0 x .0005 x 2
.7854x .11 78 )(.l 5
.15
2
= 0.864 N shn!
'" 0.864 x 10 = 1i.64]loisc. An s.
Problem 1 . 18 Two Illrgt' phll/P sIIr/tln's (I'f' 2.4 rill apart. Thl' Spllcr b(,/I<'I'I'II ,h" slIr/act's is filln!
willi giY("f'rilll'. II'lul/ fora 13 rrq"irnllo drag /I 1't"Y Ihill plllll' of sur/act' (If('ll 0.5 squarr /lleI,."
b"I»'''''''
/\1'd largl' plm,,' sII ,fa,.. s til II sprn/ of O. 6 m/s, if:
I",.
(i ) lill' 1i1ill [,/alr is ill lh" middll' oflhl' 11>'0 p!t11l1' surfllCl's. mul
fii) IIII' Illill 1'1(1/(' is at rl disliIllCl' of 0.81'111 frO/II 011(' of Ille plllll<' sIOrfll(,('3 ? Take IIII' dYliamic
"ism sil), ofgl),urill(, '" 8.10 x uri N .!1m?
Solution. Given:
Distan ce Iwtw", ... n two large ~urfac es '" 2.4 em
Ar~a of th in plate.
A = 0.5 [11 !
Velocity of thin plate.
Viscosity of g lyce rine.
/I
= 0.6
mf.~
!-! = 8.10 X 10- 1 N shn l
C:tw I. Wh en the thin plate is in the middle of the tw o plane
surfaces 1Re fer to Fig. 1.7 (n)1
FI '" Shear force on the upper s id e o f the th in plate
""2 = S h~ar forc~ On th~ low~ r s id~ o rth e th in pl~te
"" = Total force required to drag Ih~ plate
Then
The , hear
I I
S1r~ss
1.2 em
2.4 em
.,
1.2 em
Fig . 1.7 (a)
F",F1+F1
(t l ) on th e upper side o f the thin plate is give n by equation,
Ii
~ I
IL
116
Fluid Me<hanics
tl '"
~C~:~),
where 1111 '" Relative ve locit y between thin p late and upper large plane surface
'" 0.6 mlsec
d y '" Distance between thin plale and upper large plane surface
'" [.2 em '" 0.012 III (p late is a thin one and he nce thickness of plate is neglected)
11=8.IOx 10 x(.012
0·')=40.5Nllll1
I
Now shear force.
FI '" Shear stress x Are a
= t l
xll=40.5xO.5= 20.25 N
Similarly sllear st ress (t 2) on the lower si de of the 1hin plate is given by
12 =1.1
(~)
dy !
= 8.IOx 10 I x
(~)
0.012
=40.5 N/ml
Sliear force.
Fl '" t 2 )( A '" 40.5 x 0.5 '" 20.25 N
Total force.
F = FI + F1 = 20.25 + 20.25 = 40.5 N. Ans .
CaSt' II. When the thin plalc is at 11 distanceofQ.8cm from onco r
the plane surfaces [Refer to Fig. 1.7 (b)].
Lei Ihe lltil1 plate is al a distance 0.8 em from Ih e low er pia""
Ulem
surface.
Th<"n
Thc
I
disl~nec
sh~ar
of thc plalc from lh e upper
2.4 em
pl~nc sun~c<"
I
:2.4 -0.8: 1.6"m=.016m
(Negltt ting thkkllcss of the plate)
forc~ 011 th~ upper side of lhe thin plate,
FI = Shear stress x Area: I I X II
[(Iy"")
"" The shear force
011
xll:S.IOxIO
I
.,
0.8 em
•
Fig. 1.7 (b)
- -) xO.5:15. ISN
,x (0.'
0.016
the lowe r s ide of the thill plme.
:IUOx 10
I X (
0.6
0.81100
) xO.5= 30.36 N
Tot al fo rce required = PI + P z : 15. 18 + 30.36 '" 45,54 N. AilS.
Problem 1 .19 II "p nil'al gal' 2.2 rm ",id~ of infinile "X/rill ""lIIaills a fluid of \·is.'osiIY 2.0 N JIm!
aud sp('djir gr<ll'ily 0.9. IIIIIPwllic plal(' 1.2111 X 1.2 m xO.2 elll is 10 be liji('d uI' wilh (/ "OIISlmll
w'locily I)fO.15 mist'<", Ihrough Ihe gap. if Ih(' plm(' is ill Ih(' middle of Ih(' gap, jim! tllP fo ra rI'ql.ir('(l.
'/7", '''''igl'' of //", pia'" i .• 40 N.
Solu t ion, Givcll :
Width of gap
Sq. gr. of nu id
I I
" 2.2 em. viscosity, Jl" 2.0 N s1m 2
" 0.9
Ii
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IL
Prope rties of Fluids
,
Weight density of fluid
= 0.9 x 1000 '" 900 kgflm J
'"
I
900 x 9.81 N/m J
(.: I kgf=9.81 N)
Volume of plale
= 1.2
III
x 1.2 I1IxO.2 em
'" 1.2 x 1.2 x .(X)2 Ill) = .00288 Ill)
Thickness o f plate
Veloc it y of p late
171
"
= 0.2 em
= 0.15 mf.'\Cc
= 40 N.
'"
"" ,. 0.2cm
Weight of pla te
When plale is in the m iddle of the gap. the distance o f tlie plate
from vertical surface of the gap
Fig. 1.8
= ( Width of gap- T; iCkn ess of Plate)
=
(2.2 - 0.2)
2
=lcm= _Olm.
Now the shear force On the left side of tlie nw tallic plale.
FI '" Shear stress x Area
=" (:!:!.],
(f )'
."J
(O
.01
xArca=2.0x - - xl.2xL.2N
(.; Area = 1.2x 1.2m 2)
= 43.2 N.
Similarly, Ihe shear force on Ihe Tight side of the metallic plale,
(0."]
F, = Shear stress x Area = 2.0 x - - x 1.2 x 1.2 = 43.2 N
•
.01
Total sheaf force
= FI + F z = 43.2 + 43.2 = 86.4 N.
In this case lh~ weight o f plate (wh ich is acting verti cally dOwnward) and upward thrust is also to be
taken into account.
The upward thrust = Weight o f nuid displaced
= (Weigh t density of nuid) x Volume of nuid displaced
=9.81 x900x .00288 N
( .: Volume of nuid displaced = Volume of plat~ = .(0288)
The net
forc~
= 25.43 N.
acting in the downward di rection duc to weight of the plate and upward thrust
= Weight ofplatc - Upw ard thrust = 40 - 25.43 = 14.57 N
Tot al force required to lift the plate up
= Total s hear force + 14.57 = 86.4 + 14.57'" 100.97 N. Ans.
... 1.4
THERMODYNAMIC PROPERTIES
Fluids consist of liquids or gas.:s. Bul gas.:s are compressible fluids and hence thermody namic
prope r1i es play an important role. With the cha ng e of pressure and I~mpcrature. the gases undergo
I I
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IL
118
Fluid Me<hani cs
large var iati on in de nsit y. T he rel ati onship be twee n press ure (absolut e), s pecific vo lum e and
temperatu re (absolute) of a g as is give n by the equatio n of state as
p,;/= RT or'£' = RT
.. .( 1.5)
P
wh ere p = Ab so lu te pressure o f a gas in Nlm '
'<i = Spcdn e vo lum e =
I
p
R = Gas L'Olls tam
T= Abso lute temperatu re in oK
p = Densi ty of a g as.
1.4. 1 Dimension of R. The gas constant. R. depe nds upon th e pan icul a r gas. The d imension of R
is obtained fro m c qu~ti o n ( 1.5) as
R= ~
pT
( 0 In MKS units
R=
kgf/ m 2
ii5C'--
kgf -m
kgoK
(~} K
(ii) In 5 1 un its, p is ex pressed in Ncw ton/m ' or N/m!,
R = N/ m'
kg
~X K
=
For air.
Nm
louie
-- = -kg -K
kg-K
lJ oule = Nm I
J
kg- K
R in MKS = 29.] kg f m
kg c K
R in SI = 29.3 x 9.81
~ = 287
kg¢K
J
kg- K
1.4.2 botherm~1 Procen. If Ihe c hange in de nsity occu rs a1 co nstanl temp erature. the n the
process is called iMlthe nnal and relatio nship be twee n pre.'i.~ ure (p ) and densit y ( p) is give n by
J.!... == Constant
p
...( 1.6)
1.4 . 3 Adiabatic Process. If the c hange in d~ n s ity occ urs with no heat exc han ge to and fro m the
gas. the process is c all ed adiabatic. And if no h<":at is ge nerated within th e gas due to fri ct ion. the
relatio nship be twttn press ure and densi ty is give n by
...!!.... == Constant
p'
... (1 .7)
where k == Rati o o f specific hea t of a g as at co nstant pressure and constan t volum e.
== 1.4 for air.
I I
Ii
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IL
Prope rties of Fluids
1.4.4
191
Universal Cias Constant
Le,
1/1"
Mass of a gas in kg
V '" Volume of gas uf mass
1/1
J! == Absolute pressure
T= Absolute temperatu re
Then. we have
/IV == mRT
.. .( 1.8)
where R == Gas eonM:mL
Equation (1.8) can be made universal. i.I' .. applicable 10 all gases if it is expressed in moll'·bas is.
If
== Number of moles in volul11e of a ga s
'T/
== Vo lume o f th e gas
M ==
"c'c"o'coof,;,,'ho'C""c'co"'oOol'«"'olc'c'
M ass of a hydrogc ll atom
1/1" Mas;; of a gas in kg
Then. we have
II x M == III.
Substituting the val ue o f 1/1 in equation ( 1.8), we get
...( \.'))
1''V=nxMxRT
The produl'! M x R is called universal gas <:OIlStanl and is equal to 848
-=-
_.k" ',"' ' '" cc in MKS units
kg -mo le o K
and 8314 l/kg -rnolc K in 51 units.
One kilogram mole is defined as the product of One kilogrlllll mass oflhe gas and its molecular weight.
Problem 1.20
A gns wl'iglis /6 NI",J lU 25°C
Wid fII WI
IIbsolu/f prl'SslIrt' of 0.25 NIl/III':. Dnfr-
111;111' Ille glls ("OIIS/(/1/I (l1II! dflls;ry of IIII' gas.
Solution. Given:
'" '" 16 N/m~
I " 250C
T ", 273 + I"" 273 + 25 '" 2RROK
Weight den s ity.
Temperature.
P" 0.25 Nlmm~ (a bs.) '" 0.25 x 106 N/m 1 '" 25 x 104 Nlm 1
(i) Using relation", "" pg. density is obtained as
16
w
P = - '" - - '"
g 9.8 1
(ii) Using equation ( 1.5).
J.(i .~
,
kg/m· . "ns.
.!!.. = NT
P
p
N = -p-T =
Problem 1.21
A ("y/ind", of 0.6
25xlO"
CI.6~3~XC2~8"'8
"
S.l~.SS
Nm
-.
kg"
AilS.
mJ ill "01111111' ("01l/a;1IS air a/50°C allil 0 J Nlmm! lIbsollllr
"rI'HI"I'. '1111' lIir is ("omprnsed 10 0.3 IIr'. Find (i) prl'.,wrl' in~· idl' till' ("yli"dl'r ".'.IlImi"8 i.IlII/lI'rlt!(1I
pro'·I'.'.' (11"/ ( ii) prnsllrp ",,,lll'Inpnfllllrf' ('.,3,,,,,in8 ",li"ba/i' ·I,,"'·n.<. T"k" k = 104.
Solution. Give n
Initial
I I
vo lum~.
Ii
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IL
120
Fluid Me<hanics
Temperatu re
I,:SO°C
1', '" 213 + 50 '" 323°K
Pressure
P, == 0.3 N/rnm 2 '" 0.3 x 106 Nlm 2 '" 30 x 104 N/m 2
'If 2 == 0.3 m}
Final voluille
k= 1.4
(i) Isolhe r lllJj[ Ilroccss :
~ '" Constant or p'V '" Constant.
Using equation (1.6),
p
1','<1, "'1'/</:
1','rI j 30 X 10" x 0.6
'" 0.6 X 106 Nlm l '" 0.6 N/mml. AilS.
1': = - - =
0.3
'\f 2
(ii) Adiabatic
Ilroccs_~
:
-I'. '" Constant or I' V• '" Constant
p
Using equation ( 1.7),
I' I '\f~ "" 1'1'\f~.
f'2:1":~ =30XI04x(~~r' :30x104x21.4
:0.791 x lOb N/m l ", 0.791 N/ mm !. Ans.
For temperature. using equ.uiull (1.5), we gel
plt;j '" 81' and also f' Vi '" Constant
NT
NT
P '" -;;;- and -;;- x
RTv.
I '"
'<t* '" Constant
Constant
r '\fl ~ 1 '" Constant
T 'It i _ I _ T '\tk- I
I
I
(.
R is also tonstanll
1
T2=T,(;:r-
2
2
I: '" 426.2 - 273 '" IS.l.rc. AilS.
Problem 1 .22 (',,!cu/wi' tile pr('ssllfI' (,Xf'rI~d by 5 I.:g ofllilrogell gas 11/ (l1l'IIIPl'''III''l' of 100e if
Ihe \'0111111(, i5 0.4 III "'. Molnll/(lr "·e iglll of uilrogl'u is 28. A ssuml'. irinll 8"5 III\\'5 (lrl' "I'P/ iC(lblt'.
Solu t ion. Given:
Mass of nitrogen
Temperature_
'" Skg
1= lQoe
T=273+ 10=2S3°K
Volume of ni trogen.
Molecular weighl
'V = 0.4 m l
= 28
Using equat ion (1.9). we have p'V '"
I I
tJ
X M x RT
Ii
~ I
IL
Prope rties of Fluids
where M x R" Universal gas cOnstan'" 83 14
21 1
Nm
,----''-'''-c"C"
kg-mole 0 K
and one kg-mole = (kg -ma·,s) x Molecul ar weight" (kg -mass) x 28
,
8314
Nm
R for nnrogcn = - - '" 296.9 - - 28
kg " K
The gas laws for nitrogen isp'V = mRT, whe re R = Characteristic gas conSIant
or
pxO.4 =5x296.9x283
5 x 296.9 x 283
(1=
... I .S
0.4
" 1050283.7 N/m 2
"
LOS N/mml. Ans.
COMPRESSIBILITY AND BULK MODULUS
Compressibility is Ihe reciprocal of the bulk modulus uf
dasticily. K which is defined as Ih", ratio of compressive stress
v
•
-- d'V' J-o
PISTON
10 \'olurne\ric strain.
Consideracylindcr fi lled wi th a piston as shown in Fig. 1.9.
LeI 'V" Volume of a g as cnclo.\.Cd in the cy l inder
,
(J '" Pressure of gas when vo lume is V
LeI the pressure is increased to f! + d/!. the volu me o f gas
decreases frolll V to V - lil:l.
Then increase in pressure
= dp kgflm 2
CkcreaSt! in volume
= dV
CYLINDER
Fig. 1.9
dV
Vulumetric strain
- ve sign mea ns th e volume decreases with increase of pressu re.
Bulk modulus
Increase uf pn:ssure
K ~ ~~~';"""C"
Vol umetri c !)Irain
_ dp _ - dp V
- - dV-dV
... ( I . I 0)
V
I
Compressibility
K
... ( 1.1 1)
Relationship between Bulk Modu lus (K) and Pressure (p) for a Gas
The relationship betwee n bulk modulus of elastici ty (K) and pressu re for a gas for two different
processes of compression arc as:
(,) For Iso thermal Proces!l. Equation (1.6) gives the relationship between pressure (1') and density
(M of a gas as
J!..
p
I I
= Constant
Ii
~ I
IL
122
Fluid Me<hanics
1''1", Const:lII\
Diffcrc",i~ti n g
this equation. we gel (I' and '1/ both arc
pdV + 'ltdp '" 0
or
vari~blcs)
l'd'v '" - \:tlfp
Substituting this value in equation ( LlO). we get
...( 1.12)
K= p
(ii) FUT Adillha til.: I'rcKt'SS. Usi ng equat ion (1.7) for adiabatic process
-li'" Constant or f'
p
Differentiating, we gCl l'd(Vl) + 'cil(d!,)
or
pxkxV·- I,/'<I+Vldp=O
or
ph/V + ViiI' '" 0
'I;Il '" Constant
=0
ICanceliing \;It-! to both s id es ]
1'/;(1 '1", - ';ltlp
or
'It dl'
I'k=-
dV
Hence from equation (1.10), we have
K= pk
where K
= Bulk
...( 1.13)
modulus and k = Ratio of spc<:ific heats.
Problem 1 .23 Dnl' nnilll' IIIf' bllik moliulus of elasticity of a Iiquitl. if IIII' (",,!SUrf of lilt' liquid is
inf'rl'ased from 70 Nkm 2 /0 130 Nkn/. nit' "O/lIIlIe of ,II" liqllid durNlus by O. /5 I'er a/ll.
Solu llon. Given:
Initial pressure
= 70 Nfcm!
Pina l press ure
" 130 Nlc m 2
,/1' '" In crcasc in prcssure " 130 - 70 '" 60 Nlcm 2
Decreasc in vo lum e
"O.IS'l>
,/ "i
O. IS
= + -V
100
Bulk modulus. K is given by cqu31io n ( 1.1 0) as
K:~,,60N/cm2
d "i
V
.IS
100
,,60x l00
.15
"' 4xIOJN/cm2. ATl.~.
Problem 1.24 IV/Ill I is I/U' bll//I; 1I/0(/lIlu5 of "/lI5/icitv of II /iqllid ",/lid, is ('olllprnsed ill (I (' }'/il/d,'r
frolll (/ \'olwIU' of 0.0125 111 3 (1/ 80 Nklll! prnSllrf 10 (/ \'011,1111' of 0.0124 1/1" (1/ 150 Nklll! I'rnSllrl' ?
Solution. Given:
Initial \·o lum e.
Final vol ume
,,0.0 124 m)
Decrcasc in vo lum e. ,/"i" .0 125 _ .0124 '" .0001 10 3
I I
Ii
~ I
IL
Prope rties of Fluids
d'::l
';J
Initial pressure
=
23 1
nOOl
.0125
= 80 Nlcrn l
Final pressu re
= 150 Nlcrn'
Increase in pressure. dp = (150 - 80) = 70 N/cm 2
Bulk modulus is given by equation (1.10) as
lif!
70
2
K= ---;t;j= .0001 = 70 x 125 Nlcm
.0 125
= 8.75 x 10- N/cm '' . Ans .
';J
... 1.6
,
SURFACE TENSION AND CAPILLARITY
Surface tension is defmed as the te nsile force acting on the su rface of a liquid in contact with a gas
or on the surfal:c between twO immis.:iblc liquids such that the contact surfa"c behaves like a
membrane unde r tension. The I1wgnitudc of this force per uni t icngth of Ih.: free surface will have the
same val ue as the surface energy per unil area. It is denoted by Greek Idlef 0 (calkd sigma). In M KS
units. it is cxprcs~d as kgflm while in SI uniLs as N/m.
The phenomenon of surface tension is explained by
Fig. 1.10. CO rl sider three mol~cules A, ll. C of a liquid in a
mass of liquid. The molec u le A is allmcled in all directions
~'{jually by the surrounding molecules of Ihe liquid. Th us the
resultant force actirlg on the molecule A is zero. But the
molecule IJ. which is situat~d ncar the free surface. is acted
upon by upward and downward forces which arc unbalanced.
Thus a net result.1r1t fo rce on molec u le II is acting in the
downward direction. The molecule C. situated on the free
su rfa.:e of liquid. does experience a resultalll downward force.
Fig. 1.10 S"r/au unsion.
All the molecules on the free surface experience a downward
force. Thus the free surfa.:e of the liquid acts like a very thin film under tension of the surface of the
liquid act as though it is an elastic membrane under tension.
-- - -- - ---------~~~~~~~~~~~~~~~-
1.6. 1 Surfilce Tension on Liquid Droplet. Consider 11 small spherical droplet of a liquid of
radi us .,. On the enlire surface oflhe droplel, the tensi le force due 10 surface tension will be acting.
Let 17= Surface tension of the liquid
p " Pressure intensity inside the drople t (in excess o f the ou tside pressure intensi ty)
(/" Dia. of droplet.
Let the droplet is cut into two halves. The forces acting on one half (~y left half) will be
(J) (ensile force due to surface tension acting around the ci rcumference of th~ cut portion as shown
in Fig. 1.11 (b) and this is equal to
'" cr x Circumference
=crxTld
I I
Ii
~ I
IL
124
Fluid Me<hanics
(ii) pressure force on Ihe area
~ll ! = l'
x
~ ti!
as shown ir>
4
4
Fig. Lli (e). Thc~ two forces will be equal and opposite
under equilibrium co nditions. I.P ..
(9) DROPLET
(b) SURFACE TENSION
f'X ~ ,r=aXT(,!
4
(J
x
40
Ttll
...(1.14)
"
(e) PRESSURE FORCES
Fig. 1.11
Foret's on dropln.
1.6 .2 Surface Te nsion on <II Hollow Bubble . A llOliow bubble like a soap bubble in air has 1WO
surfaces in co ntact wilh air . one illsidc and other outside. Thus two surfaces arc subjected to surface
lensioll. III suc h cao.c. we have
"x.::. rr=2x(crxrrd)
4
2 and
Scr
.::. d!
d
p= - - - = -
... ( 1. 15)
4
1.6.3 Surface Tension on <II Li quid Jet. COil sider a liquid jet of diameter
shown in Fig. 1.12.
Lell' '" Pressure intensity inside the liquid jet aho ve the outside pressure
(J = Surface tension of the liquid.
.I
Consider the equilibrium of the semi jet. we have
= p x :Irea of semi jcl
==pxLxd
Force due 10 surface tension == a x 2L.
Equating the forces. we have
Force due
10
pressure
pxLxd==ax2L
a x2 L
f' ==
Lx"
,
•
... (1 . 16)
"tf
and lenglh 'L' as
•
•
•
•
•
•
•
••
•
•
•
•
•
••
••
••
••
•
•••
••
••
••
•
•
I"
Fig. 1.12 Forcts 011 liquid jn
Pro blem 1 .25 1111' sHrfll("(' IntsiOIl of ""iliff ill ("011/(1("1 will! lIir (1/ 20°C is O . 0725 NIl/!. Till' prl'Ssllr<'
ilrshlr II drop/("/ of ""fila is /Q h" 0.02 Nkm 1 grNiI"r (//(III IIII' ouuid" pre Hllr". Ca/ru/(I/(" IIII' I/imll("/(" r
of IIII' l lw/,In of "·Il/ff.
Solut ion. Given:
Su rface tensioll.
(J
I"
= 0.0725 Nlm
Pressure intensity. I' in exc ess of outside pressure is
,
• N
p = 0.02 Nlcm '" 0.02 x 10 - ,
m-
1/ = dia. of the droplet
I I
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IL
Prope rties of Fluids
2s 1
Using equation (1.1 4), we gctp" 4cr or 0.02 X 104 ", 4 )(0.0725
If
d
4 x 0.0725
d",
" ,00145 III " ,00145 x 1000" 1.45 nUll. Ans.
0.02 X(lOt
Problem 1.26
Fim/lhr sllrfu("/' II'IIS;OIl ill a soap bllubl" of 401111/1 dilllll<'la wll"l1 lilt' imide
pnssurl" is 2.5 NIIII" abm'(' (!III/osp/wrie prnsllre.
Solution. Gil'en :
(/=40111111=40 x IO- l m
Dia. of bubble.
Pressure in excess of outside./,,, 2.5 Nlm 2
['or
3
soap bubble. using equation (1. 15), we gel
80
d
p=
8xo
2.5= ~~"
40xlO 1
25)( 40 x 10- )
8
Nlrn '" 0 .0125 N/m . AJL~.
Problem 1.27
TlIi'
I'rnSllrf
VUlilid" IIii' dropli'/ O[wlII'" of diamfiPr 0.04 "'m is 10.32 Nlrl// ((1/.
-<II'f(l("1' /1'11 _';01> i.! girt'1! (IS
mospltt'rjr prns"'.-). C(II,u/(l/p /Ilf p'l'SSlI re witilill II", tlrop/l'I if
0.0725 Nlm "j ....lIIPr.
Solution. Gil'i.:n :
Dia. of droplet.
Pressure outside the droplet
0.04 mill '" .04 x 10- 1 111
,,10.32 Nlcm l " 10.32)( 104 Nlm l.
(f",
Surface tension.
(J " 0.0725 Nlm
The pressure inside the droplet. in excess of outside pressure is given by equation (I 14)
4 x 0.0725
,
:
J ,,7250Nlm-= 7250 N = 0.725 Nkml
,I
.04 x 10
10' cm ~
Pressure inside the droplet = p + I'ressure outside the droplet
4(1
P" -
= 0.725 + 10.32 = 11.045 N/em !. A il S.
Capillarity. Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a
small!U~ relative to the adjacent gcnerallevcl of liquid when the tube is held vertically in the liquid.
The rise of liquid surface is known as capi lla ry rise while the fall o f the liquid surface is known as
capillary depression. [t is ex presscd in Te rms of (;m or mm of liquid. Its
" ..
9 1"
val ue depends upon the specific weight of the liquid. diameter of Ihe
tube and surface tension o f til.: liquid.
E)(IJress ion fur Cljllillary Risc. Consider a glass lube of small
diameter 'd' opened at ooth ends and is inserted in a liquid . .~y water.
The liquid wi ll rise in the tube aoove the level of the liquid.
L<:t II" heig ht of the liquid in the lUbe. Under a state ofeq uilihrium.
: LIQUID
the weigh t of liquid of height II is balanced by the force at the surface of
The liquid in the tube. But the force at (he surface of the liquid in Ihe
F ig. 1.13 Capjflary rilc.
tube is due to surface Tensioll .
1.6 .4
,
Let
(J
= Surface tension of liquid
9" Ang le of contact betwee n liquid and glass tu be.
The we ight of liquid of heighT II in the tube" (Area of tube x III x p x g
I I
Ii
~ I
IL
126
Fluid Mechanics
=~,PxlJxpxg
.(1.1 7)
4
where p '" Density of liquid
Vertil:al \:ump<)llcm of the
surf~cc
tensile fuTl'C
'" (0 x Circumference) x cos 0
=crx/tdxcosO
•..( 1. 1H)
For equi librium. equating (1.17) and (1.18), we get
It
,Pxllxpxg=axlf.I/xcosO
4
crXltl{XCOSi}
II '" ";-"'''--'-'=-O
Il
,
4 d - xpxg
4 crL'OsO
.. (1. 19)
pxgxd
The value of 0 betwee n water and clean g lass tube is approximately equa l to zero and hence cos
equal to unity. Tlicn ri se of water is given by
II =
e is
40
---''''--c
... ( 1.20)
pxgX('
EXll rcss ion for f:a llili llry Fall . Iflhe g lass lube is dipped in mercury. the level of mercury in Ihe lube
wi ll be lower than the general level of the olllsidc liquid as shown in Fig. 1.14.
U:i II '" Height of depression in tube.
Theil in equilibrium. twO forces arc a<:ling un the mercury inside the tuhc. First one is due to surf~ce
lension ~cting in Ihe downward direction ,md is eqn~1 to (I x 1((/ x cos a.
SCl"Ond force is due to hydrostatic fo rce acting upward and is equal 10 inlensity of pressure al a
deplh '11' x Area
'" P X
1(
,
1( ,
IF '" pg )( h )( - d" 1 .. P '" pgll)
4
4
-
Equal ing the IWO. we gel
•
O"X1(dxcos9=pghx
II
Value of 9 for mercury and
= -'4cO':"eOc'~'
pgd
gl 3 s~
lube is 128".
... ( 1.21)
"'
,"
¥ ,"\'
\
MERCURY
Fig. 1.1 4
Problem 1 .26 C{I/m/III,. Ihl' copillllry riS/' ill (I glass /UbI' of 2.5111111 dimlll'lrr ",hell illllll,.rsed
\'f'rlicoll y itl (II) wllil'rlmd (b) mff("w, '. Take surfllce lellsiollS 0'" 0.0725 Nlm for II"O/,.r IIl1d 0 =- 0.52 Nlm
for ml'rmry ill ("011/(/("1 lI"illl lIir. Ti,e 51'''(" iji(" g "wiry for mnmry is gil"<'l1 1/5 13.6 mill IIl1g/e of ("011/1/("1
= 130".
Solution. Given:
Dia. of tu be.
,/ '" 2.5 mill '" 2.5)( 10- 3 111
Surface lension. 0" for water
= 0.0725 N/Ill
fo r mercury
Sp. gr. of me rcury
0"
~ I
= 0.52 N/m
=- 13.6
I~
~ I
IL
Prope rties of Fluids
271
'" 13.6 x 1000 kglln ·1.
Density
«(I) Capillary ris(' ror water (9 '" O ~)
Using equa1ion (1.20), we gd II =
40pXgxd
'"
x 0.0725
'"'=--'''''''''''''-___
=,
lOOOx9.8 1x 25xlO l
4
'" .0 118 m = 1.18 em. Ans.
(b) For mereur}'
Angle of conlact between ulcrcury and glass lu be, 9 ", 130
Q
4a cos6
4 x0.52xcosI30"
= ,,-;-~",,"''''Ci-C':-';'''';-;;0
pxgxd 13.6xlOOOx9.Rlx2.Sx lO
= -.004 III = - 0.4 em. Ail S.
The negalive sign ind icmcs the capill:lry dcpn:ssion.
Usmg equation ( I 2 1), we gel Ii '"
Problem 1.29
ill (i)
ill " glaH IlIbl' of 4 1111/1 ilialllf'/a, 1>'11('11
(ii) III"'CIO,)'. Till' /('IIIIN'wlI>rl' ofll,rlilfllill is 20 0 IIIU/Illt' I'll/III'S a/llw
slIrf{/("(' (rllsioll Ofll"(l/fr I/m/ IIII'rCllry til 20"(" ill ('Oll/ac/ w;lh lIir (Iff 0.073575 NIIli (llIti 0.51 Nlm
rnpl'Cliw'/y. Tltf augl" of rOIl{(j(,/ for 1\'(1/'" is :I'rO 11/1(/1/1(1/ for mnTury i~ 130°. Tllia' drllsity of ....me r
m 20°C liS rqUIII to 998 kgll,,-'.
illlllll'r....d
CalC/,/(I/(' III,. Cflpill<ll)' ''ifni ill mil/iII/Nfl'S
e
lI'aln, (///(/
Solution. Given:
Dia. of tube.
(1=4m m ",4x 10- 3 11\
The cap illary effect (i.e .• capillary rise or depressio n) is given by equation ( 1.20) as
where
11= 4(J oosa
pxgxd
(J == surfucc tension in N/m
a"" anglc o f c()nt:lct. and p "" density
(i) ClI l)illary efTect for wattr
(J = 0.073575 Nlm. e '" 0 "
p '" 998 ~g/m '\ 31 20°C
I,,,, 4 x 0.073575 x cos 0° ==7.5 1 x 10
998 x 9.8 1 x 4 x 10- 3
J
m ",7.51 mm. Ans.
(ii) atl)illary effect for mt'rcury
(J '"
0.51 Nlm. e", 130° and
p '" sp. gr. x 1000 '" 13.6 x 1000 '" 13600 kg/m 2
The
n~gative
sign
indica t~s
4x05lxcos130o
It '" C-;Oc;:'=;:';;;~C'=C"'"
13600x9.8lx4xIO
the capi llary depression.
'" - 2.46 x 10-
3
m '" - 2.46 mm. AilS.
Problem 1 .30 T/u' rapilhlry riu in II", glass llIj,r i .• ,wi /() ".H·rN/ 0.2 mm "fwlUrr. DNrrm;'IP
milli",,,,,, si~". gil'''n that sIIrface 1""Si,," for wain ill umtart with air'" 0.0725 Nlm.
il~
Solution. (j iven :
Capi lI ary rise.
Surface tension.
I I
I,,,, 0.2 mm '" 0.2 x 103 III
(J '" 0.0725 Nlm
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IL
128
Fluid Mechanics
" ,/
Let dia. of tube
The angle e for water
" 0"
Dt:nsily (1') for water
" 1000 kg/Ill)
Using equatiun (1.20). we get
40
="=
I," p xg x ii
or 0.2 x 10-) "
:;;~4iXCOc·,0;.72C5
, ':-,
lOOO x9.81xd
4"X,::°C·O~7~2C5c::",J ,,0.1 48111 == 14.S em. Ans.
(/= -;:",c
.
IOOOx 9.8 1x.2 xl O
Th u ~
minimum diameter o f the lube shou ld be 14.8 em .
Problem 1.31
Filld U!/f II",
minimu",
S;:;I'
of glllH
IIlI' l:" l'illary riu iu 1/'" IIIlu' is W he rPslril'tt'd 10
wilh nir a~ O.U7J575 NIl/!.
Solution. Given:
Capi Ilary risco
Surface tension.
if
II = 2.0 lllm = 2.0 x 10-) 111
a = 0 .073575 Nfm
Lei dia. of tube
The angle
/UbI' I/UII "'/II hi' U$nllO IIINuur" II'mn Il'wl
2 111m, COl!sitll'y .lllr/ucr 11'11S;"" "iw{l/p r in ,'O/lUwl
=d
e for wa ter
au
==
The densi ty for wakr,
p = 1000 kg/m l
Using cqualioll ( 1.20), we get
40
or 2.0 x 10- 1 =
,,= .,-,,"'--;px g x d
4 x 0.073575
if= -;-~""':",'="-''c~
lOOOx 9.8 l x2xlO J
Thus minimum diameter of the
l u~
shou ld
~
4 x 0.073575
lOOO x 9.8 1 x d
= 0.015 III = 1.5 em. AilS.
1.5 ern.
Problem 1.32 All oil of l'is ("Osil), 5 l",i3l' is l.s<'(1 for IIIU,i",lioll Ul'lln'l'lI II slw/l wltl £1,,<'1'1'. Tilt'
diwlI"I'" of III<' $1111/1 i:l" 0.5 m lIml il 'Ol/lles m 2()() '.p.m. ClI/cH/mp IIII' po,,"'" losl ill oil for (/ sll'el'e
Ip1IgIII of I()() """. Tile IIIirk1l1'S.1 of oil jil'" is 1.0 "''''.
Solution. Given:
Vi~osi ty.
~
= 5 poise
Dia. of shan.
Speed o f shan.
S I ~e\'c lengt h.
/. =
I{)() rnm: I{)() X IO- J III : 0.1 III
Thick ness of oil film .
I : 1.0mrn: I X IO- J m
Tangc mi a l veloc ity of shan.
I I : 11:
Using the re lation.
I I
2.
= 0.5 N s/rnl
10
D=O.5 rn
N = 200 r,p.m.
=
t:
DN :
60
II x 05 x200
60
= 5.235
Ill/S
""
~ -
Ii,
Ii
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IL
Properties of Fluids
wltcrc .
29 1
ClilIngc of velocity = 11 - 0 = Il = 5.235 tnls
dy'" Cllange of distance = I = 1 x 10-3 III
dll '"
05 5.235
'" """""i~
=
x
x
1 10
This is lhe
sh~ar Slre.<i.~
J
26175 N/m"
on the shaFt
Shear forct;! o n [he shaft. F = Shear stress x Area = 2617.5 x nO x I.
= 2617.5 x It x 0.5 x 0.1 = 410.95 N
Area = nO xL)
D
0.5
= 4 10.95 x - = 102.74 Nm
2
,
T= FOl"<:c x -
Torque on th e sitar!.
Power* lost'" T x ro Walts = T x
2rr N
60
w
,
2n x 200
= 10_.74 x
60
= 2 150 W == 2. IS kW.
.. 1.7
(":
AilS •
VAPOUR PRESSURE AND CAVITATION
A cbange from Ih e liquid slale to lhe gaseous Slale is know n as vaporiza tio n. The vaporization
(whkl! depends upon lhe prev ail ing pressure and temperature C{)udition) occurs because of continuous
escapinll of lhe molecules lh rough llie free liquid surfac".
Consider a liquid (say waler) whic h is confined in a c losed vesse l. Lellhe temperature of liquid is
20°C and pre.'\S urc i~ atmospheric. This liquid will "aporise at JOO°c. When vaporization lakes place.
Ihe molecules escapes from Ihe free surface of Ihe liquid. These vapour molecules ge t accumulated in
Ihe space between th e free liquid s urface and top of the vessel. These accumula ted vapou rs exen a
pressure on Ihe liquid su rface. This pressure is know n 3S vapo u r press u ~ of the liqu id or This is Ihe
pressure al wh ich Ihe li quid is convened into vapours.
Agai n \'onside r the same liqu id al 20°C at almosphe ric pressure in the c losed vessel. If the pressur"
above th e liquid surface is reduced by so me means. the boi ling Temperature will also reduce. If Ihe
pressure is reduced to such an exten t that it bccotnes equal to or less than the vapour pressure. Ihe
boil in g of the liqu id wi ll stan. Though th e temperature of the liquid is 20°e. Thus a liquid may boil
even at onlinary lcnlperature. if the pressure above the liquid surface is reduced ,' \0 as to h.: equal or
less than the vapou r pressur~ of the liquid m that tempe rature.
Now consid~r a flowing liquid in a syste m. If The pressure at any point in this flowing liqu id becomes
equal to or less Ihan the vapour pressu re, th~ vapori za tion of Ih e liquid starts. The bubbles of these
vapouT5 arc carr ied by the flowing liquid into th e region of high pressure where they collapse. giving
risc to high impact pressure. The pressure developed by the collapsing bubbles is so high that the
material from The adjoining boundaries geTS eroded and c aviTi es are formed on them. This phenomenon
is kno wn as ( " ,·lIalioll.
Hence the cav itat ion is the phenomenon of fornwtion of vapo ur bubbles of a flowing liquid in a
region where Ihe pressure of the liquid falls be low the vapour press ure and sudde n co llapsing of Ihese
vapour bubbles in ~ region o f higher pressure. When Ih e v;lpour bubbles collnpsc, n very high pressure
is created. The metallic surfaces. above which the liquid is flowing. is subjeCTed to these high pressures.
which c~usc pilling action on the surfnce. Thus cavi ti es arc fonned on th~ metallic surface ~nd hence
the name is cav itati on.
• Power in case of S.l. Unit
I I
~
,
2rtNr
2rr.fff
60
60.000
.
1 X (0 or - - Watts or - - - kW. The angular wloclty
(0 ~
2nN
-60 .
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130
Fluid Mechanics
HIGHLIGHTS
I . The weight density Or sp<.>eific weight of a fluid is equal [0 weight per unit volume. It is also cqualto,
w"'pxg.
2. Specific volume is the reciprocal of mass density.
,I, The shear stress is proponiOllal 10 the velocity gradient
""
"'
4 . Kinematic viscosity V is giwn by V ~ ~ .
P
5. Poise and sto~es are the unil .• of viscosity and kinematic viscosity respccti,-cly.
6. To convert the unil of viscosity from poise to MKS units. poise should be divided by 98.1 and 10 convert
poise i!l1o SI unils. the poise should be dividcU by 10.51 unit of "iscosity i< Nslm l or Pa s, where Nlm l
.. Pa '" Pascal.
7. For a perfect gas. the eq ualion of state is
!!... = RT
P
.
kg f-Ill
where R .. gas conSI:m! :md for aIr .. 29.3 - - - ., 287 Jlkg oK.
kgo K
8. For isolhcmlal process,
9.
Bul~
J!.. '" Constant whereas for adiabatic process,
modulus of elasticity is given as K
~
- lip
(":r
10. Compressibility is the reciprocal of bulk modulus of elasticity or"
II . Surface tension is
.. constanl.
p
c~presscd
~.
in N/m or dyne/em. The relation between surface tension (0 ) and difference
of prcssurc (p) betwcen the inside and outside of a liquid drop is givcn as p _
80
For a S(lap hubble.
p" ,/
For a liquid jet.
P~d '
40
"
20
02. Capillary risc or fall of a liquid is given by /, m
40eos9
"'''''!-'
""
The value of a for water is taken equal to zero and for mercury equal to 128°.
EXERCISE
(A) THEORETICAL PROBLEMS
I. Define the following fluid propenie, :
Density. weight density. specific volume and specific gravity of a fluid.
2. Differentiate between: (i) Liquids and gases . (ii ) Real fluids and ideal fluids. (iii) Specific weight and
specific volume of a fluid .
•1. What is the difference between dynamic viscosity and kinematic viscosity ? State their units of
mcasuremcnlS.
I I
Ii
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Properties of Fluids
311
4 . Explain the len". : (i) Dynamic vi>cosily. and (ii) Kinema!ic viscosity. Give their dimensions.
5. State the Newton's law of viscosity and Kjve examples of its application.
6 . Enunciale Newton's law of ,·;""osi1y. Explain the importance of viscosity in n uid motion. W hat ;s the
effect of temperature on viscosity of water and that of air?
7. Define Newl<mian and Non-Newtonian nuids
8. Whal do you understand by tenns : (I) lsothennal process, (ii) Adiabalic process. and (iii) Universal-gas
constant
9. Define compressibility. I'ro,'c that compressibility for a perfect !las undergoing isothcnnal compression is
,
-
while for a perfect gas undergoing isenlropi<" compression is - .
"V
p
Ill . Define surface tension . l>rovc thallhc reblionship between surface tension and pressure inside a droplet of
40
liquid in exee .. of outside pressure is given by p '" -
"
.
II . EXplain the phenomenon of capillarity. Obtain an expression for capillary rise of a liquid.
12. (1/) Distinguish betwcen ideal fluids and real fluids. Explain thc importance of compressibility in fluid
flow.
(b) Define the t'nTIS , density. specific volume, specific gravity. va.uum p""sure, mmp""sible and
incompressible fluids.
(R.G.P. Vishw(lI"i(/Y(l/a)"(I. IJlwp(l/ S 200Z)
13 . Deflne and explain Newton 's law of viscosity.
14. Conver! I kgls-m dynamic viscosity in poise.
15. Why does the viscosity of a gas increases with the increase in temperature while that of a liquid decreases
with increase in temperature ?
16. (a ) ~Iow doe, viscosity of a fluid "ary with temperature "
(b) Cite examples where surface tension effects playa prominent role. (J.N.TV.. Ifyilaahad S ZOOZ)
17 . (i) Develop the expression for the relation between gauge pressure P inside a droplet of liquid and the
surface tension.
(ii) Explai n the following:
Newtonian and Non-Newtonian fluids . vapour pressure. and mmpressibility.
(R.G.P. V.. Bhopal S ZOOI)
(B) NUMERICAL PROBLEMS
1. One litre of crude oil weigh' 9.6 N. Calcu late ilS specific weight. den,ity and .<pccifi. gravity.
[Ans. 9600 N/ml. 978.6 kgltn l • 0.9781
3
2. The "elocity distribution for flow oVer a flat plate i. gi"en by II ~
where u is the point
'2 Y_111.
3.
4.
5..
6.
I I
"elocity in metre per second :n a dislan'"" y metre ab<we the plate. Detennine the shear stress at y ~ ') em.
Assume dynamic viscosity as 8 poise.
(No8pur Unil'ersity) [Ans. 0.839 N/m' l
A plate 0.025 mm distant from a fixed plate. mO"cS at 50 clllis and requires a force of 1.471 NIm'to
maintain this ~peed. Detennine the fluid viscosity between the plates in the poise.
[Ans. 7.357 X 10' 1
Dctennine the intensity of shear of an oil having viseosity • 1.2 poise and is used for lubrication in the
clearance between a 10 em diameter shaft and its journal bearing. The cleamnec is 1.0 nun and shaft
rotates at 200 r.p.lll.
IAns. 125.56 Nlm' l
Two plates arc placed at a distance of 0 .15 mill apan. The lower plate is fixed while the upper plate having
surface area 1.0 Ill' is pulled at 0.3 m/s. Find the foree and power rC<:[uired to mai1l1ain this speed. if the
fluid >wpamting them is havin,!! visrosity 1.5 poise.
[An s. 300 N. 89.8 W I
An oil film of thickness 1.5 nun is used for lubrication between a square plate of size 0.9 m x 0.9111 :md an
incli ned plane having an angle of indination 20~. The weight of thc square is 3'n.4 Nand it .• lides down
the plane with a unifonn velocity of 0.2 mi•. Find the dynamic viscosity of the oil . IAn .•. 12 .42 poisel
Ii
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132
Fluid Mechanics
7. In a stream of glycerine in 1I101ion. at a certain point the "ciocily gradient is 0.25 metre pcr sec pcr melre.
The mas~ density of fluid is 1268.4 kg pcr cubic melre UTl d kinematic viscosity is 6.30 x 10 • square metre
pcr St-><:ond, Calculate lhe shear StreSS al1he poin!.
[An •• 0.2 NIm' 1
8 . Find the kinematic viscosity of an oil having density 980 kg/m' "hen al a certain point in Ihc oil. Ihe shear
stress is 0.25 Nfm' and velocity gradie1ll is 0.3/5.
[
Ans. O,IXlO849 rn' or 8.49
SlOkCS]
"'
'I. Determine the spttitic graYi1}, of a fluid h,wing viscosily 0,07 poise and kinematic viscosity 0.042 stokes.
IAns. 1.667[
10. DClcmlinc 1he viscosily of a liquid having kincmalic ";seosily 6 slokes and specific gravity 2.0.
[Ans. 11.99 poise]
II . If the velocity distribution of a fluid over a plate is gi,'en by u ~ (3/4) Y -;. where u is the velocity in melre
per scwnd al a dislance of )' melres above the plale. delennine Ihe shear slress at y • 0,15 metre. Take
[A"".. 3.825 X 10--1 kgJlm' j
dynall1ic "is:cosity of the fluid as 8.5 X 10 3 kg_sec/m Z,
12 . An oil of viscosity 5 poise is used for lubrication belween a shaft and slecve. The diametcr of shaft is
0.5 m and it rotates at 200 r.p.m. Calculale the power losl in Ihe oil for a sk'"CYe lenglh of 100 nnn, The
thickness of the oil fihn is 1.0 mm.
jAn s.. 2, 15 kWI
,
U . The ,'elocity disuibulion over a plate is gi"en by u ~ ~ y - ; in Which u is the "elocily in mlsee at a
distance of y m above the plaTe. [kterminc the shear stress at y., 0, 0.1 and 0.2 m. Take ).l = 6 poise.
IAn s.. 0,4 .. 0.028 and 0, 159 N/ml j
14. In 'luestion 13.. find Ihe dislance in melres above Ihe plate .. al which Ihe shear stress is zero.
IAns .. 0.333 ml
15 .. The ,'elocily profile of a viscous fluid m'er a plate is parabolic wilh venex 20 cm from Ihe plate. where Ihe
velocity is 120 cm/s. Calculale the "clocity gradic111 and shear stress at distances of O. 5 and 15 cm from the
plate. given the viscosity of the fluid = 6 poise. IArrs. I2Is. 7.18 N/m': 9/s. 5.385 N/m ' : 3/., 1,7<)5 Nfm' j
16. The weight of a gas is given as 17.658 N/m l at 30°C and at an absolule pre>sure of 29.43 Niem', \)elcr. J. 8k S 539.55N ml
kgo K
[ A ns' --;;;-r-'
mine Ihe gas constant and also Ihe density of the gas,
17 . A cylinder of 0.9 Ill} in volume conlains air at O°C and 39,24 NI~rn l absolule pressure. The air is
cornpres.""d 10 0045 ml . Find (i) Ihe pressure inside the cylinder a~suming isothemwl process.
(ii) pre,sure and lempem ture assumins mJiabalic proce<s. Take k ~ I A for air.
(Ans. (i) 78.48 N/em ' • (ii) 103.5 Ntm!. 140~CI
18 . Calculate the pre>.~ure exe" ed by 4 kg lna>< of nitrogen gas at a lemperature of IYC if Ihe volume is 0.35 ml .
Molccular weight of nitrogen is 28.
[Ans. 97.8 Nlem' j
19 . The pressure of a liquid is increa«ed from 60 Nlem' 10 100 Nlcm' and volume decreases hy 0.2 per cenL
Detennine the bulk modulus ofelasticily.
IAns. 2 X 10" Nleml l
20. DCle"nine Ihe bulk modulus of elaslicity of a fluid which is compressed in a cylinder from a volume of
0.009 ml at 70 Nlcm l pressure to a volume of 0.0085 m l at 270 Nlcm l pre .. ure . [An s. 3.6 X 10' N/cm l l
2 1. The surface lension of watcr in contact with air at 20"C is gh'cn as 0.0716 N/m. The pressure inside a
droplet of waler is 10 be 0.0147 N/em l grealer than the outside pressure. calculate Ihe diameter of the
<lroplelof waler ,
IAns. ) ,94 mml
22 . Find Ihe surface Icnsion in a SD.1p bubble of 30 mm diameter wheTI the inside pressure is 1.962 N/n/ above
almosphere,
IAns. 0.00735 N/ml
23. The surface tension of Waler in eont.1C1 with air is given as 0.0725 Ntm. The pressure oUlside the droplel of
water of diameter 0,02 mm is atmospheric [10,32
water.
I I
-!::,) ,
em"
Calcu lale the pressure "'ithin the droplct of
[An s. 11.77 Nlcm' l
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33 1
Prope rties of Fluids
24 . Calculate the capillary risc in:l glass tube of 3 .0 nlln diameter when hnll1cr>ed vcnically in (,,) waler. and
(1)) mercury. Take surface len.ions for mercury and water as 0.0725 N/m and 0.52 Nirn respectively in
contact with air. Spcdfic gravity for mercury is given as 13.6.
IAn •. 0.966 em. 0.3275 eml
25. The ~apillary risc in the glass lube used for mc"suring waler level is not (0 exceed 0.5 min. Dctennin" ils
minimum size. given thaI surface lension for water in conlae! wilh air,. 0,07112 Nlm.
[Ans. 5.8 em]
26 . (SI LTnits). One lilre of crude oil weighs 9 .6 N. Calculmc its specifIc weight. density and specific gravity.
[Ans. %00 Nlm'; 979.6 kg/ln ' : 0.971\61
27. (SI "nits). A piston 796 mill diameter and 200 nun IonS works in a cylinder of 800 mill diameter. If the
annular space i. filled with a lubricating oil of viscosity 5 cp (c.·nti -poi,.,), cakulate the speed of de>cem
of the piston in vertical po~jtion. The weight of the piston and 3 .• iaII03d are 9.1\1 N.
[,\ns. 7.84 m/s[
2!1, (SI lTnit~). Find the capillary risc of water in a wtlc 0.03 em diameter. The surface tension of water is
0.0735 Nlm.
[A tls. 9.99 eml
2', Calculate the specific weight. density and specific gravity of two litres of a liquid which weight 15 N.
[AilS, 7500 N/m l . 764 .5 kg/mI. 0.764 1
30. A I SO mm diameter vertical cylinder rotates concentrically inside another cylinder of diameter lSI mm.
Both the cylinders are of 250 mm height. The space between the cyl inders is filled with ~ liquid of viscos·
ity 10 poise. Detennine the torque required to rotate thc inner cylinder at 100 T.p.m. IAns. 13.S7 Nml
.~ 1 . A Shaft of diameter 120 nlln is rotating inside a journal bearing of diameter 122 mm at a speed of 360 r.p.m.
The ']XIce between the Shaft and the bearins is f,lIed with a lubricating oil of viscusity 6 poise. Find the
[Ans. 115.73 WI
power absorbed in oil if the length of bearing is 100 mm.
.\2. A shaft of diameter 100 mm is rotating inside a journal bearing of diameter 102 mm at a space of J60
r,p ,m. The space between the shafl and bearing is filled with a lubricating oil of viscosity 5 poisc. The
[Ans. 111.58 W[
length of the t1c:.ring is 200 mm. Find the power absortlcd in the lubricating oil.
.\,\ . Assumi ng that the bu lk modulus of elasticity of water is 2.07 x 106 kN/m l at standard atmospheric
conditions. detcnnine the increase of pressure necessary to produce I % reduction in volume at the same
tcmperature.
[l1int . K
~
•
,
2.07 x 10 kNlm
Increase in pressure (dl')"
-dV
;~
0 -
1
100
.0.01.
Kx ( - ~V) ~ 2.07 X10" x 0.01 = 2.07 x 10' kNhn 1.]
.\.4. A square plate of size 1 m x 1m and weighing 350 N slides down an inelined plall e with a unifonn velocily
of 1.5 mls. The inclined plane is laid on a slope of 5 venicalto 12 horizonlnl and has an oil film of 1 mm
thickness. Caiculutc the d)l11amic viscosity of oil .
[J.tV. T.V., lIy,lembm/. S 2002[
'
5 =Be
[ Ihnl.
A= I x I ~ 1m l . W_3SON.u", 1.5m/s. tan 0 .. 12 AB
Component o f weight along the plane = W x sin (I
3C _ _'
where sin (I .,
AC
[, ..
13
I' = W sin (I .. 350 x
Now
r.c:o-:,
.. JI2~+52 . 13
1
A
"
12
W
~ '"' 134.615
8
E
350N
Fig. 1.15
13
,,"dy whcre du _ u_O = u = 1.5 mlsandd.v = 1 mm= I x 10
T=).I~.
:. 1-1
I I
AC"'JAB1+BC~
c
' m
F dy 134.615 IxlO '
Ns
= -x - =
x '.";"i--= 0.0897 -:::T" 0.S97 poisei
A
du
1.5
ttl
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Ii
.. 2 . 1
fLUID PRESSURE AT A POINT
Consider a sma ll area liA in large mass of fluid. If tile fluid is stationary. tllen tile force exerted by
the surrounding fluid on tile area IiA will always be P'Crpcndicular to the su rface (LA. Let ,IF is tile force
<IF
acting on the area dA in the normal direction. Tllen the ratio of -
dA
pressure or simply pressure and this ratio is reprcsented by p. Hence
point in a fl uid at rest is
p =
is known as the intensity of
math~lllatically th~
pressure at a
dA
If the forl"C (1-1 is uniformly distributed ove r the area (A). then pressure at any point is given by
F
p
Force
Area
= -A =""'''-
Force or pressure force. F = I' x A.
The unils of pressure ar~: (i) kgffm" and kgffclll 2 in MK5 units, (i,) N~wtonfrl1 lOr Nfm l and
Nfnl1n 2 in 51 units. Nfm 2 is known as Pascal and is represented by Pa. Other common ly used units of
pressure arc :
kilo pascal = 1000 Nfm ~
'
100 kPa '" 10 Nflll-.
,
.. 2.2
PASCAL'S LAW
II states that the pressure or intensity of pressure at a point
in a .'\Iatic fluid is equal in all directions. This is proved as :
Tile flu id clement is of very small dimensions i.P .. l/X. tly
andl/$.
Conside r an arbitrary fluid clement of wedge shaP'C in a
fluid mass al rest as show n in Fig. 2.1. Let the width of tile
cle ment P'Crpendicular to the plane ofpaP'Cr is unity and p~.
P. · dy .,
>0,
•
,
p) .L\)(.\
Fi g. 2. 1 ForcN on a flll id dement.
35
I I
Ii
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136
Fluid Ml>chanics
f\. and fI: arc the pressufCs or inlcilsity o f pressure acting on Ihe face AB. AC and Be respectively. Let
LABe",
a. 1'lIcll
th e fo rces acting Oil the clc ln elll arc:
L Pressure forces nonnallo the surfaces. and
2. Weight of c lemen t in Ihe vcnical direction.
The forces 011 the faces arc:
Force on the face All
'" fI, x Area offacc AlJ
""p" xdyX I
Simi larly force on Ihe face AC '" p,. x d.f X I
Force on the face UC
'" fI , x ds x 1
Weighl of clement
'" (M ass of element) x g
'" (Volume x p) xg '"
(AB; AC x I) x px g.
where p '" density of fluid.
Resol ving Ihe forces in .r-direction. we lI ave
p, xdyx I - p(dsx I )si n (90 Q - 9) "0
p" xdyx l - f!: d~' X 1 cosO '" O.
0'
BUl from Fig. 2. 1.
ds cos e '" All "" lIy
p, xdyx l - p,- xdyx I "0
...(2.1 )
P."'Pz
or
Similarly. resolving the forces in y-dircction. we get
p,Xdxx l-p, XtlJ'X I
cos(900 _e) _~dc·'~'cdCY'xl
2
XpXg=O
tI_(dy
P-, x dx - P, tis s in e - - , - x p x g = O.
But ds sin e = d.l :md also the elemem is very small and hence weight is negligible.
P/u--p , xdx=O
or
P. =P,
From equ:uiOlls (2 I) alld (2.2). we have
..(2 _2)
... (2.3)
p" = p" = P,
The above equation s hows that the pressure aI any po int in x.)' and Z directions is equal.
Since the choice of fluid element was co mple tely arbitrary. wh ich means the pressure aI any poim is
the same in all direc tions.
~
2 .3
PRESSURE VARIATION IN A FLU ID AT REST
The pressure at :1I1y poin! in a fluid at rest is obtained by the Hydrostntic Law which slates thnt the rate of increase of pressure in a ,"e n ically dow ll ward direction must be ~-qu al to the SJXocifie weight of the
fluid at Ih at point. This is prt)l'ed as:
Consider a small fl uid cleme n! as shown in Fig. 2.2
Let M= Cross-sectional area of element
~Z'" Height of fluid c lement
P '" rres~ure on face AS
Z", Distance of fluid elemellt from free surface.
The forces acting on tile fluid clement are:
I I
ri::::-
{ REE SURFACE OF FLU ID
------------- -- ---- J ---- ---
:::::::::::::~
' ~:(::::_
B .,--_-_-_-_-_-_-_-------------_"_-_-_-_-_-_ ·tJZ
_-_-_-_
-_-_-_
-_-_-_-_-_- A -
------ --------:-:-:-:-:-:'
? -ilp-£t:-::-:-:-:-:-:
---------- ----------.{ p+
-_-_-_-.-_" -:k
Fig. 2.2
tJZ)M.--------___-_-.-_-_-
Forces on a fluid e!emeT/l.
Ii
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Pressure ilnd its Mea surement
37 1
L Pressure forc e on AB '" P x All. and acting perpendicu lar to face Ali in the downward dircClioli.
2. Pressure force on CD =(1' +
~~ llZ) x dA. actin g perpendicular 10 face CD. w rti cally upward
direction.
3 . Weight of fluid c lement = Density x g )( Vo lume = p)( g x (All. x AZ).
4. Pressure forces on surfaces IJC and AD arc equa l and opposite. For equilibriu m o f nuid
c le men!. we have
pM -
(1'+ ~~IlZ) M
I'M - I'M -
0'
~
+ pxgx (M
x~
=0
1l7..&1. +pxgxM xZ =0
-~~~+PXgX.MllZ=O
ap = p x II [cancelling MAZ on both
:~IlZ6A = p Xg xMIlZ or
Jz
"I'
aZ
... (2 ...1 )
(':pXg=w)
=PXg=w
sid~sl
where w = Weigh! density of fluid.
Elj uati on (2.4) Slat es Ihal rate of im;rcasc of pressure in a vert ic al
density of the fluid at that poin t. This is Hydrostatil: La w.
By integrating tile above equation (2.4) for liquids. we ge t
dir~'(; tjon
is equa l to we ight
{dp={pgdZ
= pgZ
...(2.5)
whe re II is the pressure iIOO\'e mmosphcrie pressure imd Z is the height uf the point frum free
surfaces.
p
From equation (2.5). we have
Z'" _ 1_'-
px,
..•(2 .6)
Here Z is called ]Irl'ssurl' hl'ad .
Problem 2.1 A hydrmdie prt'ss lUIS /I W/II of 30 elll dilllnt'lrr mul a pillnger of 4.5 e/ll dillllll"lrr. Find
Ihe !>'t'ighl fijlt'd by lilt' h)'llftllllie prt'JJ ",ht'll Iht' foret' applinl (1/ lilt' plllllger is 500 N.
Solulion. Give n :
Dia. of ram.
/) '" 30 em = 0.3m
Dia. of plunger.
d", 4.5 em = 0.045 m
Force on plunger.
Find weight lifted
"'",5OON
"W
Area of ram.
Are a of plunger.
I I
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138
Fluid Ml>chanics
Pressure il1! cnsity due to plun ge r
'"
Force o n pl un ger
A rea of plun ger
F
500
2
= - = - - - N/m
.0 0 159
(l
PLU NGER
Due to Pasc al's law . the inte ns it y o f pressu re will be
eq ua ll y trans mi u ed in a ll di rcl:1 io ns. He nce the pressure
inte ns it y at the ram
500
::: 3 14465.4 Nfrn 1
.00 159
: -:-_W""";'~'hC'_ '" ~ '"
But pressure intensit y a! ram
Area of ram
w
.07068
A
•
Fig. 2.3
_'_V_ N/m 1
.07068
= 3 14465.4
We ight
= 3 [4465.4 x.07068 = 22222 N '" 22.222 kt'l. Ans .
Problem 2.2 A Il)'liralllir pre SJ bas (/ ralll of 20 011 ,Iimll" fPr (11111 (I pi IIl1ga of 3 Til! (liw/lFler. It is
usn i for liftillg (/ wi'iglll of 30 liN. Pilld lilt' fOITe rt' qllirn/ mille p/1I1I8e r.
Solution. Give n :
/) '" 20 em = 0.2
Di a. of ram.
Area of ram.
Dia. of plunger
Area of plun ge r.
Weig lll lifted.
A '"
III
~ /)2 '" ~ ( . 2) 2 '" 0.03 14 111 2
4
4
ll=3cm =O.03 m
"'" ~(.03) 2 = 7.068 X 10- 4
rn !
4
W =30 kN = 30x IOOON =30000 N.
See Fig . 2.3.
Fo rce
F
Pressure int ensi ty develo ped due 10 p lun ge r ", - - == - .
Area
"
By Pasca l" s La w. th is pressure is tralls mi ltcd equa ll y in all d irect io ns
Hence press ure tran sm itted at th e ram ==
Fo rce actin g o n ram
!...
" inte nsity X Area o f ram
= Pressure
F
= - x A ",
But force al:tin g o n ram
"
F x .03 14 N
7.l168 x 10-4
= Weig ht li fK"d '" 30000 N
30000 :
Fx .03 14
7.068 x 10-4
F: 30000 x 7.068 X 10-4
.03 14
'" 675.2 N. Ans .
Problem 2 .3 Ca/n'/(/((' f/If prnsllr(, (/II(' to a CO/UIIIII of0.3 of(n ) w(/(a. (b) (III oil ofSI'.
(c) Ilwrcur )' of sp. gr. 13.6. Til/.;(' i/('Il$ify ofw(lfa. p '" /000 kg/IIIJ.
Solution. Give n :
Height o f liq uid co lumn.
Z=O.3 m.
I I
,~ r.
0.8. (II ld
Ii
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Pressure ilnd its Measurement
39 1
The press ure al an y point in a liquid is given by equation (2.5) as
p'" pgZ
«(I) For
wal~r.
p
= 1000 kg/Ill)
I' = pgZ '" 1000 x 9.&1 x 0.3 '" 294) Nlm 2
-_ ['(j'T
294] Ntem !-_ O"94\NI
! A liS.
.
._ .
{fll.
(b) For oil of sp. gr. 0.8.
From eq uation ( 1.IA). we know lhat Ihe densit y of a fluid is eq ual to specific grav ity of Iluid
multiplied hy density of water.
Density of oil.
Now pressu re,
Po = Sp. gr. of oi l x Dens it y of wakr
= 0.8 x P '" 0.8 x HXXl '" 800 I;g/m3
(Po= [k nsi ly of oil)
1'=PoxgxZ
= 800 x 9.8 1 x 0.3 = 2354.4
~
= 23s:.4
10
Ill '
= 0.2.\54
~.
em'
~. AilS.
,m
(c) For mercury. sp. gr.
'" 13 .6
Fro m equa tion (l. IA) we know that tile den s ity of
J
fluid is eq ua l to specific g ravi ty of fluid
multiplied by densit y o f water
Densi1y of mercury.
p. = Spcl:ific grav it y of mercury x Density of water
= 13.6 x 1000 = 13600 kg/m l
P=P.xgxZ
'" 13600 x 9.81 x 0. ] " 40025
~
40025
- - , - " 4.002
10
N
--!'
em
N
~
,
111 -
Ans.
Problem 2.4 Till'" prnSlln illlnlsily (1/ (I poim ill (lfluid is gil"nl 3.924 Nlnl/. Find Iltf ("orrnpolll/illg Iti'ighl offluid ...hi'llilti' fluid i~ ; ( II) W(I(Fr. lind (b) oil OfS!" gr. 0.9.
Solullon. Given:
N
, N
Pressure imensity.
1''' ].924 - , "].924 x 10 ~,.
cm -
m-
The cU lTcsponding heigh1. Z. of 1hc fluid is givcn by eq ua1i u ll (2 .6) as
z== - " -
I'X ,
(u ) For watcr.
p " 1000 kghn·l
z== ~"
I'X,
(b) For uil. s p. gr.
:. Density of oil
lOOOx 9 .81
- 4 III uf wa ter. An s.
" 0.9
l
Po" 0.9 x 1000 " 900 kg/m·
Z
I I
3.924 x 104
__ _ ,_, _ __ 3
o~
92~4~X~'~O~'
-:;
== 4.44
Po xg
900 x9.81
11\
uf uil. Ans.
Ii
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140
Fluid Ml>chanics
Problem 2.5
All oil of sp. gr. 0.9 is
("OIlWillt'd
ill
(I
"<'asd. AI
<I
poim 11Il' heig/II of oil is 40
tIl.
Find
Ihe corrt'Sl'olUlillg ileighl ofwO/n at lil" porlll.
Solution. Given:
Sp. gr. of oil,
So = 0.9
Hciglll of oil.
Zo=40m
Po = Sp. gL of oil x Dt:nsily of walcr = 0.9 )( 1000 = 900 kglm 3
N
p = Po x g x 7-0 = 900 x 9.81 x 40 - ,
Density of oi l.
lntcnsi ly of press ure.
'"
Corresponding liclght of water'" -;::c~_cIC' _ __
De nsityo f water xg
~
9OO)(9.8 I x40
"",:C""',:,"
= 0.9 x 40 = .\6 10 of w ~tc r. Ans .
1000)(9.8 1
Problem 2.5 All 01'1'11 {(Ink romulus W(l/er IIpto (I df/NII of 2 III (md (lboI'<' il (III oil of SI'. gr. 0.9 for
d"plil of J m. Filut 1/,1' 1'("('8811'1' ill/fluiry ( jim /11i' inlf rfowl' "llIw ""0 liquills. llIui (ii) (If /Iw hOI/om
(I
of 'h" "IIlk.
Solution. Given:
Hciglll of water.
Height of oil.
Sp. gr. of oi l.
Density of water.
Density of oi l.
Pressure intensity at
~ny
ZI",2m
Z2'" 1m
So'" 0.9
PI '" lOOO kg/m '
Pl'" Sp. gr. of oil x Dens ity of water
'" 0.9 x 1000== 900 kg/m'
poim is given by
Fig. 2.4
p =pxgxz'
(i) At interface. i.l' .. at A
P"'P2 x g X1 .O
",900x9.81 x 1.0
N
8829
•
'" 8829 - , ~ - ,- = 0.8829 Nfcm -. Ans.
III
lO
(ii) At Ihe bottom. i." .• at B
l' '" P2 X gZ2 + PI X 8 X ZI = 900 x 9.81 )( 1.0 + [000 x 9.81 x 2.0
1
28449 ,
•
= 8829 + 19620 = 28449 N/1I1 = ~ Nkm' = 2.8449 Nfcm -. Ans.
Problem 2.7
TI", dilllllt'/prs of (I sm(ll/ pi.<lOII Imd II /flrgp piMO" of (I hydraulic j(lck {lfP 3 C/ll lII"l
10 cm rp,!ppclil'dy. AftJrf'P (Jf 80 N is flpplil'l{ "" II", SIII,,/l piSI"", Filll{ Ih .. 1(J(I(llijtl'l1 by II", [flrg"
pis",,, ".hpII :
( a ) Ilw piSIOllS afF ,,/ Ih .. SIIIII" In'pl .
( b) SI1I<1I1 piS/Oil is 40 elll IIbo\'(' Iltl' large l,is/OII.
The (1I'usil), of lit" liquid iu /lle jack is g;"1'1I 1M 1000 kglllr'.
Solution. Given:
Dia. of small pislOn.
Area of small
I I
pi~IOn,
(1=3cm
1'1',2
II'" -
4
(,
11
,
= - x (3)' = 7.068 e lll
4
1
Ii
~ I
IL
41 1
Pressure ilnd its Mea surement
Dia. of large piston.
IJ=
A", P )( (1 0)2 '" 7&.54 cm 1
4
Pc: 80 N
Are a o f larger piston.
Force on small piston.
,
= W.
When thl' Ilistons arc 31 thl' same It'HI
Ld the load
(a)
10 em
Prcs~urc
lin~d
~W~~~;i'G'
P ISTON
~.-.-.-.-.-.-.
in[cllsity on small piston
F
80
II
7.068
------.
.-----__ . _-.
-----_
-. -. -. -. -. -.-
This is Iransmincd equ all y on thc large piston.
Pr".'\Sure intensity on Ih" large piston
Force on the large piston
s~,
PISTON
f i g. 2.5
= -7.068
'" Pressure )( Area
80
= - - )( 78.54 N '" 8tlX.'J6 N. AilS.
7.068
(b) Whl'n the small Ilision is 40 em abo ve th(' tar~(' I)iston
,
Pressure intens ity on Ihc small piston
F
80
(I
7J168
I
Pressurc ill1cnsity at sectio n A-A
F
'" - + Pr~ssurc i ntensity due [0 height o f 40 ern of liquid.
"
But pressure intensity due to 40 cnl of liquid
=pxgxh= IOClOx 9.8 1 xOANlm 2
1000 x9.8 1 x.40
Nlt:m! = 0.1924 Nlcm!
10'
Pressure
int~nsity
w
------"
--_._--------------------------.-----------------------
at sect ion A -A
Fig. 2.6
80
= - - + 0.3924
7068
= 1l.J2 + 0.3924 == 11.71 Nlcm 1
Pressure intensity transmitted to the large pislOn = 11.71 Nkm 2
Force on the large pislOn = Pressure x Area of the large piston
=11.71 x A == 11.71 x 78.54 = 9 19.7 N .
.. 2 .4
ABSOLUTE, GAUGE , ATMOSPHERIC AND VACUUM PRESSURES
The pressure on a fluid is measured in twu different syste ms. In one system. it is measured above
the absolute zero or complete vacuum and it is called the absulute pressure and in other sys tem.
pressure is measured above the atmospheric pressure :md it is called gauge pressure. Thus:
I. AI.soluh' 1H"t'sS UH' is defined as the pressure which is measured with refe rence to absolute
va;:uu1t1 pressure.
2 . C,lIU/tC l.rCSSUrl' is defined as th e pressure whid is measured with the help uf a pressure measuring instrument. in which the atmospheric pressure is wken as datum. The atmospheric pressure on
the s<:a le is marked as zero.
I I
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142
Fluid Ml>chanics
.\ . Vacuum IIrl'ss un' is defin ed as the pres- III
sure be low the atmos phe ric pressure.
The re lati onship bctwcc ll lhc abso lut e ptcS1;urc.
g au~c pressu re and vac uum pressure arc shown in
Fig. 2.7.
Mathe mati call y:
(i) Ahso lute pressure
== Atmosph eric pressure + Gauge press orI'
•
~
1
,
- I ./
f'"'
GAUGE PRESSURE / ~TMOSPHERIC
/"
PRESSURE
~r VACUUM PRESSURE
ASSOlUTE
PRESSURE .....
II., = P.. ," + p~",V'
"' (ii) Vac uum pressure
,
ABSOLUTE ZERO PRESSURE
Fig. 2.7 Relationship bl'llllUn pr/'S!IIrt'S.
== Atmosph eric press ure - Absol ute pressure.
Not e. (;) The atmospheric pressure at sea level al 15°C is 101.3 kNlml or 10.13 Nlcm 1 in 51 unit. In case of
M KS units. it is equal to 1,033 kgf/em 1,
(ii) The atmospheric pressure head is 760 mm of mcr<;ury or 10.33 m of water.
Problem 2.8
IV/WI (Iff liff gllUgf prrsslIrf lind {/bsollllf prrs5" ff (If II poi", 3 III bflow Ilw frt'f
511 rf ll('f (If II liquhllUl\'ill/l II dnl5i1y of J.5J x /lY k,~/1Il3 if IIw UllIl05p/wrir prrssurf i5 fqllil'lIlml 10
7501111/1 ofmnrllry? 17lf Silfrijir g fll l'i1y flfllwrrlln' i5 13,6 (II1t1 dfllsily of WUlff = lOOO kg/m 3.
Solution. Give n :
Dt: ptll of liq ui d.
Dt:nsi ty of liquid.
2 , ,, J m
At mosph eric pressure head .
20
where
mO
P," 1.53 X 10' kg/m'
750 mm of Hg
750
" WOO " 0.75 In o f Hg
"
Atm osph e ric pressure.
P..on" Po x g x Zo
Po" Densi ty of Hg " Sp. gr. o f me rc ury x Density o f water " 13.6 x 1000 kg/m'
Zo" Pressure Il ead in te nn s of merc ury.
/''''m'' (13.6 x 1(00) x 9.81 x 0.75 Nfm2
(.,'
Zo" 0.75)
" 100062 Nfml
Pressure at a poin t. which is at a de pth of 3 m fro m th e free surface of the liquid is give n by.
p=plxgxZ,
,,(1.53 x 1(00) x 9.8 1 x 3" 45028 Nlm l
G au ge pressure.
Now absol ute pressure
... 2 .S
11= 45028 N'm ~. AilS.
" Gauge press ure + Atm osplle ric pressure
= 45028 + 100062 = 145090 N/m 2 . Ans .
MEASUREMENT Of PRESSURE
The pressure of a fluid is measured by [he fo ll ow in g dev ices:
I. Man o me[e rs
2. Mec han ic al Gau ges.
2.S. 1 Milnom~ten . Mano meters are defi ned as [he dev ices used for measuri ng th e pressure at
a point in a fluid by ba lancin g th e co lu mn of fl uid by tile sa me o r anothe r co lumn of the fl uid . T il ey are
d assified as :
( b) Di ffe renti al Manomders.
(tl) Simpl e M:lllometers.
I I
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IL
Pressure ilnd its Measurement
431
2. S.2 Mechanical Gauges. Mechan ic al ga uges arc defmed as th e dev ices used for me as urin g
Ih e press ure by balanc in g llie nuid co lumn by Ih e spring or dead we ig ht. The commonly used mcc hani (;al pressure g aug es arc:
(a) Di:lphrag m pressure gauge.
(b) Bourdol1 1Ubc pressure gauge,
(r) Dead-we ight press ure gauge , and
(d) Bellows pressu re gauge.
to 2.6
SIMPLE MANOMETERS
A s im ple manometer cu nsiscs of a gl ass tube having one of its e nds connected 10 a poinT where
pressure is to be measured and mh er end re main s open to atm usphe re. Co mm on types of simple ma nometers arc:
I . PiczomclCr:
2. U· lubc Man ome ter. :mll
3. Single Column Man ometer.
T
2.6. 1 Piezometer. It is Ihe s implest form of mano meter used fOf
measurin g gauge pressures. One end of this mano meter is cu nn ected to
the point where pressure is to he measured and ot her end is open to the
atmusph erc as shown in Fig. 2.8. The ri se of liquid g ivcs th e pressure
head at that point. If at a point A. th e height of liquid say wate r is II in
piezometer tu he. the n pressure at A
N
=p x 8 xlI - , .
m-
Fig. 2.8
PinomfuT.
2. 6 .2 U-tube Manometer. It cons ists o f g lass tube bent in U-shape. one e nd uf which is
cunnected to a point at whi ch press ure is to be measured and other e nd re main s upe n to th e
atmosphere as shown in Fig. 2.9. T he tuhe gene rall y contains mercury u r any oth er liquid whose
specific gravity is grea te r than th e specific gravity uf the liquid whu se press ure is to he me asured.
T
1
(al For ga uge pressure
(b) For vacuum p<8SSUf8
U·tube Manometer.
(</) .' ur (;au!;c P!"('ss urc. Let B is the point at which pressure is to he
The datum line is A-A.
ill == Height of li ght liquid above the datum line
11 2 '" He ight uf heavy liquid abo ve the datum lin e
.';1 == Sp. gr. uf li ght liquid
PI == De ns ity of li ght liquid == 1000 X.';I
.';2 == Sp. gr. uf hcavy liquid
P2 == De ns it y uf heavy liquid '" I(X)O X S2
Fig. 2.9
I I
m~asured.
whose va lue is,).
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IL
144
Fluid Ml>chanics
As th e pressure is th e same for the horizontal surface. Hence pressure above th e horizontal datum
line A -A in the Ieli co luilln and in the right co lumn of U-Iube manometer shou ld be same.
Pressure above A-A in tile left co lumn
" P + PI X g X hi
Pressure above A-A in the right column
" Pl x g x II,
p + Pig/II'" p,-g l,!
Hence equating th e two prc!iSUrCS
P" (pUS/I, - P I X g x III)'
.. .(2 .1 )
(b) ~-or V"cuum P ....SS UI'f'. For rn ~ asuri n g vacuum pressure. the k vc l of th e heavy liquid in the
mallomC lc r wi ll be as shown ill Fi g. 2.9 (b). T hen
Pressure above A-A in the left co lumn
Pressure h"ad in Ihe ri ght column arovc A-A
"P!8"2 + Plgh l + f!
" 0
P181!2 + Pig/II + II" 0
...(2.8)
fI" - (P;oKiJ 1 + Plg" I)'
Tile right limb of a ,~imple U-/Ilbe mOI/Olllet'" CQllfaillitlg mercury is ope,1 to Ille
almO.fpllere while Ihe left limb iI connected 10 a pipe in which a fluid of sp. gr. 0.9 i.f flowing. Tile
cenlre of IIII' pipe is J1 cm below IIII' lere! of mercury in IIII' riglll limb. Find IIII' pre.l.wre of fluid in
III<' pipe if IIII' difference of mercury lel'e! in Ihe two limbs is 10 cm.
Solution. G ivcn :
Sp. gr. of fluid,
5, = 0.9
Densi!y of fluid.
p , =5, x 1000=0.9x 1000 = 900 kglm 3
Problem 2.9
Sr. gr. of me rcury,
Densi ty of mercury.
5~
r
= 13.6
"
j
p~ = l3.6 x 1000 kgfmJ
Difference of mercury level,
II~ = 20 c m '" 0.2 m
Heigh! o f fluid from A-A.
h, = 20 - 12 = Scm = O.OS m
Let P = Pressure of fluid in pipe
,
Equating Ihe pressure above A-A. we gel
"'
P + p ,Sh, = P ~gl' 2
P + 900 x 9.81 x 0.08 = 13.6 x 1000 x 9.81 x .2
Fig. 2. 10
P= 13.6x IOOOx 9.81 x.2-900x9.S1 xO.OS
= 26683 - 706 = 25977 Nfm2 '" 2.597 NIemI. ADS,
Problem 2.10 A simple lj·llIbe mOl/omeler cOilwiliing mercury is connected 10 a pipe in wllich 0
fluid of sp. gr. 0.8 lIlIIflnn·ing l'lICUUIII pressure is flowing. The OIher em f of lhe mllltollieler is open 10
almosphere. Find IIII' melwm pl"l'Hllre ill pipe, if the differellce of mercury 1{"I·el ill IIII' two limbs i.f
40 cm w,d the heighl of fluid in IIII' left frolll Ihe celllre of pipe iJ 15 cm below.
SolutIon. Givcn :
Sp. gr. of fluid.
Sp. gr. of mercury.
Density of fluid.
Cknsity of mercury,
S, '" 0.8
Sl= 13.6
PI '" SOO
P2 '" 13.6 x 1000
Difference of mercury kvel. hl = 40C1O = 0.4 nl. Height of liquid in Idl limb. I"
= 15 cm = 0. 15 m. Lei Ih e pressure in pipe = 1'. Equat ing pressure above datum
line A ·A . we ge l
I I
1
,,=
T
, "-'- ,
Fig . 2.11
Ii
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IL
Pressure ilnd its Measurement
45 1
I P28i'2 + Plg h ,]
Ii '" -
= - 11 3.6 x 1000 x 9 .31 x 0.4 + 800 x 9.81 x 0 .1 5 ]
= - [53366.4 + 11 77.2 1 = - 54543.6 N/m2 '" _ 5.454 Nie m I, Ails.
Problem 2 .11 A U-Tube m(mum eler is used /0 measure Ihe preS~'''re oj ....aler in II pipe /ine, .... hiel!
is in <'... cess of IlIIrIOJ'pileric press" re. The righl lim" of tile mUllome/a COll la;ns mercury and iJ' open /0
almasp/wu:. The con/ac/ be/ween woler (lIId mercury is ill Ihe left limb. Delermine Ihe pressure of
,nUer in the main lille. if 111<' difference in leI'e! of mercury ill IiiI' limbs of V-Illbe is JO em Ulll/lhe
free surface of mercury is in Jerel willi Ihe eenlfe of rhe pipe. If Ihe preSSIIrI' of warer in pipe line is
redllced In 98 10 Nlml, cu/cu/(lie the 1Iell" difference i ll Ihe le,-e! of mercury. Sketch Ihe arrangements
ill bOlh ea.ies.
Solution. G ive n :
Diffe rence o f merc ury'" 10 em '" 0 .1 m
The ~rran ge men t is show n in Fig. 2. 11 (a )
1st I'art
u:t p). " (pressure o f wa te r in pipe line (i.e .. at po int A)
The poi nts B and C li e o n th e sam e ho ri zonta l line. Hence pressure ,11 B s hould be equ,lil o pressure
at C. Bu t pressure at B
'" PreSl;ure at A + Pressure d ue 10 10 c m (or 0. 1 m)
RIGHT LlMB _
of wate r
;~ER
"'p~ + pxgxll
where
p '" 1000 kgl m
3
and II '" 0. 1 m
({T -
'" 1'). + 1000 x 9.8 1 x 0.1
2
"/I). + 98 1 N/m
...( i)
Pressure at C", Pressure at D + I'reSl;ure due 10 10 em o f merc ury
"'O+PoxgX/lo
where Po for mercury '" 13.6 x 1000 kg/Ill}
and
110 '" 10 em '" 0. 1 111
Press ure at C", 0 + ( 13 .6 x 1000) x 9 .&1 x 0. 1
= 1334 1.6 N
...(ii)
But pressure at B is equa l to pressure at C. Hence equat ing the equati ons (il and (ii). we ge t
1'.. + 9 8 1", 13341.6
fI).::
-
-LEFT lIMB_
,
O
1
'T
C
MERCURY _
-
Fig . 2.11 (a)
13341.6 - 9 81
N
= 12360.6 -2 . An s.
'"
lin d Pa rt
Gi ve n. 1'.. = 9 8 10 N/m 1
Find ncw di ffe re nce of merc ury level. The arrangemcnt is shown in Fig. 2. 11 (b) . In this case the
pressure at A i5981 0 Nlm '"w hie h is Jess lhanlh e 12360.6 N/ml. He nce merc ury in Je fllimb will ri se.
The ri se o f merc ury i n Jeft lim b will be equ a l 10 l he fall o f mercury in righl lim b as lhe tOlal vo lu me of
merc ury re mai ns sam e.
Lel .1 '" Ri se of merc ury in left linlb in (;m
Then fall of mcr(;ury in rig ht limb ", x cm
Th e poi nts S , C and D show lhe initial cond itions wlwre as poi nts 8 *. C* and D* show lhe
fin al conditi ons.
I I
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146
or
Fluid Ml>chanics
The pressure at 8* '" Press ure at C·
Pressure at A + Pressu re duc to (10 - xl e m of WJ\cr
'" J>rcssure at J)* + Pressore due to
(10 - 2x) ern of mercury
1910 + 1000 x 9.8 1 x
i'OO
("""'10-")
..--1+"' ~
to em
(10-2.,)
",0 + (13.6 x I OOO) x 9.8 1 x
or
or
--------
-- -- - -
100
J,j ,-r
( 10 _ 2.)
11,. __ ,,- _L
~B
- -1:0,- -
Di vid illg by 9.81. we get
1000 + 100 - lOx = 1360 - 272x
272x - l0-,= 1360-1100
262x'" 260
or
2 6()
,= -'" 0.992 em
262
New diffcfCIlCC of mercury '" 10 - 2 l em =10 - 2 x 0.992
Fig. 2.11 (b)
'" IWI6 em. Ans.
U-t"""
Problem 2.12 Fig. 2.12 .\110;"$ "colliC(IIl'l'Sse/ /I(I\'illg iu 011//1'1 til A /0 wllirh fI
mlUlOllli'll'r
i.1 rOlllll'("/I'l/. H", rNII/ill8 oflhl' /II"HomNer g;,'''" ill IIIl' figllre S/Wlt'" wlwllll", ""$.W'I is ('/"1'1)'. Filld 1/'"
rl'{I{lillg "1'/'" ""IIl"III('/l'r Wlt,," Ih" w ssd i. (""",[,Indy filled ,..ith .m/('T.
Solu tion. Vcssd is elllll' )'. Given:
Difference of mercury level
Le t ill "" He igh t of water abo ve X·X
Sr . gL of mercury,
Sr . gr. of watu.
Sl = 13,6
SI
= 1.0
Density of nU'rcury.
P2 '" 13.6 x lOoo
Dcnsity o f wmc r.
PI '" 1000
Eq uatin g the pressure above datum line X-X. we have
P2 xgxil 2 ""Plxgxil l
13 .6x IOOOx9.8 1 x 0.2"" lOoox9.8 1 xiii
or
V~ssel
,
,M
I
1
I
"
1
III "" 2.72 111 of water.
is full uf wa te r. When ves.w l is full of water. the
Fi g . 2. 12
pressure in the right limb will increase and mercury leve l in the rig ht limb will gu down. Let the
distance throu gh which mercury goes down in the right limb be . ), cm as shown in Fig. 2. 13. The
mercury wi ll rise in the left hy a d ist an ce of)' Clll. Now the datum line is Z-z. Equat ing the pressure
aoove th e datum line Z·z.
Pressure in left lim b"" Pressure in righ t limb
13.6 x WOO x 9.81 x (0.2 + 2)"1100)
'" lOOOx9.8 1 x(3+II I +yI I 00)
I I
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IL
Pressure ilnd its Measurement
13.6 x (0.2 + 2y1lOO) '" (3 + 2.72 + yllOO)
2.72 + 27.2yf1OO '" 3 + 2.72 + )'/ 100
(27.2)' - y)l100 '" 3.0
26.21' = 3 x 100 = 300
300
y ~ - - = 11.45clIl
26.2
The d ifference of mc n; ury le\'e l in twO li mbs
= (20 + 2y) e m of men:ury
47 1
hi = 2.72 e rn )
-- -1
3m
1
1,
= 20 + 2 x 11.45 = 20+22.90
.J
1- -
- T
1x (2{l + 2y)cm
i - -'-
T
1",
F T'
"',
= 42.90 em of mercury
Re ading of manometer'" 42.90 em. AilS.
Problem 2.13 A pressure gmlge cOll sis /s of ","0 c)'lilUlrim/ bulbs IJ /1111/ C r(/cl, of /0 sq. em crossseclio/l11/ oretl, which ""- cOIIIIl'Cled b y /I U-lUhl' Willi va/jell/limbs ('(wl, of 0.25 sq. em cross-sUliO/wl
/lrn l. A rnl/iqllid of sl'l'cifk gral'il), 0.9 is filled ill/a C 1/1111 c/('(l f Will'" is jilin/ imo B, IIII' S II'ftirF of
Jf'/Ulrtlliu/l bring ill Ihl' limb '1I/Ilci1n{ 10 C. Filii/II/(' ili SplaCf'1II1'I1I of 1/11' sur/art' of sepllrm ;011 will'll IIII'
pressure 011 lilt' sur[net' ill C i s gr,.fllu 1IIfIllilru i illlJ by wr WlIOIIIIII'qUlliio / olr Irpm/ o[II'/II,.r.
Solution. Give n :
Ar~ a o f eac h bulb Band C,
Ar~ a
of eac h v.:rtical li mb ,
1\= IOc rn 1
(I
= 0.25 ern 1
Its d~n si l y = 900 kg/m3
= 0.9
Sp. gr. of red liquid
X·X = In ilial se parali on leve l
Lo<
11(." = Heighl of red liquid above X-X
11[1 = He ighl of watc r abo l'c X-X
Press ure above X-X in
lh~
Idllimb = 1000 x 9.81
X "B
Press ure above X-X in the ri g ht limb = 900 x 9.81 x Ire
Eq uatin g the lWO prcssure . we get
IOOO x 9 .81 xll ll ",900x9.81 Xll c
1111 = 0.9
.. ·(0
II C
lJ40
Wh en the pressure head o\'er lh e surface in C is
inc reased by I cm of wale r. let the separat ion leve l
falls by an amo unt equa l 10 Z. Th en Y- Y b<:co m ~s lh e
fi nal separat ion le l'''l.
Now fall in surfac" IeI'd of C mul tip lied by crosssec li o nal area o f bulb C mUSl be equ a l 10 the fall in
sc parallon level mullipli ed by cross-scclion al arc a o f
limb.
FI NAL.
SEPARATION
LEVEL
Fa ll in surface lel'.:1 of C
= Fa ll in separalion kl"eJ x a
A
I I
lJ40
",
- , 1:1--- -
INITIAL
SEPARATED
LEVEL
Fig . 2.14
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148
Fluid Ml>chanics
z X0.25
Z
10
40
A
Also fall ill surfncc level of C
'" Rise in surface level of H
Z
40
The pressure o f I Cln (or 0.0 I In) of water = pgh = 1000 x 9.81 x 0.0 I'" 98.1 N/m2
Consider final separation lev e l Y-Y
Pressure above Y- ¥ in Ihe Jeft limb '" [000 x 9.81
( + "e + Z)
+
!)
Z
Pressure ahove y. Y in the rig ht limb = 900 x 9 .81 ( Z
40
Ire -
+ 98.1
Equating Ihe two pressure, we get
]000 x 9.81 (Z + 11 8 +
! )'" (2 +
he -
~ ) 900 x 9.S! + 98.1
Dividing by9.81, we get
1000 (Z+h8 +
! ) '"
Dividing by 1000. we gel Z + lIy +
Btu from cqu3lion (i),
~
900 ( Z+hc -
! )+
! )+
'" 0.9 ( 2 + he -
10
0.01
"N '" 0.9 he
Z+O. 9h c + Z '" 39Z x O.9+0.9h c + O.QI
40
40
4 !Z
39
--"-x.92+.01
40
40
"'
Z(:~ - 394~·9) = .01
_
Z
=
Of
40xO.OI
z(41~t·J) = .01
= O.067H m = 6.78 ern. Ans.
5.9
2. 6 . J Single Column Manometer. Single colulnn manometer is a 1I1odificd form of a U- IUOC
manometer in wllich a r,,~ r voir. having a large cross·scctional area (about ]00 limes) as compared to
tile area of the tube is connl....:kd 10 one of tile limbs (say left limb) of the manometer a~ shown in Fig. 2. 15.
Due to Jarge cross-sectional are a o f the reservoir. for any va riation in pressure. tile change in tile liquid
level in the reservoir will be very s mall whicll may be neglected and hence tile pressure is given by thc
height of liquid in the other limb. The other limb ma y be vcnical or inclined. Thus thcre are two types
of single column manometer as:
I. Ven icaJ Sin,gle Column Manometer.
2. Inclined Single Column Manom eter.
I. Vertical Single Column Manometer
rig. 2. IS s hows tile vertical single column manometer. Lei X-X bc the datum line in the reservoir
,md in tile right limb of the manomete r. when it is not connected to the pipe. When the manom eter is
I I
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IL
Pressure ilnd its Measurement
49 1
conn~ctcd to the pipe. due to high pressure at A. tile heavy liquid in the reservoir will be pushed
downward and will risc in the right limb.
Let 1'>11 '" Fall of heavy liquid in reservoir
r
I,! '" Rise of heavy liquid in right limb
h,
= Heighl of centre of pipe above
X-X
T
/I" = Pressure at A. w hich is \0 bll measured
",
A = Cross-section!11 area of the reservoir
(I '"
1
Cross-sc(;t;onal area of Ihe right limb
T
S, = Sp. gr. of liquid in pipe
$1
_,r!;~'::'~_:'1R~Yo~:_=~"'Qj
y
y
'"
= Sp. gr. of heavy liquid in reservoir and right limb
Fig. 2.15 V"lical si'lgie coll/mn
PI = Density of liquid in pipe
malwmeur.
Pl = Density of liqu id in reservoir
Fall of heavy liquid in rc!,crvoir will c ause a rise of heavy liquid kvcl in lh", right limb.
A
X ~,
= (/
llh =
X
(1)(
hl
112
... ( i)
A
Now consider the datum line Y-Y as shown in ri g. 2.[5. Then pressu re in the right limb above Y-Y.
Pz x g x (llh + /'2)
Pressure in lhe [eft limb above Y-Y = p, x g x (1'111 + Il,) + IJ"
=
Equaling these pressures. we ha ve
Pl xg x(l'1h + " 1) = p, Xg x {l'1h +/i,)+p"
1'" = P!8 (t'1I! + "0
"'
-
p,g(tlh + ",)
= I'1h lPlg - p,g] + "ZPlg - ",p,g
But from eq uation (i),
I'1h = a x "1
A
.. .(2 .9)
As the area A is very large as compared 10
fl.
hence ratio!!... becomes very s mall and can be
A
neglecled.
Thcnp,,="lP!8 - ",P,g
... ( 2. 10)
From equation (2. 10), it is c lear lhal as ", is known
and hence by knowing /'1 or rise of heavy liq uid in the
right limb. lhe pressure at A call be ca lcu laled .
2. Inclined Single Column Manometer
Fig. 2.16 shows the incli l1ed sin gle (o lumn mal101l\e\Cr. This manometer is more sensitive. Due to in(lination th r dis\ance moved by the heavy liquid in the rig ht
limb will be more.
I I
Fig. 2.16
Inclined single coll/mn
manometer.
Ii
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IL
Iso
Fluid Ml>chanics
Let
L", Length o f licavy liquid moved in right limb from X -X
e = Inclination ofrighllirnh Wilh horizontal
11 2 '" Vertical rise of heavy liquid in ri gh t limb from X-X ", L x sin
e
From equ ation (2.10), Ihe pressure at A is
II... = "~P2g - "IPlg·
Substituti ng the val ue of 11 2, we gCI
e
... ( 2 . 1 I )
/I,., = sin x P28 - h lPlg·
Problem 2.14 A single coluIIIn manometer is cOIlller/ed /0 (I pipe comaillins a liquid of sp. gr. 0.9
as slloll'n ill Fig. 2./7. Find lite pressure ill the pipe if the orea of Ihe reserl'Oir is 100 limes Ihe area
of lite rube for IIIe mmwme/er reading ShOIl"1I in Fig. 2.17. The specific grol'ity of mercury is 13.6.
Solution. Gi ve n:
Sp. gr . of liquid in pipe.
St = 0.9
I"
PI = 900 kglm )
Density
Sp. gr. of heavy liqu id.
S 2 = 13 .6
Density.
p, '" I1.6 x 1000
f
1
Area of reservoi r '" ~ '" 100
Area of right limb
Hcigtlt of liquid.
Ri se of mercury in right limb.
a
hi::: 20 ern '" 0.2 rn
f".g . 2 .17
h Z =40un=0.4m
Let
Using equation (2.9). we get
PA = Pressure in pipe
"
PA = -A IhlpzR
- PIS! + hop,S
• - - "IPIS
1
: X 0.4[13.6 x 1000 )( 9.81 - 900 x 9.81 J + 0.4 x [3.6 x 1000 )( 9.8 1 - 0.2 x 900 x 9.81
100
== 0.4 [1 3341 6 _ 8829 ) + 53366.4 _ 1765.8
100
= 533.664 + 53366,4 - 1765.8 Nlmz = 52 [34 Nlmz = 5.21 Niem I.
.. 2 . 7
A ilS •
DIFFERENTIAL MANOMETERS
Diff.:rcmial manomcters arc th e d.:viees used for mcasuring th e differencc of pressures betwee n
two poims in a pipe or in two differelll pipes. A differential manometer l'Ons iSls of a U· lUbe. coma in·
in g a heavy liqu id. whose two ends are con nected to lhe poinls. whose difference of pressure is 10 be
measured . Most commonly types of differential manometers arc:
!. U·tube differential lIlano lll eter and
2. In ve n ed U·tu bc diffe rent ia l manometer.
2 . 7 . 1 U - tube Differential Manometer. Fig. 2.18 shuws the differential manumeters of
U·tubc type.
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Pressure ilnd its Measurement
T
,
1
,
t
, -- J" ,
t
T
(a)Two rHpes at differenllevels
Fig. 2.18
511
'1-- '
(b) A aJ'ld B ale at the sa me Ieve1
U-wbe difftr..nlial manomtll"l'$.
In Fig. 2.18 (ll). the tWO points A and 8 are at different level and also contai ns liquid s of different
sr. g r. These point~ are C{lonecled 10 the U -luhe differential manometer. Le t th e pressure al A "fld B
arc I'A and PB"
Le\
,,=
Difference of mercury level in the U -tune.
Y'" Distance o f th e cCl1m: o f B, from the mercury level in the right limh.
x'" Distall(;c of the l-emrc o f A. from thc mercury level in the right limb.
PI
= Density of liquid at A.
P 2 = Density of liquid al B.
p & = Density of he avy liquid or m<:rcury.
Taking dmum lin e m X-X.
Pressure abo ve X-X in Ihe left limb '" Plg(1J + x) + 1',\
wherc p,\ '" pressure at A.
Pressure above X-X in the rigllt limb '" P~ x g X II + P~ X g X Y + 1'8
wllere 1'8 '" Pressure 3t 8.
Equ ating tile two pressure. we Ilave
Plg(1! + x) + I'll '" PJ X g x I! + P28Y + P/I
P.t - pB '" P, x g X II + P!G)' - Plg(1I + x)
'" It x g(P J - PI) + P2g), - Plgx
Difference of pressure at A and 8 '" I! x g(PJ - PI) + Pig)' - Plg.t·
... (2. 12)
In Fig. 2. 18 (b), the two points A and lJ are at the same le ve l and contains tile s.ame liquid of den sity
PI' Tlien
Pressure aboveX·Xin rigllt limb '" p.xg xl, + PI xg xX +PB
Pressure above X -X in le ft limb
'" PI x g x (II + xl + p ...
Equati ng the two press ure
P, x g X II + Plgx + 1'/1 '" PI X g x (II + xl + JIll
I'" - JIB'" p. X g x I! + p ,gx - p,g(1I + x)
... ( 2 . 13)
'" g x lI(p~ - PI)'
Problem 2.15 A pip .. COIII{lillS (11/ oil of sp. gr. 0.9. A diff"'''l!Iia/ 1I/1II1<I1II .. I..r mllllfC/e(1 <1/ /11 .. IWO
poillls A (I//(I Jj SIIOWI <I diff~ '<'II("" ill lIIe ,nltJ' 1"1"1'/ {U 15 CIII. Filllllh .. diff~ '<'II("" of Jlr..ssl"" III lit .. /11'0
poims.
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152
Fluid Ml>chanics
Solution. Given:
Sp. gr. of oil.
Density. PI "" 0.9 x I()(K)" 900 kg/rn
3
I,: lSClll=O.ISm
Difference in rncn:ury level.
Sp. gr. of rncffury.
Ss = 13.6
Density, p~ = 13.6 x [000 kglm J
The diffcl\)ncc of pressure is given by equation (2. 13)
f',t - PB" g X h{P, - PI)
'" 9.8! x 0. 15 (13600 - 9(0) '" 186K!1 N/m!. Ans .
Problem 2.16 A difJrTl'1IIiai m"""",,,'l''' i .• (',,""N"/I'll (l/ flIP Iwo Iminl s A ",,,18 "f I ... " l'ipl's {I$
SilO,.." ill Fig. 2./9. Til .. p'iN' A . 'omaills a Ii'll/itf of 31" gr. = 1.5 ... /,i/" [}ip" lJ collwillS 1I liquid "I
w- gr. = 0.9, Til" prnmrn a/ II mnlH (If" I kgflrm ! IIlIff /,80 kgJ7m/ rl'.!prrlil"r/y. Fi"d I/U'
dijJrrf'IICf ill mereu,)' /","/'I ill Ih" (Iiffrrnllitd """!(Jllli'lrr.
Sp . gr." ' .5
PA =1 kg/1mi'
SolutIon. Given:
Sp.gr.ofliquidatA, SI = I.S
..
PI"" 1500
Sp. gr. of liquid at /J. 52 "" 0.9
Pl"" 900
Pressure at A.
PA = I kgflcm 1 = 1 x 10· I.:gffm 2
= 10"x9 ,81 Nfm!('; 1 I.: gf=9.81 N)
Pressure al 8,
1'8'" 1.8 ~gflcm 2
2.0m
,
~
= 1.8 x 104 kgffrn l
lknsity of mercury
[
~
=1.8xlD"x9.81 Nfm2 (": Ikgf = 9.81N) x
= 13.6 x 1000 kgfm J
Sp, gr.=0.9
Pe = 1.8 kgflan'
r
Taking X-X as datum line.
Fig. 2.19
Pressure above X-X in the left limb
= 13.6 x lO00x9.8 1 xh+ 1500x9.8 1 x(2+3)+/J,\
'" 13.(; x 1000 x 9.8 1 x II + 7500 x 9.81 + 9.8 1 x 104
Pressure above X-X in the right limb '" 900 x 9.81 x (11 + 2) + Pfl
=900x9.8 1 x(h+ 2)+ 1.8 x 10"x9.81
Equ at ing the two pres.<;ure. we get
13.6 x 1000 x 9.81// + 7500 x 9.81 + 9.81 x 10·
'" 900 x 9.81 x (II + 2) + 1.8 x 10" x 9.81
Dividing by 1000 x 9.81. we gel
13.6/1 + 7.5 + 10 = (h + 2.0) x.9 + 18
"'
13,6/1+ 17.5=0.911+ 1.8 + 18=0.9h+ 19.8
{l3.6 - 0.9)11 '" 19.8 - 17.5 or 12.711 '" 2.3
23
,,= -=0.181
12.7
III
= 18.1 em .
AIlS .
Problem 2 .17 A dijJfrl'mi(l/lIwlwllwli'r i~ ("",,,w("li'd (If IIU' 1"'0 Imims A (/Jut H (IS sho"'" ill
Fig. 2.20. AlII "ir prnsurl' ;,' 9.8/ NiemI ((11M). find Ihl' "b,<oll1/l' prl'S$llrf lI/ A.
Solution. Given:
Ai r pressure al
"'
I I
II '" 9.8 1 N/cm l
1'8=9.81 x lO"N/m l
Ii
~ I
IL
Pressure ilnd its Measurement
IXnsity of oil
'" 0.9 x 1000 '" 900 kg/m 3
Density of me rcury
'" 13.6 x 1000 kg/fill
Let the pressure at A is P,1
T:lking datu m line ~t X·X
Pressure above X-X in the right limb
= 1000 x 9.81 x 0.6 + I'll
'" 5886 + 98100 = 103986
Pressure above x-x in Ihe Icfllimb
'" 13.6x I OOOx9.81 xO.1 +900
x9.81 xQ.2+PA
= 1334 1.6 + 1765.8 +
53 1
Oil OF
Sp. gr. ~O . 9
,
I'"
Equating Ihe two pressure head s
103986 = 1334 1.6 + 1765.8
+ PA
p,\ '" 103986 - 15107.4 '" 88876.8
Fig. 2.20
2
88876.8N
N
, = 8.887 -~,.
lOQOOc m "
e ll) "
A = 8.887 Nf(m l , Ans.
p", '" 88876.8 N/rn =
Absolute pressure at
2. 7.2 Inverted U-tube Differential Manometer. [[ consists of an inverted U-Wht:,
cont ainillg a light liquid. The two ends of lh~ tube arc connected to the points whose diffacncc of
pressure is to be measured. h is used for me asuring difference of low pressures. Fig. 2.21 shows an
invc n ed U-tube different ial manometer connected to thc two point!; A and 8. Lct the pressurc at A ,s
more than the pressure at 8.
Let
II, = Height of liq uid in left limb below the datum line X·X
II! = Heigh t of liquid in right limb
II == Differe nce of light liquid
p, == Density of liquid al A
P 2 = Densi ty of liquid al /J
P, == Densi ty of light liquid
p" '" Pressure at A
PB = Pressure at /J.
Taking X-X as dat um li ne. Then pressure in the left limh below X-X
,
, ~
,
==p,,- p,xgxiI,.
Fig. 2.2 1
Pressure in the right limb be low X-X
"'PB - P2 xg X /1 2 - P.xg X II
Equating the two pressure
1'" - p, x 8 X II, == I'B- p! X 8 X /12 - P. X g X II
. .. (2 . 14 )
I',,-I'B = 1', xg X " I - P! xg x1l 2 - p. xg x II.
Problem 2.18
m(l1lOm~l~r Iwvillg '111 fJii fJfsp. gr. 0.8 is CO"'''''·I(>(/. Th~ p'f'.""'~ /"'(1(/ ill 1/'" pip~ A i .• 2 /11 of»"fIIf".
jim/I/lf' pressure ill Ilu' pip" II fo r Ih" ",UllfJmNl'r re(l(lillg.1 a.1 sllo»"11 ill Fig. 2.22.
Solulion. Given:
"'
Pressule head at
A == p" =2 III ofwal~r
pg
1',j==jJxgx2= IOOOx9.81 x2= 19620Nlm~
Fig. 2.22 shows the arrangement. Taking X-X as datum linc.
I>rcs;;urc below X-X in the left limb '" p" - PI X g X /' 1
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IL
154
Fluid Ml>chanics
6''''>o, OIL
of
Sp. gr. 0.8
'" 19620 - 1000 x 9.81 x 0.3 '" 16677 N/m!.
Pressure below
x-x in
the
ri~hl
-,
limb
=PB - 1000x9.81 xO.I - 800x9.81 xO. 12
=PR - 98 1 - 941.76=PR - 1922.76
Equati ng Ih e t wO pressure. we gel
16677 =PR-1922.76
or
I'll '" 16677 + 1922.76 ", 18599.76 Nfrnl
or
I'll '" I.KS\l9 NIcOl!, AilS.
WATER
Fig . 2.22
Problem 2.19 III Fig. 2.23. (III im'fr/n/ (flffarmio/liumOIlJl'lfr i~ romlfclnJ IQ ''''0 pi"t'S A /IIullJ
",/lid, COIIW)' "'tUa. TIll' jlI,id ill IHmWlllflFr is oil ofsp_ gr. 0.8. Ff" 1/,(, IIWIWl/lnU rl'(IIIill85 showll ill
,II .. fig"'''. jiml IIU' l'rf $S/lrf difJrrl'lu'f' hp/wi'('11 It (111(/ B.
Solution. Given:
,-
= 0.8
Sr. gr. of oit
OIL of
Difference of oil in Ihe two li mbs
Sp. g'. 0.8
I:
-x
H
=(30+ 20) - 30 = 20cI11
Taking datum lin e 31 X-X
11
Pressure in th e left limb tJ..,]ow X-X
=PIl - IOOOx9.81 xO
Fig. 2.2.3
'" p" - 2943
Pressure in the ri ght limb be low X-X
"'p,, - IOOO x9.8 1 xO.J - 800x9.8 1 xO.2
Equating the twO pressure 1',,- - 2943
"'I',, - 2943 - 1569.6"'p" - 45[2.6
'" 1'" - 4512.6
p" - p,, '" 4512.6 - 2943 '" 1569.6 N/m ~. A"-~ .
Problem 2 .20
Fimf 011/ Ilw dijJl'rfllli(ll rnulillg 'iI' of 1111 im'erletl U-Iube 1I1lI1I01II<'11'r collwillillg oil of
sl",cijic grm-ity 0.7 <IS III<' m(l/lOmelric j/I,ili Wilfll COIIII<'clnl <lrrOS5 pipes A 111111 JJ (IS 5i1owII ill Fig. 2.24
below. cO/wfyillg liqllids ojW('("ijic grlll-itin 1.2 will 1.0 111111 immiscible wilh 1IU1/lOlIwlric j/llill. Pil><'5 A
IIIIIILJ lire lorflled III Ihe 5WI/<' lerel lIIul a55W"" Ihe 1'11'551111'5 (/I A (IIu/ B 10 be eqlla/.
Solution. Given:
Fig. 2.24 shows th e arrangement. Taking X-X as datum [Inc.
p,,- '" Pressure at A
1'" '" Pressure at B
Density of liquid in pipe A
'" Sp. gr. x 1000
'" 1.2 x 1000
'" 1200 kglm
Density of liquid if] pipe B
Density of oi l
2
'" I x [000 '" 1000 kg/m)
'" 0.7 x 1000", 700 kgfm3
,
&"""' j'.".~.'
-r-
,
gr·t .O
"
~-
w=
~-
Fig. 2.24
II
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IL
Pressure ilnd its Measurement
55 1
Now pressure below X-X in the left limb
"p" - 1200 x 9.81 x 0.3 - 700 x 9.81 x"
rrc~urc
below
x-x in
the right lilnb
=P8 - 1000 )(9.81 x (h + 0.3)
Equati ng the two pres,ure. we get
1'.-I-1200x9.81 xO.3-700)(9.81 Xh=Pn-1QO(}x9.81 (11+0.3)
p", = 1'8 (given)
But
1200 x 9.81 )( 0.3 - 700 x 9.81)( II = - 1000 x 9.81 (II + 0.3)
Dividing by 1000)(9.8 1
- 1.2 xO.3 - 0.711 = - (II + 0.3)
0.3 x 1.2 + 0.711 = " + 0.3 or 0.36 - 0.3 = Ir - 0.7/1 = 0.311
"'
0.06
II = 0.)6 - 0.30
"--m
0.30
0.30
I
1l1=~X
5
100 = 20 em. Ans.
Problem 2.21 All illw'rll'd U-II/be 1/1(1/10111('/'" i~ colIIlN"/ed /0 /11'0 lror;:ollllli pipn A /lml IJ
Ihrollgh ...flich Wain is flowitlg. Thl' W'rrical (/iSlalln' bnwl'fll Ihl' (/.\"t'S of tllest' pipn is 30 ('III. II'hl'lI
WI oil ofspeciftr grm'il)' 0.8 is 1151'(/ (U (I glluge f/llitl, I/U' "l'rlicollwiglw ofwlI/a ro/mlllls ill I/W two
limbs of Ow ill"erln! m(IHOIllflU
( ""WI! /IINI.,urn/
from Ihl' rl'Sl'f("I;"(' {'('mr('
lillI'S
of Ilw ,,'PI'S)
lin'
fmll/(Ito hI' SlI/IIf (IIu/ I'(I,wl to 35 Oil. Dnrrmillf till' l/ijJfr('ll{,(, of prnSllrf hl'n"ffll till' pipl'S,
SolutJon. Given :
. gr. =O.B
S..,.,cific gravity of measuring liq uid "" 0.8
The arrallgelll<'nl is shown in Fig. 2.24 (ll).
leI
PA = pressure at A
PI!" pressure at R.
The poilUs C and /) lie on th e same horizomal line.
Hence pressure at C should be equal to pressure at D.
But pressure at C
""fl,, - pgh
And pressure at /)
'" 1'" - 1000 x 9.81 x (0.35)
'" PB - p,gh, - P28 11 2
"" jiB -
WATER
1000 x 9.81 x (0.35) - 800 x 9.81 x 0.3
BUI pressu re at C", pres;;urc at D
B
Fig . 2.2~
JiA - 1000 x 9.&1 x .35
fa)
"'p/i - lOOOx9.81 xO.35 - 8oox9.81 x 0.3
or
"'
I I
8OOx9.&1 X 0.3 "'PB - PA
JiE - PA'" 800 x 9.81 x 0.3 == 2J54.4
N
on
-I.
Ans.
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156
Fluid Ml>chanics
.. 2.8
PRESSURE AT A POINT IN COMPRESSIBLE FLUID
For compress ible fiu ids . densit y (p) changes wilh tlie c han ge of pressure and te mperature . Such
pro blems arc e nco un te red in aeronautic s. ocean og raph y and meteoro logy where we arc con ce rn ed
with atm os ph eric * air where dens it y. press ure and temperature changes with e le vati o n. Thu s for fluid s
wilh varia bl e densi ty. equatio n (2.4) canno t be integ rated. unless th e relatio nship betwee n p and p is
kn ow n. Fo r g ~ scs th e equatio n of state is
1!.. '" NT
... (2. 15)
P
-
p
p - NT
dp
-
No w equat io n (2 .4) is
ilZ
=w=pg = -
P
NT
Xg
"I' = ...1...,/Z
... (2.1 6)
P
RT
In equat ion (2.4 ). Z is meas ure d ve rticall y dow n ward . But if Z is me asured ve rti call y up , th en
d[1
-
"l
"" - pg and hence e quation (2. 16) beco mes
= .=L tlZ
... (2. 17)
I'
RT
CotSI' I. [f temperature T is constant which is true for isot hC'rmalllroctip
2 .S. 1 Isothermal Process.
tOSS. equ ati o n (2 .1 7) ca n be ill1cgral~ d as
p
I
P.
til' '"
_jZL(fz= _L
1" RT
f'
I'
log Po
~
-g
-
JZd:
RT Z.
IZ - 2;. 1
NT
whcrepo is th e pressure where he ight is Zo. Ir the d atum lin e is Hlke n at
th", pressure at d:llulil linc.
log L
Po
=
Zo. th en Zo = 0 and 1'0 bo:co mcs
::.!. z
NT
L = ('- ..11FT
p,
or pressure at a hciglit Z is give n by I' = PrI'- S7/KT'
• • •( 2.18)
2 . B. 2 Adiabat ic Process. If tc m pc ralllrc T is not constant bu t the process follows adiabatic
law th en the relati on be twee n pressure and density is g i ve n by
I'
-t
= Conslanl = C
...(il
P
• The standard atmospheric pressure. temperawre and dcnsily referred 10 STP at {he sea-level are :
Pressure .. 101.325 k.N/m l; Temperature", 15°C and ();:,nsily = 1.225 I;:glm J.
I I
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57 1
Pressure ilnd its Measurement
where k is ratio of specific constant.
".
. .. ( i i)
Then equation (2.4) for Z measured vert ically up becomes.
til'
III
tip
(~l
Inlcgrating. we gel ) "
em I'
Ilk
=_pg= _(,,)II. g
C
_
=-gdZur e
III
til'
~I~
_
=-8'/Z
pc
III: ill' = j Z_glfZ
'"
"
".
[C is 3 constant. can be laken inside I
But from eq uation (;),
Substituting this va lue of
above. we gel
]
'
[ -""'
, ,.
[ p~l'• = - g[Z - ZoJor [-'~ !C1P
p
'"
-~
p
or
e llt
lZ - Z-o J
= - g-
p 1
- - +1
l
!
/.:-1 P
P •
".
= - g[2 - 201
~
' [£,_PO] =-gIZ-ZoI
k-l
P
Po
If datum line is taken :Il 4" when: prcs.<;ure. temperature and density arc Po. To :md Po. then
_'~ [.f.._PO] = _ gZ
k-IPP o
or
~I
or
4
= O.
l!.. _ ES!..= _ gZ(k-l)
PPo
k
l!..=PO_gZ(k-l)",ft [ l _k-l gZfu]
P Po
k po
k po
I~
~ I
IL
Iss
Fluid Ml>chanics
i!.. xPn
0.
P Po
But from equ at ion (il,
=[1+/::- 1gZfu]
Po
l' '" Po
?p~
Substituting th e va lu e o f
£.£.
... (iii)
k
[&.)k
'" ElLporP
Po " ( "'-)'"
P
I'
or
in equation (iii). we gel
P
P
]
(P )'" [
]
[ )-'" [
(Ll'-i"[L)';' =[, _'-:c',z&.]
/::- 1 - Po
" 1- - , - gZ 1'0
Po
I'a x
-
p
Po
0.
, -
P
/:: - 1
P
k
Po
,, 1- - - gZ.....2..
l'o
Po
l'o
.l!...."
k po
,
[1_/::-1 gZPo]N
Po
k
Po
Press ure at a he ight Z from ground leve l is g iven by
p '"
In equat io n ( 2.19), P o '" pressure
al
,
Po[1 - I fu]G
gZ
k
Po
gro und leve l. where 1 0 '" 0
k-
... (2 . 19)
Po '" densiTy of air at ground leve l
Equation of Slale is
Po : R1 0 0r££.: - ' Po
Po
RIo
Substitutin g the va lu es o f Po in eq uatio n (2. 19), we ge l
Po
2
,
8- ]'-'
1'=1'0 1- [ '-k-' -RTo
Temper<lture at any Point in Compressible Fluid.
... (2.20)
2. B. 3
For the adiab at ic process. the
lemperature al any height ill air is \' alc ulm cd as :
Equ ati o n of stale at gro und leve l and al:t heig ht Z from grou nd levd is wri llc n as
Po -- Ri'0
-
Po
3n dJ!... -
Ri'
P
Dividing thcse cqualio ll s. we get
" O)+J!...=RTo = l n
( Po
PRTT
".
I I
!..IQ
or
-Po x -P =-To
Po
=&. x J!...=..E.... x Po
po
P Po
I'
T
.. .(i)
P
Ii
~ I
IL
Pressure ilnd its Measurement
But
L from ~qualion (2.20) is given by
P.
591
•
L== [l_lr.- l /1Z ]i-=i
k
Po
Also for adiabatic process
4 = p~
or
P Po
RTo
(!2]t
== ['0
J!
P
P; . [I;:]'.
0'
[:J'
. [1-' - 1,2](: ,].( ;) . [1-' - 1,z]
k
Substituting the values of
L
RTo
£..2.. in cqu<ltion
:md
Po
1'0
·
·,
R ro
(I). we get
P
~
k
,
=[1_ /.: - 1gZ ]t-i X[1_ k-1 /lZ ]-t:1
k
RIo
k
RTo
' [l _~gZ ]i=I-i-=i = [J_k-J
k
Kin
k
k
/lZ ]
K/0
1 2]
T= "' "o [ I- '-;-!To
...(2.21)
Lapse-Rate ( L) . [\ is defined as the rate at which the te mperature
2. 8 .4 T~mperiiture
changes with elevation. To obtain an expression for Ihe tcmpc ralllrc lapsc-ralc, Ihe IC lll pcrJllIrC given
by cqumioll (2.21) is diffcrcmimcd with respect to Z as
:~: ,;~ [To (l_k;J :~J]
==
where To_ K, II and R arc constant
X70= -:(";1)
-1I (k-l)
;~~ = _ k;IXR~o
The t<!mpcraturc lapse-rate is denoted by L and hence
I. = -tiT . . dZ
/{
... (2. 22)
(JT
In cquation (2.22). if (i) k '" I whIch mcans isothermal process. '" 0, which means temperature
dZ
is constant with height.
(ii) If k > 1. the lapse· rate is negative which mca[lS temperature decreases with the increase in
heigh!.
In atmosphere. the value of k varies with height and hence the valu~ ofte11lpcrJlure I<lpse·rate also varies.
From the sea·level upw an elevation of aboul 11000 III (or II km), the teillperalure of air dec~a>cs
unifonnly at the ratc of 0.0065"Chll. from 11000 1\1 10 32CXJO Ill, the telllp"rature rellla ins ronst3nt
at - 56.5°C and hence in this range lapse·rate is :tero. Temperature rises again after 320Cl0 10 in air.
I I
-k
Ii
~ I
IL
160
Fluid Ml>chanics
Problem 2 .22 (SI Units) if rhe a/1II()Jpilere pres,~ure (1/ .~ea /el'e/ is 10. 143 Nkm ! , derami/Ie Ihe
pressure m a ''''igM of 2500 III aS$lwling Ille pressure mriarirm follows (i) Hydros/a1ie lim', and
J
(ii) isolhermal law. The density of oir is gil'en as 1.208 1r.g1m •
Solution. Given :
Press ure at se a-level.
Po = 10.1 4 3 N/c1111 = 10. 14 3 x I O~ N/rn l
2 =2500 111
1
!knsi ty of air.
Po = 1.208 kg/nr
(I) Press u re by hyd rostatic h, w. For hyd ro static law. p is assumed constallt and hence p is g ive n
.
dp
by equalloll = - pg
dZ
Height.
In teg ratin g . we get
fPdp '" f - pg dz= - pg f .Zd Z
Po
0'
For da\tJlll lin c at sea- le ve l.
70
P - Po= - pg IZ - Zol
Zo = 0
P - Po =- pgZ or P=Po - pgZ
= 10.143 x Itr - I.208x9.81 x 2500 [ ": P=Po= 1.208]
N
m
7 ]804
10
'" 10 1430 - 29626 = 71 804 ---y or __,_ N/c m 2
= 7. 18 Ni em I. Am .
( ii) I'rcss ur e by Iso therm a l Law. Pressure at an y h ~ i g ht Z by isoth cnnal Inw is give n by equ ation
(2. 18) as
= 10. 143 x 1 0~ e
-~
r.
-~
= 10. 143 X 10 4 e I'l>
= 10 . 143 x I 0 4 e (_ lloOO .
) ,1(IB x 9,8) JllO.)43. )If'
= IOl4 30xe .291 = 10 1430x - ' - = 7574 3 Nlm2
1.339 1
75743
=~
Nlc l11
1
,
= 7.574
Z
N/cm • An s.
Problem 2 .23 The barOlllelric p,,·s,m re ar .fea lel'e/ ;,5 7601111/1 of lIIercury ..... hile Ihm on a 1II00WIlIin
J
lOp i.f 735 111111. If lite d<'fu;ry of air is a.f.fljllled COl/SWill m 1.2 kg/III , ..... har ;s Ihe eiel'arion of Ihe
IIIOll1l1ain lOp?
Solution. Gi ve n :
Pressure· at s.c a ,
Po = 760 111111 of Hg
= 760 x 13.6 x 1000 x 9.8 1 N/m2 = 10 1396 N/ml
'000
• Here pressure head (Z) is given as 760 mm of Hg Hence (pfpg) .. 760 mm of Hg. The density (p) for mercury
'" 1J.6 x 1000 ~glmJ. Hence prcssure (p ) will Dc cqual to p x g X Z i.f .. 13.6 x 1000 x 9.81 x : : NI1111.
~ I
I~
~ I
IL
61 1
Pressure ilnd its Mea surement
Pressure al mount ain.
p '"
735
11lI1i
of Hg
735
,
= - - x 13.6 x Hloo x 9.81 = 98060 N/m-
lOoo
Iknsity of ai r .
,,=
p = 1.2 kg/ml
Let
Height of the mountain from sea-level.
We know Ihm as the elevation above the sea-level increases. the atmospheric pressure dc.::rcascs.
He re the density of air is given conSlallt. hence the pressure 31 any heighl .". above the sea-Icvel is
given by the equation.
1'=Po- pXgxh
It = Po - II", 101396 -98060 = 21B ..B m . ADS.
pXg
1.2x9.8 1
Problem 2.24 Ca/r"/ll/(' Ih" prnsllrt' at {/ 11I'iglll of 7500 II! nbol'f 51'(1 Inri if IIIf' 111lI1osplwricpress",.. is JO.14.i Nkm~ lind IfillpUaillre is 15°C (1/ IIlI' se a ·/{'\·e/. (lsslImillg (;) tli, is illcompn'Mib/(',
(ji) pnSSllrl" WIr;m;OIl follows iso/hal/wi /m>l. mllf (iii) f"fS5Jf 1"1" \'(,,;111;011 /01/0,,"5 (II/;lIb(l/;(' low, T nkf'
Ihe tlnlsit)' of <lir (1/ III(' snI -lr\'e/ (IS r"l/ollo 1.285 kg/llr'. Nrglrcl \'Orilllioll of g wilh Illlilll<"'.
Solution. Given:
Z", 7500 III
Height aoove sea-level.
4
2
Pressure at sc a- lel'el.
1'0" 10. 143 N/cml" 10.143 )( 10 N/m
Temperature at sca-level.
Density of air,
10 '" 15°C
To" 273 + IS" 288°K
P = Po = 1.285 kg/ml
(I) Prn.wn ,../1f/! (lir is i,u'OIll/'U'Hibll':
"p
,'Z = - pg
f Pdp = - f ZpgdZ
'0
4
1'''l'o-pgZ
or I' - ['0 = - pg[Z - lol
[.:
lo"datumline"O)
= 10.143)( 104 - 1.285)( 9.81 )( 7500
= 101430 - 94543 = 6887 N/rn l " O.6~~~ , Ans,
em "
(ii) f'nsslITl' l"(lrimioH follows iSOlIiI'rIll(lI/ow:
Using equation (2.18), we have
/, = PrI'
Jl/II.T
= f'o e ,lPn'po
= 101430 ,,-
{-:
~: = R T
:.
~: =
;A
Jl.(>rjPo = 101430,,-7500~ 12'!5~9811101~.\(I
= 101430 ('- .9.120 = 101 4 30 x .39376
= .W9.W N/m~ or .t99 .~ N/em! . Ans.
(iii) Prl'.f$Urp mrim;,m fl,lI",,"s ",/i,,/wli(" I"",,: [k = 1.41
'I
Using equation (2. 19), we have
I I
, where Po = 1.285 kg/m
l
Ii
~ I
IL
162
Fluid Ml>chanics
p = I 0 1430
[
1-
,.
(1.4 - 1.0) x9.81)( -,( 7"5"00,,X
.:,',.,2:::8,"
5) J' .
_1.0
1.4
101430
'" 10 1430 II - .2662 1' ''''· == 10 1430 X (.7337) 3.5
,
N
'" .\4.'10 Nfm - or .'.431 - -, . AIl~ .
,m'
Problem 2.25 Ca/ndllll' lilt' prnsllre (//1(/ dn/sit)' Of air (1/ II Iwight of 4000 III from 5I'1I-/PI'l'lwl!frt'
prf ss,,,/' (I/UI {('III/N' ralllrf' of /111' (lir (/rl' 10. 143 Nkm- (jill! J5°(.' rnpaliw/),. TIll' Il'lIIpff(llllrl' 1111'S('
raIl' is g;1'I'1I as OJXJ65°CIIII. Take <I""5il), of oir <1/ Ull-/(,I'ei (''1IIl1/IO 1.285 kglllri.
Solution. Given :
~lc ig h l.
Z=4000 m
,
,
N
Pressure 31 sea-level.
1'0= 10. 143 Nlcm-= 10. 143x 10 = 10 1430 - ,
m
Tc mpc rmu rc at sea-leve l.
10= 15°C
To'" 273 + IS == 288° K
tiT
L= = - O.0065° Kfm
dZ
Po::: 1.285 kg/ml
Te mperat ure lapse- rate.
Using
cq u ~t ioll
(2.22). we
=_1.(' -')
L= tiT
h ~l VC
dZ
R
k
_ OJI065= _ 9.81 ( k - ' ).WhcrcR= ~ =
R
PaTo
k
10 1430 =274.09
L185x 288
x(k-')
- 0.(1065: -9.81
274.09
k
k - ] = 0.0065 x 274.09 =0. 181 5
k
9,8]
kl]- . 18 15]: I
- ' - '" 1.222
.1815 .8 184
This means that the va lue o r powe r ind ex k'" 1.222.
(il I'r~ss ur~ at 4000 m he igh t i ~ givcn by cq uation (2. 19) as
k=
k- '
pJ,"
P"'l'o I - - - ,Z - o
[
k
,w herck= 1.222 and Po'" 1.285
1'0
un
P
'" 10 1430 [ , _ ( 1.222 - 1.0 ) x 9.&1 x 4000 x 1.285 ]1.222 - [.0
1.222
101430
'" 1 0 1 430 [I - O.0915.~ ", I0 1430x.595
,
N
'" 60350 Nfm '" (j.O.~5 - • . Ans.
cm -
I I
Ii
~ I
IL
Pressure ilnd its Measurement
63 1
(ii) I)" nsil y. Us in g eq uation of s tate. we get
,
.e = RT
f' == Pressure at 4000 111 heigh t
P == Density al 4000 III height
T", Te mperatu re at 4000 III lic igh t
Now T is calc ulated from temperatu re la pse-ralc as
where
<IT
x 4000 = 15-,(1065 x 4000 = 15-26=-II °C
<12
T=273+1=273 - 11 = 262QK
lat4000m
= 10 +
-
P
Densi ty is given by
60350
3
3
P == RT = 274 .09 x 262 kglll1 '" O.K4 k glm . AilS.
Problem 2.26 All II1'rUpillllt' is flyillg III WI {,I/;Im'" of 5000 III. C,lIn,/ale IIlI' prf'SSIIrf' tlrol/llI/liI"
Ql'rop/mw, g;\'1'1I liIl' IlIpSI'-rll/(' ill IIIl' atlllosphrrl' liS OlXJ65° Kim. N('gll'c/ "'tria/ioll of g wilh III/iIUdl'.
TakF puss/lfe 11m! umprr(l/ilre (1/ grol/lld lewl (IS 10.143 Nlclll~ (///(I 15°C (llid dellsity of air (IS
1.285 kg/ow'.
Solution. Given:
Height.
Z=5000m
,rr
Lapse-rate.
1. == = - .0065 KJm
dZ
1'0 = 10.143 x IO~ N/ml
15°C
To = 273 + 15 = 288" K
Po = 1.285 kglm)
Pressure at ground leve l.
Q
'0=
Density.
Te mperature at 5000 m hright " To + af x Height" 288 - .0065 x 5000
elZ
,,288 - 32.5" 255.5°K
First find the va lu e or power indcx k as
From equat ion (2.22). wc have
__ .If =
_ _ 1.. (k- l )
1_ _
(rz
H.
k
_.0065 =_ 9!I ( k ~ l )
or
where
H.
",...EE.....",
Po To
10 1430 ,,274.(]9
1.285 x 288
_ .0065=_~(k-l)
The pressure is giv~n
274.09
k", 1.222
by equation (2.19) as
k
k-I
, ]1.',)
"=1'0[ 1--- gZ~
k
I I
1'0
Ii
~I
IL
164
Fluid Ml>chanics
l,ll2
== 10 [430
[I_( 1.222 -
1.0 ) x 9.8 1 x 5000 x 1.285] '-:12-
1.222
'-0
10 1430
l.lll
'" 101430 [ 1- .222 x9.81 X 5000 X 1.28S j ,222
1.222
10 1430
S
'" 101430 [I - Q.112S81 ,S<J '" 101430 x 0.5175 '" 52490 Nlm 3
= 5.249 N/cm !. AIlS.
HIGHLIGHTS
I . The press ure at any point in a fluid is defined as the force per unit area ,
2. The Pascal's law ,tates that intensity ()f prc.. ure for a fluid .11 rest is equal in all direction •.
~ . l>ressure variation at a point in a fluid at reSt is given by the hydrostatic law which siate, thai the rate of
incre:''''' of pres,ure in the vertically downward direction is equal 10 the specific weight of the fluid.
tip _w*pxg.
dl
4. The pressure at any point in a incompressible fluid (liquid) is equal to the produ(,! of density of fluid at
Ihm point. :lcccleralion duc to grnvity and verlical heigh! from free surface of nuid.
p .. pXgxZ.
S. Absolute pressure is the pressure in which absolute vacuum pressure is taken as datum while gauge
pl'C'ssure is the pressure in which the atmospheric pressure is taken as datum.
p_ = p _ + 1'_1<
6. Manomeler is a device used for measuring pressure at a point in a nuid.
7. Manomelers arc classified as (n) Simple manometers and (b) Differentia l manometers.
8. Simple manometers all' used for measuring PIl'SSUIl' al a point while differentialmanomelers all' used for
measuring the difference of pressures belween the IWO poims in a pipe. or two different pipes.
9. A ~ingle column manomeler (or micromeler ) is used for measuring small pressures, where accuracy is
required.
10. The pressure al" poim in SIalic compressible fluid is oblained by combining Iwo equ~lions. i.e .. equal ion
of Slale for a gas and equ~lion given by hydroslalic law .
I L The pressure at a height Z in a stalic compressible fluid (gas) under going isolhennal compression
p " Po c
where
I'~ "
rZlRT
Absolute pressure at sea-Iewl or al ground le\'el
Z .. Height from sea or ground le"el
R .. Gas conSlam
T ~ Absolute lemperalure.
12. The pressure and lempcmlurc al a heighl Z in a SIalic compressible fluid (gas) undergoing adiabalic
compression (plpi = consl. )
,
,
[ k-I . P"j" [k-I gZ ]I,'i
I'=Po l - - , - gZ 1'0
II
= /'oI- - '-RT
o
Ii
~ I
IL
Pressure ilnd its Measurement
T ~ To
and temperature,
65 1
'Z-l
[1- -Hk- -RTo
"here Po. To are pressure and tcmpcmture at sea-level k" 1.4 for air.
13. The ratc at which the tcmrcr~ture changes with elevation is known as Tempermure Lapse-Ratc. It is
given by
-'R ('-')
,
L .- -if
(i) k
E
I. temperature is Zero.
(ii) k > 1. lcmpemturc decreases with the increase of height.
EXERCISE
(A) THEORETICAL PROBLEMS
I . Define pressure. Ohtain an expression for the pressure intensity at a point in a fluid.
2. Slate and pro"c Ihe Pascal's law.
J. What do you understand by Hydrostalic Law ?
4. D ifferentiate between: (r) Absolute and gauge pressure. (;/) Simple manometer and differential nmnom c!cr. and (iii) Piezometer and pressure ga uges.
5. What do you mean by vacuum pressure ?
6. Whm is a manometer ? How arc they dassified ?
7. What do you mean by single colum n manomClers ? Ilow are they u>ed for the mea,urement of pressure ?
8. What is the differencc between U·tube differential manometers and inverted U-lUbe differential
manometers? Where arc they used ?
9. Distinguish between manometcrs and mechanical gauges. What are tlie different types of mc.:hanical
pressure gaugcs ?
10. 1)erive an expres.<ion for the pressure m a height Z fmm sea -Ie,·el for a static air when the eompre."ion of
the air is a.smned isothennal. The pressure and temperature at sea-Ie,·els arc Po and To respectively.
II . Prove that the pressure and tem perature for an adiabatic process m a height Z from sea-level for a static air
are :
I'U ~PO [I_k-1
*
gZrlland T",TO[I_k-1 gZl
*
RTo
whcre 1'0 and To are the pressure and temperature at sea-level.
12 . What do you understand by the tcnn. ·Temperature L1pse -Rate ? Obtain an expression for the
temperature Lapse-Rat" .
13. What is hydrostatic pressure distribution ? Give one example where pressure distribution is
non -hydrostatic.
14 . Explain hriefly the working principle of Bourdon Pressure Gauge with a neat sketch.
[(fo
(J.N.1:U. , Hyderabad. S
lool)
(B) NUMERICAL PROBLEMS
J. A hydraulic press has a ram of 30 em diameter and a plunger of 5 em di:unet,·r.l'ind the weightlificd by
the h)·draulic press when the force applied althe plunger is 400 N.
[Ans. 14.4 kNI
2. /I hydraulic press has a r~m of 20 em diameter and a plunger of 4 em di~lI1eter. It is used for lifting ~
weight of 20 kN. find the foree required al the plunger.
I I
[Ans. SOONI
Ii
~ I
IL
166
.\ .
Fluid Ml>chanics
Cakul~tc th~
pressure due to a col umn of 0.4
In
of (u) wuter. (b) an oil of >1>. gr. 0.9, and (coJ mercul)' of!>p. gr.
"
13.6. Ta~e density 0( .....:l1cr. p _ 1000- 3 ,
m
jAns. (a) 0.3924 Nlcm', (b) 0.353 NiemI, (el 5.33 Nlemll
,
-" The pressure imensily at a point in a fluid i.< g;,-en 4,<) Nlcm', Find the corresponding height of fluid when il
is ,(n) water. and (I,) an oil o f sp. gr. OX
I AI~~. (n):; m o f waler. (b) 6 .25 In of oill
S. An oil of sp. gr. 0.8 is contained in a vess.cl. At a point the height of oil is 20 m. Find the correspondi,,)!
height of ....:!tcr at that point.
jAilS. 16 ml
6. An open I,mk contains waler UplO a depth of 1.5 m and above it an oil of sp. gr. 0.8 for a depth o f 2 m. FioJ
the pressure intensity: (il at the interface of the two li'luids. and (ii) al Ihc bonum of the (a"k.
IAn_,. (il 1.57 Niem I. (ii) 3.04 Nleml j
7. The diameter> of a ...mal l pislon and a largc pi'ton of a hydraulic jacl are 2 em and 10 cm respectively. A
fo("(;e of60 N is applied On the small piston . Find the lo.1d lifted by the large piston. when: (tI) the pistons me
at Ihc salHe level. and (I,) slllal] pislon is 20 Cllt above Ihe large pislon. The density of Ihe liquid in the jack is
given as
II .
':I .
10.
II .
12.
1.1 .
".
1000~.
jAns. (,,) 1.500 N. (v) 1520.5 NI
m
Oclcnn;ne the gauge and absol ute pre"ure at a point which i'i 2.0 III below lite free -,urface of walcr. Take
almo. . pheric pressure as 10.1043 Nlem'.
IAns. 1.962 Nlcm l (g.1uge). IU)66 N"-1nl (abs .) I
A simple manometer is used 10 measure Ihe pressure of oil (51'. gr. '" 0 .8) flowing in a pipe line. lis righl
limb is open to the atmo'iphere and lefll imb;s connecled to Ihe pipe. The centre of the pipe is 9 cm be low
the le'·cl of mercury (sp. gr. 13.6) in the right limb. Jfthe differcnce of mercury level in Ihe two limbs is 15
~m. dclcnnine the absolute pressurc of the oil in the pipe in N/em".
IAns. 12.058 Nlem'l
A simple manomet.:. (U·wbe) ~ontaining mercury is connected to a pipe in which an oil of sp. gr. 0.8 is
flowing. The pressure in the pipe is vacuum . The other end of the manometer is open to the atmosphere.
Find the vaCUUIII. preS'iure in pipe. if the difference of IIIcrcur)' le,·el in the Iwo limbs is 20 cm and he ight
I,\IIS. _ 27.86 Nlcm1 j
of oi l in tbe left limb from the cenlre of the pipe is 15 em below.
A single column vertical manometer (i.e .. micrometer) is conn«ted 10 a pipe containing oil of sp. gr. 0.9.
The area of Ihe reservoir is 80 times the area of Ihe manometer lube. The reservoir C()Iltains lIIercury of
51'. gr. 13 .6. The level of mercury in the reservoir is at a height of 30 em below the centre of the pipe and
difference of mercury levels in the reservoir and right limh ;s 50 CIII. Find the pressure in the pipe.
IAlls. 6.474 Nlcm1 1
A pipe contains an oil of sp. gr. O.g. A differenlial manometer connected al tbe two points A and B of tbe
pipe shows a differcnce in mercury level '15 20 cm . Find the difference of pressure at lhe lWO points.
IAlls. 25113.6 Nlml ]
A V·lube di fferential manometcr connects 11>'0 pressure pipes A and E. Pipe A contains cmbon
tetrachloride baving 3 specific gmvity 1.594 under a pressure of 11 .772 Nlcm 1 and pipe R contains oil of
'1'. gr. 0.8 undcr a pressure of I ].772 Nkm". The pipe A lies 2.5 m abo"e pipe 8 . Find the difference of
pressure measured by merc"ry as fluid filling U·tubc.
IAlls. 3 1.36 cm of mercury1
A diffcrcmi al manometer is conn«tcd at the two point.' A and R a., shown in I' ig. 2.25. Al fJ air pre"ure
is 7.848 Nkml (abs .j. fi nd the absol ute pressure at A.
IAns. 6.91 Nlcm1 j
OIL Sp.
gr. ~O.8
, -'"
,';-1
"'=
,~
i
,,~
1
-M ERCURY
Sp. g'.En.6
Fi g . 2.25
I I
OIL Sp .
T
"''' =
I
""
T
.
'
~i~=
1.
~ ~Ei
~
WATER
Fig . 2.26
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Pressure ilnd its Measurement
67 1
15. An in"crled differential manomcl~r containing an oil of sp. gr. 0 .9 is connected to find the diffuence of
pressures at twO poinB of a pipe comaining waler. If the manometer rcading is 40 em, find the difference
of prc«urcs.
IAn_s, 392.4 N/m'l
16. In above ''is' 2.26 shows an in\'ened differential rna""mc!e, connected 10 lwo pipes A and B containing
water. The fluid in manometer is nil of sp. gr. O.H. For the manometer readings shown in the figure. find the
di fference of pressure head octween A and B.
[,\US. 0 .2 6 m of waterl
17. If the atmospheric pressure at sea · lc"cI is 10.143 Nlcm l , dClcnninc the pressure at a height of 2000 m
assl!ming that (he pressun: variation fol lows: (i) Hydrostatic law, and (U) lsothennallaw. The density of
air is given as 1.208 k8/mJ.
[,\ns. (i) 7.77 Niem I. (ii) 8.03 Nlem Jj
III, Calculate the pressure at a height of 8CXXl m abo,·c sea-level if the atmospheric pressure is 101 .3 kN/ml
19.
20.
2 1.
U.
2.1.
24 .
ZS .
Z6_
~ I
and tempemlUre is 15°C at the sea-level assuming (i) air is incompressible. (ii) pressure variation follows
~diabatic law. and (iii ) pressure vari"tion follows i>othermailaw. Take the dcn.ity of air at the sea -Ic,'el a.
equal to 1.285 kg/m). Neglect "ariation of K with altitude.
jAilS. (i) 607.5 Nlml. (ii) 31.5 kNlm l (iii) 37.45 kNlml ]
Calculate the pressure and density of air at a hcight of 3000 m above sea-leve l where pressure and temperature of the air are 10.143 Niem I and IS oC res""clively. The lemperalUre lapse-rate is given as O.()(I6S"
KIm. Take dCl1Sily of air at sea-level equal to 1285 kg/111 ).
jAl1 S_ 6.896 Nfcml . 0.937 kg/1n J I
An aeroplane is flying at an altitude of 4000 m. Calculate the pressure around the aeroplane. given the
lapse-rate in the atmosphere as 0.0065°KJm. Neglect varimioll of 1/ with altitude. Take pressure and
temperature at ground Ie"el as 10.143 Nlcm l and 15°C resp«tivcly. The density of air at ground level is
gi"en as 1.285 kglnl '.
[,\ ns. 6.038 NiemI]
The atmospheric pressure at the sea"level is 101.3 kN/ml and the temperature is 15°C. Calcul"te the
pressure 8CXXlm abo.'c sea-Iewi. aSSum ing Ii) air is incompressible. (ii) isothcnnal .·"riation of pressure
and density. and (iii ) adiabatic variation of pressure and density. A,sume den,it y of air at sea-level a,
1.2!\5 kg/1n J . Ncgl""t varialion of '8' with ~Ititude.
]Ans. (i) 501.3 Nlm ' , (ii ) 37.45 kN/m '.liii) 31.5 kN111l1]
An oil of sp. gr, is 0.8 under a pressure of 137.2 kN/ml
(i) What is the pressure head cxpressed in melre of water ?
(ii) What is the pressure head cxpressed in metre of oil?
[AtlS. (i) 14 m. (ii) 17.5 m]
The atmospheric pressure at the sea-level is 101.3 kN/m l and tC11l""ralUrC is I soc. Calculate the pressure
8000 m above sea-level. assumi ng (i) isolhennal variatio!1 of pressure and dcnsity. and (ii) adiabatic
variation of pressure and density. Assume density of air at sea-level as 1.285 kg/m J . Neglect variation of
'g' with altitude .
Derive the fonnula that you may use.
IAII". U) 37.45 kNI11l'. (ii) 31.5 kN111l'1
What are the gauge pres.,ure and absolute pre.",ure at a point 4 m below the frec surface of a liquid of
specific gravity 1.53_ if atmospheric pre.<>ure is equivalent 10 750 mm of mereury.
[,\Ils. 60037 Nfm ' and 160099 N/m ' ]
Find the gauge pressure and absolute pressure in Nltn ' at a point 4 In below the free surface of a liquid of
sp. gr. 1.2. if the almospheric prcssure is equivalent to 7~ mm of tnCrcury
[Ans. 47088 NI11l '; 147150 NI11l ' j
II tank contains a liquid of specific gravity 0.8 Find the absolute pressure and gauge prcssure at a point.
which is 2 m below the free surface of lhe liquid. The atmospheric pressure head is equivalent 10 7(1) mm
oi" mercury.
[Ans. 11 7092 Nf11l' ; 15696 Nfml j
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... 3. 1
INTRODUCTION
This chapter de~ls with the fluids (i.e .. liqu ids ~nd gases) <l1 rest. This mea ns thm there will be no
relative motion between ndjilcent or ne ighbou rin g fluid Inyers. The velocity gudient. which is eq ual to
Ih e change of velocity between two adjacent fluid layers div ided by the di~tnnce between the layers.
will be zero Of dll '" O. The shearstrcss which is equal to IJ
d)"
all
a)"
wil l also be Zero. Then the forces actin g
on the flu id particles wi ll be :
I. due to pressure of fluid normal to the surface.
2. due to gravity (or
... 1.2
sclf~weigh t
of fluid particles) .
TOTAL PRESSURE AND CENTRE OF PRESSURE
Total pressure is defined as Ihe force exerted by a sta ti c fluid on a surface either plalle or curved
when the fluid comes in contact with the surfaces. This force always acts nonnal to th~ surface.
Ce nt re o r pressu re is defined as the point of application o f the total pressure on the surf~ce. There
are four ,",~ses of su bmerged surfaces on which the total pressure force and ,",entre of pressure is to be
dctemlined. The submerged surfa,",es m,ty be :
I. Venical plane surface.
2. Horizontal plane surface.
3. Inclined plane surface, and
4. CUfved surface .
... 1.1
VERTICAL PLANE SURFACE SUBMERGED IN LIQUID
Consider a plane venical surface of arbitrary shape immersed in a liquid as show ll in Fig. 3.1.
Let
A", Total area of the surface
II '" DiMaiKe of C.G. of the area from free surface of liquid
G '" Centre of gravity of plalle surface
P '" Centre of pressure
h· '" Diswnce of centre of pressure from free surface of liquid.
69
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170
Fluid Mechanics
(a) Tot a l P ress u ~ (1') . The total pressure 011 tlie surface
may be determined by dividing the emire surface into II numbe r
of small parallel strips. The force 011 small strip is then calculated and the tOlal pressure force on the whole area is calculated
by illtcgrating Ihe force on smnll Slrip.
Consider a strip of thickness til! and width /J al II depth of II
from free surface of liquid as shown in Fig. 3.1
Pressure intensity On Ihe strip.
I' == pgll
(See e quation 2.S)
Area of the slrip.
dA",bxdl!
Total pressure force on Sirip.
dF=pxArca
=pghxbxdh
FREE SURFACE OF lIaUID
"'T""'" ,-,',',',','"'"''1''1' '
,
1/ '\
"
"L
J,
!
,.
,
b
_
A
G.
Fig. 3.1
Total pressure force on lite whole surface.
F= f dF= jpghXbXdh==pg j bXhXdh
""'
J bX/,Xdh '" J hXdA
'" Moment of surface area about the free surface of liquid
= Arc~ of sorfa<:e x Distan<:e of e.G. from free surfa<:e
=Axll
F", pgAh
... (3 . 1)
For wmcr th e value of p = 1000 kglm 3 alld g '" 9.81 mfsl. The force will be ill Newtoll.
(b) C CIl I rt' of P ress u re (h t ). Centre of pressure is cal<:ulntcd by using the "Principk of Moments",
which states thatlhe mOlllent of the resultant force about an axis is equal 10 the sum of mom en IS of the
components about the same axis.
The resultan t force F is acting at P. at a distance h· from free surface of the liquid as shown in
Fig. 3.1. Hence moment of the force'" about free surface o f the liquid'" F X /1*
...(3.2)
Moment of force dF. acting on a strip about free surface of liquid
I ': dF=pghxbxdh!
'" dF x II
"'pghxbxdllx/i
Sum of mom e nts of all such forces about free surface of liquid
'" f pgllXbXdllXII=pg f bxhxlldh
",pg f bh~dh = pg f h ~dA
B",
(.:
bdh",dA)
fll ldA '" f billdh
'" Moment of Inertia orthe s urface about free surface of liquid
'" '0
Sum of moments about fn"C surface
= pglo
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...(3 .3)
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Hydrostatic Forces on Surfaces
711
Equatin g (3. 2) and 0 .3), we get
F x /!*=pg ' o
B",
F = pgAl1
pgAl1 X h*
= pg fa
1,·:= pg'?..
pgAh
0'
=~
...( 3 .4 )
All
By Ihe th eorem o f parallel ax is. We have
I O= IG + Ax h 2
where I e "M omc rll of ln cnia of ar(! 3 about an axis pass ing through the C G . of Ihe are a and parallel
\0 th e fre e s urface of liquid.
Substitutin g 10 in equat ion (3.4), we get
,_
•'o
,',-,:
+~A~I<_' ~ + -II
... ( 3 .5 )
AI<
AI<
In equat ion (3.5), h is the di stan ce o f c.G. o f the area o f Ih e ve nica l surface from fre e surface of
Ih e liqu id. He nce frolll equati on (3.5), it is c lear that :
(i) Cent re of pressure (i.e .• 11*) lies be low the centre of gra vit y of th e ve rti cal surface.
(il) The di stance of ce ntre of pressure from free s urface of liquid is independe nt of th e de ns it y of Ihe
liquid .
h* =
Tab le 3 .1
The m o ments o r in.orlia a nd olile r
Plime ,"utfOi'e
g~'tIm etric
CG. frum lilt
propcrliC5 or so m e im port an t p la ne surfaC<.'!l
Arto
,=,
I
Moment of inerTia
abowl ott axis passing
through CG. (mil
Moment of
inenia aboUi
hal'/' (101
parallel to hme (ld
Rectangle
I
T
--- -,
I
r
i 1
,.
G
d
x ~ -
2
bd
>d'
--
>d'
-J
,
bh'
-36
-
1--' ----0<
"
Triangle
,/1\i
. ~
x~
•
-
3
-bh
bh'
"
COllld ..
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172
Fluid Mechanics
Plull/'
c.G. from lhe
sur/au
1M",
I
Area
Momem of i"alia
/IIu"",,,,1 of
I'lbouIIITI axis lUlSS;IIK
i,jfrtiu aboul
rhrough C G. lInil
base (lu)
puralleilO b" se
(/01
3 Circle
)/
,
Gj 1
~
j"
.w'
d
--
.1'=-
2
4
.w'
-64
-
4. Trapezium
fzL~IG~
I
"
x_(2l1+b)~
IHb
3
(u +b )
---d
2
1
1
(u +4ab +b )Xh1
36(o+b)
-
If
Pro blem 3 .1 A ,ec/(lIIg/liar plane .mrface is 2 III wide ami 3 In deep. II lie.! in I'errical plane in
W<lIer. Dererln;/I!' rile 100al pre.!.lure and position of celllre of pressure 0/1 II,e plane surface lI'ile/l its
upper edge is horizollIal and (a) coillcides ",ill, waler surface. (b) 2.5 m below II,e f ree water surface.
Solution. Given :
Width of plane surface.
b=2 01
d=) 111
[)cplll of plane surface.
(a) Uppu edg e coincides with water s urfa ce (Fig. 3.2). Tot:,1 pressure is ~ i\'c n by equati on (3.1 )
f" = pgAI.
wh ere
P == 1000 kg/m
A ", 3 x 2 = 6
J
,
FREE WATE R SURFACE
11 == 9. 81 I11 /S1
,h
111 •
"' "2I (3) '"
15 m.
F = lOOOx9.81 x 6 x 1.5
'" 88290 N. A ns .
Depth of l'C rme of pressure is give n by eq uati on (3. 5) as
I"
AI<
-
11· =~ + ll
where
'G= 1'.1 .0.1. about e.G. o f the area of surface
Fig. 3.2
J
J
bd
2x3
4
" - " --=4.5m
12
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Hydrostatic Forces on Surfaces
73 1
4.5
h· == - - - + 1.5 == 0.5 + 1.5 == 2.0 m. " ns.
6 x 1.5
(b) Uppe r l..t g e is 2.5 III bel ow water s urface ( Fig. 3.3). T01aJ prcssur~ (F) is given by (3. 1)
F == pgAh
wliere
ii ==
==
WATER SURFACE
Distance of CG. from free surface of wate r
2.5 +
'23 =4.0 III
F = IOOOx9.8 1 x6x 4.0
= 235440 N. Ans.
I
-
Centre of pressure is given by 11·==-4r+1l
Ah
whe re IG == 4.5. A == 6.0,
Ii
--- -.=- -..
11 °'-0'-'-- 1
.-,
i
2.5m
L:J=0G
3.0m
~O'--:-i J
== 4.0
f--2 m------l
--;;4~.5';-;, +4.0
11·= -;6.0x 4.0
Fig. 3.3
== 0.1875 + 4.0 == 4.[875 == 4.1H7S m. Ao s.
Problem 3 .2 Determine the 10101 pressure on (l circu/(lr p/ale of di(lmelu /.5 m which is pluced
reTfieally ill wuler in sitch II way lillI/ Ihe cenlre of 111/: plale is J In hdow I/ll! free ~'urface of ,.."Ier. Find
the pUJ'ilion of en,lre of pressure (I/SO.
FREE SURFACE
rU
---umuum
------ -- - j
---------
Solution. Given: Dia. of plate, II == 1.5 m
Area.
.-
It = 3.0 III
Total pressure is give n by eq uati on (3. J).
F = pgAh
= 1000 x 9.81 x 1.767 x 3.0 N
3.0 m
GOL\r - - o ,
+
<-
= 52002.11 1 N. Ans .
Position of centre of pressure (11·) is g iven by equation (3.5).
I"
-
h·=~ + h
Fig. 3.4
Ah
4
whe re IG
rtd
It X 1.5
=-=
64
64
1----1.5 m --..j
4
= 0.2485 Ul~
0.2485
1. 767 x ].0
+ 3.0 = 0.0468 + 3.0
'" 3.04611 m . ADS.
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174
Fluid Mechanics
Problem 3.3 A reC/lwgl/lar .fluice gme is silUllleil QII lite I'u/ieal \\'01/ of a lock. Til<' I'errical side
oJfhe sluice is 'd ' metres ill lellglh ami deplh ofcel1lroid of tile area is 'p' III below Iile W<lIer surface.
Pro"e Ilwl Ille depth of pre:>5j,re is eq"ai /0
Solution. Gi ve n :
[kplli of vc n kal gale
lei (h e w idth of gate
1
FREE SURFACE
' .
p + -d (
/2p
11
t--- b
i Ll=
~
P
h'
r ij
= lI m
",b m
Are a.
A=b x d m 2
[krill of e.G. fro m free surface
Fig . 3.5
h =p lll .
Le i 11* is the depth o f ce ntre of pressure from free surface. whic h is give n by equati on n.5) as
bd l
I,· '" ~ + h . where I e '" - '2
AI>
I
_
[bd'/
11·= -
12
/J x,/ xl'
1+ P= - - +'
d'
12p
d'
p + - - . Ail s.
12,
Problem 3 .4 A circular opening , J //I diameter. in (/ ref/rcal side of a lan k is closed b.y a disc of
J //I diameler y,·/tich cem fo/ule about (I 1I0ri2O"I(,' diameter. Calculale :
(i) Ihe fOTn~ UI! Ihe disc. alld
(ii) Ihe IOrque required 10 maiMail! Ihe disc il! equilibrillm ill Ihe "erlie,,1 POSilioll whell the head
of"''''er abol'e Ihe IwriWIII(II diameter is -I m.
Solution. Given :
Dia. of openin g.
Are a.
d=3m
"
A = - x3 = 7.0685 m-'
4
Depth. of e.G..
1I = 4m
(i) ['orce on (he di sc is given by eq uation (J .I) as
f' =
pgAh = 1000 x 9.8 1 x 7.(l685 x 4.0
= 277368 N = 277.368 kN. Ans.
(il) To find th e torqu e requ ired to ma inta in th e di se in equilibrium , first c alculate th e poi nt o f
appli cati on o f force ac ting on th e di sc. i.e" ce ntre of press ure of th e force F. Th e depth o f cent re of
pressure (//. ) is give n by equatio n 0 .5) as
I"
-
.!!.- d"
c<64",
· __
h·= ~ + ll = -;;
All
d'
: d l )(4.0
+ 4.0
J~
: ~'--o,,+ 4.0= ~'--o" + 4.0 = 0.14+ 4.0 = 4. 14 m
16x4,0
16 )(4.0
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Hydrostatic Forces on Surfaces
B
U
75 1
----------'1- TT
------------------------------ - --0_::-- 'm
".-
10 ,
,-
3m -' -
1..
,
fig. 3.6
The for(;c F is ad;"g at a distance of 4. 14 m from free surface. Momcll1 uf this fOKe about
horizontal diam eter X-X
'" F x (h ·-ii) " 277368 (4. 14 - 4.0) '" 3HIBI
Nm. A IlS.
Hence a torque of )883 1 Nm mu st be applied o n the di sc in Ihe c loc kwise di rect ion.
Problem 3 .5 A pipe line II'lIic/1 is <I m in diameter conraillJ a gille roll'c. The pressure at rhe celllre
of 'he pipe is 19.6 Nkm!, If rhe pipe is filled wilh oil of sp. gr. 0.S7. find I/Ie force exerted by the oil
"PO" rhe gme alld POSilioli of c.'lIlre of pressurt!.
Solullon. Given:
Dia. of pipe.
"=4 rn
OJ
19.6
Fig . 3.7
Area,
A=~x 42 =4 1ln1 2
4
S = 0.87
Density o f oiL
Po = 0.87 x 1000 = 870 kg/m "'
Weight dcnsily of oil.
"'o"Po xg,,870x9.8 1 Nfm 3
Pressure at Ih~ ct;! ntre of pipe. P " 19.6 Nfcm 2 " 19.6 x 104 Nfm2
Sr . gr. of oil.
Pressure head al Ih e
cenlr~
p
19.6 x I0·
11'0
870 x 9.8 1
=-:
'" 22.988 m
The heig ht of equiva lent free oi l surface from the ce ntre of pipe" 22.988 Ill.
The depth o f e.G. of Ih e gate va lve from free oi l surface II = 22.988
(i) Now Ihe force exef(",d by Ihe oil on Ih e gate is given by
nl.
F = pgAh
where p = densily of oil = 870 kg/m 3
F = 870 x 9.8 1 X 4n x 22.988 = 2465500 N " 2.465 r.I N. Ails.
(ii) Posi li on of !;enlre of pressure (".) is given by (3.5) as
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176
Fluid Mechanics
'" 0.043 + 22.988 '" 23.031 m. Ans .
Or celltre o f pressure is be low the ce lllTC o f Ih e pipe by a di st anc e of OJl43 1ll. AilS.
Problem 3 .6 Determine tile lowl pressure atld celllre of pre.uure all WI i,I'osedes Irilmgular plale
afbase" In and allilude -I In wllell il is immu.ied I'erlically ill WI oil of 5p. gr. 0.9. Tire base of Ille plme
coincides ",ill, Ihe free .m rface of oil.
FREE OIL t-- 4 m-------i
Solution. Given:
Base o f plale.
Height of pl~IC,
Area.
Sp. gr. of oi L
,
b=4m
11=4111
~
A: bxh '" 4)( 4 =8.0 m2
2
2
s= 0.9
Fig. 3.8
l
DenS it y of oil.
P '" 900 kg/m ,
The dis tance o f e.G. from free surface of oil.
I
I
II = -xh=~)(
3
3
4
'" 1.33m.
Total pressure (F) is give n by F = pgAT.
'" 900 x 9.8 1 )( 8.0 x 1.33 N '" 9597.6. N. AilS.
Ccmrc of pressure (11* ) from free surface o f oi l is given by
II+= 'G..+h
AI<
where fc = M.0.1. of triangular section about its e.G.
bI/ 4 x4 l
= - - = - - - =7. 11 m4
36
II. =
7.1 1
36
8.0 X 1.33
+ 1.33 = 0.6667 + 1.33 = 1.99 m . AilS.
Problem 3 .7 A "ulical sluice gale is used IQ CQ l'er an opl'liilig in a dam. The opnlillg is 1 m wide
and 1.1 m /Iigll. 011 111l' IIp_lIream ofl/l e gale. I/Ie liquid of sp. gr. 1.45. 1il'.1 "1110 {/ Iwig/II of 1.5 m
abO"e I/Ie lOp of I/Ie gate. "'/Ierea.lon I/Ie dO"'II_lIream .fide Ihe waler is ami/able IIplO {/ heiglll 101lch·
illg Ihe lap of Ihe gate. Filld Ihe resulumr force acring on Ille gate alld pQ.lirioll of celi lre of pressure.
Find also Ihe force aCling horizolltally at rhe fOp of Ihe gale .... hich is capable af opening ir. Assume
Ilwl rile gille is hinged all/U' bOl/om.
Solullon. Given:
Width of gale.
b = 2 III
[kplll of gate.
d=1.2 m
Area.
Sr . gr. of liquid
I I
A =b xd= 2 x 1.2= 2.4m 1
= 1.45
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Hydrostatic Forces on Surfaces
PI = 1.45 x 1000:: 1450 ~ gJllI '
De ns it y of liqui d .
The force P, is g ive n by
where
77 1
FI
= Poree exerted by the fl uid of sp. gr. 1.45 un ga te
F~
= POfce cxen cd by wa ter on the gate.
F(:: Plg)( A x
h,
PI'" 1.45 x 1000 '" 1450 kgl m2
h , '" D epth of e.G. of gate f rom free surface o f liqui d
:: 1.5 +
21. 2 :: 2. 1 m.
FREE SURFACE OF LIQUID
UQUIOOF
Sp . gr.=1.45
FREE SURFACE
F OF WATER
UPSTREAM
DOWN STREAM
HINGE
Fig. 3.9
F,= 1450 x9.8 1 )(2.4 x2. 1 =7 169 1 N
Similarly.
l
where p! '" 1.000 kglm
F2 :: p~ . A;;l
/,!:: Depth oren . of gmc fro m free surface uf watcr
",.!.xl.2=O.6 m
2
F 1 :: 1000 x 9.81 x 2.4 x 0.6:: 141 26 N
( i) Rl."sultanl fo r ce un th e gule '" F, - F2 = 7 169 1 - 14 126 = 57565 N. A ns.
(ii) Pos ition of ce ntre of pl't'ssure of ...,s uUan! fo rce. Th e fo rce PI will be acting al a de plh of
11, * from free surface of liqu id. g iven by the rel ati o n
hl* =
I,
-
~ +h ,
Alii
where
hl*=
.288 +2. 1 =0.057 1 +2.1=2.157 1m
2. 4 x 2.1
Di sta nce of FI from hinge
= ( 15 + 1.2) - h 1* '" 2.7 - 2. 1571 = 0.5429 m
The force F2 will be actin g at a dep th of 11 2- f ro m free surface of wate r and is given by
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Fluid Mechanics
4
-
IG = 0.288 rn , /'1
= 0.6
JIl.
II!- '"
A = 2.4
2
10 •
0,288
+ 0.6 = 0.2 + 0.6 = 0.8
2.4 xO.6
III
Distance of F1 from hinge
'" 1.2 - 0.8 = 0.4 III
The r~sul ta nt force 57565 N will be acting at a distance given by
= ,7,1069clcXc.o54~2,9~-ci-l4cl,206cXcO",.4
57565
=
3392 1- 5650.4
m above hi nge
57565
= 0.578 m abo ....~ the hin ge. An s.
(iii) Force at th e 1011 of gale ",hich is capa ble of o pe nin g the ga le. Let F is tile force required
on Ihe lOp of the gmc to open ;1 as shown in Fig. 3.9. Taking the moments of F. 1', and F1 about Ihe
hin ge. we get
Fx [. 2 + 1'2)(0.4 =F, )(.5429
F=
or
=
f; x .5429 12
,.~
-
x 0.4
7169 1 x .5429 - 141 26 x 0.4
38921 - 5650.4
1.2
1.2
= 27725.5 N. Ail S.
Problem 3.8
A caisson [or closing tile elllrance 10 Ii dry dock iJ' oflrapezoidalform 16 III wide a/ Ille
rop wId 10 III wide a/ lire bOl/om and 6 III deep. Filltlille lotal pre~'SlIre allli celltre of pres.~I1'" Otl Ille
caissoli if IIII' water 011 IIII' o"lsidf' is j"~'1 /ne/ ...ill> IIII' lop and tlock is emply.
Solution. Given:
Width al top
= 16 m
Width al bottom
=lO m
Depth.
tI=6 m
Are a of tr apezoidal ALJCD.
WATER SURFACE
,..._ _ _ t 6m
A
,
=:\
' 1
'0
A = ,I~
B C,-;;+~A~D") x If
2
Fig. 3.10
"
180 + 36
78
= 2.769 m from water surface.
(i) Total P....,ss ure (F). Totill pressure. F is given by
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79 1
Hydrostatic Forces on Surfaces
,.. '" pgA h
'"
1000 x 9.8 1 x 78 x 2.769 N
'" 2118783 N '" 2. 11 87113 MN. A ns .
(ii) Cen trt' of ]'ressurt' (ht). CcmTe of pressure is
II~ '"
I
~iven
by
(3.5) as
cq u ~lion
-
c:.. + II
AI,
where fG = M.O .I. oftra[X:zoidal ABeD about its c.G.
'c;, '"
M.O. I. of rectangle FBeE about its c.G.
IG, '" M.O.1. of t wO tos ABF iHld ECD
bd J
'r;, '" [2=
Then
IOx6
12
~bout
its CG.
~
J
= 180
In
1<;, is the M. O.1. of the recltmgle about the axis passing through G,.
M.O. I. o f the rectangk atJ.oul the axis passing th rough the e.G. oflh.e trape zoida l 1(;, + Area of
I1'c[ang le x .f,l
where .r , is distance between the CG. of rectang le and e.G. of trapezoidal
'" (3.0 - 2.769) '" 0.231 III
M .0.1. of FIJCE passing through CG . of trapezoi dal
'" 180 + 10)(6)«0.231 )2 ", 180+3.20= 183.20m4
Now
IG, = M.O.1. of MBD in Fig. 3. 11 about
=
(16 - 10) X6 3
3b
=36 m
G~
=
bd'
J6
~
The distance between the e.G. of triangle and e.G. of trapeZOidal
'" (2.769 - 2.0)" 0.769
M.O. I. of the two ~~ about an axi s passing through C.G. of trape zoi dal
== la, + Area of triangles x (.769)2
6 x6
= 36.0 + - - x (.769)
I
10~64
Ie
== 36.0 +
= 46.64
= M.O.1. of trap.:widal about its e.G.
= M.O.1. of rectangle about the CG. of trapezoidal
=
F. E
A == 78.
Ii
I ,
-
-4. + II
Ah
0
;---T----;~
2
G.,.----- ____L
I
j
6m
+ M.O.1. of triang les about the e.G. of the trapezoidal
4
183.20 + 46.64 = 229.84 m
II" ==
where
A
_
"C
Fig. 3.11
== 2.769
h* "
78x2769
+ 2.769" 1.064 + 2.769" 3.833 m . Ans .
Alter nate Met hod
The diSlance of th e C.G. of the trapezoidal channel from surface AD is given by (re fer to Table 3.1
on page 7 1)
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Fluid Mechanics
x=
=
(2(1 + /1)
(a + b)
II
x-
3
(2xlO + !6)
(10 +1 6)
6
)( 3
a= IO,/>= 16andlr=6)
(.:
'" 36 x2=2.769In
26
This is also equal ro th e di stance of III" CG. o f the trapezoidal from rn:" surface of water.
Ii '"
Total pressu re.
2.769 m
F = pgAiI
(":
A = 7S)
'" 1{lOO x 9.81 x 78 x 2.769 N '" 211H783 N. AilS.
CCIHrc o f Pressu re.
Now Ie; fro m Tab le 3 .1 is g ive n by.
(a 1 + 4ab + b 1)
} (10 1 +4 )(IO XI 6 + 16 1)
l
IG",
36(a + b)
xii '"
36 (10+ 16)
x6
'"
(100 + 640 + 256)
36x26
x216=229.846m ~
229.846 + 2.769
78 x 2769
'" 3.1B3 m. AilS.
A lrapezoidol channel 2 III wide at Ille borrom (lnt! I m deep lla.! side slopes I : I.
1/. '"
Problem 3 .9
De/ermine:
(i) Ihe lOW/ pressure. and
(ii) the cenlre of pressure 011 the renieal gme closing rhe chalille! whell il is fil II o/water.
Solution. Gi ve n :
Width al bonom
!kplh.
Side s lopes
.. Tupwidlh.
Arcil uf rectangle FBEe,
=2 m
t - --
d= lm
'" I : I
.
m
-
A WATER SUR FACE
-<
,
AD==2 + 1+1==4m
AI'" 2 x l '" 2 m!
(4 - 2)
2
,
Area of TWO Trian g les ABF and ECD. A, '" - - - x I '" I 111-
.
Area uf Trapezo idal ABeD ,
A == A, + A1 == 2 + I == J m 1
DepTh uf e.G. of rl'CT angl e FBEe from waTer surface ,
..
/1 1
I I
Fig. 3.12
I
"' "2 ", 0.5 m
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Hydrostatic Forces on Surfaces
811
IXplh o f CG. of two triangles AUf' and ECD from water surface.
I
I
II I " "3 x I '"
III
"3
Depth of CG. of trapezoidal ABeD from free s urface of wmcr
-
-
A,xh , +A , x/1l
II ==
•
(AL +A!)
2 x 0.5+ I x 033333
" .44444
(2 + I)
(i) Totu! Press ure (Fl. TOlal pressure F is given hy
F == pgA h
= IOOOx9.81 x3.0 x 0.44444 = 13079.9N. Ans.
(ii) Cenl ..., of Pressure ( h*). M.O .I. of rectangle FB eE aboul its CG.,
bd J
2x] J I
fo = -- = -- = - Ill~
I
12
12
6
M .0 .1. of FBeE abou t an axis passi ng through the e.G. of trapezoidal
'c,· 'c,
==
+ AI x [DiSlancc he tween CG. of rec tangl e and CG.
of lra p.:zoidalf
I
,
'" - + 2 x 10.5 - .4444 1- " .1666 + .006 182" 0. 1727
6
1'.1.0.1. orl he twO triang les ABF am.! ECD about th eir C.G .•
M.O.I. of the
IWO
triangle s aboul1hc CG. o f trapezoidal.
I e: == I e, + A2 x [Distan ce bet wee n CG. of triang les and e.G.
of lrapczo idall l
'"
1~+ IX [h-h2r "'~+IX [.4444 -*r
'"
l~
+ (.1 111 )1",0.0555+ (. 111l)1
'" .0555 + 0.01234 '" 0.06789 m 4
M .a.]. of lite lrape zoi dal aboul il~ e.G.
' o '"
10,
. + 10 , • '" . 1727 + .06789 '" 0.24059 m
4
Tlten centrc of pressu re (h*) on lltc vcrtica llrapczoidal.
11* '" "'- + Ii
Air
'"
0.24059 + .4444 '" 0.18()4{, + .4444 '" 0.6248
3x.4444
'" 0.625 m. AilS.
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182
Fluid Mechanics
Altt'rnate M eth od
The diswncc oflhe C.G. oflhe trapezoidal channel frum su rfa ce AD is given by (refer to Tab le 3.1
on page 7 1).
x=
(2a + b)
«(I +b)
II (2)<2+ 4) I
x -=
x-
3
(2+4)
( .;
3
a=2,b=4andh= I )
== 0.444 m
h =X=
0.444 III
F=pgAh= 1000><9.81 >< 3.0 x.444
Total pressure.
(":
A
= 3.0)
= 13079 N. Ans .
'"
h*=-b+I,
AI,
Cemre of pressu re.
where 1(1 from Tab le 3
is given by
Ie;'"
(a l + 4I1b + b'-)
36(II+b)
0.2 407
3.0 x.444
3
xII =
(Zl + 4xZX4+4 l)
36(2+ 4 )
3
52
4
x l = - - - =01407111
36x6
+ .444 '" 0.625 m. Am .
Problem 3.10 A sqlwre oper7llre in the rerrical side of a ((III/.: Iws Dill' diagonal I'errical and is
camp/cleiy col'ered by a p/alle plate Ilil/ged a/ollg olle of Ihe upper sides ollhe aperlllre. The diagonals
of Ihe oper/ure (Iff! 2 III /ollg and II,e Ilmk COll10illS Ii liquid of specific gral'ily 1./5. Th e cerllre of
aperture is /.5 /II heloK' r/,,,, f ree surface. Co!c,,/rlle the thrusl exerled Oil Ille p/ale by 1111: liql/it! (lnt!
posililm of ib aI/Ire of pressu re.
Solution. G ive n : Diagonals of aperture. AC", lJD", 2 m
Area of square aperture. A = Area of ~ClJ + Area o f 6ACD
ACxlJO
2
ACxOD
+
2
Sp. gr. of liqu id
= 1.15
Density of liqu id.
p: 1.15x 1000= 1150kgfm 3
Depth of centre of aperture from free surface.
Ii '"
- .,
' . '. ' . '_
1.5 Ill.
,
.. -" ' . ' , '
SQUARE
.6.PERTURE
1.5m
Sp . !/f.,1.15
Fig. l.U
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Hydrostatic Forces on Surfaces
83 1
(i) The thrust on th e plate is give n by
F", pgAh = 1150x9.8 1 x 2 x 1.5 = 33844.5. Am.
(ii) Centre o f pressure (/1·) is given by
Ie
I,· '" ---4"
+ II
AI<
where
'G= M.O.I. of ABeD about diagonal AC
= M .O.l . of triangle ABC about AC + M.O.! . of triangle ACD about AC
ACxOB l
=
="",,'-+
12
2 X 13
="""'ACxOOJ
2 x] J
(.: M.a .l.of a tria ngle aooulits base '" b/;J)
12
I
1
12
12
6
6
~
I
= --+--=- + - = -
3
h· '" - '-
III
2x l.5
+ 1.5 '"
3x2xl.5
+ 1.5", 1.6 11 m. An s.
Problem 3.11 A /allk cO/lwills waler lipta a heighl 0[0.5 m obare fh e base. All immiscible liquid of
sp. gr. 0,8 is filled an ,lie lop a/ willer "plo I m height. C(l/cuhlle ;
(i) lollli presJ'ure Oil Olll' ~'ide ofille Illllk,
(ii) the pOSilioli of C,,"lre of prcn'ure for 011t' 'iide vf Ihe /,mll., which is 2 In wide.
Solution. Gi ven:
[krIll of wate r
'" 0.5 m
", 1m
De pll! of liqu id
Sp. gr. of liquid
'" 0.8
[kosity of liqu id,
P I '" 0.8 x 1000 '" 800 kg/m l
1
[kosity of water.
P2 '" 1000 kg/m
",2 m
Width of tank
(i) Tota l press ure on on e side is ca lculaK'd by drawing pressure diagram. wh ic h is shown in Fig. 3.1 4.
Intensity o f pressure 00 tOp.
P.. '" 0
Intensity of pressu re on D (or DE). PD'" Pl g.h l
", 800x9.81 x 1,0=7848 N/n/
,
,
Fig. 3.14
Intensity of pressure
Now force
00
hasc (or
no.po '" PlgiJ l + P2E
x 0.5
'" 7848 + 1000 x 9 .8 1 x 0.5", 7848 + 4905 '" 12753 N/m 2
F t '" Area of MDE x Width of tank
I
I
'" - x AD x DE x 2.0 = - x 1 x 7848 x 2.0 '" 7848 N
2
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184
Fluid Mechanics
F 2 = Area of rectangle DI1fT x Width of lank
Force
=O.5x 7848 x 2 = 7848 N
F J = Area of !:.EFe x Width of lank
'" "2I )( EF x Fe)( 2.0 '" "2I x 0.5 x 4905 x 2.0 '" 24525 N
F = F, + F, + F}
:. Total pressure.
'" 7848 + 7848 + 2452.5 '" 18148.5 N. AilS.
(ii) Cen tre of Pressu re (h. ). Ta kin g the Illoments of all force abou t A, we get
Fx
,,* '" F, x ~3 AD + F,- (AD + ~2 80)+ F,[AO+
~ 8D1
"
3
18148.5 X /1* = 7848 x
~
x 1 + 7848 (1.0+
O~5 ) + 24525 ( LO + ~X.5)
= 5232 + 9810 + 3270 = 18312
•
183[ 2
== 1.009 m from top. Ail S.
18148.5
Problem 3.12 A cubicol /(Ilik l/as sides of 1.5 III, II coll/oins '\"(Iter for Ihe lower 0.6 In depllt. The
upper remaining pori is filled wilh ai/ of specific gT(I\';I), 0.9. Cu/cr/Ii'/e Jor aile rerticol side of I/Ie lallk:
II =
(u) tolill pT(!SSrtre, Ulld
(h) /l05;lioli 0/ c<,,, lre of preSJ'" reo
Solutio n. Given:
Cubi<.;al wnk of sides 1.5
I:kpth of wa ter
I:kpth of liquid
In
means the dimensions of the tank :Ire 1.5 m x 1.5 m x 1.5 tn.
Sp. gr. of liquid
I:knsity of liqu id .
=O.6m
= 1.5-0.6 =0.9 m
= 0.9
PI = 0.9 x 1000 = 900 kglm J
I:knsity of water.
Il l '" 1000 kg/m>
(a) 1'01,11 pressure on one veni cal side is ca lcu lated by drawing pressu re diagram. which is shown
in Fig. 3. 15.
A
OIL OF SP. GR . ~O . 9
,
--
.
- -
- -..
--- - ---- - ~- .
I:
I'
,
,_
7946.1
IF 5686
1383i .t
,
I
:,
Fig. 3.15
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Hydrostatic Forces on Surfaces
85 1
hucnsity of press ure a1 A. p~ = 0
Intensity of pressure ~t D,PD = P ig X II '" 900 x 9.81 x 0.9 '" 7946.1 N/m 2
Intensit y of pressure al 8, Po = P igill + P2gill '" 900 x 9.8 1 x 0.9 + 1000 x 9.81 x 0.6
= 7946. 1 + 5886 = 13332. 1 Nlm l
Hence in pressure diagram:
DE: 7946.1 Nlml . BC = 13332. 1 Nlml, FC= 5886 Nlm l
The press ure diag ram is split into Iriangle AD£. rectangle BDEF and trian g le EFe. The total pres·
sure force consists of (he following components:
(i) Force
FI = Area of lrian glc ADE x Width o f tank
= (t XADx DE) x 1.5
==
(t x 0.9)( 7946.1) x 1.5 N = 5363.6
(,:
Width ", 1.5 m)
N
,
This for<:e will be acting at the CG. of tile triangle ADE. i.e .. al a di>!ancc of ~ xO.9 = 0.6 m below A
(ii) Force
Fl = Area of rectangl e BDEF x Wilhh of tank
= (BD x DE »( 1.5 = (0.6 x 7946. I) x 1.5 ", 7151.5
Th is force will be acting at the CG. of th e rectangle RDEF i.e .. at a di s tance of 0.9 + 0.6 '" 1.2 m
2
oclow A .
( iii) Force
F3
",
Area of triangle EFC x Width of tank
'" (t x EFx
FC)
x 1.5 '" (t xO.6x58&6) x 1.5 '"
2648. 7
This force will be acting at the CG. of th~ triangle EFC, i.e .. at a di stance of 0.9 +
N
"32 x 0.6 '" 1.30 m
below A.
Total pressure force on one ve ni caJ face of the tank.
F", F, + f'2 + FJ
'" 5363.6 + 7 151.5 + 2648.7 '" 15 163.H N. Ans.
(b) I'os ition of ce ntre of pressure
Let the lOtal force F is acting at a depth of 11* from th e free surface of liquid. i.e .. from A.
Taking the mom ent s of all forces :Ioout A. we gd
Fxh* = FI xO.6 + F!x 1.2 + F3X 1.3
0'
11* '"
FI x(t6+ F: x 1.2+ fj x 1.3
F
5363.6)( 0.6 + 7 151.5 x 1.2 + 264& 7 x 1.3
15163.8
'" 1.005 III from A. An s .
=
.. 3.4
HORIZONTAL PLANE SURFACE SUBMERGED IN LIQUID
Consider a plane hori7.ontal surface immersed in a stalic nuid. As every point of the surface is at the
same depth from the free s urface o f the liquid. the pressure intensity will he equal on the e ntire s urface
and equal 10. {J '" pgll, where II is depth of surface.
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186
Fluid Mechanics
FREE SURFACE
A", Total area of surface
'-'F'-'-'-'-'-'-'-'-'-,-,-,-,-'F
Then total force. F. on the surface
h
1\'
"'pxArc3= pgx/! xA '" pgAh
where II '" Deplh orCG. from fr~c surface of liquid == /,
also
Depth of centre of pressure from free surface", II.
Problem 3.13 fig. 3.17 5//011'5 a /(lIIkfull a/water. Find:
(i) Towl pressure on rhe bOI7OI11 of tank.
I,' '"
Fig. 3.16
(il) Weight oj lrlljer ill the I(I/Ik.
(iii) Hydros/alie parado;r be/wee" 1/11' THlllls of (i) (md (i;). lVidlil of /(mk is 2 III.
Solution. Given:
OAm
[Rplll of Wala on bonom of
tan~
----
h,,, 3 + 0.6 '" 3.6m
Width of lank
'" 2 m
Length of tank at bottom
'" 4 rn
Area 31thc bonom.
A : 4)(2" 8m 1
(i) Total pressure F, on the boltOlli is
F:pgAI!" (000)(9.81 x8x3.6
I·
= 282528 N. An s.
(ii) Weight of water in tank", pg x Volume of tank
•I
'm
I 1
'm
Fig . 3. 17
'" 1000 )(9.81 x P x 0.4 )( 2 + 4 x.6 x2]
= 1000 x 9.&1 12.4 + 4.&1 '" 70632 N. Ans .
(iii) Fro m the results of (i) and (ii). it is observed tlial the total we ight of water in tlie tan~ is much
less than the toml pressure at the bottom of the tank. Tliis is known as Hydrostatic paradox .
... l .S
INCLINED PLANE SURFACE SUBMERGED IN LIQUID
Consider a plane surface of arbitrary shape immersed in a liquid in sucli a way that the plane of the
su rface mak es an angle 9 witli the free surface of the liq uid as sliown in Fig. 3.1 &.
FREE LIQU ID SURFACE
Let
Fig. 3.18
A '" Total area of inclined surface
Indined immersed
~urface.
" '" Dcptli ofCG. of inc lin ed area from free surface
". '" Dist ance of centre o f pressure from free surface of liquid
9 '" Angle made by the plane of the surface with free liquid surface.
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Hydrostatic Forces on Surfaces
87 1
Let th e plane of the surface. if prod uced meet the frce liquid surface a1 O. Then 0-0 is the axi s
perpendicular to the pl3n c of the s urface .
y = distance of the e.G. of Ihe inclined surface from 0-0
y. = distance of th e centre of pressure from 0-0.
Consider a s mall strip of area dA at a depth 'II' from fret: surface and at a distance y from tlie axi s
0 -0 as shown in Fig. 3.18.
Pr~ssure ill1cnsil y 011 th e strip_
p = pgh
Press ure force. dF, on the strip. dF = p x Area o f slrip = pglt x dA
Total pressure force o n the who le area, F",
But from Fig. 3.18,
J J
tlF =
pghdA
~ = ~=~ :sine
)'
y
y.
h=ysin 9
F = j pgx yxSin9xdA : pgsin9 f Yi/A
jFM=AY
B"
whae)'
= Distance orCG. fro m axi s 0 -0
F = pgsin9 yxA
= pgA l1
(.;
h= y sin 9) . .0 .6)
Cenl .... of Press ure (h·)
Pressure force o n lhe strip.dF = pglldA
= pgy sin dA
Moment of the force. IIF. ahout axis 0·0
= IIF x Y = pgy si n e dA x y = pg sin
Sum of mom e nts of all such forces about 0-0
e
J
= pg ~i n
B"
J/ dA =
e yl dA = pg sin eJi
y*
Now
= 'P~'c'c;'~'cBc ,.
f
F
".
e]
e lilA
dA
M .O.1. o f th e surfac.: about 0-0
Sum of moments of all forces about 0-0 = pg sin e 10
Mom.:nt of th.: total force. F. aoou t 0 -0 is a lso given by
= Fx y*
where y. = Distance of centre of pressure from 0-0.
Equati ng the two va lues give n by equatio ns (3.7) and (3.8)
Fx y"= pgs in9,o
III = y sin
= 10
..•(3.7)
... (3.8)
...(3.9)
y. = - - . F= pgAI!sin 9
and ' 0 by th e theorem of parallel axis = IG + A~, l .
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188
Fluid Mechanics
Substituting llicsc va lu es in eq uatio n (3.9), we get
~ -_ PKSine[1<) + A.>C'j
sin
e
pgAII
si n ~ e
-,
I,· = --_[Ie + Ay - ]
AI,
I<
B",
==si n9
y
I,· '"
or
I,
y= - si n
""'o[
~
AI<
e
e1
IG + Ax - .-h',Sm
•
Ic si n ' S II =
+ 11
"'
... (3 . 10)
AI<
If
e '" 90 °, equatio n (3.10) hecomes same as eq uation (3.5) which. is app licab le to vert ic all y plane
submerged surfaces.
In equation (3.10), Ie '" M.O.1. of inc lined surfaces about an ax is passing th rough G ami parallel to 0 -0 .
Problem 3.14 (a) A fec/ung,,{ar pit", .. J"ur!"ce 1 m wide and 3 m d"ep!ies in ,mler iii J'uc/! tI w"Y
tiull il$ plmle makes ' m (lIIgle of 3~ " Will, tile free SlIr/llce of waler. Determine Ihe total preSSlue and
1'0,\';1;011 of [enlre of pre.ISllfe when lile Ilflper edge is /.5 III below IIIl' free water surface.
Solu t ion. G iven:
FREE WATER SURFACE
Width of plJne su rface. b = 2 m
Depth.
11= 3m
Angle,
= 30"
Di~tance of upper edge from free wa la surface = 1.5 m
(,) Total pre._~ ure fo rce is gil'en by equa li o n (3.6) as
e
o
".
F= pgAIi
where p = 1000 kglm 3
z
A=bxd=3x2=6m
Ii
= Depth of c.G. fmm free wate r surface
'" 1.5+ 1.5 sin 30°
Fig. 3.19
'" 1.5+ 1.5xt",2.25m
F '" 1000 x 9.81 x 6 x 2.25 '" 132435 N. Ans.
(ii) Centl't' or press ure (h·)
Using eq umion (3. 10). we ha ve
bil l
2X]3
where I e = - - " - -.- = 4.5 m4
12
12
" C',,30,-' +.5=
22
/r*= -,40.5C'C'O'C
1
4.5x 4
6x2.25
6 x 2.25
= 0.OS33 + 2.25 = 2.3333 m. An s.
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+ 2.25
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Hydrostatic Forces on Surfaces
89 1
Problem 3.14 (b) A rec/angI4/ar pia/ie surface J III wide alld 4 III deep lies ill water ill sudl a way
(lwl ils plalle makes atl ollgle of 30" willi Ille free SlIrface a/water. Determine rhe lotal pressure force
and position of cenlre of pressure. when the upper edge is 2 III below II,e free surface.
Solution. Gi ve n :
Free sufface of water
b", 3 m . d == 4
Ill.
e" 30°
---:".-.
Dist an ce of uppe r edge from rrce surface of wa ter == 2 III
(i) Tolal press u re fon:e is give n by eq uati o n (3.6) as
F= pgAh
where
and
V_ """""I
J
p = 1000 kglm ,
A :bxd = 3x 4 =
h '"
to plate
[2 m 2
Depth of e.G. of platc fro m
rrce wate r s urface
=2 + 3 1:::2 + Il Csi n 0
=2 + 2 ~in
30o= 2 +2x ~ = 3 m
2
Fig. 3. 19 (a)
F", 1000 x 9.8 1 x 12 x 3 '" 353167 N" 353. 167 kN . An s.
Using equat ion (3. 10). we have h· =
where
'e=
,
bd '
3 x41
~
- - = - - - = 16 m
12
12
""""7'""_',,-9 + II
A"
•
16xsin 2300
16x 4
II = "''C;=-''''~ + 3: - 36 + 3: 3.111 rn. An s.
12 x 3
Problem 3.15 (a) A circ14/af plate 3.0 m diameter is immersed in Il'IIter ill such a Imy ,Iwt its
gre(llesl and lea~'1 deplh be/ow Ihe free ~'urface are 4 m alJd 1.5 m fHpecli)'ely. lJelt'rm;IIe '''e loMI
prHSIlre 011 0111' fat'e of Ihe pla/e (lilt! pOS;/;Oll of the celilre of pressu re.
Solution. Giv" n :
FREE WATER SURFACE
Dia. of plale.
Area.
Distance
E
<I:3.0 m
A :
i
d2 :
0
1,5 m
~ (3.0)l: 7.0685 m l
,
DC= 1.5 m.BE= 4m
Dist an ce ofeG. from free s urface
: h :CD +GCsin9 : 1.5+ 1.5sin9
Om
AB
s in 9 =
BE - AE
-8-C= ""-iB~C""- =
4.0 - DC
3.0
=
4.0 - 1.5
3.0
'" 2.5 '" 0.8333
3.0
;; '" 1.5 + 1.5 x .8333 '" 1.5 + 1.249", 2.749 m
I I
Fig. 3.20
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190
Fluid Mechanics
(i) 1'01,,1 pNss ure (F)
F", pgA/'
'" 1000 x 9.81 x 7.0685 x 2.749 '" 190621 N. Am.
(ii) Ct' litre of press urt' (h *)
Using eq uation (3.10). we Ita ve 11· '"
It
/(
4
l G '" 64 (t '" 64 0) '" 3.976
where
4
III
I, . '" 3.976 x (.8333) x.8333
7.0685 x 2.749
+ 2.749 '" 0.1420 + 2. 749
= 2.89 ] m. AilS.
Problem 3.15 (b)
If in the (lbol'l! problem. Iile gi)'en circu lar plale is hu\'ing a COIla'II/ric circular
hole of diameter /.5 m, Ihen calculi"e Ihe lOla/ pressu'e , md positioll of Ihe CClil r /! of pressu'e OIl Oll l!
face of the /llare.
Solution. Gi ven: IRefer to Fig. 3.20 «(1 )1
d=3.0rn
Dia. of plate.
.
1(
,
Area of solid plale '" - d-
4
1t
,
View normal
10 plate
,
= - (3)" = 7.0685 Ill4
Dia. of hole in th e plate. do'" 1.5 III
Area of ho le
Area of the g ive n plate. A '" Area of solid plate - Area of
'" 7.0685 - 1.767 1 = 5.30 14 rn 1
Distance CD '" 1.5. BE ", 4 rn
Distance ofCG. frorn tile
fr~e
Fig. 3.20 (a)
surface.
/'",CD+GCsine
= 1.5+ 1.5sin e
Bm
sine = AB",BE - AE = 4 - 1.5=2.5
BCBC
33
25
h:I.5+1.5X 3" = 1.5+ 1.25= 2.75 rn
(i) Total press u N fOITt' (F)
F: pgA h
'" 1000 x 9.8 1 x 5.3014 x 2.75
'" 143018 N '" 143.018 kN. An s.
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Hydrostatic Forces on Surfaces
91 1
( ii) Pos ition o f centre of press ure (h*)
Usin g equation (3 .1 0 ). we ha ve
/'
I
tr
~4
fc si n2e
'"
Ah
1t
_
+ 11
444
where I c= 64 [d - d o l = 64 [3 - 1.5 [m
- ' [d'
A-
4
sin a ",
2~5
-
d'
[ -- ' [3'
0
and
4
-
15'[
.
!l1 ,
h = 2.75
+ 2.75
+ 2.75 '" I x 11.25 x 6.25 + 2.75
16x 2.75x9
= 0.]77 + 2 .75 = 2.927 m. An s.
Problem 3.16 A circular pll11e 3 metre diameter is submerged ill WaferaJ shown ill Fig. 3.21. Its
wealeH and leasl deplhs are be/ow Ihe surfaces being 2 metre and I metre respeClil'eiy. Find: (i) Ihe
lolal pressu re 011 f rom face o/rhe plole, amI (ii) Ille position of cemre of pressure.
Solution. Gi ve n :
Dia. o f plate.
Are a.
Distance.
Water surtace
d=3.0 m
"
A = - (3.0)" '" 7 J)6R5
4
DC= 1 m.llE=2 m
III
I ,
<-Jit
,
t
- f-- --',' m
,
All IJE - AE BE - DC 2.0 - 1.0
sin9 = =
=
AC
Be
Be
3.0
3
The ce ntre o f gra l'i ty of the plate is at the middle of BC. i.<1 .• at a di stance 1.5 m from C.
The d istance o f ce rlt re of gra vit y fro m th e free surface o f th e wat er is give n by
]nMBe,
-
I
" : CD + CG sin e : 1.0 + 1.5 x 3
(":s in O:!)
'" 1.5 Ill.
(i) Tota l pressure o n the front face o f the plate is give n by
F: pgA"ii
: 1000 x 9.81 x 7.0685 x 1.5" 10401 3 N. An s.
( ii) Let the di stance of th e cerltre of press ure from th e free surface of th e wmc r be II·. T hen usin g
equ ation (3. 10). we have
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Fluid Mechanics
where
It tt
IG '" 64
It
4
'" 64 OJ . A",
'4It d-,,
-
II '" 1.5
j
.
III
and Sill e '" _
Substituting the values. we get
, d , x (')'
1
- - + 1.5
, , 3 + ! 5= -16x -9xl.5
-
d~
64
-4 "" x 1.5
3'
= ~--"~~+ 1.5 ", J1416 + 1.5", 1.5416 m. Ails .
16x 9x l 5
Problem 3.17 A rec/alls"lar 8m .. 5 In xl m is /I;nged at its bll$e "IIlI indi/led ar 60" 10 IIIe horizon (a/lIS shown i" Fig. 3.22. To keep Ihe 8<1(e in "Mable position. a coulller II'ciglll of 5000 kgfis a ((ached
aT Il,e upper elld of
8a /e as sholt'/I il/jigll re. Find Ille depll! ofwa rer {I/ which Ihe gllle begins 10 fall.
Neglect Ihe weight of the gtlle and /ricliOIl at rhe Ilinge and /mlley.
Solution. Given:
Lenglh o f g ~le
= 511\
Widlh of gale
=2 m
II,,.
""". ,
9", 60 0
Weight.
--~~ -
IV", 5000 kgf
'" 5000 x 9.81 N
= 49050
(': I kgf = 9.8 1 N)
As the pulle y is fric ti on less. Ihe force actin g at H= 49050 N. First
find th e 100ai force F actin g o n Ihe g ate All for a give n de plh o f
N
"
"
C
A_
-
60
HINGE
Fig. 3.2.2.
wate r.
From figure,
,-
,
AE
II
5
211
AD", sine '" sin 60o"'""J3i2=T3
211
411
2
Area of gate imme rsed in water. A = AD)( Widt h x ""7:' x 2 == ""7:' m
..;3
..;3
"
Also deplh of the e.G. o f th e immcrs.::d area'" -h '" - '" 0.5 h
2
. .
411
II
19620 ,
Tot al force F IS gIven by F = pgA" == 1000 x 9.81 x ""7:' )( ""7:' = ----r.:-- II-N
..;3
,,2
,,3
The centre o f pressure of the immersed surface. h· is give n by
1
lG sin 9 +ii
AI,
where
IG'" M.O.1. of Ihe immersed area
=
I I
bx(AD'J
12
2 (2" y
12)( .J3)
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Hydrostatic Forces on Surfaces
~
II '"
J3)'
[
--- x
+ - =- - + - = - + - =- - " 2
9x13
~x~
J3
3h l
18Ir 2
II
4h l
2
II
II
II
h+3/1
211
2
6
2
6
3
93 1
2
2/,
CH= ,,- '" - 3 . LCDH= 60°
Now in the !J.CHD.
CH '" sin 600
CD
CD=.....0:!...-=~= 211
sin 60° si n 60° 3 x .fj
AC= AD- CD '"
211
411
2
61! - 4h
2h
---"'7ii"
- - '"
./3 3./3 J../3 3./3
Taking the moment s aoou l hinge. we gel
49050)( 5,0 = F x AC =
19620
~
,,3
1
2h
II x-:::-;;:
3,,3
.245250 = 39240 Itl
0'
3x3
III = 9 x 245250 = 56.25
39240
II
= (56.25) 1/3 = 3.S3
m.
Ail S.
Problem 3 .18 A" inclilled re"/{1II8,,/ar J/uice gme All, 1.2 111 by 5 III size as sho wn in f"ig. 3.23 is
iustal/ed 10 cOlll'Olllle di.{Clwrge o/wmer. The end A is/lillged. Determitll' Ille force norma/to lite gMf!
" f'piied ul 8 Iv open il.
Solution. Gi ve n :
A = Area o f ga le = 1.2 x 5.0 '" 6.0 111 2
Dt:plli of e.G . of the gale from free surface of the wate r =
= DC '" 8C _ BE
= 5.0 _ BG si n 45 0
== 5.0 - 0.6 x
h
FREE WATER SURFACE 0
C
0
I
Ji
== 4.576 m
The IOtal pressure furce (1) acting on the gate.
F=pgAII
== l000x9.81 x6.0x4.576
== 269343 N
This force is actin g at H. where the depth of If from
Fig. 3.23
free surface is g iven by
,,* = "'O'-';';'i"_'~. + "AI,
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194
Fluid Mechanics
1
I G" M.O.!. of gate", _bd
_
where
'" 5.0 X 1.2 ) ,,0.72
12
III
12
l
Depth of centre of pressure 11*"
n Ul from Fig. 3.23 (a),
",
0.72 x sin 45°
"'''''''0:::",'''+ 4.576 "
6x4.576
.013 + 4.5 76 '" 4.589 m
__ = s in 45 0
OH
11*
Oi~t3nce.
4.589
OH= - - ' 1 - " 4.589
sin 45°
x.fi" 6.489 In
T2
Distance.
80 = - '= 5)(.fi" 7.071
sin 45°
Dista nce.
8H = 80 - OH = 7.071 - 6.489 = 0.582
Distance
AH
= AD
- BH
III
= 1.2 - 0.582 = 0.618
III
111
Taking the moments about the hinge A
P xA8= F x (AH)
where P is the force norma l to the gate applied a1 B
p x 1.2 = 269343 x 0.618
p= 269343)(0.6 18" 1J8708 N. An s.
L2
Problem 3 .19 A gale .\"upporring Waler i.~ JlloWII ill Fig. 3.2-1. Find the fleig/II II afille \Va/a so I/WI
the gall' lips abolll Ihe hinge. Take Ille ..... idtl! of fhe gale as unit)'.
Solution. Given:
e = 60°
AC = __" _,, 211
sin 60°
Distance.
J3
where II = Depth of water.
The ga te wi ll start tippi ng ahout hinge 8 if the resul tant pressure force acts at B. If the resultant
pr.:ssure force passes through a point which is ly ing from /J to C anywhere on th.: gat.:. the gat.: will tip
ov.:r the hinge. Hence limi tin g case is when th.: resultant force passes through B. But the resultant force
passes through the centre of pressure. Hence for th.: given position. point IJ becomes th e centre of
pressure. Hence depth of centre of pressure.
I,· '" (II - 3) m
fREE WATER SURfACE
But h· is also given hy
Taking wid th of gat.: unity.
Area.
Th~n
211
-
II
A=ACxl=~xl:II= -
"1/3
2
2/,
Ix ( "0/;'
"3
12
I I
)3
Fig.3.24
gil '
211 ,
"'12x3xlf,"'9xJJ
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Hydrostatic Forces on Surfaces
95 1
Equa ti ng the two v al ues of 1/*,
2h
3
11 - 3= -
or
11 -
2/,
3
=3
~3 '" .3
or
h=3x3=9m
Height of water for tipping the gate = 9 m . An s.
Problem 3 .20 A reCllmglflar sluice gale An. 1 In wide Wid 3 m 10llg is hinged at A as ,~lIo\\'n ill
"-ig. 3.25. II is kepI closed by a weiglllfl.I"<,J 10 Ille gare. 11w IOwl weighl ofrlw gale {/tulwl'igll1jixed 10
Ihe gmt' is 3-13350 N. Find Ihe height of tile water 'II' wlliell .... ill jlw eOfue 1111" gale to open. Ti,e anlre
olgral'il}, ofrhe ...eight and gale is a/ G.
Solution. Given:
Width of g,nc.
Area.
b '" 2 m: Length or gate L= 3
A=2)(3=6m
Weight of gate ami
W.:: 343350 N
Angle of j"clination.
e '"
III
1
45°
Let /, is the required height of wale r.
Depth of e.G. of the gate and weight = II
From Fig. 3.25 (a).
h '" h -
ED '" II - (AD - AE)
'" II - (AB sin
e - EG Ian 9)
AE :. AE =
{ ... I,m& = EG
EG I~n&l
= II - (3 sin 45° - 0.6 Ian 45° )
::/! - (2.121-0.6)::(IJ-1.521)m
The 10lal pressure force. F is given by
F:: pgAii :: 1000 x 9.81 x 6 x (II - 1.521)
:58860(h - 1.52 1)N.
The IOlal force F is <I<:ling allhe cen u c of pressu re as shown in Fig. 3.25 (b) al H. The d cp lh of H
from fR..., surface is given by It' which is eq ual 10
sin ~ & -,
bd )
2x))
54 4
'
I,· :: ic;, All
+1.w hcrci c = - -= - - = - =.5m
12
12
12
' cX":::"'c'c4",'o + (II C
1.52 1)
C'
I, ' = ~4
6x(1t
I I
cO,.3c7'o'~ + (II -
1.521) = --;
(h
1.521)
<0 I) m
1..,_
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196
Fluid Mechanics
FREE WATER SURFACE}
--.-.-.-.-.-.-.-.--
~
•
1,
'".
-
" t-e '" 45"
, "
-_ -_ -. K-. --
e
:: F~ -
(b)
(a)
Fig. J .ZS
Now laking rn om,," \s about hin ge A. We gel
343350xEG",FxAH
343350
X
[
AK
0.6 '" F x - sin 450
rro m AAKH. Fig. 3.25 (b) AK '" AU sin ij ", AH sin
=
AK=
Bw
Bw
45~ ,', AH '" ~l
sin 45
0
58860 (Il - 1.521) x AK
343350 x 0.6 x sin 45" 03535 x 7
'"
58860 (" 1521)
(II 1.52 1)
A K = h· - AC =
~c3"7c5= + (II _
... ( i)
1.52 1) - AC
(1i - 1.52 1)
... ( ii)
AC = CD - AD", II - AS si n 45° '" II - 3 X sin 4:;0= 1,- 2. 12 1
Substitut in g th is value in (;,). we get
5 C" +(h - 1.52 1) - (1I - 2.12 1)
AK: -:-~J"-7,,,
11-1.521
=
375
-:-'7CC:
+ 2. 11 1 Ii 1.521
152 J '"
375
-:-'7CC:
II 1.521
+ 0.6
.. ,(i ii)
Equating the two va lu es of AK from (i) and (iii)
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Hydrostatic Forces on Surfaces
971
03535 x 7
0375
+ 0.6
11 1.521
"1.52 1
0.3S35 x 7" 0.375 + 0.6 (II - 1.521 )" 0.375 + 0.6 II - 0.6 x 1.521
0.611 '" 2.4745 - .375 + 0.6 x 1.521 == 2.0995 + 0.9 126 " 3.0121
01
5.
. 2m. Ans.
0.6
Pro blem 3.21 Find Ihe 1010/ pressure and position of celilfe of pressure 0/1 a Irhmgular plale of
base 2 m and heiglll J m which is immersed in Waler in SUcil (I way rlta/IIle plane' ofrhe plale makes an
angle 0[60" wilh the free su rface of Ihe water. The b{/~'e oftlw plate is parallel to water surface and m
{/ depOt of 2.5 III from Wafer surface.
FREE WATER SURFACE
Solution. Given :
- "=--:;.~t - '1 '60~j
Base o f plate.
b '" 2 III
h 25:
]'Ieiglll of plate,
II '" 3 III
h'
~
3.0 12
I,= -1 :
I
Ar~a.
A:bx"=2x3=3m2
2
L
(J-1
<'Ill..,..
2
Incl ination.
Dcptll of ce ntre of gravity from free surface of water.
h=2.5+AGsin60°
=2.5+
1
3
fi
x3xT
Fig . .1.26
{-: AG '" ±Of hdg ht of triangle }
'" 2.5 + .866 m = 3.366 m
(i) T o ta l p ress ure
for~
(1-')
F = pgA/' '" 1000 x 9.8 1 x 3 x 3.366 '" 9906 1.38 N. Am.
(ii) Ce nt re o r p ress ure (h "'). Dept h of ce ntre of pres.wrc from free surface of water is gil·en hy
le 1,· = Icsin
, + II
All
where
2x 3J
3
4
= - - = - = 1.5m
36
36
2
bill
Ic = -
l
n f:iJ'
I,· " 1.53 xX si3.366
+ 3.366" 0.111 + 3.366" 3 .477 m . AIlS.
to- 3.6
CURVED SURFACE SUB -MERGED IN LIQU ID
Consider a curved surface AB, sub-merged in a s tatic nuid as shown in Fig. 3.27. Let dA is the area
of a small strip at a depth of II from water surface.
Thcn pressure intensity on the area dA is "pgil
and pressure forcc.
Iff"" P X Area" pgll X dA
... (3 . 11)
This force dF acts nomlalto the surface.
Hcnce IOlal pressure force on the cu rved surface should be
F=
I I
f
pghdA
... {3. 12)
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198
Fluid Mechanics
WATER
SURFACE
,
C
E4~
--------- .....----- --::::=::~:: dF,::--:.,::--,
dA cos {I
AREAdA
(')
(.)
FIg . 3.27
Bul here as Ihe direction of Ihe forces on tbe small areas arc 110t in the same direction. but varies
from point 10 ()Oint. Hence in1egralion of equation (3.1 1) for curved surface is irnpos~ible. The problem
can. however. be solved by rc!\Olving the force dF in two components dF, and <IF,. in the x and )'
directions respectively. The total force in the .r and y direct ions. i.e., F$ and F" arc obtained hy
inlcgraling dF, and lIF,. T hen 100al force on tile curved surface is
2
F=JF, +F'
,
· .. 0 ·13)
;md inclin~tiol1 of rcsuitanl wilh horizonwl is tan
4' ==
F
... (3.14 )
-2:.
F,
Resolving the force dF given by equation (3.1 1) in x and)' directions:
dF, = rlF sin e = pgildA sin €I
and
<iF" '" dF cos 8:: pglldA cos 8
TotJI forces in 'he of Jnd }' dirediun ~re :
F , ::
and
F, =
J
J
dFA
'"
dF, '"
J
J
p.!:luiA sin 8 '" pg
pgllllA cos 8 "" pg
J
J
I': dF=pghdAI
II/fA sin 0
... (3. 15)
IIlJA cos 8
...(3. 16)
Fig. 3.27 (b) shows the cnlarged area dA. From this figure. i. e .• n.EFG.
EF=dA
FG::dAsin8
EG=dAcos9
Thus in equation (3.15). dA sin 8= FG = Vcrtical projec ti on of the
pg
J
~rea
dA and
h~nce
the expression
repr~scnts the total pressure force 011 the projected area of the curved surface on Ihe
IldA sin e
vertical plane. Thus
Fx :: Tutal pressure force on the projected area of the curvcd surface on vertical plane.
. .. (3. 17)
Also dA l'OS = EG = hurilOntal projecliun of dA and hence /IdA cos is the voluUle o f the liquid
e
e
contained in the elementary arca dA UplO free surface of the liquid. Thus
volume contained between the curved surface
Hence pg
J
I1dA cos
e~tended
J
/ldA cos
upto free surface.
e is the 10lal weighl suppo l1ed by the curved surface. Thus
F. "'pg
JhdAcosO
'" weight uf liquid su pported by the curved surface UplO free surface of liquid.
I I
e is Ihe total
. .. (3. 1&)
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Hydrostatic Forces on Surfaces
In Fig. 3.28. the curved s urface AD is not support in g an y !luid. In
sucll cases, F, is eq ual to th e weigh t of th e imag inary liquid ~upponcd
by AB UplO free su rface of liquid. The direction of F), will be taken in
upward dircdion.
Problem 3.22 Comp"te 1/11' lroriZO/llal m,,! ('ertical components
of the lolal force aCling on a CU')'ed SIlr/llce AB. "'hieh is i'l IiiI' form
of l! 'I"ail"m( of a circle a/ radills 2 III 115 sho",,, ill Fig. 3.29. Take IiiI'
wiilll, of II!I' gate as IlIIil),.
WATER SURFACE
0 ;- - -
.. . .. ______ _
----------------------------- -_._--------- --------------
A _ _ • __ _
">,,,;;;-;;,,;---~
-- --;;,,;
--";,."",,-
Solution. Give n :
Widlh of gale
Radius of the gale
Fig. 3.28
= 1.0 In
= 2.0 III
Di stan<:c
AO = OB = 2 rn
Horizomal force. Fx cxcT1cd by waler on g ale is given by l
1.5 m
equa tio n (3. 17) as
F , = Total pressure force on the projccted area of curved
surface AB on vcnica l planc
== Total pressure force on OU
{projected area of curve d surface on vertical plane = OB x I}
==
99 1
o
FREE SUR FACE OF WATER
_ C
-- -- --
,
Fig. 3.29
pgAh
== 1000 x 9.8 1 x 2 x I x ( 1.5 +
f)
{ .:
Ii =
F, == 9.81 x 2000 x 1.5
Areaof08==A=ROxl==lxl=1.
Depth o f e.G. of 08 from free surface = 1.5 + .,. }
= 49050 N. An s.
The point of application of F, is g il'en by Jr . =
I
-
~ + 11
Ah
l
bd
lx2 l
2 4
where Ie == M.O.I. of 08 about its e.G. == - - = - - == - 111
12
12
3
2
1,. == __3_ + 2.5 = - '- + 25 III
2x2.5
7.5
== 0.1333 + 2.5 = 2.633 m from free surface.
Venical forcc, F y • cxerted by wa ter is given by cquation (3.18)
Fy " Weight of wate r s upponed by AB upto free surface
= Weight of portion DABOe
'" Weight of DAOe + Wei gh t of water AOB
'" pg [Volume of DAOe + Volume of AOB]
== I OOO x 9.8 1 [ ADXAO XI +;(AOfxl ]
I I
Ii
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IL
1100
Fluid Mechani cs
= 1000 x 9.81
[L5X2.0 Xl+~X 21 XI]
== 1000 x 9.81 [3.0 + nlN
= 60249.1 N. An s.
Problem 3 .23 "-ig. 3.30 .lhows II gme I1m'ing II qlladroll/ shape of wdilIJ 2 111. "-ind Ihe re.mltal1l
fo rce dlle 10 \l'ater per metre lellg lll of Ihe gme. Find also Ihe lingle ar ....hieh Ihe IOtal fo rce \\'ill ae/.
Solution. Given:
R,I<Jiu s of g~w
Width of ga le
Hori"ttln tal
=2 m
= Im
For~
F,
= Force o n the projected area o f Ihe
curved surfa,!' on vertical plane
== Fo rce on
no = pgAI!
Fi g. 3.30
,I
where A =ArcaofB O=2x 1 =2 m , 11= - x 2= 1m:
2
P, = 1000 x9.8 1 )( 2)( 1 == 19620 N
This wi ll 3CI al a depth o f
~
3
)( 2 =
~ 111 froln
3
free surface o f liq uid.
Vertica l Force , " ..
F ,":= Weight of water (imagi ned) supported by AB
=pgxArea of A OU x 1.0
"lOOOx9.Rl)( !: (2)4 )( 1.0 = )08 19 N
4
This will act a1 a disi:l11cC of 4H = 4 x 2.0 = 0.848 m from OB.
3rt
Res ultant force. F is given by
311
F = JCFo,:-+-::F,'-'
= .j19620+30819
.j384944400 + 949810761
= 36534.4 N. Ans.
The ang le made by the resultant with horizontal is give n by
Fy 30819
Ian 9= - = - - " 1.5708
F.
e " Ian
19620
1.5708 = 57' 3 1'. Ans.
Problem 3 .24 Find tile magnitude and direclion of II,e re.illitwlI force due 10 water acting on (I
roller gale of cylindrical form of 4.0 m diameter. wl,ell the gate is placed all tile dam ill such a way that
,,·Clter is just goillg to spill. Toke Ihe lenglh of the gClte os 8 III.
Solution. Given:
Dia. of gale
=4 (11
R =2 rn
Rad ius.
Len gth of gal~.
1= 8 rn
I I
Ii
~ I
IL
Hydrostatic Forces on Surfaces
lI o rizont;oI
F,
where
ror~,
101 1
F, acting on the gate is
= pgA!J = Force on projected area of curved surface
,,
ACB on vertical plane
'" Force on vertica l area AOB
A = Area of AOB = 4 .0 x R.O = 32.0 Ill l
;0
T
1
4 .0m
h = Depth of e.G. of AOB = 412 = 2.0 III
F,'" 1000x9.81 )(32.0x2.0
= 627840 N.
Fig. 3.3 1
Vertical fo ...,e, Fy is given by
F ," '" Weight of water enc losed or supponcd (actually or ima ginary) by
the curved surface ACB
'" pg x Volume of portion AC/J
= pg xArcaofAC8x/
IT
,
IT
2
'" l ()()()x9.8 1 x-(R )" x8.0=9810x-(2) x 8.0=493 104 N
2
2
will be acting in the upward direction.
[t
F= JF,~ + F.} - J62784fJ+ 493104 '" 798328 N. AilS.
Rcsulwrn force,
o
•
,
•
Dm:clIon ofrcsultarn force ISgIVC" by tan 9=
493 104
f:F, '" 627840
0.7853
e = 3 1" 8', An s.
Problem 3.25
F illd the /wriZOlilul and I'f!r/ica/ campanelli of W(ller pressu re (lc/i' ,g all lite face of II
rainIer ga/eo/90° ~'eClorofrlldi"s 4 m (l~' si,o ,,,,, in Fig. 3.32. Take widlll ofgllte ullily.
SolutIon. Given:
Radius of g~te,
R",4rn
Horizontal cornponenl of force acting on the gale is
Fx'" Porce on area of gate
projected o n I'enica l plane
'" Force on area ADlJ
whe re
= pgA l1
A", A8 x Width of gate
",2xADxl
(": AlJ =: 2AD)
=: 2 x 4 x sin 45° =: 8 )( .707 =: 5.656 rn !
[':AD=4sin4·n
Fi g . 3.32
-h _
_ _An __ 5,656 _ 2 828 m
~~
2
_
.
2
F, = 1000x9.81 x 5.656 x 2.828 N = 156',)11 N. Ans.
Vertica l co mpone n'
Fy = Weight o f Winer suproncd or endosed by the curved surface
'" Weight of water in ronion ACHDA
'" pg x Area of ACBDA x Width of gate
'" 1000 x 9.81 x [Area of sector ACBOA - Area of Il.ABO[ x
I I
Ii
~ I
IL
1102
Fluid Mechani cs
I ': dAOE is;\ right angled]
'" 98 10 x ['::' 42 _ 4 x 4 ] '" 44796 N. An s.
4
2
Problem 3.26 Ca/cu/rue Iile IlOriZOn/u/ lind "n/icil/ components of Ihe W{lter pressure exerted on
r(lilller gale of radius 8 II! (IS shown ill Fig. 3.33. Take ,..idlh of gale "IIily.
Solution. Th e hori7.0nlal compone nt of water pressure is gil-e n by
II
F , '" pgA /, '" Fo rce on th e area proj ected on ve ni ca l plane
"" Force on the vertical area o f
whe re
BD
WATER SURFACE c
A", IJD x W idth o f gate", 4 .0 x I '" 4.0 III
-
'.Lm
I
h=-x4=2m
2
F, = lOOOx9.8 1 x 4 .0x2 .0= 78480N. An s.
Fig . l,lJ
Vertical rompone nl of Ih e wale r pressu re is give n by
F, '" Weight of Wale r su prmncd or enclosed (imagi nary) by c urved
surface
en
'" Weight of water in the portion eBDC
'" pg x [Area of portion CBD C ] x Wid t h of gale
'" pg x (Area of sector eBO - Area of the trian g le BODI x 1
= 1000 x9 .81 x
[~x 7tH! ~ BDx DO ]
360
2
= 9810 X [ -.!... 7t x 8l ~ ~
4.~
O ~X~8~.8~OO='=3~O_O]
12
2
1': DO" 80 cos 30° = 8 x cos 30° 1
'" 9810 x 116.755 ~ 13.856] = 28439 N. An s.
Problem 3.27 A cylindriClIl gllle of 4 m diameler 2 m long has ...lIIer on ils bolll sides as sllo ...n in
Fig. 3.3 4. Delermine Ihe magnilllde. 10culioll wId direClion of Ihe resuillml force e.wrted b)' Ihe ...lIler
on Ihe gille. Fillli aiso Ihe /easl "'eiglll of Ille cylinder so Ihlll il may 1101 be lifted 11 ... (1)' f rom Ihe floor.
Solution. Given:
WATER SURFACE
A
Dia. of g ale
=4m
Radius
,,2 In
WATER
(i) The forces acting on th e left Sidc of the cy linder aTe
F
4m
_ _ _ _ ~Q ___ _ D SURFACE
B
c::~~~::::::
The ho ri zontal co mpone nt. F. ,
x,__~_ ~-where
Force of wate r on area projected on ve rtlcal -=-_:_::::::~=~ ~~ _:__ FY, : Fy,
i :t
f"", "
._:_:_:_:._:_-=-_ :._:._-=-_:._.
p l an e
= Force o n ,uea AOC
" pgA h
" looox9.8 1 x8x2
where
A"ACxW idth ,, 4 x2
1
"" 8 m
'C
-:_
'm
Fx
Fig. 3.34
" 156960 N
I I
Ii
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IL
Hydrostatic Forces on Surfaces
103 1
F)", '" weight of water e nc losed by AlleOA
'" 1000 x 9.81 x
[~Rl ]
x 2.0 = 9810 x
%x 22 X 2.0 = 1232.76 N.
Ri ghi S id". of Ihe Cy linder
F" = pgA2'1l = Force on vcni ca l area CO
'" l000x9.81
x2x2x~J A,= COXI=2XI=2m l.l;l=~= 1.0)
2\
-
2
= 39240 N
F11 = Weight of water enclosed by DOeD
=pgX[~ R l] xWidlhof ga te
= 1000 x9.8 1 x ~ x2 1 x2 = 6 1638 N
4
Resulta nt force in tile direction o f x.
Fx = F" - F" = 156960 - 39240 = 11 7720 N
Resultant force in the direction o f y.
F ., =F,• I +F,, = 123276+6 1638= 1S49 14 N
(i) Res ulhmt fore", F is give n 3S
F = ~CFC,"+C-;F,~2 _ J(1 17720)l + (1849 14)2 = 2 19206 N. An s.
(ii) IJiredlon of res ultant force is given by
tan
F . 184914
e = --.L
=
Fx
11 7720
= 1.5707
e = 57° 3 1'. An s.
(iii) Location o f Ih e r ..sultan! force
rorcc. F" acts
OIl
ac ts at a distance of
a distance of 2; 4 = 2.67 m from the tOp su rface of water o n left side. while F.,
t
x 2 = 1.:\3 III from free surfa<:e on the right side o f the <:y linder. The rcsu lWnt
force F, in the d irection of.r wi ll act at a distance of y fro m the bonom as
F,
x)'
== F" [4 - 2.67 1- F" [2 - 1.33[
117720 x)' = 156960 x 1.33 - 39240 x .67 == 208756.8 - 26290.8 == 182466
Y=
182466
= 1.55 m from the bonom.
117720
Force F a<:ts at a disl<Ince 4R fro m AGe Of at a distance 4 x 2.0 == 0.8488 m from AGe IOwanls
JI
31t
31t
left of AOe.
Also F), aets at a distance 4R == 0.8488 m from AGe towards the right of AGe. The resu ltant force
,
3,
FJ will act at a distance x from AGe which is give ll by
II
Ii
~ I
IL
1104
"'
Fluid Mechani cs
Py )( x = FYI x .8488 - Fy, x .8488
184914 x x == 123276 x .8488 - 61638 x .8488 = .8488 [123276 - 61638 1 = 52318.4
523 18.4
0.2829 m from AOC.
18491 4
(i,-) uasl weight of ,,'Iinder. The resultant force in Ihe up ward direction is
Fy = 184914 N
Thus the weight of 91indtr should not be less lh:1I1 the upward force F),. Hcnl"t: least weight of
cylinder should be at leas!.
= 184914 N. A ilS.
Problem 3.28 Fig. 3.35 il'llow:; the crU!iJ"J'eCliu n of" /lllikfull of ",,,Ier ""der pre5I>''''''. The /eng/h
of lire 'a"k is :I m. Ali emply cylinder lies along the lenglh of Ille [(",k on 01!e of ils corner {IS S"O"'".
Find the horiWIl/a/ fwd renic,,/ comflom:nlS of Ihe force oc/i"g Oil 11u: cun'ed iJ'urface ABC of the
cylinder.
Solutio n. Radius.
R==lm
Length of lank .
1==2 m
-----Pressu re .
p '" 0.2 kgf/cm 2 '" 0.2 x 9.8 1 Nkm 2
-----------4
--- - -2
-----'" 1.962 Nlcm = 1.962 x 10 Nlm ~
4
11= L= 1.962xl0 =2m
pg
IOOOx9.8 1
Pressure head.
Free surface of water will be at a height uf 2 III from
the lOp of the lank.
Fig. 3.36 shuws the equi va lent free su rface of water.
----------- ----------------------------------------- ----- ----------------- --- --- --- Fig . 3.35
:::::::::;f~'"
(i) lIurl lo ntnl C umpunent uf Fure..
F, =pgA/'
where A == Area projected on vc.rtical plane
= 1.5 x 2.0 = 3.0 1l1 ~
~~~IM~~~~~~
, ' , ' , ' , ' , ' f--G
::::::::-:t-:a
15
"=2+
=2.75
_:_:_:_:1~5 ~ .:
2
F~ =
l000x9.81 x3.0x2.75
= 80932.5 N. An s .
(ii) Vt'rtical CO llll)Ollent of Forcc
:0T
:
2.5 m
,- _ ,
~ _Ol..; __
'\
H ,0
:::::}:r:::~
,:,:,:,:,:,:,:,:,:,:,:::~
------------------------------------
F, '" Weigh! of water enclosed or supported
a(;lually or imaginary by (;urvcd surface ABC
== Weight of water in the portion CODE ABC
= Weight of water ill COOFflC - Weight of water in A£FB
But weight of water in CODFBC
== Weight of wata in [COB + ODFBO[
= pg
[It:"
+ BOXOD] x 2 '" 1000 x9.81
== 64458.5 N
Weight of water in AEFB '" pg [Area of AEFB[ x 2.0
I I
Ii
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IL
Hydrostatic Forces on Surfaces
105 1
'" 1000 x 9.81 [Area of (AEFG + AGBII - A ll/I)] x 2.0
AI{ 0.5
sinO ", ==0.5
AO
1.0
EH", 80- HO = LO- AD cos e = 1.0 - J x cos 30°", 0. 134
In MHO.
ABH == Ar;)3 ABO - Area AHO
Area.
30 _ AHXHO
360
2.0
= rtR1x
Weigh! of
WOller
nRl _ 0.5 x .866 = 0.0453
12
2
in AEF8
= 98 10 x [AE x AG + AG x AI/ - 0.0453[ x 2.0
= 9810
x 12.0 x . 134
+ . 134 x.5 - J14 53 1 x2.0
== 98 10 x [.268 + .067 - .04531 x 2.0 = 5684 N
Problem 3.29
Fy '" 64458.5 - 5684 '" 58774.5 N. An s.
Find file magnitude and direction of the resu/wlil water pressure (Ictillg
,
011
a CllrI'ed
face of (I dam which is shaped uccordili g /0 Ihe f e/Mioll Y = ~ as silow" ill Fig . 3.37. Th e heighT of the
9
"'mer retained by III<' dam is 10 m. Consider 1/11' widlll of the dom (IS Imit)'.
Solut ion. Eq uat ion of curve AB is
,
.( 1
y= -
or
,
x' ",9y
9
.I '"
J9Y '" 3JY
Hcig tll of waler.
11= 10m
Width.
b=l m
The hori zonta l compone nt , f'~ is g ive n by
Fx '" Pressure due to water on the c urved area
'" Pressu re on area BC
' '''J'''''" o n vertical plane
'" pgAIi
whe re
A : BCx I: lOx I m 2.1i:! x 10:5 m
F,: l ooox9.8 1 x IOx5= 490500N
Vertical component, F,. is given by
Fy = Weight o f water su pported by the c urve AB
'" We ig ht of water in th e portion ABC
'" pg lArea of ABC ) x Width of dam
'" pg
[J~~t X dY ] x
1.0
{Area of strip '" .ldy
1'"
: 1000 x 9.81 x 0 3JY dy
=29430
[)~/~'" l" =29430x f[/'2 1~O =19620(I0Jl2(
0
: 19620 x 31.622 '" 620439 N
I I
Ii
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IL
1106
Fluid Mechani cs
RcsuJtam water prCS5ur~ on dam
F'" JF/ + F/ '" J (490500)2 + (620439)2
'" 790907 N '" 790.907 kN . An s.
Directio n of the resultant is given by
F1 620439
=
'" 1.265
F, 490500
tan9= -
9 ", 51 ° 40', An s.
Problem 3 .30
A dam has" p"rabolic shapl! y '" Yo
(c<..)2 a~' shown ill Fig. 3.38 below IUn'illg xQ '" 6 m
.',
{lnd Yo '" 9 m. The fluid is water with density '" /000 kg/m.!, Compule the "ori2Omal. !'Imical alld Ihe
resu/Ullil thrust exerted by ",mer per melre lellglh of the dam.
Solullon. Given:
Equalion of Ihe curve OA is
P)'o
"'
.t
2
(;:r =9(~r
=9 x ;: '"
1
x:
'" 4)'
.J4Y '"
of '"
2ylr.'
Width of dam.
b: lnl.
(i) lIorizontnllhrus t t'xerled by w:ller
F, '" Force c"cned by wate r on venica l surface
OB, i.e.. the surfa(;c obtained by projc(;ling
the curved s urface 011 vertical plane
Fig . 3.38
"' pgA;;
'" 1000x9.81 x(9x I)
Xl9
== 397305 N. Ans.
(ii) Vertical thru st exerted by water
F1 == Weight or water sup()Orl cd by curved surfa!;e OA upto free surface of
water
== Weight of water in th~ portion ABO
== pg x Area of OAR x Width o f dam
== 1000 x 9.81 x [ f rx
== lOOOx9.81
X[I: 2/12 XilY]x 1.0
I
=19620 x [(~:i~ ) =
= 19620 x
I I
dY]x 1.0
,
19620 x ~ [9.\12 1
3"2 x 27 == 353HiO N. AilS.
Ii
~ I
IL
Hydrostatic Forces on Surfaces 107 1
(iii) Res ult a nt th rus t exe rt ed
b~'
wa te r
F" JF} + F} - .J397305 + 353160 = 53 1574 N. Ans.
Direction of resultant is given by
tan
e=
F, = 353 160
F,
= 0.888
397305
9 = (an -I 0.888 = 41.6 3°. An s.
Problem 3 .31
A cylinder 3 m ill diameter (lnd., III 10llg retaitl.l Imler 011 one ,I'id,'. Ti,e c)'litlder is
supported as sllo .....tl ill Fig. 3.39. Determine lilt' horizolltal reaclioll III A and rhe I'erlical reaction ar B.
The cylinder weighs 196.2 kN. Ignore friction.
Solution. Given :
Dia.ofcyli[l(lcr
WATER SURFACE
--~-~
~- ~-~-~-~-~-=
_~_~_~ _~_~_~_:_:
= 3 11)
lA:ngth of cylinder
= 4 In
Weight of cy linder.
IV= 196.2 k.N = 196200N
HoriwnwJ fOT..,c exerted by water
F , = Force on vcnical area SOC
: pgAh
whe re
C
---
----
3m
l
:,
F.
L
'0
~
0" --- + ~ --A
tFy
,
f ig. 3.39
, ~
I
A =BOCx/= 3x4= 12m-.iI = - x3= 1.5m
2
F,= l000 x9 _81 x 12x L5= 176580N
The ve ni cal force e~ened by water
F, = Weight of wate r enclosed in BDC08
=pg><
(~R l) ></=
1000><9.8 1 ><~ X(1.5)lX4= 138684N
Force F, is acting in th e upward direction.
For the equilihrium of cy lind e r
Hnri7.onta l reactio n at A = Fx = 176580 N
Vertical reac tion at
B '" Weight o f cyl inder - F,.
:: 196200 - 1386M::S75 16 N. Ans.
... 3.7 TOTAL PRESSURE AND CENTRE OF PRESSURE ON LOCK GATES
Lock gates arc the devices 1I',ed for changing the water level in a cana l or a rive r for na viga ti on.
Fig. 3.40 shows plan and elev ati on of a pair o f lock gates. Let All and IlC he the two lock ga tes. Each
gate is supponed Orl twO hinges Ii~ed on their top arid ool1om at the ends A and C. In the c losed
position. th e gat~s meet at 8.
Let
F = Resulta nt force due to water on the gate All or IlC acting arc right angles to the gate
R = Reaction at the lower and upper hin ge
P = Reaction at the oommon oontact surface of the two gates and acti ng perpendic ular to the
con tact surface.
Let the force P and F meet at O. Then the reactio n R must pass through 0 as the gate All is in the
equilib rium under the adion of three forces. Let e is the inclination o f the lock gate with the normal to
the side of the lock.
I I
Ii
~ I
IL
1108
Fluid Mechani cs
In LAllO.
LOAD", LAlJO '" a.
Resolving all forces along the gale All and puning cquallU zero. we get
R cos a~ Peose '" Oor R '" I'
... (3. 19)
...- HINGE
~"'~ WATER SURFACE
ELEVATION
DOWNSTREAM
SIDE
'CAN
Fig. 3.40
"'
Resolving forces nonnailO the gale All
Rsin9+Psin9 - F=O
F", R sin €I + P sin
21' sin
e '"
e
p. _ F_
1":R=P)
... 0.20)
2 si n e
T o .,alcul"l ~ P .. nd R
In equ at ion (3.20), I' can be calculated if F and e arc I;nown. The value of 9 is calculated fro m Ihe
angle between the lock gales. The angle between the two lock gate is equa l to IRO" - 29. Hence 9 ca n
be calculated. The value of F is calculated as :
II I '" He igh! o f water on Ihe upstream side
11 2 '" Hc igh.t o f water u n tllc downstream side
F j '" Water pressure on the gate on upstream side
F1 '" Water pressure on the gate on dowlIslream side of the gate
J '" Width of gate
Now
F\ '" pgA\h\
H
",pgXH\XJX - '
2
H'
Similarl y,
Resu ltant force
'" pgJ - '
2
_
H , pgJH '
F1 "'pgA 1 h, = pgX(HlxJ)X 2"'~
F = F\ _ F _ pgIH\' _ pglHi
2-
2
2
Substituting the val ue o f ij and F in equation (3.20). the value of P and R e~n be
Reaction s .. t th e tol' li nd bott om hin ges
Let
R, '" Reaction o f the top hi nge
I I
c~kulated.
Ii
~ I
IL
Hydrostatic Forces on Surfaces
109 1
Rb ", Reaction of Ih.e bonum hinge
R=R,+R b
Then
The rcsuitam water pressure Fads norillal 10 the g ale. Half of the value of F is resisted by Ihe
hi nges of one lock gates and other half will be resisted by the hinges of01h.c r lock gate. Also FI acts at
a distance of
~
3
from bollOOl while F, acts at a di stance o f
-
H~3
from bonoill.
Taking moments about Ihe lower tJingc
R
.
X Sln
I
Fi HI F, H ,
&x H = - x - - - - x -2
3
2
3
.• ,(1)
where H = DiMallcc between twO hinges
Resol ving force s horiZOntally
.
.
fj
F,
R,sme + Rb slne= - - .....:...
...(ii)
2
2
From equations (0 and (ii). we can find R, and Rho
Problem 3.32 Each gare of a lock i,y 6 In high Wilt is supported by two hillges placcd ()II IIIe lap
alld bOl/om of lile gille. W/lell Iile gale.! (lfe c/o.!ei/. 111e)' make GIl angle of 110°. The widlll of lack is
5 m. Ifrlie ",/rer (e"e!s lIr,' -I m lind 2 III Oil 1101' upsrream and dOll'lIsrr.-am sides reSl'eC(iI'ely. derermine
rlie Imlg,lirl,,!e of rile forces 011 rlie liillges dill' ro lI'arer pressllre.
Solution. Given:
:: 6m
Heig tu of loc k:
Width of lock
:: 5m
Width of each lock gate =A B
HINGE
1'm
r -:-;:;?;-:-
10
"'
Angle belwee ll gales
AD
cos 30°
't
2.5
0
cos 30°
ELEVATIQ
:: 2.887 m
:: 120°
eo
I ,~~'
C
~~~ ~
180·- 120'
2
wa1~r
'"
R R
''''.oT- L
"'fm
PLAN A
HI = 4m
TOlal
INGE
~
DOWN$TRE
Height of wa ter on upslream side
and
''''
-
Hl=2 m
press ure on upstream side
30"
Fig . lA I
f") = pgA1ii , . wh~re AI = II I x! '" 4.0 x 2 .887 m ~
(H
4
iii,
=-t=2'= 2.0m
= 1000 x 9 .81 x 4 x 2. 887 x 2 .0
I
= 22657 1 N
F'orce
F I
will be acting at a distance of
Similarly. to tal
wal~r
fl
4
= - :: 1.33 m fro ln boltom.
3
3
_ I
pressure on th e downstrea m side
-
Fl = pgA~IIl. whe re
A~
= Hl X I = 2 x 2.887 III
= l000x9.81 x2x2.887x 1.0
I I
,
II
2
"l = .:..:.l.=
_ = 1.0m
2
2
-
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1110
Fluid Mechani cs
'" 56643 N
F, will act a1 a distance of H, = ~ = 0.67 III from bonom .
3
3
Resultant water pressure on each gale
F == FI - F, '" 226571 - 56643 '" 169928 N.
Le t .r is heig ht of F from the bottom. then taking moments of Fl' F 2 and F about the bollom. we hal'c
Fxx= FI X 1.33 - F 2 x 0.67
or
169928 xx: 226571 x 1.33 - 56643 x 0.67
.r =
22657 1 x 1.33 - 56643 x 0.67
169928
301339 - 37950
= 1.55111
169928
From equation (3.20). P ~ _ F_ = 169928 = 169928 N.
2sin e 2 ... i030
prom equation (3.19). R = P == 169928 N.
If R T ~ n d R8 arc the rCJ"Iions at the lOp <lnd bonom hinges. then NT + R8 = R = 169928 N.
Taking movemen ts of hinge rcac tiolls RT • He and R aboullhc 0011010 hinges. we have
RTx6.0+RexO=Rx 1.55
RT = [69928 x 1.55 '" 43898 N
60
RIJ=R - R r = 169928 - 43898= 126030 N. Ans.
Problem 3.33 Ti,e end gilles ABC of <l lock are 9 In IIigil and when closed include an angle of
120°, The width of the lock is /0 m. Each gate is supporled by hro hinges located at I m and 6 m
"bol'e lite bottom of Ihe lock. Ti,e depths ofwaler all Ihe two sides (Ire 8 m (!"d.$ m re~peclirely. Find:
(i) Resuillml "'("er force on each 8011'.
(ii) ReaClio" helwee" Ihe galeS AB amlBC. and
(iii) Force Oil eadl !ringe, co"sidering Ihe rellClion of Ihe gMe ac/ing ill Ihe same Ilorizon/lli plane
as resullat,l WIlier pressure.
Solution. Given:
Hdglll of gale
Inclination of gate
180°- 1200
2
'" 30
Q
A
,-~-~-~-~--- L- HINGES
,
'm
•
,
(3)
PlAN
(b)
ELEVATION
Fig.J.42
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Hydrostatic Forces on Surfaces
111 1
: 10m
W id th of loc I.:
5
" cos 30° or l= S.773 m
Width of e ach lock
0':1'111 of water o n upstre am side.
U I " 8 III
Depth of wate r o n dow nstream side. 112 " 4 III
(i) Wa ter pressure on upstrea m s id e
F I '" pgA,ii,
w here
-
H
8
A I '" I X H I " 5.773 x 8 " 46. 184 Ill. ' 11 '" _I = - = 4.0 III
2 ,
F I = 1000 x 9.81 x 46.1 84 x 4.0" 181 2260 N = 18 12.26 k N
Wate r pressure o n dow nstream sid e.
1'1 = pgA1'1!
where
A l = I X tl2 = 5.773 x 4 = 23.092
111.
";! ="24 = 2.0
FI = 1000 x 9.81 x 23.092 x 2.0 = 453065 N = 453.065 kN
Resulta nt water pressure
= " I - F2 = 181 2.26 - 453.065 = 1359. 195 kN
( ti) Reaction be tween the gales All a nd He. The re acti on (P) betwee n th e gmes All and Be is
give n by cqumi o n (3.20) as
F
F: -. ,
•
~In
0,
1359.195
.3no = 1359.195kN, An s.
X Sm
v
( iii) Force o n each hin ge. If NT and Nil arc the reactio ns at th e top and bono m hinges th en
Rr + Rs"' R
But fro m equation (3.1 9), R = P = 1359.1 95
Rr+ R s '" ])59. 195
,
.
HI 8
H,
4
The fu rl'c FI IS aetmg at - '" - '" 2.67 III from bonu m and F, at -" '" ~ = 1.33 m fro m bonum .
3
3
3
3
The resuhant fo rce F will act a1 a di stance.r from bo tlo m is give n by
Fxx= F I X 2.67 - F z x!.33
0'
.r =
c'c."X"'."67c-,c-F'LX
" ,,
1.3,,,
3
F
lS I2. 26x2.67 - 453.065x 1.33
1359.195
4838.734 - 602.576 3
3
'" .1I6= . . lI m
1359.195
Hence R is a lso aC iillg ai a d istance 3, 11 m from bol1om.
Tak ing mo ment s o f Rr and R aboll1 the bo no ll1 hinge
'"
RTx 16.0 - 1.01= R x (x - 1.0)
RT =
RX(I' - I.O)
1359.195x2.11
5.0
5.0
= 573.58 N
Rs"' R - RT'" 1359.195-573.58
'" 785.6\5 kN. A ilS .
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1112 Fluid Mechani cs
.. J.8 PRESSURE DISTRIBUTION IN A LIQUID SUBJECTED TO CONSTANT
HORIZONTA LNERTICAL
ACCELERATION
In c hapl ers 2 and 3 . the containers which co ntains liquids. arc assumed to be at resl. Hence the
liquids are also at res!. They arc in st:uic cquilibri ulll wilh respecI to con lainas. Bul if Ih e container
containing a liq uid is mad e (0 move Wilh a COn S[an l acceleration. the liquid panicles initi all y will mOVe
relmive 10 each o lh er and after some tim e. there will nO! be an y relative motion between the liquid
part icles and bound ari es o f Ihe conta in er . The liquid w ill tak.e up a new position under th e effect of
acceleration imparted to ils con tainer. T he liqu id will come \0 rest in this neW position rda tive 10 \ll<)
container. Th e en li re nu id ma ss moves as a si ng le unit. Since the liqu id aFter anainin g a new position
is in sta tic condi tion relative to the cOlllainer. the laws o f hydrostatic can be applied to determine the
liquid pressure. As there is no re lat ive motion between the liquid panicles. hence the shear stresses and
shear forces be tween liquid panicles will be zero. T he pressure will be normal to th e surface in co ntact
with the liqu id.
The following are the important c ases under consideration:
(i) Liquid (.'Ontainers subject to constant horizontal accele ration.
(i,) Liquid cont ain ers subjl'ct to ronstant vertical acceleration.
) .8. I Liquid Containers Subject to Constant Horizonta l Acceleration. Fig. 3.43 (<<)
shows a tank conta inin g a liqui d upto a certain depth. The ta nk is s tationary and free surface of liqu id
is horizontal. Let this tank is moving with a ronstant accelerat ion 'n' in the horizo nta l direction towards
right as shown in Fig. 3.43 (b). The initia l free su rface of liqu id which was horizontal, now takes the
"hape as shown in Fig. 3.43 (b). Now AB represcn ts the new free surface of th e liquid. Thus the free
surface of liqu id due to horizo ntal acceleration wi ll become a downward slopi ng inclined plan". with
the liquid ri sing at the back end. the liqu id fallin g at the front end . The equation for the free liqu id
surface c an be derived by considering the equilibrium of a fluid elcmelll C lying on the free su rface.
The forces acting o n the clement Care:
Free surface of
l;quOd
Originat liquid
sorlsCfl
,
Moving
c
Tank
e
- - - --
horizontat
C
mX8
""Rear
or
,",
end
•
lW)
(.,
•
•
•
(0'
(0'
Fig. 3.43
(i) the pressure force P exerted by the s urround ing fluid on the cle melll
C. This force is normal to
the free surface.
(it) the weig ht of the fluid cle melll i.e .. m x g acting vertically downward.
(iii) accelerating force i.e .. Nt x n aC ling in horizontal direc tion.
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Hydrostatic Forces on Surfaces 113 1
Resolving th e forces horizontally, we get
Psi n 6+m)(a=O
or
Psin8= - mll
... (r)
Resolving Ihe forces vertically. we get
Pcos6 - mg=O
or
Pcos8=mxg
. ..(ii)
Dividing (I) by (ii), we gel
tan
e" -
~
( or ; NumeriCallY )
... O .20A )
The above cqUillion, gives the slope of the free surface of the liquid which is contained in a lank
which is subjec ted 10 horizon tal CO nsta nt acceleration. The [cn n «(jIg) is 11 constalll and hence ta n will
be constant. The -ve s ign shows that the free surface of liquid is sloping downw ards. He",:e the free
surface is a straighT pla ne indilled down al:1n angle e along the directio n of :Iu;clcrmioll.
Now let us fi nd the ex pression for the pressure al any point D in the liq u id mass SUbjected to
hori zon tal ac.::clcralion. LeI the point D is at a depth of '''' from the free surface, COl1 sider all
e lementary prism DE of height 'Il' and cross·scctional area dA as show n in Fig. 3.44.
e
, ,
01 constant
/'JI:=:;,. LInes
pressure
"
' ~M::l:
f -pgh,
Fi g, 3.44
Consider the equilibrium of the eleme nta ry prism DE.
The forces acting on this pri~m DE in the ve nil:al direl:tion arc:
(i) the atmospheric pressure force (Po x riA) althe top end of the prism ac ting downwards,
(ii) tile weigh t of th e clemen t (p x g x" x dA) at the e.G. o f Ihe c le ment acting in Ih e downward
d irection. and
(iii) tile pressure force (p x <fA) althc bottom end of thc prism acting upwards.
Sincc there is no venka l acccleration givcn to the tanl.:. he nce net force acting verticall y should be
zero.
P x <fA - Po x riA - pgll riA
"'
p - Po- pgll=O
or
or
or
P=Po +pgll
P - Po =pg/r
gauge pressure at point D is given by
P
or
=0
pressure head at point D.
= pglr
- I' =/1.
PS
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1114 Fluid Mechani cs
From th e above cqual io n, it is c lea r thai press ure head al an y poim in a liqu id subjected to a
consta nt llOri zo ntal acce lerat io n is eq ua l (0 th e he ight of th e liqu id co lum n above lli at point. Therefore
the pressure distribu tio n in a liquid su bjected to a constant ho rizontal accclc rmi on is sa me as
hydrostat ic pressure di stributi on. T he pl a nes of co nst ant pressure arc therefore. para ll e l 10 th e indincd
surface as s hown in Fig. 3.44. T hi s fig ure also shows the vari al ioll of pressure o n th e rear and f ront
end of Ihe la nk .
II( = Depth of li qu id at Ihe rear end o f Ih e ta nk
II~ = De pth o f liquid at the fro nl end of I h~ tank
F I = Total pressure force exerted by liq uid 011 the rear sid e of th e tank
If
F z = T ot al pressu re force exert ed by liq uid o n th e fronl
FI
th" n
= (A rea of tri ang le
s id ~
of th ~ lank .
AM L) x W idth
I
I
= ( -rxLM X AM Xb)=-rxpglllxIlIXb=
~. b.h ~
2
F2 = (A rea o f tri a ngle DNO) x Widt h
an d
I
= ( .....•
x BN x NO)
I
= .....•
x It, x pgh, x b
•
-
=
pg .b .hi
2
where b = Widt h of tank perpend ic ular to the plane of th e pa per.
T he valu es o f FI and F2 can al so he o btained as
[R efer to F ig. 3 .44 (a )l
T-.--------------- 1
t
1
-----------
=P Xg x(hl x b) x
Itl
_
=-2I
2
A,
md
pg. b.
= h,
-
", ;[[[g{;~[,,-..... .
"I",
I"~
=-=- 2
xh and h,
I"~
~pXgx(",xb) x ......:..
-
I
:.:.:.:.:.:.:.:.:
F.-·:·:~'..
••••••••- •• -:. --.-.---••--•••
~
h
- ,~~~::~~~:::~:
2
,
Fig. 3.44(a)
= "2 pg.bxI12-·
It ca n also he prove d that th e diffe re nce of th ese two forces (i.e .. FI - FJJ is eq ual to the force
to acce lerate the lIla~s o f the liq uid co ntained in th e ta nk i.e ..
r~ qui red
f'1 - F 2= M xa
whe re M = T o ta l m ass o f the liqu id con tained in the tank
II
= Horizon tal
co nstant acceleratio n .
Note : (I) If a tank completely filled with liquid and open at the top is subjected to a COIIstam horirontal
acceleration . then some of the liquid will spill out from the tank snd new free surf""e wit h il< slope given by
equation tan 9 = _!!. will be de'·eloped.
g
(ii ) If a tanK par1ly filled with liquid and open at the top is subjected to a constant horizontal acce leration.
spilling of the liquid may taKe plae<.: depending upon the magn itude of the acceleration.
(iii) If a lan k complelely filled wilh liquid and closed althe top is subjecled 10 a con,tam horizontal accelera tion.lhen the liquid wo uld not spill oul from the tan k all d also there wil l be no adjustment in Ihe surface cle"alion
"g
oflhe liquid. Bullhc eq uation tan e = - (i" ) The example for a
ail1'l~n"
I I
during
t~ke
tan~
i< applicable for Ihi' case also.
with liquid .<ubjected 10 a constam horizontal acceleration. i, a fuel tank on an
off.
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Hydrostatic Forces on Surfaces 115 1
Problem 3 .34
A reC/angular /ank i.1 mOI'iIlS IlOrizOIl/(lI/), III lilt! directiOIl of ils '<'ngrll Willi U
CQIIS/atll acceleratio/l of 2.4 m/i. The length. ",idth and deptll of Ille /(IlIk are 6 III. 2.51/1 and 2
respecril'l'iy. If the deplh of !l'mer in III,> /(Ink is I In and {(lilt is open a/ the lOp then calcu/are :
In
(I) tile Iwgle of the \l'ater $Iu jace /0 the IlOrizonial.
(ij) Ihe max im/WI and lIIinilllum flres.lure in/ensilie,1 (j{ Ille bo l/om.
(iii) Ihe /0/(1/ forc e due 10 ...tller (lelillg 0/1 eaeil end of flu! tank.
Solution. Gi ve n
F..... surface t'
Constant ;!(;ce lcrmion.
a = 2.4 mls l,
Le ngth" 6 III : Width = 2.5 m and depth '" 2 Ill. 2
Dc ptll o f Wa!U in tank . II = I m
(I) The a ngle or the ... aler s urface tn the
h o rb;onlul
~
Let 9 ", tile an g le of wa ter surface to th e hori zontal
0.2662 m
Usi ng equ ation (3.20 ). we get
Fig. 3.45
"
2.4
tan • " - - ,, - - - " - 0 .2446
g
9.81
(Ihe - ve sign shows that Ihe free surface o f waler is s lo pin g dow nward as sho wn in Fig . 3.45)
Jan
e"
0.2 446 (slope downw~rd)
fJ " lan- I 0 .2446" 13.7446" or 13° 44.6'. An s .
( if) Th e maximum and minimum press llrt' intens ities a t the bottom of th e tank
From th e Fi g. 3.45.
Depth o f water at th e front e nd,
irl" 1- 3 t3n fJ " 1- 3 x O.2446" O. 2662 m
Depth of wate r at th c re ar e nd.
it1 " I + 3 tan fJ " I + 3 x 0 .2446 " 1.7338 111
The pressure inte nsity will be IllJ ximum at the oouo m. where depth o f wat er is max imulll .
Now the maximum pressure inte ns ity m the OOuo m will be at point A and it is g iven by.
fI ....," p xgxl,!
" 1000 x 9 .8 1 x 1.7338 N/m 1 " 17008.5 N/m 2 • An s.
The minimum pressure intensity at th e OOllo m will be a t point B a nd it is g iven by
Pmin" p xg x l'l
,, 1000 x9 .8 1 x O. 266l" 2611.4N/ m 2. Ans.
(iii) Thl."" 101,,1 for.:I."" dul."" 10 wull.""r "cling on eu.:h I.""nd of Ihe lunk
Let
F I "to tal force actin g
On
the fmnt side (i.e .• on face BlJ )
F 1 " tota l fo ree acting on the rear side (i.e .. on face A C)
Th en
rnd
F I " pgAl h l, where AI " BD x width of lank ""I x 2.5,,0.2662 xl.5
,-11 __ BD
_!!L _
_
_ 0.2662 _
_ 0 . 133 1 m
2
2
2
" 1000 x 9.8 1 x (0.2662 x 2.5) x 0. 133 1
" M8.95 N. Ans.
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1116 Fluid Mechanics
F 2 ", p.g.A 2.hl. w herc A2 '" AB x width o f mnk '" 112 X 2.5 '" 1.7338 x 2.5
and
'ii, '"
•
AB '" III = 1.7338 '" 0.8669 m
2
2
2
= 1000 x 9.81 x (1.7 338 x 2.5) x 0.8669
= 3686 1.8 N. An s.
Rcsul! ant force'" F, - "'2
c
= 36861 .8 N - 868.95
liquid (water)
= 35992.88 N
Not .... The difference of the for~cs acting on the lWO ends
of the lan k is equal 10 the force necessary 10 acccicraic the
liq u id maS,. This can be provc<i as shown below,
Consider the '"Oolrol volume of the liq uid i ..... control vol·
ume is ACDHA as shown in I'ig. 3.46. The nel force acting on
the cont rol volume in the horizontal di recti on must be equal 10
the pl'<)duct of ma.% of the liquid in CQntml vol um" and accel -
F2
__ . '. '••••••••••••••• '••• ".
----'-- ,:::::~:::::::::::::::::::::: ~:: -:-. .
D
-----------------------------------------------------------------------------------------------------------------------------~I.-------"m ------~'
eration of the liquid.
F ig. 3.46
(F, - f',) 5!>1 XlI
= (p
X
, "
,
volumc of cont rol vol umc) x
II
.. (1000 X Area of tl BDC£x width) x 2.4
5
. 1x2.4
2
xA8xwldth
[lOOO x ( AC"D)
[-: Area of trapezium = ( AC;
IJI))x tlB]
) x 6 x_.
~5 x •~ .4
.. '000 x (1.7 338 +0.2662
2
= 36000 N
(": AC = h, " 1.7338 Ill. SO = h, = 0.2662 tll. and tlB = 6 Ill. width," 2.5 m)
Th e abovc force is "early the same as thc difference of the forces acting on thc two ends of the tank. (i.e,.
35992,88 :< 36000)
Problem 3.35 Tile reClang"lar lallk of lite abOl'" probl"m contaill S ...aler 10 a deplll of 1.5 III. Fin d
Ille I,orizon/{,l acceleralion wllicll m"y be im(!llfled 10 lite IlI/lk ilt lite direCliOIl of ilJ' lengllt J"O Ilwl
(i) Ihe spillillg of ..."ter f rom Ihe 1'lIIk is jllst 011 Ihe !"erge of latill g pl(!ce.
(il) Ihe f r0l11 bo/lom co mer of Ihe ,,,"k iJ' j ll S! e_.posed.
(iii) lite bOl/om of I/Il' /(IIIk is eXpOJ'ed upla ils mid·poinl.
AI,w c,dClllate Ihe tolal force.! exerted by II,e ...mer on each end of lite lank ilt eaclt caJ'e. A 1$0 pro!"e
Illat IIII' di/fer('/tce belweelt Ilte,le forces is elflw l to lite fo rce "eCeJsa ,)" 10 accelerale lite mass of ...ater
wllk.
Solution. Given:
Dimensions of the tank from pre vio us problem.
L '" 6
I I
Ill.
width (b ) '" 2.5 m and depth '" 2 III
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Hydrostatic Forces on Surfaces 117 1
/,=
IXplh of water in tan k.
Horizo ntal """elcrnlion Imparted 10 the lank
(0 (iI) IV/len Ille splllwg of'Huer from Ihe tonk IS
JUII 0/111/1'
"
,age oftak./Ilg place
d'
I
---
l .5m
C
000aco>
I ~:::...
HA
T__
u_.
_ _ 3",
-, -,,-. ". 8
I
":. 3.. ::._
"-:.::"j,":
:_:::_:::':::':::_:':~_:::_: _. _: ._:
..xl a '" reqUiTe ..omo nla aeee cral10n
When the sp lllm g of water from thc lank IS JUS1 o n th e 15m . ~::: :_:'~. ~:.~: ~:_~~~:- ~:::_
\ ergc of lakmg pi ace. tlie water wou Id rise u PIO the rear
' .' : -,::, .:_ . _ . _.: _' - .:_.:::.:__
tOp corne r of the tank as sliown 111 FIg. 3.47 (a)
E,"- - - - .-,-•. , .-, •. , - ,-,-- - - ,.
I · ---' m --~
~
' I
Ian 9= AC '" (2- 1.5) '" 0.5 =0. 1667
3
AO
But frolll equation (3.20) Ian
(l
a = -"g
Fig. 3.47
3
F
(J) Spilling of 'IlNlur is just on
tht tJerge of taking place.
(Numerica ll y)
= g x tan e = 9.8 1 x 0.1667 = 1.635 m/s 2 • Ans.
(b ) TOIIlI forces e.u:rled by water on each end of IIII' tallk
The furce exerted by water un the end CE uf the tank is
FI = pgA ,/',. where A, = CE x width of the tan~ = 2 x 2.5
CE 2
/11 = - = - = 1111
2
2
= 1000 x9.81 x(2 x 2.5) x I
= 49050 N. Ans.
The force exerted by wa ter un the end PD of the tank is
F1 = pgA 1 x
/'1.
where A1 = FD x width = 1 x 2.5
(.: AC= BD=0.5 111.
= l ooox9.81 x(J x2.5)xO.5
FD=Bf' - BD = 1.5 - 0.5= I )
-
FD
I
I" = - - = - =0.5 III
,
2
2
= 12262.5 N. An s.
(c ) DijJere1> ce of IIII' forces is eqlla/to the force 1>eCeSI(lry 10 (lcce/emle IIII' mllSS ofw(lier in Ille /(lllk
Difference o f the forces = FI - Fl
= 49050 - 12262.5 = 36787.5 N
Volume of water in the tank before acceleration is imparted to it '" L x b x dcpth of water
=6x2 .5x 1.5 =22.5m 3.
The force neces-"llry to accelerate the mass of watrr in th e tank
= Mass of water in lank x Acceleration
= (p x volu me o f watrr) x 1.635
= 1000 x 22.5 x 1.635 [Thcre is no spi llin g o f water a nd volu me of
wate r = 22.5 mO[
= 36787.5 N
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1118 Fl uid Mechan ics
Hence Ihe difference between the forces on the 1WO ends of the tank is equal to the force necessary
\0 acccierate tlie mass of water in the tank.
Volume of water in Ihe !>mk elll1 also be calculmcd as volume = ( CC+
2 FD ) x EFx Width. [Refer to
Fig. 3.47 (a) 1
=
F"I'8$urtaoe
(22+1 ) x6x2.5=22.5m.,
Original l"",
aHer ~e<.1ioo sunace
(Ii) (iI) Horizontal acceieral;on ... I,en tile frolll
bOl/om cortler of Ihe Iallk
i,~
just e.rpoud
Refe r 10 Fig. 3.47 (b). In this c ase the free su rface of water in the tank will be along CD.
leI 1I '" required hori zontal acceleration.
In th is <:ase,
2
6
CE
I
3
,
1309 = - = - = -
ED
BUI from equatioll (J.l7),
tan
•
e = ~ (Numerically)
,
I .---- 'm ------~
"
.I
Fig. 3.47 (b)
g
a=gxtan8=9.81 xi =3.17m1s!. Ans.
(b) Tolal jorces exerted by waler 011 eaell end of Ihe l(llik
The force exerted by water on the end CE of the tank is
F, "'pg xA , x
whe re
h'
A,=CExw idth= 2x2.S",'; m"
CE 2
: - = lm
2
2
h, = -
= 1{lOOx9.81 x5x 1
= 49050 N. Ans.
The force exerted by Waler on the end BD of the tank is zero as there is no water against the face BD
Fl =0
Difference of forces = 49050 - 0 = 49050 N
(e) Difference offorces is eqlw/IO rhe force neeeHary 10 accelerate rhe lIIass ofwarer ill rhe
Volume o f water in the tank = Area of CEO x Width of tank
= (
CEXED)
2
x 2.5
( ...
/(llik.
Width of tank =2.5 m)
= 2;6 x2.5= 15 m3
Force necessary to accele rate the mass of water in th e tank
= Mass of water in tank x Ac(;e lcratiun
= (1000 x Volume of water) x 3.27
= 1000 x 15 x3.27 =4,)050 N
Difference of two forces is also = 490S0 N
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Hydrostatic Forces on Surfaces
Hence d ifference be twee n Ihe forces on the two e nds
of the ta n~ is eq ual \0 the force necessary to accelerate
the mass of wate r in th e tank.
(iii) (a) Horizon/al (lac/era/ioll wllell the bOl/om of
till' lallk is eXJlosed lip/a its mid-poiru
Refer 10 Fig. 3.47 (e). hi thi s casc the free surface of
W31cr in the tank w ill be a lo ng
mid -po int of ED.
CD'. where D * is the
Let (/ '" required horizontal acce lerat io n from
Fig. 3.47 (e), it is clear that
C£
2
ED'
3
Free >urf"""
olle, accew.ralion
119 1
Original I, ..
If
i ....... 7. '
r
•
:-:-:-:-:':':':-:-.
lane= -- ~ ­
But from cquiltion (3.20 ) numerically
,
1ane=~
,
a=gxlan 6=9.81 x -
3
,
=6.54 mls. Ail S.
(b) Tolal forces exerted by ,m/er on cod, end of Ihe lank
The force cxcrwd by water on the end C£ of the tant is
F I=pxgxAlx/IJ
where
A 1 =CExWidlh=2x2.S=Sm!
CE 2
1,,= = - : lm
,
2
"lOOOx9.81 x5x 1
'" 49050 N. AilS.
The force exen cd by waler on lhe end 8D is zero as lh ere is no waler againsllhe face BD.
Fl '" 0
Differe nce o f lhe forces'" FI - Fl = 49050 - 0 = 49050 N
(c) Difference of rhe 1»"0 force.! is equtl/ fa rile force neces.lary fa aealerare rile Intls.f of lI"aler
reI/wining ill Ihe Iallk
Volume o f waler in lhe lank:: Area CED* x Widlh o f tank
C£xED*
2x3
2
X2.5" - , - X2.5=7.5 m)
Force necessary
10
accele rate th e mass of water in the tank
" Mass of water x Acce leration
= p x Volume of water x 6.54
= 1000 x 7.5 x 6.54
=49050 N
This is th e same force JS the difference of the lWO forces on the lWO ends of the wnk.
Problem 3.36 A reclallgular wnk of lellglll 6 m. widlll 2.5 m (md lleighl 2 In is completely filled lI"ilh
K·(ller when al resl. The tUIlk. 6· opell 01 Ihe lop. The 1(JIlk is subjected 10 0 hor;zoll/o/ COlis/mi l {illear
accelemlioll of 2..1 mli ill Ihe directiOl1 of ils /ellglh. Find Ihe ,·olume of ,,"ala spilled from 1111;0 lallk.
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1120
Fluid Mechani cs
So lution. Given:
L "" 6 Ill . b " 2.5 111 and height, H", 2 m
2
Horizontal acceleration.
1I '" 2.4 rnls ,
The slope of Ihe free surface of water after the lank is subjected to linear comaa,,1 acceleration is
given by equation (3 .20) as
Ian
.
e = -" (Nullwrlcally)
g
'" 2.4 '" 0.2446
9.& 1
From Fig. 3.48,
8e
tan6= -
AB
BC=ABxtan8
=6)(0.2446
I'
(": AB '" Length = 6 III ; Ian
e = 0.2446)
8m
'1
Fig . 3.48
= 1.4676 III
Volume of water spilled = Area of ABC x Width of tank
=(+XABxBC)
x 2.5
=+)(6)(1.4676)(2.5
'" i 1.007 nrl,
(':
(,:
Width = 2.5 10)
IJC= 1.4676111)
Ail S.
3.8.2 Liquid Container Subjected to Constant Vertical Acceleration . Fig. 3.49 sliows
a tank containi ng a liquid and the tank is moving vertically upward wilh a constant 3(;ccicrmion. The
liquid in the tank will be su bjected to the same ve nic al accckration. To obtain Ihe expression for the
pressure at any point in the liqu id Illass subjec ted to venical upward accele ration. consider a vertical
elementary prism of liquid CDFE.
Free surface
•
1
I
1--01- ""-1
f-
pgh(1 · t ) - 1
Fig. 3.49
LeI dA '" Cross-sectional area of prism
I,,,, Hdghl of prism
/10 " Atmospheric pre5.';u re acting on Ihe face CE
p " Pn:ssure at a depth II acting on the fat"'! DF
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Hydrostatic Forces on Surfaces
121 1
The forces acting on tile clcmcnlary prism arc:
(0 Pressu re fOrl:c equa l 10 PI;) x dA acting on the face CE vcnically downward
(ii) Pressu re fOrl:c equal !O p x dA acting on the face OF vertically upward
(iii) Weight of the prism eq llal to p x g )( dA )( II acting through C.G. of the clement vcnicalJy
downward.
According to Newton"s second law of motion. the nel force acting on Ih" clement must be eq ual 10
mass multiplied by accekration in (tIC same di rection.
Net force in \'cnically upward di rection'" Mass x accelerat ion
p x ciA - Po x <fA - pgdA . 11 '" (p)( dA x 11) x a
(": Mass'" p x dA x 11)
or
p - Po - pgil '" ph x a
(Cancelli ng dA from bolh sides)
or
P-Po=pgll + plw
= pglt [I
,1
...
+~]
But (p - Po) is the g~uge pressure. l'len<:(; gauge pressure OIl any point in the liquid
constam vertical upward ~lcceleration , is given by
m~ ss
P~ ==pgh[I+;]
(3.11)
subjec ted to
... (3.22)
... (3 .22A)
== pgh + plw
where P, == P - 1'0 == gauge pressure
In equation (3.22) p. g and a are constant. Hence variation of gu~ge pressure is linear. Also when
h == 0.1'8 == O. This means P - Po == 0 or I' = Po' Hence when II == O. the pressure is equa l to atmospheric
pressure. Hence free surface of liquid subject~d to constant venical acceleration wi ll be horizon tal.
From equation (3.22A) it is also c lear that the pressu re at any poi nt in the liquid mass is greater than
the hydrostatic pressure (hydrostatic pressure is = pgil) by an amount of p x II x a.
rig. 3.49 shows the variation o f pressure for the liquid mass subjected to a constant vertical upward
acceleration.
If the tank containing liquid is mo ving vertica ll y downward with a constant acceleration. then the
gauge pressure at any point in the liquid at a depth of II from the free surface will be given by
(P-Po)=pgh[ I-~] =pgh-Pha
... (3.23)
,fo
The above equation shows that the pressure at any
point in the liquid mass is less than the hydrostatic pressure by an ~mount of plla. Fig. 3.50 shows the variation
of pressure for the liquid mass subjected to a constant
vertical dow nward acceleration.
If the wnk cunt aini ng liquid is moving downw ard with
a constant ac(;c!eration equal to g (i.e., when a == g). then
equation reduces to P - Po == 0 or P = Po' This means the
pgn
9
pressure at an y point in the liquid is equal to surrounding
atmospheric pressure. There will be no force o n the
Fig. 3.50
walls or on the base of the tank,
Note. If a lank containing a liquid is 'Ubjectl'{/ to a constant acceleration in the inclined direction. then the
acceleration may be resolved ~long the horiront~l direction aTTd ,"ertkal direclion. Then each of these cases
may be separately analysed in accordance with the aoo"e procedure.
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1122 Fluid Mechani cs
Problem 3.37 A limk cOlllailling ,water upla a dep/II of 500 mm is moring I'erlically upward WillI
a conSWlI1 acceleralion of 1.45 mh-, Find IiiI' force exert,'d by Waler 011 Ihe side of II,e wnk. Also
mlcu/are II,e force on Ihe side of Ihe /allk wh"11 the wid/II of tank i.! 1 111 and
(i) {(1IIi: is mOI·ing I'erlically downward wilh II conSlalit accelermion of 2.45 m/I, and
(ii) Ihe limk is IW( moving oj all.
Solution. Given:
[krIll of waler,
/, = 500
Vcnil:al acce leration.
a = 2.45 fills!
Width of tank.
b= 2m
n11n
= 0.5
III
I
"
To find the force exerted by wate r 011 the side
of the tank when moving vertically upward. Ie! us
first tlnd the pressure al the honom o f th e tank.
The gauge pressure at the bonolll (i.e .. at point B)
for this case is give R by equation as
JIB == pgl!
(I+~)
= IOOOx9.8 1 xO.5
This pressure is re prcscl11cd by line
Now the force on the
~idc
Fig. 1.51
2.45)
- - ==6 131.25Nfm ,
(1+9>
'
Be.
AB = Area of triangle ABC x Width of tank
==(txA8X8C) Xb
= (t x O.5x6131.25) x2
(.: BC=6131.25 and/!= 2 m)
== 3065.6 N. AilS.
(r) Fo rce On the s ide or th., honk, wh.,n honk Is m ovi ng nrlio:u ll y do wn ward.
The pressure variation is shown in Fig. 3.52. For this easc. th" pressure m th" bottom of the tank
(i.e., at point 8) is given by equmion (3.23) as
PB== pgli
,
(1 -~)
2.45)
( ,.,
==1000x9.8IxO.51- - -
"
== 3678.75 Nfn/
This pressure is represented by line
Be.
Now th" force on the side AB == Area of triangle ABC x Width
==(tXAJJxIJC)X b
= (t X 0.5 x 3678.75) x 2
Fig. 3.52
('c Be. "78.75. b. 2)
== 1839.37 N. Am .
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Hydrostatic Forces on Surfaces 123 1
~'o rc" 0 11 th e s id e or Iht' lank, ... ht'li tunk Is st..lionary.
The pressure at point B is given by.
fl8'"pgh = lO00x9.81 x05 = 4905 Nlrn 2
(ii)
This pressure is represented by line BD in Fig. 3.52
Force on the side An
:: Area of triangle ABD x Width
=(t XA B xBD)X b
= (i-xO.5x 4905)x2
(': ED :: 4905)
= 2452.5 N. AilS.
For this case. the force on An can also be obtained as
"AIl= pgA.;;
wllerc
A:: AB x Widlll = 0.5 x 2 = 1 m 2
AS 0 .5
It:: - : =0.25111 = 1000 x9.81 x 1 xO.25
2
2
= 2452.5 N. AilS.
Problem 3.38 A umk coll/oins »'a/er "plo (j depth of 1.5 m, The leng/II amI wid/I, of /lIe lank (lrl!
-I m (lIld :z m respec/indy. Th e wnk is mo\'ing lip "" jllc/ined p/(Jlle with (I cons/,w/ (lccderu/iol! of
-I ",Ii. The inc/inaliOll of Ihe p/mw will, the /u)riZOJ1la/ is 30° as 11110 ...11 in Fig. 3.53. Find.
(i) Ihe ''''gle made hy Ihefree J'u rf"ce ofll'lIler will! Ihe horiWII{(,/.
(ii) Ihe pressure III Ihe hOI/om of I/le lallt (1/ Ihe fro n/ ami rear ellds.
Solution. Given:
Dcptli of water. 11 == 1.5 10 : Lengt li. L == 4 III and
Width. b == 2 Ul
'm
Constant acceleration along th e inclined plane.
II '" 4 Ill /5
1
Inclination of plaoe. 0: == 3D"
a
Let == Angle mad~ by the free surfac", of w~ter
after tlie acceleration is imparted to the tank
p~
== Pressure ,lithe bo1l0111 of the tank :It the from ",nd
and Pv == Pressure at the bottom of the taJlk at the reaT
end.
Fig. 3.53
This proble m can be donr by resolving the give n acceleration along thc horizontal direction and
vertical d irection. The n eac h of these c ases Ill ay be separately anal ysed according to the sct procedure.
Horizontal and vertical eomponcnts of the acceleration arc:
2
a" '" II cos Cl " 4 cos 30° '" 3.464 lll/s
,
{/ == a sin Cl == 4 sin 30° == 2 m/s 2
When th~ tnnk is station~ry on the inclined plane, free s urfac", o f liquid will be along EF ~s shown
in Fig, 3.53, But when til", tank is moving upward alo ng the indi Jled plane th e fr"'e surface of liquid
will be along ne. When the tank containing a liquid is moving up an inclined plane with a constant
accderation. thc angl e madc by the frce surface of thc liquid with thc horizontal is g iven by
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1124 Fl uid Mechan ics
_ "_,_
tan , '"
1I., +g
3.464
2+9.8 1
"" 0.293)
= tan - I 0.2933 '" 16 .346
Ans.
"'
Now lei us first find the depth of liquid at the front and rear end of the tank.
{I
G
IXrlll of liquid at front end", II I '" AB
[krill of liquid at Tear ~nd '" 11 2 '" CD
From Fig. 3.53, in triangle eOE, tan
a = ~~
CE = EO tan
"'
e = 2 x 0.2933
(,: EO", 2 m. tan
= 0.5866 III
CD = 112 = ED + CE = 1.5 + 0.5866 '" 2.0&66
e = 0.2933)
m
hl=AB=AF-BF
Similarly
= 1.5 - 0.5866
(": AF = 1.5. BF = CE = 0.5866)
'" 0.9134 m
The pressure at the bouom of tank 3l Ihe fear end is g iven by.
(I
+ ';)
PD= pghl
= 1000 x9.81 x Ul866 ( , + _ 2_) '" 24642.7 N/m z. Ans.
9$ '
The pressure at the boU01n of lank at the front e nd is given by
Pit '" pgh t ( I + (:')
'" 1000 x9.81 xO.9134 ( , + -'-) '" 10787.2 Ntml, Ans..
9>,
HIGHLIGHTS
1. When the nuid is at rest. Ihe shear stress i. zero.
2. The force exerted by
~
Sl:J.tic fluid on a vertical. horitontal Or an inclined plane immersed surface.
/' '" pgAh
where
p • Density of the liquid.
A • Area of the immersed surface. and
ii .. Depth of the centre of gravity of the immersed .urface from
free .urface of the liq uid.
J. Centre of pressure is defined as the point of application of the resultant pressure.
4. The depth of centre of pre,sure of an immersed surface from free surface of the liquid.
for vertically inlinCrsc<i surface_
for inclined immcr>ed surface.
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Hydrostatic Forces on Surfaces 125 1
S. The centre of pressure for a plane ,'crtical surface III's al a depth of two-third the height of the
immersed surface.
6. The t01a1 force on a curved surface is given by F.
where
f~ ",
K
Hnriwntal force on cun"cd surface and is equal to total pressure force on the projected area
of Ihe curved surface on the ,-"nieal plane.
pgAh
F, '" Vertical force on sub-merged curvcd surface and is equal 10 Ihc weight of liquid actually
or imaginary supported by the curved surface.
and
F
7. The inclinalion of Ihc result,ml force on curved surface with horizontal. tan e = .2....
F,
8. The resultant for • .:: on a sluice gate. F = FI - Fl
where
F, = Pressure force On (he upstream side o f (he sluice !laIc and
F : '" Pressure force on the downstream side of the sluice gate.
9. Po. a lock gate. the reaction ""'tween the two gates i. equal to the r~action at the
hing~"
R" I'.
Also the reaction between the two gate •• p .. __
F_
2 sin {I
where F .. Resultant water pressure on the lod gate .. F, - F2
and
{I
= Indination of the gale with the nonnal to the side of the lock.
EXERCISE
(A) THEORETICAL PROBLEMS
I . What do you understand by 'Totall'Tessure' and "Centre of Pressure" ?
2. Derive an expres,ion for the force exerted on a sub·merged vertical plane surface by the static liquid
and locate the position of centre of pressure.
J . Prove that the centre of press ure of a completely sub·merged plane surface is always ""' low the centre
of gravity of the sub·merged surface or at most coincklc with the centre of gravity when the plane
surface is horirontal.
4. Prove th~tthe tOl~1 pressure Herted by a static liquid on an inclined plane sub·merged surface is lhe
""me a< the foree exerted on a vertical plane surface as long as the depth of the centre of gravity of the
,urface is unal!ered.
5. Deri~e an expression for thc deplh of centre of pressure from free surface of liquid of an inclined plane
surface sub·merged in the liquid.
6. (u) How would you detcnnine the horiwntal and wrtical components of the re,uUant pres,ure on a sub·
merged curved surface?
(h) Explain the procedure of finding hydrostatic forces on curved surfaces.
(f)elhi Un;\·er.<ity. Hec. 2002)
7. l: xplain how you would find the resultant pressure on 3 curved surface inllHersed in 3 liquid _
S. Why the resultant pressure on a curved sub·merged surface is dctcnnincd by first findin~ horizontal
and ycrtical forces on the curved surface? Why is the same method not adopted for a plane inclined
surface sub·merged in a liquid?
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1126 Fluid Mechani cs
!I. De .. ribe briefly with sketches the various methods used fo r measuring pressure excncd by fluid s.
10 . Prove that the "cr1ical component of the resu ltant pressure On a sub-merge<J curved surface is equal to
the weight of the liquid supponcd by the curved surface.
I I . What is the difference between sluice gale atld lock gate?
,.
11. Prove 1ha1 the reaction between the gates of a lock is equal to the reaction at the hinge.
u.
Derive an expression for the reaction between the gates as I' '" _ __
2 sin e
where"" Resultant water pressure on lock gale, (I '" inclination of the gate with nonnal to the side oflhe lock.
1-1 . When will centfe of pressure and centre of gmvity of an immersed plane surface coincide?
IS. Find an expression for the force cxened and ccnlre of pressure for a complclely SUb-merged inclined pi:me
surf~ce. Can Ihe same melhod be applied for finding Ihe resull~nl for.:e On a curved surface innners..'<l in
Ihe liljuid ? If nOI. why?
16. Whal do yo u undersland by Ihe hydroslalic equal ion ? Wilh Ihe help oflhis equalion derive Ihe expressions
for lhe IOlal Ihrusl on a sub·merged plane area and Ihe buoyam force aCling on a sub·merged body.
(B) NUMERICAL PROBLEMS
I . DClcnninc Ihe 10lal press ure and dcplh of cenlre of pressure on a plane re<:I:,ngulur surface of 1 111 wide
and 3 111 deep when ils upper edge is horizonlal and (al coincides wilh Waler surface (b) 2 111 below Ihe
[Ans. (a) 44 145 N. 2.0m. (b) 103005 N. 3.71 4 ml
free waler surface.
2. Dctennine the IOlal pressure on a cir.:ular plale of diamcler 1.5 m which is placed verlkally in walcr in
such a way Ihal ceTllrc of plale is 2 m below Ihe free surface of waler Find Ihe posilion of cenlre of
pressure also.
IAn< . 34668.54N.2.07rnl
3. A re<;langular sluice gale is situaled on the ,·crlkal wall of a lock. The verlkal side of the sluice is 6 m
in lenglh and depth of centroid of area is 8 m below the Water surface. Prove Ihat Ihe depth of cemre of
pressure is given by 8.475 m.
4. A circular opening. 3 m diamelcr. in a vertkal side of a tank is closro by a disc of 3 m diameter Which can
rotate about a horizontal diameter. Calculate (i) the force on the disc. and (iI) the torque required 10
maintain the di", in equilibrium in the .. ertical position when the head of water abo,·e the hori,ontal
IAns. (i)416.05 kN . ( ii) 39005 Nml
diameter is 6 m.
5. The pressure at the centre of a pipe of diatneter 3 m is 29.43 Nkml. The pire contains oi l of sp. gr. 0.87
and is filled wilh a gate valve . Find thc force cxene<:l by the oi! on the gate and posilion of centre of
pressure.
IAns.l.OIl MN ..016 m below cenlre of pipe]
6. Determine the tota l pre.<>ure and ccntre of prc<Sute on an isosceles triangular piate of ba,e 5 m and
allilude 5 m when Ihc plale is immersed verlically in an oil of sp. gL 0.11. The base of the plale is I m
[An •• 261927 N. 3.19 ml
below the free surface of waleL
7. The opening in a dam is 3 m wide and 2 III high. A ycnical sluice gale is used to cover the opening. On
the up,lream of Ihe gale. the liquid of sp. gr. 1.5. lie, uplo a heighl of 2.0 m above the lop of Ihe gale.
whereas on the downSlrcalll ,ide. Ihe waler is available uplO a height of the top of Ihe g~le. Find the
rcsuUanl force aCling on Ihe gale and posilion of centre of pr~ss~re. Assume Ihal Ihe gale is higher al
Ihe bononl.
[Ans. 206010 N. 0 .964 '" above Ihe hingel
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Hydrostatic Forces on Surfaces 127 1
K. A cOlis>')" for closing thc enlrance [0 a dry dock is of Impel0idal form 16
'I.
10 .
II .
I Z.
13.
14.
In
wide at the lOp and 12 m
wide at the bonolll ,md 8 111 deep. Find Ihc 10lai pressure and cenlre of pressure on Ihc caisson if the
water on the outside is 1 m below the lOp le"cl of thc caisson and dod is empty.
[Ans. 3.164 MN. 4.56 m below water surface I
A .Iiding gale 2 III wide and 1.5 In high lie, in a venical plane and has a co-effocicnt of friction of 0.2
between itself and guides. If Inc gate weighs one tonne. find the vertical force required to raise thc gate if
its upper edge is at a depth of 4 m from free surface of wmer.
IAns. )7768.5 NI
A lank conuin, WOller upto a height o f 1 In above the ha"", An imm;",,;ble liquid of _'po gr. 0.8 is (,lied on
the lOp of waler UplD 1.5 m height . Calculale , (il total pre«ure on One side of the lank. (iiI Ihe position of
cenlre of pressure for one side of the tank. which is 3 In wide.
IAn •. 76518 N. 1686 In from lopl
A reclangular tank 4 m long. 1.5 m wide cont";,,s water Uplo ~I height of 2 m. Calculale Ihe force duc 10
waler pressure on Ihe base of Ihe lank. Find also Ihe deplh of cenlre of pres,ure from frcc surface.
IAns. 117720 N. 2 m from free surface I
A reclangular plane surface I m wi<k and 3 In deep lies;n water;n such a way Ihal;t,; pl:!n" makes an angle
of 30" wilh (he free surf"ee of water. Dele"";ne lhe (Olal pressure and pos;(;on of eenlre of pressure when
lhe upper edge of (he plale ;s 2 m below Ihe free waler surface.
IAns. 80932.5 N. 2.318 Illi
A circular pla1c 3.0 In diameter is immersed in water in sllch a way lha( 1he plane of lhe plale makes an
angle of 60° wi1h (he free surface of waler. Dete""ine the (otal pressure and posilion of centre of pressure
when lhe u!'per edge of the plale is 2 m below the free water surface.
IA lls. 22l!.69 kN, 3427 m from free surface I
A rectangular gale 6 m)( 2 m is hinged at its base and inclined at 60° to the horizontal as shown in Fig. 3.54.
To kee!, (he gale in a stable position. a COUnler weighl of 29430 N is a!1ached at (he upper end of lhe gate.
fi nd the depth of wmer at which the gate begins (0 fall. Neglect (he weight of the gale and also friction a1
Ihe hinge and pulley.
IAns. 3.43 ml
WATER SURFACE
Fig..H4
Fig.3.SS
15 . An inclined rcct:lngular gale of width 5 m and depth .5 m is installed to comrol the discharge of waler as
shown in Fig. 3.55. The end A is hinged. Dcle""inc Ihe force nonnal to Ihe gale applied al H to open it.
{An s. 97435.8NI
16. A gale supporting wa(er is shown in Fig. 3.56. F;nd lhe heigh(
'h' of (he wa1er so tha( lhe ga(c begins to (ip aooul 1he hinge.
Take the widlh of lhe gale as unilY.
IAns. 3)(./3 "'I
17. Find the tOlal pres,u re and depth of cen1re of pressure on a
Iriangular plate of base 3 m and heighl 3 III which is immersed in
waler in such a way lhal plane of lhe pla1e makes an angle of 60"
Fig. 3.56
wilh the free surface. The ba<e of Ihe plale is parallel 10 waler surface and at a deplh of 2 III from water
surface.
[Ans. 126.52 IN. 2.996ml
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1128 Fluid Mechani cs
Ill . Find the horizontal and vertical components of lhe 101al force acting on a curved surface AB. which is
in lhe fonn of a quadrant of a ,-irde of radius 2 m as shown in Fig. 3.57. Take lhe width of Ihe gale 2 111.
IAns. F, = 117.72 kN. F, = 140.114tNI
WATER SURFACE
-------------
,
Fig. 3.S7
Fig. 3.58
19 . fig. 3.58 shows a gale having a quadrant shape of radius of 3 Ill. rind Ihe resultant force due to waler pcr
metre length of the gale. rind also the angle at which the total force wil l acl. [Ans. 82.201 kN, 0 = 57° 31 ' I
20 . A roller gate is shown in I:ig . ).59. It is cylindrical fonn of 6.0 In diameter. It is placed on lhe dam. I'ind
lhe magnitude and direction of lhe reSUllan! force d ue 10 waler aCling On the gale when the waler is just
going 10 spilL The I
1 of [he gal" is gi"en 10m.
Fig. 3.59
( Ans. 2.245 M N.e~3l!" g·1
Fig. 3.60
21. Find Ihe horizontal and vertical components of Ihe water pressure exerted on a tainter gate of radius
4 rn as shown in Fig. 3.60. Consider widlh of thc gate unity.
(Ans. F, _1962 kN. F, _ 7102.44 N I
22. Find thc magnitude and dirtttion of thc resu ltant water
pressure act illg on a cun'cd face of a dam which is shaped
",
according to Ihe relalion y '" -
WATER SURFACE
as shown in I' ig. 3.61. The
height of water retained by the dalll is 12 Ill , Take the width
IAns. 970 ,74 kN.
43 " 19']
of dalll as un ily
23 . Each gale of a lock is 5 III high and is supporled by two
hinges placed on Ihe lOp and bollom of Ihe gale. When Ihe
e ..
Fig. 3.61
gales are closed. Ihcy make an angle of 120". Tn e widTh of The lock is 4 Ill. [f The depths of water on the IWO
sides of the gates are 4 m and 3 m respectively. o;ietem,inc , (I) the magnitude of resultant pressure on cach
gate. and (ii) magnitude of the hinge reactions. IAns.(i) 79,279 kN. (ii) Rr • 27.924 kN. R~. 51.355 kN]
24 . The end gates ARC of a lock arc 8 m high and when closed make an angle of 120°. The width of lock
is 10 m. Each gate is supJXlrlcd by IWO hinge s located at 1 III and 5 m above the bollOIll of the loc\;, The
depth of Waler on the upslream ami downstream side, of the lock are 6 m :md 4 HI respccti~ely . Find :
(I) Rcsulta1l1 water foree on each gate.
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Hydrostatic Forces on Surfaces 129 1
(ii) R""c(ion between the gates AB and BC, and
(iii) f orce on each hinge, considering the reaclion of the gate acting in Ihe same horizomal plane 3S
re sultant water pressure. [A ns. 566.33 i:N. (iiI 566.33 kN. and (iii) R, _ ! 73.64 kN. R8 _ 392.69 kNI
25. A hollow circ ular pl~IC of 2 m external and I 111 internal diameter is immer>ed ""l1i'31Iy in waler such thm
the cenlre of plate is 4 111 deep from water surface. Find the to(al pressure and depth of centre of pressure.
[Ans. 92 .508 kN. 4.078 ml
26 . A rec~~ngular opening 2 111 wide and I 111 deq, in the vertical side of a tanl; is closed by a sluice gate of the same
size. The gUll: can turn abom the horirontai ~en\rOidal axis. Dclcrminc : (.) !he lotl! prc,;surc on the slu;"" gale
and (il) the torque on the ~Iuice gatc. 11Ie head of wale, alxwe the upper edge of the gale is 1.5 111.
IAns. (il 39.24 IN. (il) 1635 Nml
,
27 . Octennine the total force and kx:ation of .'entre of pressure on one face of FREE SURFACE OF L1aU ID
the plate shown in Fig. 3.62 immersed in a liquid of specific gravity 0.9.
IAn • . 62. 4 kN. 3.04 ml
28 . A circular opening. 3 m diameter. in the "er1kal side of water tank: is cloSC<l
by a disc of 3 m diameter which can rolate about a horizontal diameter?
Calculate: (i) the force on the disc. and (ii) the torque re«uired 10 maintain
the disc in equilibrium in thc "cnical position whcn the head of waler
above Ihe horizontal diameter is 4 m. IAn •. (I) 270 kN. and (ii) 38 kN till
29 . A penstock made up by a pipe of 2 m diameler contains a circular disc of
Same di:nneter 10 act as a ,·al,·c which controls Ihe discharge passing
through it. It can rotate aoout a horiwnlal diameter. If the head of watcr
Fig. 3.62
above ilS cefilre is 20 111. find thc lotal force acting 011 Ihe disc and the
IOrque r'-"quireJ 10 maint'lin it in Ihe "cr1ic,tI posilion.
30 . A circular drum 1.8 m diameter and 1.2 111 height is submerged with its axi, vertical and il< upper end at
a depth of 1.11 m below water lewl. Detennine :
(i) tOlal pressure on top. bonom and cU"'ed surfaces of Ihe drum.
(ii) rcsuhall1 pressure on Ihe whole surface. and
(iii) depth of centre of pres,ure on curved -,urface.
31 . A circular plate of diameter 3 m is imme=d in watcr in such a way that ils least and greatest dcpth from
the free surfacc of walcr arc 1 m and 3 m re .• pectivc ly. For the front side of the plate. find (i) tolal force
IAns. (i) 1386S4 N : (ii) 2.1 25 ml
exerted by walcr and (ii) the POSilion of ccntre of pressure.
3 2. A tank contains watcr upto a hei ght o f 10 111. One of the sidc< of the tank is inclined. The anglc between
free surface of water and inclined side is 60". The widlh of the tank i. 5111. Find: (i ) the force exerted by
waler On inclined side and (ii) position of centre of pressure.
IAns. (,)283.1901 IN. (if) 6.67 ml
33. /I circular platc of 3 m diameter is undcr " 'ater wilh ils plane making an angle of 30" wilh the water
surface. If Ihc lOp edge of the plate is I m below the waler surface. find the force 011 one side of the plate
and ils loculion.
(J.N.T. U.. !fyi/em/mt! S 20(2)
".30·
IlIl nt. i/. 3 111.9.30". heighl of top edge. I 111. h • I + 1.5 X sin 30"
'" 1.75
-
(' ,] X 1.75", 121.J5kN.
F=pgAh '" l000x9.81 X "4 x3 "
h' •
+ 1.75 .. 0.08 + 1.75 .. 1.83 m.1
Fig. 3.63
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CHAPTER
.. 4. 1 INTRODUCTION
In th i~ c hapte r. the equ ilibrium Ml he Oo alin g and sub-me rged bodies will be co nsidered . Thu s Ihe
c hapter will inc lude: I. Buoya ncy. 2. Ce ntre of buoy ancy. 3. Mclace ntre. 4. Metace ntri c hf.> ight.
5. Ana lytica l method fo r deta minin g mdacentric heig ht. 6. Condition s of equilibrium o f a fl oatin g
and sub -merged bod y. and 7. Expe rim ental met hod fo r metacc ntri c he ight.
.. 4.2
BUOYANCY
Whe n a bod y is imm ersed in a fluid_ an up ward force is exe n cd by the fluid o n the bod y. Th i~
upw ard forec is e qual to the we ight o f the fluid displaced by the body and is c al k d th e force of
buoy an cy or s impl y bu oy ancy .
.. 4.3
CENTRE OF BUOYANCY
It is de fin ed as the poi nt. th rough whi ch th e force of buoy ancy is supposed to ac t. As tlw force of
buoyancy is a ve rtica l force and is equal to th e we ight of th e Ou id d isplaced by th e body. the centre of
b uoy ancy wi ll be thc ce ntre o f grav it y of th e fluid di splaced.
Proble m 4.1 Fittd lite I'Olume of Ille lI'aler displaced alld pwiliOlt of cenlre of buoyancy for a
lI'oodell block of wid III 2.5 m and of depth 1.5 m. whell il floals horizollwl/y in Waler. The densilY of
woodell block i.f 650 kg/III ) {Illd ils lenglli 6.0 m.
Solution . Gi vc n :
Width
" 2.5 10
Ikplh
Le ngth
Vol ume of Ihe bl oc k
" 1.5 111
" 6.0 111
= 2.5 x 1.5 x 6.0 = 22.50 ln 3
p = 650 kgfm}
Iknsi ty of wood.
Wf.> ig ht of bloc k
= px g x Volum c
= 650 x 9 .81 x 22.50N= 14347 1 N
jw
G
2.5 m
Fig. 4.1
131
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1132
Fluid Mechani cs
For equilibrium Ihe wei gh l of water di sp la ced'" We igh.t of wooden block
= 14 347 1 N
Volum~
o f W31.-:r displaced
Weig ht of wa ter di sp laced
We ight de nsit y of water
14347 1
~~"':'cc= 14.625 ml, AilS.
1{)()() X 9.8 1
(': We ight density of wmcr '" 1000 x 9.81 N/ml)
Pos it io n uf Ce ntre of Uuoyancy. Vol ume of woode n b lock in water
or
'" Vo lume o f water d ispla(;cd
25 x II x 6.0 '" 14.625 m) , w here /, i s depth of wooden block in W3Ie r
II '"
14.625 '" 0 .975
2.5 X 6.0
III
0.975
Cent I"C of Buoyanc y '" - 2- '" 0.4875 m from base.
Ail S.
Problem 4.2 A ...oodell fog 0/0.6 m diameter and 5 m lenglll is j7<w(illg ill ,;,"er waler. Pind Ille
depth of Ille ,,"OOdl'l! log ill \\'(l/er when lile !Ip. gf",-;Iy of flU! log i,y O. 7.
Solution . Given:
Dia. of log
=O.6m
L:5m
Length.
Sp. gr ..
S '" 0.7
Dens ity of log'" 0.7 x 1000", 700 kgfm}
We igh! dcnsi!y of log .
w =
,
,
o
o
px g
Fig. 4.2
=700x9.81 Nfm 3
Find depth of immersion ur II
W~igtl! of wooden log
::
'" Weigh! densily x Volum e of log
,
,
=7 00x9.8 1 X 4 (DrXL
=700x 9 .81
Fur equi librium.
Weigh! of wooden log
X~(.6)2 X5N=989.6X9.81 N
= Weigh! of wa!er displaced
= Weigh! densi!y of water X Volu ,nc o fwatcr di splaced
Vo lume u f wa!cr displaced '" 989.6 x 9.8 1 = 0.9896
l000x9.81
m3
(': We igh! densi!y o f wat~r '" 1000 x 9.81 Nfm3)
Let" is the dcpth o fi'nmc rs ion
Volume of log ins ide wa ter'" Area of ADCA x Len gth
= Area of ADCA x 5.0
But vo lum ~ of log inside water == Vo lum e of water displaced == 0.9896 m}
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Buoyancy and Floa tation
133 1
0.9896:: Area of A DCA x 5.0
:. Area of ADCA
__ 0.9896 __0. 1979
5.0
Bm area of ADCA
:: Area o f cu rved su rface ADCOA + Are a of MOe
:: It?
)·-"l
[3(j(360"
:: It?
[l - ~l
IS(r + , 2 OOS. 9 .sin 9
0. 1979 = It
.00157
or
e-
III '
+
"21 r (;Os e x 2r si n e
(.3)~ [ l- l ~.] + (.3)~cos 9sin e
0. 1979:: .2827 - .{1()157 e + 0.91:0s e si n e
.09 cos e sin 9:: .2827 - . 1979 :: 0.0848
, .,
.0848
, - -.09
- - cos S10 :: - - .00157
.00 157
e - 57.32 cos e sin e:: 54.01.
or e - 57.32 ("(IS e si n e - 54.01 = 0
For 0:: 6(}0,
60 - 57.32 x 0.5 x .S()6 - 54.01'" 60 - 24.81 - 54.01:: - llUn
For
e:: 70°,
70 - 57.32 x .342 x 0.9396 - 54.01 '" 70 - 18.4 - 54.01 :: - 2.41
For
e::
72 - 57.32 x.309 x .951 - 54.01 = 72 - 16.&4 - 54.01 = + 1.14
72°,
For e:: 7 1°,
e:: 71.5",
T hen
7 1 - 57.32 x .325 x .9455 - 54.01 :: 7 1 - 17.61 - 54.01:: - 0.376
7 1.5- 57.32x .3 173 x .948- 54.0 1 :: 7 1.5- 17.24-54.0 1 = + .248
/,=
r + rl'Os7 15 °
= 0.3 + 0.3 x 0.3173 = 0.395 m . Am.
Problem 4 .3 A slOlIe weighs 392.4 N ill air and 196.2 N in .mler. Compule Ihe )"olume of slon e
and ils specific grm·ily.
Solution . Given:
Weigtlt of SlO ne in air
Wdglll of Slone in water
= 392.4 N
== 196.2 N
ror equilibrium.
Weigtll in air - Weight of sto ne in water = Weigh t of w~H er displa<:ed
or
392.4 - 196.2 == 196.2 = 1000 x 9.81 x Volume o f wata displaced
Volume o f wa ler displaced
196.2
t
t
= ~~~~== _ rn 3 == - x 1 0~ c rn 3 == 2 x 10' em l . Ans.
looox9.8 1 50
50
= Volume of Slo ne
Volume o f Slo ne
I I
== 2 x 10· cm 3. An s.
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1134
Fluid Mechani cs
Specific G rav it y or Ston e
'" Wcig tll in air '" 392.4 = 40 kg
g
9.8 1
Mass in air '" 40.0 kg
Mass o f stone
=
Densi ty o f stone
Vol ume
~rn J
kg
= 40x50=2000 -
rn
l
5()
De nsi ty of Slone
Sp. gr. uf stolle
'"
2000
'"
'" 2.0. AilS.
Density o f wak r
1000
Problem 4 .4 A body of dimensions 1.5", x f.O '" x 21M, weighs /962 N in ....afer. Find ils I\"eight
in air. IViI,,/ will be ilJ' JJlecific gra,·;ly ?
Solution . Gi ve n :
Volum e of bod y
'" 1.50 x 1.0 x 2 .0 '" 3.0 Ill l
Wcig tll of bod y in wa!c r
'" 1962 N
Vol um e of rhe wa ter di splaced", Vo lume o f the body '" 3.0 Ill]
Weight of water d isplaced = 1000 x 9.8 1 x 3.0 '" 29430 N
For th e equilibrium o f rhe hod y
Wc igtn o f body in air - We ight of water di sp laced'" Weight in water
\¥.. r - 19430= 1962
IV..,'" 29430 + 1962 = 31 392 N
= Weight in air '" 31392 '" )200 kg
Mass of body
9.81
g
=
Den sit y o f rhe body
Sp. grav ity o f th e bod y
=
M ass
Vo lu me
1066.67
'" 3200 '" 1066.67
3.0
'" 1.067. Ans .
1000
Problem 4.5 Find rlre den .~ily of a melilllic hody which /loms at Ihe inlerface of merCllry of
Sf'. gr. 13.6 alld Waler silch 1/1(1140% of ils \·olume is sub-merged ill mercllr)" and 60% ill wal"'.
Solution . Le t th e vo lume of the body '" V m 3
The n vo lume of body sub-me rged in me rcu ry
40
= - V", 0 .4 Vm 3
100
Vo lum e of body
sub - m c r~ed
in wate r
""
~_ )( V",0.6 Vm 3
I()()
For the equilihrium of the bod y
Fig . 4. 3
To ta l bu oy ant fo r"e (up ward force) '" We ig ht of the body
Bu t to tal buoy ant force '" Forc~ of buoy an cy du ~ to wa ter + Force of buoyan cy d ue to merc ury
Force of buoyancy d ue to water '" Weig ht o f wa tn d isplaced hy IxHl y
'" De ns ity of water )( g )( Vo lume o f wate r d isplaced
= 1(00)( g )( Vo lume of body in water
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Buoyancy and Floa tation
135 1
= 1000 x g x 0.6 x V N
and Force of buoya llcy du e to mercu ry
= Weig ht of me rcury di splaced by body
= g x D ensi ty o f mercury x Vo lume of mercury displ ace d
= g x 13.6 x 1000 x Vo lum e o f body in merc ury
We igh t of the bod y
wtwre p is the de nsit y of th e bod y
For equilibriu m. we ha ve
= Densi ty x
=Ex [) .6x I OOOx O.4 VN
g x V olume of body
=Px
To tal buoyant force
= Weight o f th e body
I OOOxgxO.6xV + 1),6x 1000xg x .4 V=pXg x V
or
p = 600 + 13600 x.4 = 600 + 54400
De ns it y o f the body
'" 6040.00 kg / m ) . An s.
Problem 4 .6
g x V
= 6040.00 kgfm l
A flolli WIll'" regulates I/le flow of oil of sp. gr. 0.8 illio "ciSlem, The JpileriClil float
iJ' 15 em ill diameter. AOlJ is {/ weighlleS!>' link carrying lile floal al anI' <'11</. Will {/ m/"e at Ihe OIlier
end wllich closes Ille pipe Iltmugll wllich oil flows illla IIII' cis/cm . The /i,,1: i.1 mounled in (I friClionle.H
liillge m 0 alld llie (llIgle AOB i.f 135°. Tlit' lellglh ofOA is 20 em, (l1II! Ille disrtillee belweell Ille eelllrt'
of Ihe flom mId II,e hillge is 50 em. WIIt'll Ihe flow i.l slOpped AO will be I·ulieal. The \"G/re is 10 be
pressed all 10 II,e se(ll WillI a force of 9.81 N 10 co mplerel)' slop Ihe flow of oil inlo Ille cistern. It W(lS
obserl'ed IIwl the flow of oil i')' slopped whell Ihe free SlI rfrlCe of ai/ ill Ille cislem is 35 em bela,,· Ille
hillge. Delermil1f: Ihe weiglll of Ihe floo/.
Solution . Gi ve n :
0"
Sp. gr. of oil
SUPPLY
'" 0.8
Po = 0.8 x 1000
Densit y of oil.
0:
800 kglm'
D o: I5cm
Dia. of flo at.
LAOB = 135 Q
OA=20cm
Force .
P=9.8 1 N
O B = 50 cm
Find tile weig llt of [h e float. Le t it is eq ual [ 0 W.
Wile n til e flow of o il is stopped. the ce ntre of float is s how n in Fig. 4.4
Tlic leve l of o il is al50 sllown. Til e ce ntre of fl oat is be low th e Icve l of o il. by a de pth '/r' .
From tJB OD,
sin 450 0: 00 " OC + CD
OB
OB
35 + II
50
50xsin45° ,,35 + h
II = 50 x
I
J'i - 35 '" 35.355 -
35 " 0.355 Ctn " .00355 tn.
Th e weig lu of floa t is ac ting th ro ugh 8, but Ih" upward buoya nt force is acti ng throug h th" CCl)tre
o f weight o f o il displa(;ed .
Vol um e uf oil disp l;lccd
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1136
Fluid Mechani cs
'" t
Buoy ant force
x
It
X {.075)3 + .{)(}355 x
It
x (.075) 2 '" 0.000945 111 3
'" We ig ht of oil di splaced
'" Po x g X Volume of oil
: 800x9.8 1 x.000945 = 7.416 N
The buoyall1 force ~nd weight of Ihe floal passes through th e sa me vertical li ne. pass ing through B.
Let the we ig ht of float is W. Th en net vertical force o n 110.11
'" Buoyalll fo rce - Wc igtu of flo at'" (7.416 - IV)
Tak ing mo ments about the hin ge O. we gel
or
Px 20 '" (7 .4 16- IV) x 8D = (7.4 16 - IV) x 50 x \:0545 0
9.8 1 x 20 '" (7 .4 16 - IV) x 35.355
16 - 20x9.8
1"=7.4
T
35.355 1 = 7 .4 16 - 5.55= U!66 N. Ans .
.. 4.4
META -CENTRE
It is defined as Ihe (Xlint about wllich a body Slarts oscillatin g when Ih e body is tilled by a small
an gle. The 11lc1a-ccntre may also be de fi ned as the point at which Ihe li ne of act io n of th e force of
b uoya ncy will meet th e nonnal axi s of th e body whe n th e body is give n a s mal! angulardisplaccmcnt.
Consider a body fi oating in a liq uid as s hown in Fig. 4.5 (a). Let the body is in equilibrium and G is
th e ce ntre of grav it y and B the ce ntre o f buoyancy. Fo r equilibrium. both the points li e on the norma l
axis. whic h is ve n ic al.
J-... ANGUlAR
NOj MAL AXtS
- "'-!-".
!;
M DISPlACEME
•. . . :.. t...
~
-~
I
,, ,
I"
/
I
-;;'~?';-'
I'.
NORMAL AXIS
(b)
(a)
Fig. 4.5
M na-<:f'nlr t
Le t th e body is given a small angular displa~"t!ment in the clockwise di recti on as Show n in Fig. 4.5 (b).
The I:entre of buoya ncy. wllich is Ille I:entre of gravity of Ill e di spl al:ed liquid or cenlre uf grav ily uf the
ponion of Ihe body sub-11Ie rJ(cd in liquid. will now be Shifted towa rds righl from Ihe nurm al axi s. Lei
iI is al B[ as show n in Fig. 4.5 (b ). The line o f action of lhe force of buoyancy in Ih is new posi tion. will
in tc["Sectlhe [lOTinal a~is of th e hody at SOllie point say M. This point /If is called Met,,· IT n lre .
.. 4 .S
META -CENTRIC HEIGHT
The d islanc e MG. i.e .. Ih e disl anl:e belween th e mcl a-ce ntre of a n oal in g body and Ille cemre of
grav it y o f the body is called IllCla-centric heig ht.
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Buoyancy and Floa tation
.. 4.6
137 1
ANALYTICAL METHOD FOR META -CENTRE HEIGHT
Fig. 4.6 (0) shows the position of a noming body in equilibrium. The location of centre of gravity
and centre of buoyallcy in Ihis position is al G and 8. The floming body is given a small angu lar
displacement in the clocl;wisc direction. Tliis is shown in Fig. 4.6 (b). The new centre of buoyancy is
al 8 1" The vcnica l linc through 8 1 cuts Ihe normal ax is at M. Hence M is the meta-centre and GM is
mew-centric height.
----.
ANGULAR
M
A
ISPLACEMENT
"'.
S'
S
Go
S
Gi--(+~I:--"C
,
'. C
T
101
(
l (c)
PLAN OF BODY AT WATER LINE
~.E-".
Fig. 4.6
Mela-Cf'nlre height of f/Qdfillg body.
The angular displacement of the body in Ihe cloc kwise direction causcs the wedge-shaped prism
B08' un Ihe rigtu of the axis 10 go inside the water while tile idemical wedge-shaped prism reprcscmed
by AOA' emerges OuI of tile water un tile len uf tile axis. These wedges represent a gain in buuyam
force on the right side and a corresponding loss of buoyant force on the len side. The gain is
r~pr~scnted by a vertical force dFJj acti ng th rough the c.G. of the prism BOB' while th e loss is
represcnt~d by an equal and op]Xlsite force dFo ac ting Vertically downward through the centroid of
AOA'. The coup le due to these buoyant forces <IF/! tends to rotate the ship in the counte rclockwise
direction. Also the moment caused hy the displacement of the centre of huoyancy from B to Bl is also
in the coullterclocl; wisc direction. Thus these two coup les must he equal.
Co uille I) ue to Wed ges. Co nsider towards the right of the axis a sma ll strip of thicl;ness <Ix at a
distance x from 0 as Shown in Fig. 4.5 (b). The height of strip x x LBOB': x x e.
! ... L B08'" LAOA'" 8MB,'" 6)
:= Height x Thickness", .r x 6 x <Ix
Area of strip
If L is the lenglh of lhe floating body. then
Volume of strip
:AreaxL
"xx6xLxdx
Weight of stri p := pg X Volu1\le:= pgx fJL dx
Similarl y, if a small Slrip of thickn~ss dx at a distance .l from 0 towards the left of the axis is
considered, the weight of Slrip will he pg_.fJ L d.L The two weights are acting in the opposite direction
and h~nce constitute a coupl~.
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1138
Fluid Mechani cs
Moment of ttJis couple
'" Weight of each stri p x Distance between Ihes<: two weigh ts
=pgx fJLdxlx+_t]
= pgx aL dx
x 2x = 2pgxl 9L dx
Moment of the couple for the whole wedge
"J 2pg_~ aL dx
.. ,(4 . 1)
MomcnI of coup!c duc 10 shirting of cemrc of buoyancy from B 10 HI
:FyxBBI
= FHXBMxfJ
('0" BB I
"
BMx fJif9is "cry
" WxBMx9
~m alll
I': Fe" WI ...(4.2)
BUl these two l'Oup lcs arc the same. Hence equating equations (4.1) and (4.2), we get
x HM x e" J lpg_'.! e ulx
IV x 8M x e" 2pg9 J:h_Ax
IV
IV x 8M "" 2pg
J_'~Ldx
Now LAx == Elem en tal area on the wmcr line shown in Fig. 4.6 (c) and == dA
II' x 8M '" 2pg
j .,.2dA .
BUl from Fig. 4.5 (el it is clear lhm 2/ x 2 dA is lh", second moment of area of the plan o f Ihe body
at water surface about the axis Y~Y. Therefore
2
Wx JiM", pgl
lwhere I '" 2 f x dAl
8M = pgl
IV
IV" Weight of the body
""'
= Weight of the fluid d isp laced by the body
= pg x Volume of the fiuid displaced by th e body
= pg x Volume of the body sub-merged in water
= pgx V
BM = pgx'=i
pgxV
.. .(4 .3)
V
I
GM=BM - BG= - - BG
Meta-ccntric height
Problem 4.7
I
•
= GM = - -8G.
•
... (4 .4 )
A rec/(Ingu/ar prwlOafl is 5 /II 101lg. J m ...ide and 1.20 m lIigll. The deplll of
immerJ'iOll of Ihe pOllloOll is 0.80 m ill se(l waler. If /lle cell/re of grtll"ily is 0.6 '" abo\'e Ihe bollom of
(h e pOll/OOll. delermill e lite meia -celllric heighl. The dellJ'ily for Sell waler = 1025 kg/", J.
Solution. Given:
Dimension of pontoon
!kpth of imm ersion
I I
=5mx3mx l. 20m
= 0.8 m
Ii
~ I
IL
Buoyancy and Floa tation
AG", 0.6
Distance
Distance
AB
=
t
1-'-
III
x Depth of imlllersion
'" -t. x
.8=0.4 III
Iknsity for se a Waler
'" 1025 kg/Ill '
Meta -ce ntre height GM. given by equat ion (4.4) is
GM= -
where
f'" M.D.
IlIcni~
I
•
3 m -----.,
",,-r-:t;
G
,Tl+.
, ...&,
.. s ' 4 6m 1mO.6m
T
,
A
T
-BG
of th'" plan of the poll1oon about Y-Y allis
S.Om
45 m~
12
4
'<i = Volume of the body sub-mugcd in water
'"
139 1
1
~ x 5)( 31 m~ '"
,
=3)(0.8)(5.0= 12.0ml
PLAN AT WATER SURFACE
BG = AG - Ali = 0.6 - 0.4 '" 0.2 111
45
GM = -
4
Fig . 4.7
I
45
x - - - 0.2 '" - - 0.2 '" 0.9375 - 0.2 = 0.7375 m. Ans.
12.0
48
Problem 4 .8 A wllform hody of size 3 m long X 2 m wide x I 111 deep jlOlllS in ImIN. IV/WI is the
weighl of I/le botly if depth of immersion is 0.8 //I ? De/ermine the mew-centric fleig/II (I/$().
Solution. (jiven :
Dimcnsioll o f body
= 3x2xl
Deptli of imm ersion
= 0.8
-t--. --+-
III
Find (i)Wcigh t of lxKly. IV
(ii) Meta-centric lIeiglit. GM
(i) Wt:ight of Bod~' , W
j.- 3.0m ~.J
== Wcigllt of water displaced
== pg x Volume of water displaced
-~c4'C!tft--~-
== lOOO x 9.81 X Volume o f lxKly in water
== H)OOx9.8 1 x J x 2 xO.8 N
== 47088 N. AilS.
A
ELEVATION
Fig. 4.8
(iI) Ml."h,-cl."ntrk !lelght . GM
Using equation (4.4). we get
I
GM== - - BG
whe re
f
•
= M.O.I about y. y axis of the plan of the body
!
3
]x2 J
~
= - x]x2 = - - =2.0 m
12
12
V == Volume of body in wmer
=3x2xO.8=4.8 m3
BG= AG - AB =
~_
2
0.8 = 0.5 - 0.4 =0. 1
2
GM = 2.0 _ 0.1 = 0.4167 _ 0.1 = 0.3 167 III. A" s.
4.8
~ I
I~
~ I
IL
1140
Fluid Mechani cs
Problem 4 .9 A block of wood of specific gm!'il}' 0.7 j1oalJ' ill Waler. Determine Ihe mew-celllric
hciglll of the blod:. if its size i.! 2 III X I III x O.S m.
Solution. Given:
=2 xlxO.8
Dimension of block
1
2.0 m
leI depth of immersion
'" II
Sp. gr. of wood
'" 0.7
Wcigllt of wooden piece
'" Weight density of wood· x Volume
In
:
'CAN ,
= 0.7 x IOOOx 9.Rl x 2 x 1 xO.8 N
Wc igllt uf water displ<lccd '" Weight density of water
X Volunw of the wood sub-me rged in
-
wa ter
I
-'-
-.
~!
A
~_8';;- ~ c
1
~
=1000x9.8Ix2xlxhN
Fig. 4.9
For eq uilibrium.
Wciglll of wooden piece
'" Weighl of watl'r displaced
700 x9.8 1 x 2 x 1 x O.ll '" 1{lOO x9Jll x2 x I x II
/,=
100x9.81x2x l xO.8 :O.7xO.8=O.56m
lOOOx9.8 1x2x l
Distance of ccnlre of Buoyancy from 001l001, i.e .•
AB =
and
I!
0.56
- '" - - '" 0.28 1lI
2
2
AG '" 0.8/2.0 '" 0.4 m
UG =AG - AB '" 0.4 - 0.28= 0.12 In
The Inew -ce ntril: height is given by equation (4.4) or
GM: -
where
1
\
1
4
f
•
-BG
/: - x2x1.0· : - m
12
6
'I;f:
Volume of wood in wate r
:2x 1 xll",2x I x.56: 112m l
I
I
GM: - X - 0.12 '" 0.1 488 - 0.12 '" 0.0288 tn. Ans.
6
1.1 2
Problem 4.10 A solid cylillder of diollleler -1.0 III has (I heighl of 3 melres. Filld Ille melli-centric
/!eiglll of Ihe cylillder whell il is floil/illg ill Wil/er wilh ils axis I"eflical. Ti, e sp. gr. of Ihe cylillder
'" 0.6.
Solution. Given:
Dia. o f cylindcr.
D", 4.0 m
Height of cylinder,
II '" 3.0 III
• Weight density of wood
I I
.. p x g. where p .. density of wood
.. 0.7 X 1000 = 700 kg/m ). Hence ... for wood = 700 x 9.8 1 Nlm l.
Ii
~ I
IL
Buoyancy and Floa tation
Sp. gr.
or cylinder
'" 0.6
j-o-
,
4m -
[kpIII of im mersion of cylinde r
'" 0.6 x 3.0 '" 1.8
1.8 09
AB"'T'"
.
141 1
o<
III
III
'CAN
3
AG=2'" 1.5m
I
T
~l, 'r
~
~
BG: AG - All
= 1.5-0,9=0.6111
Now the meta-centric he ight GM is given by equation (4.4)
GM= -
"0'
/
~
Fig. 4.10
-nG
"
J = M.O.!. about Y-Y ax is of the plan of Ih.-: bod y
= ~ 0 4 : ~ X (4.0)4
64
64
md
'V '" Volume of cylinder in wata
= .::. D! x Depth of inllllcrsion
4
GM=
.
~x ( 4.0)·
64
,
_ 0.6
- x (4.0 )" x 1.8
4
I
= 16
4.0'
,
x - - - 0.(, = 1.8
I
1.8
- 0.6 = 0.55 - 0.6 = - 0.05 m. AilS.
- vc sig n 1I\eans 1ha1 IllcI3 -cemrc. (M) is below the c.;,olre of gravity (G).
Problem 4.11 A hotly /W.I lilt! cylintlricalllpper ponion of 3 m diameter alld I.S //I deep. nle lower
POrllOlI is a cun'cd one....hich displaces a )'oilllne of 0.6 /IIJ of wa/er. The cenrre of blw)'oncy of ,he
cun'ell pOrlion is at a diswnce of 1.95 m below the lop of the cylinder. Th e celilre of gwriry of Ihe
whole body is 1.20 m below the lOp of the cylinder. The lolal displacement ofwatl'( is 3.9 1011llt'S. Find
Ihe mela--cenlric heiglll of the body.
Solution. Given:
Dia. of body
[kptli of body
" J.O
III
" 1.8 III
Volume disp laced by curved portion
= 0.6
1
m o f water.
Let 8 I is tlie cemre of buoyancy o f the curved surf:lce and G is tlie centre of gravity of the whole
body.
I I
Ii
~ I
IL
1142 Fluid Mechani cs
CB, '" 1.95 III
CG:I.20m
Then
Total wCigtll of water displaced by body = 3.9 lonncs
== 3.9 x 1000 == 3900 kgf
+-.-+
= 3900 x 9.81 N = 38259 N
Find mela- ct'nlric heighl of the body.
Lei tile heig ht of the body abov" th e water s urface x 111. Total
weight of water displa\:cd by body
= Weight density of water x [Vulume of water displaced I
'" 1000
x 9.81 x IVolu nw uflhe body in water1
= 9810 [Volume o f "ylindrical pan in wmcr + Volume
of curved portion I
'" 98 [0
,
ELEVATION
Fig. 4. 11
[~x D" xDe pth ofcylindrical part in water
+ Vo lume di spl aced by cur.... ed port ion ]
"'
38259
=98 1 0 [~ (3)l x(1.8-X ) +O.6]
) + 0 .6= 38259
-' 3
() ' xO. 8
-x
- =3.9
4
98 10
,
'4
,
x 3 )( (1.8-x)= 3.9 - 0.6
1.8 _.( =
= 3.3
3.3 x 4 '" 0.4668
Itx 3 x 3
x = 1.8 - .4668 = 1.33 m
li:l B~ is the centre of buoyancy of cy li ndrical part and B is the centre of buoyancy of thc who le
body.
Then depth of cylind rical part in water = 1.8 - x = 0.467 111
CH 2 "'.r +
.4~7
= 1.33 + .2335 = 1.5635 m.
Th" d is tanc" of the ccntm o f buoyancy of the who l" body from the top o f the cylindrical part is
given as
CB", (Vo lume of curved portion x CHI + Vo lume of cylindrical part in water x CB l )
+ (TOlal volume of wmcr displaced)
=
0.6 )( 1.95 + 3.J)( 1.5635
1.1 7+5. 159
'" 1.623 m.
'"
(0.6+33)
39
Then
BG = C8 - CG= 1.623 - 1.20= .423m.
Met a-centric heig ht, GM. is givcn by
/
GM= - - BG
•
I I
Ii
~ I
IL
Buoyancy and Floa tation
whe re
f'" M.O.I. or the plan of llie body al water surface about
)'[
';j
4
)'[
143 1
y.y axi s
44
= - xD = - x3 III
64
64
'" Vo lum e ort he body in water '" 3.9 Ill l
IT
3'
x - - .423 " 1.0 19 - .423 '" 0.596 m. Ans.
64 3.9
GM'" -
.. 4.7 CONDITIONS Of EQUILIBRIUM Of A fLOATING AND SUB-MERGED
BODIES
II. sub- Ill crj;cd o r a floating body is said 10 be stable if it w ill es back to its original position afl cr a
slig ht disturbance. T he relative position of the ce lllfC of grav ity (G) and cemfC of buoyancy (Ill) of a
body de termines the stability of a sub-ilwrgcd hody.
4 . 7 . 1 Stilbility of iI Sub- merged Body. The position of ce lltre of grav ity and centre of buoy ancy in case of a complete ly su b-merged hody arc fixed. Consider a balloon. which is complete ly su bmerged in air. LeI the lower portion of the balloon comai ns heavie r IHalerial, so thaI ils eemfe of
gravity is luwe r thnn its wntre of buoyancy ~s shown in Fig. 4.12 «(I). Let the weight uf the balloon is
W. The weight W is acting through O. verticall y in the downward direction. while the buoyant force FB
is acting vertica ll y up. Ihrough n. For the equilibrium of the balloon IV= Fl!' If Ihe balloon is given an
angular di splacement in the clocl.:wisc di r~ction as shown in Fig. 4.1 2 (a) . then IV and FIJ const itute a
couple act in g in the ant i-c lockwise direction and brings the balloon in the orig inal pos ition. T hu s the
balloon in the position. show n by Fig. 4.12 (a) is in stab le equ ilibrium.
G
,
(,)
w
(b)
STABLE EQUILIBRIUM
Fig. 4.12
UNST ABLE EQUILIBRIUM
o
«)
NEUTRAL EQU ILIBR IUM
Stabiliries of mb-merged bodies.
(a) Sta hl e Equillhrium. When IV = Fe and poin l n is above G. the body is said 10 be in stahk
equilibrium.
(b) Unstahle Equilibri um . If IV = F o' bUI Ihe ce ntre o f buoyancy (8) is h~low ee nlre of gravity (a),
Ihe body is in unstable equilibrium as shown in Fig. 4. 12 (b). A ~light disp lacement 10 Ihe body. in Ihe
clockwise direc tion. gives the co upl e due to W and I'IJ also in [he c lock wise dirt."ction. Th us Ih e body
docs nOI relurn 10 its origi nal po~ilion and hence Ih e body is in unslable equi lihrium.
(c) Ne utml Equi lib rium . If F/I= Wand IJ and G arc at [he same point. as shown in Fig. 4.12 (c). the
body is said to be in neutral equili brium.
4 .7.2 Stability of Floilting Body. The sWbility of a nual in g body is determined from [he positiun of Meta-ce ntre (M). In C3SC of fi(Hlting body. the weig ht o f the bod y is equ al to dIe weight of liquid
displaced.
I I
Ii
~ I
IL
1144
Fluid Mechani cs
(a) SlabJe EqullibrlullI.
[flh~
Jl'Oilli M is above G. the floalillg bod y will be in stab le equilibrium as
shown in Fig. 4.[3 (a). If a slight angular displacement is given to Ihe flo atin g body in Ihe clockwise
direction. the centre of buoyancy shifts from 8 to 8 1 such thm the vcnicallinc through 8 1 cutS al M.
Then Ihe buoyant force Felhroug h 8, ,IIlO weight IV lhrough G constitute a couple aCling in the anticlockwise dircclion and thus brillging th e floaling body in the original position.
DISTURB ING
----.. COUPLE
,
tal Slablc equilibrium /If is above G
Flg. 4.13
(b) UnSlablt equilibrium /If i< below G.
Stability offloating bodin.
(b) Unstab le Equilibrium. If the point hi is below G. the floating body will be in unstable equilib·
rium as shown in Pig. 4.13 (b). The disturbing co upl e is acting in tht clod:wis.: direction. The coup lt
due to buoyaru force F/:I and IV is also aCling ill the cloc kwis.: dirc.::tion 311d thus oven urnin g the
floating body.
(c) Ncutr:ll Equilibrium. Jfthe poim M is at Iliecel1treofgravily oflhc body. 1he floa1ing body will
be in neu1ral equilibri um .
Problem 4.12 A solid cy/inda of diameler -1. 0 '" ha~· a lreiglll of -1.0 m. Find Ihe meta-cenlric heigill
of Ille cylinder if rile specific gflll"ily of Illl' material of cylinde r'" 0.6 ami ir is floaling in l..arer lI"illl ils
a.{is I·urical. Slale II"herher IIIe eqw·librillm is ~·I,jbll' or unswble.
SolutIon. Given:
D=4m
Heigh!.
11=4m
Sp. gr.
!kptll of cylinder in water
1 4·~ 1
+- .
'" 0.6
=Sp.gr.xh
'" 0.6 x 4.0 '" 2.4 m
PLAN
y
Distance of cent re o f buoyancy (B) from A
AD=
24
i'" 1.2m
-------:.~:.
4.0m 2.4 m
Distance of centre of gravity (G) from A
"'
AG =
,
~ '" 4.0 =2.0 m
2
2
BG '" AG - AB
, " -".1:
,G
-L 1
Fig . 4 .14
= 2.0 -
1.2
= 0.8 III
Now 1he 111em·ccruric height GM is givcn by
I
GM"'"r/-BG
I I
Ii
~ I
IL
Buoyancy and Floa tation
w ltc rc
145 1
f = M ,0.1. o f Ihe plan of tile body abo ut Y- Y ax is
= ~ 04
",
64
2:.
X (4 .0)4
64
'1 " Vo lume o f cyli nder in w ater
= ~ x D2 X I)"pl h of cy lind er in wat er '" ~ X 4 2 x 2.4 m 3
4.0
4
~ )(4'
]
4!
1
= ~64>'c-_ _ = ~ x = =0.4167 m
'r:I
;X 4 1 )( 2.4
16
2.4
2.4
I
GM = 'r/ - BG '" 0.41 67 - 0. 8 '" - O.3H33 m. Ail s.
- vc sig n me ans thalthc meta-centre (M ) is be low the centre o f grav ity (G). T itus th e c ylind er is in
un stabl e equilibrium . Ails.
Problem 4.13 A solid cylinder of !O em diwneler alld 40 em /ollg. COIIS;SfS of nm parrs made of
differel17 mll/erin/s. Theftrsl pari GIllie base is 1.0 em long and of specific gral'iry '" 6.0. The Oilier pari
of Ille cylinder ;s mode of tlie mareria/ IWI'il1g specific gWl'ily 0.6. Siale, if if call flom \'enically in
Irma.
Solution. G ive n:
Length.
Len gth o f I st pa rt.
Sp. gr ..
Densit y of 1st part.
Le ngth o f 2nd pan,
Sp. gr ..
Dc n~i t y of 2nd part ,
D : Wc m
L ", 40 cm
1.0 u n
SI == 6.0
'I:
PI'" 6 x 1000 '" 6000 kg/m
, ", 40- 1.0=39.0cm
l
l
S2 '" 0.6
Sp.gr.
=0.6 -
T
P2 '" 0.6 x 1000 '" 600 kgfml
The cy linde r will floal l'c n ic all y in W31c r iF its meta-ce ntric he ight GM is
1.0cm
posi tive. To find me ta -cc ntric hc ight. find the location of ccntrc o f gra vit y
(G) and ce nt re o f huoy all cy ( 8 ) of th e combined so lid c ylind er. Th e di s tance
of the ce nt re of gra vity o f th e so lid cy linder fro m A is given as
Sp.gr = 6.0
A G : [(Weight o f 1st pan x Distance ofCG. o f 1st pa rt from A)
F ig . 4.15
+ (Wcight of 2nd pan of cy lindcr
x Di stance uf CG. of 2nd pan fro m A) I
+ IWc ight uf 1st part + wcig ht of 2nd p~rtl
U.
T LJ.-l
,
==
(~O l
1
X1.0 x 6.0 x 0.5) + (~O l X39.0 x 0.6 x (1.0 x 3912 ))
( : 0 " x 1.0 x 6.0 + : 0 " x 39 x 0.6)
=
1.0 x 6.0 x 0.5 + 39.0 x.6 x (20.5)
I.Ox6.0+39.0xO.6
Cance l : Dl in th e Numerator and De nu mina to r ==
I I
3.0 + 479.7
482.7
4 = - - == 16.42.
6.0 +23.
29.4
Ii
~ I
IL
1146
Fluid Mechani cs
To fi nd th e ce lllfC o f bu oyancy of th e co mbined two parts o r of th e cy lind er. de term ine Ihe dcptll
of imme rsion of the cyl in der. Let the de pth of imm ersion of the cy linde r is II . T he n
Weigh t of the cy lin de r " We ight of wa ter displaced
It
, 39 .0
11:
,
1.0
It
,
II
- )( (. It x - - )(600 )( 9.81 + - Cit x )(6000 )(9,8 1 = - (. 1)" x x 1000 x9.8 1
4
100
4
[00
4
100
I ": h is in cml
I:a ncc llin g
~
, cIOOO
=:~X~9~.8,,1 th rougho ut. we get
( I) x
100
39.0xO.6+ I.Ox6.0 =11 or 11=23.4+6.0=29.4
Th", d istance of the cen tre of the buoyancy B, of the cylinder from A is
29.4
AB = 1!12 '" - , - = 14.7
BG
= AG - AS '" 16.42 - 14.70 = 1.72
C Ill .
Meta -ce ntric heig ht GM is give n by
GM= ~ - BG
wllerc
•
1 = M.O.1. of pla n of the body about Y- Y ax is
'" ~ D~= ~ (lO)~c m ~
64
.64
'It = Volume o f cy linder in water
=
,
,
1
'4If D-xh=
4"It (10)-x29.4
111
~
'" ~ (10)4/ .:!.( 1O !
'V
64
4
)
x 29.4 '"
-.!... x
[6
10 ' = -,-'~OO
O;;;-c = 0.2 12
29.4
19 x 29.4
GM = 0.2 12 - 1.72 '" - 1.50S e rn
As GM is - ve. it mea ns th at the Me ta -ce ntre M is below the cen tre o f gra vi ty (G). T hu s tlw
cy linder is in uns tab le eq ui librium and !\O it ca nnot fl oat ven ica ll y in water. An s.
Problem 4.14 A rectangillar POII/OOII 10.0/11 long. 7 m broad (lnd 2.5 m deep weighs 686.7 kN. /I
carries Oil ils "pper deck (In empty boiler 0/5.0 m di",,,eter weighing 588.6 kN. The celi lre 0/ gw!"il)'
o/Ihe boiler alllilile POIi/OOII are (1/ Illei r respectil'e relltres alollg a l"erlicallille. Filld tlte //Iela-celllric
height. lI'eighl dellsity a/sea lrater is /0./04 kN1mJ.
Solution. Give n : Di me nsiu n u f pon tuon = [0 x 7 x 2 .5
Weig ht o f pont oon.
Dia. of boile r.
Wr ight o f boile r.
D=5.0 m
11'1 '" 5SS.6 kN
= [0. [04 kN/m 3
\\' fu r sea wate r
To find the met a-ce nt ric heig ht. first dc tc nni n ~ the co mmo n ce ntre of gravi ty G and cOl11 mon ce ntre of buoya ncy B of the ooi la and
po ntOO n. Let G I nnd G , nrc the ce ntre of grav iti es of pontOOn and
boiler r.:sj)Ccli ,·e ly. The n
I I
f
11'1'" 6S6.7 kN
IT
IG,
lG
"t
G
B
I
,
A
T
5.0 m
gm ~
1.
1 -7.0m- l
Fig. 4.16
Ii
~ I
IL
Buoyancy and Floa tation
AG 1 ",
2.5
2 '"
AG1 ", 2.5 +
1.25
147 1
,
III
I
25.0 "" 2.5 +2.5 '" 5.0 III
The distanl'C of co mm on (;cnt rc of gravity G from A is given as
1\1 x AC, + IV, X AG,
AG '" ~---.~o.'--~
\~ + IVz
686.7 x 1.25 + 588.6 x 5.0
(686.7 + 588.6)
to.Om
'" 2.98
111.
Let!J is th e depth of immersion . Then
Total weigh t of polUoon and boiler'" Weight of sea water displaced
(686.7 + 588.6) = Ii' x Volume o f th e pomoon in water
= 10.104 x I. x h x Deplh of immersion
I·
7.0
~ ------l
J
Fig . 4. 17 Plan of tbe body
a/ water·line
1275.3", 10.104 x 10 x 7 X II
1/: -;-;~~1~27~5~.3~~ : 1.803m
10 x7 x 10.104
The di stance of the com mon cen tre of buoyancy IJ from A is
All=
~==
1.803 == .90 15 III
2
2
BG == AG - AB '" 2.98 - .9015 '" 2.0785
Mct3 -ccmric height is given by GM '"
where
.!.... V
In -
2.078 111
BG
1== M.O.1. of the pl'lll of the body at th e water leve l along Y-Y
=...:..
x [O.Ox7J: IOx 49x7 m4
[2
';j
12
== V o lume of Ille lxxl y in water
== Lxbx II == [0.0 x 7 x 1.857
J :
';j
GM "
IOx49x7
[2x[Ox7xI.857
I
';j -
CC~4~9~ =2.[9&111
12 x 1.857
= ~
BG = 2.198 - 2.078 = 0. 12 m.
Mela·t...,nlric lIeighl o f bolh Ih e ponloon 1lnd boiler == 0.12 Ill. An s.
Problem 4.15 A lI"oodell cy/illder of sp. gr. = 0.6 lmd cirCli/ll f ill cron·· seClioll i~· reqllired 10 flo (ll
in oil (sp. gf. = 0.90). Fint/ Ihe LID ralio for Ihe cylint/er 10 floal will, ils /Ongilut/inal a.ris ,·erliC(l/
in oil ....IIue L is Ih e Ireiglll of cylinder fwd 0 is ils diam eler.
Solution. Gi\"~n :
Dia. of cy linder
Heighl o f cyli nder
Sr . gr. of cylinder.
I I
Ii
~ I
IL
1148 Fluid Mechani cs
Sp. gr. of oil
52'" 0.9
Let the depth of cylinder immersed in oi l '" II
ror the principle of buoyancy
Wcigtlt of q lindc r '" WI. of oil displaced
11:
It
,
C' ~
,
,
- 0- x L x 0.6 x 1000)( 9.81: - 0- x II x 0.9 x 1000 )( 9.81
4
4
h",O.6)(L = ~L
0.9
3
L
,
0
1•
LxO.6 : /1)(0.9
0'
'-----'
- "4
lG
,
°
Fig . 4.18
1
The distance of centre o f grav it y G from A, AU '" -
2
The dist ance of ccnlT" o f buoyanc y B from A.
The meta -ce ntric height GM is given by
GM"' ~- BG
where
I ",""::'"
M
D~ and ';j
•
'"
Volum e of cylinder in oil '" ~ 0 2 X "
4
!....=(~ D./~ D2")=-.!...
V
64
4
16
Ol =_ D,,'-;-_ =3Dl
II
16
x,
2L
32 L
{ h=H
3Dl
L
GM: - - - .
32L
(,
For Mable equilibrium, GM should be +vc or
GM>O
3D'
or
32 L
L
- >0
6
1
3D
L
-- > -
0'
32L
6
1
t
18
- <OJ
32
0'
9
16
~ <~ =%
IJD < 3/4. A. ns.
Problem 4 .16
511011' /lw / a cy/;ndrit'(I/ buoy of I m diameter Illld 2.0 m heighl \\'eiglling 7.848 kN
wil/ 110/ floll i "ulial/ly in sea waler of densily f()JO kglm J• Find Ihe force necessary in a "erlica/
chain a((a("/It:d al Ihe cell ire of base of Ihe buoy Ihal will keep il ,·ulica/.
I I
Ii
~ I
IL
Buoyancy and Floa tation
149 1
Solution . Gi ve n : Dia. o f bu oy. 0 '" I III
= 2.0 III
Hciglil.
H
Wci~1I1.
IV = 7.848 kN
Density.
= 7.348 x 1000", 784 8 N
p::: 1030 kg/ml
(i) Show th e cylinder will not l1oa1 ve rticall y.
( ii) Find the force in the c hain.
Purl I. Th e cy linder will no t fl oat if me ta·centric heig ht is - 'Ie.
LeI the d~pll1 of immersio n be II
---,-_.
2.0 h
Then for equilibriuln , Wei ght o f cy linder
'" Wcig lll of wa ter displaced
"" Density x g x Volume of cylinder in water
7848 = 1030 )(9.81 x~ Olxh
Fig. 4.19
4
,
,
1-,, -1
,
::: 10 104.3 x - (It)( II
4
4 x 7848
10104.3)(11
The di stance of centre of buo ya ncy
::: 0.989
n from A,
AB::: ~ ::: 0.989 = 0 .494
2
Ill.
Ill.
2
And the distance of ccmre of gra\'ily G. from A is AG :::
22.0 :::
1.0 III
8C::: AG - AB::: 1.0 - .494 ::: .506 m.
Now meta-ce nlri c height GM is given by
where
and
1=
It;f
GM:::'!"" -BG
V
~ D~= ~ X(I)~m4
== Volume of cy linde r in wa ter == ..::. D I x " == ..::. I I x .989
,
4
J
~X 1 4
-, x I'
64
V " ~ D 2 xh
4
6'
"::'X I ~ X.989
4
== ~ xtlx _'_ =
16
.989
=0.063 11\
16x.989
GM = .063 - .506 == - 0.443 m. Am.
As the m ~la·ce nlri c hdghl is - ye, Ih e poinl M ties below G and he nce Ihe cy linde r will be in
un sta bl e equilibrium :lI1d hence cy linder will nOl float vertica ll y.
I I
Ii
~ I
IL
1150
Fluid Mechani cs
Purl II. Let tlie force applied in a ve rtical chain 3nachcd aI the centre .,.,.,J-t-1~,,-~
of the base of Ihe buoy is T (0 ~ccp the buoy venical.
:0
Now find the combined position of ccmn: of grav ity (G') and cen tre of
buoyancy (B'). For the combined Centfe uf bUUyJIH':Y. let
/J' = depth of immersion when the force T is applied. Then
TQlal downward force = Weight of water displaced
or (784S +
= Density o f wate r x g x Volume of cy lilldcr in water
n
~ D')( Ii' I
= lO30x9.81 x
,
II'
1 [ 7848 + T ]
AB=T="2
7935.9
where II': depth ufimmeThion I
7848 + T
10 I ()..I.3+ T
I Ol04.3X~Xl l
7935.9
m
,
A
Fig. 4.20
7848 + T
= 15871.8 m.
The combined cenlTC of gravity (G') duc 10 wcigtn of cy linder and d uc to tension T ill the chain
from A is
AG'" IWI. of l:Ylindc r x Distance of CG. of cylinder from A
+ T x Distance of e.G. of T from A] + ]Wcig hl of cylinder + T]
" (7848 x
B'G'
~ + T x 0)
2
= AG' _ AB' =
+ ]7848 + T] ==
7848
(7848+ T)
The meta-centric height GM is given by GM ==
7848 m
7848 + T
(7848 + T)
15871.8
Vf - B'G'
== ~XD4== ~ X]4=~nl4
where
64
V
= ~ D~
64
X II'
=~
4
64
X ]1
4
x (7848 + T) ==.!:. x 7848 + T
7935.9
4
7935.9
,
f
• " =oi'64~",
(7&48+ T )
4 7935.9
It
=
(7848 + T )
GM = -:-C~79,"3:"'.,-9
16
71\48
[ (7848 + T)
T) ]
(7848+
15871.8
For stable equilibriunl GM shou ld be positi ve
GM> 0
7935.9
16 (7848 +T)
I I
[
7848
(7848 + T)
(7848 +
T)] >
15871.8
0
Ii
~ I
IL
Buoyancy and Floa tation
7935.9
16 (7848 + T )
151 1
7848 + 7848 + T , 0
(7848 +T)
15871.8
,-79"3;;',.;9,,-~'6;-':..;7~84,,,8 + (7848 + T)
16 (7&48 + T)
>
0
15871.8
~-"I~'7C;6~:n",+ (7848 + T) > 0
J6 (7848+T)
15871.8
(7848 + T)
:>
15871.8
,,-~''~7~63~2""
16(7848 + T)
(7&48 + n 2:> 117632 x 1587 1.8
0'
16.0
:> 11 6689473.5
:> ( 10802.))!
7848 + T:> 10802.3
T:>
10802.3 - 7848
:>
2954. 3 N. A ns .
Th.e force in lhe chain IllUS! be at least 2954.3 N so that the cy lindric al buoy can be !;cpt in
vert ic al position. An s.
Pro blem 4.17 A solid coile floats in warer ",itl, irs apex downwards. Determill e Ille lew/ apex ollgle
of cone jar .liable equilibrium. Tlw specific 8WI';I)' of lile material of the coile is gil'el l O.S.
Solullo n. Given:
Spo gr. of cone
== 0.8
P == 0.8 x 1000 == SOO kg/m l
DensiTy uf I:one.
/
D == Dia. of the cone
d == Dia. of cone ,H water leve l
Co<
29
+
,
CAN 0'
C ONE AT
= Apex angl e o f cOil e
w" ER LINE
H == Hei ght o f C()nc
D
I'
•I
II = De pth of cone in water
For tllc I:onc. thc
di st~ncc
G'" Centre of gravi ty of the cone
B = Centrc uf buoya ncy of Th c conc
of ce nTre of gravity from the ~[lcx A is
AC = t height of conc '" tH
also
AB", t dcpth of cone in wa1cr '"
Volume of water di splaced '"
Volume of cone
= 1-
:. Weight of cone
=800 xgx tx TtR'" xH
Now from JiAEF.
Similarly.
I I
t It? x II
X TtR 2 X II
EF
Ian 9= -
t"
"
,
Fig. 4.2 1
il
R
=-
EA
H
R=Hta n 9
r",htan6
Ii
~ I
IL
1152
:.
Fluid Mechani cs
Wei~hl
of cOile
t
.. Weight of water displaced", ]()OO x g x
x II'! X II
. .000"""x""'
'" lOOOxgx tXI[(htan 9) , X/I= -1
For equilibrium
Wcigllt of l"OIlC
3.0
'" Weight of water displaced
3.0
3.0
800xlt=
or
'X'O'ITfX~h" _'""'''_',>.
1000
x
II)
1000 X11.1or!!'" = (1000)'"
H3 =
800
/,
800
For stable cljuilibrium. Meta -centric height GM should be positive. But GM is given by
GM= -
/
•
-8G
where 1= M.O.! . of cone al wateT-lil'" = ~ d~
64
.
In,
V = Volume of con" In water =- - d" x h
34
/
IT d ' l "3)("4
' IT d ,- X "
V=64
I x3 d 2 3d'
3
' 3r'
=x - ==x(2, r = - 16
=
II
161,
1611
4 II
3 (/l lan6)1
4
I':
1/
r=I,[an91
=t illa ,,2e
and
BG '" AG - AB = fH - til'" f (H - II)
GM =
til tanl e -
t(ll - II)
For stable equilibrium GM should be positive or
t ht3n!e -1 (1I - 11»O
or
htan 2e>(H _ h)
or
ur
htan!e - (If - h»O
htan 2 e + h>H
hltan!a+ 11>11
So<
!!.. = ('OOO)iIJ = 1.077
h
800
,
I
scc~ e > 1.077 or cos' 0 > - - '" 0.9285
cos
e > 0.9635
1.077
0> 15° 30' or 20>31 °
Apex angle (20) s hould be aileasl 3 1°. Am.
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153 1
Buoyancy and Floatati on
Problem 4 .18
di(lmeler
f)
A cOile of Ifleciftc gra\'it)' S. is floating ill wmer willi its apex dowlI.",ard,~. II has a
(md l'erlin,/Iu:ighl H. Show Ilwl jor stah/e equilibrium of the COli I' H <
!.. [V i. 5;;; ]"1,
2 2
S
Solution. Given:
Dia. of oone
=D
Hciglll o f cone'" H
Sp. gr. o f cone = S
Let
G", CemTC of gravity of cone
B '" Ccmrc of
PLAN OF
CONE AT
WATER LINE
bUOYHIlCy
~ D _I
29 = Arcx angle
T
= Apex of the cone
II'" Depth of immersion
J = Dia. of cmw at Waler surface
A
Then
"
1
,
AG:2H
4
fig. 4.22
3
A8= - II
4
Also weight of cone =Weigh t of water displaced.
I'
I
,
lOOOSxgX,lTWX
H : IOOOxgx-rrr'xh
3
or
'2
sn-H=,h
SR!H
II = - ,-
B"
R" ,
t~ne = - = -
H
"
N=Hmne.r=ltlan8
SX(Hlan6)l xH
II '"
(1/ tan
er,
SX II l xtan 1 exfI
SH J
"
: __
It" tan " e
I,,,, (SH1)1fJ = Sill H
I,:
Distancc.
,,1
or h 3 =SH 3
... (1)
8G '" AG -An
J
4
3 3 (H - b)= -3H(- .1/3
S /I)
4
4
4
=-H - - b= -
'" 2 HI I _SI Il I
Also
... (2)
4
I", M.O. Inenia of the plan of body at wate r surface
'" 3... It
64
.
V", Volume o f cu ne III water'"
I I
,
In: ,
3In:
x 4 x iI + x h '" 34 iI+
IH.S
III
I
Ii
~ I
IL
1154
Fluid Mechani cs
-
J
V
=
~d'
64
.----o"'c----c
1
x
3
It
d
H.S'1l
16.H.slll
4
Now Mcl<l-ce ntric height GM is given as
I
3d "
3H
GM = - _ BG '" ~~'"'- :.-- II _ S'/3 1
'rJ
16.fI ,S"J
4
GM should be +ve for swblc equi librium or GM > 0
... (3)
Also we know R = titan
e and
R
r
r= h tan
e
H
D
= - =II
d
If '" DII '" D X HS' fl '" DS' fJ
H
H
Substituting the value of di n cqum io n (3), we get
3(0 5
1
1))1
3H
~::':;--"i, > -
1.6.H.S'IJ
4
.lt3
(l - S
) or
D J .S'/J
H2< 4 ( 1_ 5"1)
"'
II>
4.8
EXPERIMENTAL METHOD Of DETERMINATION OF META -CENTRIC
HEIGHT
The mcla-cc ntric heigh t of a noati ng vessel can be determined, provided we know th e (emre of
grav it y of the Ooaling vc~scl. Lei"", is a known we ig hl placed over the ce ntre of the VCS1;e J as show n in
Fig. 4.23 (iI) and the vesse l is floating.
w
(a) Floating body
(b) Tilted body
Fig. 4.23 M l!l d-<:f'rltric bright.
I I
Ii
~ I
IL
Buoyancy and Floa tation
Let
IV '" Weight of vessel including
155 1
WI
G '" Celltrc of gmvity of the I'essel
B '" Celllrc of buuyancy of the vessel
The weight 11" 1 is 1Oo\'Cd across Ihe vessel towards right through a di Sinn/;c x as shown in Fig. 4.23 (b).
The vessel will be tilled. The angle of heel is measured by means of a plumblinc and a protractor
attached on tlie vesse l. The neW cenlrcof gravity of the VC~!ic J will sh ift to G 1 as the we ighl WI hasbcen
moved towards Ihe right Also the centre of buoyancy will change to B] as Ih" vcss<:l ha.~ lih~d. Under
equilibrium. the moment caused by the movement of lhe load WI through a distance x must be eq ual to
Ihe rnOlTIcl\t caused by the s hifl o f the relltrc of g ravity from G to G I' Thus
e
= GG I X IV=
movement of WI = 11'1 X X
The momenl due to change of
The moment duc to
IVx GM tall
G
e
11'1 '\ '" IVGM tan e
GM = cc'"~"~'cc
...(4 .5)
IV tan
Problem 4 .19 A J'l!ip 70 III 10llg {md 10 III bro{{d h{{s a displacemenl of 19610 kN. A weight of
343.35 kN is mOI'l'd across Ille deck: IhrOllgh a tii.,·I{Wce of6 III. The .IMp is lilled Ihrollgll 6 ". The
momelll of inertia of Ille ship m wmer-line abOli1 ils fore Wid afl axis is 75% of M.O.I. of I/Il'
circlIIIIJ"cribing reClmlg/e. Tlw celllre of buoyancy is 1.15 III belo", ",ala-lille. Find tile meta -cen/rit"
IJeigllt lIlIIl PO$ilioli of celilre of gr""ity of J'hip. Specific weight of sell '\"lIler is 10 I 04 Nlm J •
Hence
e
Solution. Given:
Length of ship.
Bread th of ship.
L=7UIlI
b = to m
IV= 19620kN
Displacement.
Angle of heel.
", .0.1. of ship at water-line
". for sea- water
e" 6°
" 75% of M.O,I. of circumscribing rectangle
= 10 I 04 N11lI 1
"
l
10.104 kN/m
w, " 343.35 kN
Movable weight.
Distance l\loved by "'I'
Centre o f buoyancy
Find (I) Meta-ce ntri \: height. GM
.\ =6 m
= 2.25 m below Waler surface
(ii) Position of centre of grav ity. G.
(i) Meta- cc ntric height . GM is given by equation (4.5)
GM =
"
"'IX
IV tan e
343.35 kN x 6.0
19620 kN x lal16°
34335 kN x6.0
19620 kN x .105!
'" 0.999 m. Ans.
(ii) Pos ition of Centre of Gra vit y, G
GM=.!.... - BG
where
~ I
•
1== M.O.l. of the ship al water-line about Y-Y
I~
~ I
IL
1156
Fluid Mechani cs
.-WATER LINE
2.25
~10m ~
Fig. 4.24
'" 75% o f
lUlU
-.!....
[2
x 70 X 101
Fig. 4.25
.75 X...!... x 70 X 101
",
12
V" Vu lume of ship in wme r '"
.!... '"
'rJ
",
4375
~h ip
We ight of
-:::c-:'c'-'C'':'''-'O''--:-:c:
Weight df.>nsily of wate r
4375
'" 2.253
111 4
19620 '" 1941.74 m1
10. 104
111
1941. 74
GM", 2.253 - IJG or .999 '" 2.25] - BG
BG
= 2.253 - .999 = 1.254 m.
From Fig. 4. 25. it is c lear lhal Ih e di stan ce of G from free surface o f the wa ter '" dist ance o f B
from water surface - BG
" 2.25 - 1.254 ", 0.996 111. An s.
A pOIl/QOI! of 15696 kN di~placemt'1lI iJ floating in water. A »'<'igh! of 245.25 kN is
IIwred 1/"014gh a diSllIllce of 8 m across the deck ojpOIl/OOIl, II'hic// lills the pontoon l!lrOJlgll all angle
Problem 4 .20
·r.
Find mela-anlric height of the pontOOIi.
Solution. Given:
Weig ht of (lOntoo n = Di splace ment
"' MOl' abl e we ight.
W = 15696 1; N
245,25 kN
Distan ce moved by weig ht "'I ' x == 8 m
Angle of hee l.
6 == 4 °
Th.;, nwta-ce ntri c heig ht. GM is giv.;, ,, hy eq uati o n (4.5)
11'( '"
GM == ~w",x'-o
1\1 tan 6
"'
245.25 kN x 8
15696 kN x tan 4°
--;:""",,96,,2;.-;;:,;;;;
== -;-:
== 1.788
15696 x 0.0699
~
4.9
III . '\ IIS.
OSCILLATION (ROLLING) OF A FLOATING BODY
Consider a fio at ing body. which is tilted thro ugh an a ng k by a n ol'e nurnin g co uple as s how n in
Fig. 4. 26 . Let the o\'Crtu rn ing co upl e is suddenly re moved . The bod y will start osc ill ati ng. T hu s. the
I I
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IL
Buoyancy and Floa tation
157 1
body will be in a Siale of oscillation as if suspended allhe mcta·CCnlrc M. This is similar 10 the case of
a pe ndulu lll . The only force acting on the body is due to Ihe restoring couple duc to the weight Wof
the body force of buoyancy FR"
,
•
Fig. 4.26
Restoring couple
'" \I' x Distance GA
= II'x GM sin 9
. .. ( i)
This couple tries 10 decrease Ihe imgle
Angular 3(;celcralion of the body.
(l
d1 S
= - - ,- .
d,
- ve sign has been introduced as Ihe restoring couple tries to dcr rcasc the angle B.
Torque due \0 inenia
= Moment of 11lcrlia about Y-Y x Angular acceleration
,[- -dtlr"J
= I y_y x
IV ,
f r' r = gK
where
IV = Weigh t of body, K = Radius of gyra tion about r·Y
_ '" K' ( _
[nenin torque
g
d1,' J__ _~ .,2 dle,
g
dt "
"
dl '
... (U)
Equating (I) and (il). we ge l
IVxGMsin9=
For sma ll angle
K'
Dividing by -
,
e.
, we gel
_~ K 2 d2?
g
dr '
"'
sine - e
d1?+ GM Xfx e ",0
dl-
K-
Th" above equalion is a differenlial equat ion of second deg ree. The solUlion is
. JGM.g
~
e ", C ,S l n
I I
XI + C2 COS
JGM.g
xI
Kl
...(iii)
Ii
~ I
IL
1158
Fluid Mechani cs
where C t and C2 arc constants of integration.
The values o f C 1 and C 1 arc obtained from boundary conditions whic h arc
(i) 311==0,6==0
(ii)
all=
~,e =o
where Tis Ihe time period o f one complete o!;(:ilimion.
Suhstituting the 1st ooundary condition in (iii), we get
(': si06=0.0:0s6= 11
O=C 1 xO+C1 X1
C1 == 0
Substituting 2nd boundary conditions in (iii), we get
Bm C 1 cannot Ix: equal to
sin
1.Cro
and so tile olher ahcmalivc is
JGM.s:
T
- - - x -=O=sinlt
Kl
(.: sinlt=OI
2
T=2n~ Kl
...(4 .6)
GM .g
Time period of oscillation is give n by ~qualion (4.6).
Problem 4.21 Til e
rat/illS of gy,III;OIl of {/ sllip is S III
la/e Ihe lime period of o.\'ciliatiQlI of Ihe Jllip.
Solution. Given:
Least radiu s of gy ration. K = 8 III
Mcta -cc nlric height. GM == 70 em == 0.70 111
The tim e period of oscillation is given by equation (4.6).
"',1.11
T: 2n
j£,'
(/111/
meta-centric height 70 CIII. Caleu-
8x8
- - : 2n , 1 ;cOCC~~ : 19. IH sec. Ans .
0.7x9.81
GM.g
Problem 4 .22 The lime period of rof/illg of a ship of weighI 29430 k.N ill sea waler is 10 seconds.
T/l e cellire of buo}'(l/Icy of Ihe ship is /.5 m be/ow II/e cen/re of grlll'it)'. Fifl(llile radius of g)"",liOl/ of
Ihe ~IIip if Ihe moment of inerlia af Ihe ship al IIIe lI'aler Urle aboUI fore and aft luis is I()(x) m~. Take
j'l'ecijic weiglll af sea Imler as = 10 I (X) Ntm J•
Solution. Given:
T : 10 s«c
Time period.
Distance between centre of buoyancy and centre of gravity. BG : I.S 111
Mome nt of Inertia.
I:
10000
m~
WciglH.
IV" 29430 kN '" 29430 x 1000 N
Let the radi us of gyration: K
PiTSt calculate the meta-cenuie height GM. whieh is given as
GM: BM - BG '" -
I
•
I I
- BG
Ii
~ I
IL
Buoyancy and Floa tation
wlterc
.,d
159 1
1 = M.O. llicnia
V = Volume of water displaced
Weight o f ship
-
29430 x 1000
- ~~"";;';~' = 2912.6
- Sp. we ight of sea water -
10 104
rn
3
10000
GM = - - - 1.5 = 3.433- I.S = 1.933
29 12.6
Using equation (4.6), we get
T= lit
~
'
GM xg
K'
2,K
1.933x9.81
JI.933 x 9.8 1
10= 211 ,I~~~~
"
Ill.
J~I'~3c3C'C'C',,-1 = 6.93 m . An s .
K = clO
- c,-,-
"
2,
HIGHLIGHTS
1. The upward fOrl'e cxcncd by a liquid on a body when the body is immersed in the liquid is known 8S
buoyancy or force of huoyancy.
2. The point through which force o f buoyancy is _,upposed to act is called centre of buoyancy.
J. The point about which a body slanS oscillating when the body is ti lled is known as mcla-ccntre,
4. The distance between the mela-ccnlre and centre of llravilY is known as mcla-Ccnlric height
5. The meta-ccnlric height (OM) is gi,'en by GII1 =
where
I
~
,
V
-IJG
Moment of lnenia of \he floating body (in plan) at water surface nboutthe axis Y-Y
'!j .. Volume of the OOdy sub-merged in wuter
BG '" Distance between
~entre
of grnvity and centre of buoyancy.
b. Conditions of equilibrium of a floating and sub-merged OOdy are :
Equilibrium
(i) Stable Equilibrium
(ii) Unstable Equilibrium
{iii} Ncutr.. 1 Equilibrium
Flooting Hody
Sub·merged /JOlly
111 is above G
M is below G
M and G coincide
/J is abo,·" G
/J is below G
Band G coincide
' oc
7. The value of meta-centric height GM, experimentally is given as GM "' -;c,'
"'"'
Wtan6
where
11',
~
of ..
Movable
w~ight
Distance through which 11'1 is mowd
W .. Weight of lhe ship or flooling body including
1<',
6., Angle through the ship or flo.1ling OOdy is lilted due to the movement of 11',.
8. The time period of o!'Cillation or rolling of a floating OOdy is given by T", 21f
,,'here
v~
c;;;t;g
K .. Radius of gymlion. GM .. Mela -cenlric height
T", Time of one complete o!'Ciliation.
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1160
Fluid Mechani cs
EXERCISE
(A) THEORETICAL PROBLEMS
l. Define the lenn, 'buoyanq"
~nd
'cenlre of buoyancy' ,
2 . Explain the terms 'meta-<:entrc' and 'meta-centric heighI'.
3. Derive an expression for the mela-ccmric height of a floating body.
4 . Show thai the distance
bctw~..,n
the mcta-ccntre and centre of buoyancy is giw" by Hili
~
-
,
V
where I • Moment of incnia of the plan of the float ing body at water surface about longitudinal axis.
V _ VOlume of the body sub -merged in liq uid.
S . What arc the conditions of equilibrium of a floallng I>ody and a sub-merged body ?
6. How will you deten"inc the meta"':c"l,;C height of a floating body experimentally ? Explain wilh ncal
sketch.
7. Sclectlhc correct Staiement:
(a) The buoyant force for a floating body passes through the
(i) centre of gravity of the body
( iii) mcta-.celllre of the body
(ii) centroid of vol ume of the body
(il·) cCllIre of gravity of the suh.merged pan of the body
(,,) ccntroid of thc displaced volume.
(b )
A body
sub-mer~~"<.l
in liquid is in equilibrium when;
(i) it' meta-.centre is above the centre of gravity
(I;) its mcta...:cn!rc is above the cenlre of buoyancy
(iii) its centre of gravity is above thc centre of buoyancy
(i" ) its centre of buopn,y is abo,·e the centre of gral·ity
(,,) noneoflhesc.
IAn s. 7 (II) (" J. (b ) (i1')1
8. Dcri,·c an cxpression for !he lime pcriO<l of lhe oscillation of a floating body in lcnns of radius of gyrmion
and mela-.centrie hcight of the floating body.
9. Define the termS meta-<.·entrc. ccntrc of buoyancy. mcla-centric height. ga uge pres,ure and absolute
pressure.
10 . What do you understand by the hydroslatic equation ? With the he lpof this equation. derivc the expression
for the buoyant force acting on a sub·merged body.
II . With neal sketches. explain lhe conditions of equilibrium for fiOal ing and sub-merged bodies.
12. Differenlime between:
(i) Dynamic viseosi1y and kinematic ,·iseosity. (ii) Absol ute and ga uge pressure (iii) Simple and
differenti al manomelers (i" ) Cent re of gr~vily and cenlrc of buoyancy.
(Del/,i Ulli'·t"rsiry. Dec. 10M)
(B) NUMER ICAL PROBLEMS
1. 1\ wooden block of width 2 m, depth 1.5 til and lenglh 4 til floats horizontally in water. f ind the volume
of water displaced and position of centre of buoyancy. Thc specific gravily of thc wooden hloc k is 0.7.
IAns. 8.4 til ! . 0.525 m from (he basel
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Buoyancy and Floatation
161 1
2 . A wooden log ofO .8m di,,,"c(er and 6 III length is noaling in rivcr Waler. Find the depth of wooden
log in water when the sp. gr. of the wooden log is 0.7.
[Ans. O.54m ]
3. A Slone weighs 4SlO.5 N in air and 196.2 N in waler. Determine the volume of Slone and its specific
gravity.
IAn s. 0.03 Ill J or 3 x W' eml, 1.671
4 . A body of dimen,ions 1.0 III X 1.0 m x 3.0 III weighs 3924 N in water. I:ind its weight in air. W hat will be
its spedfic gravity?
[Ans. 62784 N. 1.06671
5. A metallic body floats at the interface of mercury of sp. gr. 13.6 and water in such a way that 30% of its
volume is sub-merged in mercury and 70% in water. I:;nd the den<it y of the meu!!ic body.
[Ans. 4780 kglmJI
6 . A hody of dimensions 0.5 III )( 0.5 III x 1.0 m and of sp. gr. 3.0 is immersed in water. Delenni"e the leaSI
(An .•. 49()5 N I
force required to lift the body .
1. A "'''tangular pontoon is 4 rn long. 3 10 wide and 1.4010 high. The depth o f immersion of the pontoon is
1.0 m in sea -wmer. If the centre of gr~"ily is 0.70 m above the boltom of the ponloon. detemline Ihc meta (Ans. 0.45 ml
cent ric height. Take the densily of sea-Water as 1030 kg/m J •
8. A un ifonn bod~ o f size 4 m long X 2 m wide x I rn deep noats in water . Whal is the weight of Ihe body if
depth of immersion is 0.6 m? Detennine thc mCla-ccntric height also .
(,\n s. 4 7088 N. 0.355 IllI
9. A block o f wood of specific grJvity 0.8 no.,ts in water. Detenninc the meta-centric height of thc block if
its size is 3m)( 2m x I 111.
(Ans. 0.316 IllI
10 . A solid cylinder of diameter 3.0 m has a height of 2 m. find the meta-centric heighl of tne cylinder when
(AilS. 0.1017 ml
it is noaling in water with its axis ,·wical. The sp. gr. of the cyhnder is 0 .7.
II . A body has the cylindrical upper poniOll of 4 m diamete r and 2 m deep. The lower p<mion is a curved one,
which displace .• a volume of 0.<) m1 of water. The cenlre of buoyancy of the curved ponion is at a distance
of 2.10 III below Ihe lOp of the cylinder. The centre of gravity o f the whole body is 1 50 m below the tOP
of the cylinder. Th e totnl displacement of water is 4.5 tonnes . find the Illeta-centric height of the body.
(Ans. 23871ll1
12. A solid cylinder of di~meter 5.0 10 ha .• a height of 5.0 10, Find the Illeta--centric he ight of the cylinder if
the spe"ific gravity of the material of cylin<.lcr is 0.7 and it is n=ting in water with it> axis vertical. State
whether thc equilibrium is Slable Or unstable.
IAR •. - 0 ,304 m. Unstable Equilibriuml
13. A solid cylin<.lcr of 15 cm diameter and 60 Crn long . consists of tWO parts rna<.le of different materials. The
f,rsl part al the base is 1,20 cm long and of specific gravily '" 5 ,0. The olhcr parts of the cylindcr is made
of Ihe material havins specific gravily 0 ,6, Slale. if il can noat \'er(ically in water.
(Ans. G/I1 _ - 5.26. Unstable. Equilibriuml
14. A rectangular pontoon 80 to long. 7 m broad and 3.0 m <.leep weiShs 588.6 kN.lt carries on its uppcrdeck
nn empty boiler of 4.0 to diameter weigh ing 392.4 kN. Th e centre of gravity of the boiler and the pontoon
mc m their respective centres along a vertical line. Find the mcta--ccntric height. Weight density of sea water is 10104 N/ml .
(Ans. 0.325 ml
I S. A wooden cylinder of sp. gr, 0.6 and circular in cross-secti on is require<.l to nOOt in oil (sp gr. 0 .8 ). Find
the /.JO ratio for the cylinder to noot with its longitudinal a~is ,'ertical in oi l where L is the heigh t of
cylinder and 0 is it, diameter,
IAns. (UO) < 0. 81641
16. Show that a cylindrical buoy of 1.5", di~meter an<.l 3 10 long weighing 2.5 tonne> will not n<Xlt vertically
in sea-water of density 1030 kg/",'. Find Ihe force necessary in a ,'crtical chain all~chc<.l at the centre oflhe
b",;e Oflhe buoy Ihal will keep il vertical.
(A ns. 10609.5 N I
11 . A solid cone nOMS in water its apex downwards. Dc-Iennine Ihe least apex angle of cone for slable equilib·
rium. The specific gravily of thc materia l of Ihe cone is given 0.7.
(Ans. 39° 7'1
18 . A ship 60 m long and 12 tn broad has a <.lisplacement of 19620 kN. A weight of 294 ,3 kN is moved across
the deck through a distance of 6.5 m, The sh ip is ti lted through 5". The momenl o f inertia of the sh ip at
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IL
1162 Fluid Mechani cs
walcr line aboUl its fo",,, and aft axis is 75% of mOment of inertia the circumscribing rec(,mgle. The ccnlre
of buoyancy is 2.75 Tn below watcr line. Find the mcl~.ccnuic height and position of centre of gnlvity of
ship. Take spc<:ific weight of sea walcr = 10104 Nlrn J ,
IA ns. 1.1145 m. 0.53 m below walcrsurfaccl
19. II pontoon of 1500 lnnnes dispiacClllCnl is !looting in watef. II weight of 20 lonnes is moved through
a distance of 6 m across the deck of poilloon. which tilts the pontoon through an angle of 5", Find
Illcla-<:cntric height of the pontoon.
lA ns. O.9145ml
20 . Find the lime period of rolling of a solid e;"",lm cylinder of radius 2.5 Tn and 5.0 m long. Th,- specific
gravity of the cylinder is 0.9 and is floating in water wilh its axis vertical.
IAn s, 0.35 secl
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CR~ER
A. KIN EMATICS OF flOW
~
S.I
INTRODUCTION
Kinematics is dctincd as that branch of sc ience whic h deals with motion of particles without
co nsidering the forces causing the motion. The ve locit y at any point in a flow field at an y till1~ is
studied in this brandt of fluid ll1~chanics. Once the velocity is known. (h e n the pressure distri bution
and hence forces ac tin g on the fluid can be dClemlincd. In this <:hapter. the methods of determining
velocity and acceleration arc discussed .
.. S .2
METHODS OF DESCRIBING flUID MOTION
The fluid motion is described by two methods. The y arc - (i) Lagrangian Method. and (ii) Eulerian
Method. In the Lagrangian method. a sin gle fluid particle is followed during its nlO1ion and its
yelocity. acceicration, density. clc .• ilre described . In case of Eulerian melhod. lhe velocily. ilcceleration. prcssure. dens it y etc .. arc described ,,( n point in flow field. The Eulerian method is common ly
used in fluid mechanics .
II> S.l
TYPES OF FLUID FLOW
The fluid flow is classified as:
(i ) Sleady and unsteady flows:
(ii) Un ifo rm and non·uniform flows:
(iii ) Lami nar and turbulent flows;
(i v) Comp ressible and in compressible flows:
(v ) Rotational and irrOlationaJ flows; and
(ri ) One. two and thrt,e -dimensional flows.
S.l . I Steady and Un steady FloW5. Steady flow iSdefi lled as thilttypc of flow in which the fluid
charac teristics Ii!;e ve loc ity. pressure. density. etc.. at a point do not change with time. T hlls for
steady flow, 1I1athematically, we have
163
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1164
Fluid Mechanics
,.
- 0 -P
- . ( dill )
•. J.....
~
('P) "" ,.",
=0 -
'at
=0
where (.(0- )'0 - 4 ) is a fixed poim in fluid field.
Unsteady flow is that type of now, in which the velocity, pressure or dcnsily ;n a point changes wilh
respect to time. Thus. mathematically. for uIl5tcady flow
(~~)
.~o.(~:)
... ·'0· '"
.... y• • ""
~OCIC
.
5. 3. 2 Uniform and Non -uniform Flows. Uniform flow is dcfillcd as lhal lyPl' of flow in
which Ih~ velocity at an y given time does nOI cllange with respect 10 space (i.e .• length of di~ct ion of
the flow ). Mathematically. for uniform flow
(~~L~I =0
where
av = Change of ve locity
as = Length of flow in the ui rcction S.
Non-uniform flow is lhal type of flow in which the veloci ty al any given time ctl311gCS with respect
\0 space. Thus, mathematically. for non -uniform flow
('V)
as ' _ron"",
.0
.
5. 3. 3 Laminar and Turbulent Flows. Laminar flow is defined as that type of flow in whieh
the fluid panicles move along we ll·deflned paths or stream line and all the stream-lines arc straight and
parallel. Thus the panicles move in laminas or layers gliding smooth ly over the adj acent layer. This
type of Ilow is also called stream-line Ilow or viscous flow.
Turbulent now is that type of flow in which the fluid panicles move in a zig-zag way. Due to the
movement of fluid particles in a zig-zag way. the eddies fonnation takes place which arc responsible
VD
for high energy loss. For a pipe flow. the type of flow is determined by a non-dimensional number-
"
called the Reynold number.
where D = Diamckr of pipe
V = Mean vcioci ty of now in pipe
and
t' = Kinematic viscosity of fluid.
If the Reynold number is less than 2000. the now is calkd laminar. If th~ Reynold number is more
thaI14000. it is called turbulent now . Irthe Reynold number lies between 2000 and 4()()().thc now may
be laminar or turbul~nt.
5.3.4
Compressible and Incompressible Flows. Co mpressible flow is that type of flow in
which the density of the fluid c hanges from point \0 point o r in other words the densit y (p) is not
constant for the fluid. Thus, math~matically, for compressible flow
p ... Constant
Incompressible flow is that type of flow in which the density is constant for the fluid flow. Liquids
arc generally incompressible while gases arc compressi ble. Mathematically. for incompressible flow
p = ConslanL
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Kinematics o f Flow and Ideal Flow
165 1
S. 3. S Rotational and Irrotational flows . ROlalional flow is 1lial type of flow in wllich the
fluid panicles wll ile flowing along stream-lines. also rolate about thcir own axis. And if the fluid
panicles white flowi llg along stream-li nes, do not rotale abou t their own axis then ttJallypc of flow is
"ailed irrotational flow.
5.3.6
One- , Two- ilnd Three- Dimensional Flows.
Onc·dirn cnskmlll n ow is thai lyp" of
flow in which Ihe flow parameter such as ve locity is a function of lime a nd One space co-ordinate on ly_
say.T. For a steady one-dimensional flow. the I'doc ity is a func tion of onc-spacc-co-ordinatc only. The
variation of veloc itie s in ol her 1WO mutually perpendicular directions is assumed negligible. H ence
mathematically. for one-dimensional flow
II =j(x).,·= Oand w= 0
where 11_ V and W <Ire velocity components in x. y ,lnd z directions respe\:tively.
Two-dim e ... ion a l n ow is thaI type of flow in which the veloci ty is a function of time and two
r~ctangular space co-ordinates say x and y. For a steady two - dimen~iOllal flow the ve locit y is a func lion
of IWO space C{}-ordinales only. The variation of veloci ly in the third dircction is negligihle. Thus.
mathematically for two-dimensional flow
II
= f,( .... y ). v = f 2(x. y) and w :: O.
Th ~ _dimcn sional fl nw is that Iype of flow in which the veloci ty is a function of time and th ree
mutually perpemlkular diredions. BUI for a sle:,dy three-dimensional flow the fluid p<trameten; arc
functions of three spa ce co-ordinates (x . y and z) only. Thus. mathematically. for three-dimensional
now
II::
.. 5.4
fl(x . y . z) . ,' :: fl (. y . z) and
11'::
fix . y. z) .
RATE OF FLOW OR DISCHARGE (Q)
It is delined as the quanti ly of a fluid flowing per second through a seclion of a pipe or a chanrlCl.
For an incom pressib le fluid (or liquid) Ihe rale of flow or discharge is expres.scd as the vo lum e of fluid
flowing across the s.cclion per second. For compressibl e fluids. Ihe rale o f flow is usually expressed as
the we ight of fluid flowing across Ihc section . Thus
(i) For liquids Ihe uni ls of Q arc mJfs or iilres/s
(ii) For gascs the uni ls of Q is kgffs or Newtonfs
Consider a liquid flowing through a pipe in which
A :: Cross-sectional area of pipe
V:: Average ve loc it y of fluid a(;fOSS the section
...(5.1 )
Q= A xY.
Then discharge
.. S .S
CONTINUITY EQUATION
The eq uat ion based on Ihe principle of conservation o f mass is called contin uit y eq uati on. Thus for
a fluid flow in g through th e pipe at all the cross-sectiun. the quantity of fluid per secund is constant.
Consider IWO ",ross-sections of a pipe as show n in Fig. 5. 1.
Let VI:: A vera ge velocity at cross-sedion I- I
PI = Density at section I- I
A I = Area o f pipe at section I-I
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1166
and
Fluid Mcchanics
V2. P2'
A~
<D
arc correspo ndin g valu es al section. 2-2.
~~
The n rate of now
section 1- 1 "" p ,A, VI
Rate of flow at "ce uon 2-2
== P ~!Vl
,_",.l~'"'.'"'.'"'.'"".'"'."11.,"".,.,."
i
'
,
7""
DIRECTl ON~
-
Acco rdin g 10 law o f co nse rva ti on of 1ll,ISS
OF FLOW
Rate of flow al seclio n I - I
== Rate of fl ow al sec lio n 2-2
= PzAIVl
@
i
+.
b"""'" " ',~"",..,.
... (5 .2 )
Eq u:u ion (5.2) is appli ca ble 10 the co mpressi ble as we ll as inCQlIl - Fig . 5.1 Fluid flowi ng through
pn: ...s iblc fluids and is callcd Contlnull}' Equation. If the fluid is in a pi~.
compressi ble. Ihc n P, '" p, and continuity eq uati on (5.2) redu ces \0
or
p, A I V,
... ( 5.3 )
Problem 5.1
The diameters of a pipe atlhe seclions I (/lId 2 lire /0 em mid 15 em respeclil-ely. Find
rhe disc/wrge Ihrough lilt' pipe if the relocily of W<lIer flowing through rhe pipe ar sec lion 1 is
5 mls. Determine also IIIe )'e/oeify al sl'clioll 2.
Solution. G iven:
<ZJ
CD
At sc(;ti on 1.
D,,,, lO cm ,,,O.1 1ll
!
AI
At scc tio n 2.
"'..::.
4
(0 , 2)
:"::'
4
(. I) l " 0 .007854 Ill l _
V, '" 5 m/s.
O 2 ,,, 15cIll ", 0.1 5
A2
III
="4" (. 15t =0 .0 1767 m-,
, D,=1Ocm
V," 5m./s.ec
Fig. 5.2
(/) DiSCharge through pipe is g iven by equa tion (5. 1)
or
Q=Al x V,
= 0.007854 x 5 = 11.03927 III lls. A/l s.
Using equ.u io n (5.3). we ha ve A,V, " A 2 V1
( ii)
A, V,
,O~.=
OO 78~54
,=-
V 2 = - A,
-,- " - 0.0 1767
)( 5 .0 = 2.22 m'-~. Ans.
Problem 5 .2 A 30 em diameler pipe. coIII'eying warer. branches illlo 111'0 pipes of diametl'Ts
20 em and 15 em respecti)'e/y. If IIIe (II'erage ~'e/oeir)' in II,e 30 em diamerl'r pipe is 2.5 mls. find the
disc/mrge ill this pipe. Also derermille rhe ),elocity ill 15 em pipe if tile (II'ew ge I'e/oeify ill 20 em
diumeler pipe is 2 mls.
Solution. G iven :
V," 2.5m/ sec
0, "3Ocm-
(j)
Fig. 5.3
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Kinematics o f Flow and Ideal Flow
D, '" 3Dcm =0.30
11"
1
167 1
III
11"
,
,
A, '" - V , = - x .3" =0.07068 m "
4
4
V, '" 2.5 In/s
D2 '" 20 ern = 0.20
Al
=~
III
(.2)2 = ~ x .4
= 0.0 3 14 OI l ,
V2 =2 rn /s
DJ = 15cm = O.15m
,
II
A):-
Find
,
,
If
,
(. Is r =- x 0.22S '" 0.01767 m -
(I) Discharge in pi pe I o r Q ,
(ii) Veloc ity in pipe of di a. 15 (Ill or V)
Le t QI' Q1 nnd QJ are disch<lrgcs in pipe 1. 2 and 3 rcs[X:ct ivcl y.
Then accordi ng 10 continu ity equ ati o n
Q,=Q2 + Q3
.. .(1 )
(,) Th~ dl~c hu rgc Q, in pi pe I is g ive n hy
Q, = A, l', = 0.07068 x 25 lO lls = 0.1767 rn l/s . A n~.
( ii) Value o r V.l
Q l = A1V1 = 0.03 14 )( 2.0 = 0.0628 m3/s
Substitutin g th e values of Q, a nd Q1 in cqu~t io n (I)
0.1 767 = 0.0628 + Q3
QJ'" 0 .1 767 - 0.0628 '" 0. 1139 m1ls
Q3 = A1 XV3 =O.0 1767XV)
or
O. I 139=O.O I 767 XV3
0.1 139 = 6.·U m/s. An s.
0.01767
Problem 5 .3 \Yarer flows ["ro ugh II pipe AH 1.2 m diameler a/ j mls and Ihell pt/SIes II!rollgh II
pipe BC 1.5 m diameter. AI C, the pipe brallches. Ilrallch CD is 0.8 m it! diameter alld carries Ollel/!ird of Ille flow in All. The flow I"elocity ill brallch CE i.~ 2.5 mls. Pilld Ihe I"olume rale of flow ill
AIJ. Ihe )"e/OCily ill BC, Ihe \"e1ocif)' ill CD alld l/ie diameter of CEo
V}=
Sol ut io n. G ive n:
Dia meter of pipe AB.
DAS= I .2 m
Veloci ty o f fl ow th rough AB. VAS = 3.0 m/s
Di a. of pi pe HC,
lJ /iC = 1.5 m
Dia. o f branch ed pipe CD.
Dcv = O.8m
Ve locity o f flow in pi pe CEo Vn · = 2.5 mls
Lei lhe fl ow rate in pi pe
AB=Q mlts
Ve loc ity o f fl ow in pipe
BC = VIIC mts
Ve loc it y of fl ow in pipe
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CD = Ve/J mts
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1168
Fluid Mechanics
,
A
B,
t
1.5
12m
,
m
I
!
V>B= 3 m/sec
~
0
~
~
,
VeE" 2.5 m/$<)C
Fig. 5.4
Diameter o f pipe
Then now ratc thrQu~h
CE", Da :
CD : QI3
CE=Q-QI3= 2Q
3
(i) Now volume flow ralc th roug h AB = Q = V,\~ x Area of An
'IliU now ralc th roug h
= 3.0 x -
1t
2
1t
2
11
2
J
(DAR) = 3.0 x - (1.2) = 3.393 m Is. Ans .
4
4
(rr) App lying co ntinu ity equatio n \0 pipe AB and pipe BC,
VAH x Area of pipe AB == VHe x Area o f pipe Be
It
,
3.0 x- (D.... /I)" == V/IC x- (Doc)
4
4
"'
3.0 x (1.2)2 == VBe X ( 1.5) 2
V nc
3)( 1.2 2
1.5
l
=
[Di Vide by
~]
1.92 m/s . An s.
(iii) The flow ratc through pipe
CD=QI= Q = 3.393 = 1.13 1 rn J/s
3
3
Q 1 '" VCD x Are a uf pipe CD
. 131", Veo
)(~
(Om)!
, )(0.8-, '" 0.5026 Veo
.
)(4
1.1 31
= 2.25 m /s. Ans.
0.5026
V ClJ : - - -
(i,') Flow rate th rough CEo
Q 2 == Q - Q 1 = 3.393 - 1.1 3 1 = 2.262 rn J/s
Q 1 = Vet- )( Area of pipe CE
2.263 = 2.5
"'
DeE::
,
= Vc£~ (Dn/
,
)("4 x (DClf
2.2(3)( 4
2.5 x Il
~5'
=,, \.1;).0" 1.0735m
Diame ter of pipe CE:: 1.0735 m. A n s.
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Kinematics of Flow and Ideal Flow
169 1
Problem 5.4 A 25 elll diallle/I'r pipe carries oil of sp. gr. 0.9 aI II \'e/ociry of 3 rnls. AI atlOt!!er
secriolllire diameter is 20 em. Filld the \'elociry a/ Ihis .Iec/ioll and also mass rate offlow of oil.
Solution. Given
at section I.
v, = 25 em =0.25 m
IT
at section
2
It
1
A , : - D , =- x 0 .25 = 0.049
4
4
V,=3 rnls
DJ '" 20 ern = 0.2 rn
2.
OJ
1
"
Al : - (D.l t '" 0.0314 111 -,
4
V2 "'?
Mass ra tc o f flow o f oi l = ?
Apply ing continuity equat io n al sectio ns and 2 .
A,V, = AlV l
0'
0.049 x 3.0 = 0.03 14 X Vl
0.049 x 3.0
V. '"
= 4.68 IIlls. AilS.
0.03[4
Mass ratc of fl ow of o il
= Ma ss de nsity x Q = p X A , )( VI
o
Density of o il
Sr . gr. of oil
Densit y of water
'" Sp. gr. of oil x Densi ty of' wate r
1
900 ~g
= 0 .9 x 1000 kgfm = - ,-
De ns ity of oil
m
Mass ratc of flow
= 900 x 0.049 x 3.0 kg/s = 132.23 kg/so AilS.
Problem 5.5 A jel ofll'aler from a 15 mm diameter nozzle is directed l'erlically up\>'drds. Anl/ming
Illat llie jet remains circular and neglerling any loss of energy, 111(11 lI'ill be llie diameler at II poim 4.5 m
abow! Ille IIOU Ie, if Ille "elodly "'ill, II'/licll Ille jel {cares rile 1I0ule is /1 mls.
,
Solution. Give n
Dia. o f nozz le.
0, '" 25 mm '" 0.025 rn
Velocity o f jet at nozz le.
VI'" 12 rn/s
Heigllt of poin t A.
II'" 45 III
Lei Ille velocily of Ih e jel at a heig ll t 4.5 rn '" V 2
Consider th e ve rlkal motion of Ihe jel from th e out ld of Ille
nozzle to Ille poilU A (neglecting an y loss of e nergy).
Ini lial veiocily.
U=
Final veloc it y.
V = V2
g= - 9.1I 1 m/s 2 and li0: 4.5m
Value o f
Using.
V, '" 12m/s
JET Of
WATE
NOZZLE
T
I
• .• m
J
- 1/ 0: 2g/l . we gel
,-12
, =2)«(-9.81 »( 45
V!
V~
Fig . 5.5
I I
Ii
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IL
1170 Fluid Mechanics
~1 21
V2
2x 9.8 1:.:4.5
Jl44
88.29
7.46111/s
Now applying continu ity cqu:nion to the ouliet of nouk and at point A.
we gel
"
,
A\c
VI-X VI Itx (0.025)2 xI2
Al = - ' _I", 4
= 0.0007896
V1
V1
4x 1.46
OJ = Di31nclcr of jd 3\ point A.
Then
.. 5. 6
Al
="2If 02 !
OJ
x4
"'V/0.0007896
It
or 0.0007396
="411
x DJ
2
= 0.03 17 m '" 31.7 mm. A ns .
CONTINUITY EQUATION IN THREE · DIMENSIONS
Consider a Iluid cle ment of Icnglhsdx. tty and dZ in the direct ion o f .(, y and Z. LeI U, I' and II' arc the
x.)' and Z directions respectively. Mass of fluid ente ring th e face ABCD
inlet ve loc it y co mponents in
per seco nd
= p X Velocity in x-d irection x Area o f ABeD
=pxux(dvxdz)
Then mass of fluid leavi ng the fa<:c EFGH pe r second = pu dydz
+~
a., (pu dydz ) d...
Gain of mass in x-dircction
= Mass through ABeD - Ma ss through EFGH per sccolld
'" pu dplz - pJI dydz - -
=-
=-
a
- (pll dydz) dx
o (pu dydz)tlx
ih-
,
ax
o
- (pu) Ii.( dydz
ax
.. dyllz is l:o nstant I
Similarly. the lI et g ain of mass in y-dircctioll
"
ay (pI') d,l.dydz
a
- (pw) d_l dydz
a,
'" - -
and in z- di rcct ion
'" -
Net gaill of masses '"
y
Fig. 5.6
-[~
(plI) + ~ (PI') + ~ (pw)] dxdyd:
O.l
oy
oz
Since the mass is neither created nor destroyed in the fl uid cle ment. the ne t inc rease of mass per unit
time in the fluid cle ment must be equal to the rate of increase of mass of fluid in thc clement. But mass
I I
Ii
~ I
IL
171 1
Kinematics o f Flow and Ideal Flow
of fluid ill the cleme n! is p. d.\". dy. tlz and its ralC of incrca!;C wilh time
is~
a,
(p dx. dy. tlz) or
. dx d)' dz.
Equati ng the two cxprcssio lls.
"'
"'
1
- [-' (pu) + -a (pI') + -a (rw)
ax
ay
(Jz
Or . dxdydz
d_w/yilz '" -
dt
ap + ~ (pu)+ ~(pL') + ~ (pw) = 0
at ,h
oy
<lz
ICancelling dx.dy.dz from both sides1 ... (5. 3.-\ )
Equation (5.3A) is the continuity cq um iun in I:arlcsian l'O-urdinatcs in its most general rOlln. This
equation is upplicablc to :
(il Steady and unsteady flow,
(ii) U niform and non -uniform flow, and
(iii) Compressible lInd incompressible fluids.
For stead y flow. Up = 0 and hence equation (5.3A) becomes as
a,
o
,
0
- (pu) +- (pl') + - trw) = 0
ax
uy
<Jz
.. .(5 .3R)
If the fluid is incompress ible, then p is COnSt31l1 and the above eq uat ion becomes as
all
vv vW
vy VZ
-+-+-~O
0.(
... ( 5.4 )
Equation (5.4 ) is th~ continuity "quation in three-dimensions. For a two-dimen sional flow. the component w" 0 and hence continui ty equation becomes as
au + a,· ,,0.
... (5 .5)
ax ay
5. 6. 1 Continuity Equation in Cylindrical Polar Co-ordinates. The continu it y equation in
cylindrical polar en·ordinates (i.e .• ~. 9. Z co-ordinates) is deri ved hy the procedure given below.
Consider a two-dimensional ineompr~ssibl e now field. The
two-dimensional polar co-ord inates are, and 9. Consider a fluid
~ · dr
element ABCD hctween the radii, and r + dr as s hown in
Fig. 5.7. The angle subtendcd by the clement at the centre is d9.
Th" componen ts of the velocity V arc "r in the radial direction
and ue in the tangential direction. The sides of the c lement ar~
having the lengths as
Side All "" rd9.BC "" dr. DC", ('+ Iff) d9. AD" Ifr.
The thidncss of the c leme nt perpendicular to the plane o r
the paper is assumed to be unity.
COllsider the flow in radial direclioll
Mass of fluid en tering the fa"e AB per unit tim e
"p x Ve locity in , -direction x Area
•
",
,."
Fig. 5.7
I I
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IL
1172
Fluid Mechanics
Mass o f fluid
Ic~ving
"'pxu,x(A LIx I)
'" p X II, X (rilO x I ) = p.
the fa<:c CD per uni t time
= p x Velocity x Arca
= px(u, +
(
=px u, +
(,: Arca '" AB x Thickness = rdO x I )
II," rdO
a;; .df) x(CDx
J:
a..
I)
(,: Area = COx I )
dr X(r+dr)dO
I': CD=(r+tlr)dO I
)
a.., d,+ -a.., (d,')' 1d O
aT ar
=p xlI,xr + u,dr+r -
[
'" r{lIrxr +u , xdr+ rJ;; .df] dO
[Th e tenn containing (11,)1 is very small and has been ncgk~ t cd l
Gain of mass in r-di rect ion per un it time
'" (M aS/; through All - Mass through CD) pcr unit time
'" p. ",. rdO - P[U,.f + 1I,.dr+ r
a;~ .lIr] dO
a..,
[
1
= p.Il ,.rdO- p.u r·r.de - p u,.dr+r dr,ilf de
''', 1
[
= -p 1I, .dr + r dr,ilT .dO
[Tlli s is writlcn in this form because
(r. de. dr. I) is eq ual to vo lum e of
ele mentl
Now COllsir/er the flow in S-direr/io"
Gain in mass in a·direction per uni t time
= (Mass through Be - Mass through AD) per unit lime
'" [p x Velocity th rough Be x Area - p x Velocity through AD x Area l
:: [P.ua.ii, x1-p (110 + i.l~~ ,de)xdr xlJ
", - p( ~ .de) drx
(': Area=drx I )
1
aU a r.J6.dr
= -P aE! '
,
[Multiplying and div iding by
rl
Total ga in in fluid mass per unit time
u,
(lU
'l . r. dr. dfJ - p-aUa- · rtia,. tir
= - P [ -;:- + -, -,
I I
ae
...(S.SA )
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IL
Kinematics o f Flow and Ideal Flow
173 1
p x Volume of fluid clemen!
=px[rd6xdrxll
Bul11lilSS of fluid clement
=:
=px,,'9.dr
Rat e o f in creasc of fluid mass in the cle ment with lime
a [p. rde.dr ] = a;'
ap rdedr
"a;
... (5.58)
(": rde . <If. 1 is th e vol ume of c lemen t and is a constant quantity)
Since the mass is neither created nor destroyed in the fluid clement. hence net ga in of mass per unit
lime in the flu id cle ment must be eq ual 10 the rale of increase o f mass o f fluid in the clemen!.
Hence equating th e two express ions given by equ ations (5.5 A) ,Old (5.5 8). we gel
1
-p [ -U, + -(111, r.dr.d9-p -dUo
rar
"'
Ja
U,
"'
rde. dr
- p[
Ju,]
---; + Jr
[u,
r
dUe
-p
Ja
ilp
rd9dr
dt
= -
r =
dP
[Cancelling rilr . d9 from bolh sides ]
at
Ju,] + p -aile . ~ __ 0
-dp +p - + r
dr
...(5.50
r
Equation (5.5 C) is Ihe continuity equation in polar co -ordinates for two-dimensional flow.
at
For steady flow
ap :
a,
i.l9
0 and hence equation (5.5 C) reduces to
u, au, ]
aUe I
p[ - + +p - . - :O
r
ae
ar
r
~+au,+ aue
.!.:o
,a, ae ',
au, aUe
u,+ '-a;+ de :0
"'
:, (ru,)
"'
+ dd
O
(un) '" 0
Equation (5.5 /J) represcnts the cuntinuity cquation in
incompressible flow,
Problem 5.5A
flow?
[.: : ,
pol~ r
(r. u,) '" r. daU; + u,]
...(5 .5D)
cu-ordin ates for two-di mensional steady
Examine whetllef Ille [o/lQ ..... illg I'e/oeity compollell ts rt'preselll a physically possible
", = r Jill O. Uo '" 2rtos O.
u, '" r sin e and ue '" 2r cos 8
For phys ica ll y possible flow. the cont inu ity equation.
Solution . Given:
a
ar (,u,) +
Now
Multiplying the
I I
asa (110): 0 should bt'satisfied.
e
abov~
",: 'sin
equation by,. we gd
fII,=?sin8
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~ I
IL
1174 Fluid Mechanics
Differentiating Ihe preceding equation W.r. l. r. we get
~
{)T
(furl"
~ (r sin
aT
8)
=2rsin8
(":
si n
e is cons tant
w.r.!. r)
Now
jje'" 2r cos e
Differentiati ng th e nbovc cqumion w.r.l . e. we gel
a
{}9 (" 0) '"
a (2,- cos 9)
as
(.:
'" 2, (- sin 8)
2,- is conSlalll W.r.I. 9)
= -lrsi n e
i.(nI,) +~ (u e ) = 2r sin
a,
ae
e - 2,- sin e = 0
Hene.: the cominuity eq U<ll ion is sn li sfied. Hence Illc given velocity co mponents rcprcsem a physi·
cally possible flow.
to 5.7
VELOCITY AND ACCELERATION
LeI V is the rcsu hant ve locit y a1 any poim in a fluid flow. LeI 1/. I' and w arc its componclll in .1', y and
Z di rections. The velocity co mponents arc functions of space-eo-ordinates and lime. M31llcmal icall y.
Ihe ve locit y co m ponents arc given as
U=ft(.I',Y·Z,I)
\' "'li·I, y, Z,
I)
w=f,(.r.), . z. t)
and
R~~ullaI11
vdoc ily.
V=
IIi + vj+ wI: = Ju l + v l + w l
Lei <1r <1 , and a, arc Ilic lot al acce lerdtlo n in x. y and Z directions respective ly. Tllen by the cliain
rulc o f d ifferc ntia tion. we lIave
du au iI.l au dy au dz au
a = - = --+--+- .- + ,
ill
dl
til
til
dl
ax
ay
az
dx
ely
dz
- =u. - = I'anO- =w
til
ill
til
du
til
dll
dx
d,'
dl'
dll
dy
du
dZ
du
a = -;II-+I'-+W-+-
,
dl
a,· d" av
ax +v -d), +w -dZ + -(II
a . = - =u )
til
Similarl y.
dw
a.= -
I
If
dW
=U -
ih
dW
+>' -
ay
dW
...(5.6)
dW
+>' - + -
azal
av '" O. wllcrc V is result ant veloc it y
a,
('or ste ad y no w. -
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Kinematics of Flow and Ideal Flow
175 1
VII =O,Jv =Oand Jw =0
Jt
iJl
U/
"'
Hence acceleration in
-I',
y and Z directions bci:OIllCS
dll
all
<II
(Ix
dl'
o - - =' -
all
+l' -
au
~,
+ Wdz
a,·
dv
dv
dll'
aw
all'
all'
dl
ax
()y
Jz
, -
... (5.7 )
a , = - =II - +I' - +W }
dl
ax
0;,
Jz
" =-=,-+,-+wt
Acceleration vector
A=ai +aj+at:
J ' +a, , +d" , .
=
1
.. .(S .&)
ii,
5. 7. 1 local Acceleration and Convective Acceleration . Local acct'lcrution is defined as
the rate o fincrcasc ofvcloci ty with respect to lime at a given poillt in a now field. In the equation given
.(JUUI,
by (5.6), the cxprcsslOn -
.-
dt dl
C(l n n:o;Ii,'C acceleration
i~
(JW,
,
or "",,\ IS known as local acceleration.
of
defined as th e ratc of change of veloci ty d ue 10 the challge of position of
Ju dv
fluid particles in a fluid now. The expressions other than -
,-
at at
Jw
afld -
at
in cqualioll (5.6) arc known
as convcclive acceleralion.
Problem 5 .6
Til e I·elocity ,'eClor ill a fl' lid flo w iJ' gil'en
V", 4.';i - IOYyj
+ 21k.
f-illd rile l'e1ocity lIml acre/amioll of a flllid particle m (2, I. 3) aI rim e I '" I.
Solution . The velocity componel11s II. v and IV arc II == 4.r'. I' '" - IO.~ y. IV '" 21
For Ihe poinl (2. 1. 3). we have x == 2. ), == I and z = 3 allime I = I.
Hence ve loc ily components at (2. I. 3) arc
II = 4 X (2» ) == 32 units
I' = - 10(2)1(1) = - 40 unils
w=2x I =2uni ls
Velocily veclor Vat (2. I. 3) = 32; - 40j + 2k
Resultant velocity =
Jil"+ ,.1 + ","
==b2l
+(-40)' +2 2 -JI024 + 1600 + 4 = 51.26 units. Ans.
Ac ce leration is given by equation (5.6)
all
all
ih
dY
all all
a =11 -+ 1' -+ 11' - + -
az at
av av av av
,. =II -ax + V-dy +W -az + -ar
.(
(l
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~ I
IL
1176 Fluid Mechanics
" ==
;:
d",
dw
Il -
(.Ix
h
-
~.
dw
Jz
+W -
d...
+ -
at
Now from velocity components. we have
au
, all == o. -au
-
= 12.,-, -
ax
dz
dY
Jy = - 20.\)" ~
ax
Jy
dW
ax
= 0,
ow
Oil
= 0 alld -
at
~
0
= - I O_~. a~ =0('11' = 0
iJz
ow
oy '" 0, az '" 0
iJ!
dW
and
at = 2.1
Su bstitutin g the valu es. the accel erati o n co mpone nts at (2. l. 3 ) at lime I '" I arc
= 4~.3 ( 12x2) + (- 10;')') (0) + 21 x (0) + 0
lIx
'" 4&x~ = 48 )( (2)~ = 48 x 32 = 1536 units
",." 4x3 (- 20xy) + (- IOx!y) (- IOx~) + 21 (0) + 0
=-
4
&ox y + l00x\-
=_80(2)4( 1)+ HlO (2)4 X I = - 1280+ 1600= 32Q unils.
<I,
"'
== 4.yl (0) + (- In;)') (0) + (11) (0) + 2.1 == 2 .0 units
II! = 15361 + 320j + 2k. An s.
Accele ration is
A = a,i + (I) +
Resultan t
A =J( 1536)l + (320)1 + (2)l units
=J2359296 + 102400 + 4 '" 1568.9 unit s. Ans.
Problem 5.7
Th e [o/lowing CUjCS represent Ihe Iwo ,-e/oe ily compOI.enIS, determine the third componnll of "e/ociry SlIdl flwr riley salisf)' rile COII/in uir), eqllllfio" :
(i) 1/ =.y2 + l + z" : "=xl-Y{] +.t)'
(Ii) \. =
= 2.\)"z.
2/. '"
Solution.
The continuity cllualio n for inco mpressible fluid is ~ivc n by equation (5.4) as
all + 0" + aw
ih
a),
=0
ih
a" =2.T
a."
a,
,
- =2x)' -z + x
Cuse I.
-
iJy
Substituting the va lu cs of
h+2q- l
I I
all
dx
and dl' in continuity equa tio n.
d),
, +x+ aw~o
a,
ow
,"\
'
- =- 3x - 2~)" + z-orow= (- 3.(- 2ty + z-) dz
a,
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~ I
IL
Kinematics o f Flow and Ideal Flow
Integration of both sidc ~ gives j dw '"
i( - 3..- -
177 1
l .ry + Zl) til
+
W '" ( - 3.>.l - 2.\)"
~: ) + Const31l1 of integration,
w here l.'O ll stam of integratio n canllot Ix: a function of z. BUl it can be a funct ion o f x and y Ilia! is f(x . y).
IV
=(
- 3u - 2:<yz +
+ f(:<, y). An s.
a,· =4)'
\' '" 2l
Case II.
z:)
-
dy
-
aw = 2.')'
a,
-
Su bstitutin g the va lu es of ~ and dW in conti nuity eq uatioll. we get
Jz
dY
dw
-
d.•
+4y+2xy=O
d,
".
-
d.•
" = _ 4.\)' _ 2)' _;1 + f ey,
[nt egrati ng. we get
Problem 5 .8
'" - 4)' - 2.\)' or dlj '" (- 4y - 2ry) dx
Z) = _ 4xy _ x 2y
+ fey, z), Am .
A fluid flow field is gi,-<'I/ by
V", xl )'; +
/;j _(2.>.)'z + YZ 1)/c
PrOl'e Ilwl il is (I case of possible slnlll), incompreJJiblc fluid jlow. Co/ell/are lire I'e/ociry and aeee/em/ioll at II,e point (2. I. J).
Solution.
For the givcll Ouid flow field"
= ..?y
I' :.lz
ow
-ax
a,·
-
ay
= lxy
:2yz
aw
az-,, - h y - 2yz.
['or a case o f possible stead y inco mpre ssib le fluid flow. th e continuity eqU3lion (5.4 ) should be
sm isfied.
i.e ..
Substituting the I'alues of dll ,
d.'
dw
,uld - . we ge t
d,
dll dl' dll'
- + - + - '" 2xy + 2)"2 dx dy d,
I I
2.1)' - 2)'2 '" 0
Ii
~ I
IL
1178 Fluid Mechanics
Hence the veloc it y fi eld
v '" .~}'i + yZzj _ (2.ry< + ),Zl)
is a possible case of fluid fiow. AilS.
j:
Ve locity al (2. 1,3)
Substituting the va lues
z= 3
.f '" 2. Y = I and
V
=
+
' .
[JI
y 1!j
in velocity field, we gel
' k
- (1.1)'< + yz')
= 2"x l i+ l'x 3j - (2x2 x I x3+ I x3')k
= 4i+3j-21k. An s.
:~4 Z + 3'- + (_ 2 If == J16 + 9 + 441
and Rcsuhant veloc it y
..}466 '" 21.587 unils. AilS.
Accd.,ratkm nt (2, 1,3)
The acccJcralio li co mpon ents (Ix' a" and 11, for Slc;\(ly now arc
1I
au
.(
au +W -i}J j
:U - + V -
ih
dy
ilv
dV
(l , =U -
)
+ V-
ax
dy
OZ
ai'
dZ
+II' -
all' h -ali' h ' -a!l'
il,= It -
-
ax
2
dz
d)'
dU
all
ai'
dy
au
l
u"'_t)'.-= l.ry. -= x 300 2
<lV
th'
dZ
00
rlv,
V=Y Z. ax = 0. {)y =2y:, dZ =y'
, i)waw
w= - 21yz-yz·.-:;-= - 2yz. -,
oX
y
, aw
= - 2.t: -:' , -,
= - 21)'- 2)': .
Z
Substituting these values in acceleration components, we gel acceleration
~t
(2. 1. 3)
a, == .\1)' (2x)') + iz (xi - (21),z + )'Z') (0)
= l,.-ll + .r"iz
=2(2)11'+2'x l ' x3 =2 x8+ 12
== 16+ 12 ==28 units
a,. == x 2)' (0) + iz (2),:) - (21)'z + YZ2) (i )
. == l.l'? - 2xlz - lz2
==2x 11 X3'-2x2x 11 X3_1 1X3'= 18-12-9 == -3units
U;.
= x")' (- 2)"z) + iz (- 2xz- Z2 ) - (ll)': + ),Z2) (- 2.\")" - 2)"z)
= - 2Xlit - lxiz' _iZl + [4Jy'z + l.ll:' + 4.,).1: 2 + l izl r
== _ 2 X 22 X l ' x 3 _ 2 x 2 X J2 X 3 2 - J2 X 3.1
+ [4 X 22 x J2 X 3 + 2 x 2 X J2 X 32 + 4 x 2 X 12 X 3 2 + 2 X 12 x 31r
= - 24 - 36 - 27 + 14K + 36 + 72 + 54r
= - 24 - 36 - 27 + 48 + 36 + 72 + 54 = 123
Acceleration
I I
=
ai + II) + {If = 281- 3J +
I2Jk. An s.
Ii
~ I
IL
Kinematics o f Flow and [deal Flo w
or
J28 l + ( 3)1 + 123 2
Resultant acce leration
179 1
../784 + 9 + 15 129
= .J15922 = 126 .1 8 units. Am .
Problem 5.9 "-ind Ille conr<'Ctil'l! acceleration allil e middle of a pipe whiel! co/ll'erg<'J uniformly
from 0..1 m diameter to 0.2 In diameter OJ'a 2 III lengtll. Tile rale off/o ..... is 20 lith. If Ihe rale of flow
c/wnge.! IlIliformly from 20 lis 10 40 lis in 30 seconds. find rhe towl acceleration ill II,e middle of Ihe
pipe a/ 151h second.
Solut ion. Given:
Diameter at section I.
D I = 0.4 m ; D2= 0.2 Ill. L = 2 Ill. Q = 20 lis = 0.02 111 3/S a~ one lilre
=O.OOlm}= 1000 ern}
Find (I) Con>'c<:!;v" ac<:clcration m middle i.e .. at A when Q = 20 lis.
(ii) Total acceleration al II when Q c hanges from 20 lis 10 40 lis in 30 seconds.
Case I. In this case. tile rate of flow is constant and equal 10 0.02 m}/s. Th" velocity of flow is in
_(-d irecti on only. Hence this is one-dimensional flow and velocity componenl~ in y and Z direction s are
lcroorl' =O,Z=O.
"'
= I' ay only
Convectivc acce Jcration
.
Lct us find the value o j
I'
all
and -
".,
.. , (i)
_
.
at a dlstance.f from Inlet
(lJ
Th e diameter (D,) at a distance _f from inlet or at section X-X is given by,
0.4 - 0.2
D, =O.4 xx
2
=(0.4-0.1 x)m
The area of cross-section (At) at sec tion X-X is givcn by.
A
"'~ D 2= ~ (0.4 - 0.1
''' -.
T r-+----'
~·T - -'
1 _ x...: x :
O.4m
O.2m
,
:""' lm - ,
;- 2m
X) l
4 '
4
Ve locity (1') at the section X -X in terms of Q (i.e .. in tenns of rate of flow)
.r
I'
=
---.SL '"
~ '"
Ar"a
A,
---.SL=
~D l
4
1.2730
= (0.4 O.I _r)!
"ax
To find -
~
It
Fig. 5.8
4Q
Il(OA - O.lx/
'
1.273 0 (0.4 - 0.1
-Irl
fills
... (ii)
. we must diffe re ntiate equation (it) with respcci to x.
au = -:;a [1.2 73 Q (0.4 - 0.1 x)--•[
ax
-
0.1"
= 1.273 Q (- 2) (0.4 - 0.1 .1)- 1x (- 0.1)
= 0.2546 Q (0.4 - 0.1 _r) ,
Substituting the val ue of
uand au
ax
[Here Q is co nstant]
... (iii)
in equation (i). we get
Convective accelerat ion = 11.273 Q (0.4 - 0.1 .I;-l ] x [0.2546 0 (0.4 - 0. 1 _rr']
= 1.273 x 0.2546 x Q2 x (0.4 _ 0.1 _IT l
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1180
Fluid Mcchanics
'" 1.27] x 0.2546 x (0.02) 2 X (0.4 - 0.1
:.Convective
acccl~ralion
x»
[-:
Q '" 0.D2 m'/sl
al the middle (where x = I 111)
= 1.273 x 0.2546 x (0.02)2 x (0.4 _ 0 .1 x 1)-3 m/s 2
"" 1.273 x 0.2546 x
(o.oll x
(O.3r3 mls 2
= 0. 0048 mls·. An s.
Case II . When Q changes from 0.02 m3/s 10 0.04 1I1 3/s in ]0 seconds. find the 10lai acceleration
at x'" I 111 and r '" 15 seconds.
Total accclcr;l1ion = Convective 'lccclcrmion + Local ac(;clcration ,11 t'" 15 seconds.
The rmc of now at I = 15 sccunds is given by
Q'" Q 1 + Qz 3~ Q1 x 15 where Q2 = 0.04 m 31s and Q1 = 0.02 1ll 3,s
=0.02+
The velocity (14) and gradient
:. Convective acceleration",
(0.04 - 0,02)
30
J
x15= O.03mfs
('"a.. ) in terms of Q arc given by equations
,I.
(Ii) and (iii) res[l<-'l:tivcly
a"
"
'" I 1.273 Q (0.4 - 0.1 xr21 x [0.2546 Q (0.4 _ 0.1
= 1.273 x 0.2546 Q2 x (0.4 _ 0.1 )( 1)-3
.(r l [
:. Convective acceleration (w hen Q = 0.03 lO lls nnd x = I rn)
: 1.273 x 0.2546 x (0.03)! )( (0.4 - 0. 1 x 1
= 1.273 )( 0.2546 )( (0.03) 2 )( (0.3)-.1 m/s2
rl
=0.0108m/s"
Local acceleration:
... (il')
i.l"
-a;
: -a;i.l [1.273 Q (0.4 - 0.1 .rr !
1
I .:
II
from equation (ii) is
: 1.273 x (0.4 - 0.1 x)
I .:
1x
II:
1.273 Q (0.4 - 0.1 X)-l ]
,Q
-
a,
Local acceleration is at a point wherc.f is constant but Q is changi ng]
Local acceleration (at x: 1 m)
, aQ
: 1.2731'.(0.4 -0.1 x I) 1'.= 1.273 x (0.3)
-2
0.02
x 30
"
[
. .'Q = QrQ I: 0.04 - 0.02: 0.D2 ]
30
30
.. .Iv J
. a,
= 0.00943 m/s 2
Hcnce adding cquations (i,,) and ("). we gc t lOtal acceleration.
:. Tmal <lcccier;,tion : Convcl'livc ac(;clermion + Local <lccelermion
== 0.0108 + 0.00943 == 0.02023 IlI/s2. An s.
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Kinematics of Flow and Ideal Flow
.. S. 8
VELOCITY POTENTIAL FUNCTION AND STREAM FUNCTION
5.8 . 1
Velocity Potential Function .
181 1
II is defined as a scalar function of space and time such
that its negative derivative w ilh rcspccl l0 any dircrtion g ives the nuid velocity in that direction. It is
defined by 0;. (Ph i). Mathematically. Ihe velocity. polcnlial is defined as 0;. '" f(x. y. z) for steady flow
sucli that
U=_dQ
d'
v'" -
dQ
ily
...(5.9)
d,
11'=- -
il,
where II, \' and I\' MC the components of velocity in .l.), and z dircdions respectively.
The velocity components in cy lindrical polar co.()rdinales in terms of vctocity potential funct ion <Ire
given by
". = ; ;
lie =
)
.!. dQ
.. .(5.9A )
'"
whe re
II, '" ve locity component in radial direction (i.c .. in , dircctioll)
and
lie = velocity component in tangential direction (i.e .• in 0 direction)
..
.
.
.
. au + -a" + -a\~ == o.
a.T ay at
The COl11lnUlty equation fur an IIKumpresstble steady nuw tS -
Substituting the values of u. \' and I,' from equati on (5.9). we gtl
... (5. 10)
Equation (5.10) is a Laplace equation.
alq.
a~9
f'or two -dimension case. equation (5.10) reduces to - , + - , '" o.
a.T - ay '
If any valu e of
9 that
. .. (5. 1 I)
s.11isfies Ihe Laplace equation. wilil"()rrespond IU some ,",<lse of fluid flow.
P ro lle rties o r th e l'o te nt !:,1 I'- unctio n. The rotational co mponents· arc gil'en by
• Pleasc. refer to equation (5. t 7) on page 192.
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1182 Fluid Mechanics
ro '"
.!. (all _dlY)
==
.!.[aw
- ~l
2 ayaZ
,. 2 dZ
U)
X
ih·
Substituting the values. of I•. I' and w from cquatioll (5.9) in the above rotational components. we
gel
'"'
a!q.
(l! Q
a!q.
a"41
(hay
dyi}x
ilzih
dxilz
If 9 is a continuous function. IlIcn - - " - - 0- - " - - : etc.
ro~ =w. =wx =O.
When rotational components <lfe zero. the flow is called irrotational. Hence the propcnics of the
poccntial function arc :
L If velocity IX'iClllial (41)
c!\iS!~.
the flow should be irrotalional.
2. If veloci ty potential (41) satisfies the Lapla(;c cqualion. it represents the possi b le steady inc01l1pressible irrotational flow.
S.S.2 Stream function . It is defined as Ihe scalar fUIl(;tion of space and time. SU(;h. Ihat its partial
derivative with rcsp.:ct to any direction gives the veloci ty component at rig ht angles to that direction. It
is denoted by IjI (Psi) and defined unly for two·dimc nsional flow. Mathcmati<:alJy. for steady flow it is
defined as IjI : / (x. y ) such thut
I
a.
Jx ",.
... {5.12)
J IjI : _ "
and
The velocity
cOl1lponenl~
'J'
in cy lindrical polar co·ordinates in ICrms of stream function are given as
I
aljl
-
-
aljl
... (S .12A)
and "a" - r
ar
where "r'" mdi al velocity and "9 " tangential velocity
II , :
. _.
ae
_ .
' , a" + -(/1' "
ax ay
The COntln ully equation for two-dimensional Ilow
I I
IS -
o.
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Kinematics o f Flow and Ideal Flow
Substituting the values of
1/
and \' from equation (5.12). we get
a( aya.) a(""J
a:;- '"
,h
+ dy
-
183 1
a'. a'.
0 or - a.•ay + (lxdy '" O.
Hence cxistclH:c of \jI means a possible case of fl uid flow. The flow may be rot ational or irrmat ion al.
'(',ax '")
The rolallonal
component W, .IS gIven
by 00, '" .
.
2
Substituting
th~
-
- -
it),
.
va lues of 1/ and I' from cqualio rl (5. 12) in the above rotational component. we gel
For irrotational flow. U1, '" O. Hence above equation becomes
a'
a
as~ + ~
ax- G1y-
=0
which is Lap!iI!;c equ ation for 1jI.
The properties of strea m function (1jI) arc:
L If stream function (1jI) exists. it is a possible use of fluid flow whic h may be rotational or
inOlational.
2. If stream function (ljI) sat isfi es the Laplace equation. it is a poss ible case of an irrotat ional flow.
5 .8 .3 Equipotential Line . A line a long which the velocity potenti al 9 is cons tant. is called
equipotential li ne.
For equipotential line
<j) = Constant
d~
=0
,.ax a.a),
<j) = f(x. y) For steady now
B"
d<j)= -
'" -
For equipotential line.
- (utfx
0'
But
5.8-4
+
an -
IIlf.f -
ay
wi)'
'" - (utf.{ + I'd}').
tf<j) '" 0
rd}') '" 0 or lIdx + rtf)' '" 0
dy '"
"
tfx
\'
dy =
d.,
... (S. 13)
Slop~ of equipoten lial
line.
Line of Constant Stream function
~
'" Constanl
d~:O
Bm
I I
d~ = aljl
ax
d.{ +
aljl
a},
dy '" + "dx _ udl'
.
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1184
Fluid Mechanics
For a line o f constant stream function
'" dljl = 0 or
dy
:
dx
IIdy '" 0
1'1/-'.' -
I'
... (5. 14)
u
But dy is slope of stream line.
d,
From cqU3tions(5.13) and (5.14) it is clear thallhc product Orlhc slope of Ihe equipolenliallinc and
the s lope of Ihe stream line a1 the point of inlers.ection is equal to - I. Thus Ih" equipotcnliallines are
othogonal to the stream lines at all points of intersection.
5.8 .5 Flow Net. A grid oblliined by drnwing a scricsof cquipotcntiallincs and stream lines is called
a now net The flow n<:1 is an illlporlanilool in analysillg two-dirn"nsional irrOlalional flow problclllS.
5. 8. 6
Relation between Stream function and Velocity Potential function
From equation (5.9),
_df
we have
andv =
"
JIjI
dltl
and v = ""'\"
dy
ax
From equation (5.12). we have u = - -
u= _ d¢= _ aIj/ 3Ild,·= _ #
Thus. we h3vc
ax
ilQ
ay
dy
= alj/
ax
'xilQ "''y"
:
)
-"'
"
" ",
... (5. 15)
:
Problem 5 .10
The "elocily pOlelllial funclion ($) is gil'en by (III e.>.pressiolJ
xy
1
,
.f y
$=---x + -
+y
1
J
3
(i) Fi/ld IIJe "e/oeily compolle/l/s ill x alllll' direclion.
(ii) Sholl' 11101 9 represelllS U possible wse offloll'.
The p3nial dcriv31il'cs of
1
3
q.
,
xy
~
3
W.r.l..l and yare
dQ
i
J.l ly
-
ax
=- -
-2 x + - -
... (1 )
d$
3xy".I J ,
: - - - + - +-y
... (2)
-
ay
I I
,
x)'
9=- - - X + - + ,
Solution. Given:
3
3
3
3
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Kinematics o f Flow and Ideal Flow
185 1
(il The velocity components II and I' arc given by eq uation (5.9)
""d.l : - [y'
- 3- 2
11= -
,
.
11= "'3
ly
]X
X+
-- ]
3
y'
"' )+ 2.r -x ' y
,
+2x -x-y. Ans.
An s .
(ii) The given val ue of <p. will rcprcSCIH a possi ble case offlow if it satisfies llie Laplace cqumion. i.e..
(fiji
a
2
<1>
- , + - , =0
ih · dy ·
From cqumion s ( 1) and (2). we have
dQ = - J'/3 - 2r+x2y
Now
ax
,f<ll
-, : - 2 + 2.ry
,hO
{)¢I
-
and
,Xl
= - .ry- + -
(ly
3
+2y
(1 141
= - lx)'+2
dy l
d1¢l
ih
J1<l>
dy
- , + - 1 = (- 2 + 2\)') + (- 213' + 2) = 0
Laplace equat ion is sa tisficu and he nce Q represe nt a possible case of flow . Ans.
Problem 5 .11 The ndocill' pOlential j,,"(lioll is gil'e" by <p = 5 (x! ((lieu/ale 'he ",doc;/y components a llil e poill/ (4, 5).
Solution .
op = 5 (Xl - i )
dQ '"
ax
lOx
i.lQ
= - lOy.
ay
Bul velocity components" and
I'
i).
arc given by t>{jualion (5.9) as
u:-d1p:-IOx
ax
a. = - (-10y) = lOy
" ,, - -
ay
Th" velocity components at the point (4. 5), i.e .. al x " 4. Y " 5
"" - 10 X 4 " - 40 units. An s.
I'" 10 x .5 " SO unit s, A ns.
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1186 Fluid Mechanics
Problem 5.12 A stream junclioll is given by
IjI '" Sol - 6)'.
C,,/cu/mf! the ..efocil)' componellts o /l d ,,/so lII(lg1l;llIile (mil direction of Ihe resu/I"/I' I'e/ocity al ,my
poill/.
Solution .
1jI =5x - 6y
-"" = 5 and -d. = - 6.
ax
dY
But the velocity l'OmponCllts II and I' in tel ms of stre am function arc g iven by equation (512) as
!I=
_ dljl = - (- 6)= 6 uni ts/sec.
d),
a.
1' = -
a.,
AilS .
= 5 un li s/sec. An s.
Rcsul1ant velocity
,. 5
lanS ", - = - =0.833
" 6
== lan- l .833 = 39° 48', An s.
Problem 5 .13 If for II two-dimensional potelltial flow, the )'elociTy pOlelitial is gil'ell by
41=..-(2), - 1)
delerm;" " Ihe I'docily III Ihe [XliII/ P (4, 5). Dekrm;IIe "bo Ihe WI/Ill.' of slreOIll "melioll IjI (II the poillt P.
Direction is give n by .
e
Solution. G i ve n
~=x(2y - l)
(i) The vc loc ity componcllIs in the directi o n of x and ), arc
a;
a
= - -ih '" - -ax Ix (2)' - 1)J =- (2)' - I J = I \'= _ i>t", _ ~ Ix(2y - I)] = - [2x] = - 2x
1/
dY
AtlhepointP (4,5 ),i.e.. alx=
u=
)' =
.:
VelocitY;11 P
Of
=
Res ul1ant veloc ity at P
2)'
dY
4.y=5
1-2x5 = - 9 unit s/sec
- 2 x 4 = - 8 uni lsiscc
- 9i -8j
J9' + 8
1
'"
~81 + 64 =
12.04 units/sec = 12.04 unit,Jse<:. Ans.
(ii) Vallie of S trea m Function a t I'
We know that
and
d1f
= - u= - ( 1 - 2)')=2)' - 1
:
=I'= - 2X
d),
Integrat ing equat io n ( I) w.r.t . 'y', we get
j dIy "" j (2y - I) dy o r 1.[1 =
I I
... ( i)
... ( i i)
+2
.1
y + Constall! of integratio n.
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Kinematics o f Flow and Ideal Flow
187 1
The const ant o f integration is not a fUliction of y bUI il can be a fUllcti on of x. Le t the val ue of
constant of integration is /e Tllen
\V '" l- y + k.
. ..(iii)
Differentiating the above equation w.r.L 'x', we get
dill '"
(J.T
But f rom equation (ii),
ak
ax
Oljl = _ 2x
0,
dljl
dk
ox
oX
Equating the value of -,;;--. we get -:;- '" - 2.1.
1ntcgratlng
. th
" ' n. we get k
IS equalto
= f - 2xd x = - -2x~
, - '" - X-.,
Substituting this value of k in equation (iii), we gel It' '" y1 _ , . _ ,,1, AilS .
Stream fun(;tion If a1 I' (4. 5) '" 5' - 5 - 4' '" 25 - 5 - 16", 4 IInits.
Problem 5.14
711,. stream /1I1II:tioll for
II
two-dimensional flow is git'en by \jI = 2xy. (aleulMe Ille
\'cloeily at IIII' point P (2. 3). Find lile I'elocity pOfelllillljullc/ion
Solution. GiVt'1l :
The ve locity componen ts
IjI
1/
1/=
At
th~
point P (2. 3).
w<"
get
$.
= 2 ly
and I' in tcrms of Itf
I'
d.
ar~
a
- - = - - (21}')= - 2x
ay
= -"" = -a
ih ax
ay
(2.1)')
= 2y.
= - 2 x 2 = - 4 units/sec
v = 2 x 3 '" 6 units/sec
1/
/, , /, , 1"'=
Resultant vcloci ty at
p=" .. + 1' _ ", 4 +6 = ,,1 6 + 36
Vel ocity l'olenli,.1 Fu n ction ¢I
We know
A ilS.
~
.J5i = 7.21
"' -Il = -(-2x) =lx
d.jl = -v=-2y
units/sec,
...( i)
... (i i)
Oy
Int egrating equation (i). we get
!d$=!2xdx
"'
,
2x'
1
., +C= x + C
... (iii)
where C is a constant which is indepe ndent of x bu t can be a function of y.
Differentiating equation (iii) w. r.t.
I I
y. we ge t
ay oc
oy
Oljl '"
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1188
Fluid Mcchanics
But fro m (ii),
~
0,
= - 2y
'0,C "' - 2.1'
-
Integrating lliis equatio n. we get C ==
2 '
J- 2y dy =- ~ . == - l
Substituting this val ue of C in eq uati on ( iii). we get q. == x~ _ yl,
Problem 5.15 Skelch Ihe s/re(lfll lines repreSented hy W == .; +
Also Jim{ QuI Ihe I'e/ocil)' lind ilJ' dir ediOIl al poin/
Solution. Given
\j1 '" ,,) +
The ve locity compon ents II and I' arc
AilS .
i.
(I. 1 ).
I
,
_ iA¥ =_i. (..-2 + I)=-2y
dy
I'
==
Al tlie poi nt ( I, 2). tlie velocity
11=
()y
,
oW = i. (.(2 + l ) = 2.1'
,h
ax
compon~nL~
4 UNITS/SEC
Fig. 5.9
arc
- 2 x 2 '" - 4 units/sec
y
\' = 2 x I = 2 uni ts/sec
Resul tant ve loci ty
==
~u! + vl_~(_4)1 +2 '
=.fiO
lane=
oed
I'
== 4.47 uniHi/so.-"C
2
,
-HH--- +--+-+-t--~
I
-=-:II
4
e = t all
I .
2
5 = 26° 34'
Res ultant veloc ity makes an an gle o f 26" 34' wilh .f- axis.
Sketch of Stream li nes
w=x2+l
Ld
Then We ha ve
1jI =
Fig. 5. 10
1.2.3 and so on.
1 =->.,l + l
2=-I) + l
3=·,J+i
and so 011.
Each equa ti on is a equm io ll of a ci rcl e. T hus we s hall ge t concemric circles o f differcm diam ete rs
as show n in Fig. 5.10.
Problem 5 .16 Th e l'e!oCily components in (I Il!"o -dimensioll(l/ flow field fo r (In incompressible
fluid (Ire liS follows:
u=
LJ
+ 2-1· _ -Ily {lnd
obtain {Ill expressioll for rhe Slream fUllc/io n
I I
I'
= -1/ _ 2y -
7!/3
v.
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Kinematics o f Flow and Ideal Flow
189 1
u " if] + 2\" _ _(2y
= .1)'2 _ 2y _ x 1/3 .
Solution. Given:
I'
The velocity components in terms of stream function arc
a; = 1' = .l:l -2y - ~f3
ihv
OJ'
Int egrating (il w.r.l, x. we gd
···(il
= - 11= - 113 - 2\"+.,-2,)'
IjI
... (; i)
= f (xi- 2y - ..:"3) dx
~
1
~
q - -2xy- -"- + ' .
.: 2
4 x3
. .. (iii)
where k is a COnSl3J11 o f integrat ion which is independent of x but can be a function of y.
Differentiating equation (iii) w.r.l. y. we gd
l
Oljl == 2.t y _ 21" +
~
dljl
Bul from (ii),
-
ay
at = .f \' _2.1 + ~
2
~
J
1
~
= - yf3 - 2 l +XY
,
dljl'
<lk
. we get x ")' - 2x +- == - y 113 - 2.1 + .I- y
,
Comparmg the value of -
oy
oy
dk= _ 113
oy
k =
Integrating. we gel
,
f (- 113)
,
-=.L = -y
dy =
4 x3
12
Substituting this value in Uii). we get
1
1
••
. : -11: ·V- - 2xv - -" V
- -· - . Ans.
2
Problem 5.17
.
12
12
III a rWQ·dimensional incompressible j/ow. ,lie fluid reiociry compoll ents are gircn by
II = .l - 4YG/ld I' = - y - 4.1".
SilO»" Ihat reiocil)' pOlelilial e.(i.~IS allli de/ermine ils form . Find ol.fo lite slrcom filliCliotl .
Solution. Gi ve n:
II" .l - 4y and I'" - )' - 4x
a" :
-
a.<
1
VI'
-= - 1
ay
~+,)I'= I - I =O
ih
d)'
Hence flow is continuous and velocity potent ial exists.
41 = Velocity potential.
Let
I I
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IL
1190
Fluid Mechanics
Let veloc ity componcnls in
t~rrns
a.;.
ax
,"d
of velocity potential is given by
= - Il= - (.r - 4v)=-x+ 4y
. .. ( i)
d.jl = - I' =-(-y - 4x)=y + 4x
... (i I)
.
dy
"2
Int egrating eq uation (il, we ge t ¢ '" - -
+ 4xy + C
where C is a co nstant of integration. which is independent o f x.
This conSla nt can be a function of y.
Differentiating Ih e aoov" equatio n, i.e .• equation (iii) wi lh respect 10
..
-
dy
... (i i i)
y, we ge l
de
=0+4x + -
dy
But from equation (iii), we have
at
i)y
== y + 4.•'
Equati llg Ihe two valu es of d¢ ,we get
oy
de
de
4.l+ - =y+4x or
dy
Integrating the above eq uation. we get
c= L
where C 1 is a CO ll Man !
=y
OJ'
,
+ C1
2
of integration. which is independent o f x and y.
Tak ing it equa l to zero . we gel C == yl .
2
Substituting trw va lu e o f C in equation (iii), we ge t
-1, 2
<jl
=- -
2
)'1
+ 4 .'1+ -
2
. Ans.
Value of Stream functions
u:t \II = Stream function
The veloci ty com ponell!s in terms of SCream function arc
d.
d,
d.
-
- = I ' = -y- 4x
and
=-,,=-(.(- 4y)=-.\"+4y
... {;,,)
... ( v)
OJ'
Integrating eq uati on (1"1')
W.f.!.
\II
x. we get
4.r 1
= - )'.\"- - -
H
2
where k is a cOnStalll of illlegration which is independent of x but can be a function o f y.
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Kinematics o f Flow and Ideal Flow
191 1
Ok
Diffacnliating equ ation (ri) w.r.1. y. We gel QIjI", _ x _ o+
oy
il)'
a.
But from cqu3lion (I'). we Itave
-
ay
"oy
Equatillg the twO values of d\jJ ,we gel
-x+ -
"
= - .t+4,}'
'y"
=-x+4y
-
=4y
Integrating the above equation. we get
SultS\ilulinlllhc va lu e o f k in equat ion (I'i). we gel
IV =
... 5.9
-)'x -
2.\"2 + 2/ . AilS•
TYPES OF MOTION
A fluid pani cle whi le moving may undergo anyollc or combination of follow ing four ly pes of
disp lacements:
(i) Linear Translation or Pure Tran slation.
(il) Linear Dcformmion.
(iii) Angular Dcfornwtion. and
(iJ') R01alion.
S. 9. 1 Lineilr Translation. It is defined as the movement of a fluid c lement in such a way thai it
moves bodily from olle position 10 another posit ion and the twO axes ab and cd rcprcsemcd ill new
positions by (I'b' and c'd' lire parallel as shown in Fig. 5.11 (a).
5 .9.2 linear Defo rmati o n . II i~ddlned as the defonnalion of a fluid clemenl in lineardireclion
when lhe eleme nl moves. The axes of lhe elemenl in lhe deformed POSilion and un-deformed posilion
are parallel. bUI lheir lenglhs change as shown in Fig. 5 II (b).
,
,
.' .'
,
eI ~+~
• -t
0
0
(0) LINEAR TRANSLATtON
~
0
0
(b) LI NEAR DEFORMATION
,"
'" , .
t ,'
,
•
, • ,,' "
,
-~-
,
,
'!~
'"
"
.-"'
o.
,
(e) ANGULAR DEFORMATION
Fig . 5.11.
~ I
_4 d'
0
o
,
,
(eI) PURE ROTATION
DiJplaument of a fluid deme"f.
I~
~ I
IL
1192
Fluid Mcchanics
5.9.3
Angu lar Deformation or She af Deformation.
It is defi ned as the average change in
lhe angle conta in ed by two adjacent s id es. LCl ll.9 1 and "'6 2 is the change in angle ~ twccn two
adjacent sides of a fluid c lement as shown in Fig. 5.11 (c) . then angular deformation or s hear strain
rale
.... 6 1 :
Nuw
Angular d erorma tion '"
dV AI" ':h·
-x- · -
ax
~
S h c ll r slru]n r>lt" '" -I
ax
Ih
and l!.6,=
-
a"
6)' all
. oy ll.y dy'
-
[d6 1 + d B! 1
[a" + -a,,]
-
2 i,h
... ( 5. 16)
(ly
5 .9.4 Rotation . II is ddincd as Ihe mOVellll'nl of a flui d element in sucll a way Ihat both of it~
a)les (horizontal as well as ... crtical) rolah! in Ihe same dir"ction as s hown il\ Fig. 5.1 1 (d). II is equal
ox Jill
oy for a two-dimensional clemen t in x-y p lane. The rotational compon ents are
to~(~2
W. '
-
~(a"
2i1x
-,,,]
Jy
.. ,( 5.17)
W •
J'
~(a"
_awl
2ozox
Vorticity. It is defined as the vai lic twicc o f the rotation and h ~ncc it is given as 200.
Problem S.18 Aflilidflow is gi),en by V '" 8fi _ iOAJyj.
Find Ihe ~'he(lr slmin rale lind ~'/('Ie whelher Ihe floK' is rola/iuII(I/ or irrol(lliOllll/.
5 .9 .5
Solution. Given
V", S..-li - I Q~yj
II'"
\' '" - lOx·y.
~ h car
'"
24.,.2.
~
'" 0
a,·
,
20.1)" (ly = - lOx"
strain ratc is givcn by equation (5.16) as
=
I I
~:
,av
a.t = -
and
(i)
8..-1.
.!..(av
+ au]",..!. (-20.•y +O) = - 10sv. An~.
2axd}'2
.
Ii
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IL
Kinematics of Flo w and Ideal Flow
(ii) Rotation in
,f -
193 1
Y plane is given by equation (5.17) or
00, ==
-.!. [~- d
)= ~ (- lOX)'. _ 0) == _ tnt)'
U
2ihdyl
As rotation w, ~ O. Hence flow is rolmional. Ans.
Problem 5.19 The "e/OCil)' com/l(ments in 1I (wo -dim,," siOll,,/ flow lIre
'
, - 2y - x ' I J.
""'),'I3 + 2x - x)'lmd,'=xY
SllOw flwt Ille.le compolI<'lIrJ represelll a po.l.fible case of lIIl irrOlalional flow .
Solution. Given
II"
if3 + 2,1' - _~y
0"
-=2
-2xy
0,<
a"3y ' , , ,
-
oy = -3- - .C=Y--.I\' == xy2 _ 2)' -
:(31)
0'
-=2.l'y
- 2
oy
dV
1
3 .\'!
-== y - --=y
ax
(i)
3
" 1 0 ow. continu
" ity
r'o r a two-dmlcnSlona
Sub~liluling
au
the value of -
<'h
2
1
- ,f.
. .IS -audv
cqu3110n
+ - == O
a\ ay
dl-
and -
oy. we gel
au ai'
- + - =2 -2->')'+ 2-')'-2
ax Jy
==
0
It is a possible case of fluid now.
,
..
(ii) ROla1101l. 00, IS given by w, == -
[al' au)
I -
- -
2a.ray
1
,
,
1
'
== - [(y- - .c ) - (y - x-)] .: 0
2
Rotation is zao. which means it is case of irrotational flow . AilS.
.. 5.10
VORTEX FLOW
Vortcx flow is defined as thc flow of a fluid along a curvcd path or the flow of a rotating mass of
fluid is knowll a . VOrtell Flow'. Thc VOrtCll flow is of twO types namely:
I, ForlCcd vOrtell flow, ;lnd
2. Free vonell flow .
5 . 10. 1 Forced Vortex Flow. Forced vo rtex flow is defined as that type of vortex flow. in
which some ellternal torque is requircd to rotate the fluid mass. T he fluid mass in thi s type of flow.
rotates at constant angular velocity. (0. Thc tangential vcloc ity of any fluid particle is given by
I'= rox r
... (5.18)
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1194
Fluid ML>chanics
where, '" Radius of fluid particle from tlie axis of rotation.
I CENTRAL AXIS
V
I
_::1'__ _
=~=~=-{--::::~
VERT iCAL ....
CYLI NDER
LIQUID
:::::=t::-::
_-_-_1--_-_-
::=:::f:::::
:-:-:-£cccc
_-:-:-:t-:-:-
-=-=-}-:-=-=
•
L1
(a) CYLINDER IS STATIONARY (b) CYLINDER IS ROTATING
Fig. S.12
Forced 'OOrrex floUi.
Hence angular I'clocil y ro is g iven by
1-
,
... (5.19)
m" - '" Consta nt.
Examp les of forced \'OnCll arc:
L A vert ica l cylinder containing liquid wll ietl is rOlaled abollt its centra l axis with a constant
ang ul ar veloc ity ro, as sliow n in Fig. 5. 12.
2. Fl ow of liquid inside the im peller of a cen trifugal pump.
] . Fl ow of waier through the runner of a lUrbinc.
S. IO.2 free Vortex flow. When no cx tcmal torq ue is required to rotate the fluid mass. that
type of flow is call ed free vortex flow. Thus the liquid in case of free vonex is rotating due to the
rotation which is imparted to the fluid prev iously.
Examp les of the free \'onex flow are :
I. Flow of liquid through a ho le provided at the OOllom of a contai ner.
2 . Fl ow of liquid around a c ircular bend in a pipe.
3. A whirlpool in a river.
4. Fl ow of fluid in a cent rifuga l pump casing.
The relation between velocity and radius. in free vort ex is obtai ned by putting the va lu e uf extern,1I
torque equal to zero, or. the ti1l1e Tate of change of angu],lr momenlum. i.e .. 1I10melll of m01l1emum
must be zero. Consider a fluid panicle uf mass 'm" at a radial distance, from the axis of rotation.
havin g a tangential velocity I'. Then
" Mass)( Velocity" m )( v
Angular momemum
Morucm of mOnlemum
" Momenlum x r'" /II X I' X ,
Time rate o f change of angular momentum '"
~
a,
(lin',)
a (ml") '" 0
For free vonex Int egratin g. we get
I I
"'
1111"
'"
Constan t or (" '"
Constan t
m
'" Constant
... (5.20)
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Kinematics o f Flow and Ideal Flow
195 1
S. 10.3 Equiltion of Motion for Vortex flow . Consider a fluid clemen! ABeD (shown
shaded) in Fig. 'i. 13 rotating at a uniform velocity in a horizontal plane about an ax is perpendicular to
Ihe plane of paper and passing through O.
Let
r= Radius of the element from O.
<loa", Angle subtcndcd by the cleme nt at O.
dr'" Radial thickness of Ihe clement.
<loA '" Ar~a of cross-section of clemc"L
The forces acting on [he element are:
(i) Pressure force. 11M. 011 (he face AB.
(ii) Pressure forcc'(p
+ ~~
tJ.r) M
on llie face CD.
,
(iii) Centrifugal force, till" acting in the dirc(;tion away
,
Fig . 5.13
from the cemfC. O.
Now. the mass of Ihe clement'" M ass density x Volume
=pxMx!:J.r
",
Centrifugal force
'" piMAr - ,
Equali ng the forces in the rad ial direc tion, we get
l
ap)
,,'
a, ty L\A - pL\A = pL\A6r - ,
ap
v
a, 6r L\A = pL\A6r - , ,
p+-
l
dp
Cancelling 6r)( L\A from both sides. we get -
a,
,,1
= p -
...(5.2 t)
,
Equation (5.21) gives the pressure variation along the r:,dii.1 di rection for a forced or free von cJ\
flow in
J
horiwllml plJnc. The cJ\pression dp is called pressure gradient in the radial direct ion. As ap
ar
ar
is posilive. hence pressure increases with the increase of radius ' r' ,
The pres~ure variation in the "enical plane is given by Ihe hydrostatic law, i.e. ,
ap
az-= - pg
... (5 .22)
In equation (5.22). z is measured venically in the upward direction.
The pressure, p varies with respeci to rand z or pis J fune lion of rand Z and hence total derivative
of {! is
ap dr+ -ap dz .
or aZ
dp: -
ap
a,
Substituting the values of -
I I
frOnl eljuatioll (5.21) and
ap
aZ
front cljumion (5.22), we gel
Ii
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1196 Fluid ML>chanics
dp '"
P -" dr - pgd:
,
...(5 .23)
Equation (5.23) gives the variation of pressure of a rmating fluid in any plane.
S. IO.4
Equation of Forced Vortex Flow.
For th., forced vorlex flow. from equation (5. 18).
Wi' have
.'=",xr
where ill '" Angular ve locity'" Constant.
Substituti ng lhe value of,' in equat ion (5.23), we gel
(!) ~ r l
dp '"
,
P x- - dr - pg dz.
Consider two points 1 and 2 in Ihe fluid ha vin g forced von e,; flow as shown in Fig. 5.14.
Int egrat in g the abo ve equation for poilUs I and 2. we gel
J,~p
'" f pm!
rdr -
I~gdZ
"'
"'
p
'1
l'
='2(00-'1 - oor n - pg lz2- zd
= £b}-I',ll- pgI Zl-z dJ': V1=rof1 l
1
2
v,"' (!)r, f
Fig. 5.1 4
If Ihe points 1 and 2 lie on the free surfal'C of the liquid. then PI = /11 and
hence above equation bc(;OUlCS
0= ~ [1'/ - ",21 _ pg Izl-zil
pg
[Z1 -
zd '"
f 11'/ _
[zl-z d= -
I
2,
----...r:
_:_:...;z,
__ :..r_
",2(
-:-:1:-- -
[ "/ - "I~l.
[f thc point I lies on the axis of rotation, then \., "ro x 'I "ro x 0" O. Thc abovc
equati on becomes as
Ld
I I
I',·,
l , - l , = Z. thcn wc ha vc Z= - '-
-
2g
=
ro
,
'
Xr,-
28
Fig. 5.15
•
...(5 .2 4 )
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Kinematics of Flo w and Ideal Flow
197 1
Thus Z var ies wilh th e square o f r. He nce equation (5 .24) is an equatio n of para bo la. This means
Ih e free surface of th e liquid is a paraboloid.
Problem 5 .20 Pro,'/< Ilwl ;1' case of fo rced I'orlex, l/i e riSt: of liquid {el'e! <II Ihe ends is eq",,/Io
Ihe /111/ of liquid lewd (II the axis of rolatioll .
Sol ution. Lei
R = radiu ~ of I h" cy linder.
0 ·0 = Initial level of liquid in cy linder when the cylinder is not rolating.
:. Initiaillcighl o f liquid
'" (11 + x)
Vo lum e of liquid in cylinder" ItR 2 X Hei ght of liquid
'" TtR1 x (II + x)
.. U)
Lcl1lie cylinder is rmated al consta nt angular ve locity w. T he liq uid will risc at Ihe e nds and will fall
,H th e ce ntrc.
Let
y = Rise of liquid at the e nds from 0·0
x = Fall of liquid <11 the (e mf" fronl 0-0.
The n volum e of liqu id
'" IVolume o f qtl inder UplO level 8 -81
- IVolume of pJraboloidl
'" InR ! j( He ig h1 of liquid up10 level 8-81
,
- [2' H'
j(
He ight o f paraboloid
,
1
""'
, ~
'" nR-, x ( II + X + Y) - - , - x (.( + Y)
AXtS OF
ROTATION
Fig_ 5.16
,
,
nR l
" nR- x II + nR - (x + Y) x (x + Y)
2
,
nRl
" nR - x II +- - (x + y)
... (ii)
2
Equ atin g (I) J nd (ii). we get
,
rtR- (II
1
nR"
2
+ .f )" nR x It +- - (.l + y)
",
nR '
nRl
- - .f= - - y
or
2
Fall of liquid at ce ntre = Rise of liquid at th e e nds.
2
or
.f" Y
Problem 5 .21 All ope" firc"fa r la"t of 20 cm d;wl1t:ler 'IIId 100 CIII fOllg COllla;" 5 ...aler "1"0 a
lreiglll of 60 CIII. Tile /'IIIt is rolated {,bOUI ils I'erlical llXiJ' at 300 r.p.m., fiml Ille deplll of (lflfabola
forllled a/ lite f ree Sllrface of "'I'/er.
Solution. Given
,,20 em
Diame ter o f cy linder
Rad iu s.
I I
2. =!Oem
'. -
2
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1198
Fluid Mechanics
Hcigllt of liq uid.
II : 6Q em
Speed.
N", )00 r.p.m.
21f.N
2x
An~ular
ve locity,
Ld the depth of parabol a
w= -- =
1'()(
60
300
60
,,3 1.41 rad/scl:.
=z
Usin g eq uation (5.24),
w'R'
(31.41)' x (tOi
==
"'''''"-c''CCC'"-= 50.28 e m . An s.
2g
2x981
Problem 5.22 An open circular cyiindt'f of 15 em diameter and 100 em long conwin.i water JjfllO
a lleight of 80 em. Find ti,e maximum sp<'cd 0 1 ....lliell tile cylinder is to be rOIa/ed a/W ill liS I'ertical
axis so thm 110 ",mer spills.
Solution. Given
Diamclcr of cy linder
= 15 em
:. Radiu s .
Leng th o f cy li nd er.
Initial he ight of water
15
R =-=7.5cm
2
L= lOQcm
= 80 CIlI.
Let the cyli nder is TOMkd at an angular sP<"cd of ill radfscc , wh en Ihe water is i!hou llO spi ll. Then
usi ng.
Ri se o f liquid at e nds
BIll rise o f liqu id at ends
Fa!! of liqu id at ce ntre
He ight of parabo la
'" Fall of liquid at centre
'" Le nglh - Inilial he ight
: IOO - 80:20cm
'" 20 cm
",20 + 20",40em
Z= 40em
I!J ~ R ~
Usin g Ih e rci:lt ion.
Z= - - , we gel 40 '"
2,
w1 ",
4Ox 2x98 [
7.5 x 7.5
1!J ~{7 .5 ).!
2 x981
'" 1395.2
w", J1395.2 : 37.35 radls
Speed. N is give n by
2><N
w = --
60
x00 '" -"60~X"3"7".3"5 '" 356.66 r.p.m. An s.
60N= 2n
2 x It
Prob lem 5.23 A cylindrical \·<'.uei 11 cm in diameter and 30 cm deep is filled wirll water apro Ille
lOp. The renei is opell at Ihe lOp. Filld Ihe qualilily of liquid left in Ihe I'e.uel. wilen il is rotated aboul
ils rerlical axis wilh a speed of (a) J()(X) r.p.m .. and (b) 600 r.p.m.
Solution. Given:
Diumclcr of cy linder
'" [ 2 em
Radiu ~.
R '" 6 em
Initial height of water
'" 30 em
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Kinematics of Flow and Ideal Flow
Initial vo lume o f water
199 1
'" Area x Initial he ight of wate r
:: ~ X 12l X 30 em) = 3392.9 ern )
4
(II) Speed,
N = 300 cp.m.
,""
60
21t x 300
=31.4 1 rad/s
60
W= - - =
(jJ l Rl
(3 1.4 1)" X6 1
Hcigtu of parabola is g iven by Z= - - =
= 18.10 em.
2g
2x98 1
As vessel is initially ful! of water. water will be spilled if it is rOlll1ed. Volume of wmeT spilled is
equal 10 the volume of paraboloid.
But volume of paraboloi d
= li\rea of cross-section x Height of parabola I + 2
,
It
Z
'" - V - x-
If
424
Volume of w ater left
(b) Speed.
,
= - x [2-
18.10
x- - :: 1023.53 ern
2
l
= Initial vo lume - Volume o f water spilled
:: 3392,9 - 1023.53:: 2369.37 em l, Ans.
N:: 600 r.p.m.
,""
60
21t x 600
60
'" 62.82 radls
W=-- =
Height of parabola.
As the Itcight o f parabola is more than the Itcight of cy linder th e shape of imaginary parabola wi ll be
as showil ill Fig. 5.11.
Lei
r =' Radius of the parahQla at the boltom of the vessel.
Hei glll of imaginary parahQla
== 12.40 - 30 == 42.40 em.
Volume of wa lU left in Ihe vessel
= Volume o f water in portions ABC and OEF
'" Initial vo lume o f wala
- Volume of parabo loid AOF
+ Volume of paraboloid COD.
I" \
4cm
Now vol ume of paraboloid
42.
l \
ill~
I
"'tmSIj~ :~ :E 72.4cm
T
AOF ='
~
x Oi.
X
4
It
I
I
Height of panlbola
72.4:
'" - x 12" x - - == 4094. 12 em
4
,
For Ihe imaginary parabola (COD). w == 62.82 rad/sec
3
I' I
I
I
I
I
I
I
I
1
1GL,_jt>
I
IMAGINARY
CYLINDER
IMAGINARY
PARABOLA
I
,
\
I.
\
.
\ I '\
'
_IH
FIg. S.17
Z == 42.4 ern
r == Radius ill Ihe bottom of vessc l
I I
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1200
Fluid Mechanics
Using Ihe rdation
ro' "
Z = - -.
2,
r ==
get 42.4 '"
We
62.822
X,'
2 x981
2 )(981 x 42.40 '" 2 1.079
62.82 X62.82
r '" J2 1.079
==
4.59 em
Volume ofparabo luid COD
,
==
~ x Area at llic lOp of the imaginary parabola x Heig lll of parabola
==
"21 x It? x 42.4 "'21 x 1t x 459 2 x 42.4 '"
1403.89 eml
Volume of wate r len '" 3392.9 _ 4094.12 + 1403.89", 702.67 em J • Ans.
Problem 5 .24
An open circular (flllll/er of 15 em diameter and 100 em 1001g ( 0111<1;11$ waler 14plO
(1/ "'hiel, rile cylinda iJ /0 b,' row/ed abou i its l'erlicaiaxi.l . .\0 IlulI
II ',eigh l of 70 em. f 'jm!l/Je .~peed
the {lXial deplll becomes zero.
Solution. Given
Diameter of cy linder
Radius.
u,ngth of cylinder
Initial height o f wa ler
==
15
I
T"'"
em
15
R = -=7.5cm
==
==
2
100 ~m
70 em.
om
Tl
When axial depth is zero. the depth of paraboloid = 100 e m.
Usin g the
......... 15cm ~
rcl~tion.
Fig. 5.18
w =.~I00
~X~2~X~9.~8}1 " 442.92 "59.05 radls
75x75
.". Speed. N is given by
0'
7.5
',H
00 = - -
60
,v __
6Q x
21t
w _
,60~X;,'"9c·O",
21t
" 563.88 r.p.m . An s.
Problem 5.25 For tile problelll (5.24), find tile difference in IOtal preHlfreforce (i) at tile boltOIll
of cylinder. {lnd (ii) at tile sides of the cylinder due to row tion.
I I
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Kinematics of Flo w and Ideal Flow
201 1
Solution. (i) T he data is given in Problem 5.24. The difference in \olal pressure force al the
bonom of cy lind er is obtained by finding 100ai hydrostatic force at tbe bottom before rotation and aflcr
row lion.
Before rotation.
force'" pgAil
/f
1
l
it
22 -
1000 kg/m-. A '" Area of bonum = - D : - x (0. 15) m . II '" 70 em '" 0.70 III
where
3
rorcc= 1000x9.81
x~
4
X (0.15)1 xO.7 N '" 121.35 N
After rotation. the depth of water at the bottom is nut cunstant and hell,",c pn:ssurc force duc 10 the
height of w~ tcr. will 1I0t be l'OIISlanL Consider a circular ring of radiu s r and width tiT as shown in
Fig. 5.19. Lee Ihe height of water from tlie boUOIll of the lank uplo free ~urface of water at a radius
fil','
r=Z= - - .
2,
Hydrostatic force on ring a1 Ihe bonum.
<IF", pg x Area o f ring x Z
1
'" WOO x 9.81
TOlal
pr~ssun: forc~
al
tll~
bollom
o/r'
R
J io
'" dF'"
r·
071
'"
o
d.
9810x2xltrx--d,
2,
00 '
19620XItX -
2,
/
J
r dr
Fig . 5.19
I"'rom PrOblem 5.24.
ro '" 59.05 radls
R", 7.5 em '" .075 nl.
Substituling Ihese values. we ~cI lOla I pressur~ force
19620 x rc. x (59.05) '
2x9.8 1
'"
19620 x It x (59.05)1
2 x9.8 1
x
(.07 5)'
4
",86.62N
Dill"ercnce in pre~ure forces al the bottom
121.35 - 86.62", 34.73 N. An s.
(;i) Forces Orl Ihe sides of the c ylind er
Before rOlalion
where
~ I
'" pgA/'
A '" Surface area of Ihe sides of Ihe cy linder UplO heighl of water
'" itO X Heigl\! o f waler '" It x .15 x 0.70 1112 '" 0.33 m 1
I~
~ I
IL
1202 Fluid ML>chanics
h '" e.G.
'"
~
2
of the wcucd area of the sides
x height of wate r = 0.70 '" 0.35 m
2
!'oree on Ihe sides before rotation '" I{X)() x 9.S1 x 0.33 x 0.35 '" 1133 N
After rotation, the water is UplO llie lOp o f Ihe cylinde r and lienee force on the sides
'" 1000 x 9.81 x Wcucd area of the sides x
= 9810 x rrD x 1.0 x
~
x 1.0", 9810 x
11:
~
x Height of water
x .15 x
~ '" 2311.43 N
Difference in pressure On Ih e sides
2311.43 - 1133 ", 11 7H.4J N. Ans.
S.IO.S Clos ed Cy lind rical V ess els. If a cy lindrical vesse l is closed allhe lOp, which con tains
some liquid, lhe shape of paraboloid formed duc to rotation of the I'essel will be as shown in Fig. 5.20
for different s~d of rotations.
Fig. 5.20 (a) shows Ihe initial siage of the cylinder. when it is nO! rotated. Fig. 5.20 (b) sliows Ihe
shape of Ihe paraboloid fonned when the speed of rotat ion is WI' If the speed is increased funher say
w1• the shape of paraboloid formed will be as s hown in Fig. 5.20 (c). In this case the radius of the
parabola m the tOp of the vessel is unknown. Also the height of the parJboloid fonned corresponding
to angular S[)t....,d 001 is unknown . Thus to solve the twO unknown. we should hav~ twO equations. One
equation is
z=
lI) ;r1
2,
The second cquation is obtaincd from the fact that ror closed I'essel. I'olume of air berore rotation
is equal to thc volume of air after rotation.
Volume of air before rotation'" Volume of closed vcsscl- Volume of liquid in vessel
/tr l x Z
Volume of air after rotation'" Volu mc o f paraboloid formed '" "'-~-"
2
----y'l'--=
-- -
-- -
=:=:=t=::-:-]-:---i"
~
: --'
-_-_t::-_c
-:-1-:-:---::E
--- -:
i'l
i<1
fig. 5.20
Pro blem 5.26 II "en'd, c)'/illlhiw/ ill shape (11111 cloud at Ihe lop (llId bollom, eOlllaills .wler IIplo
a "eighl of 80 em. Th e diameter of rhe ressei is 20 em (lIId lellglh of rnse/ is 120 cm. The "esse/ is
rOlmed (II (I speed of -I()() r.p.m. ahOll1 ils "erli~'(li luis. Filld Ihe heighl of pllwh%id formed.
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Kinematics of Flo w and Ideal Flow
Solution. Given:
Initial height of water
Diameter of vesse l
:. Radius,
,,80
em
= 20 em
R = 10 em
= 120 em
N = 400 T.p.m.
Length of V('sscl
Speed.
w~
2.N
-60
~
120
60
'" 41.88 radls
When tlie vcs.sc l is rotated. let Z
'" Hei ght of paraboloid formed
11
==,=
r '" Radius of paraboloid at the top of Ihe vesse l
This is lhe ca'ie of closed vesseL
Volume o f air !x·forc rol31ion '" Volume of air MIN rotation
tt
2
Jt
2
,
T
,
1
TT
2 11 x4()()
203 1
Fig. S.21
Z
- 0 xL - - D x80=/(,· x -
"'
442
where Z = He ight of paraboloid. r = Radius of parabola.
"'
It
11
,
,
,Z
- xV- x ( 120 - 00)= - 0-)(40= It,· X -
"'
4
4
2
~X 202X40= 4000x1t=1f.rx ~
"'
,
,·xZ=
Using relation
4000xllx2
•
= 8000
... ( i)
.4C
""'';-'CX
,-,-,'_ = -.4C"C"8~'Ox",_'
w' , ! ,wcgc\Z= Z~ -2g
28
2 x98 1
=0.894 ,2
? ~ --'--
0.894
Substituting this value of? in (I). we get
"
_ Z _ xZ= 8000
0.89 4
Z2 = 8000 x 0.894 = 7 152
Z"" J7152 "" H4.56 em . Ans.
lind Method
Lc( Z, = Hcighl of paraboloid, if (hc vcssel wo uld nOi havc bcen
closed at (hc top. corrcspond ing 10 speed,
N"" 400 r.p.lll.
or
ro "" 41.88 radfs
Then
I I
ro
2
R 2
"
I,
"1
I,
•
r - -,--,
.4~L8~8~';'~L~0-_'
Z, "" - - "" 2g
2x98 1
TT
11
120
I,
:--:I- :
:-:-r-:-:
T
"
1
Fig. 5.22
"" 89.34 Clll.
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1204
Flui d Mechanics
Half of Z, will be below the initial hei ght of water in the vessel
i.e..
AO
= 2, '" 89.34 '" 44.67 em
2
2
But heig ht of pa raboloid for dosed vessel
'" CO '" CA + AO= ( 120 - 80) +44.67 em
'" 40 + 44.6 7 '" 84.67 em . An s.
Problem 5.27 For Ihe dll/a gin'lI in Problem 5.26. find Iile speed of rota/ion of Ihe ,'essel. when
tHill1 depth of Waler is zero.
Solution. Given
Diameter of vessel
'" 20 em
:. Radius.
R '" 10 em
Initial height uf wa ter
'" 80 (;111
u:ngth o f vessel
120
= 120cnI
LeI (oj is Ihe angular speed. when axial depth is zero.
When axial dcplh. is zero. the heig ht of paraboloid is 120 em and
radius of the parabola at the tOP of the vessel is r.
,,
:. Using tlie relation.
IT
'11"
, ,
Z:~ or l20=ooxr
2g
2 x980
(fi ,J '" 2 x 980 x 120" 235200
''''
1
Fig. 5.23
... (i)
Vulume of air before rOl<ltion '" Volume o f ai r after paraboloid
IIR! )( (120 - 180) = Volume of paraboloid
=
II(1)(
Z
2
"'
r
1
1 = 11 )(10 )(40)(2
~
""-"'::'C;;;"'-'"
=
r
11)(120
Substituting the value of? in equat ion (i). we get
{Ii )( 66.67 " 235200
to
Speed N is given hy
(oj
35200
66.67
= 2rr.N
60
60 )( 1,1
N 0 ~~211
"'
8000
0
-
120
= 66.67
== 59.4 radls
60)( 59.4
0
"'~"'":: == 567.22 r . p,m . Ans .
211
Problem 5 .28
The q/i",hk(.Il'(:n'c/ of IllI! problem 5.26 is r Olflled {II 700 ,.p.m. ahO"1 ils }'erlieal
luis. Find Ihe area ,mCOl'ued {II Iile bOl/om of Ille lank.
Solullon. Given
Initial height of water
Diameter of vessel
Rad ius.
u:ngth of vessel
I I
" KO cm
" 20 cm
R" 10 em
= 120cnI
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Kinematics of Flow and Ideal Flow
205 1
N", 700 T.p.lll.
Speed,
2rr.N
w ~ - - '"
60
2x
It
X 700
60
'" 73.30 rad/s.
If the tank is nOI closed allhc lOp and also is ve ry long. then th e height of
to w = 7].3 will be
73.3 l
2Xg
X
IO ~
,'--- 1----\, ------.
""
I
':
Froll) Fig. 5.24.
correspo nding
I
.
"
== 274.12 ern
2 x 980
par~bol;\
'.
""
I- r, --.t: '
A K
Of
I
r/
l
'"
x l +x 2 =174.12 - 120= 154.1 2c m . .. {i)
From the parabola. KOM. we ha ve
73.3~ x
2 )(980
I
x,
... (ii)
['or the parabola. LON. We have
73.3~ x rl!
... (iii)
2 )( 980
Fig. 5.24
Now. vo lume o f air before rotation = Volume o f air after rotation
Volume of air before rotation = rrRl x ( 120 - 80) = It X 102 X 40 = 12566.3 em J
Volume of air afta rotation
= Volume of parabolo id KOM - I'olu me o f parabol oi d LON
= ltr,
Equating (il') and
(1') .
l
x
(120+ x, )
l
-1I:r,
.
2
Xi
1
x-
..• (v)
we gel
Substituling th e va lue of 'l2 from (ii) in (I'i) . we get
12566.3 = /t x ( 120 + XI ) x, 2 x 980 x (120 +
2
73,3-
1
.:
xd - -'"O,,'- ,,+;.c:"L'
.
2
1
From (U)"I '"
2X980X ( 12o + x
(733)"
'
)1
12566.3 = 0.573 ( 120 + .l l)" _
Substituting th e va lue of XI from (iii) ill the above eq uati o n
_
21
2
2
_ 0.573 ( 120 + 2,74 '1) - 4,3 x '1 X'l
'" 0.573 [1 20 ' + 2,74 ' ,,4
,:
1 _ 4.3
. + 2 x 120 x 2.74 ,,2
.
.
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1206
Fluid ML>chanics
'" 0.573 114400 + 7.506 ,:4 + 657.6 r/ I - 4.3,:4
~5~66~.3o '" 2 1930 = 14400 + 7.506'24 + 657.6
-~J20.573
r/ (7.506 -
'l
~
-
4.3 ' 14
r/ + 14400 - 2 1930= 0
r/ + 657.6 r/ - 7530", 0
4.3) + 657.6
3.206
"'
-657.6 ± J657.6 1 - 4 X(- 7530) x (3. 206)
2 x 3.206
=_
-~6.~
S7~.6~±~J~4~3~2~
43~7~.7~6~+~9~6~
564
=.7,,
2
6.4 12
= - 657.6 ± 727.32 '" _ 21 5.98 or 10. 87
6.412
Neg at ive va lu e is not possi ble
r/ '"
Area uncovere d at [h e base '"
10.87 ern
Itr/ '"
If
2
x 10.87'" 34.149 <.: rn 2• An s.
Problem 5.29 A closed cylindrical I'esse! of diameter 30 0/1 and II<!iglll 100 em cOll taills waler
upla (I depth of 80 em. The air (,hol'e ,he Imler SlIr/oce is or a presslm~ of 5.88fJ Nkm". Tile l't'!iSe! is
To/rued (II (l speed of 250 T.p.m. u/;JOJl I ils ,'u/jeul luis. Find Ihe pressu re i,e(lil (l/111e bollom of the
I'esse! : (I) (II Ihe u n lre. alld (b)
(l/
the edge.
Solution. G iven:
Diameter of vesse l
:. Radiu s .
T
:: 30 em
R :: 15 em
Initial heig ht of wale r.
H = 80cm
Le ngth of cy linder,
L :: 100 elll
Pressure of a ir above water :: 5.886 N/cm!
"
1
P '" 5.886 x
"'
H ead du e 10
prc~s u rc.
II
= plpg
::
S(lecd .
"'"
5.886 x 10"
IOOOx9.8J
l - 30cm- l
=6 m ofw ater
Fig. 5.25
N = 250 T. p. m.
21tN
w ~ - - ::
60
2/t x 250
60
:: 26. 18 radJs
Le t xI:: He ig h! of parabol o id fo rm ed. if the vesse l is assum ed ope n at the to p and il is vc ry long.
Then we ha ve
'
w-R': : -.26.".1",8-,'"x ",
" :...'
XI:: 2g
2x 981
= 71\.60cl11
... (i)
leI 'I is the radiu s of llie actua l para bo la of hc ig hl .T2
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Kinematics of Flo w and Ideal Flow
Then
207 1
... (ii)
The volume of air before rotat ion
= rcR" (100 - 80) '"
1[
x 152 X 20 = 14 1]7 em}
Volume o f air after rotation'" Volum e of paraboloid EOF
I
= '2
,
x It"
X Xl
But volu me of air before and after rolmion is sa me.
I
,
14137 = zxltr , XX l
Btu from (ii),
1
xl '" 0.35 '1
I
,
,
14137 = 2" x 1If, x 0.35 "
,,4 = 2 x 14137 = 257 14
Il
x 0.35
" '" (25714)"4 '" 12.66 ern
Substituting Ihe value of " in (ii), we gel
Xl = 0.35 X 12.66
Pressun' head at th e bottom o f Ih e H'SS e!
1
",
56. 1 em
(a) At Ihc centre. The pressure head at the centre. i .e.. aI II = Press ure head duc [0 air + 011
[ ": OH=LH - LOI
=6.0+ (HL - LO)
HL = IOOcm =l rn
LO = Xl = 56. 1em = .561 m
= 6.0 + (1.0 - 0561)
(b) Mille edge. i.e .. at
l
= 6.439 m or wu le r . An s.
a = Pressu re head due to air + height of water above G
= 6.0 + AG = 6.0 + (GM + MA)
= 6.0 + HO + 0,786
= 6.0 + 0.439 + 0.786
== 6.0 + (HO + x,)
l'
(
.f ,
= 78.0 cm = 0.786 m}
': HO : Uf -LO : lOO -56.ll
= 43.9 cm = 0,4 39111
== 7.225 In or waleT, AilS.
A closed cylillder of wdills R alld lIeiglll H is compte/ely filled lI"illl lI"alU. II is
rolated abolll its rerlical a.lis lI"ilh a speed ofro radialls/s. Determille tile t01il1 pressllre exerted by
"'iller on Ihe rap (/111/ bottom of IIIe cyti"der.
Problem 5.30
Solution, Given
Radiu s of cyl inder
Heiglll of cylindcr
Angular sp.:cd
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Ii
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1208 Fluid ML>chanics
As llie cylinder is c losed and completely Illl ed with water. Ihe ri sc o f wate r level a1 the ends and
depression of wate r at the ce ntre due \0 rotation of llie vessel. will be preve nted. Thus th e water will
e llen force o n the complete top of Ihe vessel. Also the pressure will be cxcncd a1 th e oonom of Ihe
cylinde r.
Total )' ressure exerted on Ihe lop or cylinder. T he top o f cyl inde r
is in con tact w ith. wata and is in hori zoma l plane. Thc pressure var iation
a1 an y radius in horizontal plane is g iven by eq uation (5.2 1)
1
"
, l
up
PI'-' POO",
,
- = - - = - - " pro r [o.' 1'=CllXrl
dr
r
J
r
Integrating. we gel
0'
Consider an elementary t: ircular rin g of rlldiu s r and width dr 0111he lOp
of Ihe cy linder as shown in Fig. 5.26.
Are a of circul ar ring = 2rt rdr
Force 011 Ihe demclllary ri ng = Intensity o f pressure x Area of ring
"P X 2nrdr
-
""
e
. II ' ·
-
Fig. 5.26
" x 2nrdr.
'" '2p orr'
Total force on the top o f th e e ylinda is obtainL..:l by ink grat in g the above equat ion be tween the
lim its and R.
°
.
i"
i"'
Total force or FT ",
-(f)
r dr
P -r" x2 n rdr= -(f)P ' x2 n
o 2
2
0
... {S.2S)
Total pressure foree on the bollom of cy linder. F /I
= Weight of water in cylinder + total force on the top of cylinder
. .. (5.26)
p" Density of wate r.
Problem 5.31 A closed cylillder of diameter 200 1/11/1 alld hdglll 150 1/11/1 is comple/ely filled WillI
Wilier. C{I/culme rhe 101111 pre.!Sl jre force exerted by Wilier
is rormed aboul irs I'utical axis (1/ 200 r.p.m.
Solut ion. G iven
Dia. of cy linder
" 200 mm = 0.20 m
Radius.
R"O.llIl
He igh t of cylinder.
H" ISO mm = O. IS III
Speed.
N" 2oo r.p.m.
Angular spL'"Cd.
I I
21tN
w = -- =
60
OIl
rhe rap {llId bO/fom of tile cylillder. if it
21t x 200
" 20.94 rad/s
60
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Kinematics o f Flow and Ideal Flow
209 1
TOlal pressure force on Ih e top of Ihe cy linder is g iven by equ ation (5.25)
p
,
4 100 0 ,
4
J'[ x R '" - - )(20.94")( It)( (0.1) '" 34.44 N. An s.
4
4
Now total pressure force on the bottom of the cylinder is given by equation (5.26) as
F T ", -)( ro-)(
F B ", pg)( nR!)( H + Fr
= 1000 x 9.81 )( It x (0.1) 1 x 0.15 + 34.44
'" 46.22 + 34.44 '" 80.66 N. Ail s.
5 . 10.6
Equation of Free Vortex Flow.
For the free vo rtex, from eq uation (5.20), we have
\' x r'" Constanl '" say c
,
"'
\'
"'-,
Substituting the value of \' in equation (5.23), we gCI
dp =
t'
P-
2
r
ilr -
C
1
pg liZ = p)( - , r
X
1
ilr -
r
pg liZ = P X
C
- )
r
ilr -
pg dZ
Consider two poims I and 2 in the fluid having radiu s 'I and'2 from the centml axis respectively as
shown in rig. 5.27. The heights of tlie poill1s from bonom o f the vessel is 'I and Z2'
Integrating the above equation for the points 1 and 2. we get
,
, pel
1
Z
1,r-dp"' J I- - , ) d'- Jpgd
I
[ -"']',
, ,
"'PC" - -2
pc,
-pg [z l - zd " ' -
- 2
"' -fH - vn - pg[Zl - ZI]
'" f [VIl - I"ij- pg [Zl - ZI]
Dividing by
pg. we gel
I
Pl - P,
L
pg
"'
/ +ZI
.E.!... + _I
pg 2g
I,
...(5.27)
Equation (5.27) is Bernoulli' s equation. Helice in case of free vortex
flow. Berlloul[i"s equation is applicable.
Fi g. 5.27
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1210 Fluid ML>chanics
Problem 5.32 In a free cylindrical !'Or/ex jlow. at a point ill 111<' fluid at a radii/.\' of 200 ml/) and
at (/ height of 100 mm. rhe \'e1ocil)' and pre.l.mres are JO mh alld 117.72 kNlm' absolute. Find Ille
pressure at a radi/IS of 400 mm alld III a IIdg/1I of 200 min. Ti,e fluid is air IIa)'ing dell.!!t)' eqllal to
1.24 k,g/m J •
Solution. Al Poin! I : Give n :
RJdiu s.
'I'" 200 mill '" 0.20 m
Heigh t.
ZI'" 100 mm '" 0. 10 III
Velocity.
\', '" 10 m/s
Pressu re.
P, '" 117.72 kN/rn ! '" 11 7.72 x 10 1 N/m"
At Puint 2 :
' 2 = 400 10m '" 0.4 111
Zl=
200 10m '" 0.2 III
1'1 = press ure at point 2
p '" 1.24 I; g/ml
For the frce vortex from equation (5.20), we have
\' X
r = constant or
1'1', '" I'l'l
X ' I lOxO.2
v1 = - - =
'" 5 IIl/s
\' 1
Now usin g equation (5.27) , we ge t
117.72 X [o j
1.24 x9.8 1
"'
+
10 2
2x9.8 1
0.4
"
+ 0. 1= !!1.. +
pg
f!.l.=
pg
"
2 x9.81
+0.2
1
117.72x IOl + 10
+ 0.152
- 0.2
1.24x9,S I
2x9.SI
2x9.8 1
"" 9677.4 + 5.096 + 0. 1 - 1.274 - 0.2 '" 9676.22
Pl "" 9676.22 x pg = 9676.22 x 1.24 x 9.8 1
= 117705 Nfm 2 = 117.705 x 10) Nfl1l 2
= 117,705 kNfml (abs.) = 117.705 kNfm l. " " s.
(8) IDEAL FLOW (POTENTIAL flOW)
to .s . 11
INTRODUCTION
Ideal fluid is a fluid which is incom pressible and inv iscid. Incompressible fluid is 3 fluid for whi ch
density (p) re main s conSlanl. ln viscid fluid is a fluid for which viscosity (1-1) is zero. Hence a fluid for
whic h densilY is constanT and viscos it y is zero. is kn own as an ideal fluid.
The shear s tress is give n by. t = 1-1 dll . Hel1l'e for ideal flu id th e shear Slress will be zero as 1-1 "" 0
dy
for ideal fluid. Also the shear force (w hi ch is equal to s hear stress multiplied by area) will be ze ro in
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Kinematics of Flo w and Ideal Flow
211 1
case of ideal or polclUial flow. The ideal fluids wi ll be mo ving with uniform ve locit y. Alillic fluid
panicles wi ll be movi ng w ith llie s ame veloc ity.
The l'Onccpt of ide al fluid si mplifies the typical mat hem ati ca l anal ysis. £lluids such as Waler and air
hav e low visl:os it y. Alw when the speed of air is apprec iabl y lower than that of sou nd in it. the
rom prcssibili ty is so low Ihat air is assumed to be incompress ibl e. Helice und er certain co ndit ions.
c~rtain rea l fluid S such as wa ler and aiT may bc t,-.,atcd like idea l f lu ids .
.. S. 12
IMPORTANT CASES Of POTENTIAL FLOW
The following arc th e import ant cases of pot ential flow:
(i) Uni form flow,
(ii) Source flow,
(il·) Free-vo rtex flow.
(iii) S in k flow.
( v) Superimposed flow .
.. 5. 1)
UNIFORM flOW
In a uniform fl ow. th e velocit y remains eonstall1. All the fluid panicles are movi ng with th e same
veloci ty. The uniform flow may be :
(i) Par<llieltox·axis
(ii) P<lral le l to y-ax is.
5 . 1) . 1 Uniform Flow Parallel to x - Axi s. Fig. 5.27 ((I) shows th e uniform flow paralld to
...·axis. In a unifo rm flow. the ve locity re main s co nstant. A ll the fluid panicles are movi ng with tlw
same veloc it y.
Fig. 5.27 (a)
•
•
•
•
•
U" Ve locit y which is uniform o r constall1 along x-axis
II and V" Co mpone nts of uniform ve loc ity U along x and y-ax is.
For the uni form flow. parallel w x- axis, thc ve locit y components II and v ~ rc givcn ~s
I I " U and ,. " 0
... (5.28)
But th~ ve loci ty II in term~ of st rea m functio n is give n by.
Let
1/"
o~
Oy
and intcTms o f ve locit y potcntial thc vc locit y II is givcn by.
,= -a,
a.,
<l¢
,= -a.
oy =-a,-
Similarly, il ca n be shown
lh~t
v=_
... (5.29)
o~ "" o¢
,h
oy
.. .(5,29A)
But II = U fro m equatio n (5.28). Substituting u = U in equation (5.29). we hav e
I I
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1212
Fluid Mechanics
u '"
dljl "" <lQ
dy ax
u= -a.
...(5 .30)
#
ity
a nd also U= -
a.I'
Pirs1 pa rI g ives tI\¥ '" U tly whereas seco nd pan gives dip '" U d.\" .
Jm cgrati on of th ese pans gives as
\11 == Uy + C 1 and 9= Ux + C2
where C ! and C 1 arc constant of integrat io n.
Now leI us plm the stream lin es and p01 cn lial lin es fo r un ifonn flo w pilrallc l lo x- ax is.
Pl otti ng or Stream li JlL'!j. For stre am Ji nes, the eq uati on is
Ij/ =Uxy +C t
le I IjI '" O. w h ~ rc Y '" O. Substitutin g lh es.e values in [h,:, above equatio n, we gel
O=U x O+Ctor Ct",O
Hence the equati on of st ream lin es occo mcs as
Ij/ :U. )'
... (5.1 I)
Stream tines
The stre am lines arc straight lines paral le l \0 x-axis and at a
di Sla nce y from th e .I'-axis as show n in Fig . 5 .28. In equat io n
(5. 3 1), U. Y re prese nts the vo lum e flow ra te (i.e., m3/ s) be~ U x4 ~4U
twee n x-a xis and th at stream line al a d islance y.
3
N ull'. The thickness of the fluid stream perpcndkulm to the
y" 3
'1';1. mU x 3 ~ 3U
plnne is assumed to be unity. Thcn y x I or}" represenls thc area of
y~2
~ U x 2~2U
flow . And U. )' represents tne product of velocity and area. Hence
p '
... a U X \EU
U. )" rcpresenls the volume flow rate.
,
,., Ii
,
,
,
,
•
"
y~lJ
.,
•
,
Plotti ng lJf potcntiH I lin es. For pote nti a l lines, th ~ ~quat i on is
Fig. 5.28
1\1 == U . x + C2
... (5 .32)
Lei ~ == O. where.f '" O. Substituting these va lu es in the above equ alion. we gel C2 == O.
Hence equat io n of po te llli a l lines becomes as
~ = U . .t
The above equat ion shows that polc ntialli nes arc slrai ghtlin es parali c I to y· axis and at a di slance o f
.f from y- axis as s how n in r ig . 5.29.
Fi g. 5.30 shows th c plol of strea m lines and polenliallines for uniform flow para li c I to x-ax is. The
Slrca m li nes a nd pote llli a l lin es interseci each other 31 ri g hT angles.
,
,
"•
••
~
"'
•
••
• • "• ~
• • • •
0
,
~
-
V
,
PolenUat lines
,"
5'
• •
• •" •"
o
3
Po1ential
-?1 ~"'"
;;
,
Fig. 5.21
I I
o
•••
•
tream lines
•
e-- 1/, ,.
II
,
V
"
,."
,-,
,
Fig. 5.30
Ii
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Kinematics o f Flow and Idea l Flo w
213 1
S. I 3.2 Uniform Potential flow Puallel to y· Axis. Fig. 5.31 shows the uniform potclUial
flow parallel to y- axis in which U is the uniform velocity along y- ax is.
Fig . 5.3 1
Th" velocity components u, y a long ,(-axis and )~axis are given by
Il=Oand,' =U
... (5.33)
These velocity components in Icnns of strcmn functi on (1jI) and ve locit y polcmial function (9) arc
given as
ow
II" -
.(5 .34 )
,, = _ dl¥=Jq.
Jx dy
>0'
But from equation (5.33),
ilQ
0 -
ay ax
"==
... (5.3 5)
U. Subst ituting
U=_ dlj/ = dQ
dY
ax
I ' '"
or
U in eq uati on (5.35), we gel
U=_dW 3ndaL<;oU=
ax
at
iJy
First p~rl gives iI'lI = - U dx whe reas scl:o nd pan giv~s dq. = U d),.
Inte gration of Ihese pans gives as
1( = - U . x + C 1 and 9 = U. y + C 2
•.. (5.36)
where C 1 and C 2 are constant of integration. Lei us now plot tlte stream lines and potential lin es.
Pl utt lng or S lreum lin es. For s tream lines. tlt e ~q uatjon is \jI '" U .. I' + C 1
Let \jI '" 0. wlt ere x '" 0 , Tlten C 1 '" 0.
Hence tlt~ equatio n of stre am lin es becomes as \jI '" - U.X
... (5 .37)
Tlte allo\'~ equation sltows that stream lines are slTaigltt lines paralle l to y~axis and at a distance of x from
tlte )~ ax i s as shown in Fig. 5.32. Tlte -ve sig n sltows that tlte stream lines are in tlte downward direction.
,
Stream lines
o
•
>
o
., . ,
, ,. , ,. ,
,,, ,
>
;'
;' 5
potenUat
, ,
N
"• "
0
. /,
I
"•
•
•
"•
I
•
•
"
"-
-
Fig. 5.32
I I
~""S
' -0
•
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1214
Fluid Mechanics
Pl otting of Potential lin es. For potentia! lines. the equation is <fl = U.y + C 2
Let q. = 0, wllerc), = O. Then C z = O.
Hence equation of potential lines becomes as 9 = V.)'
... (5 .38 )
The above equatiun shows that potential li nes J rc straight lines parallel 10 .I-axis and at a diswnce of
)' from the .{-axis as shown in Fig. 5.32.
to 5 . 14 SOURCE FLOW
The source flow is the now coming from a point (source) and
mov ing uut r:ldial1y in all dir~tions of a plane at uniform Tmc.
Fig. 5.33 shows a sour(.'C flow in whi<:h the point a is the source
from which the fluid moves radially outward. Tlw strength of a
source is defined as the volume flow rate per unit depth. The unit
of strength of soun;" is m 2ls. It is rcprcsemcd by q.
Let u , = radial veloci ty of flow at a radius r from the source 0
If = volume flow rat¢ per unit d¢pth
r = radius
The radial velocity II, at any radius (is gil'¢n by.
"
.. . (5.39)
", = 21((
o
Fig.S.33
SoIlrr:t{lIJW(Flvwa'way
fro m wll rct")
The above equation shows that with th¢ increase of r. the radial velocity dec re3S<:s. And at a large
distance away from the source. the veloc it y will be approximately equal to zero. The flow is in radial
direction. hence the tangenTial velocity,,& = O.
li:t us now find the equation of stream function and velocity potenTi,tl function for the source flow .
As in this case. /Ie '" O. the equation of stream funct ion and velocity potential function will be obtained
from 11,.
Equation of Stream Function
By definition. the radial velocity and tangential velocity compone nt s in tenns of stream function arc
given by
/I
"m
I aW
=- andll e = -ihv
-
ar
' rae
(See equation (j. llA)]
(Sec equation (5.39) ]
u, = 2!r
_ "" = --.!L
r dO
21(r
d'V= r. -
'/
21((
.d6= -
q
2 11
de
InTegrating the above equatiun w.r.1. 6. we get
Ill '"
..!L x e +
2,
IjI = O. when 6 = O. then C t = O.
Ct. where C t is constant of integration.
Let
Hence the equation of stream function becomes as
I I
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IL
Kinematics of Flow and Ideal Flow
1jI",.iL. e
215 1
.. . (5.40)
2,
In the above equation. q is constant.
The above cqumion shows Ihal stream fUllction is a function of a. For 11 given v~luc of a. the
stream function IjI will be l'()l1stant. And this will be a radial line. The stream lines \:an be ploued by
hav ing different values of Here is taken in rndians.
e.
't'-t
e
Plotting of 5tream lines
"••,
When 9 =O, IjI = O
,
9=45° = _
4
9=90Q=~ ra d "I:ms.
2
31'[.
135" = -
"
.~;'/>_, Stream lines
units
4
IjI '" -
q
21t
radlans,ljI =
-
.
!!..",!1
2
4
q3Jt3q
21t
.- = 4
o· ,
unil~
8
are radial
S 9_0
.
units
The stream lines will be radial lines as shown in ri g. 5.34.
'I' - ~
Equation of Potential Function
Fig . 5.3~
By de fi nition. the radial and tangential components in
terms of velocity funct ion arc given by
source flow .
a. andua = -I ·-a.
It,= But from
~quation (5.39). ", =
Equati ng
th~
[See equation (."i.9A)[
, iJO
"2!r
Strea m lin e for
two values of "" we get
dQ = ...!L or <I¢ = ...!L
ar
2nr
2nr
<If
Integrating the above equation. we get
Jd9 = J...!L.dr
2"
2 f -;-dr
q
1jl=
=
n
q
2,
Potential tir>8S
are cirele
q
[ : 2n
is a eonstant term]
,. 2
log,
r
,. ,
... (5.4 1)
In the above equation. q is constant.
The above equation shows. thatth~ vclocily potential function is
a function of r. For a given value of r. the velocity function 9 will be
C()nstant. H~nce it will he a circle with origin at the source. The
velocity potential lines will bc circles wilh origin at the sourc" as
shown in Fig. 5.35.
Let us now find an expression for the pressure in terms of
radius.
I I
~I _"'"''''>o\:.t
o. / -v
s
Fig. S.l S
'0
Pottwtiallinn for souru.
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1216
Fluid Mechanics
Pressure distribution in a plant source flow
The pressure distri bution in n pl~nc source flow can be obtained with the
tion. Let us assume that the plane of the now is horiwnt,d. In that case the
for two poin ts of now.
LeI p" prc~sure at a (lOint I which is at a radius r from Ihe source al
II,: velocity al point I
Po" pressure al point 2, whirl! is al a large dist ance away from Ihe
be zero at point 2. [Refer \0 equation (5.39)]
Apply ing Bernoulli's equation. we get
(p- Po)
pg
,
p .u,
help uf Bernoulli's equadmum head will be S<11llC
point 1
source. The velocity will
=_ u;
2g
({) - I'o)= - - -
0'
2
But from cqu ation(5.39),
II
q
,
=-2 ltr
Substituting the value of II, in the .(bove equation. we get
=-
..
8~(1
(5.4 2)
In the above cqumion. p and II arc constants.
The abovc cqualiOll shows lhat the pressure is im'ersdy pro)Xlrt iooal to the square of Ihe radius
fro m the source .
... S.15
SINK FLOW
The s i o~ flow is the flow in which fluid mov es rad ially
inwards tow ards a )Xlint where it disappears at a consta nt ratc.
This flow is JUSt opposite tu the source flow. Fig. 5.36 shows
a si nk flow in which the fluid moves radially inwilrds towards
point 0, where it disappea rs at a constant rate. The pattern of
strealll lines and equipotential lines of a sill k flow is the sallie
as thai of a source flow. All the equations derived for a '-Ource
flow shall hold to good for sin~ flow also except that in sin~
flow equations. q is 10 be replaced by (- q).
Problem S.33
Plor Ihe srream lilies for
II
o
uniform flo ... of:
(i) 5 IIIls pllralld to Ihe positi!"e direcIioli of the x-a.>.is alld
Fig . 5.36 Sillk flow
(ii) 10 IIIls pUTIJlIeito Ille pO~·ili>·e direCI;OIl of the }"-(lxis.
(Flow toward centre)
Solution . (i) The stream func tion for a uniform flow parallel to the positive direction of the
.. - a~is is givell by equation (5.31) as
If== Ux)'
The above equation shows that stream lines are straight li nes p<trallel 10 the
from the x-ax is. Here U == 5 IIIls alld hence above equation heroInes as
I I
x-a~is
al a distance )'
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Kinematics of Flow and Id eal Flow
IjI
stream function IjI
For), '" 0.2. stream function IjI
For y'" 0.4. S!Tcam function IjI
["or )' '" 0,
217 1
'" 5)'
'" 0
'" 5 x 0.2 = I unit
'" :; x 0.4 '" 2 unil
The other values of S1rcam function can be obtain ed by substituting the different values of y . The
Slrcam lillcs arc horizontal as shown in Fig. 5.36 «(I).
,
,.",
'+' '" 4
Y E 0.8
,.
....,
\II '" 3
Y E 0.6
y
,.,
z
0.4
,
y: 0.2
,
,.,
p'
,
Fig. 5.36(11)
(ii) The stream funcTion for a uniform flow parallclto the positive direction of the y- axis is given by
equatiun (5.37) as
1jI = -UX .l
The above equation shows thal stream lines arc straight lines parallcl 10 the y·axis .11 a distance
from the y-axis. Here U", 10 mts and hence the ahove equation hecornes as
.l
1jI= - lOxx
The
For
For
For
For
The
stream
negative sign shows Ihm the stream lines arc in
.l" O.
the stream fU"':tlon '¥" 0
.I = 0.1.
the stream function '¥" - I 0 x 0.1
x = 0.2. the stream (un ction '¥ = - lOx 0.2
.' = 0.3. the stream function '¥ = - lOx 0.3
oth~r valu~s o f stream function can be obt ain ed
l i n ~s arc vertical as shown in Fig. 5.36 (h).
,
•
•
..•
•
the downward di roxtion.
= - 1.0 unit
= - 2.0 unit
= - 3.0 unit
by substitutin g the different va lues of .1". The
7
..
•;0
M
o
•
o
O.t
o
..
..
•
..
..
•
•
0.2
0.3
Fig. 5.36 (b)
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1218 Fluid Mechanics
Problem 5.34
D elerllline (lie \'clocily a/Jlow
(1/
radii 0/0.11/1,0.4 1/1 {lin/ 0.8 m. Id,ell Ille water
II source (1/ a slrength of 12 1/1"/.1.
Solution. Given
Strength of so urce. q '" 12 111 '/s
The rad ia l veloci ty II, at any radius r is give n by equa tio n (5 .39) as
is flowing radially QIII\\'ard in a /wrizonlal pia/Ie from
" , '" .!!...2 1(,
Wh en r =
O.l m.
=
12
= 'J,SS m/s. An s.
21txO.2
II, '"
12
'" 4.77 m/s. An s.
lit x 0.4
II,
Wh en r = 0.4 m.
12
= 2.31\ mls. Ail S.
2n x O.8
Problem 5 .35 T wo di,KS (lrc placed in II hQriZ(lnld/ plane, onc OI'a lire OIlier. Tile lI'aler elllUS M
tile centre of the lower disc oml flows radially oil/ward from a source of sirength 0.628 1/1"/.1. The
Wh en r= 0.8 !l1 .
II,
=
pressure, at a radius 50 mm. is ZOO kN/m' . Find:
0) pre;'sure in tN/III' lIl" ",dillS of 500 mm mil/
(ii) .{{rcwo JuneliOIl (1/ anglCl" of 30" tlllt! 60" if \I' = 0 at 0 = 0".
Solution . Gi ve n
So urce Mre ng th .
q = 0.628 m 21s
Pre~surc a t radius 50 mm . P, '" 200 kN/m2 '" 200 x 10) N/m2
( i) Prcs!iIIre III (I radil/J 500 mIn
p, = pressure at rad ius 500 111m
Let
(ur), '" \'eloci ty at radiu s 50 nlill
(u r), '" velocity at radiu s 500 mm
The rad ial ve loci ty at any radius r is give n hy equat ion (5.39) as
II, '"
....!1....-
2"
When
r = 50 mill '" 0.05 111.
(Ur), =
When
r = 500 111111 = 0.5 Ill.
(urh '"
0.628
= 1.998 IIl/s '" 2 IIl /s
21t X 0.05
0.628
21t xO.5
'" 0.2 IIl/s
Apply ing Berno ull i's equati on at radiu s 0.05 III and at radiu s 0.5 III.
P,+(u,)~
P' +(u,):
- - = _.
-_
.
pg
2g
pg
2g
P,
-
I I
(u,)~ _
p,
(u,):
+ ---- + -p
2
P
2
Ii
~ I
IL
Kinematics of Flow and Id eal Flow
=2()()'7,;'~'~0,-' + _2 1
'"
2
]()oo
21 9 1
_1'_'_ + _0,_2_'
1000
2
p,
200 + 2= - '- +0.Q2
1000
1 ~ '" 202 - 0.02 = 201.98
p, = 201.98 x 1000 N/ml
(ii) StrewlI fUllctiol1,~ at
= 201.98 kN/m l . Ans.
0", 30" Gm! 0", 60"
For the SOUT(;C flow. the equation of stream functi on is given by equation (5.40) as
\11 '"
.1....0. where e is in
2,
radians
0.628 30)(
.' --,-lit
180
e '" 300 '" ]0)( It
it
radians)
ISO
'" 0.052J m l/s. AilS.
0.628
6011
21l
180
- - , - - '" 0. 1046
.. 5. 16
1
111
Is. Am .
fREE · VORTEX flOW
Frcc·vo ... c~ flow is a <:ircul:llory flow of a fluid suc h Ihm its stream lines arc conccmric <:ircles.
For a free-vortex flow, u& x r '" constalll (say C)
Also. circulation around a Slrealll Ii"c of an irrolalion vonc~ is
r",2rrrxue=lnxC
(":rx"e=C)
where ue '" tangential velocity at any radius r from the centre.
r
jje=-27fr
The circulation r is takcn positivc if the frec vortex is anticlockwisc.
ror a frcc -vortex flow. thc vclocity cumponents are
r
lie = - 2"
and
1/,
=0
Equation of Stream Function
By definitiun. the stream function is given by
-a.
"e=-a,
and
I ",
" , =de
--
(See equ ati on (5. I2AlJ
In casc of freC-I'ortel flow_the radi al velocity (u,) is zero. Hence equation of s tream fum:tion will
be obtained from t:lngential velocity. I/e' The value of lie is given by
r
I/e = - -
2"
Equating the two values of "e- we get
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1220 Fluid Mechanics
d.
r
- -- =
d,
2 lt r
or
d'¥ = - -r- d,
2"
Integrating the above equation. we get
-
"'
r .IS a consta nt (e nn )
2,
The above equat ion shows 1li al stream function is a fUliction o f
radius. For a given value of r. the stream function is constant. Hence
the stream lines arc concentric circles as shown in Fig. 5.37.
Equ a llon of p otential fun ction. By definition. the potent ial
function is given by.
I
u~=
Here
II,
=
d,
--
d e
0 and
and
... (5.·B )
, -"'" Stream
lin<!s
[Sec equation (S.9A)[
r
ue = - - , Hence. the cquntion of polcnlial
2"
Fig. 5.37
function will be obwincd from ue'
Equaling Ihe two values of us- we gel
~ d$ = r
d~
ae
r
2 tt r
Integrating the above equation. we get
fd¢= f;ll d9
or
= r.
r
2"
r
.<19= 2,
de
!II = ~ J d9 = ~. e
2n
21l
... (5.4.f)
The aho ve equatio n shows thai velocity potenti al function is a function of 9. for a g iven value or9.
potential function is a constant. Hence equipotenti al lines are radial as shown in Fig. 5.38.
Veloci!y pote!1t~t
tines
~
•
••
•
Y-
o '
,.,
•
Fig. 5.38
~ I
POlenlialli"~1 ar~
radial.
I~
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Kinematics of Flow and Id eal Flow
.. 5. 11
221
I
SUPER-IMPOSED FLOW
The flow pancms due to uniform flow. a source tlow. a sink. flow and a free vortex flow can be
supe r-imposed in any lin ear combination to ge\ a resultant flow whic h close ly resembles th e fl ow
around bodies. The rcsultan! flow will st ill be potential and ideal. The following arc the importan t
supcr-im(X>scd flow:
(i) Source and sin k pai r
(ii) Doublet (specia l case of source and sin k combination)
(iii) A pl;lnc source in a uniform flow (flow past a half body)
(i,') A soun:c and sink pair in a unifonn Flow
(I') A doublet in a uniform flow.
5.1 7. 1 Source and Sink Pair. Fig. 5.39 shows a source and a s ink of s tre ngt h q and (- q)
placed at A and lJ respectively a1 equa l distance from tile point 0 on the x-axis. TtIUS tile source and
s ink are plal'Cd symmetrically o n the x-axis. The sou rce of s tre ngth q is placed at A and sin k of
strength (- q) is placed al 8. The combination of the source and the si nk would result in a !lownet
whe re stream lin es will be circ ular arcs staM ing from point A and end ing at poi nt IJ as s ho wn in
Pig. 5.40.
A
o
/1-1-'--- . ------II-- ' ------I
Source (q)
Fig. 5.39
Sou ~u alld sillk pair.
Stream
line"
,
~-~ Sink
Fig. 5.40
StrMm lillo for lvuree-lillk pair.
Equation of stream function and po tential fun ction
Ld P be any point in the resultant !lownci of source and sink as shown in Fig. 5.41.
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1222
Fluid Mechanics
e
= Cylindri cal co-ordinates of point P wilh rcs(lCct 10 origi n 0
x.Y'" Corresponding co·ordinates of point P
',_ 8 , " Position of point P with respect to sourcr placed at A
'2_ 6 2 '" Position of point P Wilh respect to sink placed a1 n
a", Angle subtcndcd at P by the join of source and sink i.e .. angle APE.
LeI us find th e equation for the resultant stream function and velocity potential function. The
leI
r,
equation for stream function due \0 source is given by equation (5.40) as III , '" If· 8 , wllcrcas due 10
2,
sink it is given by IV, '"
-
(- qe 2 l. The equation for resultant stream function
2,
(\II) will be the sum of
these twO stream function.
. U
~
[
...
- q.U
... (5 ...1 5)
The eq ua tion for potential function due 10 source is given hy equation (5.41) as 4>,
'" .!L log " , and
2,
due to sin~ il is g iven as 4', = - q log,r,. The equation for re~ultanl potential function (9) will be the
- l /t
sum of these two potential function.
4'=4'1+4',
(-q)
=.!L log, r l +
log, r 2
l/t
l /t
I I
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Kinematics of Flow and Ideal Flow
'" ..!L
21(
110&,' 1 - log,'21""
.!L log ,
(.'L)
223 1
...(5 .46)
21 t ,l.
To pron' that n'Sult lml strt'!lnl lines will be circular arc pass ing through source and s ink
The resultant stre am fUllcti o n is give n by equ ati on (5.45) as
-q.a
.~ -­
"
For a given stream lin e l¥ '" co nstant. In the above eq uati on tile te rm .!L is al so constant. T liis
"
mea ns th ai (9 2 - e,l Of ang le 0: will also be constant fo r vario us positio ns o f P in the pl ane.
To s,uisfy ttJi s. th e locus of P mu st be a (;ircJc with AB as chord . hav ing its ce ntre on y- ax is. as
shown in Fi g. 5 .40.
Consider th e cquali on (5.45) ag ain as
- q
21\:
-q
(EI1 - EI,)
2/f
. ~ - "~ -
v
= ...2...(6 1 - O
(e, -a!)=
"
2 lTIjI
q
Tak ing tange nt 10 both s ides. we ge t
1= lan (2:1tf )
HIn (9 1 - 9 2
B",
tan 9 1 '"
-
and
'-
x +a
Suhstilut ing Ih e values o f Ian 9 1 and Ian
- )-' - - - '(x+ a)
(.I· - a)
= Ian
1+ -' - .--'(.1+ (1) (X- II )
9~
tan 9,
or
-
tan
0;
,".( i)
1+l an9 , .1an8.
ta n 9 , = - Y• x- a
... (S A M )
in equ al ion (i).
(2'.)
,
q
y (x-o))' (.1'+0) = Ian ( " . )
,
l
x'
a + .vl
,
,: I
"
0'
,"0 '
I-- ,
x·, - a"" +
I I
y- = -
2a)" COl
q
(".)
---t=.:..:'-=1,__
F ig. 5.41 (d )
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1224
Fluid Mechanics
, "
.\--a + y" + 2(1},l:ol
.l!+ /
(2"'
q 1')
=0
COl (2:W) -(/ = 0
,[2+ / + 2ay COl(2:'V ) + al COIl (2:"') _al
+
2a)'
CO,2
(2:1JI ) _al= 0
[ Adding and subt ract ing I' ! cot! (2:1¥)]
... (5.4 7)
The above is the eq uat ion of a cin:lc· with centre on y·axis M a distance of± a COl (2:"' ) from
Ih e origi n. The radius of the cin:lc will be a cusc<: (2:'41 ).
Sim ilarly, il Can be show n th al Ihe poten tial lin es for Ihe sou rc e-si nk pair will bc eccentric nonimcrse(;,i ng cirdcs with th eir centres 011 the x- axis as shuw n in Fig. 5.41 (b).
PoIent;alline.
Sink
Fig. SAl (b)
Polentia/linn for SQuru sink pair (potentia/lina are «untric
non-inlerlt'Ct ing circlN with Ibrir t:t'ntrn on x-axis).
°The equation.C + y' ~
I I
(II
is the equation of a c;rcle wilh centre a1 origin and of radius ',,-,
Ii
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Kinematics of Flow and Id eal Flow
2251
Problem 5.36 A source atilt II sillk of slrcng lll 4 m!ls and 8 m!h are IOC{/led at (- I, 0) lind ( I. 0)
respeC/ire/y. D<'lermillt' tile I'e/aeily and stream fllllNion (I{ II paiM P ( I. I j which is lying 0/1 Ihe
jlulI'llei of Ille resu l/mil stream lille.
Solution. Gil'en
Source Sl re ng lh.
'I,
= 4 In l{s
'I, = 8 m 2{s
Distance of Ihe source and sink from origin. (I = I unit.
The position o f Ihe source. s ink and point P in the flow field is s hown in Fig. 5 .42.
From Fig. S.42. ;1 is dear that angle 6 2 will be 900 and angle 8 1 can be ca kulatcd fro m right angled
tri:lIlglc ABP.
Sink strength.
The eq ua tion for stream func tion duc to sou rce is g ive n by equ ati o n (5.40) as
. k·
w h.:reas ...' lIe to Sin
' II ·IS g .i ven by 1jI ,
-=
-'I'- x e,• . The rcsultam
,-,
Mrc~ m
fUllction
IjI
1jI,
= ", x 0 , .
is givell
2,
~s
,
p (• . y)
(1 , 1)
,
Fig. 5.42
+(- l/l X6 l )= l/, X6,
21t
21t
... ( i)
Le i us find llie values o f 6 , and 6 2 in radian s. Fro m th e geomelry. it is cl ear tlim llie traingle ABP is
a rig hl angl ed Irian gle wilh angle 0, '" 90° '" 90 x
•
180
AI<;o
It"
~
2
radian s.
liP
1
Ian Ell :: - : : - " 0.5
2
AB
,
8 1 :: tan
,
0.5" 26.56° " 26.56 x 180 radians" 0..463
Subslituting Ih ese values in eq uati on (i).
1jI "~ x0.463 - {Il x~
21t
21t 2
I I
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IL
1226
Fluid Mechanics
8 ,
xO.463 - - x l it
In 2
' -
,
= 0.294 - 2.0 = - 1.706 m !/s. Ans.
To lind the veloc it y m th e poin t P. let us tirst fi nd the stream function in terms of
o rd inates. TIle stream function in terms of 9 1 and 9 2 is giv,, " by equ ati on (i) above as
.l
and y
(.X).
1f= rJ,x9 1 _(h X9 l
211
21'(
The values of 9 1 and 9 2 in terms of x. y and a arc give n by equation (5 .4611) as
tan 9 1 =
- '-
9 1 ""
13n-
"'
Substituting th ese
va lu c~
1
- Y-
(.[ -a )
and
H'
- ,-
9 2 ", tan
(x - a )
of 9 1 and 9 2 in eq uat ion (iI, we gel
, _ Y_ _ q1 [an 1_ '_'_
1jI =.!l.!.. t:l n
2,
The velocity com ponen t
Ian 9 2 = - y-
and
x +a
,, -a.0,
/I
=
\" + a
~nd
lit
x- a
I' = _d\v .
"
d",
0,
=
1an
~
or [!!J...
21l
q
'"
2~ x
, _ ,__ (h tan
x + a 21(
1_'_]
.t -
{l
q.
I
x +a
Ifl
I
[
(Y J' x -( _,- +-,-) - -2-', x 1+ (Y
J' x -(,- - -,-)
1+ - -x -a
(x + ,, )2
I
if'
(x -a )!
, -,-, '(.-, '+-0")"+
~,,, x -(_'-+-0- ) - -,-; x 7(_,~-oC),C+~y2'X -(,- _- ,-,)
(x+a)
(h
(.t -a )
= 2; (.\ +(1)1 + / - 21l (x_all + vl
q,
At th e poilll P( 1. I). the componelll "is obtained by substituting .f = I and )' = I in th e above
equa ti on. Th e value of" is also ,;,qual to o ne.
q,
u= -
I+I
2 lt(1 + 1) ;: +1 ~
I I
(12
(I - I )
2ft(l _ 1)"+ l l
Ii
~ I
IL
Kinematics o f Flo w and Ideal Flow
Now
I' '" _
2:1.7 1
iilv
ax
'" - -
, [q,
-
;:h 21t
=_ [!!J..
- - - - " lall
21t
x +a
1
y
--
-I'-a
, x y(- I), Xl _iL X
I
l it 1+
q, -,
_I)'
""
(_Y_)"
(.I + af
2 1t
x+a
, x Y(- I), XI ]
I + (~ )- (.\ - a)"
I
x
a
(12
Y
2 n (.f _a)l + /
At th e point P( I , I).
'I,
I
til ~_',-_~
I' " -2n x (I + III + I! - -21t x (I _ 1)1 + ]~
(': a= I)
'I'
I
= -q, x -J - -"
x-
2n
5
l it
1
= _, _, x .!.. - _q_l = _4_ x .!. _ _8_ = 0.1272 _ 1.272 = _ L 145 mlsl
21t 5 2n 21t 5 21t
T he n:sulta n! vc IOl.'it y, V =
Ju 1 + v" _ ~rOc2c5c44:c;,-+-(C_-IC,C4C,::T)1 = I 174 m/ s. AilS.
For Ihe abo,'" problem, de/ami" " Ihe preswre al P( I, I) if Ihe pressure al infinity
is zero Wid dellsily o/fluid is 1000 kg/m J.
Problem 5.37
Solut ion. Given:
Pressure at in fini ty,
Po = 0
Dcn~ily of fluid ,
p = 1000 kgfm 1
The ve loc ity· o f fl uid aI infi nity w ill be zero, If Vo = ve loci ty at infi ni t y. thcn Vo = 0,
The result ant veloc it y of fluid at P( I , I) = 1.174 mfs (c alc ul akd above)
or
V= 1. 174 m/s.
LCl p = pressure al P( I . I )
A pp ly in g Be rno ull i's tlleorem at point m infi ni ty and m poin t P, we get
Po
Vo'
P
V"
-+= -+pg 2g
pg 2g
0'
0'
V'
0 + 0= L+ _
pg
2g
or
~= _ V' =_ 1.174 '
P
2
-------'--
p
V '-
pg
2g
0= - + -
or
p
V'
2
0= - + p
(':
V", 1.1 74 mls)
2
* From equation (5,39), the velocity at a distance 'r' from sourcc or sink is g;"cn by u," ~ , At infinity,
2lfr
r is very very !:\fge hene" velocity is Zero,
I I
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1228
Fluid Mechanics
1.1 74'
1.174 l x 1000
= _ 689. 14 N/ml. A il S.
2
S. 17 .2 Doublet. It is a spec ial case of a source anti sin k pair (bo th of them arc o f equ al strength)
when the 1WO approach each other in suc h a way that the "islancc 2a between them approac hes ze ro
and the product 2a . II remains constant. This product 2(1 . q is kno wn as doubl e! strength and is
P= - - ,-
deno ted by
x p =-
1-1.
Doublet strength.
Jl '" 2a . "
... {5 .4 8)
Let q and (- q) may be th e slrenglh of the source and th e si nk respectively as show n ill Fig. 5.43.
leI 2(1 be the distance between 1hel11 aud P be any poin t in Ihe coulbincd field of so urce and sin k.
Fig . 5A)
Let e is the angle made by P al A whereas (9 +
Now the stream funct ion at P.
Ifa
Be) is the angle at B.
q
If
(9 + 00) ': - - 39
... ( 5 .4 9)
21t 21t
21t
From B. draw Be J. on AP. Le I AC '" or. CP = rand AP = r + or. Also angle HPC: 09. Tile angl~
is very sma ll. Tile distance BC can Ix: taken equal to r x 00. In triangle ABC. angle BCA = 90 0 and
h~nce distanc~ BC is al so equal to 2<1 . sin G. Eq u ~ting the two values of Be. we get
• = -
- -
oa
fxOO=2<1.s inG
SG= 2a.si nG
,
Substituting the va lue of
oa in equation (5.49). we get
q
l it
21t sin
J.l
si n
2,
,
W': - - x
e
r
e
= -- x --
[.,'
2n. q = J.l
from equa tio n (5.48)1 ... (5.50)
In Fi g. 5A3, when 2(1 -t O. th e a ngle 09 sub tendcd by point P with A and B becomes very small .
Also or --+ 0 and AP hecornes ~qua l to r. Then
s inG: PD =2'.
AP
,
Apl '" A02 + P02 or r2", .( 2 +
Also
Substituting the va lue of sin G in equation (550). we get
I I
)'2
Ii
~ I
IL
Kinematics of Flow and Ideal Flow
\jI :
~
)'
I
I-iy
--X-X- = - - ,= 21t r
r
21lr-
I-'Y
"
2rr (_t- + y'
229 1
)
... (S.SOA)
.~+l=-....!2....
2,.
The above equati on can be written as
.r2+i+2X YX-"+[-")' -(_
, _)2=0
41t1jl
41t'l'
41f\jl
or
.t
[Addi ng and subtra<:ting
1+ (Y + - "-)' = ( - "- )'
41t1t/
(4~\I r]
... (5,51)
4ltlj1
The above is Ihe equmion of a circle with centre (0. - " - ) ,md radius - "- . The centre o f Ihe
41t1jJ
circle lkson
41t'll
y -a~is at a distance of - " - from x-ax is. As Ihe radius of the circle is also equal 10 - "- .
41l1j1
41l0/
hence the circk will be tangent to the x-axis. Hence str.:am lines of Ihe doublet will be the fam ily of
circles tangen l to (he x- axis as shown in Fig. 5.44.
Stream
~"""
are circles
tangent 10 . -aX;S wittl
centre on y-axiS
Stream lines
y lines
Fig. 5.44
Potent i ~1
fu" c:t ion at P
Refer to Fig. 5.43.
Th ~
potential function at P is give n by
$=
I I
SITI'am /inff for II doub/fl .
;~ IOg , (r+or)+(-;~) log ,r
[Refer to equation (5. 41 )]
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1230
Fluid Mechanics
q Iog '= q '0,
: - q Jog (r +or) - l it
'
l it
'
:'L[" +("]' X~+ ...l
211
r
",..!L
or
2,
l it
'
(<+,,]
q Iog (1+
"]
-- : -'
r
l it
r
2
r
[ "
.
("]'
As ---; IS a sllIall quamu y. Hence ---;
,
'
BUI in fi g. 5.43. from triangle ABC, we gel
or",
Or '" cos e
2"
e
2" cos
Substituting lhe value of 15r, we gel
. 1
becomes neglig ible
OP '" ...!L )( 2" cos 9
2,
,
Jl
cos e
: - X- 11l
r
1-: 2{/
x q '" Jl from equation (ill ... (5.52)
oe
In Fig. 5.43. when 2a ~ O. Ihe angle
becomes very sma ll.
Also or ....... 0 and AP becomes equal to r. Then
AD
.r
AI'
r
cos9= - - : -
l
Also Ap l '" AD2 + pD l or ? '" _~ +
Substituting [he valu e o f cos 9 in equatio n (5.52), we gel
41= L
x (~]x!=J:..x
-~
21[
r
r 2)[ , -
.~ +l", Lx~
lit 0;.
The above equation c;m be wrinen ~
~- ilt i+( 4~r -(4~~r +l= o
or
(.I" - 4~4>
r
+i =
(4~9
r
The above is Ihe equation of a cin:]" wilh cenlre
circle lies on x-axis at a diM3nce of
l
4,.
(-"-.0]
4mp
[Addingand subtracting(4~9r]
.
(5.53)
and radius ( -'-'- ]. T he centre of the
41'(4)
frotH y-a xis. As the radius of the circic is equal to the
distanw of the cent re of the circle from the y-axis, henc e the cirdc will be tangent to the y-axis.
*
I I
Expansion of log, ( I + xl '" x +
2"""
" +,
Ii
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Kinematics of Flow and Ideal Flow
231
I
Hence tile polclUiallincs of a doub let wi ll be a family of ci rcles tange nt to the y- axis with their centres
on the x-axis as sliown in Fig. 5.4'5.
Potential lio&S
POlentia l lines
Of
, lines are ardes
w ith centre on x_axis
£.::::~Q'-'""
:; 1 13og,,01 10 y-axis
Problem 5.38
Fig . 5. 45 Potenlial linn fo r a doublet.
A paim P(O.5, I) is SilUaled ill II,e flow field of
II doubler of streng/II 5 mlls.
CII/cu/ale IIII' velocity al Iilis point and "/50 Ihe "olue of Ihe slreom fu nction .
Solution. Given: Point 1'(0.5. 1). This means of = 0.5 ,md y == 1.0
Strength of doublet.
J.I " 5 111 2/s
(i) VelOcify at point P
The velocity at the given point Can be oblain"d if we know the stream funclion (1jI). But stream
function is givell by equation (5.50A) as
1jI=- ~ X
21l
The velocity componc llts
1/
y
(Xl + /)
and I' arc obtained from the st ream funct ion as
d,, [-2;X(X2+l)
" ,]
u=ay-=(1),
"-:, ;y[(>'~r')]
~ is a consw nt ternl)
2,
"-;, [(:,':)~>l
[ ... ~'[!' ('<'+r') }y[-Jj[.<'+"'l ' [2y] +(>' +,.' )
_ 2),2
(.ol +l) ,
I I
.I
1
+ (X-'
+)' ')
-
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~ I
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1232
Fluid Mechanics
= irr ;, [i" ~ y' )1=irr [ix;:~;)' 1
Su bstitutin g the va lu es o f )l = 5 m'ls, x" 0 .5 and y" 1.0, we ge t tile veloc ity co mpone nts as
u= _L[ (XlXl +-if
i'l = _ ~[ O.5l -ll'1= _ 2- 0.7~
+ IT
2 J't
and
2 1t (0.5'
= - 0.382
2It 1.25"
\'" ;n [(x::"/ )!]'" n[(O~5~ :'~l)l[ ] '" ~[~] ,,- 0.509
5
2
Res ultant ve loc ity,
V=
Ju
1
- ~(- 0.382 i
+ v"
+ ( - O.509 )" = O.63fi m/s. ADS.
( ii) Value of strewn lime/ion m poilll P
,
I
l it
1.25
-- x == _ 0 .636 m1/s. A DS.
Solution in polar co-ordinates
e
The above qu estion ca n also be done in r. (i.e.. polar ) co-ord inalCs. The s tream func tion in r. 9
co-ord inates is give n by equatio n (550) as
)l
si n 9
1jI =
-- x --
2rr
--.(i)
,
and vel oc it y contlx mc nt s in radial and tangC llli a l di rec ti o ns arc given as
" : ~ x ,)IjI= ~ ~ [_ LSi lle]
,
ae rae
r
'" ..!. x
r
,,'
21t
r
(_L] x ..!.~(sin
9)
J9
lit
r
[ .: L2rr
is a constant te rm and also r is conSlanl w. r. t. e]
=- -x....,
"
I cos e
21t r·
... (Ii)
"" '[" ';0 ']
ue : - - : - - - - - ar ar 21t r
= _ ( _ _11
21t
sin e]~ [~]= ~ sine (-I ) . -
ar '
21t
, !
[.:
~
sin e
21t
,.
= - - ' -,-
I I
~~i;eiSaCOnSlanl w.r. I. ']
... (iii)
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Kinematics of Flow and Ideal Flow
233 1
1.",5
~.f " + Y _ ~0.5 " +1 _",""
y '
894 and cos , x
O.5
sin , =-:~=O.
=-:~= 0 .44 7
Now
r=
r
Substituti ng the values of
r,
r
,,1.25
"1.25
a and cos e in above cqumions (i), (ii) and (iii), we gel
J.I sin e
5 0.894
z
' l I = - - - - = - - x r.= =- O.636 m Is. Ans.
sin
21t
r
J.I
21t
I
21t
If,'" - - x ----,- x cos
,"
,, 1.25
e '" - - 5
I
x - - x 0.447 = - 0.2845 IlI/s
21t (1.25)
J.I
sin 9
5
0.894
lie'" - - x -,- =- - x - - = - 0.569 rnls
lUlU
21t
r
211
1.25
Resu ltant velocity.
= J(-Q.2845)! + (-0.569)2 = 0.636
S. 11.3
m'.~.
AilS.
A Plane Source in a Uniform flow (flow Past a Half-Body), Fig. 5.46 (d) shows
a unifonn flow of velocity U and Fig. 5.46 (b) shows a source fluw of strength q. When this unifonn
flow is flowing over the sourc<; flow , a rcsuhant flow will b<: obtained as shown in Fig. 5.46. This
resultant flow is also known as the fluw past a Ilalf-body. Let the source is placed on The ori~in o.
Consider II point P(.\". y) lying in The resultant now field with pol~r co-ordin<lfes r <lnd a as shown in
Fig. 5.46.
,
,
+
•
(al Uniform !low
'01
(b) Soufce now
Point P(x . y)
(f. &)
Hall body
,
"
fG
~
\
( \&
0
,'
",
~
,"
,
0
, . 0
1--'. _
Fig.5.4-6
I I
FlO'W pall"n reuliling from Iht combinalir)n of a uniform flow and a $Our("t.
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1234
Fluid Mechanics
Th e stream fUllction (\V) and pot ential fllllCiion (IJI) for the rc suhant flo w arc obtained as give n
belo w :
It' " Stream func ti on du c \0 uniform flow + stream func tio n du c to source
= u . y + .1... e
2,
... ( 5 .54 )
=U. r sin9+!Le
(":
2,
and
9" Vdocity po te ntial
y =
r
sin Ell ... (554A)
function due to unifonn flow + Ve locity potcr)[i al function due to source
= U . .1" + - '- Iog , r= U . rcos 9 +.!L log , '
21(
... (5.5-1.8 )
211
The follow in g arc Ihe importa nt po ints for tile resultant flow panc rn :
(i) SraglUuiQlI poi/I I. On th e lefl side of the source. at the point S lying o n the x- axis , th e ve loc ity o f
unifo rn! flow and Iha! due to source are equal and opposite 10 each othe r.
H~ncc
the ne t ve locit y o f the
combined fl ow fi eld is ze ro . Thi s point i s know n as sta gnation point and is dcnOlcd by S. The polar co-
ord inat es o f the stagnatio n point S arc 's and It, where 's is radial diSl31lCC of point S from O.
The nd veloc it y (or resultant veloc ity) is zero at the stag nation point S.
", =
=
(v.
.!La)
+.!L] v. ...!i.....-
..!. altl =.!. ~
r aa
..!.
[v.r
r
r
ae
sin a +
21t
=
cos a +
21t
2 J(r
At the s tag nati on point. f) = II radian s ( 180G ) and f = fsand net ve locit y is zero. This mean s u, = 0
and 1'& = O. Substituting these values ill the abo ve equati on. we get
r
cos a
0= V. cos 180° + - ' 2J(f S
=- v+ - ' -
"'
2J(rs
q
's=2,U
...(5 .55)
From th e abo ve equation it is d e ar that positio n of st agnati on point depends upon th e f ree stream
ve loci ty V and source s trength 1/. Al th e Slngnatio n point. the value o f strea m fundi on is obtained
fr011l equat ion (5.54A) a~
If= V .fsina+ .!L. 8
2,
For th e stagnatio n point. the above equation becomes as
ItI s = V .T, si n ] 80 0
( .:
. 0,1
2 = 12
+.!L2,
x 8
At sta gnation point. 8 '"
It
radian s = 180 0 and r '" fs l
... ( <<6)
~ ..'
The abo ve re lat io ll g ives the equatio n o f stre am line passin g throu gh sta gnation point. We kn ow
that no fluid mass crosses a stream line . He nce a stre am line is a I'irlua l solid surface.
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Kinematics of Flow and Ideal Flow
235 1
(ii) Shape of resul/atll flow. 1\( tlie stagnation point S. the net ve loci ty is zero. The fluid panic les
lhal isslie from tlie source ca nn ot proceed further to the left of stagnation poin!. They arc carried
along the contour BSB' Ihm separates the source flow from uniform flo w. The curve BSB' ca n be
regarded as ttte so lid bound a r y or a rou nd nosed body suc h. as a bridge pie r arou nd wll ic h th e
uniform flow i ~ forced to pass. The contour RSB' is ca ll ed the half body. bcc auS<) il ha s o nl y th e
leading point. it trails [() infinity al down "tream e nd.
The value of stream fUll ction of the strc~m line passing through stilgll,Hion poin t Sand p<lssing over
the so lid boundary (i.e .. curve 858') is ltIs'"
!l.
2
Thus the composite fluw cunsists of :
(1) flow over a plalle half-body (i.e., flow over curve BSB')
out~idc
IjI
= '1 alld
2
(2) source flow wi thill the plane half-body.
The p lane half·body is described by the dividillg stream lille,
IjI
=
%.
But the stream function at allY poi nt ill the combined flow field is !!ivell by equatioll (5.54) as
IjI=U.y+..!L{}
If we take
IjI
=
"
%in th e above equation. we will!!et the equalioll of the divid illg stream line.
EquatiOIl o f the dividing stream line (i.e .. equation of curve 8SB') will be
'1 = U
2
. Y+
.!L . {} or
2lt
U . Y = 'i
2
_.!L
2lt
6
=
" "--(I
2U -Ill
"'
... (5.57)
It
From tlie above eq uation, tlie main dimensions of tlie plane lialf·body may be obt ain ed. From tliis
eq uation. it is clear that y is maximu m. when 6", O.
Hence At 6 '" O.
Y is maximum and y ...., '" 2(~
y=
At6=lt,
At
6
3lt
=2'
y
2(~ ( 1 -
= -.!L
2U
T' ~) '" 4~
(I- ~l '" 0
It
)' = 2(~ (I- ~: ) =- 4~
... tlie maximum ordinate
... tlie ordi nate above tlie orig in
... tlie !cadiug po int of the half-bod y
... the ordinate be low the origill.
The main dimensions arc shown ill Fig. 5.47.
(iii) Re.m ltll/!/ !"dodt)' III Ilny poi/lt
Tlie vclocity components at any point in the flow field are givcll by
iI,=
I I
~ dljl = ~ ~[u.rsine +.!Lel
r de
r dfJ
2lt
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1236
Fluid Mechani cs
==
.!.[u.rcos a +.2....] '" V.cos e +..!!...r
2lt
Leading
peOn!
2M
-"-
Y_ -- 21)
-"-
'"
Origin
Fig. SA7
The above equati on gives the radial veloc it y at any point in the now field. Tllis radi al ve locity is due
,,,
tn uniform flow and due \0 sou rce. Due \0 source the radial veloc ity is~. lience the velocity due
10 source dimillishcs wilh im:rease in radial distance from the source. At lar~c distance from the
source the contribution of soure", is negligible ,md hence free stream uniform flow is nOl influenced
by Ihe prescncc of source.
'" - IV. sin
Rcsul1aol velocity,
V
e + 01 '" -
U si n
.2.... e is cO lman! w. f . \.
e
2.
r]
= ~u; + uJ
(il') Locatioll of swgnafion poilll
At the stagnati on point. the velocity compone nt s are lero. Hence equating the radial and tan gent illl
velocily componen ts to zero, we gel
or
11,=0
,.U
0'
rl:os e = - - q-
,.,
ucuse + ~=o
BU1
0'
fl'OSe=X
x= - -q2nU
Wh~n
"6= 0 or
e = 0 or
-Usine=O
Hence Stagna tion point is at
I I
or
nBu1 y=rsine
(- ~. 0). the
,.U
e :: 0
:. y'" 0
si n
as U canno t be zero
leading point of the hal f-body.
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Kinematics of Flow and Ideal Flow
237 1
(I') Pressure ar (IllY point ill flow field
Let
Po" prc~so re at infinity where velocity is U
p == pressore at any point P in the fiow field, where velocity is V
Now applying the Bernoulli>s equation at a point at infinity and at a point P in the now field. we gel
1
Po
U
P8
2g
- +-
V~
28
P
: - +pg
"'
_v _' _ _V_1 =.l!....._!!..2...", P-Po
pg
2, 2, pg pg
The press ure l"()-cm"icnt is defined as
c=P-Po
,
1
"2
PU
,
V' V']
[
: ----'T"--:c'''-''
P8
-
Zg
2g
!pu 1
2
... (:'i.58)
Problem 5.39 A III1i/orm flow .... irh a I'e/ocity of J mh is flowing ora a p/(lIIe source of slrell8111
30 m!k Til<' uniform jlow lind $Ollrce flow are in IIIC sallie plwle. A palm P is siwaled i/I the flow
field. The dimmee of Ihe /loim P from Ihe source i.! 0.5 III ond il is III a/l angle of 30 0 10 Ille ulliform
flow. De/ermine: (i) Slreamj,m CliOIl a/ point P, (ii) resuil'IIIll'docil), of jIow <II P allli (iii) location
of ,~Iag,wlion poinl from lite ,~ource.
Solution. Givcn : Unirorm veloci1y. U = 3 m/s : sourcc streng1h. If = 30 m1fs : co·ordin;i1cs or
point Pare r = 0.5 m and (} = 30°.
(i) Siream fimcliOIl at poinl P
The stream run<:tion at any point in the combined now ncld is given by equntion (S.S4A)
IjI=U,rsin(}+.!La
2,
at poi nt p, r = 0.5 m and (} = 30° or
30
180
x 1t radians.
,
,
----l
1.59m
I""~­
Fig. 5.48
Stream runction at point P.
1jI=3x0.5xsin30o + 30 x
21t
I I
(30
X1t)
180
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Fluid Mechanics
'" 0.75 + 2.5 = 3.25 m I,s. Ans.
(ii) Resu/I(lI'/l'e/ocity II/ P
The velocity componen ts anywhere in the flow arc given by
"r=..!. dljl :.!.~[u.rSin 9+ ..2.... 9]
r ae rae
2rr
'" ..!.[u.rcos 9
r
+.!L]
= V.cos e + l
2 1t
2rc r
.-::3~O~
:]xcos300 + -::;21txO.5
'" 2.59& + 9.55 '" 12. 14
,II1U
lie =
-:,ljI = - ~: [u.rSine + ;~. 6]
= - U sin
e
+ 0 '" - U sin
= - ]xsin30 = - 1.5
e
Q
Rcs ulwm velocity.
v=
Ju; + u~
= J12.l 4 1 +(- LSl '" 12.24 m/s . An s.
(iii) LoctUiOlI of stagnation poilll
The horizontal uistancc of the
st:'~l1alion
poilll 5 from tlie source is given by equation (5.55) as
's= -q- = - ]0
-- =
1.59 m . An s.
lit x 3
The stagnation point wi ll be at a distance o f 159 III to tlie lert side o f Ihe source on the x- axis.
Problem 5 .40 A wli/o'm jlOII' Willi" ,'eiOCil)' of 20 mls is ]/0"" " 8 orer " J'ou,ce of Slrength 10 111 21$.
Tlu~ uniform flow Glllt -Wlm; e flow are in Ille same pili/Ie. Oblain III" eqllMi011 of Ille dil'idil1g Siream
/in e ""'/ skelCh Ihe flow p"l/em.
Solution. Given: Un iform 1'~ loc ity. U :: 20 m/s : Source strength. q :: 10 m 2/s
(i) £,/lIl1lion of Ille dil'iding SIri'lim line
The st re31l1 function a1 any point in th~ combined now field is given by equation (5.54A)
2/tU
IjI
= U. rsi n e+..2....
2rr
e
::20x rsine+..!.Q.e
2rr
Th" value of th" stream function for the d ividing stream line is 1jI::
2.. He nce substituti ng IjI '" 2.
2
2
in
the above equation. we get the equation of the dividing stream line.
1. :: 20r sin e + ..!.Q. e
2
"'
I I
2rr
10
10
- =20rsin6+ - 9
2
2rr
c·
q:: 10)
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Kinematics of Flow and Ideal Flow
.
10
10
5 = 20rslll 9+ - 9 :20y + - ,
2n
lit
"'
(":
239 1
rsin8= y)
10
20)'=5 - - '
2,
"'
e
e
5
10
- - x - = 0,25- 20
l it
20
4n
The above relation gives the equation of the di viding stream line.
From llie above equmion. for d iffe re nt values o f e the value of}' is obtained as :
y= -
Value of 9
0
,
2
,
3,
-
2
"
Value of y from 0)
0.25
... ( i)
Remarks
Max. ha lf width of body
III
0.125 m
T he + vc ordin ate above the orig in
0
T he leading point
- 0. 125 m
The - vc ordinalc below the origin
- 0.25
The max. - \'C ordin ale
III
(ii) 5kelci! of flow pal/ern
For ske tc hin g Ihe flow pattern. Jet us first find the location of the stagnation poilll. The horizomal
distance o f the stagnation p<linl S from the source is given hy th e equation.
', = -q- :
10
'" 0.0795 III
2rr.U lit x 20
Hen,;:" Ihe s tagnat ion poin t lies on the x- a~is a1 a di~lanc", o f 0.0795 m o r 79.5 mm frorn Ih'" source
tow'lrds left of th e so urce. The flow p'l\1Crn is shown in Fig. 5.49.
-~§:,,~
. ";~
J}
79 .5 mm
Fig. 5.49
Problem 5.41 A uniform flow wi,1I a I'e/ocity of 2 m/s is flowing ol'er a source placed III ,lie
origin. Tile s/agl1a /ioll poill' occIlrs (II ( - 0.398. 0). De/ermille "
(i) Slrellg/II of Ille source.
(ii) Maximum widlll of Rallkine half-body allli
(iii) OIlier pril1cipal dimellsiolls of Ihe Rallkine Imlf-body.
Solution. Gi ve n
Uniforrn ve locity.
u= 2 rnls
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1240
Fluid Mechani cs
Co-aminalcs o( stagnation point'" (- 0.398. 0)
This means,. " 0.398 and stagnation point lies on x-axis al a distance 0(0.398 IIl lowards left of
origin. The soum: is placed al urigin.
(i) SI":tIgll! of tile J'ource
Let q " strength of tlie source
We know thm
or
r == ~
,
2 lf U
q= 2ltUx ' . " 2ltx 2 x 0.398 = 5.0014 m1/s '" 5 m l/s. Ans.
(it) Maximum Iddtl' of Ril/lid/le half-body
The main dimensions of the Rankine half-body arc obtained from equation (5.57) as
y=
.!L(l-!)
2U
,
...
(1)
The value o f )' is maximum. when G = O.
Y""" '"
2~ ( I - ~) '" 2(~ '" 2 : 2 '" 1.25
III
Maxim um width of Rankine body '" 2 )( Y ...... '" 2 )( 1.25 = 2.5 Ill. AilS.
dillWII$iOIlS of Rallkin<' lIal/-body
Using equation (5.57), we gel
(iii) Olher Principal
y = -"2U
y
(I --"-),
-,,- [Ji)1= -"-[1_-'-]=-"- =- '
=2V
It
2V
2
4U
4x2
=0.625 m
The above value gives the upper ordinate at the origin. w here source is placed.
Width of body at origin == 2 x 0.625 == 1.25 III
At the stagnation point. th e width of the body is zero.
0.625m
5
0.398
Origin
Stagnation
point
(Soon;<! is placed here)
Rankine ha ll·body
Fig. 5.511
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Kinematics o f Flo w and Ideal Flow
241 I
.5 . 11 .4 A SOUTce and Sink Pair in a Uniform Flow (Flow Past a Rankine Ova l Body).
Fig. 5.51 «I) shows a unifonn now of velocit y U and r ig. 5.51 (b) shows a source sink pair of equal
strength. When this I,mifoTm flow is flowing over the source sink pair. a resultant now wil l he obta ined
as shown in Fig. 5.51 (e). This resultant now is also known as the now past a Rankine oval body.
Sink
"'""'
·;te'
•
Source
(a) Unif"",, flow
•
Sink
,
I-
~{'
•
0
I·
' --I
(b) Source and sink pair
"
,
Rank;"" Oval8od~
Pix.
~)
"'
,
-
,'
'.
0
•
•
•
"
"
•
•
x,
••
,,'
Fig. 5.51
U == Velocity of uniform flow along .I-axis
If" Strength of source
(- q)" Strc ngt li of si nk
20" DiSlancc between source and sink whid is alun g x- axis.
The origin 0 of the .(-y co-ordinates is mid -w ay between source and sin k. Consider a poilU P(x. y)
lying in the resu ltant flow field. The stream function (III) and veloc it y potential function (q.) for the
resultant now fie ld are obta ined as given below:
III == Stream function due to uniform flow + stream function due to source
+ stream fu nction due to sin k
Let
== 1jI..;{orm now + 1jI><>I/fCe + 1jI..""
q
(- q)
2lT
=Uxy+-',+--xe,
2n
(where 9 1 is the ungle made by P with source along x- a;o;is ilnd 9! with sin k)
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1242 Fluid Mechanics
" u x y + (Jf.l , _ qfJ) "u)( y + !L
21t
2rr
(G, _ G,)
21t·
(... )' =, "0 9) ... (5.59)
.p" (I'Otcll1iaJ
and
fUllction duc to uniform flow
+ potential fUllction duc 10
source + potential function duc co sink
'" U x r cos
e + !L
[log, '1 2,
" U )( r cus
e+
;'
_It
log,'21
(.:
[10g,.2..]
-,""reos6)
... (5 .60)
')
The following arc the importam poil11s for the resultant flow p.dtcrn :
(a) The re will be two stagnati on points 5, and 52' one to the left o f the source and other to the right
of the sink. /\11hc stagnat ion points. llie resu ltant ve locity (i.e .• velocity duc 10 uniform flow. ve locity
duc to sou rce and ve locity due to s ink) wi ll he zero. The stagnation point 51 is to the left o f th e source
and st agnat ion point S2 will be to tlie rig ht of th e sink on the x-ax is.
Let ).'s = Distancc of tile slaguation points from origin 0 along .r-axis.
Let us ca lcu lat<) this distane<) " 5'
For the stag nat ion poin! 51'
(i) Velocity due to uniform flow = U
(ii) Velocity due 10 source
The ve locily a( any radius due 10 source == ~l
= cc,,~q,-,
2n (xs a)
r"Or .':II' the radius from source == ("'s
-"
(iii) Velocily due 10 sink '" ~,-'!.,--,
Al 5 1'
2n ("s + a)
th~
- a)
2nr
radius from sink'" ("'5 + a))
At poi nt 51' the velocity d ue to uniform flow is in the positive .l -direction whereas due 10 source and
sink are in Ihe -ve x-direction.
The resultant ve locity at 5 1 = U - cc-,'
::;
'' --,
2n ( x.~ a)
(- q)
Butthc resu ltan t velocit y at stagnation point 51 shou ld be zero.
uU=
II
2n (x s -a)
q
2n(" 5-a)
I I
-
+
q
=0
21l" (xs+ll)
q
2n(xs +a)
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Kinematics of Flow and Ideal Flow
q [
'" l it
(xs
243 1
I
"'
(1+:~U )
... (5.61 )
Th" above eq uatio n gives th e locati o n of the stagnation point on the .(· axis.
(b) The s tream line passing thmugh th e stagnation points is having zero ve loc it y and hence ca n be
re p laced by a so lid body. This so lid body is having a s hape of 0'131 as show n in Fig. 5 ,5 1. T here wilt
be two flow fields. o ne wit hin the ova l contour and the Olh er outside the solid bod y. TIle flow fi el d
within Ihe ova l co ntour w ill be due 10 source and s ink whereas tilt! flow field outside the body w ill be
du e to unifonn flow onl y.
The shape of so lid body is obtained from Ih e s tream lin e ha vin g stream function eq ual 10 zero. But
th e stream function is g iv en by equat io n as
IjI
= U x r sin
a + .!L
(9 , _ 0,)
2]'(
•
For th e shape of so lid body.\jI = 0
0 = U x r sin a +
"'
U)( rSin
.!L (9 1 _ e ,)
21'f
•
e ", _ .!L (ti l _ e,)",.!L (e, - ( 1)
2rr
•
21t
•
(el - 6, )
if
'" -2rr
... (5 .62 )
Usi n 6
Prom th e abo ve equation. th e distances of the s urface o f th e so lid body from the origin call be
obcai ncd or th e shape of th e so lid body can be obtained. The maxi mum width of the body (Y""",) will
be equal to OM as shown in Fig. 5.52.
M
"
s,
A
"
]<- '
,
0
,
- +- .----<
s,
F ig. 3.32
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1244
Fluid Mechanics
From triang le AOM. we have
OM
tane = - -
,
AO
OM == AO tan 6 1 "
"'
Y......
"' LeI us lind th e va lu e of 6 1,
Whe n lhe point Plies 011 M. then r == OM.
and
{/
tan 6 1
= a tan 6 1
(.,. OM " Y,,..,) ...(5.63)
e" 90
Q
== ;
6 2 = 180 0 -6[ =1t-6 1
I":
AM=BM
[R efer \0 Fig. 5.521
Ang le ARM" Ang le BAM == 6 1J
Substituting th ese va lu es in equatio n (5.62), we gel
OM
~(I_
'_
- '-",,)-,-',,-,)
= _"
-" It
U
.
Sill
I'f
2
Iwhere OM "
"'
or 2rr. UYm, ==
It _
y"",, 1
2 (1 1
q
"
_
_ I -
"'
It
_ l ItU)'.,,,
q
"'
Substituting thi s value of 6 1 in cqualiOIl (5 .63). we get
Y.m == a Ian
[% _rru;m>x ] " a WI [ 1t U~;m~ ]
... ( 5 .6 4 )
From th e above cqumion.lhc valu e ofy ..., is obtained by hit and triallll cth od till L.H.S." R.H.S . In
this equation
( J'[ u~..., ) is in radian s.
The kngl h and w idth of the Rank i ne oval is obtained as :
Length.
L=2
X .l S
"2xo
and Width.
B=2 x
=
(,+"1
,0u
y"""
2a (nu;-. )cot
... ( :'i .66)
Prob lem 5.42 A Iwifarm flo ... of I'eloeity 6 mls is flo ..... ing {liong x-{l.fi,~ 01'(' , {I SOilree {lnd a sink
...hicil are silllmed along .(-axis. The strengtli of source and sink is 15 m"/s alld they are m a distallce
of 1. 5 III alwrt. Determille :
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Kinematics o f Flo w a nd Ideal Flow
245 1
(i) wCaiiOIl of srtlgnali(1II points,
(ii) unglh lind width of Ihe R(mkil1c oml
(iii) Eqllation of profile of Ihe Rankine body.
Solulion. Given: Uniform fl ow velocity. U'" 6 In/s
Strength of source and sin k.
q'" 15 rn ~15
Distance bt,',twccn source and sink,
2,,: 1.5 III
II'"
1.5
""2
= 0.75
III
(Refer 10 Fig. 5.51)
For Finding tlw localion of the Slagll31iOH points. Ilic equation (5.61) is used.
(I) wnllioll of sl(lgnmioJi poillls
x. =a
J(I+rwUq )=0.75 [1+ IlxO.75x6
15 1'" 1.076rn
Ttw above cqu ali oll gives lhe di,13nce of Ihe stagnation points frolll the origin. There will be two
stagnation poi nts.
The diswnce ofslllgoation poims from the source and sink'" x, - II = 1.076 - 0.75 = 0.326 m. Ans.
(Ii) Lenglh olld ",idlll 0/ fhe Runkine 0>'01
Length.
L = 2 x x, = 2 x 1.076 = 2.152 nl.
Width.
B '" 2 x y"""
... (i)
Let us now lind Ihe value of Y"",.
Usin g equ3lion (5.64). we gel
y
Y"",,"'<ICOI ( 'Uqm ..
= 0.75 COl
1
=0.75col
(" 6X" )
15'"''
= 0.75 COl (O.4Il Y....,)
(0.411 Y..., x I~}
[ .; (OAIl y ..... ) is in rad ians an d hence (OAII Yo.,,) x
I~
will be in degrees]
'" 0.75 cot (72 )( y ..... )O
The above equation will be sol\'ed by hit and trial method. The value of x, = 1.076. But x, is equal
10 length of major axis of Rankine body and y""", is Ihe length of minor axis of Ihe Rankine body. The
length of minor axis will be less Ihan length of major axi s. Le i us flrsl assume Yma.\ = 0.8 ]I). Then
y"""
[" H.s.
R.H.S.
0.75 COl (72 x 0.8)° = 0.75 COl 5l N '" 0.475
07
07
0.75 cot (72 x 0.7)° '" 0.75 cot 50.4" = 0.577
06
06
0.75 cot (72 x 0.6)" '" 0.75 cot 43.2" '" 0.798
0.65
0.65
0. 75 cot (72 x 0.65)" = 0.75 cot 46.8° '" 0.704
0.67
0.67
0.75 COt (72 x 0.67)° = 0.75 cot 48.24° = 0.669 '" 0.67
f.r0111 above it is clear that. when You, '" 0.67. then L.H.S. '" R.H.S.
0.8
0.8
y ..... '" 0.67 II)
Substituting this value in equ3lion (i). we get
Width. 8", 2 x Yn"" = 2 x 0.67 '" 1.34 rn . Ans.
(iii) Eqlwlion of profile of Ille Ronkilte body
The equation uf profile of the Rank ine body is give n by eq uJ tion (5.62) as
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1246
Fluid Mechani cs
,= -q la, -a,)
211: U si n e
5. 17.5
A Doublet in
ill
15 (8 2 - 9,)
0.398 (9 2 - 6, )
=
. A ilS .
21l 6 x si n 9
Uniform Flow (Flow Past
ill
sine
Circular Cylinder).
Fig. 5.53
(a)
shows a unifonn flow of velocity U in the positive .H lircction and Fig. 5.53 (b) shows a doublet at the
origin. Doublet is a spcdal case of a source and a sink combination in which bolh of equal strength
approad each olher such that distance between them tends to be zero. When the unifonn flow is
flowi ng over tlie doublet. a resultant flow will be obtained as shown ill Fig. 5.53 (c). This res ultan!
now is kn ow n as th e n ow llas l a Rankine ova l or equ;'tl :,xes o r n ow p as t a circular cylinde r .
/
H
f--'
(a) Unif"",, flow
\\
"
(b) Dout»el
Poten~al l >nes
Stream lines
.., ~~,m .,
s,
$,
,,'
Fig. 5.53
The stream function ( 1jI) and velocity potenti al functioo ($) for the resul tant flow is obtained as
given below:
I¥ " SITe,lm function due 10 uniform flow + stream fUlH:lion due to doublet
,,, a)
"uxy+(-J.l
si n
[ Stream function due to doublet is given by equation (5.50) as'" -
~ I
,,,
L
sin
e]
I~
~ I
IL
Kinematics of Flow and Id eal Flow
L
=Ux rxsin6 -
(";
sin6
2rcr
247 1
y=rsin6)
~(u xr-L) sino
... (5.6 7)
2"
(jI '" Pote ntial func tion due to unifo rm flow
=UX .l+ -
).1
2,
~"()s
+ potential fU llction due to do ubl et
e
x --
,
[ From
COS6]
eq uation (5.52). potent ial function due to do ublet ", - ).l x 21t
II
cos
e
'" Uxrcos 6+ - x - 2,
,
r
(':x=rcos 6)
~(U x,+L)cos e
...(5.68)
2"
Shape of Rankine oval of equal axes
To get the profile of th e Rankine ov al of equal axes. the stream lin e IjI is taken as zero. He nce
substitutin g IjI = 0 in equa tion (5.67), we ge l
O~ (u xr- L)Sine
2"
This mean s either
. e =0 or UX·, _ ..1:...-_o
sm
e
(i) If si n = O. [h en 9 = 0 and
hori zontal line is the x-ax is.
(ii) If
Uxr - L
2n:r
2"
± 1'( i.e.• a tlOri zonl al tine th ro ug h lh e origi " of the douhlet Th is
=0, then
Vxr= Lor ?", - "2ltr
211:U
,-- J).1
= a constant
2n:U
as).l and U arc constant.
Let 1his eo nswm is equal 10 R.
r=
~2~tU
=R
This givcs that the close d body profile is a circular e ylimler of T:ldius R wi1h ce ntre on double t. The
di vidin g strcam linc corrcsponds to 'V = O. Th is strcam linc is a circlc of radius H. The stream line'll = 0 has
twO stagnat ion poims SI and S2. At SI . the un iform fl ow sp li ts into two stream s that !low along the
c ircle with radius R "
~ 2n:U
P . the twO branches mect again at the stagnation poin! S, and th e flow
-
continucs in the downward diredion. The uniform !low occurs outs ide the circle whereas the fl ow
fi eld du e to doublet lies entirel y wi thin the ci rclc. Th e strcam function for the co mposite flow is g ivc R
by equation (5.67) as
~ I
I~
~ I
IL
1248
Fluid Mechanics
.' (u xr -l-it,"- )Sin
=U
9= U ( , - - "- )
2ltU,
R' ]sine
(r--;-
sin e
- P- , R
2,U
2) ... (5.69)
V .... oclt y Co m po ne nts (u , a n d ( 9)
Th" I'elocity components at any point in Ihe flow fie ld are given by.
R'].sinS1", -I U ( , - -R
'l cuse
","'-,IJ.
-ae =- [U( , - - r
," (}9
r
r
... (5. 70)
wd
11 0
a.(lr
a [U (, - -R
'].sin e1=-u (1+-,
R'].sma
=--= - -
(Jr
f
, "
.. .(5 .7 t)
. .. ( 5.72)
Resu ltanl veloc ity.
On IIII'
surfac~
of the cy linder. r
=R
I l r=U[I - :~ ] cos e
1-"
In equal ion (5.70),r=R]
,0
[ R'l
"0'
u9= - U 1+ R'
sinfl=-2Usinfl
... (5 .73)
- VI' sign s liows the cloc kwise direction of tangential ve locity at that point. The value of lie i~
maximum. when e '" 9(10 ,md 270°.
At e = 0 " or 180".lhc value o f lie = O. Hence on lhe surface of the cylinder.l hc resultant ve locity
is zero, when 0", 0" or ]lIDo, These two points on the surface of cyli nder li. e .. at 0 '" 0 0 and 1&0°1
whae resultant velocity is ze ro. are known as stagn3lion points. Th~y ar~ denoted by SI and S2'
Stagnation poin t SI corresponds to e "" 180° and S1 corresponds to e '" 0°.
Pressure dis tribution on the surface of the cyli nder
Let
I I
Po =
pr~ssure at a point in the unifonn flow far away frolll Ihe cy linder and towards the
left of Ihe cylinder [i.e" approaching uniform flow ]
U '" velocity of uniform flow 31 that point
/I = pressure at a point on the surfa<.:e of the <.:ylindc r
V '" resu ltant vc locity at th at point on the surface of the cy linder. T his velocity is equ al to
lie as ", is zero on the surfacc of th~ cylinder.
Ii
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Kinematics of Flow and Id eal Flow
249 1
V=u a = - 2Usin9
Applying Bernoulli's equatio n at the above two points.
Po
-
pg
UI
+-
II
VI
: - +28
pg 28
~ + ~ == ..E... + [-
"'
pg
28
2Us in
pg
at
[-:
2g
v=
09=
- 2U sin 91
-p" + -U" = -P +-,4cUC'...,'C"C'_'C
p
"'
2
P
2
p
But P - Po is a dim,,"sionle_~s term and is kn ow n as dimensionless pr~ssure co-effi cie nt and is
1 U'
"
denoted by
Cpo
Cp = ~-p~ = 1_4 sin 2 e
,
- pU'
Va lue of preuure co-efficient for di fferent values of
Vulue of B
o
e
Value of Cp
1 _ 4sin 2
1
e
= 1- 0 = 1
GY =1-~ :
30·
1 _ 4 sin 30o : 1 - 4x
90·
1 _ 4 sin 1 90o = 1 - 4x I == 1- 4= - 3
I - I =0
I - I =0
1 _ 4 sin 1 ISOo = 1- 0= I
AI
Al
Al
The
sh own
I I
9 = 0 a nd 180°, th ere arc stag nati on points 52 and 51 respectivel y.
30 Q and 150 0 , the pressure co-efficient is zero.
9 = 90¢, lhc prcs~urc cO-1>'ffici<' nl is - 3 (i.e.,leaM pressure)
va riation o f press ure co-efficie nt along th" surface of th ~ c ylind er for diff~renl valu<,s o f
in Fig. 5.54.
e=
e are
Ii
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IL
1250
Fluid Mecha nics
The positi ve pressure is actin g normal 10 th e surface and IOwards the surface o f th e cy l inder
whe reas th e neg at ive pressure is acting normal to tile surface and away from th e surface o f the
cylinder as sllow n in Fi g. 5.55.
' f'
.i\
o· ",.
w·
I-
\
-,
,
,
00"
-
"-
V
. / 150~180·
0
/
F ig. 5.54
~
/ --,r--<:- - pressu r"
,",~- pressure
".
Fig. 5.55
Problem 5.43 A IIl1iform jlow of 11 m/s is flowillg ora a doublet of strengtll II:!
is in Ihe line of the ulliform flow. Determine:
(i) slrope of Ihe Rankine oml
(iij radius of Ihe Rmlkine circle
•
I/I"/S.
'I'Iledoublel
(iii) )'(1111<' of .l'lrealll litle fimcliOl1 (1/ Rankine circle
(il'J re~·"II<"'1 re/ocily at (I poinl 011 Ille Rankine circle a/ lm angle of 30° from x·,uis
( I' j \'{jille of II/{uimum ,·e/oi.'iry 0' 1 the Ran/due circle (lnd lQ(;lIliOlI of Ihe poilll wllere "e[o<:iry is maT.
Solution. Given
U = 12 m/s: J.I = 18 1I1 2/s
(i ) Shape of Ihe Rankine 0,",,/
When a uniform no w is Flowing over a do ubl et and doublet and uni foml n ow are in lin e. then the
sh ape of the R ankine ova l w ill be a circle of radius =
I I
J2,U
I-l •
A DS.
Ii
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Kinematics of Flow and Ideal Flow
251
I
(ii) RIUIiII$ of Ille Rallkille circle
R=r=
J)l
2nU
=
~
V~
= 0.488 m. AilS.
(iii) Value of stream lille /,melion al /I,e Rallkin e ('irc/e
The va lue o f strea m lin e function (\11) at th e Ranki ne c ircle is zero i.e .. \jJ = O.
= JO
On the s urface of the cyl inder, the radial velocity (u,) is lero. The langenlin] veloci ty ("a) is give n
by equation (5.73) as
ua=-2Usin 6=- 2 x 12 x sin 300 =- l2m/s. AilS.
(;v) Rend/mil ndocily 011 the sur/ace of 1/"" rir dl!. ""hell
e
Q
- ve sign shows the clockwise direction of tangential velocity a1 thai poi n\.
Rcsul1alll ve locity.
(,,) Maximum "e1ocity and ils foell/ioll
The rcslllwnt ve locity at an y point 011 Ih e surface of the I:ylindcr is equal to I'a" But "6 is give n by,
" a=-2UsiIl6
This velocity wi ll be maximum. when = 90°.
e
Max. ve locity
" ~ 2U" - 2 x 12 = ~ 24 m/s . Ans.
Problem 5.44 A uniform flow 0/ JO /Ills i$ flowillg OI'er a doublel of j'lrenglll 15 m 2/$. Tile doublel
i.~ ill IIII' lille of Ihl' ulliformJIow. 1"11<' polar co-ordilWles of a poim P ill IIII' flow field are 0.9 III and 30°.
Find: (i) Slrealll line/ul1Cliol1 lIml (ii) /Ile "'j'ullllll/ "docil)" a/lire poini.
Solution. Given: U = 10 I11 ls: ~ = 15 111 2/s: r = 0.9111 and = 30°.
Let us fi rst find the radius (R) of the Rankine c ircle. This is given by
e
R=
~
Jl5
P =
=0.488m
21!U V~
The polar co-ordinates of the point /' are 0.9 111 and 30°.
Hence r '" 0.9 111 and
30°.
As the va lu e of r is more than th e radius of the Rankine circle. hence point P lies outside
cylinder.
(i) Vallte 0/ Siream line/wne/ioll III III<' poil1( P
The st ream line function for the cOllljXlsite flow al any point is given by equa ti on (5.69) as
e '"
th ~
,
P{O .9 m, 30' )
-
Fig. 5.56
I I
Ii
~ I
IL
1252 Fluid Mechanics
,= 0.9 Ill. R = 0.488 and 9 '" 30")
I
,
= 10(0.9 - 0.2646) x - '" 3. 177 m Is. An s.
2
(ii) He,mllalll \'e1ociry al lilt! poilll P
The radial ve loc ity and tangential veloc ity at any point in tlie flow field arc g iven by equations (S.70)
and (5.7 1) respectively.
[ 7")
"r=U 1-
(0488')
cos9= 10 1-
~.9 l
CQs300 =6 11 m/s
+vc sign s how s th e radial \'clocit y is o utward.
ue =- U
and
[1+ 7")
sin9=-10
[ 0488')
I +~
si n 30 0
= - 6.47
mf~
-vc sign shows the clockwise direction of tangential ve locit y.
R csul1~ nt velocity,
= J6.l 12 + (- 6,47)' = .)3733 + 44,86
= 8.89 m /s. A ns .
HIGHLIGHTS
1. If the fluid characteristics like velocity. pres.ure. density etc. do not change at 3 point with ",spc<;t to
lime. the tluid flow is called steady flow. If they change w,r.l. time . the fluid now is called unsteady
now.
(~;)
= 0 for sleady flow
and
(~,)"" 0 for un.leady flow.
2. If lhe velocily in a fluid flow does nm change with re'pee! to space (length of direction of flow). the
flow is said unifoml otherwise non-unifonn. Thus.
(~) '" 0 for unifoml flow and (~;) c#- 0 for non-unifonn flow.
J. If the Reynolds number in a pipe is less than 2000. the flow is said to be laminar and if Reynold number
is more Ihun 4000. the flow is said 10 be turbulent.
4. For compressible flow. p"" constant
For incompressible flow. p '" constant.
5. Rate of discharge for incompressible fluid (liquid). Q ~ A x I·.
6. Continuity equation is wrillen a, A,", '"
A,,·).
A,'·. '"
I I
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253 1
Kinematics of Flow and Ideal Flow
7. Continuity C{Juation in di ffcrcnliul fonn.
(Ju
(n'
ilx + dy +
flu
<rn'
a;'" 0 for Ihrce-<limcnsional flow
i'l,'
a( + iJy
.. 0 for two-dimensional flow.
II. The components of acceleration in x. )' and Z direction are
(,
au
'
dU au
iJX{)yi.lZ(J1
()u
",u-+,·~+
.. '~ + -
;n,
;),0
i),. a,·
+ " - + II' - + axiJyilz dt
II : U-
'
al<'
all'
: ,h
a", a...
" " - H -()y +W -at ' -i)I
9. The components of
~elocit y
u__
in x. )' and
l dir~""lion
'
in Icnns of velocity polential
(~)
are
"*."
K_'*ily and", _ _ ()q>dZ .
d.t
10. T he stream funclion (1jI) is defmed only for Iwo -<;l imensional flow. The ""Ioc i!y CQln ponems in
y direClions in {CnTIS of stream function are u • _
a'll
;ry
and """
~
~
J
and
.
II. Angular defonnation or shear strain rate is given as
Shear strai n rale _
~ [().. + au]
2 ax
ily
12. Rolational components of a fluid par1icie are
13. Vonicity is two times the value of rotation .
14. flow of a f1uid along a curved p;tth is known as vortcx f1ow . If the part icles afl' moving round in curved
p;tth with Ihe help of some external torque the f10w is called forced '"Ortex f1ow. And if no external
torque is req ll ircd to rotate the nuid p;tnicles. the f10w is called free -vortex f1ow.
IS. T he rclation beNeen tangential velocity and radi us:
for forced vortex, \. ,. to X r.
for free vortex, I' X r = con,tall!.
16. The pressure variation along the radial direction for ,-ortex now along a horizont:ll plane.
and pressure variation in the venical plane
,,2
For the
where
I I
W 2,l
iJp _ _ P8.
a,
wl R l
for~e<.l
vortex now, z,. - ' " - - . - 28
28
28
Z .. height of p;trJboloid fanned
w = angular velocity.
Ii
~ I
IL
1254
Fl uid Mechanics
18. For D forced vonex flow in a ope" mol.
Fall of liquid level at cemre Rise of liquid level at the ends.
%
19 . In case of clo"'" cylinder. thc volume of air before rotalion is equal to Ihc volume of ai r after rotation.
20. If a close cylindrical vessel completely filled wi th waler is rolated about its "cMical axis. the IOU!
pressure forces acting on lhe top and OO1(om arc
Fr"
and
where
FT=
F 8 ..
1Il ..
R=
..e. oillK
4
F 1/"' FT + weight of wale' in cylinder
Pressure force on lOp of cylinder
Pressure force on the bonom of cylinder
Angular vclocity
Radius of the vessel
w
p .. Density of fluid,. - ,
g
21. For a free vonex flow the equation is
.£L -+- ,·f
pg
2g
~
+ ZI .. .!!.l... +
-+pg 2g
Zl'
EXERCISE
(Al THEORETICAL PROBLEMS
1. What arc the methods of describing fluid flow ?
2. bplain thc lcnns:
(i) Path line,
(iii) Stream line, and
( ii) Streak line,
(i") Stream tube.
3. Distinguish between
(i) Steady flow and un-steady flow. (ii) Unifonn and non-uniform flow,
(iii ) Compre,siblc and incompressible flow.
(i") Rotati onal and irrOlational flow. (v) Laminar and turbulent flow,
~ . Define the following and give one pmctical example for each
(I) Laminar flow.
(ii) Turbulcnl flow.
(iii) Steady flow. and
(i") Unifonn flow.
S. Define the equation of continuity. Obtain an expression for continuity equation for a three-<limensional
flow.
(R-G.P.V. S 2(02)
6 . What do you understand by the terms
(,) TOIal acceleration. ( ii) Convective acceleration. and
(ii,) Local accclcr~lion ?
(Dellii Unit·u,'il),. Dl"c. 2(02)
7. (<I) Define the tenos :
(ii) Stream (unelion .
(I) Velocity potential function. and
(1)) What arc the co nditions for flow to be irrotational ?
K. What do you mean by equipotenliai line and :t line of "onstant st ream function ?
!J. (<I) Describe the uSC and limitalions of the flow nets.
(b ) Under what conditions can one dmw flow net ?
10 . Define Ihe tcmlS :
(i) Vortex flow.
(ii) r~rced vortex flow. and
(iii) f ree vortex flow.
II . Differentiate between forced vortex and free vortex flow.
I I
Ii
~ I
IL
Kinematics of Flow and Ideal Flow
255 1
12 . Dcrive an e~pres,ion for the depth of paraboloid formed hy the surface of a liquid cOnlaincd in a
cylindrical lank which is rotated at a constant angular vel ocity (II ahou! ils venical axis.
U . Derive an expression for the difference of pres,ure between two points in ~ frce "orln flow. Does the
difference of pressure sat isfy Ucmoullrs equation? Can Bemoulli"s equation t>c applied to a forced
,'one" flow 'I
14 . Dc,;"c. from first principles. the condition for irrola(ional flow. I'row that. for potential flow. OOlh the
stream fu nction and velocity potential function ",-[isfy the Laplace equation.
15 . Define "elClCity potential function and slream funnion.
16 . Under Whal conditions Call one trcat real fluid flow as irrolationa] (as an approximation).
17. Define the following:
(I) Steady now.
(Ii) Non -unifomt now.
(iii) Laminar now. and
(i") Two·dimcnsiona l now.
Ill . (a) Disting uish between rotational flow and irrotational flow. Give One example of e.leh
(b) Cite two examples of unsteady . non-uniform flow, How ~"n thc Unsteady flow be transformed to
steady flow ?
/J.N.T. Vlli,·asil),. S 10(1)
19. Explain uni fonn flow with souree and sink. Obtain expressions for stream atld velocity potential
functions.
20 . II poi1ll source is a point where an incompressible fluid is imagined 10 he created and senl OUI e"enly
in al l directions . Detenn;ne its velocity potential and Slream function.
21 . (I) Explain doublet and define the strength (If the doublet
(ii) Di stingui'h between a source and a sink.
22 . SketCh the fl(lw pattern of an ideal fluid flow past a cylinder with circulation.
23 . Show that in case (If forced vortex flow. the rise of liquid IC"e! at the ends is "'Iual to thc fall of liquid
le"el at thc axL. of rotati(ln,
2~ . Differentiate between
(i) Stream function and "elocity potential function
(ii) Stream line and streak line and
(iii) Rotalional and irrolational fl(lws.
(B) NUMERICAL PROBLEMS
1.
2.
3.
4.
S.
~ I
The diameler:s of a pipe .11 the seclions I and 2 me 15 cm and 20 cm respecti,·e ]y. Find the discharge
through the pipe if "ciocily of water at section I is 4 mIs, Dctemline also the ve locity at section 2.
IAns. 0.07068 m'/s, 2.25 m/sl
A 40 cm diameter pipe. conveying water. branches into two pipes of diameters 30 cm and 20 cm
respectively. If the averagc veloci ty in the 40 em diameter pipe is 3 m/s. Find tne discharge in this pipe.
lliso detennine the velocity in 20 em pipe if the average velocity in 30 em diameter pipe is 2 m/s.
IAns. 0.3769 ml/s . 7.5 mlsl
II 30 em diameter pipe carries oil of sp. gr. 0.8 at a velocity of 2 m/s. lit another section the diameter is
20 Cm. Find the velocity at this >cetion and also mass rat~ of flow of oil.
IAn s. 4.5 m/s. 113 kg/sl
The velocity "cctor in a nuid now is given by V = 2~ 1 - 5.rYJ + 4/k.
Find the velo<'ity and accelemtion of a fluid panicle at ( I. 2. 3) at time, / ~ I.
(Ans. 10.95 units. 16.12 unitsl
The following case .. represent the two velocily components. delenninc the thin! component of velocity
such that they sa(isfy (he continuity equat ion
I~
~ I
IL
1256
Fluid Mechanics
(Il u., 4,,-2, \' = 4.I)'Z
(ii) u = 4Xl + 3.,)" ", ..
[An•. (i) ... _ - 8<z -
t - 4.')' -
2y< _
2xz1 + I( x. y)
(ii) I' = -!UY
>" + 3}"l '' + I(x, t)]
- "2
Calculate the unknown "clocily components so (hm they satisfy the following cqualions
(II i t . 2.-2, v. 29'<.'" _?
(ii) ,,_ 2K + 2sy. 1<'.
- 4xz + 2yz .•'.?
[AilS. (ill<' _ - LIz
r
6. A fluid now is given by : II =
7.
H.
9.
10 .
11 .
_ x'zi
xy'i - 2yz'j _ ( 2),' - 2t) k.
Prove that it is a ca5e of possible steady incompressible fluid f1ow_
Cakulatc the velocity and acceleration at the point [1. 2. 3].
[A ns. 36.7 units. 874.50 unils]
Find the convective accclcr.tlion al1hc middle of a pipe which co,,,'crges unifonn ly from 0.6 III diameter
to 0.3 m diameter over 3 111 length. The Me of flow is 40 litis. If the rate of flow changes unironn l), from
40 liUs to 80 hils in 40 second,. find Ihe IOlal a"elcmlion al the middle of the pipe al 20lh second.
lAos• .04<)<) tnls' : .1 IS74 mis' I
T he velocily potential function. , . is given by, '" x' - /. Find the "cloc ity components in x and y
direction . AI"" show thaI, rep rcscms a possible case of fluid flow.
IAns. u = 2x and I' ~ - 2yl
f or Ihe "elod ty polential function. <l>"
fi nd the velocily components at the point (4. 5).
[Ans . u = S. ,. = - 10 IInitsl
II stream function is given by: 'I' _ 2r - 5)". Calculate the "elocity components and also magni1Ude and
direction of the res ultanl velocity at any point. [Ans. u .. 5. 1'" 2. Resultant .. 5.384 and e .. 21 0 48'1
If for a two-dimen,ional potential now. thc velocity potential is given by : , = 4.t(3)' - 4). deteonine the
velocity at the point (2. 3). Dctennine also the value of stream fun ction 'I' al Ihe poinl (2. 3).
.r -/'
[ Ans. 40 units. '1''' 6.r' - 4
(f)'l - 4 Y). - IS]
12 . The stream function for a Iwo-<limen,ional flow is given by 'I' = !!.'Y. calculate Ihe velocily al the point
[All' . Ux - 32 unils. I' ~ 40 un il •. ¢I = 4i _ 4x' l
p(4. 5). Find Ihe w locity potcmiul function ¢I.
13 . Sketch Ihe stream lines reprcsemed by 'I' '" .•y. lliso find out Ihe "e!ocit y and its di=tion al point (2 . 3).
[AilS. 3.60 un il< and e = 56 18.6' or 123" 42'1
14 . f or the "cloci ty components given as : I< .. flY sin .'Y. I' _ ax sin xy.
O blain an expression for Ihe velocily polcntial function.
[Ans. ¢I_ (J cos.I)' l
0
1 S. II fluid now is gh'cn by : V _ I o..-Ji - 8.1'\;'
f ind the shear strain rale and state whether the flow is rotational or irrotational. [AilS. - &ry. rotational I
16 . T he vc!ocity components in a two-<l imensiona l flow arc
u ..
&r), - ~ }'l and v .. _ &xl + ~ r.
J
3
Show that these veloci ty components represent a possible case of an irrotational flow .
[
An s.
au + a,· 0=0. ro . 0=0]
ax ay
.
17 . An open circular cylinder of 20 em diameter and 100 cm long contains water upto a hcight of 80 em. It
is rolated about it.. ,"enieal axis. I' ind the .• peed of rotation when
(I) no Water spills.
(iil uial depth is zero. [AilS. (I) 267.51 r.p.111 .• (ii) 422.98 r.p.m.]
18 . A cylindrical \"Cs~1 15 cm in diametcr and 40 em long is complelely filled with watcr. The ves~el is open
at the top. Find the quantity of watcr left in the ve,,,,l, when il is rotated about ils ,'ertieal axis wilh a
[Ans.4566.3cm ' ]
speed of 300 r.p.m.
~I
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Kinematics of Flow and Ideal Flow
257 1
19 . An open circular cylinder of 20 em diameter and 120 em long contains water upto a height of W em. It
is rotated about its vertical a~is at 400 r.p.m. Find the difference in lotal pressure force (il at the OOno111
of the cylinder. and (Ii) al the sides of the cylinder dilC 10 rotation.
[Ans. (i) 14 .52 N. ( if) 2465,45 N I
20 . A c1o<;ed cylindrical \'e,sel of diameter 15 em and length 100 em contains water urlO a he ight of 80 em.
The \'esse l is rotated at a speed of 500 Lp,m, about ils wrtical axis. Find the height of paraboloid
formed.
[Ans. 5606 em]
2 1. For the data gi"cn in question 20. find the speed of rotation of the vessel. when axial depth is zero.
[Ans. 891.7 r,p .m.1
22 . If the cylindrical vessel of question 20. is rot<ltcd al 950 •. p.m . about its "crtical axis . find tll c ~rea
unco"crcd at the base of thc tank.
[An s . 20.4 em' ]
23 . A closed cylindrical vessel of diameter 20 cm and height 100 cm contains water upto a height of 70 cm. The
air above the willer surface is at a pressure of 7S,4 8 kN ImI. The vess.,1 is rotated at a speed of:lOU Lp.m.
about its ,'crtical axis, Find the pressure head at the bottom of the \'essel ; (a) at the centre. and (b) at
the edge.
[Ans. (a) 84485 m (b) 8,95 15 m[
14. A closed <'ylinder of diameter 30 cm and height 20 "n is completely filled with Water. Calculate the IOtal
pressure force exerted by water On the lOp and oottom of the cylinder. if it is rotated about its .'ertical
IAns. F T ", 392.4 N, F~ '" 531 N ]
axis at 300 .. p.m.
25. ]n a free cylindrical \,ortex flow of w~ter> ~t a point al a rad ius of ] 50 nun the velocity and pressure arc
5 mls and 14.7 15 Nkm l , Find the pressure at a radius of 300 nnn ,
[Ans. 15,65 Niem I ]
26. 1)0 the follow ing "elocity components rcpre""nt physically possible flows ?
II .. .r' + l' + 5. ,... / + z'. 1<' .. 4.tyZ.
[Ans. No.[
27. State if the flow represented by" .. 3x + 4y ana \' .. 2r - 3)' is rotational or irrotational. [Ans. Rotational[
28. A Yes.e!. cylinarical in shape ana dosea at the top and bOllom. <"ontain. water upto a height of
700 mm. T he di~meter of the vessel is 200 mm and length of vessel is 1.1 m. Find the speed of
rotation of the ves",] if the axial depth of water is '.ero.
29 . Define rotational ana irrolational flow. The stream fUn <'I;on ~nd veloc;ty potenti al for a flow are given
by,
'lI 3 2 (),.¢/ 3 X< - j .
Show that th~ cond il;on' of continuity and irrotational flow arc sati.fied.
30 . f or Ihe steady incompres .• ihlc flow. are the following val ues of u and .' possible ?
(I) " z 4.ry+ / .,· .. 6.ry+3xand
(ii) u=2.C+ / ••'=-4.l)·.
[Ans. (i) No. (ii) Yes[
J] . Dcfinc two ·dimensional stream fun ction and velocity potential. Show that followi ng stream function
'lI .. 6x - 4y+7xy + 9
represents an irrOiational now, I;ind its velocity potential.
(A n.•. 41 = 4.< + 6y - 3.5.<1 + 3.5/ + C]
32 . Check if oil '" .•' - y' + y r{'presents the >'C locity potential for 2--dimensional irrot:ltional flow . If it docs.
thcn dctcmline the stream function 1(1.
[Ans. Yes. 1(1 = - ] .ry + x l
3J . If stream function for steady now is given by 1(1 _ li - ...-l). dctenninc whether thc flow is rouuional or
[Ans. IrrotJtional. 41" - 21y + CI
irrotational. Th en detcrmine the velocily potential 41.
.l4 . A pipe ( I ) 4SO mm in diameter branc.hcs into two pipes (2 ) and (3 ) of diameters 300 mm and 200 mm
respectiYely as shown in Fig. 5,57. If the 3YCrage .'elocity in 450 mm diameter pipe is 3 mls. fin<l
(i ) discharge through 450 mm dia. pipe :lnd (ii) velocity in 200 mm diameter pipe if the ave rage
velodty in 300 nlln pipe is 2.5 m!s.
(J.N,T.V .. Hyder"!;>,,,!. S 200])
[Hint. Given:
~ I
tI , =4SOmm=0.45m.tI,,,,300mm=O.Jm
tI J .. 200 mm = 0.2 m. VI = 3 mIs, V.! .. 2.5 Illis
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1258
Fluid Mechanics
(i)
It
~
J
0 , "A,V, "'"4 (0.45)x 3 '" 0.'177 m Is.
(j)
-
d ,_ 450mm
Fig. 5.57
_ ; (.3 1 ) x 25., 0. 176 rrhs
(ii)
OJ •
IJ ut
0 , '" OJ + QJ
Also
O J "'A l X V1
A l V:
QJ '" 0 , - Ol = 0 477 -0.176= 0.301
" (0.2-), X V1
"' "4
Q,
V, • - _.-
~(O_ 2~
4
I I
D.lO l
• -0.0314
~
11.6 m/s. ]
Ii
.. 6. 1
INTRODUCTION
In the previous c hapter. we stud ied the ve locity and acceleration 31 a point in a fluid flow. without
tak ing into consideration the forces causing the flow. Th is chapter includes the study o f forces causing
fluid Flow. Thu s d ynamics of fluid flow is thc st udy of fluid motion with the forces causing flow. The
dy namic behaviour of thc fluid flow is anal ysed by Ihe Newlon'S second law of molion, which relates
Ihe accele ralion wilh Ihe forces. The tluid is assumed to be incompressib le and non- viscous .
.. 6.2
EQUATIONS OF MOTION
According 10 Newlon's seco nd law of motion. Ihc nel force F , acti ng On a fluid elemenl in the
direClion of X is eq ual 10 mass In o f the fluid clement multiplied by Ih e acceleration (/. in Ihe x-direction.
Th us mathemati cally.
...(6.1 )
F, = m.(/,
In Ihe fluid flow. Ihe following forces are pre.'lCnt:
(i) F g• gravity force,
(ii) Fr Ihe pressure force.
(iii) F .. force due 10 vis.:osi ty.
(iI') F,. force due 10 lurbulence.
(I') Fe' force due 10 compre"sibilily.
Th us in equa tion (6. 1). the net force
F, = (F, )" + (Fp) , + (F) , + (F,)x + (Fe),'
(0 If the force due to compressibility. Fe is neg ligible, the resulting net force
F,
= (F, )" + (Fp) , + (F,.) , + ( F,),
an,j equ31ion of motions arc called Reynold' s ell uation s of motion .
(ii) For flow, where (F,) is negligible. th e resulting equatio ns of 1Il0t ion are known as
Na"ier-Stokes Equation.
(iii) If the flow is aSl;umcd \0 be ideal. viSl:ous force (F.) is zero and equation of 1Il0tions arc
known as Eule r 's elillation of mo tion.
259
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1260
Fluid Mechanics
EU LER'S EQUATION OF MOTION
... 6.l
This is equation of motion in which the forces due to gravity and press ure arc take n il1m consideration. This is derived by considering the motion of a fluid cle ment along a s trea m- line as :
Consider a strcam-lin<l in which flow i s takin g place in s-d ircction as sltown in Fig. 6.1. Consider a
cylindrical clcrncllt of cross-section <fA and length lis. The forces acti ng on the cy lindrical clement arc:
I. Pressure force pdA in the direction of flow.
2. Pressure force
(p + ~ dS) dA opposite 10 Ihe direction of flow.
3. Weight of clement pgdAds.
Let
e is the angle between the d irect ion of flow and the line of action of the weig ht of clement.
The resultant force on the flu id c lement in the direction of
clement X ae.::cleralion in the dire.::tion s.
pdA - [p +
= pdAds )(
where
ap
~'
must be equal to the mass of fiuid
dS) dA _ pgdAds cos
d,I'
s
a
... (6 .2)
<l,
is the acceleration in the direction of s.
<l,
til'
Now
" . '" - . where v is a function of sa nd r.
,"
=
av ds + av
a,~ dt
at
I'dl' + ai' (.;
a~·
at
=
Id'" '"
I')
a,· = 0
",
If the flow is steady. -
I~I'
= --
it
,
"'
Substituting the value of II, in equa tion (6.2) and simpl ify·
ing the equalioll. we gCl
ap
- -
as
pd,j'dA. -
pas
Forcl!S on
(b)
<l
fluid l'Ieml'rll.
ap
ill·
pas
as
- - + gcos 9+1' - =0
a= ~
d,
lip
<lz ",h,
- +g - + =0
p lis
d.!
ds
1
I I
Fig . 6 .1
"""',
a~·
But from Fig. 6.1 (b). We have eos
0'
(a )
a = pdA<I~' x -:;oS
ap - 8 cos e =I'al'-
dsdA - pg dAds '::OS
Dividing by
0'
d,'
or
lip
-
p
+ gdz + \"lit, = 0
dp
- +gdz+"d,,=O
p
Equmioll (6.3) is known as EuJc(s equmion of mot ion.
.. .(6.3)
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Dynamics of Fluid Flow
.. 6.4
261 1
BERNOULLI'S EQUATION fROM EULER'S EQUATION
Bernoulli's eq uati on is obta ined by integrating tlie Euler's equatio n of motion (6.3) as
f d; + Jgdz + J
I'dl' '"
constant
If flow is incompressible. p is oon,13n1 and
I
\ .:
p
2
L + gz + )
I ,l
-'- + <+ P8
2g
'" constant
'" constant
l
-P + -v +z=constant
"'
pg
...(6--')
28
Equation (6.4) is a Bernoulli's equation in which
~ '" pressure energy per unit we ight of fluid o r pressure he~d.
pg
,?-ng '" kinetic ene rgy per unit weig ht u r kinetic head.
l '"
.. 6 . .5
p01cmial energy per unit weight or potential head .
ASSUMPTIONS
The following arc the assumptions made in th e dcri vati on o f Bernoulli' s equat io n :
(il Th e fluid is ideal. i.e., viscos ity is zero (U) Tile flow is Meady
(iiI) The flow is incompressible
(iv) Tile fl ow is irrOlalional.
Problem 6 .1 Water is flowing Ihrough a pipt! of 5 cm diameler under 0 pressllre of 29.43 Nkm ?
(gouge) lind lI'ilh lI!ellll I'e/oeily of2.0 mA Filld Ille 100a/lleml o r rollIl ellergy per utilI weiglll oflllt!
water m a cross-seclioll. which is 5 m IIbol't! Ihe dmum lille.
Solution. Given:
Diameler of pipe
= 5 el11 = 0.5 m
2
Pressu re.
p '" 29.43 Nkm '" 29.43 x
v=2.0mfs
VelocilY.
Datum head.
z =5 m
Total head
'" pressure head + kille ti c head + datum head
Pressure head
:
~
:
pg
1,2
Kindic head
Total head
:
2,
:
29.43 x 10"
=30 m
1000 x 9.81
2x2
2x9.81
{ p for wa ter =
IOO(L.~~}
l
m
= 0.204 III
: -p + -" + Z'" 30 + 0.204 + 5 '" 35.204 m. Ail S.
pg 2g
Prob lem 6 .2 A pipe. 111f00'gh ... "ieh Wilie r isj/owill8. is /1(II'illg diameters. 20 cm (IIul10 cm allhe
cross-seCliOIl S I alld 2 respeClil"e/y. The "elOCily of waler al secliull 1 i~' givell 4.0 mls. Filld Ihe veloCily
IWlld II/ secliolls I alld 2 IIl1d also rale of discharge.
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1262 Fluid Mechanics
Solution. Given
D,,,, 20cm=O.2 m
/flit
A,,,,D, = - (.2) ! =O.031 4 m1
Area.
4
VI = 4.0 111/5
4
_
D,~20cm
V, ., 40 mlsec
D1 =0. 1 III
r
A l '" ..::. (.1)1 '" .00785 m 1
4
Fig. 6.2
(i) Vclocily head at section 1
=~= 4.0x 4.0 '" 0.815 m . Ans.
2g
2 x 9.8 1
(iiI Velocity hC~ld at .s«tion 2 = V/12g
To rind VI_ apply cont inuity ~-quntion at 1 and 2
VI
VI
Velocity head at SCi:lion 2 '" ---.L =
2g
A,V,
.03 14
Al
.00785
= -- = ---
16.0 x 16.0
2x9.81
x 4 .0
= 16.0 m/s
'" 83.047 m. A il S.
:: A ,V, or A 1V2
'" 0.03 14 x 4.0 = 0. 1256 ml/s
= 12S.6I1tNs/s. r\,,-~ .
Problem 6.3 SIMI' Bem ol/IIi 's tileorem for steady flow of "n ;nco mpTHsib/e fluid. Deril'C (In
c.<pression fo r Bern oulli 's c'I,wrion from first principle {lilt! Slale Ih e a~'~'umplions made for ~'uch a
deril'ation.
Solution. Stat ement of ll e rn oulli 's Theo ~m _ It stales th at in a steady. ideal n ow of an incom pressible fluid, Ihe 10t,11 e nergy at any puint uf the fluid is constant. The total e nergy cunsists of
pressure e nergy. kinetic e nergy and po tential energy or datum e nergy. These energies per unit weight
of the fluid are:
(iii) Rate of discharge
Pressure energy"
.1!....
pg
,
,,'
Kindie energy" -
2,
Datum e nergy" Z
Thus mathematically, Bernuulli's theorem is wnnen as
p
"t
- + - + z " Constant.
pg 21(
Derh'ution or lh-rn oulll 's theONIII. ror derivation o f Bernoull i's thL'Orem, Aniclcs 6.1 and 6.4
should bt;: written.
Ass umptlons arc given in Article 6.5.
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Dynamics of Fluid Flow
263 1
Problem 6.4 Ti,e WIlier i,~ flowing (I/fougll a
pipe IlaI'ing dhmreters 20 em and 10 em (1/ seCliolis f
alld 2 respec /il'e/y. Tile rate of flow IhrOi,gh pipe
h 35 Ii/fo/s. The seC!iOlI I is 6 m "bol'<' dlllum alii!
iJ'eclioli 2 is./ In abo,'e dll/Ilm . Iftl", prnyu'e III sec~
lion I is 39.24 Nkm 1, find the ill tensify of preSSllre
(II iJ'eclioli 2.
Solution. G i ven:
A I sec/io n I.
D 1 ",20clll =O.2 m
Fig. 6.3
"
A, '" - (.2) '" .03 14111-'
4
P, '" 39.24 Nk m!
'" 39.24 x 10 4 N/m '
z,=6.0 m
AI secti on 2.
D2
:=
0. 10
111
"
4 (0. 1)" '" .00785
A, '" -
,
III
,
Zl=4 m
po'" '1
Rat e o f n ow .
Now
Q" 35 li tis'"
.l2... = .035 m 3/s
1000
Q = A1V I = A1V,
v,=iL=
A,
.035 = 1.114mls
.031 4
V, = -Q.Q35
= - - - : 4.456 mIs
A!
.00785
and
Apply ing Bc mu ull i's equ ation al sc(;l ions I and 2. we gel
4
oc
39.24x 10 + (1.1 14 )1 + 6.0 ",,,,,;"PL',,,,, + (4.456)1 +4.0
IOOOx9.81
2 x9.81
l 000x9.81
2 x9.81
40+0 . 063 + 6.0=~+ 1.0 12+ 4.0
9810
oc
46.063 =....!!L + 5.0 12
9810
~ = 46.063 - 5.0 12 '" 41 .051
9810
Pl ", 41 .051 x98 1ONlm 1
'"
I I
4 1.05 1x9810
10'
Nlc m 1 '" 40.27 Nlcm 1. Ans.
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1264
Fluid Mechanics
Problem 6 .5 Warer i,~ flowing Illrough II pipe iIarillg diameter 300 mill alii/ 200 111m III Ille boftom
and upper <'lid rcspecli\'e/y. The illtensi1), of pressure ar Ihe halTom t1nd i.! 24.525 Nkml and flw
pressure a/ Ihe upper end is 9.81 Nlcm l. D elumine Ihe differe' lce in darum /wad if Ille mIl! of flow
through pipe is 40 lilk
Solution.
Gi\'~n
:
Sect io n 1,
DI = 300 mm = 0.3 111
P I'" 24.525 Nlc rn ! '" 24.525 x 104 Nlm!
Sect io n 2,
D1 '" 200 111111 '" 0.2 m
,
40
004
'I
"'
m
P2 "9.81 Nlcnl
\
ill
Z,
'" 4 0 litis
Q " lOOCl " ' .
0 ," 200 mm
1
4'
Pl=9.8 1 Nkm-=9.81 x 10 Nfm-
Rate of n ow
I
D,"'300mm
.-
",
.__ .
2
P1 '" 24.525 Nlcm
DATUM LINE
S
Fig. 6.4
Now
.04
.04
,,0.5658 mfs
V,"'-" - - "
~ D'
4 '
0.566 m/s
AI
0::
v,,, _"'_ " ~.04",-_ "
-
Al
..!:. (D,)l
-;,0",.04,-- _ 1.274 III/s
~(O.2)2
4 '
4
Applying Bernou lli 's equation al secti ons ( I) and (2). we gel
~l
V!
EL+-'- +z, '" P! + ---.L+ z,
pg2g
24.525 X 10
4
l000 x 9.81
+
.566x.566
2x9.81
pg 2 g "
+ ZI =
9.8 I xI0 ·
rOOOx9.81
+
{1.274)1
2x9.81
25 + .32 + ZI = 10 + 1.623 + Z2
25.32 + Zl = 11.623 + Zl
Zl - Zl '" 25.32 - r 1.623 '" 13.697", 13.70 111
Differe nce in datu111 head = Z2 - ZI = 13.70 m . Ans.
Problem 6 .6 The water is jlowillg Illrough a taper pipe of /e/l glh IOQ 1/1 IUII'ins diometers 60Q mm
(lllhe upper elld ,md JOQ mm (II Ihe 10l<'er end. allile role of 50 lilres/s. The pipe lUIS a slope of I in 30.
2
Find Ihe press lire 01 Ihe lower elld if Ihe pressure III Ihe higller lerel is 19.62 Nkm •
Solution. Given:
u,ng th o f pipe,
L= 100m
Dia. at the upper e nd.
D I '" (lOO 111111 '" 0.6 111
"'
"'
Arc~.
1t
2
1t
,
AI= - DI = - x ( .6) -
4
4
= 0 .2827 111 2
c'"
PI'" pressure at upper end 0 ~.yP./= 19.62 N/cI11-,
'
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Fig . 6.5
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265 1
Dynamics of Fluid Flow
4
'" 19.62 X 10 N/m2
Di a. at lo we r e nd ,
D2 '" 300 111m '" 0.3 111
: . Arca.
A , '" - D,' '" - (.3) - '" 0 .07068
•
4 •
4
/t
,
I(
,
Q = ralC o f now = 50 lit rcsls =
Ld
lh ~
datum Ii Jlc
III
..22....
'" 0 .05 m31s
1()()()
th ro ugh (h e ce nl re o f the lowe r end.
p a~s
T hen
Zl = 0
As slope is 1 in 30 mean s
ZI'" -
Also we know
Q=A 1V1=A1V1
1
30
10
)( HlO = -
3
III
VI = Q = 0.05 = 0 . 176& m/sec = 0. 177 Ill/s
A
.2827
V, '" .Q. =
ami
+
Al
~
.07068
= 0.707 4 m/sec = 0.70 7 Ill/s
Apply ing Be rno ulli 's eq uati on a1 scc li o lls ( 1) and (2) , we gel
Pl
-
,
V,
+-
pg2g
"'
0'
,
P' V,
+ZI = - ' + - '- + ,.
pg
2g
+
19.62 )(10'
.177l
10
Pl
.707 2
+0
+
+ - "' - +
looox9.8 1 2 x 9.81
3
pg 2x9.81
20 + 0.00 1596 + 3.334 '" Pl + 0.0254
pg
23.335 - 0,0254 = -;;:""
-;- PL'"'"
looox9.8 1
Ill '" 23.3 x 98 10 Nlm 1 '" 228573 N/m 1 = 22.857 N/cm 1. Ans .
or
... 6 .6
BERNOULLI 'S EQUATION FOR REAL FLUID
The Be rno ulli 's equat io n was de ri ved o n th e assumpt io n thm fluid is invis.:id (non-\' is,",ous) and
the refore fricti o nl ess. But a ll the rea ll1uids Jre viscous and he n,",e o ffer resistan ce to now. Thus there
arc alwa>'s so mc losses in fl uid flows and hence in the applic ati o n of Bc rno ulli' s equati o n, thcse losses
hJ\'e to be ta ~ e n into conside ration. Thus the Bc rn oulli 's equ ati o n (o r re al flu ids betwee n points I and
2 is g il'c n 3S
,
,
..b.+~ + Z l = P2 + ~
pg 2g
pg 2g
where
I I
+Z2 + !JL
.,.(6.5)
"L is loss of ene rgy betwee n points I and 2,
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1266
Fluid Mechanics
Problem 6.7 A pipe of diameter 400,mm carries water,III (/ I'elocity of 25 mls. The pre.~,~ure.f lU Ihe
points A and B are gil'ell as 29.43 Nkm- "lilt 21.563 Nk",- respec/iI-ely ...hile IIII' da/lun /,e(l(/ {/( A and
B "re 28 m lind 30 m. Find I/,e IOH' of head ber".een A lind B.
Solution. Given:
Dia. of pipe.
D = 400 mm '" 0.4
Vclocily.
V=25mfs
At point A ,
111
4
<'~
I'll '" I' '"
25
,
~9""'~
ZB
_ _ _ _D
o ATuM l LINE
=
fig. 6.6
29.43 x [ O~
25'
+
+ 28
IOOOx 9.81
2x9.8 1
'" 30 + 3 1.85 + 28 '" 89.85 111
PH'" 22.563 Nlcm' '" 22563 x 104 Nlm l
zo",30m
\'/1 '" I ' '"
Total Crlcrgy at B,
~?l
""?~-~
Ill/s
Total energy at A.
At Ilo ini B .
•
"" = 29.43 Nlcm 2 :: 29.43 X 10 N/m 2
~
z" = 28 III
1:P ((\(I'I~\C~
V" '" 25 mls
P
1,2
P8
2g
Eo= ....!. + ....!. + ZB
4
'" 22.563xt0 + 25'
+30=23 + 3 1.85+30=84.85m
JOOOx9.8 1 2x9.8 1
Loss of e nergy
= Ell - EIJ= 89.85 - 84.85 '" 5.0 m. Ans.
Problem 6.8 A conical lube of lengrh 2.0 m is fixed I'erlicallv wirh ils smaller end upwards. The
~'elociry offlow ar fhe smaller elld is 5 mls wllile at rhe lower end if is 2 mls. The pressure head ar fhe
l'mllller<'IId is 2.5111 ofliqw·d. The loss of liead ill file IIIbe
i,~ 0.35(\,/ -i'll: . wllere 1'1 is file I'docify at
2,
Ille SlIIlIlIer end lIlid v: {II lile lower end relpeCfil·dy. D ef<'fmim: Iile pfeSJ'llfe ilead af Iile lo ..... er end.
Flow lakes pl(lct' in lile dOIl'nW(lrd direcfion.
Solution. LCllhc smalle r clld is rcprcsclllcd by (I) alld lowcrcild by (2)
Givell :
U:llgth of lube.
L=2.0m
1' 1" 5 nils
P1/pg = 2.5 m of liquid
\'1 '" 2 mls
Loss of head
'------+1-----' ®
Fi g. 6.7
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Dynamics of Fluid Flow
0.35 ]5 - 2]'
"
28
0.35 x 9
'"
2 x9.8 1
267 1
:=U. 16 m
{/, "'?
Pressure he ad.
PH
App ly ing Be rno ulli 's equ ation at secti ons ( I) and (2). we gel
",
lg
,.2
pg 2 g - -
12. + _1 + z, ,,, Pl + -'- + z, + h/
pg
Le t the datu m li ne passes th ro ugh see ti o ll (2). Theil Z2 '" 0 , Zj '" 2.0
5'
2.5+~',-,,, + 2.0=
2 x 9.8 1
2.5 + 1.27 + 2.0 '"
Problem 6.9
P1
pg
2'
+~o",, + O+O. 1 6
2)( 9.81
l!.l.. + 0.203 + .16
pg
p,
- - '" (2 .5 + 1.27 + 2.0) - (.203 + . 16)
pg
'" 5.77 - .363 '" 5.407 III of nuid . Ans.
A flipclille ca r,-yillg oil of .~pedjic gral'iry 0.87, chatlge,~ ill iliameler from zoo mm
dioll!efer (I{ (/ 11O.lllioll A /0 500 mil! diameter at a po.filioll B which is 4 melres at a hig/IeT lel'el. If lile
fI' f'Swres at A all</ n arc 9.81 Nlem' and 5.886 N/cm' respec /i"ely a",1 IIIe disdwrge is ZOO IiIresls
de/ami'le the loss of II<' ad Wid direc/iOIl of flow .
Solution . Disc hargc,
Sp. gr. of oil
Q = 200 li tis = 0.2 m 31s
= 0.87
kg
p for oil
=.87x 1000= 870 - ,
T
m
=200 111111 =0.2 m
Givcn: AI s l'(! lion A .
DII
Area.
All =
'm
~ (D~ ) 2 = ~ (.2)2
4
4
= 0.03 14 m 2
PA = 9.S1 Nlcm"
=9.Sl x 1 0~N/m l
Fig. 6.8
If d atum li ne is passi ng throu gh A, th en
ZII = 0
VII =
JL = ~ = 6 .369 Illis
AA
0.031 4
At s rtliun fl ,
DIJ = 500 mm = 0.50
Arc ~ .
AB= "4DB ="4 ( .5>-=0.1 963 111 -
If
l
/!
III
'
,
PB = 5.886 Nlcm-, = 5.886 x 10" Nlrn -
I I
Ii
~ I
IL
1268 Fluid Mechanics
2/1 '" 4.0
III
Q
02
Arc ~
.1963
vB = - - " - - = 1.018111/5
V'
=E~=&+.....:1.. + Z...
pg
28
Total e nergy al A
'"
Total e ne rgy at B
9.8 1)( 10 '
870x9.8 1
+
(6.369)1
2x9.8 1
p~
vi
pg
2g
+ 0= [1.49 + 2.067: 13.S57 m
=E8 = - + - + z,
=
5.886 x 10"
+
(1.0 18):
+ 4.0 = 6.896 + 0.052 + 4.0 = 10.948 m
870x9.81
2x9.8 1
(i) Directio n a rn ow. As Ell is more Ihan E8 and hence flow is taking place from A to 8. An .•.
(ii) Losso fh ead =IIL=E,, - E 8 = 13.557 - IO.943 = 2.609rn . An s•
.. 6.7
PRACTICAL APPLICATIONS OF BERNOULLI 'S EQUATION
Bernoulli' s equatio n is appli ed in all problems of incomprcssi bk fluid now where ene rgy cons;(I erations are in vo lve d. But we s hall consider its applicat ion 10 the follow ing measuring devices:
l. Vemurimctcr.
2. Orifice meter.
3 . Pilot-lube.
6 . 7 . 1 Venturi meter. /I. vc nturimctcr is a device used for measuring the rate of a flow o f a fluid
flowing through a pipe. It co nsists of three pans:
(i) A s hort converging parI. (ii) Throat. and ( iii) Di ve rging pan. II is hascd on the Principle of
Bernoulli' s eq uati on.
Expression for ute of flow through ve nturi meter
Consider ~ vc murim eter fllted in a hori zoillal pipe through whil:h a fluid is flowing (say water).
show n in Fig. 6.9.
Let til = diame ter at in le t or at SCl:tion ( I).
PI = pressure at section ( I)
1'1 = ve locit y of fluid at sec tion ( 1).
<I S
1
_ T
- '.
" "I
.
a == area at section
(I ) == 1.
4
and
tl2 _ P 2' "2' £l2 arc correspondin g va lu es at secti on (2).
Apply ing Berno ulli' s equation at sect ions ( I ) and (2). we get
,
HROAT
,
F Ig. 6.9
Venlur;meUr.
.£!.. + s... + ZI== P2 + ~+Zl
pg
pg
2g
As pipe is horizonwi. he nt"e ZI ==
PI
2
1'1
2g
Zl
Pl
vi
- +- " - +pg 2g
pg 2g
I I
or
2
P I ~ PI _ I'; _ 1'1
.
-
pg
2g
28
Ii
~ I
IL
Dynamics of Fluid Flow
BUI
PI-P2
pg
is Ihe difference of prc:;sure head s m scc~ iulls I and 2 and it is equal 10 " or
269 1
Pl - P2
pg
== II
S ub~ti!uling Ihi~ val ue of PI - PI in Ih e abo ve equation. we gel
pg
,,2
/
2g
28
11= -'- -'
..,(6.6)
Now appl ying cont inuit y equatio n al sec tion s I and 2
(I,V
I
vl= ~
Substituting this value of 1'1 in equatio n (6.6)
" [";:'r
" = -' 2g
.[ '] vi'["d,· -, (/; ]
"i
"--';f::"-28
2g
1 _ (j~
(/,-
=
28
(Ii
x
~2 gh
"'
Discharge.
Q '"
==
" 2" 2
<1 2
II I
Ja~ ai
)(
~2gh"
(I I":
Ja~
a;
...(6.7)
Equation (6.7 ) gives lh e discharge ullder ideal condition s and is ca ll ed . theoretic al disdtargc. Act ual
discharge will be less than th eoretical discharge.
Q=
Cx
"'"
where
d
"I'"
I'
'lU i
7
,
ai
x
2,,,
.J2ih
...(6.8)
Cd == Co -efficie nt o f vc nlurimdcr and irs va lu e is less than I.
Value of ' h' ,iven by differential U· tube milnometer
Case I. Let the differential
throu gh thc pipe. Lct
manom~ te r
contai ns a liquid which is heavier than
th~
liquid flowing
Sp. gra vi ty of thc ht.> a l·ier liquid
So '" Sp. gravi ty of the liquid flowing throu gh pipe
.\" '" Differe nce o f th e hcavier liquid column in U-tubc
S h :=
T hen
"=.,[,,s" -,]
...(6 .9)
Cllse II. If the dilTerc nt ial manometer contai ns a liquid which is lig hter than th e liquid nowing
through the pipe. th e value of II is give n by
~ I
I~
~ I
IL
1270 Fluid Mechanics
... (6. 10)
whe re
5, = Sr. gr. of lighter liquid in U·1Ube
So = Sp. gr. of fluid flowi ng through pipe
.r = D ifference o f Ihe lighter liquid columns in U -lube.
Case III . Incl ined Ventu r lrn eler wi th mrr.,renilu l U. lu be mu norn" le r. The above two cas.cs are
given for a horizunlal \'CnmrirnClcr. This case is rdated \1.1 inclined vcnturimClcr having d iffcrcmial
V-lube Inanometer. LClth., diffcrcIltialllHlrlOIllCler contains heavier liquid [hell" is given as
,,= (;~ + ll) - (~ + Zl)
= .r
[~: -I]
... (6.11)
Case IV. Similarly. for inclined \,cnlurimeli:r in which differential IllallomCler contains a liquid
which is lighter Ihan lhe liquid flowing through Ihe pipe. lite value of II is given as
II = (;~ +Z, ) - (~; + Zl) =x [I-
:J
... (6. 12)
Problem 6.10 A horizolltal relllllrimelf'r wilh inlet and Il"oat diameters 30 em and 15 em respectinny is I/Sed to mellsure Ihe flow of 'WlIer. Ti,e rellding of differenti,,' manomeler conllecled to II,e
inlet and Ihe rhrom ;.1 20 em of mercury. Dererlllille rile flUe of flow. Ta/(e Cd '" 0.98.
S olution. Given:
Dia. at inlet.
Area at inlet,
a,
,
= ~/f <f , -=
-/f
4
d 2 = 15cm
Dia. at throat.
4
,
,
(30)" = 706.85 em-
a,,,,~xI51"'176.7cm2
.
4
Cd '" 0.98
Reading of different ial manometer'" of '" 20 cm of mcrcury.
Difference of pressure head is given by (6.9)
where Sh = Sp. gral'ity of me rcu ry'" 13.6. So '" Sp. gral'ity o f water =
1
=20 [ -136
, - -I
=20 x 12.6cm = 252.0 em ofwatcr.
The discha rge through \'cnturimcter is givcn by cqn. (6.8)
Q=C
II , a,
, J'a ,
'" 0.98 x
-
,x
.Jfiii
28/'
"2
706.85 x 176.7
x ~2 x 9.& 1 x 252
J(706.85)2 _( 176.7)2
~ I
I~
~ I
IL
Dynamics of Fluid Flow
&606759336
=~~~=
'/499636.9 3 1222.9
'" 125756
C1Il
3
/s '"
1~:6
271 1
86067 593.36
684.4
litIs", 125.756 litis. An s.
Problem 6.11 An oil of sp. gr. 0.8 is flowing !hrougll (I l'en7llrillleler /1(I)';n8 illier diameter 20 em
and IhrOal diameter /0 em. Th e oi/·mere",)' differential mOllomeler sllOWS ( 1 reading of 25 em. CalcuIme lite discharge of ai/through IIII' horizol1lal 1'l'lIlUrimeler. Toke Cd '" 0.98.
Solution. Given:
Sr . gr. of oil.
So '" 0.8
s~ " 13.6
R eadin g of differential manometer. x:= 25 em
Sp. gr. of merc ury.
DifJcrcncc of pressure head. II"
.f [
~:
I]
-
1
13.6
:25 [ - -I cm o f o il =25 [! 7 - 1J = 400c lll ofoi l.
0.8
Dia. nt inl et.
rr
,
,
"4 x 20" '" 31 4 .16 c m d 2 '" IQem
"1 '" ~ X 102 ", 78.54 cm 1
4
CJ = 0.98
Th e discharg.-:
Q is g i wn by Cq U31io rl (6.8)
Q=C
(/ 1(/,
dJ
2
1
" I - U,
'" 0.9& x
~I
Xv"'!;"
314.16 x 78.54
Jc.~ 14.1 6)1
2 1421375.68
J98696
6 168
x
J2 X 98 1 x 400
(785 4 )1
1/
2 1421375.68
"
3().1
(; 111
S
" 70465 CI111/S = 70.465 1itn' 'iI.~. An s.
Problem 6.12 A horiZOl1tlll l"el1turimeter .... ilh il1lel di","eler 20 e ll! allli Ihroat diameter 10 em is
used to ",e"sure Ihe flow of oil of Sf!. gr. 0.8. TIl e discharge of oi/through "e1lluri",eter is 60 litres/s.
Fil1d 'he re(U/ili g of Ille oil·ml'rCliry diJferellfill1 mO/Jometer. Tuke CJ = 0.98.
Solution . Gi vc n :
<I, = 20 cm
rr
,
,
", = - 20- = 31 4.1 6 cm4
<ll =lOc m
I I
Ii
~ I
IL
1272 Fluid Mechanics
a, '" ~ x 102 ", 78.54 em!
•
4
Cd == 0.98
Q = 60 IilTCsls = 60
x
IQOO .::rn 1ls
Using the equati on (6.8).
60
"'
314.16x78.54
x 1000 = 9 .81 x c.;;;;;;:;:,;~~..".
x .J2 x 981 x II
J(314.l6)~ _(78.54)2
=
I07 1068.78Jh
304
IT
304 x 60000
"h = 107 1068.78 = 17.029
"'
II '" (17.029)" = 289.98 nn of oil
B~
II= '{~: -I]
wh ere ::i/o = Sp. gr. of me rcury = 13.6
50 = Sr. gr. o f oil = 0.8
.f == Reading of manometer
239.9S=x [ 13.6_ 1] = 16.<
0.8
289,9 8
X= - - -
16
= 18. 12 em.
Rcadin g o f oil-mercury differential manom eter", 18.12 em . Ans.
Problem 6.13 A IlOriWl!lai I'cl1lUrimelcr )I"rlll ililet diamerer 20 em alld Ihro(l/ diameter 10 em is
u!led 10 measure 'he jlow of w(I/er. The pressure (1/ ;,,/el is /7.658 Niem i alld the "('euum pressu re (1/
the IhrO(l/ is JO em a/mercury_ Fi" d the disclwrge ojlt/(lier through ",mlllri",cler. Take Cd '" 0.98.
Solution. Gh'cn :
Dia. :II inlet.
al
Dia. at throat.
,
'" -
.
,
x (20t '" 3 14. 16 em"
4
d1", IO em
.
"
a, '" - x 10' '" 7&.74 em '
. 4
PI'" 17.6$8 Nfem z '" 17.658 x 104 Nfm 1
p for wate r
'" 10<Xl
[.;: ~
m
and
p
17.658 X 10·
pg
9.81 x 1000
_, '"
'" 18 1ll of wata
p,
- - '" - 30 cm of mercury
pg
'" - 0.30 m of lI1~rcu ry '" - 0.30 )( 13.6'" - 4.08 m of
I I
wat~r
Ii
~ I
IL
Dynamics of Fluid Flow
= II = Ji _ P, '" 18-(-4.08)
pg
pg
'" 18 + 4.08 '" 22.08 III of w ater
DilTcrc nt ial head
'"
2208
273 1
em o f wat<'r
The di!\Chargc Q is given by eq uation (6.8)
'" 0.98 x
3 14. 16 x 78.54
x .jr"C,"9ii'''IC,C2n2'''0''
J(3 14.1 6) 1 - (78.74)'
'" 50328837.2 1 x 165555 cm 3/s
=:
165.555 litis. Ans .
304
Problem 6.14
The ;II/el 'IIld I/"oal diameters of II horiWllla/ l'"n(urimt'ler are 30 em ,,,,d
10 em respeClil'('ly. Tile liquid flowing l11rollgll Ihe meIer is water. The flrCn"llft' i,llen)"ily at i" lel i.~
f 3.734 Nkm l wllile I/'e mel'"m pn<sS14re head (II the Ih ro(l/ is 37 em a/mercury. Find fhe mit' of flow.
Assume liIal 4 % o/Ihe differential head is lost be/ween IIII' inlel "lid Illroal. Filla 1,Iso l/ie ..a/ue of Cd
for Ihe I'fmll>rimeler.
Solution. Given:
Dia. at in let.
"
", '" - (30r = 706.85 cm-,
4
Dia. at throat.
d 2 =lOcm
"
"1 =="4 (10) == 78.54 eln-'
Pressure.
Pressure head.
P, = 13.734 Nfcm 1 = 13.714 x 104 N/m 2
p
--.!..
pg
p,
~
pg
==
13.714 x 10'
'C2;:':'ii'c
= 14 m of water
IOOOx9.8 1
,,- 37 em o f mercury
= -37 x 13.6
100
Differentia l head.
In
of water" - 5. 032 m of water
II '" !'I/pg - pipg
= 14.0 - (- 5.032) " 14.0 + 5.032
== 19.032 m uf w ater = 1903.2 em
Head lost.
4
1I/ =4%ofI1= 100 x 19.032=0.1613 m
Cd "
~h-hl
"
I I
==
19.032
.76 13
19.032
== 0.98
Ii
~ I
IL
1274
Fluid Mechanics
Discharge
'" 0.98 x 706.85 x 78.54 x ./2 x 98 1x 1903.2
J(706.85) ' -(78.54f
[05[32247.8
3
" ",;::~"'~~'" '" 149692.8 cm /s '" 0.1 4969 mJ/s. ,\IIS.
,1499636.9 6168
PROBLEMS ON INCLINED VENTURIMETER
Problem 6 .15 A 30 em x 15 em )'emurimeler is iliseY/cd i/l {l >'crlic,,/ pipe nlfryillg "'(lIer, flo K'illg
in Ihe upward direc/iOlL A different;,,/ mI'TCIl'Y mU/lomder C01mccll'd /0 'he inlel (md IhTool girl'S II
Teadillg 0[ 20 em. Find Ihe discharge. Take Cd := 0.98.
Solution. Given:
Di3. at inlet.
d,=30cm
{/, = !:. (30)2 = 706.85
Dia. at th roat.
cm 2
4
d l = 15cm
"
a,= - ( 15)-= 176.7cl11
-
4
,
[.s~ -1]= 20 [13.6 _LO] = 20x 12.6 = 252 .0 em of wate r
II "' .r
So
1.0
Cd = 0.98
Discharge,
,
Q= C
(1 , (/,
/,
"a
l
, x ,/28/'
{/ l
706.85 x 176.7
= 0.98 x [:~'§¥~~~ x ./2 x 981 x 252
~(706.85)1 (176.7 )1
8606759336
" rtiiii~~
./499636.3 31222.9 "
86067593.36
684.4
= 125756 cm 3 1s = 125.756 litis. Am.
Problem 6.16 A 20 CIII X 10 cm renlurimel/!f is i"sel·'ed in a rerlin,l pipe earning oil ofsp. gr. 0.8.
Ib e flo .... of oil is in "fI'nm/ direClion. Tbe tlijJerenn! of /el"e/s hel"·een Ille II'TOal and in/el J·eclion iJ·
50 cm . The oil merc"ry differenlial manometer Ril"es a reading of )0 em of mercury. Find li'I!
tlisc/wrge of oil. Neglectlon·es.
Solution. Dia. at inlet.
d , = 20 em
"
a , = - (20) - = 3 14. 16cm
Dia. at throat .
I I
4
d z =lOem
,
Ii
~ I
IL
275 1
Dynamics of Fluid Flow
{/! '"
Sp. gr. of oil.
Sp. gr. o f mercury.
'4" (10)' '" 7854 em ,
So '" 0.8
S~ = 13.6
Differential mano met er reading. x '" 30 e rn
=30
13.6 [ ~0.8
1
1
=30(17 - 1]= 30x 16 =480crn of oil
T he discharge.
'"
'"
1.0 x 3 14.1 6 x 78.54
)( J2 x 98 1 )(480 c m3/s
J(3 14.16)1 - (7854 )1
23932630.7
J
•
= 78725.75 em Is", 78.725 llt resls. Ans.
304
Problem 6.17 I" (I "erlic,,/ pipe cOlll'eying oil of specific 8m)'il), 0.8, Iwo pressure gllllge:; hU\"I:
been illS/ailed (II A and n where 1/", diameters are 16 ClIIllnd 8 em respectire/y. A is 2 melres abo",: 8.
The preSJure glwge readings 1/(/\'/: s/loll'n Illallhe pres.illTe al B is greater t!ian III A by 0.98/ Nlem",
Neglectillg all losses. cu/cu/a/e I/!e flow f(!le. Iflhe glluges (II A alUi B aTe rep/aced by lUbes filled wilh
Ihe Jllllle liquid (//1(1 collnec/ed 10 a lj·lllbe COIIWilling mercllry. ClI/cli/aie Ih,. difference lIf lel'eI of
mercllry ill Ihe two limbs oflhe U-lIIb,..
Solution. Given:
Sp. gr. of oil.
S,, '" O.S
Density.
P '" O.S x 1000 '" SOO
DA
Dia. at A.
",
k.~
16cm
m
16 cm",0. 16m
-'
®
AI=~(.16)''''0.020Im'
Area at A.
Dia. at 8.
D IJ '"
8 e m'" 0.08
(i) Differe nce of pressures. PR - p", '" O.9S1 Nlc m'
'" 0.98 1 x
Ht
9SlON
Nlm ' '" - ,-
®
f-t-LG\
, ~ _I
m
Fig. 6.9 (,,)
Differe nce of pressure head
p
p
8 -
pg
I I
'm
m
A, '" -" (.OS) - '" 0.005026 m ,
• 4
Area at 8.
t--+----f'-"':)
A '"
98 10
;;;:""~7 '" 1.25
800 x 9.8 1
Ii
~ I
IL
1276 Fluid Mechanics
Applying Bernoull i's th eore m at A and B and takin g the refere nce line passing through sec tion B.
we gel
PA
pg
PB.
P,
- - - -ZA~Z8=
P"'p~PB
(
]
vi v;
- - 2g
2g
v'v'
+2.0-0.0= 2; - 2;
v"
Vl
2g
2g
-1.25+2.0:....!..- --<i..
V"
V~
O.75=....!...-.....:L
2g
28
... ( i)
Now appl ying conti nuity equation at A and B. we gel
VA xAI = V8 xA!
L
!
®
0'
/
,
Substituting the va lu e o f VI! in ctjualion (i). we gel
®
0.75 = L6V,;' _ V; " 15V;
28
V" '"
Rate of nuw.
28
2g
O.75X2X9.8 1
J
15
_
'" 0.99 m/s
,
J
,
T
.L
Fig. 6.9 (h)
Q=V... XA 1
'" 0.99 xO.OlOI '" 0.01 989 m'/s. Ail S.
(ii) DifFcrcn;;c of le ve l of mercury in the U-lu be.
u,1 II "" Differe nce of mercury level.
h=X(~:-I)
Then
where
""
"'"
(~; + Z... J- (~; + 2 8) = ,p"'Cp-C8'P~8 + Z... -
Z/I
=-1.25+2.0-0
PB - P... ==
" 0.75
pg
1
1.25]
136
- - 1 =xx 16
0.75=_1' [ -
0'
x'" 0.75 '" O.(l4687
III
= 4.687 e m . Ail S.
16
I I
Ii
~ I
IL
Dynamics of Fluid Flow
277 1
Problem 6.18 Find Ihe diRharge of wmer flowing II/rough a pipe 30 em diameter placed in WI
inclilled POSilioll where a )'{mll/rime/a is itlserled. /wI'ing a th roat diameter of 15 CIII. Ti,e difference of
pressure belween lilt: main mid Ihroar is mcas" red by Ii liqu it! of sp. gr. 0.6 ill an jm'a/ed U -/" be which
girl'S {/ , e(liIing of 30 Clil. The lo.H a/head be/ween Ihe lII{1illlllU/IIIfQa/ is 0.1 limes Ille killerle head of
Ih e pipe.
Solution.
Dia. at ink1.
til'" 30 I:In
" (30)
(/1'" -
4
Dia. al throat.
= 706.85 em
,
II
til'" 15cm
"
a,= - { 15t= 176.7cl11',
,
4
Rc~ding of differenti al manometer. x = 30 em
Difkrcnce o f pressure head. II is give" by
(~~ + Zl ) - ( ~; +Zl)=11
A lso
where 51 " 0.6 and So " 1.0
= 30 [ 1- 0.6 ] = JOx.4 = 12.0c1I1 ofwa!cr
1.0
Fig. 6.10
"
Loss of head. ilL = 0.2 x kinetic head of pipe = 0.2 x - '
2g
Now app lying Bernoulli' s equation at sections (I) and (2). we get
(:~ +'1) - (~ + Zl) +;: -;~
0"
(:~ + Zl ) - (~ + Zl) =11= 12.0cm o f wmcr
and
\'1"
12.0+ -
28
12.0 + 0.8
~2g
I I
, .;
- -'
2g
1
=0.2 X VI
,.; .0
2g
28
... (1 )
Ii
~ I
IL
1278 Fluid Mech a nics
Apply ing continuit y eq uation a1 sect ions (I) and (2), we gel
(1\,,\ '"
al".
tt
,
4 (15) ~l
If
4
=~
4
(3of
Substituting tliis valu e of 1', in eq uation (I ) , we gel
1'; ["'
16- 11== 0
v; '" 0
28 4 - 2~
12.0 + 0.8 (",)"
or 12.0 + 2~
-";'- [.05 - 1[ =- 12.0 or 0.951'i == 12.0
2,
2,
2x98 1x12.0
0.95
Disclia rgc
'" 157.4 emfs
= a 2 "l
'" 176.7 x 157.4 cm 31S:o 27800 cml/s
= 27.8 lil rt."Sls. Ail S.
Problem 6.19 A 301.'111 x 15 em \'tmmrimeler is pro l'ided in a )'eflim/ pipe lille carrying oil of
specific grlll·ily 0.9, Ihe flow being upwurds. The difference in eiel'alion of the I/Iroa/ sec/ion and
e/!lrance sectioll of the I'emurimeler is 30 1.'111. The differential U-lIIbe mercury mallometer shows a
gauge deflectioll of 25 cm. Calculate:
(i) Ille di.~c1/arge of oil. lind
(ii) the pressure differellce between rile elllrance secriol1 and rile rlima! secriol1. Take rlie
cQ-efficiellr of discliarge lIS 0.98 (mu ~·pecific gra,·iry of mercury as 13.6.
Solution. G iven:
Dia. at inl et.
ti, = 30 em
Area.
Dia. at throat.
Area.
a , = ..:: (30)2 = 706.85 c m 2
4
d 2 = 15cm
tt
,
a,= - (15) = 176.7 em 2
,
4
u.,t section ( I) rcprescil is inle t and sect io n (2) represents throat. Then Z2 - Z, = 30 cm
S" = 0.9
S~ = 13.6
Read ing of diff. manometer. .l = 25 cm
The differen tial head. II is g ivcn by
Sr· gr. of oil.
Sr. g r. o f mercury.
/,=
(.E.!.
+ZI) - (!!.pgJ. . +Z1)
pg
=X
[' 1 [".6 1
~- 1
So
I I
=25 - - - I = 352.77 Clll o f oi l
0.9
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IL
279 1
Dynamics of Fluid Flow
(i) Th e discharge. Q of oil
0.98)(706.85)( 176.7
12 98
3'
'" ~(706.85)
="
x
Ix 5_.77
1 _ (176.7)l
'" 101832219.9", 148790.5 cl11 3/s
684.4
= 148.79 lilrl's/s. AilS.
(ii) Pressure diffucnce betwee n en trance and throat seeli on
I,: (ll.+
ZI)-(l!l.pg +Zl) =352.77
pg
1
!/
,
,
!<D
"'
"m
r"O
li _ Pl)_ 30 '" 352.77
( pg pg
,
~
~
ttf
Fig. 6.11
.Ii _.!!.J... '" 352.77 + 30 '" 382.77
pg
pg
(PI - Pl) '" 3.8277 )(
ern of oi l = 3.8277 m or oil. AilS.
pg
= Sp. gr. o f o il x 1000 kg/Ill l
But dt nsi ty of o il
'" 0.9 )( 1000 = 900 kg/em'
N
(PI - p,) '" 3.8277 )( 900 x 9.8 1 - ,
-
111 -
33795
1
1
: -, - Nlc m '" 3.J7~ Ntcm . AilS.
10
Problem 6.20 Crude oil of specific gral-it)' 0 .85 flo .... s up ..... ards ar a roll/me rale offloll' 0[60 lilre
per second II"Of/gil a I'err;cal I'ellwrimeter .... illI lUi inlel diameler of zoo mm and a Ihroal diameter of
100 mm. Tile co-efficielll of di.~c1w'ge of Ille rel/llfrimeler is 0.98. The I'ulical di.ltance belween Ihe
pre.~.mre tappings is J(}() 111m.
(i) If Iwo preJ'sure gauges are cunnee led 01 Ihe lappings sIIch Ilwl liley are posilioned 01 'h e (el'els
of Iheir corre,"pondillg lapp;'lg poinls, determine Ihe difference of readings in Nkm 1 of Ihe two pfeSJ'ure glmges.
Iii) If a men""y differelilial m(lIiomeler is COlinected. in fllace uf preSJ'ure gauges. 10 tile tappingJ'
J'uch Ihal Iile cOli/leCling lube IIfllO merc'lf)" are filled wilh oil. delermine Ihe dif/erenCl' iii 111<' level of
'he mercury CO/Willi.
Solution. Gi ven:
Specific gra vi ty of oil.
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1280
Fluid Mechanics
Densit y.
Discha rg",
p - 0.85 x 1000 850 ~ g/Ill J
Q - 60 litrels
60
,
· - - =O.06 m /s
1000
Inle t dia.
tI, '" 200
a, "' "4" (.2)
Arca.
Throat d ia .•
0.2
lllill '"
I
~
IJ
III
'" O.03 14m
,
til'" 100 mill '" a. lm
4"
a, = - (0. 1) '" 0.00785
Area.
•
Ill ''
,
JOO mm
..L
Val ue of Cd
'" 0.98
Let section ( 1) represents in let and sect ion (2)
re presents throa t. Then
I
(j)
!
(il Difference of readings in Nfc ml of tile /h'O prl'SSlIre gauges
The discharge Q is given by.
,~
(/,(/,.
'
(I,
0'
-
,
(11
x
.Jfih
2,1,
0.06 = 0.98 x 0.Q314 x 0.00785 x
~O.03 1 4 z _ 0.00785 '
'" 0.98
\=
I,
1
Zl- Z, = 300 111111 = 0.3 In
Q=C
--
~I
!
T,
~
...
F ig. 6. 11 (a)
..j'2c,",".8"'c,c,',
x 0.00024649 x 4.429 ,fh
0.0304
.[h =
O.U6 x 0.0304
= 1.705
0.98 x 0.00024649 x 4.429
II'" 1.7051
BIH for a venica ] vC llturimc tcr. II =
2.908",
(;~
= 2.908 m
+:,) -(;;
+ l, )
(li zl)_(fi Zl) . (l!J.. _li)
pg
+
pg
+
pg
pg
+ z, -
Zl
= 3.208 III of oil
p, - Pl =pgx3.208
'" 350 x 9.8 1 x 3.208 N /m ~ '" 850 x 9.8~ x 3.208 N/c m ~
10
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IL
Dynamics of Fluid Flow
(ii) Difference
ill
rhe /el'el.l of mereul)'
The v alue of 10 is g i ven by.
COIIIIllIl ,\'
281 1
(i.e .. x)
iI=_r [~: -I]
2.908:x[ [3.6 - I] :x I1 6 - 1)= I S.l
0.85
2.908
.r '" - - '" 0. 1938 m '" 19.JS em of 011. Ans.
IS
Problem 6.21 III II 100 mm diameter iJoriZO/lfa/ pipe (/ l'elllUrimeler 0/0.5 cOlltractioll ratio /1</.1
bee:nfi):ed. Ti,e Ileail afwater 011 Ille melre w/ltm Illere ;.f no flow is 3 III (gauge). Find lire rate afjlalt.'
fur ".hiel, Ihe Ihroal pressu re wm be 2 melres ofwlller ahsolule. The co-efficien{ 0/ diSC/llIrge is 0.97.
r "ke Mlllospileric preSJ'llfe ileml = fO.3111 o!wlIfer.
Solution. Gi ven
Dia. of pi pe.
il, '" 100 mm '" 10 em
Area,
Dia. at throat.
il1 '" 0.5 X il, '" 0.5)( 10 '" S em
Area,
"
(/, '" - (S) '" 19.635 em
•
4
,
H ead of water for no flow
'" i!..L '" 3 m (ga uge ) '" 3 + 10.3", 13.3 In (abs.)
p,
Throat pressure he ad
p,
= - ' = 2 III of wa~er absolute.
pg
-.
P, p,
DI !fere nec of press ure lIe ad. II '" - - - ' : 13.3 - 2.0: 1 1.3 m : 1130 e m
pg
Ra~cofflow.
pg
Q is~iven by Q= Cd
:0.97)(
78.54 )(1 9.635
xJ2x98 1x ll 30
J(7&54)' - (19.635)'
:
22273 18.17
76
3
•
: 29306.8 cm Is : 29.3Ofi hlres/s. Ans.
6. 7.2 Orifice Meter or Orifice Plate. It is a device u.',cd for measuring thc rak of flow of a
flu id tllrough a pipe. It is a cheaper devicc as co mpared to I'cnturimd cr. It also worb on the Same
pri ncip le as that of vcnturimetc r. It co nsists of a flat circular plate which has a ci rcular sharp edged
hole ca ll ed orifice. which is coocen tri c with the pipe. Tile orifice diamder is kept ge nerally 0.5 times
the diameter of Ihe pipe. though il Ill~y vary from 0.4 10 0.8 tim es th e pipe dimncter.
A differemial manomete r is connected Jt stttion (I). which is at a distan"" of aboul 1.5 to 2.0 tim<,s
the pipe diameter upstream from the orifl"e plate. and at section (2). whid is at a distance of about half
the diameter of Ihe orifl"e on the downstream side from Ihe orifice plate.
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1282 Fluid Mechanics
Let PI" pressure at section (I).
VI" velocity at section (I).
U("
area of pi pc al section ( II, and
<DI
PIPE
OR IFICE METER
12>
--
DIRECTION OF FLOW
.i
,
.-- D IFFERENTIALMANOMETER
T
Fig. 6.12. Orifiu m('ler.
Pl>
)'1' "1 arc corresponding values al section (2). Applying Bernoulli 's equation al sections (I) and
{21, we gel
1
PIl'I
- +pg 28
'I
pg
28
+',
(li+ZI)-(li+Zl) 28"i _ ~28
(Plpg ll) '" h"
( lpgi +
pg
3m
+
!
p,
",
== -" + -"-
==
pg
Zl)-
+
Differential head
Now section (2) is al the vcna·co nlracta and
the area of orifice then. we have
<12
represents [he area al the vena -cont racta. If "0 is
",
Cc = -
".
where Cc == Co-effident of contraction
"l="OXCc
... (ii)
By colltinuity equation. we lIavc
...(iii)
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Dynamics of Fluid Flow
,
283 1
(""J'
\', =2gl1+ ~
\', ==
J2gl1
'-[::1' c',
Substituting this va lue o f C< in equatio n (II'). we get
Q="o XC d
c',
'-[""1'
",
x
C;
,-[""
",]' ,-[::]'
CJuoat M
,- [",]'
",
~2g"
Ja~ -l/&
... (6. 13)
where Cd == Co-efficient of discharge for orifice meter.
The co-efficicill of discharg e for orifice mClCf is Illuc h smaller than that for a venturi meter.
Problem 6.22 All orifice lIIeler will! orifice diameter {O em is iliseT/cd ill a pipe of 20 em diameler.
The pressure gauges fifled upstream wId down.nream of Ihe orifice meIer gil'e.! readillgs of
19.62 N/cm 2 alld 9.81 Nlem z respeC/iI'ely. Co-efficietll of discharge for Ille orifice meier is gil'ell {IS
0.6. Filld Ille di,~clllI'ge of Waler IflrOllgll pipe.
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1284
Fluid Mechanics
Solution. Given
Dia. of orifice.
do'"
Area.
aO
] 0 em
"' "4" (10) '
'" 78.54 em
,
til'" 20 em
Dia. of pipe.
Area,
(11=
~ (20)1= 314.16c01 '
1
4
PI'" 19.62 N/(;m = 19.62)( 10 N/rn'
19.62)(104
P
---.!. '"
pg
-:-;O:::-""'
C- = 20 III of waIi: r
1000)(9.81
9.8 1x I0·
"" 10 m of water
1000 x 9.8 1
Similarly
II =
~ - p~ '" 20.0 - 10.0 = 10 m of water = 1000 em of wate r
pg
pg
Cd = 0.6
The discharge, Q is given by equalion (6.13)
'" 0.6 x
c~7~8".5"4",,,.;,3,,14~.l~6~ x .}2 x 981 x 1000
J (314. 16)' -(78.54)1
'" 20736838.09 '" 68213.28 cm 3 /s = 68.Z1 liires/s. Ans.
304
Problem 6.23 All orifice meier wilh orifice diameter 15 em is inserted ;11 IIp'pe of 30 elll tlir,meler.
Tile pressure diffeTe"c/! meaSllred hy II mercury oil differential m(mameleT Oil lire /"'0 sideJ' of lile
orij"'" meier gi.'es" reading of 50
of mercury. Find Ihe rMe offlow vf oil of sp. gr. 0.9 .,.hen the co efJicil'1lt of disch"rge of the orifice mcter = 0.64.
Solution. Giwn :
Oia. of orifice,
" 0 =15cm
"m
"
a o = - ( 15)"= 176.7 cm-,
4
til = 30 cm
Area.
Oia. of pipe,
"
(30) = 706.85 cm
4
Sp. gr. of oil.
So = 0.9
cm of mercury
Reading of diff. ll1anOlneter. x =
Area.
Oiffcr~l)l i al
I I
III
head,
=-
,
50
h= x [S,So - I] = 50 [13.6
-I] cm of oil
0 .9
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285 1
Dynamics of Fluid Flow
'" 50 x 14.11 '" 705.5 em o f o il
Cd '" 0.64
Th e rat e o rlh e fl ow. Q is given by eq uation (6. 13)
",0.64 x
176.7 x 706.85
xJ2x981x 705.5
~(7 06.85)2 _ (176.7)2
'" 94046317.78 '" 137414.25 cm 3/s= 137.414I11resJs. Ans.
684.4
6 . 7 .3 Pitot- tube. 11 is a d~vicc uscd for measuring the
velocity of now al all Y poil1l i n a pipe or a channel. It is baso:d on
th e principle Ihm i f the ve locit y of flow m a poill1 becomes le ro.
th e pressur.: there is in<:rcased due w the (;unvcrsion of til..., kine!i"
energy into pressun: energy. In its simp iesl ronn, Ihe pilot-mix:
consists of a g lass lube. bent at right ang les as shown in rig. 6. 13.
T
" :-C-,
~:":":":":":":":""""""""""":__;:-""
H
--q>---------------------.------------ -"l'1..-=,..!!
.i
The lower end. wlli ch is bent th rough 90 Q i s directed in [h e up -
--------------------------- . - '
stre3111 direction as show n in Fig. 6. [3. The liquid rises up in the
---------------------------------tu~ du~ to th ~ co n v~ rsion of kinetic e nergy into pressure energy.
The veloc it y is determ ined by meas urin g the risc o f liquid in the lU be. Fig. 6. IJ PiIQt-tl/be.
Consider two po ints (I) and (2) mt he sa me level in suc h a way that poim (2) is j ust as the inl et of
th e pitot-tube and point ( I) is far away f m m th e tube.
1'1: intensity of pressure at point <I )
\'1 = ve loc ity o f n ow at ([ )
P2 = pressure at poin t (2)
\'2 = ve loc ity at poin l (2), wllich is zero
H = d~rllh of tube in the liqu id
I,: ri se of liquid in the tube above th e free su rfa(;e.
App[y ing Bernou[[i' sequmion at points ( I ) and (2). we get
).
).
,-"+"
v,
- ' - + '1: - '
pg2g
pg 2 g p,
But ZI =
Z2
",
p,
as points (I) and (2) arc On th e sa me lin c and
.!i '" pressun: he ad at
pg
1'2
= O.
(I ) '" H
Po : pressure head at (2) : (II + Il)
pg
Substituting these values. we get
1,2
H , -'- =(h +/f)
2,
1,1
I,,, _ l
2,
or
1'1" J2g/r
This is th eore tica l ve loc ity. Actual veloc ity is given by
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IL
1286
Fluid Mechanics
(I',) "" '" C. J2g/r
where C , = Co-efficient of pilot-lube
Ve locity al an y point
... (6 . 14 )
"'" C• .j2gh
Velud ty of n ow in a Pi llt' by pitot-tu be. For findin g the veloc ity J i an y point ill a pipe by piwltube. the following arrangements arc adopted:
I. Pilot-lube al ong wilh a >'c l1ira] pie zo mder lube as sllown in Fi g. 6 .14.
2. Pitol-Iulk connected wilh pic;o;omc!er tube as shown in Fi g. 6.]5.
3. Pilot-tu be a nd vertica l pi ezo me ter \tJhc conn ected wilh a diffe re nt ial V- tube man o meter as
sh own in Fig. 6. 16.
PIEZOMETER .....
TUBE
1
"
if
V-PITOT.TUBE
~
-_
f~:: ~ =::=::=:=:=:=:=:=:=:=:=:
------------
= __ _-----------_
::."'" ----.
IT
... _-_._--
.. ., _- ......___--..
....
<
::::~ -:-:-:-:-:-:-:-:-:-:-:
':..-'10
-----------
---------------------
Fig. 6.16
4 . Pi lOt-stat ic lube. wllich co nsists o f two circu la r concel11ric lubeS o ne inside the ol her wilh some
annular space in between as show n in Fig. 6.1 7. The outlel of lh ese lwo lubes are co nn ec ted 10 the
diffe rcl11 i;ll mano meTe r whe re Th e d iffe rence of pressure head 'h' is measured by kn ow in g the
difference o f Th e levels o f Th e mallo mcter liquid sa y.\". Thell h
"'.f
[~:
-I}
Pro blem 6 .24 A pilOt-.I/atic lUbe placed in the cemre of a 300 mm pipe lille has one orifice pointing
upslream mid olher perpelli/icu/ilr 10 il. Ti,e meall I'e/oeily in II,e pipe is 0.80 of the cell/TtlI I·e/ocil)'.
"-ind II,e discharge Ihrougll the pipe If the pre.l$llre difference bet .....een the 11>'0 orifices i.l
60 mm af ..... ater. Take the co-effieielll of pitot IIIbe as C. '" 0.98.
Solu t io n. G iven:
Dia. of pipe.
d", 300 mm '" 0.30 III
Di ff. o f pr~ssure head.
I,,,, 60 mm o f wa ter '" .06 m of wate r
C. '" 0.98
Mean veloc iT y.
V", 0.80 x Central ve loci ty
Central ve locity is give n by eq uatio ll (6. 14)
'" C • .J2g11 = 0.98 x .J2 x 9.8 1 x.06 '" 1.063 m/s
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Dynamics of Fluid Flow
287 1
V "" 0.80 x 1.063 = 0.8504 m/s
Discharge.
Q = Ar~a of pipe x V
j'(
,_
== 4d- x V ==
1t
"4 (.30)
1
.
J
x 0.8504 == 0.06 m Is. Ans.
Problem 6.25 Find the I'e/oeify ofl/I(' flow of GIl oil Ihrough <l pipe. ",1'1'0 IIII' difference of mercury
lerel ill a differential V-lube mil/wmeler COllnected /0 Jhe two tappings of the pilOHube is 100 mm.
Toke co-efJicietll of pilor-wbe 0.98 alld sf!. gr. of oil" 0.8.
Solution. Given:
Diff. of mercury level,
X= IOOmlll=O.lm
Sp. gr. of oil.
5,,= 0.8
Sr. gr. o f me rcury.
S~ '" 13.6
C. = 0.98
Diff. of pressure head.
Velociiy o f flow
II = x
[~:
-I]
=. 1
[_'g_.:__ I]
""
].6 III of oil
'" C. J2gl1 = 0.98 .J2 x 9.81 x 1.6 '" 5.49 m/s. Ans.
Problem 6.26 A pilol -Slalic (ubI' is u~'ed /0 ",ea~'ure Ihe ..docily oflt'(l/er in a pipe. The ~'Illg'illlio"
pressure head is 6 II! (I"d Sialic pressure head is 5 m. Ca/culllle Ihe "e/OCily of flow aSJ"umillg Ihe co I'jficil'nl of 'uhe equulla 0.98.
Solullon. Given:
Stagnation pressul\! liead,
11, =6111
SIalic pressure head.
",=5m
11=6 - 5= I m
Velocity of now.
V = C. JZg" = 0.98 JZ x 9.81 x I = 4.34
III/S.
Ans.
Problem 6.27 A sub-marille mOI'es IlOrizollilllly ill seo Imd I/Os ils oxis /5 m bdow 1/1/' sllrface of
"·Iller. A pilo/./ube properly placed jusl ill frOlll of Ihe suh -marille olld (l/oll g ils axis is cOimecled 10 Ihe
IWO limbs of II U-lIIbe COIlIO;Ilillg mercury. The differellce of mercury lel'el is found ro be 170 mm. Filld
l/ie speed of Ihe SUb'lIlllrille kllowin8 1//01 Ihe sp. gr. of mercury is 13.6 and 1//01 of sea· Willer is 1.026
willI re.lpect offre.lh Willer.
Solulion. Given:
Diff. of lIIercury level,
.r = 170 111111 = 0. 17 11\
Sp. gr. of mercury ,
Sj= 13.6
Sp. gr. of !.ea-watcr.
SQ= 1.026
[S'l
lI=x -
So
['3.6 1
- I =0. 17 - - -J
1.026
=2.083411\
V = J2gh = J2 x 9.81 x 2.0834 = 6.393 Ill/s
=
6.393 x 60 x 60
1000
kmlhr = 23.01 km/hr.
An.~ .
Problem 6.28 A pilOI-lube iJ' ill$er/t"d ill a pipe of 300 IIlm diameler. The J'Ia/ie pressu re ill pipe is
tOO 111m of mercury (mCllllm). The slllgllaliOli pressure aI Ihe e<:llire of IIIe pipe, reco rded by Ihe
I I
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IL
1288
Fluid Mechanics
pilOI-lUbe is 0.981 N/cm<. ea/clliare Ille rale of flow of Wilier II/rough pipe, if Ille meatl \'elocity of
flow is 0.85 limes rhe cemral 1't1/ociry. Toke C. '" 0.98.
Solutio n. Given:
Dia. of pipe.
d" 300 mm '" 0.30 III
Area.
SIalic pressure head
a = ~ <1 2 '" ~ (.3) 2 ", 0.07068 Ill l
4
4
" I()() nUll of mercury (vac uum)
100
= - - - x 13.6= - 1.36 III of wate r
1000
Stagnation prc,,-w rc
Stagnation pressure head
= .981 Nfcm l = .981 x 10' Nfrn l
=
.98 I x IO· = .98 1 X 10' = I m
WOO x 9.8 1
pg
II " Stagnation pressure head - Slatic pressure head
= 1.0 - (- 1.36) = 1.0 + 1.36 = 2.36 III of water
Veloci ty at centre
'" C. J2g/1
= 0.98 x ...;RjC,c9'g"C'''2.'36'' 6.668 tnls
Mean velocity.
Rate of fl ow of water
~
6.8
V = 0.85 x 6 .668 = 5.6678 Illis
= V x area of pipe
= 5.6678 x 0.07068 rn l/s = 0.4006 1Il 3/s. An s.
THE MOMENTUM EQUATION
It is based on the law of conservation of momentum or on the mOllie ilium principle. which states
th~t the net force "cling on ~ fiu id mass is cqual to Ihe ch"ngc in moment um of now p.:r unit time in
that dircction. The force acting on a fillid mass '/11' is givcn by thc Newton's second law of motion.
F:/II:><o
where a is the acce leration acting in the sallie direction as force F.
8",
d,'
F= m -
,"
,"
'" d(m •. )
F ==
d(IIII')
d'
!m
is constant and can be taken inside the diffe re lll ial }
... (6. 15)
Equation (6.15) is known as the momentum principle.
... {6 . 16)
Equation (6.15) can be written as FAt", d(ml')
which is known as the illlpu{se'lIIomenllllll equation and states that the im pulse o f a force F ac ting on a
fiuid of mass /II in a shon interval ofti mc dt is equal to the ch~nge of momentu m d(mv) in thc direction
of force.
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Dynamics of Fluid Flow
Force exerted by
~
fl owing fluid on
OJ
289 1
pipe bend
The impulse-momentum l"<luation (6.16) is used to deter mine the rcsu[(anl force exerted by a
flowing fluid on a pipe be nd.
Consider two sections (I) and (2). as shown in Fig. 6. 11\.
leI
VI'" ve locity of flow at seclion (I),
PI'" pressure intensity at section (1).
AI'" area of cross-section of pipe al sedion (1) and
1'2- Pl- Al '" corresponding va lues of velocity. pressure and area al section (2).
Lei F. and F, be Ihe components o f tile forces exerted hy [lie flowing fluid o n the bend in x-and
y-dircctions r..:spcclivcly. Then Ihe force cxcn cd by the bend on the fluid in Ihe directions of x and y
will be equa l \0 F. aut! Fy bul in the opposi te di rections. Hence component of lhe force exe rted by
bend on the fluid in the ,[·direction = - F, and in the direction of y '" - Fr The other externa l forces
actin g on the fiuid arc PIAl and py\~ on the sections (I) and (2) respectively. Then momentum
equation in x-di rection is give n by
"•
P,A"_-ll-_
(0)
(0)
Fig. 6. 111
For~
011 bend.
Net force acting o n fiui d in the direction of x = Rate of change of momentum in x-direction
PIA I - Pt\~ cos 0 (Mass per sec) (change of ve locit y)
'" pQ (Final velocity in the direction of x
- Initial ve loc ity in the direction of .f)
...(6. 17)
=pQ(V2CO~O-VI)
.(6. 18)
F , '" pQ (VI - V2 cos 0) + PIAl - PY\l cos
Similarl y the momentum equation in y-dircction gives
... (6. 19)
Pt\~ sin
F. = pQ (V2 sin
0)
...
(6.20)
,..,. '" pQ (- Vl sin 0) - p0~ sin 0
Now lhe resultam force (FR ) 3<:ling on lhe bend
"'x'"
e
o-
e-
e-
"
'" 'Fz +Fl.
,
And the angle made by lhe resultant force with horizontal direction is given by
,.
tanO"'~
F,
... (6.2 1)
... (6. 22 )
Problem 6 .29 A 45° reducillg belld is connecled ill a pipe line. ,be diamerers ar 'he inlel and OIl/lei
of,he bend being 600 mm Imd 300 mm respec/ire/y. Filld lire force e:terled b), warer 011 II,e bend if/be
in/ellsil), of pressure ar illiel 10 belld is 8.829 Nkm! lind rale of flo ..... of ..... arer is 600 lirres/s.
I I
Ii
~ I
IL
1290
Fluid Mech a nics
Solution. Given:
9 = 45"
Angle of bend.
Dia. at inl c!'
DI '" 6CKl1ll1ll = 0.6 III
f[
It
:
2
Area.
A1='4D1 ='4(.6)
Dia. al outle!,
= 0,2827 m2
Dl = 300 mill '" 0.30
ill
,
- - ""'1----/ /
_P2~ ___ ~ ~'
III
Fig. 6.19
:. Area.
"
, 8.829 x 10 N/rn"
PI'" 8.829 Nlcm-:::
Q = 600 litis", 0.6 mJ/ s
Pressure at inlet.
Q
0.6
VI"" - = - - =2. 122 m/s
AI
.2827
Q
0.6
= - - - '" 8.488 mls.
Al
.07068
V, '" +
Apply ing Bernoulli's equation a1 scc ti o lls ( 1) and (2) , we gel
,
.£.!..+~+Zl=
pg
B",
2g
V'
II) + ....L+zz
PI!
28
11'" 12
,
,
l!J...+~ = III + V1
pg 28
pg 28
or
8.829 X 10
4
1000x9,81
+
2.1 22"
/I,
8.488'
=~
+
2 x9.8 1
pg
2 )(9.81
9 + .2295'" PtPS + 3.672
P! '" 9.2295
~ 3.672 '" 5.5575 m of wmcr
"
li z = 5.5575 x 1000 x 9.81 Nfm 1 '" 5.45)( 104 Nlm 1
Forces on lht: bend in x- and y-dircclions arc given by equ ation s (6. 18) and (6.20)
a~
F, '" pQ IVI - V1 cos 91 + PI A l - P0 2 cos a
= 1000 x 0.6 [2.122 - 8.488 cus 45°1
+ 8.829 x 10' x .2827 - 5.45 x 104 x .07068 x cos 45"
'" - 2327.9 + 24959.6 - 2720.3 '" 24959.6 - 5048.2
'" 19911.4 N
,..,. '" pQ 1- V1 si n
md
- vc sign means
r,.
sin
e
. '" 1000 x 0.6 [- 8.488 si n 45° [_ 5.45 x 104 x .07068 x sin 45°
'" - 3601.1 - 272 1.1 ", - 6322.2 N
is acting in tile down ward direction
Res ultant force.
I I
e] - PY\ l
,.. I? '"
Jr} + F/
Ii
~ I
IL
Dynamics of Fluid Flow
==
291 1
~(1991 1 .4)Z + ( _6 322.2)2
= 20890.9 N. An ~.
The ang le made by resultan! force with x-ax is is given by
equa tion (6.22) or
F,
Fi g . 6.20
= 0.3175
tan9= -'- = 6322.2
199 11.4
F,
9= tan - I .3 175 == 17° 36' , An s.
Problem 6.30 250 IilTes/s of ,Wiler is jlOWillg ill a pipe Illn'itlg a di"meler of 300 mill. If tile pipe is
bellI by 135 0 (IIml i.1 c/range from illilial 10 filial "iree/iOIl i,1 I J 5 °J. find I/Ie mag/lilude amI <Ii reCliOIl of
l/ie reSII//{ml force 011 Ihe bend. Tlw pressure of ,,'aler flowing iJ' 39.24 Nlcml.
Solution. Givcn :
Pressu re.
1'1 = 1'2 = 39.24 Nlcrn 1 = 39.24 x 104 N/m 2
Disch~rgc.
Q == 250 litrcsls == 0.25 m3/s
Dia. of bend at inlet nod o ullet. VI = D1 = 300 111m = 0.3 m
l'I:
Area.
Veloc it y of
l
ll
l l
A 1=A!=4"D 1 =4X.3 =0.07068 111
wat~r at S~"(:lioJ\s (1) and (2), V = VI = V, = ~ :
V2 sin 45'
~
.1'
p;/"
~
cos 45'
"- ",
I")
-
V
0.25 : 3.537 m/s.
.07068
Area
o
,
p' $ -----IDI
~
V, cos 4 5'
''I
135'
Lx
Fig. 6.21
Force along .(·axis
: F, : pQ[V I,- V~,J + PI..A I + P;o..Al
initia l ".. Iocily in th .. direction of.>:: 3.537 m/s
V~, " final ve loci ty ill the dircctioll of x = - V~ cos 4 5° = - 3.537 x .707 1
PI.< : pressure at s,:,ction ( I) in x-dir.. c ti o n
: 39.24 N/cm ~ " 39.24 x 104 Nlm 2
Pl, : pressure at section (2) in x-dircction
: p~ cos 45°: 39.24 X 104 X .70 71
F,= 1000 X .25[3.537 - (- 3.537 X .7071)[ + 39.24 X 104 X .07068 + 39.24
X 104 X .07068 X .707 1
: 1000 X .25[3.537 + 3.537 X .7071 [+ 39.24 X 104 X .07068 II + .70711
V lx :
I I
Ii
~ I
IL
1292 Fluid Mechanics
" 1509.4 + 47346 = 48855.4 N
'" Fy " pQIYI , - V!yJ + (PIA 1),_+ (P:01) ..
= init ial velocity in y-d ireClion = 0
V 1., = final velocity in y-dircclion = - V! si11 45" = 3.537 x .707]
(PIA,), = pressure force in y-dirccli on = 0
whe re
VI ."
(P02) , = pressure force at (2) in y-dircction
- = - P0 2 si n 45 6 = - 39.24 X [0 4 X .07068 x .7071
F, = 1000 x .2510 - 3.537 x .70711 + 0 + (- 39.24 x IO~ x .07068 x .7071)
= - 625.2 - 19611.1 = - 20236.3 N
- vc sign means "'., is acting in the downward direction
Rcsuhant force.
,
,
F = J F,1 + Fl
F,=4980.1
• "'CJi-='------,
°
= 528110.6 N . Ans .
Th" direction of the resultant force F H' wi th the x-axis is given as "
~
Ian
e = F,. =
Fs
LL"
" t _______________ _
20236.3 = 0.4 142
48855.4
Fig. 6.22
e" 22" 30 ' , Ans.
Problem 6.31 A 300 "'''' diameter pipe C(lrries W(1/er under a Ilead of 20 me lres will, a ndociry of
3.5 ",/~·. if Iile a.tis of Iile pipe lu rtlS Illrougil 45 °, find Ille mag" il"de and dire/;Iioll of Iile rcsu/lllJIl farce
allll<' bend.
Solu t ion. Given:
Dia. of bend.
D = DI = D z = 300 mm = 0.30 m
1t
Area.
A=A , =A 2 = -
4
Velocity.
2
1t
D = -
4
2
x.) =0.07068m
2
V=V , =V1 =]·5m/s
e = 45°
Q = A x V = 0.07068 x 3.5 = 0.2475 m 3/s
Discharge.
= 20 m of water
Pressure head
or
...E.... = 20
pg
III
of water
p=20xpg=20x IOOOx9.81 N/m 2 = 196200Nlm 1
Pressure intensity.
Now
p = P, = P2 '" 196200 N/m!
= 3.5 1l1/S, VZr = V1 oos 4 5 0 = 3.5 x .707 J
V ly = O. V 2y = V2 sin 45" = 3.5 x .7071
V"
(PIA I )~ '" PIA I = 196200 x .07068. (PIA I),. = 0
(P;oAl)~ '" - p0!
Forte along
I I
.{-{I.ri.l.
cos 45°, (P;oAl),' '" - p0! sin 45°
pQl VI., - V2,1 + (PI A ,)~ + (P:0- 2)~
= 1000 x .247513.5 - 3.5 x .70711 + 196200 x .07068 -
F~ '"
p;oA~
x tOS 4SO
Ii
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IL
Dynamics of Fluid Flow
293 1
'" 253.6& + 196200 x .07068 - 196200 x .07068 x 0.7071
'" 253.68 + 13871.34 - 9808.()..I = 4316.98 N
F,,: pQ [ VIY - Vy'l + (PI A I)\' + (1']"\2),.
'" 1000 x .247510 - 3.5 x .707 11 + 0 + [- Pfo2 si n 45°]
'" - 612.44 - 196200 x .07068 x .7071
Force along Y'{/XiI,
'" - 612.44 - 9808 '" - 10420.44 N
F II '"
Resultant fore\'
JF.1 + F/ '" J( 4316.98) 2 + (10420.44 )2 '" 11279 N. ,\IIS.
"~
A'""''' .'
V.sin45"
q'
I7A,cos 45"
V,
45· V.COS 45·
,
/
- - ~- -
(j)
L
'.
Fig. 6.23
The ang le made by F/I with .r-axis
lane", F, '" 10420.44 = 2,411
4316.98
F,
tan - I 2.411 '" ii7 c 28'. An s.
belli! {/ rcell/llgular (lir duel of f m 2 cross-secliQIWlllrell iI grll<ludlly
a",
Problem 6.32
In {/
-15°
reduced 10 0.5 m l area. Find the magllirude oml direc/ioll of the jorce required /0 Iwld Ihe duel ill
position iflhe I'e/Orit)' offlow ai/he I m ! seClioll is /0 mIs, and I'rej'sure is 2.9-/3 N/cml . Take dellj'ily
of "ir as /./6 kg/mJ.
Solution. Give n :
Arca at section ( I).
Area at section (2) .
Velocily at section (1 ),
Pressure at section ( I ).
Density of air,
A I" lm
1
A 1 " 0.5 m
1
VI "lOm/s
PI" 2.943 Nfcm 2 '" 2.943 x 104 N/m 1 " 29430 N/m 1
P '" 1.16kgfm 1
Apply in g continuity equ ati on at sections (I) and (2)
A1 V I ",A 1 V1
AI VI
1
V1 ", - - o - x 10" 20 m/s
Al
~ I
0.5
I~
~ I
IL
1294
Fluid Mechanics
Discharge
V, sin 4 5 '
V, cos 45"
/ )\f7L
p,A, cos 45'
".
Fig. 6.24
Apply ing BCrrloulli' s cqualioll al sections (I) and (1)
P,
V,~
P!
¥l!
-+-=-+pg
2g
pg 2g
2.943 x 10' + -;;,-J~
O '= "" P! +
"'
1.1 6x9.8 1
pg
2x9.8 1
20'
P2 = 2.943 X 10' +
pg
1.16x9.81
2x9.8 1
2 x9.8]
"" 2586.2 + 5.0968 - 20.387 '" 2570.90 m
p, '" 2570.90 x 1.l6 x 9.81 '" 29255.8 N
Force along .I·axis.
whe re
A I,
'"
F~
= pQ IV" - Yhl + (p,A,). + (PY\l),
10 mls. Vb"" VI (;OS 45° '" 20 x .707 1.
(PIA, ), '" PIA , == 29430 x I '" 29430 N
and
(1'02), = - P02 cos 4.5° = - 29255.8 x 0.5 x .7071
F, = 1.16 x 10110 - 20 x .7011 J + 29430 x 1 - 29255.8 x .5 x .707 1
'" - 48.04 + 29430 - 10343.:n '" 0 - 19038.59 N
Similarly furce (1/0118 )"-luis. F. := pQ[V h, - V2,.1 + V',A,)y + (p;Y\:J ),
wh~ re
V,), = 0, V2r = V1 si n 45 ° = 20 x .7071 = 14.142
(PIA,),_= 0 and (pyl.!»), = - PY\l sin 45° = - 29255.8 x.5 x ,707 1 : - 10343.37
F,, : 1.16 x 10 [0 - 14.142 [ + 0 - 10343.37
: - 164.05 - 10343.37: - 10507.42 N
R,,~ullanl
force,
FK: ~F} + F/ : ~( 1 9038.6)~ + (10507.42 )"
: 21746.6 N. An s.
The direclion of FII Wilh x- ax is is givcn as
lanG:
10507.42 :0.5519
19038.6
t
8 : tan - .5519: 28 c 53'. Ans.
Fy
:
Fx
I I
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Dynamics of Fluid Flow
295 1
PI< is (h e force exerted on bend. Hence th c fo rce required 10 hold th e duct in position i s eq ual lO
2 1746.6 N but it is actin g in th e opposite direction of FR' AD S.
Problem 6.33 A pipe of 300 mm diameter con "eyillg 0.30 ", JIJ' of ,mler hus a righlllllgied be.rd in
II horiZOliful pilll/c. Find Ihe rC$u/tallf jorce e.fer/ed on Ille bend If IIIC pro.fllre at i, 111'1 lind QU lld of Ille
bend are 24.525 N/cll/ ami 23.544 Nkm 1.
Solution. Given:
0;3. of bend.
D '" 300 Imn '" 0.3 111
"
Area.
A=A 1 =A 2 = - ( .3) =0.0706810
4
DiM:hargc.
Q" 0.30 rn J/s
Ve locity.
v=
VI" V2 ",
~
==
.O~~8
~-
/
p,A,
=4.244 Ill is
e - - <2!
v
(j)
v,
,
y
L,
Fig. 6.25
a == 90°
PI = 24525 Nfcm 1 = 24.525 x 10· Nlm 2 = 245250 Nf m l
1'2'" 23.544 Nfcm! " 23.544 x 104 Nlm 2 " 235440 N/m!
Angle of bend.
['orce
whe re
DO
Fx '" pQ lVI" - V2, 1 + (PIAl)" + <,,02),
p = I(X)(}. VI> = VI = 4.244 mls. V~, = 0
bend along x-axis
(PIAl), = PI A l = 245250 x .07008
(P:A2) ,
=0
,..~
= 1000 x 0.30 [4.244 - 0] + 245 250 x .07068 + 0
= 1273.2 + 17334.3 = 18607.5 N
Foree on bend alon g y-axis. Fy = pQ lVI , - V2)· [ + (PI A l), + (Pi'-2)y
where
VI , = O. V2, = V2 = 4 .244 Ill is
(PIAl),. = O. (P:A:U,. = - P0-2 = - 235440 x .07068 = - 16640.9
F,. '" 1000 x 0.30]0 - 4. 244] + 0 - 16640.9
= -1 273.2 -16640.9= -1 79 14.1 N
Resullanl force.
and
F,,=
~F,' +F,'
F,
lana : - . 17914.1 = 0.9627
18607.5
F,
a = 43
I I
= J(l8&J7.5)' +(179 14 .1 )' = - 25829.3 N
0
54'. Ans.
Ii
~ I
IL
1296 Fluid Mechanics
Problem 6 .34 A tlOU!<- of diameter 20 mm i.1 jilled 10 a pipe of diameter ~O mill. "';lId lite force
"'.I'erled by Ille noule 011 the water ..... hich is flowing 1III0llgii I/Ie pipe m lilt! rale of 1.2 111 3/lIIlllIIle.
Solution. Giv~n :
Dia. of pipe.
0 , = 40 mm = 40 x 10-1 m ", .04 m
Area.
Dia. of nozzk.
VI '" 20
0.02
m ill ",
111
Area.
Discharge.
i
(j)
Fig. 6.26
Apply ing continuity equalion at scrtions (I) and (2 ).
A,V,=A 1Y2=Q
VI =
~I
=
.,d
'(Xl~'~56
= 15.92 rnls
0.2
'" 63.69 rn ls
.0003 14
Applying Bernoulli's equ ation
<II
sec tions (1) and (2). we gel
V'
Ie '
1:l + _'_ + z, '" ~ + - '- ' "
pg
2g
Now
pg
28
Z, '" Zl' 1'1 '"
pg
mmosphcriC pressure", 0
-PI + -VI' : -V}
pg
28
28
1:l =
(1S.92')
V,' _ V,' = (63.69') ""~~
= 206.749-12.917
pg
28
2g
2 x9.8 1
2 x9.81
= 193.83 m of wate r
N
N
In '
rn '
p, = 193.83 x 1000 x 9 .81 - , = 1901472 - ,
I I
Ii
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Dynamics of Fluid Flow
297 1
Let the force exerted by Ihe nozz le on water'" F,
.l '" r~tc
Net force in the di rection of
PI A l - P;tAl
of change of momentum in the direction of x
+ F, '" pQWl -
VI)
where 1'1 '" at mosp heric pressure", 0 and p '" 1000
1901472 )( .00 1256 - 0 + F, '" 1000 x 0.02(63.69 - 15.92) or 2388.24 + Fx'" 916.15
F , '" - 2388.24 + 916. 15 '" - 1472.09. AilS.
- VI.' s ign indicates lhal the force exe rted by Ihe noa k on waler is acting froln right to left.
Problem 6.35 The diameter v/a pipe grodual/y ~educeJfro'" I III /0 0.7 III G J' sho ..." ill Fig. 6. 27.
Tlw pressure ;nlensily II/Ihe C<'IIlre-/i",: of I III sec/ion 7.848 kNlm! (1111/ Tale offlow a/woler Illrougil
th e pipe is 600 li/res/s. Fill d Ihe ;n'CIIsity of pressure a l Ihe cen/re·/ille of0.7 m seclion, Also delermin/!
rhe force au/ed by flowing Imrer 011 rratlJirion of rile pipe.
Solution. Given:
Di~ . of pipe at scl:lion 1.
D I '" I In
"
Are a.
A I ",-(I),,,,0.7854m
4
,
~
GJ
I
3r'f '+
'"
;_~ll""'""v;_-"A,!""---1
1
A
CD
@
Fi g. 6.27
Dia. of pipe at scction 2.
,
"
A, '" - (0.7)' '" 0.3848 III
•
4
1
2
PI'" 7.848 kNlm '" 7848 N/m
Are a.
Pressure at scction I.
Q'" 600 litres/s '" 600 '" 0.6 rn'ls
Discharge.
1000
Apply ing continuity equation.
A I V I ", A ZV2 ",
Q
Al
VI'" -
~
Q
O.
- - - '" 0.764 IlIls
0.7854
V,,,,.Q.= ~ ,,,
1.55m/s
• A!
3854
Apply ing Bernoulli' s equation at scl:tiuns (1) and (2).
PI
-
pg
7848
1000 x 9.81
I I
VI!
+-
2g
~ P2 V}
- - +pg
2g
!...
pipe is horizontal.
+ (.764)2 '" Ii + ",(Ic:'",S);C'
2 x9.8 1 pg 2x9.81
Ii
~ I
IL
1298 Fluid Mechanics
",
,( . ,,,,64
,,,Ic,-'
2 x9.8 1
-'-"- == 0.8 + 7"
pg
(J.55)'
2x9.8 1
== 0.8 + 0.0297 - 0.122 == 0.7077 111 of water
1', "" 0,7077
x 9,81 x
1000
== 6942.54 N/ml or 6.942 kN/ml, Ans.
Let
F~
== the force cxcrlcd by pi pe lransition on the
flowin~
water in the dircdion of flow
Then net force in the direction of flow == ralC of change of momentum in Ihe dirc<:!ion of flow
= p(V2 - VI)
7848 x .7854 - 6942.54 x .3848 + F, = 1000 x 0.6]155 - .764]
6163.8 - 2671.5 + F, = 471.56
F, "" 47156 - 6163.8 + 2671.5" - 3020.74 N
PIAl - P~l + F,
or
The force exu[cd by water on pipe transition
= - F, " - (- 3020.74) == 3020.74 N. AilS.
to 6. 9
MOMENT OF MOMENTUM EQUATION
MomcnI o f rnomclltum equation is derived from moment of 1ll0lllCnlum prillciplc which states that
the resulting torque acting on a rotating fluid is equal 10 the rate of change of moment of momentum.
Let
VI == velocity of fluid at section I.
'1 == radius of curvature at section I.
Q == rate of flow of fluid.
P == density of fl uid.
and
V z and '1 = velocity and
radiu~
of curvature nt sec tion 2
Momentum of fluid at section 1 = mass x velocity == pQ x Vl/,f
MOlllent of 1Il0tllcntUill per second nt sectiOn I.
=PQXV1x'l
Similarly moment o f momentum per second of fluid at section 2
=pQxV!x,!
Rate of change of moment of momentum
= pQV!'2 - pQVl'1 = pOlY!'! - VI'tI
According to moment of momentum principle
Resultant torque == rate of chan ge of moment of momentum
... (6.23)
Equation (6.23) is k.now n as moment of momentum equation. This eq uati on is applied :
I. ror analysis of flow problems in turb ines and centrifugal pumps.
2. ror fmding torque exert ed by water on sprink.ler.
Problem 6 .36 A lawn Jprlllkler II'Ilh MO nou{ej of diamele," mm each Is cotlnecteil aCfOSJ a lap
of Wilier as shown ill Fig. 6.28. The 1I0zzies are at a distance of 30 cm alld 20 Cllifrolll Ihe cellrre of IIIe
tap. Tile rale of flow of waler Ihrollgll lap is 120 cmJ/.i. The 1I0zzies discll<lrge Wilier ill Ihe dowlllI'lird
directioll. Determllie Ihe lIIlgu{ar Jpeed at II'lIicll Ihe ,Iprin kle, will rotate free.
I I
Ii
~ I
IL
Dynamics of Fluid Flow
r-
Solution. Given
Dia. of nozzles A and B.
D- D~ - D8
A",
Area.
(.004)2 ",
"I
_ /0 ____
30
cm---J
ro t
III
.00001256 111 2
4
F Ig . 1>.28
Q'" 120 em}'s
Discharge
Assuming
2:
4mm -.OO4
20cm
299 1
(h~
di.<;ehargc 10 be equally divided bclWc.:1l the two nozz les. we have
Q 120
3
6 )
Q,,=QI/= 2 =T=60c11l Is=60X IO 11I Is
Velocity of
Wal~r
at the ou tld of each nonle.
Q
...
whe re
fA'"
8
=_
6OxlO --<>
A '"
= 4.777 m/s.
A
.0000 1256
The jet of water coming oul from nozzics A aJld 8 is havi ng veloc ity 4.777 mls. These jets of water
will exc n force in llle opposite direction, i.e .. force exerted hy Ihe jets will be in Ihe upward direction.
The torque exerted will also be ;lIlhe opposite directiun. Hence IOrquc at B wi II be in the ami-clockwise
direction imd al A in the dockwis.: di rection. But torque at B is 1001\: (him the torque at A ;md hence
sprin kle. if free. wi ll rot ate in the anti·clockwise di rection as s hown in rig. 6.28.
lA:t (0 '" angular veloci ty o f the sprin kl er.
Then absol ute velocity of wata at A.
V '" V
VI = VA + (Ox 'A
distance of nozzle A from the centre of tap
" 20
elll
= 0.2
IW x
III
'A=
tangential ve loc it y d ue to rota tion I
VI = (4.777 + 00 x 0.2) lilts
Here 00 x ' A is added to VA as VA and tangential ve locity d ue to rotation (00 x fA) are in the sallie
dire<.:tion as shown in Fig. 6.28.
Similarly. absolute velocity of wate r at B.
V 1 ", VB - wngential velocity due to rotation
=4.777 - wx
(where 'B= 30 em = 0.3 III)
'IJ
'" (4.777 - 00 x 0.3)
Now applying eq uation (6.23), we get
I Here,z" 'B"I='A
T" pQ(VZ'2 - VI'I(
'" pQAIV 1f B - Vlr,, 1
= 1000 x 60 x 10
6
Q'" QA '" Q R
((4.777 x 0.3 (0) x .3 - (4.777 + 0.2
(0)
x .21
The moment of momell1 UIII of the fluid entering sp rinkl er is given zero and also there is no external
torque applied 0 11 the sp rinkler. Hence resultant external torque is zero. i.e.. T", 0
1000 x 60 x 10 6 [(4.777 - 0.3 (0) x.3 - (4.777 + 0.2 (0) x .21 = 0
(4.777 - 0.3 (0) x 0.3 - (4.777 + 0.2 (0) x.2 '" 0
4.777 x.3 - .09 00 - 4.777 x.2 - .04 (1)= 0
0. 1 x 4.777 '" (.09 +
.4777
w" - -
0.13
I I
.(4)(0 '"
. 13
OJ
= 3.6746 rod/s. An s.
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IL
1300
Fluid Mecha nics
Problem 6.37 A h,WII sfJrillkler shown ill Pig. 6.29 lIa.1 O.S elll diameter 'lOuie ar lile end of a
fOlilIing arlll alllt discharges Wafer ill the rate of to mls \'e/ocily. Defemlim' Ille torque required /0 hold
Ihe rofaling "rm s/alio,wry. Also determine the conJ'tan' spud of rOI,,'iOIl of Ille
If free 10
'If"'.
t10
rO/lIIe.
Solution. Dia. of each nOlzle = 0.8 elll = .008 III
20 em
Are a of each nozzle
Velocity of !low al each nozzic = IU m/s.
Discharge through each noak.
I
25 em
'9
II'j
1
mise.;
liA
Q '" Area x Velocity
10 mlsec
Fig . 6.29
'" JXlOO5026 x 10 = .0005026 rn 3/s
Torque cxc-fled by wate r coming through nozzle A on the sprinkler = moment o f momentum of
water through A
=
f" X P x Q x V~ '" 0.25 x 1000 x .0005026 x 10 clockwise
Torque exerted by water coming through nozzle B on the spr inkler
= 'II X P x Q X VB'" 0.20 x 1000 x .0005026 x 10 clockwise
Tot al torque exerted by wmcr on sprinkler
Torque required
10
'" .25 x 1000 x .0005026 )( 10+ .20 x 1000 x .0005026 x 10
'" 1.2565 + 1.0052 '" 2.26 Nm
ho ld the rotating arm stmionary '" Torque exerted by water on sprinkler
'" 2.26 Nm. Ans.
Speed of rotation of arm , if fret' to rot:.t e
Let
00 '" speed of rotation of the sprinkle r
The absolute velocity of flow of wate r at the nozzles A and Bare
VI'" 10.0 ~ 0.25 x 00 and V! '" 10.0
Torq ue exerted by water coming out at A. on sprin k ler
~
0.20 x 00
= FAX P x Q X VI '" 0.25 x 1000 x .0005026 x (10 - 0.2500)
'" 0.12565 (10 ~ 0.25 (0)
Torque exerted by water coming out at B. on sprin k ler
X P x Q X V! '" 0.20 x 1000 x .0005026 x (10.0
'" 0.10052 (10.0 ~ 0.2 (0)
Total torque exerted by water: 0. 12565 (10.0 - 0.25 (0) + 0.10052 (10.0 - 0.2 (0)
:
Tg
~
0.200)
Since moment of momentum o f the flow entering is zero and no external torque is applied on
sprinkler. so the result ant torq ue on the sprinkler must be ze ro.
0.12565 ( 10.0 - 0.25 (0) + 0.10052(10.0 - 0.2 (0) '" 0
1.2565 - 0.0314 00 + 1.0052 - 0.0201 00: 0
1.2565 + 1.0052 '" 00 (0.Q3 14 + 0.020 I)
2.26 17: 0.0515 00
2.2617
w: - - - '" 43.9 md /s . A n s.
0.051 5
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Ii
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IL
301 1
Dynamics of Fluid Flow
60 x 43.9
~-';::?'"- h
and
... 6. 10
'" 419.2 r,ll.tn. Ails .
FREE LIQUID JETS
Free liquid jet is defined a ~ th e jd of water corning oul from the noule in atmosphere. The palh
travelled hy the frcejet is parabolic.
Consider a jet coming from the nozzle as shown in Fig. 6.30. LeI the jet at A. nwkcs an ilngle with
th e horizontal dirc~lion. If U is the velocity or jet o f waler, Ih ,," the horizontal component and vcrticnl
~u11lponcm of Ihis velOCiTy al A arc U cos e and U sin 9.
Consider another point P(x. y) on the centre line of Ihe jeT. The co-ordinates of P from A arc x and y.
Let the ve locity ofjct at P in the of- and y,d ircctiolls arc u and v. LeI a liquid particle tak es lime '( to reac h
from A to P. Tllen the horizontal ami vert ical distances tra velled by thc liquid panicle in lime 'r' are;
e
V
TRAJECTORY
J":;;::::1- -..(/
.,
u
"
PATH
"
: -;:--·L f
;;
;.., \c':-:-:
'
Ucos9
. --1
NOZZLE
Fru liquid il'l .
Fig. 6.3O
x'" velocity l"t>mpone nt in x-direction x
1
.. ( i)
=Ucos €lx t
and
y = (I'enica l component in y-direction
= Usin 9XI ,.
±
g1
X
tim., _
.!. g(2)
2
2
... (,. i)
Horizontal componen! of velocity is consta n! whi le th e venka l distance is affected by gravity I
From eq uation (i). the valu e of ( is given as I = --,-"'
-; C·'"
U cos
Substituting this val ue in equation (il)
a
y=UsinOx
=.Ttan 9 _
x
UeosO
g.<
I
(
x
- xgx
2
Ucos9
1J.:
)'
.
sma
=x - - -
cos 9
gx
2U
2
cos'e
e) . .
sec' a
- '-,- = sec!
(6.24 )
2Ucos-O
Equation (6.24) gives the varia tion of y with thc squarc of.T. Hence thi s is the equatio n of a parabola.
Thus the path travelled by the free jet in atmosphere is parabo lic.
I I
Ii
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IL
1302 Fluid Mechanics
(I) l\ la xl mum h eig ht uttnlntod b~' th e Jet . Using the relation V/ - VI! '" - 2g5. we get in this
case VI '" 0 at the highest point
VI = Ill ilial vertical component
=UsinG
-vc sign on right tHmd side is taken as g is acting in the downw:ml dircdion but panicles is moving up.
0- (U sin e)~ '" - 2g x S
where S is Ihe Illnimurn vertical heig h! attained by the particle.
or
- U 1 sin1e == - 2gS
s=
Ulsin " e
... (6.25)
2,
(ii) Tim e of fli ght. It is the time taken by Ihe fluid particle in reaching fro m A to IJ as shown in
Fig. 6.30. Let T is the time of fligh!.
Using equation (iiI, we have y = U sin
when the panicle reaches at B.
e x /_ ..!. gIl
2
y'" 0 and I'" T
Above equation becomes asO '" U sin exT -
..!.g x r2
O=Usin9 - ..!.g'f
"'
1Cancelling T)
2
,
T = .2~U~si~""
,
... {6.26)
(iii) T im e 10 rea l' h hIghes t po int. The tillle to rcaeli liigliest point is halftlie tillle of niglit. Let P
is tlie tillle to reacli liigliest point. tlien
P= T = 2Usin8 = Usi n8
... (6.27)
2
gx2
g
(iv ) Horl:ton tul run g" of th " j cl. Th~ tOlal horizontal distauce travelled hy til<: fluid particle i~
called horizontal range of the jet, i.e .. the horizo ntal distance AB in Fig. 6.30 is c alled horizontal range
of the jet. LeI this range is deooted by x·.
Then
x· = velocily .::omponent in .l-dire'::l ion
x time taken by IiiI' panicle 10 rea.::h from A 10 B
= U ,::os 8 x Time uf flight
=U\'os8x 2Usin 8
g
U
1
=-
e sin 8 = -U
1
.,,(6.28)
sin 28
g
g
(v) Valu e of 8 fo r lll" x illlUnl range. The range x· will be maximum for a given velocity of
projection (U). when sin 28 is maximulll
or when
sin 28 = 1 or si n 28 '" sin 90° =
28=900or8=45°
Then maximum range.
I I
2 \'Os
U" . '
.l *rn'" '" -5111- 8
g
U
= g
l
{ '.'
,;" 90" = 1{ ... {6.29)
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303 1
Dynamics of Fluid Flow
Problem 6 .38 A I'ulica/wall is of 8 In ;" Iwighr. A jel of warer is cOllling QUI from a !lOule WillI
a 1't!locily of 20 mls. The noule i.! silllated ar (I diswllce 0[20 //I from Ill<' I'ulieal wall. Find the {/lIgle
of projection of lire llOule to Ihe //OriZOllwl 50 Illar the jet of \\"(l/er jusl clears rhe lOp of Ihe wall.
Solution. Given
: 8m
Hci gllt of wall
Velocil y of jd.
Dislancc of jet from wall.
l;:t Ihe required angle
T
'm
U=20 mfs
.P'
20
WALL:
In
=e
~i 'f---20 m------t
Using equation (6.24), we ha ve
y=x tan
e-
1,'.(
,
1
,
-"- , sec" a
Fig . 6.31
2U·
where y = 8 m. x = 20 Ill. U = 20 Illis
8 = 201:tnfl-
"'
"'
9.8lx20 2
,
2 x 20-
scc 2
e
'" 20 tan
e - 4 .905 scr 2 e
'" 20 (an
'" 20 (,III
e - 4 .905 [I + (an ! 81
e - 4.905 - 4.905 1:1112 e
e - 20 Ian e + 8 + 4.905 '" 0
4.905 ta n' e - 20 Ian e + 12.905'" 0
4.905 (an'
tan e ""
""
20 ±
J20
l
-
4 x 12.9{lS x 4.905
2x 4.905
20±JI46.81
9.81
=
20 ± Jr.400M-2~5~1C;19"
"" ",ccce,;;,:,;.",,,,,,-
9.8 1
20±12.116
32. 116
7.889
=
or
9.81
9.8 1
9.81
= 3.273 or 0.8036
9 = 73 9 0.8' or 38 9 37'. An s.
Problem 6.39 A fire-brigade Imm iI holding II fire sIred'" nozzle of 50 ",m did",eler aI shown in
Fig. 6.32. Thejel isslles 0141 wilh a reloClr)' of JJ ",h and strik.es rile willdow. Find Ihe angle orallgles
of in dill ali 011 will! ...!!ieh Ihe jel issues from lite nozzle. Wlwi ... il/ be rhe WIIO/Ull of ",,,'er [dl/illg 011 Ihe
willdow ?
Solution. Givell :
WINDOW
d,=50mm,=.05m
Di3. of nozzle.
Area.
A
= 2:(.05)! = 0.001963
10
I I
f
2
4
Veloc ity of jet.
U,= 13ml'i.
The jet is corning out from n07,7. le at A. It strikes the window
and let the angle made by the jet at A wit h hori zo ntal is eq ual to 9.
The co-ordinates of win dow. with respect to origin at A.
.\" '= 5 Ill. ), '= 7.5 - 1.5,= 6.0 m
The equation of the jet is give n by (6.24) as
,
7.5m
WALL
1
1-- 5 m---.1
Fig. 6.32
Ii
~ I
IL
1304
Fluid Mechanics
,
'"
,a
y=xI1ln9- --, sec
2U"
9.S ! )(5
,
, 11 + tan-91
2 x 13 -
6.0= 5 X (3n 9 6.0", .5
0 _ .7256 (1
(an
+
(a1l
2 EI)
e-
'" 5 tan
.7256 - .7256 tan l e
l
0.7256 lan e - 5 Ian 0 + 6 + .1256'" 0
0.7256 tan 2
5 tan + 6.7256 '" 0
e-
This is a quadratic equation in tan
e. Hence solution is
5±
I an
e '"
e
JSl - 4 x .7256 )(6.7256
"-''-'''---"""""",,,,,7-'''':'=''
2x.7256
'" s± J25
19.52 '"
~5;,:+,,2.,,34;C'
1.4 512
1.45 12
'" 5.058 or 1.8322
9= lan - I 5.058 or tan- I 1.8322", 78.8 0 o r 61.3r. AilS.
Amoun t of W31t'T f alling on window = Discharge from nou ie
'" Area of nozzle )( Velocity of jet at nozzle
= 0.001963 x U = 0.001963 x 13.0 = 0.0255 m J/s. AilS.
Problem 6.40 A "ozzle is J'il""kd at" diSlan",' of f '" "bOI"e the gmllml !e"e/ a",/ is inclined III an
angle of 45 0 10 file Ilorizonlal. The diameler of Ihe "oule is 50 111m (llid Ihe jet V/Warer from Ihe ,'ouie
s/rike5 the grOimd (1/ (l "oriZOlll(l/ diII(mce of.J Ill. Find Ihe Tille of flow ofw(IIer.
Solution. Given:
m
Distance of noule abo ve grou nd '" I III
Anglc o f inclination.
6", 45°
Dia. of nozzle.
d", 50 mill '" .05 III
1 _________
'
J~
, m
•
Area.
B
Thc horizontal distance
.I" '" 4 III
Fig. 6.33
The co·ordinates of the point B. which is on the ccmre·linc of the jet o f wate r and is situatcd on thc
ground. with respect to A (ori gin ) are
.1": 4 m 3ndy "'- I.O m (From A. poim B isvcnically down by I III I
,
Thc eq ua tion of the jet is given by (6.24) a.~)' "" x tan 9 _ ~s"c2e
2U
Substituling the known values as
-1.0=4tan450 -
9.8l~4 1
xSC(:145 0
2U
"" 4 _
7~~S x (J2y
{sec 45° "" COSI450
"" - : - ""
J2}
.fi
~ I
I~
~ I
IL
Dynamics of Fluid Flow
- 10=4 - 78.48x2
.
U'
"'
305 1
78.4&, x 2 '" + 4.0 + 1.0 " 5.0
U·
U1" 78.48 x 2.0 ,,31.39
5.0
u '" .J31.19 '" 5.60 111/s
= Are a x Velocity of jet
'" A xU ", .001963 x 5.6 1l) 3 lscc
'" 0.0 1099 .. .011 nlls. AD S.
Problem 6.41 A window. ill II ,'erlieal wal/. is a/ a disrance of 30 III abo!,t! rllt' grOlllld lel'el. A jet of
"'(!fer. issuing/rom (l /louie of diameter 50 mil! is 10 slrike Ihe wi"do .... The rille offlow o/,,"(lier IhrOl'gil
the lIou/e is 3.5 ",J/minute (lnd /lou./e is si/IUlled "/ (I dis/lweI! of I /II abo)'e ground le"f:/. Find th e
greate:;1 /Iorizonlili di~'I{mce fro m th e .ml/ of IIII' /lou.le so l/ul/ jel afwaler MrikeJ' the window.
Solution. Given:
:,;.W INDOW
Di stance of window from ground level = 30 m
Dia. of nozzle.
d=50mm=O.05m
Now the rntc of flow of fluid
, T
" ,
,
A", "4 (.05 t = 0.001 963 In-
Are a
30 em
3
Q = 3.5 m /mi nute
The discharge.
x
= 3.5 = 0.0583 m3/s
6Q
Fig. 6.34
Distance of nozz le from ground '" I n1.
Let the greatest horizonlal di51ance of the nozz le from the wa ll = x and le t an gle of inclination = O.
If the jet reaches the window. the n the poinl B on th e window is on the centre-l in e of the jel. The
co-ordinates of B with respect 10 A arc
.r=x.)'=30 - 1.0=29rn
The velocity of jcc.
U=
Dis<:ha r~e
Q
.0583
'" - =
= 29.69 m/kc
Are a
A
.00 1963
Using the equal ion (6.34). which is Ihe <"quation of jeC
)'
gx
,
' sec- e
= x lane - ~-
2V'
0",,' e
9.8h· 2
1 ""
2 x (29.69)
= .f tan 0 _ 0.0055 seclO x x l
.0055.1. 1
=.r lan 0 ,
cos- 0
29 .0 =.r lan 9 -
.. (i)
or tan e - .0055 .l leos'e - 29 '" 0
The
ma~imurn
value of or with res[)Cct to 0 is obtained, by diffcremi<lti ng th e above equ<ltion w.r.t.
dx
o and Subsliluling the value of ~
[
I I
.r sec l 9 + tan 0 x
dxl
"9
"9
'" O. Hence differe miating the eq uation (il W. r.t. O. we have
[ , ((-2»).
- 0.0055 x x
I
d' ]
~-,- (-sIl10)+ ~-,- x h~
cos 0
cos 9
dO
Ii
~ I
IL
1306
Fluid Mechanics
x tanB) = x scc
1J.: !!.-(
d9
x sec
2
2
e + tan e dx and .!:!.- [~l = x 2 .!!...(_I_)+ _ 1_, --"-Ix'l)
de de cos· e
d(J COO l e
cos · e de
2.r dx ]
,_.t·sln
' · 0
- .0055 [ oos J 0 + ----rcos O-'( 0 = 0
dx
e + tan 9 -IIG
For max imum val ue of .T. w.r.L
e,
we have d.l '" 0
dO
Substituting this val ue in the above equmion. we hav e
, ' 0 - .00 55
.I SCC
[2X1sinej_
J
_ 0
,~
.OO55 X2x l Si nB _
x
\:05"9 -
"'
"'
0
cool e
-
.\" - .011 X" iII" 9 = 0 or 1- .O ll .r lan
.I"
o
orx -.011
.1sin8 __0
cosS
X .I
e=0
I
Ian 9 = - - = 90.9
.,.( i i)
.011
90.9
... (i i i)
.r : - tan {I
Substituting this value of .r in equIItion (i), we ge l
90.9
lane
- - x tan
0
- .0055 x
(90.9)"
I
\ x ----;-:- - 29 '" 0
tan - a
cos ' s
909 - 45.445 - 29=0 or 6 1.9 _ 45.~45 = 0
.
si n ' e
sin " a
45.445
. 10 = 45.445
61.9 = -,- or sin
- - = 0 .7 341
si n "
e
61.90
sin e '" JO.7341 '" 0.8568
9=
Substitutin g this va lue of
lan- I
e in equation (iii). we gel
90.9
Ian 9
x= - = 54.76
I I
.8568 = 58° 57.8'
~
n1.
90.9
_~9~O~.9=
90.9
=
=- =54.759 In
Ian 58° 57.8'
Ian 58.95
1.66
An s.
Ii
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Dynamics of Fluid Flow
307 1
HIGHLIGHTS
I. The .tudy of fluid motiun with the forces causing flow is called dynamics of fluid flow, which i,
analysed by the Newton"s second law of motion.
2. Ilemoulli"s equation is obtained by integrating the Euler"s equation of motion. Bernoulli's equation Slales
"Fo r a steady. ideal flow of an incompressible fluid. the lotal energy which consist, of pressure energy,
kinetic energy and datum energy, at any point of Ihc fluid is ron,l;ml' . Mathematically •
..El.+ ,}
28
pg
where
li
+t
_!!.l.+
' - PB
,·i
28
+"
'-
= pressure energy per unit weight = pressur<: head
"
~
'"
"
~inctic
energy pcr unit weight., kinetic head
z, • datum energy pcr unit
weighl. da!iJ111 head.
J. Bernoulli s equation for real fluids
'il + z : [11 + "i + z, +h
L
pg2g'pg2g"
..El.+
where
hL
K
loss of energy Detween section> I
~nd
2,
4. The discharge. Q. through a venturi meter Or an orifke meter is given by
_C
Q-
°to;:
" J'lit - (/;:,
x J2gh
where {II '" area at the inlet of "enlUrimetcr.
0, z area at the throat of ,·cniurimelcr.
Cd ~ co-efficienl of \'enturimeter,
h '" difference o f pressure head in tcnns of fluid head flowing through venturimclcr.
5. The value of his gi\'cn by diffcrential U-tUDe manometer
h~ .{ [Sh
S"
-I]
h .. .,{ I - ::]
h ..
(~+ <L) - ('&; +t~ ) .. x [ ~: - I]
.
(when differentia l man01Heier contains heavier liquid )
... (when differential munometer comu;ns lightcr liquid )
... (for inclined ve1l1urimetcr in which diffcre1l1ial manometer
contains
hC~Lvier
liquid )
... (for inclined "eolurimeter in which differential man0111elCr
contains lighter liquid )
where x. difference in the readings of differe1l1ial manometer.
S. '" .p. gr. of heavier liquid
S• •• p. gr. o f fluid flowing through velllurimetcr
S, .. sp. gr. o f lighler liquid.
I I
Ii
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IL
1308 Fluid Mechanics
6 . PilOt-lube is used 10 fl nd the ,-c!ocity of a flowing fluid a1 any point in a pip" or a channel. The velocity
is given by Ihc relation
\I ~ C, J2gh
where C, z co-efficknt of PiloHuhe
h '" rise of li'luid in thc lube ~bo,'e free surface of liquid
. x[ ~: -1] (for pipes or channels).
7. The 1110menlum equalion Slales thaI Ihe nct force acting on a fluid
tum
p<."'
second in that dire.:lion. This i, gh'en HS F
M
ma~S
is equal 10 Ihc change in
mom~n­
,"
.!!..(I/II·j
The impulse -momentum equation is given by F . dl "d{m"j.
II. The force excned by a fluid On a pi"" bend in the directions of ..- and y arc given by
mass
F , '" - -
=
(Initial velocity in the direction of.r - Final wlocily in x-direction )
.;. In itial pressure force in .T-direclion + Final pressure force in
- pQlV" - V.:.. J + (p ,A,), + (1'0.:) ,
Ii, = pQ I v" - V"J + (p ,A,)" + (p,A,)r
""'
Ii" _
Resultant force.
.r-dir~"::lion
JI-~l + 1-~1
and Ihe direction of Ihe rcsullalll wilh horizonl:!1 is Ian
a.
F
---.!....
F,
9. The forcc exerted by (he nozzle on (he walcr is givcn by F, " pQW.:.. -V" l
and force exerted by the watcr on Ihe noUle is ~ - Fx ~ pQI V,-, - V.:..l.
10. Moment of momenlum equation Slales lhalthe resullanl lorque Hcting on a rotating nuid i, equal to lhe
rate of change of 1HOlnent of momenlum. Ma1hema1ically.itisgivenbyT .. pQIV••::_V•• ll.
II. f ree liquid jet is Ihe jet of water i"suing from a noule in atmosphere. The palh lravelled by Ihe free jel is
parabolic. The eq uation of lhejel is given by
grl
Y "'x Ian 6 - ~sccl{l
2U
where x, y .. oo-ordinales of any point on jet w.r .I. 10 Ihe nozzle
U .. velocily of je( of waler issuing from nozzle
6 • inclination of jet issuing from nozzle wi1h horizontal.
1 1
U sin 6
ll. (i) Maximum heighl attained by jel " "--"'-'~
"
g
(iii) Time 10 reach highesl poinl, .,... •
!.. •
2
U sin {I
18
1
(il') 1·lmimm:!1 range of Ihe jet ..• _ '"
U sin 26
g
(.,) Value of {I for maximum range. {I. 45·
~ U1'g.
(..,1 Maximum range, x-__
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Dynamics of Fluid Flow
309 1
EXERCISE
( A) THEORETICAL PROBLEM S
l. Name the di ffer':,I! forces present in a fl uid flo ..... , For the Eu ler"s equation of mOlion. which forces arc
taken in10 con,ideralion.
2. What is Eulcr"s equation of mOl ion ? How will you obtnin llemuull;'s equation from it?
J . Derive Bernoulli's eq uation for the flow of an incompressible f.iclionlcss fluid from consideration of
momentum.
4. State Bemoulli's theorem for steady flow of an incompressible fluid. Derive an expression for Bernoulli s
theorem from first principle and Slale Ihe assumptions made for such a Jcrivalion.
5. What is a "cnwrimetcr ? Derive an expression for the discharge through a "cnlUrimetcr.
6. Explain the principle of "cnwrimet.:. with a oem sKelch. Derive the cxprcSSiOl1 for the rate of flow of fluid
through it
7. Discuss the re lativc mer;ls and demerits of "c1l1urimeter with respect to oriflce -mctcr.
(Delhi Ulliwrsily. Dec. 2(02)
II . Dcfine an orifice ·melcL l>rovc that the discharge Ihrough an orificc-mc ter is given by the relation
Q~C"
"aa,
r:>7
x ,,2gh
J",l - ,<5
'" ~ area of pipe in which orifice -muter is fitl~-d
" 0 • area of orifice
(Technical Uni.-ersity of M.P.. S 2(02)
9. What is a piloc -Iube? How will you dctennine the velocity at any point wilh the help ofpilOl-tube?
(Dd"i UlliWfSily. Dec. 2(02)
Ill. What is the difference between pitot-tube and pitot _static tube ?
II . Siale the momcntum C4uation . How will you apply mOmentum equation for dClennin ing thc fo.-.:c cxcrted
by a flow ing liquid on a pipe bend ?
12. Whal is Ihe difference belween momentum equmion and impulse momentum equation.
13. Define momenl of momentum equation. Whe.-.: Ihis equation is used.
l~ . What is a free jet of liquid? Deri"e an e ~prcssion for thc path tra"clled by free jet i,suing from a nOl7.le.
I S. Prove Ihatlhe equalion of Ihe free jet of liquid is given by the expression.
where
l
gx
l
2Uwhere x. y • eO-<lrdi nales of a poinl on !he jC!
u. "elocily of issuing jet
o ~ inclination of the jet with horizontal.
Whieh of the following stmemelll is eorrce! in case of pipe flow
(0) flow I:lkes pkt~c from hiJ!hcr pressure 10 lower pressure ;
(b) flow takes pl"cc from higher velocilY to lower vel<:x,ity ;
(e) flow takes place from highcr ele"aliOIl to lower elevalion ;
(iI) flow takes pla~e from higher encrJ!Y to 10"'cr energy.
Dcrive Eulcr's equation of motion along a stream line for an ideal nuid stating clea rly the ass umptions.
Explain how Ihis is integrated to gel Bernoulli's equalion along a stream -line.
Stale l3crnoulli's theorem. fo,·lention the as>umptions made. How is it modified while applying in
practice? List OUI its engineering applications.
Define continuity equalion and Bernoulli's equation.
y ,. x Un 0 - -::--:-:osee 0
16 .
17 .
Ill.
19 .
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IL
1310 Fluid Mechanics
20 . What arc the different forms of energy in a flowing fluid? Represent s<;hematically the Il ernoulli s
equation for flow through a tapering pipe and show Ihe position of total energy line and the datum line.
21 . Write Eulers equation of mmion a long a stream line and integrate it to obtain BernouUi"" equation.
Stale all ass umptions {"ade
22. Describe wi th the he lp of sketch the construction. operation and u>c of 1';I01-slalic lube .
ll. Staning wilh Eu lds equation ormation along a stream line. Obtain Bernou lli's equation by its in!Cgralion.
List al11he a"umplions made.
24. Slate Ihe different devices that one "In use 10 meaSure Ihe discharge through a pipe and also through an
open channel. D,'scribe one of such devices with a Ilcal sketch and nplain how one can obuin the actual
discharge with its help'
25. Derive Bemoullfs equation from fundamentals.
(8) NUMERICAL PROBLEMS
1. Water is flowing through a pipe of 100 mm diameter under a pressure of 19.62 Niem I (gauge) and with
mean ~elocity of 3.0 mls. Find the total hend of the water at a cross-section. which is 8 m above the datum
line.
[Ans. 28.458 ml
2. A pipe. through " 'hich water is flowing is having diameters 40 cm and 20 Cm at the cross-sections I and 2
respectively. The velocity of wmer at sect ion I is given 5.0 mls. fi nd the velocity head m the scrtions I
IAns. 1,274 m : 20.387 m; 0.628 mJlsl
and 2 and also rate of discharge ,
J . The water is flowing through a pipe having diameters 20 cm and 15 cm at sections I and 2 respectively.
The rate of flow through pipe is 40 litresls. The 'lCCtion I is 6 m above datum line and section 2 is 3 m
above lhe datum. If the pressure at section 1 is 29, 43 Nlcm l . find the intensi ty of pressure at section 2,
[Ans. 32.19 Nlcm' l
4. Water is flowing through a pipe having diameters JO cm and 15 em at the bottom and upper end respec tive ly. The inten~ity of pre .• sure at the bottom end is 29,43 Nlcm l and the pres,ure at the upper end i~
14.715 Nlem1. Delennine Ihe difference in dalum head if the rale of flow through pipe is 50 Hlis
[,\ns. 14.618 Illi
5. The waler is flowing through a taper pipe of length 50 m having diameters 40 em at the upper end and
20 emallhelowerend.allhe rate of 60 litresls. The pipe has a slope of I in 40. Find the pressure at the
[Ans. 25.58 Nlem' l
lower end iflhe pr~ss ure al lhe higher le~eI is 24.525 Nkm',
6. II pipe of diameter 30 em carries water at a "elocity of 20 mlscr. The pressures al Ihe points A and Bare
given as 34,335 Nlem' and 29.43 Nlem' respecti,·ely. while the datum head at A and II arc 25 m and 28m.
Find lhe loss of head belweeTT A and B,
[Ans . 2 ml
7. A conical tube of length J.O m is fixed \'enically WiTh i" smaller end upwards. The velocity of flow at the
smaller end is 4 mls while al the lower end it is 2 mis, The pressure head al the smallcr end is 2,0 m of
liquid. The loss of head in Ihe tube is 0.95 \" 1 - ,·,)'ng. where "J is the velocity at the smaller end and ",
at the 10"'cr end respectively, DClennine the pressure head at the lower end. Flow takes place in downward
di re<:tion ,
[Ans. 5.56 m of fluid I
II . A pipe line carrying oil of specific gravily 0.8. changes in diameter from 300 mm al a position A 10
500 mm diameter 10 a position B which is 5 m at a higher level. If the pre"ures at A and Bare 19.62
Nlem ' aT,d 14.91 N!cm ' respective ly. and Ihc discharge is 150 Htresls. detennine the loss of head and
[Ans. J .45 m. I-l ow takes place from A to 8 1
direction of 110w,
9 . A horizontal venturimeler with in let and throat diamelers 30 em and J5 em respecl ivcly is used 10
measure the flow of water. The reading of differential manometer connected 10 inlet and throat is 10 cm
of mel"(:ury . Detennine lhe rme of flow. Take CJ • 0,98
IAn s. 88.92 litreslsl
I I
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Dynamics of Fluid Flow
311 1
10 . An oil of ~p _ gr. 0.9 is flowing through a "cOlurimeter having inlet diameter 20 em and lhroat diameter
10 em. T he oil-mercury differential manometer shows a rcading of 20 em. Calculate the discharge of oil
through the horizontal venturimcl"r. Ta ke Cd " 0.98.
IAn s. 59.15 lilrcslsl
II . A horizontal venturiml-('" Wilh inlet diameter 30 em and throat diameter 15 em;, used to measure the flow
of oil of sp . gr. 0 .8. Th e discharge of oilthroullh venturim,,!cr is 50 litTCsls. find the reading of the oil mercury differential manometer. Take Cd ~ 0.98.
IAns. 2.4!!9 eml
12. A horiwntal vcnturimeter Wilh inlet diameter 20 em and throat diameter 10 em j. u-"cd 10 meas ure Ihe flow
of water. The pressure at inle! is 14. 715 Nlcm 1 and Va~Uum pressure 31 the Ihroal is 40 em of mercury . Fin<l
the discharge of waler Ihrough venturimeter.
[.... ns. 162.539 liUs[
13. A 30em X 15 em "cnt urimerer is insened in a venical pipe carrying water. flowing in Ihe upwar<.J d irection.
A differential mercury manometer connected 10 the inlet and throal giYCs a read ing of 30 cm. Find Ihe
discharge. Take
0.98.
[Ans. 154.02 litls[
14. Ifin the problem 13. instc~<l of watcr. oil ofsp . gr. 0.8 is nowinglhrough Ihc ,·cnturimeter. detennine the
ratc of flow of oil in litres/s.
[Ans. 173.56 lills[
15. Thc water is flowing through a pipe of diameter 30 em. The pi pe is inclined and a "cnlUrimeter is inserte<l
in the pipe. The d iameter of ,'ent urimeter at throat is 15 cm. The d ifference of pressu re belween the in let
and throal o f the venlUrimeter is measured by a liquid of sp. gr. 0.8 in an inverte<l V-lUbe which gives a
reading of 40 cm. The 10" of hea<l between the inlet and throat is 0.3 times the kinetic head of the pipe.
fi nd the discharge .
[Ans_ 22.64 liUsl
16 . A 20 )( 10 em "enturimeter is provided in a vertical pipe line carrying oil of sp. gr. 0. 8. the now being
upwards. The difference in cle,·ation of the throat seelion and entrance ,,"clion of the venturimeter is
50 em. The differential V-tube mereury manometer shows a gauge deflection of 40 cm. Calculute: (i) thc
discharge of oil, and (ii) the pressure difference between the e ntrance section and the throal ",clion.
[Ans_(i) 89. 132 litts. (ii) 5.415 N/cm ' l
Take Cd . 0.98 and sp. gr. of mercury as 13.6.
17 . In a 200 nlln diameter horiwntal pipe a ven lUr im~tcr of 0.5 contr..ction ratio hilS been fixed . The h e~d of
water on lhe venturimcter ..... hen there is no now is 4 m (ga uge). find Ihc rate of flow for ..... hich the Ihroat
pressure will be 4 I1lClres of w~ter abSOlute. T"ke Cd '" 0 .97 and atmospheric pressure head'" 10.3 m of
water.
[Ans. 111.92 litl'l
18 . An orifice meter with orifice diameter 15 "m is insertl'<l in a pipe of 30 em d iameter. Th~ pressure gauges
fille<l upstream and downSlream of the orifice meter give readings of 14.715 N/cm ' and 9.8 1 Nlcm~
respectively. Find Ihe rate of flow of watcr Ihrough the pipe in litrest" Take Cd . 06. [Ans. 108.434litfsl
19 . If in problem 18. in stead of wate r, oil of sp. gr. 0.1I is flowing through the Orifice meter in wh ich the
pressure di fference is measured hy a mercury oil differential !Hanometer on the two sides of the ori fke
meter. find the rate of flow of oil when the reading of manometer is 40 cm.
[,\Ils. 122.68 litts[
20. The pressure di fference measured by the 1WO tappings of a pilot_static tube. one lapping pointing upstream
and other perpendicular 10 the now. placed in the cenlre of II pipe line o f diameter 40 cm is 10 em of watcr.
The mean ""Ioc it y in the pipe is 0.75 times the cenlml ,·clocity. Fin<llhe discharge through the pipe. Take
[An s. 0.1293 mltsl
eo-efflcient of pilot -tube as 0.98 .
C".
21. Find the vclocily of flow of nn oillhrough a pipe. when the d ifference of mercury level in a differenti al
U-tube mUnomeler ,:onneeted to the two tappings of the pilOt -tube is 15 crn. Taite sp. gr. of oil ~ 0.8 and
co-effle ien l of pitot_tube as O.9 R.
[An •. 6.72 m/sl
22. A sub -marine mOveS horizontally in sea and has ils axis 20 m below the surface of waler. A pitot -st:lIic
lUbe placed in front of ,ub· marine and along it' axis. is connecled to the two limbs of aU -tube
containi ng mereury . Th c difference of mercury IC"el is found to be 20 em. Find Ihe speed of sub-marine.
Ta ke sp. gr. of mercury 13.6 and of sea-wate r 1.026.
[Ans. 24.95l\ kmlhr.1
23 . A 45 " reducing bend is connected in a pipe line, the diameters at the in lel and outlet of the bend being
40 C111 and 20 em respectively. Find the force exerted by water on the bend if the intensity of pressure
at inlet of bend is 2 1.58 N!cm'. The rate of flow of water is 500 litres/s.
[.... ns. 22696.5 N; 20" 3.5'1
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IL
1312 Fluid Mechanics
24 . The discharge of water through a pipe of diameter 40 em
i~
4()() ]ilres/s. If the pipe is bend by I J5°, find
the magnitude and direction of the re<ultant force on the bend. The pre,.urc of flowing watcr i.
29.43 N/c", l,
[Ans. 7063 .2 N. 8 _ 22 ° 29.9' wilh x-axis clockwise[
25 . A 30 em diameter pipe carries watcr under a hca<l of 15 melres with a velocity of" m/s. If the axis of Ihe
pipe turns through 45", find the magnitude Dnd direclion of the re,uitanl force at the bend.
[Ans. 11717.5 N. 8 _ 67" 30' 1
",'!"""
26. A pipe of 20 em diameter conveying 0 .20
of water has a right angled bend in a horizontal plane.
Find (he rcsull:mt force exerted on (he bend if thc pressure at inlet and outlet of the bend are 22.563 Nlcm'!
Dnd 21.582 NiemI resf>C<;ti\'cly.
I An~. 116047 N. 0 " 43" 54.2']
27. A n07.zle of diameter 30 "nn is fitted 10 a pipe of60 nun diameter, Find lhe force exerted by the nOlzle on
the water which is now;ng through the pipe at the rate of 4.0 mJ!minute.
[Ans, 7057.7 NI
28. A lawn sprinkler with two nozz les of diameters 3 mm each is connlXted across a tap of water. The nozzles
are at a distance of 40 em and 30 em from the centre of the tap. The rate of water through tap is
100 em 3!s. Th e nozzle discharges water in the downward directions. Detennine the angular speed at which
the sprinkler will rot"te free .
[Ans, 2.83 ""tlsi
29 , A lawn sprinkler has two 'tozzles of diameters 8 mm each at the end of a rotating ann and the "ciocity of
now of water from each nOI.l.le is 12 m/s. Onc noule discharges water in the downward dirlXtion. while
the other nozzle discharges water ~ertically up. The nozzles arc at a distance of 40 em from the centre of
the rotating ann. Detennine lhe torque require{! to hold the rotating ann stationary. Also detennine the
constant .<peed of rotation of ann. ifit is free to rotate.
[An• . 5.711 Nm . 30 radJsl
JIl. A "ertical wall is of 10 m in height. A jet o f water is issuing from a nozzle with a "elocity of 25 m/s. The
nonle is situated at a horizontal distance of 20 m from the "ertical wall. Find the angle of projection of the
nozzle to the horizontal so that the jet o f water just clears the tap of thc wall. [Ans, 79° 55' or 36° 41' 1
31. /I fire-brigade man is holding a fire stream nozzle of 50 mm diameter at a distance o f 1 m above the ground
and 6 m from a vertical wall. The jet is coming out with a velocity of 15 m/s. This jet is to strike a window.
situated at a distance of 10m above ground in the vertical wall. Find the angle or angles of indination with
the horizontal made by the jet. coming out frum the noule. What will be the amount of water falling onlhe
window?
[An s_ 79" 16.7' Or 67° 3.7'; 0.0294 m'!sl
J2. /I window. in a ~enica) wall. is at a distance of 12 m above the ground level. A jet of water. issuing from
a nozzle of diameter 50 mm. is to strike the window. The r~te of now of water through the nOlzle is
40 htres/sec. The n01.zlc is situa1e{! at a distance of I m alx)\"e ground l c~el. Find the greatest horizontal
distance from the wall of the nozzle so thatjct of water strikes the window.
[,\ ns. 29.38 ml
JJ . Explain in brief the working o f a pitot -tube. Calculate the velocity of now of water in a pipe of diameter
300 mm at a point. where the stagnation pressure head is 5 m and static pressure head is 4 m. G i\'en the coefficient of pitot-tube '" 0.97.
[Ans, 4. 3 m!sce[
34. Find the rate of now of water through a vcnturimeter fitted in a pipeline of diameter 30 cm. The ratio of
diameter of throat and inlel of the venturi meter is •. The pressure at thc inlet of the vcnturimeter is
13.734 N/em 1 (gauge) and vacuum in the thro.1t is 37.5 em of mercury. The eo-effieicnt of vcnlUrimeter is
givcn as 0.98.
IAns, 0. 15 m'lsl
J5. /I 30 em x 15 em "enturimeter is inserted in a "ertieal pipe carrying an oil of sp. gr . 0 .8. flowing in the
upward direction. A differential mercury manometer connceted to the inle1 and throat gh'e, a reading of
30 cm . The difference in the elevation of the throat section and inlet section is 50 cm. Find the rate of flow
of oil.
J6. A venturi me te r is u<;cd for measurcment of discharge of water in horizontal pipe line. If the ratio of
upstream pipe diameter to that of throat is 2 : I. upstream diameter is 300 mm. the difference in pressure
between the throat and upstream is C<Jual to 3 m head of water and loss of head through meter is one·
eighth of the throat velocity head. calculate the dis.::harge in the pipe.
[A ns, O.107m)!s[
37. /I liquid of specific gravity 0 .8 is nowing upwards at the rate of 0.08 m'!s. through a venical
"cnturirncter with an inlet diameter of 200 nnn and throat diumeler of 100 mm. The Cd = 0.98 and th"
venieal distance between pressure tappi ngs is 300 nun. Find
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IL
Dynamics of Fluid Flow
313 1
(I) the difference in readings of Ihe Iwo pressure gauges. which are connected to the two pressure
tappings, and
(iil Ihe difference in the level of the mercury columns of the differential manometer which is connected
to the tappings. in place of pressure gauges .
[An s. (i)42.928I<Nfm'. (ii) 32.3 .1111
[Hin t. Q'" 0.08 mJ/.s. ti l ,,200 111m" 0 .2 m. til " 100 mill" 0.1 m,
Cd" 0.98, Z, -Z t" 300 mill. 0.3
.
II
l
•
_
11,= _(. 1 ) =O.Q07854m ",UsmsQ=Cd
-
.
,
,
m. II," _ (.2") ",0,0314 m"
4
4
<II X
n,
•
Ja,2- ai
M-::i:
x .. 2gh
Find 'h , This "3lue of
h '" 5.17 Ill.
Now use
, • [.!!L!i) .;. (ZI _ '2)' where p .. 800 kg/m!, find {PI - I'll.
P8 pg
Now use the fonnub
where h" 5 . 17 m, S," 13.6 and Sf" 0.8. Find the ~alue of x which will be 32.3 em. I
JII. A venturimetcr is installed in a 300 mm diameter horizontal pipe line. The throat pipe rates is 1/3. Water
flows through the installation. The pressure in the pipe line is 13783 Nkml (gauge) and vacuum in the
throat is 37.5 em of mercury. Negl"'ting head los< in the vemurimcter. delennine the rate of flow in thc
pipe linc .
[Ans. O. 153 m'/sec[
[llin t. ill ,. 300 mm .. 0.3 m. ill "
Hence
Value of
"3I x 300 ,. 100 mm = 0.1
•
~
.
m. P t ., 1].783 Nlcm " .. 13.783 x Iv Nhn -.
p,lp x g ~ (3 .783 x 10'/1000 x 9.81
'" 14.05 m. P/P8 '" - 37.5 em of Il g = - 0.375 x 13.6 'n of water
.. - 5.1 In of water. Hence h .. 14.05 - (- 5.1) '" 19.15 m of water.
CJ
,.
1.0. Now usc the fonnula Q w CJ
J<I,'1 "• 1 x vr;;-;:o:
2gh I
til - "1
39. The ma~imum flow through a 300 mm diameter horiwntal main pipe line is 18200 litre/minute. A
vcnturimcter;< inlroduce<l at a poi nt of the pipe linc where the pressure head is 4. 6 m of water. Find the
smal lest dia. of throat so that the pressure at the throal is never negali'·e. Assume eo-efftcicnt of meter as
unily.
[Ans. ii, = 19204 mm]
[Hint . "I '" 300 nnn .. 0.3 m. Q .. 18200 litres/minule .. 18200/60 .. 303.33 litres/s .. 0.3033 m' ls. P,lP8
fond the "alue of " , ' Thcn " , '"
.
,
"4 d i
and find d , .]
40. The following arc the data given of a change in diameter effected in laying a water .• upply pipe. The
change in diameter is grddual from 20 ern at A to 50 ern at H. l'ressures at A and Hare 7.848 Niem I and
5.886 Nkm ' respccliYcly with the end B being 3 m higher than A. If the flow in the pipe line is 200 litre/s. find :
(i) direction of flow, (ii) the head lost in friction I:>elwecn A and 11.
[Ans. (i) From A to 8. (ii) 1.015 111 ]
[Hin t. D A .. 20 em = 0.2 m. DB = 50 em '" 0.5 m. PA = 7.84 8 Niem I = 7.848 x 10"' NIm"
fiB = 5.886 Nlem 1 = 5.886 x 10" Nlrnl. ZA '" O. ZB = 3 m. Q '" 0.2 m' ,.
I I
Ii
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IL
1314 Fluid Mechanics
v'
E~ .. (r~/p x g ) .;. ....d... .;. Z~ .. (7_848)( 10' /1000 x 9 _81) ';' (6.369'12 x 9.81) + 0 .. 10.067 m
"v'
EB .. (I'i/r )( g) + ~ + ZH"' (5.886 x 10'/1000 x 9.81) + (l.DlS"12 x 9.81 ) .;. 3 .. 9.052 ml
2,
41 . II ,"cnt urimeter of inlet diameter 300 mm and throat diameter 150 mm is fi ~ ed in a vertical pipe li ne. 1\
liquid of 'po 8f. 0. 11 is flowing upward th rQugh the pipe line. II differential manometer containing mercury
Siy"," a reading of 100 mm when connected at in let and throat . The vertical difference belween inlet and
throat is 500 nllll. If Cd " 0.98. then find : (r) Me of flow of liq uid in lilre pcr seeo",' and (i') differcn.:c of
pressure between in lel and throat in Nlm '
[AIL<. (i) 100 litre/s. (i') 15980 N/ml ]
4 2. 1\ "colurimeter with a throat diameter o f 7.5 CHI is installed in a IS em diameter pipe . The pressure al the
entrance 10 Ihe meIer is 70 kPa (gauge) and il is desired Ihallhe pressure al any poinl sho uld nOI fa ll below
2.5 m nf absolute water. Detenn ine the ma~imUln flow rate of water through the meter. Take Cd " 0 .97 and
atmo spheric pressure as 100 )':Pa.
(J.N.T V .. Ifyliaab(l(/ S 2(02 )
IHln t. The press ure at the throat will be minimum . Hence ~ '" 2.5 m (abs .)
.,
pg
,
tI , '" 15cm:. A , '" 4" ( 15-) '" 176.7 em
G iven:
"
til "' 7.5 em ;. Al ", - (7 .5 ) ", 44. 175 em -'
4
p, : 70 kPa : 70x 10) Nhn l(gauj;ej,p.... '" 100 k Pa '" loox 10J Nhn l
PI (abs. ) .. 70 x 101 + 100 X 10J .. 170 X IO J Nhn l (abs .)
P
Pi:
LL ..
h ..
170 x 103
.. 17.33 m of water (abs. )
)OOOx 9 .81
.!i_ ~
pg
Q~
Now
PS
C"AI A1
z 17.33 _ 2.5 .. 14.83 m o f waler .. 1483 em o f water
JA,!- Ai
r:>7
O.97 xI 76.7 x44 .175x.J2x 98IXI483
x,,2Xh '"
J176. 71 -44 .175!
z 7S.41UI litre/s.]
43. Find the discharge of waler flowing through a pipe 20 em diameter placed in an inclined position. where a
vcnturimCler is inscrted, having a thro at diametcr of 10 cm. The difference o f pressure between the main
and Ihroat is measured by a liquid of specific gravity OA in an inverted V-tube . which gi"cs a reading of
30 CI1I . The loss of head Octween the main and throal is 0.2 timcs the kinetic he~d of pipe.
(i>elh; VIII"t.",sily. Dec. 2(02)
II
"
II
"
Illin t_ G ivcn : ii, '" 20 em :. " I '" -(20'
) = 100 II em' ; tI,_ = 10 em :. a,. '" -(
4
4 10-) ~ 25 II em-.
[ S"s, ] ( 01.0.4 )
x .. 30 em.h ..... 1- ----;- .. 30 1- -
Il UI h is
I I
~1 8 cm
.. O.1 8 1TI
al<o
Ii
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Dynamics of Fluid Flow
315 1
"'28
h L ",O.2x..:.L
Ii
From Bernoulli's C<jualion. -
PI(
m
1':"
-+- - '-
28
f'
V"
+" .. ~
-+- ---.L -I- z: + hL
pg
21(
(l!L+ZI)-(!i+
Zl) + .1.._~
= ilL
pg
pg
28 28
"
O. 18 +~ - ~ .. 0.2 "1
28
28
,
[
28
.. (12 +
pg
ZI)-(rlpg Zl) .. o.18m and
-+-
hL ..
0.2 v,l)
21(
v,z
\til
V,l
0.2
0.8 \li l V,l
0.18 + - - - ' - - - - = 0 orOIS + - - - - " - - = 0
28 28
28
28
2g
From continuity l"lUalion.
Now
(I,
V, = ",V,
O.I!! + 0.8 V,l
28
~4V,
or
_:1 .. 0 or 0.1 8 + 0.8 V,! _ (4\)2
28
0. 18 -+- 0.8 \<]1 _ 16"1
28
28
1
28
~0
28
= 0 or 0.1 8 = 16111 _ 0.8 V,l = 15.2\11
2$
21(
21(
~O:II~'~'~'~'~9~'~1 = 0.48 mls = 48 crnls
V, = ,/
IS.2
,
,
,
Q = A , V, = "4 (20" ) x 48 = 15140 em Is = 15. 14 li tre/ • . 1
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.. 7. 1
INTRODUCTION
Orifice is a small opening of any cross-section (~uch a~ circular. triangular. rectangulardC.) on the
side or al Ihe bonom of a tan k. throug h which a fluid is flow ing. A mouthpiece is a short lenglh of a
pipe which is IWO to three times its diameter in lenglh. fllted in a tan k or vesse l conlaining Ihe fluid.
Orifices as we ll as mouthpieces arc used for measuring the rale of fl ow of fluid .
.. 7.2
CLASSIFICATIONS OF ORIFICES
The orifices arc classified on the basis of their size, shape, nature of discharge and shape of the
upSlrealn edge. The follow ing are Ihe important classifiea t io n ~:
I. The orifices arc class ifi ed as s ma ll orifice or la rgt: o rifice depending upon the size of orifice
and head o f liquid from the centre of Ihe orifice. If the head of liquid from the centre of orifice is more
than five times the depth of orifice, Ihe orifice is c all ed small orifice. And iflhe head of liquids is less
Ihan five limes Ihe dept h of orifice. il is known as large orifice.
2. The orifices arc classified as (,) Circular ori fi ce. (i,) Triangular orifit""', (ii,) Rectang ula r orifice
and (iv) Sq uare orifice depending upon their cross-secliollal areas.
3. The orifices arc c l a~sified as (i) Sharp-cdged oriflce and (ii) Bell-mouthed orifice depending
upon Ihe shape of upslream edge of Ih.:: orifices.
4. The orifices are c lassifi.::d as (i) Free diseharging orifices and (i,) Drow ned or sub-merged orifices depending upon Ihe nalure o f discharge.
The sub-merged orifices arc further classified as (al r ully sub· merged orifices and (b) Parti ally
sub-merged orifices .
.. 7.3
FLOW THROUGH AN ORIF ICE
Consider a tank filled wilh a ci rcular orifice in one of ils sides as shown in Fig. 7. 1. Let I I be the head
of the liqu id above the centre of the orifice. The liquid flowing through the orifice fonns II jet of liquid
whose area of cross-section is less than that of orifice. The area of jet of fluid goes on deneasing and at
II section C·C, the area is minimum. This section is ~pproxim~tdy al a distan ce of half of dia meter of the
orifice. At this section, the streamli nes arc straighl and parallel to each other and perpendicular 10 the
317
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1318 Fluid Mechanics
plane of Ihe orifice. This sec ti on is ca ll ed Venu-contr"cla.
Beyond thi s section. the jet diverges and is anrac lcd in the
downward direction by the gravity.
Consider twO poinTS 1 and 2 as shown in Fig. 7. 1. Point 1 is
inside 111e tank and point 2 31 the vcn J-contracla. LC11hc now is
stead y a nd at a constant head H. Applying Bernoulli's equation at
points I and 2.
,
",i!!+I'i+ z,
pg 2g
-
""'
Fig. 7.1 Tank with " 11 orific~.
ZI '" Z2
,
'
P!
-PI + -VI = +I'i
-
pg
2g
!i
Now
pg
2g
=H
pg
P! '" 0 (atmosp heric pressure)
pg
I' , is very s mall ill comparison to v 2 as area of tank is very large as compared 10 the area of the jet of
.,,
liquid.
f/+O=O+ - '
2g
.J2gll
This is th eoretical ve loc ity. Actual veloci ty will be Jess th a n this va lu e.
\'2 '"
to 7.4
...(7.1 )
HYDRA U LIC CQ -EFFICIENTS
The hyd raulic co-cfflcients are
I . Co-efficient of velocity. C.
2. Co·efflcielll uf cumractiun. Co
]. Cu-efflcien! uf discharge. Cd'
7 .4 . 1 Co-e fficient of Velocity (C y ) ' It is defined as the ratio between the actual velocity of a
jet of liquid at vena-con tracta and the theoretical velocity of jet. It is denoted hy C. and mathemati cally. C. is given a.s
C. =
=
Actual velocit y of jet at vena -cont raeta
Tlleoretieal velocit y
v
r:>::U '
v2gJ1
whe re V = actual veloc ity . .)2gH '" Theoret ica l velocit y
... (7. 2 )
The value of C,. varies from 0.95 10 0.99 for different orifices. de(l<:nding On the ~ha pc. size of the
orifice and on the twad under which flow takes p lace. Generally the value of C. = 0.98 is taken for
sharp-edged orifices.
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Orifices and Mouthpieces 319 1
7.4 .2 Co-efficient of Contraction (C e) . h is defined as the rmia of Ih e area of the jet at
vena-contracta to the area o f th e orifice. It is denoted by Ce .
(I '" area of orifice and
G" = area of jet al vcna -co n lr~C l a.
C,, =
T hen
area o f jet at \'cna-con tracta
area o f orifice
'" G o
...(7 .3)
"
The va lu e of Cc varies from 0.61 100.69 dep"nding on slla pe and size of [h e orifice and tlead of
liquid unda w hich flow takes p lace. In ge nera l. the va lu e of Cc m~y be taken as 0.64.
7.4. 3 Co-efficient of Discharge (Cd)' It is defined as th e mtio of the actua l discharge from:m
ori fice to the th eoretical discharge from the oriti<:c. It isdc nolcd by CJ' If Q is actual di scharge i1 nd Q,.
is Ih" th~"(Irc tical discharge then mathematically. Cd is give n as
Cd'"
JL",
Q,h
Actual ve locil y x Actual area
Theoretical ve locit y x Th eoretica l area
I
: ",~A"
':"C'CI:'O'e""oc"'";'~Yc:: x 'CCA"'C""'O
' C,~''C'C' ::Theore tica l ve loc ity T heoretical area
... ( 7.4)
Cd",C"x C c
The value of Cd varies from 0 .6 1 to 0.65. For genera l purpose the va lu e of Cd is take n as 0.62.
Problem 7 .1 The head of W(lter o\"er (m orifice of diameter -10 mm is 10 m. Find the actlla/ dischargc ""d "elua/"e/oeily of thc jet al ,·c"a-contracta. Take Cd '" 0.6 ,md C. '" 0.98.
Solution. Give n:
H~nd.
H = 10cm
Dia. of ori fi ce.
40 mill 0.04 III
,,=
Area,
a
= 4,
=
. =.001256 ,
~ (.04 )-
111
Cd = 0.6
C. '" 0,98
-c.C
, A"':"e'C'cdC;"~""h,",~gc'=
(i)
Th eore tical di sc harge
BUI
: 0 ,6
Theoretical discharge = V'h x Area of orifice
V/h '" Theoretical veloc it y. where V'h '" J2g H = J2 x 9.S ! x JO =14 m/s
m2
,
T heoretic al discharge = 14 x .00 1256 = 0.0175SAc tua l dL.;;charge
= 0.6 x Theoretica l di scharge
= 0.6 x .0 175S = 0.01054 mlfs. Ans .
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1
320 Fluid Mechanics
-o;CA"'
:; C"'C"C"C'CICOC';"""Y;:::
(ii )
Tht:orctical \'d ocit y
'" c. = 0.98
Actual ve locity '" 0.98 x Thcorctical ve locity
'" 0.98 x 14", 13.72 m/s. Ails.
Problem 7.2 Th e head of \I"o/er o"er tile Cenlre of (ill orifice of diameter 20 """ is I m. Th e {lelul' /
d;~'c1wrge Ihrough IIII' Qrifice is 0.85 lilre/s. Find Ihe co -effieien, of discharge.
Solullon. Given:
Dia. of orifice.
d = 20 mill = 0.02 III
Area.
l/=
~ (O.02 ) 2 '" 0.000314111 2
4
Head.
H ", I
Actua l discharge,
Q = 0.85 litr.. /s = 0.00085 1I1 3/s
Theoretical ve loci ty.
III
V'h = J2g H = J2 X 9.8 1 X I = 4.429 Ill is
Theoretical discharge. Q,h = V,h x Area of orifice
= 4.4 29 x 0.000314 = OJ,OI39 ml/s
..
Co-effici e nt of discharge =
... 7.S
Actual discharge
0.00085
=
'" 0.61. AilS.
Theoret ic al discharge
0.00 139
EXPERIMENTAL DETERMINATION Of HYDRAULIC CO - EFFICIENTS
7 .S . 1 Determination of Co-e.fficie.nt of Discharge (Cd)' T ile waTer is allowed to flow
through an orifice filled to a tank under a constant head. H as s hown in rig. 7.2. The water is collecTed
in a measuring Tank for a known time. /. The heighT uf water in the measuring Tank is noted dow n. Then
acwal discharge through orifice.
Q" ,A"C'c.,",r"'"'''''="=","c,~,,=,,'k~Xc;H='=i'.'hC'CoCf~w,'=',,·'=i="~m='c·c',""=·,".,,',"='k=
Time (/)
and
thcOrc1ical disch~rge = area of orifice x ~2gH
S~Y~
WATER
I
I
'!
b"",',h, ,r
:,.·l
1
Fig. 7.2
Valu~ofCd.
. .. (7 .5 )
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Orifices and Mouthpieces 321 1
7.S .2 Determiniltion of Co- efficient of Velocity (C v ) ' Let C-C represents the ve narontracta of ,( je t of W3lCr com ing out from an orifice under constant heal! H as s ho wn in Fig. 7.2.
Consider a liquid particle which is at vc na-co ntraCIa at a ny time and takes Ih e position al P alon g the
j et intim c '(.
Let .1 '" horizontal distance tra ve ll ed by the part id c in tim e ' r'
)' '" vertic al di stance betwee n P and C-C
V", actual velocit y of j et al vc na-colltracta.
The n ho rizomal di stance ,
x'" V X I
. ..( i)
I
,
...(ii)
and vertic al di stance.
y= - gr
2
Prom equatio n (i) ,
"
1= -
V
Substituting thi s valu e of 'r' in (iil, we get
.( 1
1
Y= - g x - ,
2
V-
BU11hcorcti cai ve loc it y.
:. Co-cfficic nl of I'c loc it y. C. '"
~
= ~gXl
V",
2y
X
~ = J4yH
x'
"Z g H
"
= J4 yH '
7 . S. l Determination of Co -efficient of Contraction (C e) .
is dcle nnilled from the equat ion (7.4) as
...(7 .6 )
Th e co-cfnci ent of contracti on
Cd=Cv xCe
C = Cd
c
C,'
... (7.7)
Problem 7 .3 A jel of !l'afer. issuing frolll a sllarp-edgell rerrical orifice under a conSlall1 head of
10.0 cm. af a cerlain poim. liaS 'he horizon/aland !'erricai co-ordinates measllred fro/ll IIII' l·nla·comraC/II
(IS 20.0 C/ll Imd /0.5 elli resper/ire/y. Fi" d 'he mille of C,.. A IIO find /111' I'lllue ofCe if Cd = 0.60.
Solution. Give n :
Head.
Horizo ntal di s tance.
Vc n ic al dis tan ce .
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II = 10.0 ern
x = 20. 0 c rn
y = 10.5 ern
Cd = 0 .6
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1322
Fluid Mechanics
The value of C,. is given by equat ion (7.6) as
C = _ .'_
" .)4)'/1
'"
J4
20.0
x [0.5 x 10.0
20
- - '" 0.9759 == 0.976. An s.
20.493
The value of Cc is given by equation (7.7) as
CJ
0.6
C, = - = - - == 0.6147 '" 0.6 15. Ans.
C" 0.976
Problem 7.4 The head o/wafer ol'a (lfl orifice of diamerer 100 mm is 10 m. The waler coming Oll[
from orifice is collected in a circu/a, /mlk of diameter 1.5 m. The rist> of water iel'e/ in Il,is lank is
1.0/11 in 25 seconds. Also 1/11.' co-ordinates of (l poinT 011 1/,1.' jet. measured'rom rena-COll/wel« ar/!
4.3 In /Iorizon/ll/ (lnd 0.5 lit l"I:rlica/. FillJ tile co-efficients, Cd> C. (mlf Cr'
Solution. Given:
H~ad.
H == 10 III
Dia. of orifice.
Area ofoJificc.
Dia. of Illca5uring lank.
Area.
Rise of wate r.
Time.
Horizontal distance.
Venical distance.
d= lOOrnrn:O. lm
(I'"
,
,
,
- (.1)' '" 0.00785) m '
4
D: 1.5rn
2
It
A= - (1.5) =1.767 m
2
4
11= 1111
t = 25 seco nds
x = 4.3 III
y = 05 m
Now theoretical velocity. V,h: .J2g H : .J2 x9.81 x 10 : 14.0 m/s
Theoretical disch~rge, Q'A : V,h x Area of orifice: 14.0 x 0,()()7854 = 0.1099 m 3/s
Actu al disc harge.
Q: II
XII :
1.767 x 1.0 = 0.07068
,
Q
25
0.07068
Cd: =
: 0.643. Ans.
0.1099
Q'h
The value of C,. is given by equation (7.6) as
c= _ .' - =
,.
.J4yH
4.3
.J4 x 0.5 x [0
43
4.4 72
= 0.96. Ans.
C.. is give n by eljuation (7.7) as C( '" CJ = 0.643 = 0.669. AilS.
C,
0.96
Problem 7 .5 1V(ller discllarge (II Ihe mle of 98.2 lilresls IlmlUglIll 120 mm di(lmeler ..erliCCII
J'/r"rp -edged or iJke placed II/1du 1I COIIS(lIIll head of 10 melres. II pOilll. Oil Ille jel. meaSll r ed from II,,!
I'el!(l-COII/mcw of II,e jel I!(IS co-ordillmes 4.5 melres horiwllwl and 0.54 melres I'erli[(ll. Filld Ihe
co -eJJrcielll C,. Cc «lid Cd of Ihe orifice,
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Orifices and Mouthpieces 323 1
Solution. Given:
Discharge.
Dia. of orifice.
Q'" 98.2 litIs" 0.0982 ml/s
d= 1201l11n=O. 12m
:. Area of orifice,
O= ~ (O. 1 2)!=O.Dl13 1
ml
fI:lOm
Head.
Horizontal distance of a point o n the jet from vc na-contracta . .\' '" 4.5 m
Jnu vertical distance. }, = 0.54 In
Now theoretical velocity.
Theoretical
disc h~rgc.
V'h = J2g x H '" J2 )(9.8 1 x 10 = 14.0 m/s
Q'h'" V'h )( Area of orifice
3
'" 14.0 x 0.QI131 '" 0. 1583 m fs
. .
Actual discharge
Q
0 ,0982
The val ue o f C, IS g ive n by. Cd =
.
'" = - - - = 0.62. ADS.
,
TheorclJcal discharge
Q'h
0.1583
The value of Cc is given by equation (7.6).
C"'x=
4.5
'" 0.968. Ans.
'" J4yH
J 4x O.54xlO
The value of Cr is g iven by equation (7.7) as
Cd
C< '" -
C,
0.62
= - - = 0.64. Ans.
0.968
Problem 7.6 A 25 mm diameter nozzle di.~c1/{/rge$ 0.76 mJ of water fler minule when Ihe Ilead is
60 III. The diallleter of IIIe j,'1 i.1 22.5 111111. Delerlllin e : (i) Ihe I'allle$ of co-efflciet!/$ C". C. alld Cd ali<I
(iij Ille loss of head dlle 10 flilid resis/allce.
Solution. Givcn :
Dia. of nou:lc.
Actual discllargc.
~~3a.d~fjCt.
D=25mm",,0.025m
Qoct " 0.76 m]/minut c =
0~6
,,0.0 1267
t I3
N~"" I ~;'~~
--=~3:.=
==3)-
~:~.~nmm"0.0225m.
(i) VoilleJ ofco-eJftcicllI.l:
Co-cfficicm o f contraction (C~) is givcn by.
C _
C -
111]/S
-
.
Area of jcl
Area of nozzle
!!. d'
= _4__ "
!!. Dl
Fig . 7.3
~=
0.0225l
l
D
0.025
1
,, 0.81. Ans .
4
Co-efficient of discharge (Cd) is givcn by.
C _
d -
II
Actu al dischargc
Th core ti<.:al disc hargc
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1324 Fluid Mechanics
0.01267
===~"'===
Theo retical vel oc ity x Area of non le
=
0.0 1267
0.01267
--="';;---:J2gH X~D "
J2 x 9.81 x60 x 4 (0.025)'
4
It
'" 0.752. An s.
Co-efficient of velocity (C.) is
~ivcn
C. ""
by.
Cd
0.752
Co
0.81
- = - - '" 0 .928. An s.
(ii) Loss of ilead due to fluid feJ"iJ'/allce :
Apply ing Be rnoulli' s equation 31thc oU llet of nozz le and \0 thc j cl of water. we gel
~"
l!!.. + _,_ + z, '"
pg
28
V
1
Pl + -L + II + Loss of head
pg 28
~ '" PI", Atmosphe ric pressu re head
pg
pg
z, '" l l ' VI '" .)2811 . Vl '" Actual ve locity of jet'" C. J2g H
(.fiiii)'
=
(C, ~2,H)'
2,
+ Loss o f head
H:C;xH+Lossofhcad
0.
=1I -C;x l/ : I/( [ _ C,2)
'" 60(1 - 0 .928 2) '" 60 x 0.1388 '" 8.328
:. Loss o f head
m. Am .
Problem 7.7 A pipe. 100 ",m in diumeter, Iws (l /JOule aI/ached to il at rhe discharge elllf. til e
diame ler of Ihe IIoufe is 50 "'nt. Tlu" Tole of discharge oj"'(1/er Ihrough Ihe IIou/e is 20 lilres/s ,/lid Ihe
I'fes~'''re al Ihe base oj the nou/e is 5.886 Nlem l. Ca/cu/ale Ihe ('o -efficient oj disclwrge. An"me Ilwl
Ihe ba~'e oj Ihe nozzle ,/li d OIl1lel oj Ihe nou /e a fe at Ihe same e/el'lliion.
Solution. Given:
D= 100111111 =0. 1 m
Dia. of pipe.
"
AI= 4 (.0- = .007854 111-,
Dia. of nozz le.
d = 50
Al =
111111 '"
4" (.05)
0.05
G)
NOZZLE
c_~,,~=t:/0~'~.tT
~ipe
= .001963 III ,
III
,
BASE OF NOZZLE
AClUal discharge.
Q = 20 litis = 0.02 1I1 3/S
Pressure at the ha st:.
,
, N
PI'" 5.886 N/cill '" 5.886 x 10 - ,
PI = 5.686 Nlcm
Fig . 7.4
m'
.007854 VI'" .001963 V1
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Orifices and Mouthpieces 325 1
_ V1
4
where VI and V1 arc theoretical ve locit y at se ctio ns ( I ) and (2 ).
App ly ing Bcrno ulli 's eq ua ti on at sec tio ns ( 1) and (2) , we gel
V
I963Vl
,-_ .OO.007854
,l
p,
V,1
-PI + -V = - ' + -'pg 2g
pg 28
V, )'
4
5.8S6x 10 + ( 4"
I OOOx9.8 1
'" 0 +
v/
1.. l!.l.
'" Atmosph eric press ure =01
1
pg
2g
28
V1
V1
.,-"''cc,
= - '2g x l 6
2g
6 .0 + ,
2~
V
" [1- 16' ] = 6.0
v,'
or 2~
[15]
16 = 6.0
V1 ", J 6.0 X 2X9.8 1X :: "" 11.205m/scc
Tlicoretic al d ischarge
'" V 1 X A l = 11.205 x .00 1963 '" 0.022 111 3,s
_
C,_
Problem 7 .8
Actua l diiiCh argc _ 0.02 _ o . ~
_ - - _ .7"", Ans.
Th eo re tical d isc harge 0.022
A I(mk has Iwo ,dell/ie,,/ orifices VII one of ;I~' l'erliCa/
be/ow Ihe ,..aler $lIr!<I('(! ami/ower Oil" is 5 m he/oM' Ille waler
orifice is 0.96, find IIIe point of i"lersediuII of Ihe Iwo jelJ'.
Solullon. G ivc n :
Height of water from orifice (I ). 11 1 = 3 m
r rom orifl cc (2).
H~ = 5 m
C. fo r both '" 0.96
LCI Pis thc point of interscctio n of the twO jelS coming
from orifices ( I ) and (2). suc h that
~·ide5.
Sur/IK e.
x = horizo ntal di stance of P
)"1 '" ve rtica l di stance o f P Fro m o rifi ce ( I )
)'1 '" ve rt ic al di stance of P From o rifice (2)
Th en
)'1"')'2 + (5 - 3) ==.\'2 + 2 III
The value o f C. is given by e quat ion (7.6) as
I I
The upper orifice is J m
If the 'Oll/ue of C./or cad,
Fi g. 7.5
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1326 Fluid Mechanics
For orifice (I ).
c
For orifice (2),
C '"
",
=
.r
.,)4v,
H,
., J4y.r fl
l
=
=
l
"
.. ,(1)
.)4)', x 3.0
.,
.j4 X)"
.•.( i i)
x 5.0
As both the orifices are identical
C" '" C"
"'
B",
Y'''')'2 + 2 .O
30'2 + 2.0) '" 5)'2
2)" '" 6.0
y, '" 3.0
From (iI),
"'
0.96 '"
::'~'
~~
J4)( 3.0 x 5.0
.r ", 0.96 x ,)4 x 3.0 x 5.0 = 7.436 nl , An s.
Problem 7 .9 A closed I'essei cOllwil lS 'WI/er "(I/O a IJeigill of 1. 5 m 01111 o\'er II,e \l'llter surftlce IllerI'
is air /wl ling pressure 7.848 Nkm' (0.8 kgf/cm") lIbOl'e llimospheric f!"'smre. Alllie bOffom of Ihe
,'ej'se! Ihere is at, orifice of diameter I(){) mm. Fi" d Ihe rule of flow of ",,,'er /rOlll orifice. T"ke
Cd'" 0.6.
Solullon. Gil'en :
Dia. of orifice.
If '" 100 mm '" 0. 1 III
p " 7.848 Nlcm
Cd'" 0.6
Hdglll of water.
If '" 15 III
Ai r pressure.
p '" 7.848 Nfcm 2 '" 7.848)( 104 N/m'
Apply ing Bernoulli's eq uati on 3l sections ( I) (water surface) and
(2), we gd
II
v: 1
P
Vl
_' + _1 +ZL"'.:...l.+..:.L+Z,
pg2g
pg2g
'
Taking datum linc passing throug h (2) whic h is very clo.'iC to the
bonom surface o f the tank. Then Zl = 0, tl = 1.5 III
Also
"od
,
.............
-. --.- ---. _ .. •. -.-. -. '
I
I
L---1 !,t----'
H
T
~1.5
m
1
Fi g . 7.6
l!1.. = 0 (allnospheric pressure)
pg
..!!J..
pg
= 7.848 x 10'
1000)( 9.& 1
v;
8+0+1.5=0+ ---+0
,
2,
=8
III
of W3!cr
I Vj is negligible}
9.5 = V2
2g
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Orifices and Mouthpieces 327 1
v!
J2x9.8lx9.5 _13.652m/s
Rate of now of wawr = Cd x a 1 x V ,
n
~
3
1
:0.6x "4 (. 1) xlJ.652m/s: O.064J m /s. An s.
Problem 7.10 A closed {(10k partially filled willi wat'" UpfO (j heighl of 0.9 III hm'ing lIIl orifice of
diwlleler 15 mm a/ the bOl/om oflhe wllk. The air is pumpl!(/ inlO Ihe ljpper part of the Iilllk. De/ermine
tI!I' preJJure reqllired for a disc/mrge of 1.5 lilres/s Ihrough Ihe orifice. Takf' Cd = 0.62.
Solution. Given:
Heiglll of wa ter above orifice, If = 0.9 In
Dia.uforifice.
d= 15mm=O.O I5m
'4'[']"
t/" = '4 (.015t = 0.0001767 In-,
:. Area.
(/ =
Discharge.
Q = 1.5 litrcsls = .0015 mlls
Cd = 0.62
Let p is intensity of press ure required above water surface in Nlcm 1.
p x 10"
lOp
- - III of W3Icr.
=
"'~"''"'
pg
lOOOx9.8 1 9.81
p
Then pressure head of ail'= -
If V, is the v,",ocity at outle t of orifice, then
2 x 9.81(0.9+ lOP)
9.81
Q==CJxax J2g(H+p/pg)
Discharge
.0015 == 0.6 x .0001767 x ~r.2-X~9~.S~I"
(O~.9~+c,,~1p
- ,')
r-~---co"
2X9.81(0.9+ lOP) =
.0015
= 14.1 48
9.81
0.6x.0001767
or
2x9.8 1 ( 0.9+
IO P
9.81
)= 14. 148 XI 4.148
to II == -,14.,IC40SC'CIC4CIC4,S 0.9 = 10.202 - 0.9 = 9.302
9.81
2 x9.81
9302 x9.81
10
... 7.6
= 9. 125 Nfcm!. An s.
FLOW THROUGH LARGE ORIFICES
If the head of liquid is less than 5 times the depth of the orifice. the orifice is called large orifice. In
case of sm all orifice. the ve locity in tho.) entire cross-section of tho.) jet is consi dered to be const ant and
discharge can be calculated by Q == Cd x a x J2gh. But in case of a large orifice. th e ve locit y is not
conStant over the en tire cross-section of the jet nnll hence Q cannOl be calc ulated by Q = CJ x a x J2gh.
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1328 Fluid Mechanics
7 .6. 1
Disch.ug~
Through L.uge Rectangular Orifice . Consider a large rectangular orifice
in one side of the tank disch argi ng freely InlU atmosphe re under a constant head. H as sliow n in
Fig. 7.7.
HI " height of liquid above top edge of o rifi ce
H2 = height of liquid above txltlorn edge o f orifice
b = breadth of orifice
"',
If = depth of orifi<:c = H2 - HI
Cd = co-efficient of discharge.
Consider an ele mentary horizontal strip of d,'pl h . dll' 31 a depth of
liquid in the tan~ as show n in Fig. 7.7 (b).
'/1'
below the free surface of the
-
-
---- --
~~-.---
,.)
Fig. 7.1
'01
LarKt uctangu!ar ori/ict.
Area of strip = b x dl!
and lheordical veloc it y of watCT through strip = ~2gl!.
Oi scharge tllrough elemen tary
dQ
~lrip
is give n
= Cd x Area o f strip x Velocity
= CJ x bxdl/x J2gh :Cd bx,j2gh dh
By int~grating the above equation between tlw lim its HI and H!, tile tOlal dischar£e tlirough the
whole onli<.:e is obtained
,: Cd )( b x
.J2iJI/' Jh dlt ,: C~ x b x.J2i -3f2
"-'" ]'"
Ii,
[
1/
,
... (7 .8 )
Problem 7 .11 Find Ihe diIcharge Ihrough a reclangular orifice 2.0 m ...ide and 1.5 m deep filled 10
a w"lu lallk. The "'aler IeI'd in Ihe lallk is 3.0 III abo}'e Ihe lOp edge of Ihe orifire. Take Cd = 0.62.
Solution. Given:
Width of ori li<.:e,
b = 2.0 III
!kpth of ori fice,
"=1.50t
Hei gh t of water ahove top edge of the orifice, H I = 3 III
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IL
Orifices and Mouthpieces 329 1
He igh! of wmc r above bo tto m edge of 111e o rifi ce.
H1 ", II I + <1= 3 + 1.5 = 4.5
In
Cd == 0.62
Disc harge Q is give n by equati o n (7.8) as
,
2
~
31'
3f1
Q= - Cd xb x,, 2g 11f 2 - - 1f 1 I
==
~
3
x 0 .62 x 2.0 x
~2 + 9.S! [4.51.,
_ 3 U I mlfs
"3.66[9.545 - S.1961m 3/s '" 15.9 17 molls. '\I1 S .
Problem 7 . 12 A r('c{(//Igulor orifice. 1.5 II! wide and 1.0 III deep is discharging Wall'r frolll a wnk.
If Ihe \l'ater 11'1'1'/ in rhe /lUI " is 3.0 II! above Ihe lOp edg e of the orifice, filld Ihe discharge through Ih e
orifice. T(l ke the co-effrciem of disc/wrgil18 fo r Ihe orifice " 0.6.
Solullon. Give n :
b == 15 In
d == 1.0 III
H 1 =3.0 rn
Hl = H I + d = 3.0 + 1.0 = 4.0 111
Cd = 0 .6
Disc harge. Q is giw n by lh e equati o n (7.8) as
Width of ori fice.
Depth of orifice.
2
~
Q= ] X Cd XbX ..; 2g
,x
2
= -
IH;3f1 -HIlt2 1
0.6 x 1.5 x .)2 + 9.8114.0 u _ 3.0 u [ rn 1/s
'" 2.657 [8.0 - 5. 196] ml/s = 7.45 rn l/s. ADS.
Problem 7 .13 A recumglilar orifice 0.9 m wide alld 1.2 III deep is dischargillg water from a \"essel.
The lop edge of tile orifia is 0.6 m be/ow the waler sll rfila in the ,·esse!. Ca/cll/me (he discharge
Ih rOllgh the orifice if C J = 0.6 Imd perce ntage error if the orifice iI treall'd as a small orifice.
Solution. Givc n :
Width of uri flcc,
Dcptli uf orificc.
b=O.9rn
d= 1.2 rn
Hl = O.6 rn
111 = II I +d=0.6 + 1.2= 1.8 rn
Cd = 0 .6
Disc li argc Q is g ive n as
Q
="32 X Cd x b x J2i
=
X
IlI lll- H i~1l 1
"32 x 0.6 x 2.9 x .)2 x 9.81
11.8.lJ! - 0.63/1] mJ/s
= 1.5946 12.4149 - .4647 J = 3. 1097 III l/s. Ans.
Disc liarg in g for a s mall orifi ce
QI= C J xa x.j2gh
II
where II = II I + "2 =0.6 +
~ I
21.2
= 1.2 m and a= b x 11=0.9 x 1.2
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IL
1330 Fluid Mechanics
O.6x.9x1.2xJ2x9.8 I x l.2 = 3.1442 m3/s
Q1
'l> error = OJ - Q = 3.1442 - 3.1 097 = 0.01109 or 1. 109%. Ail S.
Q
~
3. [097
7. 7 DISCHARGE THROUGH FULLY SUB - MERGED ORIFICE
Fu lly sub-maged orifice is one which has its who]" of lhe outlet
side sub-merged under liquid so that it d ischarges ajcl of liquid into the
liquid of the same kind. It is also called totally drowned orifice. Fig. 7.8
shows the fully sub-merged orifice. Consider two points (I) and (2).
point I being in the reservoir on Ihe upstream si de of the ori fice and
point 2 being at the vcna-contracca as shown in Fig. 7.8.
Let III '" Height of water above the top of Ihe orifice on
the upstream side.
H2 '" H eigh! of water above lhe 00[[0111 of the orifice.
fI '" Difference in water level.
b = Width of orifice.
Cd := Co-efficient of discharge.
Height of water above the ce ntre of orifice on upstream side
Fig. 7.8
= ", + Hl - H, := H, + Hl
2
2
Hdght of water above the ccntre of orifice on downstream side
= H , + Hl _ H
2
Apply ing Bernoulli·sequalion at ( I) and (2). we get
p,
-
pg
Now
li",
pg
V,'
+-
2g
p.
V/
pg
2g
... ( 1)
... (2)
: - ' +-
H , + Hl , Pl :
2
Fully sub.merged ori[ir:t'.
- H and V, is n~gligible
pg
H , + Hl + 0 '" H , + Ih
2
2
V,'
H+~'~
2,
V1' = H
2,
V2 '" .J2g H
=l>x(H1 -H,)
Areil of orifice
Oischarge through orifice = Cd x Area x Velocity
:=
Cdx I> (1/ 2 - H,) x .J2g11
Q:= Cdx I>(H 1 -
I I
H,»( .J2gH.
... (7 .9)
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Orifices and Mouthpieces
331 1
Problem 7.14 find r/i,> di.lclwrge rlrmugll a fully Sllb-merged orifice of\\'itllil 2 II! if Ille diliaellc,of water lel'els OIl bOIl! sides of Ihe orifice be 50 em . Ti,e Ileighr of water from lOp and boltom of the
orifice are 2.5 III and 2.75 /II r('speCfil'ely. Take Cd " 0.6.
Solution. Give n :
Widlh o f orifice,
b=2 m
Diffe re nce of wate r leve l.
H : JO c m '" 0 .5
Hcigtn of water from top of orifice.
II , = 2.5 111
I-Ie ight uf wate r frum bottom o f o rifice. H! '" 2.5 III
III
CJ = 0 .6
Disd,argc lhrough full y suh-merge d orifice is given by equat io n (7.9)
Q = C J Xb X (1I2 - lI ,)X .J2g H
'" 0.6 x 2.0 x (2. 75 _ 2.5) x .J"2cxC9".8",cxc O".<5 m3/s
= 0.93% m 3/s. AilS.
Problem 7 .15 Find Ihe di~'ch(lrge I/l rough (l lolul/)' droll"lIed orifice 2.0 m wide (lnd I
difference of ...."ler /elds on both Ibe J'ides oflbe orifice be 3 m. TlIke Cd = 0.62.
Solut ion. Given:
Width of orifice.
b = 2.0 m
D<: pth of orifice.
d = I m.
Differe nce o f water leve l on both th e sides
H=3m
Cd == 0.62
/II
deep. if IllI!
Disc harge throu gh o rifice is Q "" Cd x Area x ,j2g11
"" 0 ,62 x b x If x ,j2gl/
= 0.62 x 2.0 x 1.0 x
,12 x 9.81 x 3
111 3/S = 9.5 13 mJ/s . Ans .
.. 7.8 DISCHARGE THROUGH PARTIALLY SUB · MERGED ORifiCE
Paninll y sub· me rged orifi ce is o ne whic h has its Outlet side
pnn iall y sub· me rged und er liquid liS show n in Fig. 7.9 . It is nlso
kno wn as pnrtia ll y dro wned orifice. Thus th e pnrt iall y
sub·m erged orifice has two pon ions. The upper po rtion be haves
as an orifi ce d isc harg in g free wh ile th e lower porti on hc ha ves as
a s ub· merge d orifice. Onl y a large or ifi ce can be ha ve as a
partiall y sub· me rged o rifice. The t0 1a1 di sc harge Q th roug h
partiall y sub· merge d o rifice is e qu al to the d isc harges throug h
Fig. 7.9
free and the sub · merged porti ons.
Disc harge throu gh th e s ub·m e rged portion is give n by
equation (7.9)
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Partially IIIb.muged
orifice.
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1332
Fluid Mechanics
Disctmrgc through Ihe free ponion is given by equation (7 .8) as
Q ~:
2
M-::
3f1
312
']C d Xbx v 2g If/; - HI I
Q = Q 1 + Q1
Tutal u ischarge
'" Cd X b X (11 2 - I I) x .J2g11
+f
Problem 7.16
A r{'(/lUlgl/lar orifice 0/2
In ..... id/I!
Cd
X
b x
/ii
[H /,,2 - H
3I2
1. ...(7. 10)
IInif 1.2 m deep isjifleil ill 0I1e J'ide o/a large
Wllk. The ",aler lel'e! 011 Olle side of Ihe orifice is 3 III abo!'e rhe lOp edge of the orifice, II'Mle 011 Ille
OJ/WT side of lire orifice. Ihe ...lIla /e"ei is 0.5 '" be/ow its lop edge. Calculate Ihe diiicilarge Ihroug"
Ihe orifice if Cd '" 0.64.
Sol ulion. Given: Width of orifi~. /,
=2
rn
Deplll of orifice. If = 1.2 111
Height
or w~tcr from
tOP edge of orifi<:c, H I '" ) 111
DiffeT"",:c of water leve l on both sides. H", 3 + O.S '" 3.5 111
Hcigtll o f Waler from the bottom edge of orific,,-
Ifl
= HI + d = 3 + 1,2 = 4.2 HI
Th" oritice is partially sub·maged. The discharg" throug h ,ob-maged ponion.
Q I = Cdx bx (H 2 - H)x .J2gH
= 0 .64 x 2.0 x (4. 2 _ 3.5) x
..j",C,c9".oS"I<=3.<5 '"
7.4249 mJ/s
Ttw discharge through free portion is
Q2 =
t
Cd
=
~
x 0.64 x 2.0 x .J2 x 9.81 [3.5"\12 - 3.0.112]
X
bx ,fii[H3J2 -
HIJ/2]
= 3.779 [6.5479 - 5. 1961] = 5.108 mJ/s
Total discharge through the orific" i,
Q = Q I + Q1 = 7.4249 + 5.108 = 12.532.9 m l/s. Am .
... 7.9 TIME OF EMPTYING A TANK THROUGH AN ORIFICE AT ITS BOTTOM
Consider a tank containing SOme liquid upto a hei gh t of HI. Let an orifice is fined at the bollOIll of
the tank. 11 is required to find the tillle for the liquid surface to faJi from the height II I to a heigh t H 2.
Let A = Area of the lank
II = Area of the orifice
Ill'" [n itial height oflhe liquid
fl 2 = Fi nal height of th~ liquid
T", Time in sL'Co nds for llie liqu id to fall from fli to H,.
Let at any time, the height of liquid from orifice is II and IClt lie
liquid surface fall by a sl1\all h"ighl till in lime tiT. Ttwn
Volume of liquid I"aving lhe tank in tim", tiT: A x rill
Also lhe theorctkal vcJOCily through orifice, V = .j2g11
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Orifices and Mouthpieces
333 1
Disctmrgc throu gh orifice/scc.
dQ = Cd x Area of orifice x Theoretical veloci ty" Cd'
(l •
,J2gl1
Discharge tltrough ori fi ce in time interva l
dT =C,j.a . .j2gll. dT
As lh e vo lum e o f liquid
I ~aving
Ihe tank i s equal 10
Ih~
vo lume of liquid Flow ing throug h orifice in
lime dT. we have
A (- dll)=C,/.a . .j2gh .dT
- vc s ign is inscn cd becau se with Ihe increase of lim e. head on o rifice decreases.
f'>::l:
- Adil=C,j.a. ,,2gll.dTordT=
By integrating the aoove equati on betwee n llie
r'"
T
Iill1it~
CJ .a .,,2gh
H I and H, othe total time. T is obtai ned as
dh
-A
J ill
)1'
h -·
.lI . .j2i - C" . a . .,ffi II,
_ All -I I!
jo d7'-- j II,, C
J
-AdII
F>7.
~
[ru
ruj -_ 2A [JH, -=
,lH,j
,H, _,H,
-2 A
Fe
Cd .(1.,,2g·
Cd . a. ,,2g
... (7 . 1 I)
For emp tyi ng th e lank comp letely. Hl == 0 and hence
lA
- Cd
T-
.[ii;
.a.J2i"
... (7. 12)
Problem 7 .17 A circu/<If la"k of diameler 4 m COJ/ fai/IS Inlier uplO {/ /wiglll of 5 m. The ({lIIk is
prodded Willi aJ/ orifice of diameler 0.5 III (j( Ihe bOllom. FiJ/d Ille lillie fakell by waler (i) I() fall from
5 m la 2 iii (ii) for co mpletely t'mplyillg the ({lIIk. Take Cd = 0.6.
Solution. Gi ve n :
Dia. of tank.
D=4111
Area.
Dia. of orifice.
Area.
Initial he ig ht of wate r.
Pinal heig ht o f wat er. (i)
Firs t ClIse. Wh~n
A=
~ (4)! = 12.566 Ill !
4
d=0.5 111
"
(1= - (. 5) =0. 1963 111"'
4
1-1 ,=5 111
I-I! = 2 III (ii) H ! = 0
fll =2 111
Using eq uation (7.1 1). we have T =
I I
2A
Jii {.{ii; ~.JH;""]
Cd . ( I . 2g
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1334
Fluid Mechanics
""
2 x 12.566
0.6)( .1963 x ../2)( 9.81
[r.
,,5- ....=1
2,0 ~conds
20,653
'" 0.5217 '" 39.58 lIt'm nd s. An ~ .
Senmd
ellS'"
When H2 = 0
2A
T",
Cd .a .
.Jii
~ '" ;;c:~2~X~I~2.~566""x~,f5",~
0.6x.1963x..)2)(9.8 1
= 107.7 seco nd s . Ans .
Problem 7.18 A circu/M /{mk of <lhll/ieler /.25 '" CO"/u;II5 waler "pia a heiglll of 5 m. All orifice of
50 """ diameter is prOl'ided (1/ i/~' hollom. If Cd '" 0.62. find Ihe Iwiglll o/waler abu,'" Jhe unfice after
1.5 minute:;.
Solution. Given:
Dia. of lank.
D:l.25m
:. Arc",
Dia. of orifice.
d=50mm=.05m
:. Area,
(I
"
= - (.05)- = .001963
,
01-
4
Initial height of w ater.
Time in seconds.
",=5111
T= 1.5x60=90seoonds
Lei the height of wa ter after 90 seconds = Hl
Usi ng equat ion (7.11 ) , we have T =
JH: -JH:]
2A
- -:;'---'---ic--'-"
Cd .a . 2g
JH:]
J'
=455.2 15 [2.236-~1
0.62 x 0.00 1963 x 2x9.8 1
.
2 x 1m[,f5 -
90:
,[iT;. '" 2.236 -
90
455.2 15
Hl '" 2.0383)( 2.0383
I I
'" 236 - 0.1977 '" 2.0383
= 4.154 m. An s.
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Orifices and Mouthpieces
.. 7. 10
335 1
TIME OF IMPTYING A HEMISPHERICAL TANK
Consider a Ilcrnisphtrical lanl.: of radius R filled with an orifice
of area '(( al its bouo m as shown in Fig. 7. 10. Th.e lank cOnlains
some liquid whose initial height is H I and in lillie T. the hciglit of
liqu id falls to H! . It is required tn find the time T.
U:i a1 any instant o f time. the head of liquid over the orifice is
hand <lllhis inswnt lel .1 be the radius orlhe liquid surface. Then
Area of liquid surface, A 0: IlIl
ORIFICE
and theoretical ve locity of liquid '" J2gh.
Fig. 7.10
lI~milpbmcaf
tallk.
Let th.: liquid level falls down by an amo unt of dh in time dT.
Volume of liquid leaving lank in lim", tiT = A x III!
= 1t1.2 x till
Also
volurn~
•.. ( i)
o f liquid nowing through orifice
= Cd x area of orifice x velocity
= Cd.a . ./2811 second
Vulume or liquid fluwing through orifke in time liT
... (ii)
: Ca.a. J2gh )( dT
From equations (i) and (ii). we get
2
7U ( - dll) '" Cd.a .
..j2gh . tiT
-ve sign is introduced. because wilh the increase or T. h will
- ltr!
de<.:Te~se
... (iii)
dl,: CJ.a . J2gh . dT
But from Fig. 7.10. for ""'OCD. we hal'c OC '" R
DO=R-II
OJ",.( ==
.\J
Substituting
~OCl _ OD 1 _ ~Rl _(R_h)l
== R'- _ (R _ It)" == R" _ ( R" + h'- _ 2Rh) '" 2RII _ Il"
x"in cquation (iii). we get
- n(2RII - IIl)dll == Cd.li. J2gh . dT
"'
dT:
1
- n (2 Rh - h )dll",
CJ .a.J2gh
The total tim e Treq uired 10 bring th e liquid lel'd from HI to H2 is obtained by integrating the above
equation between the limits H, and H2.
T-
f'-;c---,
-c'-".- (2Rll ln - "Jnjdh
JI, Ca ·a.j2i
-, fll, (2 Rh In- - II -jdll
~
In
Ca .a.,,2g II,
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1336
Fluid Mechanics
'"
[' R-,,-,,_,._I - "lll C+] ]"'
-It
~+1
cJ xax fii
C_'3 -+ ,C
2
2
=
[:: R ( H~i2 _ lIi'1) _ ~ (1I11) - 111m )]
- It
3
Cd X II X,fii
:
[.:!. R(H,l!! - H;'2 ) _ ~ (Htl) - Hi/C )] ... (7.13)
'
For comp lete ly emptyi ng lh e lank . HI = 0 and
Problem 7.19
5
Cd x a x ,ffi 3
T=
It
H,
5
h c ~ cc
fii[:: RHi'! - ~ H~12] .
Cd ,a . 2/1
3
... (7.1 4 )
5
A hemispher jw/ IW11:: of dill/nerer .,
III
CQl lfains waler lip/a
II
height of 1.5 m. All
orifice of diameter 50 111111 is prol'ided allile bOl/olll. Pilld IIII' lillie required by W<lIer (iJ 10 fail /rolll
1.51/110 1.0 m (iii for comple/ely emptying IIII' rank. Tank C~ '" 0.6.
Solution. Given:
Dia. of he mi splie ri cal tank. D = 4 III
:. Radiu s.
Dia. o f or ifice.
R ", 2.0 In
II" 50 mm = 0.05
:. Area.
(1
III
"
= - (.05) = O'()O I963
'
111 -
4
Ini tial heig h! of wate r.
111= 1.5 111
Cd '" 0.6
Fi rs l Case. H ~ = 1.0
Time T is give n by equatio n (7.13)
T-=
-=
[~ R ( H:'l _ H~ll )_ '-( HI)ll _ H;ll )]
II
Cd XIIX jii 3
5
X [~X 2.0 (1.5)12 - I.O lll )
O.6x.OO I96 3xJ2X9 .8 1
3
II
-
~ (1.5 512 - I .O~ll)]
5
-= 602. 189 [2.2323 - 0 .70221 " 921.4 second
= 15 min 21.4 sec. Ans.
SeCQnd Case. H2 = 0 and he nce ti m", T is give n by equ at ion (7. 14)
'r __
'
Cd . a .
:
I I
.J2i
[4
H lil 2H
$!l]
3
5
- R
I
- -
I
'
[~X2.0X
0.6x.OOI963J2x9.81 3
I .5Jll _ ~ X I..5m]
.5
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Orifices a nd Mo uthp ieces
337 1
'" 602. 189 [4.8989 - 1 10221 sec '" 2286.:n sec
'" 3H min 6.33 s.,.", An s.
Problem 7 .20
A hemis{!/laical cis/em of 6 III radius iii full of Waler. II is filled willi a 75 mm
dh'meler sllllrp edged orifice III "ie bollom. CalCl'/ale (I,e lime required 10 lower the le"e/ in 1/"" cislem
by :z melres. An'ume co-effie;..'", of discharge Jor lire orifice is 0.6.
Solullon. (jive n :
R~d ius
uf
hc rni ~phc rica l
ciste rn. R = 6
11\
",=6 m
Initial he ig h! of w ater.
Dia. of ori fi ce.
d = 75 mill = 0.075 III
:. Area.
0=
! (.075)l", .00441 8 m 2
4
Fall of height of wa ter
:2 m
(' i nal hei ght o f wa in, 111 == 6 - 2 == 4 III
Cd == 0.6
The ti me T is given by equat ion (7.3 1)
T=
:
[~R ( 1IIm _H~I1) _~ ( llt'2 _Hi'2)]
II
Cd x a x,fii 3
:;
,
~=;';;-c-~~
0.6 x .Q0..\4IS x
x9.81
J2
x
JI
[~x
, 6 (6.0 ! -
4.0 lll ) -
~
s (6.0$12 -
4.0" 2)]
'" 267.56 [8( 14.6969 - 8.0) - 0 .4 (88. 18 - 32.0) )
:: 267.56 153.575 - 22.472 ) sec
:: 832 1.9 sec '" 2hrs 18 min 42 sec. Ans.
Problem 7.21 A cylilldriw/ tllllk is IUII-ing a /utm isplwrical bllst'. Tilt' heig/II of cy/ilidriCi// portion
orifice of diameler 200 mm is jilled. Find the
is 5 m alld diameter is -.I m. All/Ie bottom of Ihis 10111.
,III
lim;~:~~'i ~~~ ~i~~,7~llele/y emplying
Ihe
/(111/;.
T(I/;e Cd :: 0.6.
T -:
_\~~~~{~~~~;~~~(~:_; -
Heig ht o f cy lind ri cal port ion (II ) = 5 m
=4.0 m
Di a. of lank
- :.:.:.:.:.:. _.
A :: ~ (4)!:: 12.566 m 2
Area.
"
4
tI---- 4.0 m -
<l:: 200 mm ::O.2 m
Di a. of orifi ce.
Area.
(I'" -
" {.2t '" 0 .03 14 m-,
4
- {1
z.o m
Cd:: 0 .6
ORIFICE
7
Fig. 7. 11
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1338 Fluid Mechanics
The tan~ is splittcd in two ponions. Fi rst portion is a hemispherical tank ami second portion is
cylindrical tank.
LeI T, '" time for e mptying hemispherical portion I.
T) == time for emptying cylindrical portion II.
Then lowl time T= 1'1 + T1"
For Portio n I. H I == 2.0 Ill. H 2 = O. Then TI isgi\'cn by equalion (7 .14) as
T, =
=
It
[ :.
Cd xax,fii 3
RH~l _2. Hi'l ]
5
'
[~X2.0X2.0JI2
O.6x.031 4 xJ2x9.8 1 3
_ 2.X2.0)ll]
5
== 37.646 [75424 - 2.2621 sec == 198.78 sec.
For Port io n II. H I == 2.0 + 5.0", 7.0 Ill. 112 '" 2.0. Then 1'2 is given by equation (7.11) as
2A[,fH; -,IH,]
Tl = "
Cd X{lX,ffi
2X12.566[J7-J2:0]
O.6x.oJ1 4 XJ2x9.81
To: 7', + 1'2"" 198.78 + 370.92
:. TOlallimc.
== 9 min 29
.... 7. 11
==
S L"<:,
== 569.7
sec = 370.92 sec
SL'i:
Ans .
TIME OF EMPTYING A CIRCULAR HORIZONTAL TANK
Consider a circular horizontal tank of length L and radius R, comaining liquid upto a height of H I'
Let 3n orifice of ~rea ' Q' is fillcd at the bottom of the tank. Then the time requi red to bring the liquid
level from fli to 112 is obtaincd ~s :
Lc t at any time. tlie licigli t of liquid over orifice is '/1' and in time dT. Jet tlie hciglit falls by an licight
of 'dh·. Le t at tliis time. the width of liquid surface == AC as sliown in Fig. 7.12.
I_
L
_I
Fi g . 7.12
SurFace area of liquid = L )( A C
Bo<
I
AC = 2 xAD = 2[JAO' - OB' =
~ ~H' - (R -
")'
J
= Z~R l _ (R' + Il " _ ZRII) = ZJZRII _ Il "
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Orifices and Mouthpieces
339 1
Surface area.
Vo lum e of liquid lea vin g tank in til llC tiT
==
A xdh == 2L J2RII - II " xdll
... (1)
A lso th e vo lum e o f liqu id fl ow in g th ro ugh orifice in time dT
==
Cdx Are a of o rifkc x Vclm: ity x tiT
Bm the veloc ity of liquid at the l im e cons ide red == .J2gh
Vo lum e o f liq uid flowing th rough urifi ce in tim e tiT
= CJ xax ../2g11 xlrr
... ( i i)
Equatin g (il and ( ii). we get
2L JZRh
_ 1, "
x (- dll) = Cd X II
X
,JZgh x tiT
- vc s ign is introd uced as w illi Ihe increase of
r. the
he ight II dccrcascs,
- lL.j (2R
C~
III dll
Xi/ Xlii
ITakin g
",
I
.fh COl11lll on]
",
- 2L (2R - h) - dll
To ta l tim e. T=
II,
Jii
CJ xax 2g
==.:-,,..1
x ax,f2i
- 2L
=
==
II,
Cd
II,
- 2L
III
[2K - I,]
dh
[12K - h)'" '' X(-I)] "'
~+I
C,s x a x ,ffi
2
=
=
2L
CJ x a x ./2i
",
x ~ X[(2R - h)lI1 J"'
4L
3Cd x a x ..{ii
3
H,
[(2R_H1)JIl_(2 R _ Ht )Jil ]
... O . IS)
For l'On1pl ctc ly em pt ying th e tank. Hl == 0 a nd hence
!].
J1
4L
[(2R/ 12 - (2R- Hd
... (7 . 16)
3C" x " x
Problem 7,2.2 A/I orifice of diameler f()() mm iJ filled III rhe hOllom of a boiler drllm of fe/lglfl
5 1M GIld of diameler 2 m. The drllm is IlOfizollla{ (IIld half full of warer. Filld Ille rime required 10
empl}' rhe boiler. gil'ell Ihe mlue of Cd " 0.6.
T=
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1340
Fluid Mechanics
Solution. Gi ven:
Di3. of orifice.
d '" 100 !lllll '" 0.1 III
"
a'" - (. n- '" .007854
4
Area.
Length.
L =S Ill
Dia. of drum.
D= 2 m
Radiu s.
,
nJ '
R=lm
Initial he ig ht of wa ter.
11 1= 1111
Final height o f wa t~ r.
Ill'" 0
Cd'" 0.6
f'orcompletely emptyin g the tank. Tis give n by equati o n (7. 16)
'"
4 x5.0
[(2 x])Y.! _ (2x l _ l)3r2]
3 x.06 x .007854 x ./2 x 9.81
'" 319.39 12.8284 - 1.01 = 583.98 sec = 9 min 44 sec. Ans.
An orifice of diameter 150 mm iJ filled at lile' bOl/om of II boiler ilrllm of lenglll 8 m
and of diameter J metres. 1'I,e drum ;.1 horizontal (md cOll/ain.! "'<lIer IIpSO a IIdg/1f of 2.4 m. Find Ille
lime required 10 empl}' tile boiler. Take CJ '" 0.6.
Solution. Gi ven:
Dia. of orifice.
d= 1501111n =0, 15m
Problem 7 .23
Area.
(I
"
= "4 (. 15)- = 0.01767
Lenglh.
L =8.0 rn
Dia. of boiler,
D=3.0m
:. Radius,
R= 1.5 [1)
Ini,ial height of water.
H)=2.4m
rind he ighT o f wa Ter.
Hl '" 0
,
Ill "
Cd'" 0.6.
For co mple tel y empl ying the lanl;. T is given by equ ation (7. 16) as
4 x &.0
3 x.6 x .0 1767 x
J2 x9.8 1
[(2 X I.S).lIl _ (2 x I.S _ 2 .4)"~n)
:227. 14[5.196 - 0.4647): 1074.66 sec
: 17 min 54.66 sec. An s.
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Orifices and Mouthpieces
.. 1. 12
341 1
CLASSIFICATION Of MOUTHPIECES
I , The mouthpieces arc classified as (i) External mouthpkcc o r {iil Inlc mal mouthpiece depend·
in g upon their position wi th respect 10 the tank or vesse l to which th.ey are fined.
2. The mouthpiece arc c lass ified as (I) Cy lind rical Illoulllpiccc o r ( ii) Con ve rge nt mouthpiece or
(iii) Convergent -divergent mouthpiece depending upon their shapes.
3. The mouthpi eces arc d a.%ifi ed ,IS (i) Mo uthp ieces running full or (ii) MUUThpieces running free.
depending upon the nmure of discharge ;u the outlet of th e mo uthpiece. T his d assificatiun is o nl y for
internal rnoulhpi c<:cs which arc known Borda's or Rc-cnlran t mouthpieces. A mouthpiece is said to bo:
rurming free if th e ~ l of liqu id after contraction docs no t touch the si des o f the rl1outhpiece. But if the
jet after contraction expa nds and fill s the whole mouthpiece it is known as run ning full.
.. 7 _13
FLOW THROUGH AN EXTERNAL CYLINDRICAL MOUTHPIECE
A mouthpi«:e is a short le ngth of a pipe whi ch is two o r three
times its d iameter in leng th. If thi s pipe is fitted e xte rnall y to the
ori fice. the mouthpiece is called ex ternal cyli ndrical mouthpi ece and
the discharge thro ugh orifice increases.
Conside r a tank ha vi ng an exlernal cylindrical mouthpi ece o f
cross-sectional area (i l ' allached 10 one of it s s ides as s how n in
Fig. 7. 13. The jet o f liquid e ntering th e mo uthpiece co ntracts to form
a ve na-co ntracta al a section C-c. Beyond this section. the jet agai n
expands and fiB the mouthpiece comp lete ly.
Let
II = Height of liquid above the centre of mouthpiece Fig. 7_13
v,. = Velocity of liqu id at C-C set: li on
a,. = Area o f fl ow at ve na-contraCla
"t = Velocity of liqu id al o utl el
at = Area o f mouthpicce at outlet
C,. = Co-e ffic ie nt of contraction.
Appl ying continuity eq uation at C-C and (1) -(1)_ we get
ExuTnal cylindrical
mOlflbpiecN .
ae x 1',. = arv r
" == al"1 == - "-' -
..
B",
!!.c.
",
a,.
lIJar
= C, = Co-effici ent of contraction
Taking C,. = 0 .62 . we get.:!..r.. = 0 .62
",
'"
0.62
" = --
,.
,
The jet of liljuid from section C-C s udde nly e nlarges at sec tion (1 )-( 1). Duc to s udd en e nlarg eme nt.
(",. -vd
Ihe re will be a loss of head. IlL· which is gi ve n as hi = "''-;;-''"-2,
•
I I
Please refer An. 11.4. I for loss of head duc to sudden enlargement
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1342
Fluid Mechanics
( -0'.6' 2
But " '" - -' 'c
0.62
-"1)' =;~ [O.~2- 1 j'- = 0.375 vi
2g
2,
Apply ing Bernoulli' s equati on \0 point A and ( I H I)
v1
{!
.....::!.. + ~ + Z~
pg
wher" Zot
= ZI' l'A is
2g
P
1,1
= - ' + - '- +ZI +h L
pg 28
neg li gi ble.
~ = atmospheric pressure = 0
pg
, ,1
If '" 1. 375 - '
2g
"g il
= 0.855 J2g H
1.J75
-- -
Theoretical veloc ity of liquid at outlet is
V'h
= JZglI
Co -efficient o f ve locity for mOUlhpkcc
J2iii
0.855
C '"
Actua l ve locity
'"
J2gH
'" Th eoretical velocit y
= 0 .855.
Cc for mouthpiece ", I as Ihe are a o f jet of liqu id at o l,lliet is equa l to the area of mouthpiece at o Ull et.
T hu s
Cd = Cc x C. = 1.0 x .855 = 0.855
Thus Ih e va lu e of Cd for mouthpiece is more Ihan the va lue of Cd for o rifice. and so di ;;c hargc
throu gh mouthpi c<:c will be more.
Problem 7.24 Find the discharge from II 100 mm diameler c.u emal mouthpiece. filled to a side of
(l /tlfge \'esse/ if the head ol'a the mouthpiece is../ metres.
Solution. Given:
Dia. of mouthpiece = 100 11\ = 0. 1 In
Area.
Hc.:u./,
Cd ror mo uthpiece
Disc harge
(l
= ~ (0. 1 ) 2 =0.007854 11\ 2
4
H = 4.0 11\
= 0.855
= Cd x Area x Veloc ity = 0.855 X li X.)2g H
= .855 x .007854 x .)2 x9.81 x4.0 = .05948 IIl l /s.
AilS.
Problem 7.25 An eX le",,,1 cylindriCll/ moulhpiece of diame/a 150 mm is discllllrging "."Ier u",fer"
conJHml llead of6 m. Delermine rhe disclwrge "nd "bw/"re pressure head of ,.."Ia 1I1 n'nll·conlraC/".
TlIke Cd = 0.855 alld C,. fo r \'ellll'COlllfllCW = 0.62 . Allllospilaic pressure head = 10.3 til of Wafer.
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Orifices a nd Mo uthp ieces
Solution. Given:
Dia. of mouthpiece.
d= 150 Imn:Q. 15cllI
:. Area.
a '" "::(.15) 2 '" 0.0 1767 rn 2
343 1
4
H= 6.0 III
Head,
Cd = 0.855
C~
al vcna-contracta '" 0.62
Atmospheric pressure head.
H~
Discharge
= [0.3
III
=CJ xaxJ2g H
'" 0.855 X .01767 X
.J2 X 9.8 1 X 6.0 '" 0.1639 ",lis. An s.
I' ress UN Head a t Vena-co ntractu
Apply ing Bernoulli's equation at A and C-G. we get
------.
_-------.
............. .........
p I '!
p~!
~+.....1..+ZA=_
c +-'-Hc
pg
B",
28
pg
2g
.!!..t,.. = II. + II.
pg
I'A '"
"
O.
A
I
Fig . 7.14
Bm
,~ =-0".62'1I,, =Hu+lI- ( - " ) ' X - I = "'"+ II - -,,''
.62
B",
H", 1.315
2g
28
x -I-,
(.62)
I'I!
2,
Vil = ~= 0.7272 II
28
1.375
I
He'" fl. + f/ - .7272 Hx - ,
(.62)'
'" fl. + f/- 1.89 H= Ha - ,89 H
= 10.3 - .89 X 6.0
{-: Ha '" 10.3 and 11= 6.0)
= [0.3 - 5.34 = 4.96 m (Absolute). An s.
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1344
Fluid Mechanics
.. 7. 14
flOW THROUGH A CONVERGENT- DIVERGENT MOUTHPIECE
If 3 mouthpiece converges UplO v"na-conlraCla and then di verges as s hown in Fi g. 1 15 tlle n that
type of mouthpiece is ca ll ed COllvcrgcm-Divcrgcnt " 1outhp iecc. As in this lIl oulhpicr c the re is no
s udde n clllargc men! o f llie jet. th e loss of ene rgy du e to sudd en en large me nt is elim inated. The coefficie nt of discharge for this mouth piece is unit y. Let H is th e head of liqu id over llie mouthpiece.
Applying Bernoulli's equation to the free SUrf,I CC of water in tank and sect io n C-c. we hav e
-
II~'
pg
,
+-
2g
Pc
,
Vc
+ z= - + - +,"
P8 2g
Taking datum pass ing through (he centre of orifice. we ge l
... ( i)
,. ~
~ : Il +/f - I '
28
~
I'e '"
... (ii)
c
~2g( H. + II
Conv ergentdivergent mQllt hpiece.
Fig. 7.15
Hr)
Now a ppl ying Bernoulli' s equatio n at sect ions C·C and (I )·(l)
,
,
I'"
-P'
' + ~'~+ Z =' + _1
pgZg '
pgZg
+Z I
Bm
1,2
1,2
H,. + ~'~ =Hd + -'
Also from (;) .
Zg
Zg
Hc + v/ IZg=II + H"
lid + v/IZg = II + lid
... ( iii)
"I=.j2g H
Now by co ntinuity equatio n.
tlc v e
= "I X (I I
!2. _ 3:..= JZg( H. + H He )
lI,
-
VI
.jZgH
1 + lId - l-l~
II
The discharge. Q is given as Q = a, x .jZgH
where
I I
li e
Hd
lie
H
H
- +l~ ­
...(7 . 17)
... (7 . 18)
= area at vena-conlraC la.
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Orifices a nd Mo uthp ieces
345 1
Problem 7 .26 A com 'ergel1 /·dil'ergelll mOlllhpiece /ral'ing llirom diameter of 4.0 em is discharg ing
Waler IIIlder a cOlIswlllllead of 2.0 m. deTermine II,e maximum OIlier dimneler for maximum di.IClwrge.
"-ind maximum discharge also. Take Ha " /0.3 III afwater and H ",p ,,2.5 III a/waler (absolule).
Solution. Given :
Dia. of throat.
Area.
Il c "
" (4t
= 12.566 ern-,
-
4
fI '" 2.0 III
ConSlant head.
Find max. dia. at o Ull e!, d l and
Om.,
H. '" 10.3
H"p '" 2.5
In
In
(a bso lute)
The di scharge, Q in converge nt.diverge nl mouthpiece depends 0 1> lh e are a al Ihroat.
Qm>., '" a, x ,)28 1/ '" 12.566 x .J2 x 9.& 1 x 200 '" 7H7 1.5 CIlIJ/S. Ails.
Now ratio of are as at oullet and throat is give n by eq uati on (7 .17) as
~
oc
'" I]+ 11"
V
He"
11
II + 010=.3" ,,=2.='
V
2.0
=2.2 135
,
,/'
- lr
4
I
- d
4
, '" 2.2 135 or
c
r::r
= 2. 2 1JS
ill '" 1.4877 x d~ "" 1.4877 x 4 .0 "" 5.95 em. A ns .
Problem 7.27 Ti,e tl/rom Illld exil diameters of conl'ergenl·d;"ergelll mouthpiece are 5 em Illld
10 CIIl respec/il'e/y. { I isfilled 10 Ihe l'erlielll side olillollk. cOIII(lilli"g Inlier. Find 'he lIliuim ,u" helld
of II water lor steady floK'. The maxim"m "lIcuum preS~'ure is 8 m of \l'lIler alltl lake lIlmospheric
pressure"" 10.3 m I\·(l/er.
S o lution. Give n :
Dia. al throal,
Dia. a[ exi t.
A [mo~ph e ric p re~sure
de "" 5 em
", "" 10 cm
head. H. "" 10.3 m
The maximum vacuum pressur.:: will be at a throat o nl y
Pressure h.::ad at [hroal " 8 m ( vacuum )
He "" HQ - 8.0 (ab,'IO IUle)
" 10.3 - R,O " 2.3 m (abs.)
Let maximum head o f wate r ove r mouthpi ece "" H m o f water.
The ratio of areas al out le t and th roat o f a convc rge lll-di l'erge lll moulhpiece is given by equat ion (7. I 7).
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1346
Fluid Mechanics
~1 + 1O.3
23
H
Hi
to !
-,-:
4=
,-
0'
If '"
8 or 15", -8
1+ - or 16= 1+ H
H
H
.! '" 0.5333 m of water
15
Maximum head of water'" 0 .533 m. An s.
Problem 7.28 A con",:'gf!n/-t/ij'ergenl moull/piece is filled /0 Ihe side of (I Illllk. Th e discharge
through muulilpier( "",Ier (I COll:!/,ml lier,d of /.5 In is 5 liITes/s. Th e helld loss ill Ihe dil'erg<',,1 portion
is 0.10 limes Ihe killetic IW{ld at oliliel. Find the ,"roalllnd ".til diameters. if 5cl'(I"'liOlI pressure is
2.5 In w/(I a/mo,~pheric pressure llead '" 10.3 III ojW(,ler.
Solullon. Given:
Constant head.
1/ '" 1.5 III
Discharge.
Q '" 5 rilres '" .005 111 3,s
ilL or H ead loss in divergent '" 0.1 x k i netic head at oUllet
lie or H"p = 2.5 (abs.)
If" "" 10.3 m of wmcr
Find (i) Dia. a1 th roat. d<
(ii) Dia. at oUIIc!. til
(r) Ilia. a l th roa t (d <). Applying Bernoulli' s equation to the free water surface and throat section.
we get (See Fig. 7.15).
-
P
pg
I,l
+-
2g
+ z==
P
I,l
pg
2g
- ' +-'- +<,
Taking th e centre line of mouthpiece as datum. we get
,
"
H" + O+H ==H,. + -
2,
1,2
-'-== Ha+H-Hc== 10.3+ 1.5-J.5 ==9.3 morwater
2,
I'e ==
J2 )( 9.81 )( 9.3 == 13.508 IlIfs
Now
.005 x 4
It X
13.508
" J.0CXJ47 == .0217 m
==
2. 17
CIII .
AilS.
(ii) Dia. a \ oullet (d,). App lying Bernoulli's eq uation to the free water surface and ou tlet of mouth·
piece (See !'ig. 7.15). we get
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Orifices and Mouthpieces
347 1
1,2
II~ + 0+ 1/ : H" + - ' +0 + 0.1
2,
.,2
",l
28
28
H = - ' + .I x 1'1
=
"J
= .1 - '
28
J2gH = ,p2~X~9~.8~'GX~1.5~= 5. 1724
l.l
l.l
Now
4 x .005
11: X I',
II> 7 . 1S
=
,1:04::X7:.00
",5~ '" 0.035 III = 3.5 .:m. Ans.
1f
X 5. 1724
FLOW THROUGH INTERNAL OR RE- ENT RANT ON BORDA 'S
MOUTH PIECE
A shon cy lind ric al tu be anached to an orifice ill s uc h a way Ilial th e lube projects in wardl y \0 a
tank . is c alled an inlc rn al mo uthpiece. II is also ca lled Rc-c mraru or Borda 's mouthpi ece. If the length
of the tu be is equ al to its di ame te r. the jet of liqu id co mes Oll t fro m mouthpi ece with out to uc hin g the
s idc50ftbc mbe as ~h ow n in ri ~. 7.16. T he moU!hpiccc is kn ow n as fUlI/ling/ree. But if the leng th of
the tu be is about 3 li mes its d iameter. th e jet comes out wi th its diameter equal to th e d iamete r o f
mouthpi ece a t o ut let as s how n in Fi g . 7. 17. T he llI o uth piece is said to be fUlUlingfull.
(i) Bo ni n's !\Io ut hpi l'C" Run n in g Fre.. , Fi g. 7. 16 s hows the Bonia's llI o uthpiece runnin g free.
Let
H = heig ht o f liquid above th e Ill outhp iece,
a = area of Illouthpiece,
" e " area of contracted jet in th e mo uthpi ece,
ve " veloc ity th ro ugh mouthpiece .
RUN NING FREE
Fig. 7.16
RU NNt NG FULL
Fig. 7.17
Thc n ow of fluid th ro ugh mo uth p iece is takin g place d ue to th e press ure force exc n ed by the flu id
o n th e entrance SC1:tion of the mo uth p iece. As the area of the mOllthpk'i:e is "a' hence total pressure
force o n entrance
"pg.a . 11
where II = di stance of e.G. of area ',,' from free s urface = H.
, ..(i)
=pg.a. H
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1348
Fluid Mechanics
According 10 Newton's second law of motion. the net force is equal lO the ralC of change of mOlllen ttl 111.
Now mass of liquid flowing/sec '" p x (I e )( I'e
The liquid is initially at rcst and hence initial velociTy is zero bUI final veloci ty of fluid is I'c'
Riue of change of lllomClllum = mass of liquid flowing/sec x [tinal velocity - initial velocityj
= pac x 1',lv, - 0] = pac v/
... (Ii)
Eq uating (I) and (Ii). we get
pg ,a.H. = pac' \./
... (i ii)
Apply ing Be rnoulli's equation to free su rface of liquid and section (1) -( 1) of Fig. 7.16
Taking the centre line of mouthpiece as datum. we have
z=
H.
z, '"
p
0, pg
v= 0
'" 0,
,
0+0+11=0 +-'- +0 or H= ~
"
2g
28
Substituting the valu e of
Vc
in (iii), we gel
pg . <l • H. = P . a" 2g . H
_ ')
a,_
_ _I __O .5
"
2
l!_~a c or -
.
.
U
Cu-cfficlcill of cun tTacUOIl. C<" ...L '" 0.5
"
Since there is no loss of h~ad . co-effident of velocity. C, '" 1.0
..
Co-effident of discharge. Cd" C< X C. " 0.5)( 1.0" 0.5
Discharge
Q '" Cd a.}2gff
...(7. 19)
'" O.S )( aJ2gH
(ii) Bord a's l\l oulhp iece RUllnill1l: l-' ull . rig. 7.17 shows Borda"s mouthpic(;c rUllning full.
Let
H"" height of liquid above the mouthpic(;~.
"I = velocity UI outlet or lit (1).(1) of mouth[licc~ ,
(l = ar~a of mouthpiece,
a < = area of the flow at C· C,
V C = velocity of liquid at \,ena·contrac ta or at C·c.
The jet of liqu id after passing through C·C, suddrnly enlargrs at sc£tion (1 H I). Thus therr will be
a loss of head due to ~uddc n cn l arg~1llen\.
It L
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(v<_1'd
'"
1
'-'-c;-"2,
... ( i)
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Orifices and Mouthpieces
Now f rom continuit y . we have a,. x
I'c '"
v~
0'
Substituting this va lu e of
I ' ..
~
a~
Vi'
349 1
= a t x VI
X I' I
==
--"'- = ~ = ~
aC I {' I
CC
0.5
= 21"1
in (i). we ge t hI."
(2" 1-1(1)1
"-",,...c,,-2s
_
1','
28
Apply ing Berno ulli 's equation to free surface of water in tan k and section (1).( I). we ge t
p
1, 2
pg
lg
-+ T~kil1g
PI
+z= -
pg
VI'
+-
2g
+ z, + il l
.
dmulll line passing through th e CClllre line of mo uthpi ece
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V !
O+O+ H =O + - '- + O+ -'-
28
28
2
~
2g
28
l
H=~+~: !L
g
== .JgH
He re \', is act ual velocity as losses ha ve b<:c n taken illlo consider-Hion.
\'1
But th eoretical ve loc ity.
.
.
\' ,h
==
,JZgH
l'l = J,[iii. =
I
M= 0.707
2811
,,2
As th e are a of th e jet at o utle t is equal to the are a of the mouthp iece. hence co-effi cient of
:. Co-efficlen l of ve loc1t y, C, = -
V'h
contracti on ", I
Discharge.
Cd '" Cc X C,.= 1.0 x .707" 0.707
Q" Cd X (/ x J2 gH "" 0 .707 x a x J2gH
Problem 7 .29 An in tern(ll mouthpiece of 80 mm diameter iJ" di~"C!lIIrging ....aler unde r
head of 8 metres. Find Ih e discharg e through mouthpiece .....1"'"
(i) Ti,e mouthpiece is runnillg free . alld (ii ) The mo uthpiece is rI/IIlling JIlII.
Solullon. Give n :
Dia. of mou thpiece. d = 80 10m = 0.08 III
Area.
Constant head.
... (7.20)
(I
constant
a = ~ (.08 ) 2 = .005026 m 2
4
11 = 4 m.
(i) Mouthp;"ce runu;n g f ree . The di sc harge. Q is g iven by equat io n (7.19) as
Q=0.5x(lx J 2gH
= 0 .5
x .005026 x J2 x 9.8 1 x 4.0
= 0.02226 m 3 /s "" 22.26 lit res/s. Aos .
(ii) Mouthpiece ruuning full . The discharge. Q is given by l'q uation (7.20) as
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1350
Fluid Mechanics
Q = 0.707 x a x J2gl/
'" 0.707 x .005026 x
J2
x9.81 x4.Q
== 0.03147 m 3fs '" 31.47 1I1rt'ls. A ns .
HIGHLIGHTS
1. Orifice is a small opening on (he side or al the bottom of a lank while mouthpiece is a shon length of
pipe which is 1WO or three limes its diameter in length.
2. Orifices as well as moulhpic~es arc used for measuring (he rJle of flow of liquid.
J. Theoretical "clocity of jet of waler from orifice iii giwn by
V ~ J 2gH • where If _ Height of Water from the cenlre of orifice.
4. There arc three hydraulic co-efricients namely:
c
(a) Co-efficicnl of ,-elocity.
= Actual ve loc ity at vcna-contmcla =
x
Theoretical velocity
J4 yll
•
(b) Co-efficicnl of contraction.
k ) Co-efficicnt of discharge.
C • Area of jet at vena - COIHracta
Area of o rifice
<
c =
d
Actual disdarge
... C x C
Theoretical disc harge
'
,
where.t and yare the co-<;>rdinate .• uf any point of jct of walcr from ,'ena-«lnlracta.
S. A large orifice is one , where the head of liquid above the centre of orifice is Ie,s thun 5 times the depth of
orifice. The discharge thro ugh a large rectangular orifice is
,
Q=~
where
C" xb X.J2iI H,JI1 _ H ,JI:~J
b ~ Width ofurifice.
Cd = Co-cfficient of discharge for orifice.
II J s Height of liq uid abo,'e top edge of ori fice. und
Hl = Il eight of liquid above bollom edge of orifice.
6. The discharge through fully sub-merged urifice. Q C" X b X (H l - H,) X .j2gH
where b ~ Width of orifice.
Cd .. Co-cfficienl of discharge for orifice.
Hl '" Height of liquid ,tbovc OOtlOrll edge of orifice on upstream side.
II , ~ Height of liquid aoow top edge of orifice On upstrcum side.
H = Difference of liquid h,,'els on ooth sides of the orifice.
7. Discharge through panially sub-merged ori f,ce.
K
Q-Q , + Ql
D
C,I> (H, _ H)
X.J2gH
+ 2/3 CJ>
X.J2i W ,n _ H, Jn ]
where b .. Width of orifice
C", H,. H, and H are having their usual meaning .
8. Time of emptying a lank through un orifice at its OOtlOrll is given by.
where H , = In itial heigh! of liquid in tank.
H, ~ I:inal he igh! of liquid in tank ,
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Orifices and Mouthpieces
351 1
A _ Area of lant,
a = Arca of orifice.
Cd" Co-efficicnt of discharge.
If the tan); is to be completely emptied. then time T.
T
2A.[ii
.. Cd.ll,jf;'
9. Time of emptying a hemispherical
taTI);
by an orifice fiued at its bonom.
[:. R( 111311 - Hil!) - ~ (ffl~i 1 - lIil!)]
,
Cd.a./ii)
and for completely emptying the
where
tan~.
T=
5
It
.{I ,j2i
CJ
[~ RH~12 - ~ 1I1~12]
3
5
R .. Radius of the hemispherical {anl.
If I
..
Initial height of liquid,
" l " Final heigh! of liquid,
II .. Area o f orifice, and
CJ .. Co-efficicnl of discharge.
10. Time o f e mptying a circu lar horizontal tan k by an orifice at thc bouom of the tan k,
T_
4C
}CJ .(/ . ..[fi
](2R_ H,r_(2R_ H, )Jr. 1
where L .. Length of horizonta l lant.
II . Co-efficicnl of dischmgc for.
E ~lcmal mouthpiece,
Cd" 0.855
Internal moul hpicre. f\lnning fuli,
Cd" 0.707
Interna l mouthpiece . running free.
Cd ~O ..so
Convergent or con"crgent -<Ii ,·crgent. Cd" I .0.
11. For an e~tcrnal mouthpiecc. absolute pressure head at vena-contracta
(i)
(ii)
(iii)
(i" )
H, .. Hu -O .89H
where Ha "atmospheric pressure hcad " 10.3 m of water
H '" head of liquid above the mouthpiece.
13 . For a cOl\ vcrgcnt·di,·crgcnt mouthpiece, the ratio of areas at outlet and at vena ·contracta is
'"
J
- :
1+ H. - H,
",
H
where il , '" Area of mouthpiece at outlet
il, " Area of mouthpiece at vcna-contracta
H• .. At mosp heric pressure head
If, "Absolute pressure head at vcna-contmcta
If " Hcight of liquid above mouthpiece.
14. In case of internal mouthpieces. if the jet of liquid cOmeS out from mo uthpiece without touching its sidcs
it is known as running free. But if the jetlOuchc, the sides of the mouthpiece. it is known as runn ing full.
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1352 Fluid Mechanics
EXERCISE
(A) THEORETICAL PROBLEM S
I. Defi ne an orifice and a mouthpiece. Wh at i, the difference between the two ?
2. h plain the classification of orifices and mout hpi«cs b;tscd on thelr shape. size and sharpness ?
J. W hat are hydraulic co..,ffk icnl. ? Name them.
4. Defme the following "o -dficienls (i) Co -... rr.cicnl of vciocity. ( ii) Co-efficient of contraction ~nd
(iii) Co...,fficient of discharge.
S. Derive the c~ p rcssion Cd . C. xC,.
6 . Define Ycna--<:onlracta.
7. D ifferemiatc between a large and a small ori fice. Obtai n an cxp",,,;on for d;",hargc through a b rge
rectangular orifice.
8. What <10 you understand by Ihc [cnllS wholly sub-merged orifice and panially sub-merged orifIce?
9. Prove that the expression for discharge through an external mouthpiece is gi"cn by
Q .. .855 x <1)<"
10.
II .
12.
13.
14.
where a .. Area of mo ut hpiece m outlet nnd
•• • Velocity of jet of water at outlet.
Distinguish bet ween : (i) External mo ut hpiece and internal mo uthpiece. (ii) Mouthpiece runn ing free and
momhpie<:e running full.
Obt:,in an expression for absolute pressure head at vcna-cuntmcta for an cxtcm~ 1 mouthpiece .
Whm is a conwrFnt-<l ivergem mouthpiece ? Obtain an expression for the ratio of diamcters at outlet and
at vcna-<:untracta for a con"ergcnt-<li"crgcnt "mouthpiecc' in tenos of ab>olutc pre"ure head at Vcnu COnlracta. head of liqu id above mout hpiece and atmospheric pres.''''c head,
T hc length of the divcrgeOl outlet part in a venlurimeter is usually made longer compared with that of the
converging in let part. Why?
Justify the statemcnt . '" In a con"ergcnt-<li~crgent mouthpiece the loss of head is practically eliminated"
(B) NUMERICAL PROBLEMS
I . T hc head of water over an orifice of diameter 50 10m is 12 m. Find the :tctual discharge and actual velocity
of jet at vena-COntr~cta. Take Cd ~ 0.6:md C, ~ 0.98.
IAns • .018 m'/s ; 15.04 mlsi
2. The head of water o,'cr the centre of an orifice of diametc r 30 10m is 1.5 m. The actual discharge through
the orifice is 2.35 litrcs/sec. Fi nd the co-cfficient of discharge.
IAn s. 0.613]
3. A jet of W:ltcr. iss uing from a sharp cdged "crtical orifice under a constant hend of 60 em. has the horizon tal and "enical co-ordinalcs mcasured from lhc vena-conlmcta at a cenain point as 10,0 cm and 0.45 cm
respectively , Find thc "alu~ of C •. Also find the value of C, if Cd -O.60_
[Ans. 0_962. 0 _6231
4. The head of "':tIer over an orifice of diameter 100 mm is 5 111. The waler coming 011\ from ori fice is
COlleCICd in a circular tank of diameter 2 m. The rise of ",ater level in circ ular tank is .45 111 in 30 seconds.
Also the co-ordinales of a cenain point On the jet. measur~..J from \'ena<nntracta ,ore 100 ern horizonl:]1
and 5.2 ern venica!. Find lhc hydraulic co-cffieients Cd' C, and C,.
IAn s. 0.605. O.9 ~. 0.6171
5. A tal1~ has two identical orifices in one of its "crtical sides. T he upper orifice is 4 tl1 below the water
.urface and lower one 6 III belo'" thc water surfacc. If the value of C, for each orificc is 0.98. find tlte point
[Ans. At a horizontal dislance of 9.60 eml
of intcrsection of the two jets.
6. A closed vessel contains watcr UP!O a height of 2_0 m and over the water .urface there is air having
pressure 8. ~29 Nlcm ' alx we atmospheric pressure. At the bottom of the vessel there is an orifice of diam eter 15 em. Find the rate of flow of water from orifice. Ta~ e Cd " 0.6.
IA ns. 0.15575 m' lsl
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Orifices and Mouthpieces
7. A closed laok partially filled with water Upl0 a heig ht of I
In,
353 1
having an orifice of diameter 20"'111 althe
bonom of thc wnk. Dctcmlinc thc pressure required for a discharge o f 3.0 litresls through thc orifice . Take
C" .,0.62.
IAns. 1088 N/cm' j
8. Find (he dischnrge through a rccwngular orifice 3.0 m wide and 2 In deep fined to a water tank. The watcr
Ic,-eI in the tank is4 m nixwc thc lOp edge of thc orifice. Ta~c Cd _ 0.62
IA ns. 36.77 mJ/sl
9. A rectangular orifice. 2.0 In wide and 1.5 III deep is discharging water from a lanio;. . l rlhe water level in thc
tun\: is 3.0 m above the tOP edge of the orifice. find the discharge through Ihc orifice. Take Cd 0.6.
IA ns. 15.40 Inl/sl
10. A rectangular orifice . 1.0 m wide and 1.5111 deep is dischargiTlg water from a vessel, The lOp edge of the
orifice is 0.8 m below the water surface in Ihe vessel. Calculate the discharge through the orifice if
C J ", 0.6. At"" calculale the percemage error if the orifice is treated as a small orifice. [Ans. 1.0511%[
11 . Find Ihe discharge through a fully sub· merged orifice of width 2 m if the difference of water levels on
both the sides of the orifice t>e 8(X) I1U11. The height o f water from top and bonom of the orifice are 2 .5 m
and 3 m rcspc.;tively. TaKe Cd ~ 0.6.
[Ans. 2.377 ", l/sl
12. Find the discharge through a toully drowned o rifice 1.5 III wide and I m deep. if the difference of "'ater
le"els on both the side", of the orifice t>e 2.5 m. Take Cd - 0 .62 .
IAns. 6.513 ml/sl
13. A rectangular orifkc of 1.5 01 wide and 1.2 rn deep is fined in one side of a large tank. The water le\"el On
One side of the orifi~e is 2 m abo,·c the top edge of the orifice. While On the other side of the o rifice. the
water IO"el is 0.4 m be low its top edge. Calculate the discharge th rough the orifice if Cd ~ 0.62.
IA ns. 7.549 mlls l
14. A circular tank of diameter 3 m conta ins water upto a height of 4 m. The tank is provided with an orifICe
of diameter 0.4 m at the oollom. Find the time taken by water: (i) 10 fall from 4 m 102m and (ii) for
completely emptying the tank. Take C" '" 0.6.
IAn •. (i) 24.8 s. (ii) 84.7 sl
15 . A circular lank of diameter 1.5 m contains water UplO a height of 4 m. An orifice of 40 mm diameter is
provided at its bollom. If Cd '" 0.62 . find the heighl of w"lcr abov~ the orifice after 10 minules. IA ns. 2 m I
16 . A hemispherical tank of diameter 4 m co~tains ",ater upto a hei ght of 2 .0 m. An orifice of diameter 50 mm
is provided al the bollom . Find the time required by water (/) 10 fall from 2 .0 III to 1.0 III (ii) for complelely
emptying the unk. Take Cd" 0.6
I"' ns. (I) 30 min 14.34 s. (ii) 52 min 59 51
11. A hemispherical ci'lern of 4 III radius is full of"'''ler. It is filled with a 60 mOl diameler sharp edged orifice
al Ihc bollom. Calculate the time required to lower the level in the cistern by 2 metrn. Take CJ ,", 0.6.
[Ans. I hr 58 min 45.9 s l
IS. A cylindrical tall. is having a hemispherical base. The height of cylindrical portion is 4 111 and diameter is
3 m. At the bottom of th i, lank 3n orifice o f diameter 300 mm is fitted. Find thc time required to completely
cmplying Ihe tank . Take Co z 0.6.
[Ans. 2 min 7.37 ,[
I II . An orifice of diameter 200 l1un is fitted at Ihe bottom of a boiler drum of length 6 III and of diamete r 2 m.
The drum is hori7.0l1lal and half full of water. Find the time required to empty the boiler. given the "alue of
Cd - 0.6
[Ans. 2 min 55.20 51
20. An orifl,e of diameter 150 mm is filled at Ihe boltom o f a boiler drum of length 6 m and of diameter 2 m.
The drum is horizontal "nd contain, water upto a height of 1.8 01. Find the time required to empty the
boiler. Take Cd '" 0 .6 .
IA ns. 7 min 46.64 s l
2 1. Find the discharge from a 80 mm diameler extern"l mouthpiece. filled to" 'ide of a large vessel if the head
O,'er Ihc moulhpk'<;e is 6 m.
[Ans. OJ)466 10)/5 1
22 . An external cylindrical mOUlhpiCO"e of diameter 100 mm is discharging water under a con .• tal1l head o f 8 m.
Detemline the discharge and absolute pressure head of water at vena-eontracta. Take CJ • 0.855 and Co for
vena-eont racta .. 0 .62. Take almospheric pressure head", 10.3 m of water.
IA n,. 0.084m3,s; 3.18 ml
23 . A eonvergent-<liycrgent mouthpiece having throat diameter of 60 mm is discharging waler under a con ·
Slant head of 3.0 m. Delennine the maximum oullel diameter for maximum discharge. Find maximum
discharge also. Take atmospheric pre."",e head", 10.3 m o f water and separation pressure head"," 2.5 111 of
waler absolute.
[A ns. 6.88 cm. Qm» _ 0.Q\506 ml/s l
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1354
Fluid Mechanics
24 . The throat and ~ ~ it diameter of a convergem-d;\,ergcnt lHouthpiece are 40 mm and 80 ["In re~peclively.
It is fitled to thc ""Mical side of a ImIO: . containing water. Find the maximum head of water for steady
flow. The maximum vacUum pressure is 8
In
of Wa ler. T ake atmospheric pressure head = 10.3 In of "",,(eT.
[Ans. 0.533 ml
25. The discharge through a (:{)nvergcnl-divergcnt mouthpiece titled [0 the side of a (anK under a constant head
of2111 is 7litrcsls. The head loss in the divergent portion is 0.10 limes the kinelic head at OUtlet Find the
throat and exit diameters. if separation pressure head ~ 2.5 m and atmospheric pressure head 10.3 111 of
waler.
[Ans. 25.3 nnn ; 38.6 mml
26. An i1l1cnml mouthpiece of 100 mTll diameter is discharging Water Ilnder a COIlstant head of 5 m. Find the
E
discharge through mouthpiece. when
(i) the mouthpiece is running free. ,md (ii) the mouthpiece is running full.
I Ans. (,) 38.8 lines/s. (ii) 54 .86 l itre",1
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.. 8. 1
INTRODUCTION
A notch is a device used for measuring Ihe rate of flow of a liquid through a small chan nel or a
tank. It may be del1ned as an opening in the side of a tank or a sm all ehannel in sueh a way that the
liquid surface in the tan k or channel is below the tOp edge of the opening.
A weir is a concrete or m3!\onary structure, placed in an open channel over which the flow occurs.
It is generally in the form of ven ic~1 wall. with a sharp edge at the tOp. running all the way across the
open dannel. The notch is of small size while the weir isof a bigger size. The notch is generally made
of metallic plate whi Ie wei r is Illade of concrete or masonary structure.
I. Na p pe o r Vein. The sheet of water flowing through a notch or over a weir is called Nappe or Vein.
2. C res t or Sill. The bottom edge of a notch or a top of a weir over which the water flows. is known
as the sill or crest.
.. 8 .2
CLASSifiCATION Of NOTCHES AND WEIRS
The notches arc classified as :
I. According to the shape of the opening:
(a) Rc.:: langular notch.
(b) Triangular notc h.
(e) Trape7,Oidai notc h. and
(d) Stepped notch.
2. According to the effect of the sides on Ihe nappe:
(a) Notch with end conlraction.
(b) Notch without end contraction or suppressed notch.
Weirs arc classified according to the shape of the opening. the sh ape of the crest. the effect of the
sides on the nappe and n~ture of d ischarge. The following arc importal11 classificmions.
(a) According 10 the shape of the opening:
(I) Rc.:: langular weir.
(ii) Triangular weir. and
(iii) Trapezoidal weir (C ipollcui weir)
(b) According to the shape of the crest:
(i) Sharp-crested weir.
(iii) Narrow-crested weir. and
(ii) Broad-crested weir_
(iv) Ogee-shaped weir.
355
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1356
Fluid Mechanics
(e) According to the effect of sides on the emerging nappe :
(i) Wei r with end contraction. and
(il) WeiT without end contraction.
to 8 .3
DISCHARGE OVER A RECTANGULAR NOTCH OR WEIR
The expression for discharge over a rcctallgular notch or weir is the same.
NAPPE
~~
I 7>,-'
i
'''''''j'''fJQ TI
CREST ' /
OR SILL
,' ~
1
I-- L
IT'"
----..j
(e) SECTION AT
CREST
(al RECTANGUlAR NOTCH
NAPPE
c:!,~r\
(b) RECTANGULAR WEIR
Fig. 8.1 R~ctangular notch alld 'Il;rir.
Consider a rectangular notch or weir provided in a channel carrying water as shown in Fig. 8.1.
Le!
H " Head of water over Ihe nest
L " Length of the notch or weir
For finding the discharge of water flowing over the weir or notch, cOlIsidcT an elementary horiwntal
strip of water of thickl1es~ dll alld length L at a depth" fronl the free surface of wakr as shown in
Fig. 8.1 (c).
The area of strip
"Lxdh
and theore tical velocity of water flowing through strip" J2gl!
The dischargc dO. through strip is
dQ'" Cd X Arca of Strip x Th~'Orctical velocity
... (1)
where Cd '" Co·cffleient of dischargc.
The total discharge. Q. for the whole notch or weir is dctcnn ined by imcgrating equation (I) bctwCt:n
the limits 0 and If.
=C, 'L'.j2i -'1111"
,
[2 + [
1' =C, 'L'.j2i [h- )I1]"
2
' " IHI ~.
="3Cd
X/.X.y2g
3/2
0
...(8. 1)
Problem 8. 1 Find rhe discharg e of Wilier flowillg o ra a reclangulor norell of 2 m /engrll Wllell ,he
COIISIaIll head ora rile lIorch is 300 mm. Take Cd " 0.60.
Solu l io n. Given:
Length of the notch.
I I
L=2.0m
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Notches and Weirs
Head over no tch.
II '"
300 m '" 0.30
357 1
III
Cd '" 0 .60
Di sc harge,
Q=
~ Cd XL x.J2g" [HJIl ]
1
2
= - xO.6 x 2.0 x J2x9.S r x [O.3011,s m J {s
3
'" 3.5435 x 0.1643 '" O.SSZ m' /s. ADS.
Problem 8 .2
D e/ami/Ie I/Ie /leighl of II reC/(lIIgu/M weir of I<mgll! 6 111 10 b<! buill llc mS.f II rectwlgll/ilf chalillel. Tire I/I(1Xillllll1l deptll of Wil Ier 0/1 rile lips/ream $ide oflhe weir is I .S III alld disC/llIrg" is
2(X)() l i lres/s. Take Cd '" 0.6 alllt lIegleel end cO/llme/iolls.
Solution. Given:
Length of wt'ir.
!krIll o f wate r.
DiSl:hargc.
L=6 m
H I= 1.8rn
Q '" 2000 litIs", 2 m 3/s
Cd '" 0.6
Let II is he ig ht o f Wale r above the crest of wei r. and Ill'" he ight o f we ir (rig. 8.2)
The d isch:lrgc uve r Ihe we ir is gi ve n by the equatio n (8 . 1) as
Q '"
2.0 '"
"'
"32 Cd X L x./2i
,,312
~ x 0.6 x 6.0 x ,J"2~<~9~.8"1
'" 10.623 H
X 11 312
Jn
1l3l2 = ~
Fig. 8.2
10.623
II =
He ight of we ir.
'- -0 )'" '" 0.328 m
( -10.623
H1 ", H , -H
'" De pth. of water on upstream side - /I
= 1.8 - .328 = 1.472 m. An~.
Problem 8.3 The he(ld of ,,'«Ier 01'1." ( I recl""glll(lr nolch ;s 900 mm. Tile disclwrge is 300 lilres/s.
Find Ihe /eng lh oflht, nO ld,. wilen Cd '" 0.62 .
Solution. Give n :
H~ad
over notc h.
Disc harge.
/I = 90 cm '" 0 .9 m
Q =300 litIs = 0.3 m 3 /s
Cd'" 0.62
lA:t le ngth. of nutch
Using equatio n (8 . 1), we have
"L
2
Q'" 1
I I
X
Cd X l.
x./fi X /1 )12
Ii
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IL
1358
Fluid Mechanics
0.3 =
~ )( 0.62 x L x
..}2 x 9.8 1 x (0.9) .112
= 1.83 x L x 0.8538
L:
... 8.4
7"CCO"·" '''''O = .192
1.83 x .8538
III
= 192
mm.
Ans.
DISCHARGE OVER Po TRIANGULAR NOTCH OR WEIR
The expression for the discharge over <I triang ular nOlch or weir is the same. II is deri ved as :
Le t If = head of water above the V- notch
e = angle of notch
Consider a horizontal strip OrWJtcr o f lhickncss 'dh' at a de pth of II from the free surface of Waler
as sho wn in Fig. 8.3.
From Fi g. 8.3 (b). we ha ve
e = -AC :
t;lI1 -
2
AC
;-;:C"'
"
( H II )
OC
AC=(If - II) tan
,
,
,
~ r
-z
•
Width of strip
=AB=2AC=2(H - l!)tan-
:. Area of Mrip
=2( H-h) ian 2"xdh
,
.'\t'o--'-
,.,
<0,
Fig. 8.3
2
Th~
triangular notch.
The theoretical velocity of water through strip '" ../2gh
Oischarg~ ,
tllrough the "trip.
,
dQ = Cd x Area ofslrip x Velocity (theoretical)
'" Cd
X
2 (H - II) tnn
'" 2C,, ( N - II) tan
:. Total di scharge,
,
"2 x till X
"2 x ./2gh
./2gll
X dll
jo" 2Cd (H - II) tall -,2 x ./2gh x dll
=2C , xwn -' x,,2g
(H - /J)II '"-till
re j"
,
2
"
==
Cd x tan -, x J2i j" (HI,! n
Q=
2 X
_11
2
==
2 x C x tan -,
d
I I
2
312
"
Hhlll " SI!
x,fig [ -- -3/2
5/2
)
dll
r
Ii
~ I
IL
Notches and Weirs
=2 x C" xtan
'"
!XJ2i[~ H3Ill
2
Q=.!
Problem 8 .4
,'2 '"
xO.6x I x .}2 )(9.8 1 x
15
'" 1.4 17 H5I2 ,
Frill! t/ie discharge oreT
...(8.2)
2
tan
Discharge,
15
~ Cd x Ian! x J2i x Hsn.
15
For 3 right -allglcd V-nOich. if Cd '" 0.6
II
359 1
wn
... (8.3)
Iriatlgl4/ar nOlcll of /wgle 60" ..... hell the Ilead Mer Ille
V-nore/! is 0 .3111. Anume Cd '" 0.6.
Solution. Gi ve n :
Angle of V-notch .
e '" 60"
Head over nOlch.
H =O.3 m
Cd '" 0.6
Disc harge, Q over a V-notch is give n by equation (8.2)
8
e
Q = _ xC" xtan _ x.j'fi x HSfl
15
8
2
.
60'
'" IS X 0.6 tan 2)(
.}2 x 9.8 1 x (O.3)Y.!
'" 0,8 182 x 0.0493 = 0.040 mJ/s. Ans.
Problem 8 .5 Warer flows 0I'e( II rec/allgular weir I III wide m a deptll of 150 mm and aftenI'Grds
pancs {llrol jgh a rriangular righi-angled weir. Taking C,,/or II,e rec/angulM (Uullrhlllgular weir as
0.62 (l1Ii1 0.59 resper/iI-e/y. find the deprh Ol'/'f rhe triangular ,,·{'ir.
Solu t Ion. G ive n:
For rectangular weir. length ,
t:o I m
Depth of water.
H = 150 111m = 0. 15 111
Cd = 0.62
e == 90°
For trian g ular weir.
Cd == 0.59
lA:t depth uvcr
The
I I
disch~rl(e
tri angul~r
ove r the
weir
:0
reet~ n gular
HI
weir is given by equat ion (8. 1) as
Ii
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IL
1360
Fluid Mechanics
2
Q= - xCdXLX.Jfi xHIf!
3
'" '32 x 0.62 x
1.0 x .)2 x 9.8 1 x (.15)">11 mJ/s '" 0.10635 ml/s
The same dischargc rasS<.'s through the tria ngular Tight -angled weir. But discharge, Q, is giver> by
equation (8.2) for a triangular weir as
8
15
Q= -
X
e
Cd x tan - x.[fi X H~f!
2
8
90'
0. 10635 == - x .59 x tan ""2 x
15
8
= -
15
H
Y1.
FE x H/'/2
x .59 x I x 4.429 X H
IY2=
1.3936
HI 512
= 0.10635 = 0.07631
1.3936
I
HI == (.07631)11.4 = 0.3572 m. Ans.
Problem a.SA W,,[er flows 111f0Ug/1 II (rfangulllT rig/u-ongled weir firJ"' amllhen Ol'('( {/ r<,ClanguIlIr weir of I In widill. The di.~cha'ge cQ-ef!iciellfJ of Ihe Iriangul<If and re( /angular weirs are 0.6 {lnd
0.7 respec/i,·e/y. If the deptll oflHller Ol"er Ihe triangular weir is 360 mm, find I/,e deplll o/water m"er
Ihe red'l1Igll/ar ""'ir.
Solution. Given:
rOf triangular we ir
e = 90°. Cd = 0.6. 1/ = 360 111111 = 0.36 III
For rectangular weir
L = 1 Ill. Cd == 0.7. fI==?
The discharge for ~ tria ngular weir is given by equati on (8.2) as
=
e 2g xl-/ w
Q= -8 xCd xlan -xv
15
2
8
== - xO.6x tan
15
(90°)
-
2
xJ2X9.81 x (0.36)w= 0.11 02 1ll 3/s
The same discharge is passing through the rcct.lngular wei r. But disc harge for a I"\:l:tangul ar weir is
given by equation (IU) as
2
Q="3 xCd xLx
0.1 102 ==
"'
I I
~
3
Iii xfl·Vl.
xO.7 x I x J2 x9.8 1 x fI·Vl. == 2.067 H )J2
H312 '" 0.1 102 '" 0.05:n
2.067
1-/ = (0.0533) U3 '" 0.1 41 5 m '" 141.5 mm. Ans.
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Notches and Weirs
361 1
Problem 8 .6 A f<'c/{wgular c1wlllwl 2.0 m wide ilaJ a discharge of 250 litres per .>'ecalili. whiel! is
measured by (I righl·ongled V·notell weir. Find the positioll oflhe apex of tile nOlel, from lilt' bed of the
dUlI/nel if maximum depth afwiller is no/ ro exceed 1.3 m. Td/(e Cd '" 0.62 .
Solution. Given:
Width of rccwngular chnno.:!. L = 2.0!l1
Dischilrgc.
Q = 250 litis = 0.25 Ol lis
[)cplll of wa te r in channel
= 1.3 In
Lei lhe height of wakr ove r V-notch = H
The rate of n ow throug h V- notch is given by equation (8.2) as
8
e
xcdx.[fi xtan- xll 5t2
15
2
Q= -
90·
8
X.62x.j2x9. Rl x tan 2
15
Q= -
"'
"'
xH~
0.25 '" 18 x .62 x 4.429 x 1 x HSi'l
5
.25 x 15
H Y! = ;;-c~=~~ "" 0. 1707
R x .62 x 4.429
H = (.1707):!I~ '" (. [ 707)°·4 '" 0.493 III
Position of apex of the notch from the bed of channe l
= de pth of waler in ch<umcl-hcig ht of water ove r V-notch
= 1.3 - .493 = 0.807 m. AilS.
.. 8.5
ADVANTAGES Of TRIANGULAR NOTCH OR WE IR OVER RECTANGULAR
NOTCH OR WEIR
II. triangular notch or weir is preferred to a rectangular weir or notch due to following reasons:
I. The expression for discharge for a right-angled V-notch or weir is very simple.
2. For rnc~suring low disdargc. a tri;mgu)ar notch givcs 1110re a<.:curate results th~n a rectangular
notch.
3. In case oftriangulur notch. only one reading. i.e., H is required for the computation of dischnrge.
4. Ventilation of a triangular notch is not necessary .
.. 8.6
DISCHARGE OVER A TRAPEZOIDAL NOTCH OR WEIR
II.s s hown in Fig. 8.4. a trapezoidal notd or wci r is a
combinati on of a rectangular and triangular notch or weir.
Thus the total di scharge will be eq ual 10 the sum of
discharge through a rectangular weir or notch and discharge
through a triangular notch or weir.
Let H = Height of water o\"er the notch
L = Length of the crest of the notch
F
------
e
\l\H~r
D
E
C
I - - , --.,
F ig. 8.4 The t~(lpezQida/ ,,(Itch.
I I
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IL
1362 Fluid Mechanics
c~,
= Co-efficient of discharge for rectangu lar portion ABeD of Fig. &.4.
Cd, " Co-cffkicnt of discharge for Triangular portion IFAD and BeE1
The diSCharge through rectangu lar portion ABeD is given by (8.1)
,
2
Q, '" - x CJ
'
)(
I'>::
1I2
L )( ,,28 )( H -
The discharge through twO lri~ngular notches FDA and BeE is equal 10 Ihe discharge through a
single triangular notch of angle 9 and it is given by equation (8.2) as
a
8
~
Y1
Q1 = - XCd X I<ln- )( ,,2g xH
15
'
2
Di sch,n ge through trapczoidalnoH:h or weir FDCEF= Q, + Q1
2
3
"" -
Cd L
I
'28 x fI -1I2 + -158 Cd' )( Ian en)( v"X
f2ii x H SIl.
v -'-I;
...(8 .4)
Pro blem 8 .7
Find Ihe discharge through (l Ir(lpezoiJal II oleil "'hid, is I m "'ide al the lOp Gild
0..10 m til the bulIO'" wu/ is 30 em jll height. The /W{Id of ",Mer 011 the li DICh is 20 em. A ssume C,,Jor
reCllmg"{"f portio" '" 0.62 ,,"/life/Of lTitmg,,/ar portioll '" 0.60.
, ,
Solution. Giv",n :
Top width.
liE = I III
Base width.
CD", L = 0.4
Head o f wate r.
H", 0.20
III
:........
,
,
III
-
Cd, = 0.62
,
For triangular portion.
Cd," 0.60
f0- e
WI'
-
~:-"i!i"-
For r"C!angular portion.
From l!.ABC.
, ,
--- - --
0
T
,
1
-i
have
Fig. 8.5
tan ~ " _A_8 : c( A~E~---;;
C~
D~
)I~2
2
Be
11
=
( 1.0 - 0.4 )f2
0.3
0.6/2
0.3
0.3
0.3
; --: -
:1
OiSl:liarge through trapezoidal notd is given by eq uati on (8.4)
Q= -2 Cd
3
'
X
L x.ffi x~ +
'
-8
Cd x tan - x../2i x
15
2
3
'
2
H '>r.
8
x .60 x I x ./2 x 9.81 x (0.2)Y.!
15
= - x 0.02 x 0.4 x ./2 x 9.81 x (0.2 l Y.! + -
= 0.06549 + 0.02535 '" 0.09084 m3fs " 90.84 Ill res/s. An s .
... 8.7
DISCHARGE OVER A STEPPED NOTCH
A stepped notch is a combination of rectangu lar notclies. The di .>eharge through
equal to the su m of the discliarges through tlie different rectangular notclies.
I I
st~ppcd
notch is
Ii
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363 1
Notches and Weirs
Co nsider a ste ppe d no tc h as show n in Fig. 8 .6.
Lei H I" Height o f water abo ve the ercS( of nOlc h !.
Ll
"
Length of notch I.
H1. L2 and H3_ LJ arc corre sponding v:tlucs for notdlcS
2 and 3 respect ive ly.
f-- L' --1
Cd " Co-effi cie nt of di sc harge fo r all no tc hes
Total di Sl: hargc Q '" Q 1 + Q2 + QJ
r-- l ,
I•
2
r;;:
31'1
.Yl
Q= "] xCd x LI x ,,2g IHI -Hl 1
_I
L,
•
I
Thr Itrpped notch .
Fig. 8.6
... (8 .5)
Problem 8 .8
= 0.62.
Fig. 8.7 sho ws
II
stepped lIolc/'. Find lilt' discharg/' l"rOJ/gll Ihe notch
il e d/o r 1111
sec/ion
Solution. Given:
T
L I = 40 em. ~" 80 em.
LJ = 120 Clll
"I
= 50 + 30 + 15 =95 em.
Hl '" 80 em. HJ
fl,
50cm
= SO em.
Cd = 0.62
-~
Iscm
T
_
,.
Total di scharge. Q '" Q, + Ql + QJ
where
~40cm""'1
80cm _
120 cm
. ,
Fig. 8.7
2
r->::
QI ="3 X C" X L1x .;2g
J.i2
IHI
- H!
.lI2
1
= ~ x 0.62 x 40 )( J2 x 98 1 x [95 31l - 80Jl'1 1
3
= 732.261925.94 - 7 1'i.54 1 == 154067 cm3/s '" 154.067 li tis
f'>::
1I1
1I1
2
Q2 ""3xCJ xL2 X,,2 g xlH2 - HJ 1
" ~XO.62X80 X J2X98 1
3
x [80 312 _ 50 312 1
== 1464.52(7 15.54 - 353.55( cm3/s= 5301 41 cl1I 3ls" 530.144 li tis
"
~
3
x 0.62 x 120 x
J2 x 98 1 x 50ll.l " 77677 1 cm 3,s" 776 .77 1 litis
Q = Q 1 + Q 2 + Q 3 = 154.067 + 530 . 144 + 776.77 1
= 1460.98 ]iUs. ADS.
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1364
Fluid Mechanics
.. B.8
EffECT ON DISCHARGE OVER A NOTCH OR WEIR DUE TO ERROR IN
THE MEASUREMENT OF HEAD
For all accurate value of the discharge over a weir or notc h. an accurate measurement o f he ad ove r
the weir or notch is ve ry esse ntial as the disch arg e ove r a triangular !lotch is proportional 10 HYl and in
case of rectang ular no tch i1 is proportional \0 HY2 . A s mall error in thc measurement o f head. will
affect lhe di.<;chargc co ns iderabl y. The fo llowing c ases of error in Ihe rnCa5 urcmcnt of head will be
considered :
( i) For Rectangular Weir or No tch.
(ii) For Trian gular Weir or Notclt.
8 . 8 . 1 for Rectangular Weir or Notch .
by equation (8.1) as
Th e disc harge for a rectangular weir or notch is g iven
=
~
Q= -2 xCJ xLx,,2g xH3
== KH1I2
where K
2
= "3 Cd X L )(
...( i)
.Jfi
Differe ntiating th e above equation. we get
3
I12
dQ- Kx -2 H dH
K x ~xH"2dH
dQ '" -~2~;n- '" 3 dH
Q
KH JI2
2 H
Di vidin g (ii) by (i).
...(8.6 )
Equ ati o n (1\.6) shows that an error o f 1% in mcasurin g fJ will produce 5<;<, c rror in di scharge ovcr
a rectang u lar weir or notch.
8. 8.2 For Triangulilr Weir or Notch.
by equation (8.2) as
Q'"
The discharge uver a triangular weir or nutch is g ivcn
8
e r->:
15
C d' tan T,,2 g
'" KH
xH
YJ.
SIZ
... (iii)
8
e
Cd' tan - ,J2i
IS
2
where K '" -
Differe nli~ting equ~tion
(iii). we gel
dQ= K~H3I2 XdH
.. .( i,')
2
K~
1/ 311
dll
Di viding (il') by (iii ). we get dQ '" _.2co",,~JIl
Q
K fi
Equation (8.7) show s thaI :111 error of 1% in
a triangul~r weir or notch.
I I
5 dH
2 H
III c ~ suring
H will produce 2.5% e rror in
...(8.7)
disch~rg c
over
Ii
~ I
IL
Notches and Weirs
365 1
Problem 8 .9 A recrangular 1I(1Icii 40 em hlllg is IIsed for lIIea~'llfing a dj.~c1wrge of 30 Iilre.1 per
second. All error of 1.5 mm was made. wllile medsurillg the head ol'a Ihe notcll. Calcula te IIII'
pefcelilage error in II,e discharge. Take Cd '" 0.60.
Solu tion. Given:
Length of notch.
L=40cm
Dischilrgc.
Q = 30 litis = 30000 c11l 3fs
Error in head.
dH= l.5mm = O.IScm
Cd = 0.60
Let the hdght of water over rectangu lar notch = Ii
Ttw discharge through a r;:c(angular notch is given by (8.1 )
2
Q=")XCd XLx.[fiXH 3I!
or
,
30000
"'
=..:. x 0.60 x 40 x .J2 x 98 1 x 11Y2
3
H3Il =
3 x 300Xl
2 x.6O x 40x
.J2 x 98 1
= 42.33
II : (42 .33)Z13 = 12.16cm
Usi ng eq uatio n (8.6). we gel
dQ
Q
= ~dH =~X~ =O.OI85=
2 H
2
1.85%. A ns.
12.16
Problem 8 .10 A righl ·angled V.notch is IHed Jor me(lSllring (/ di~Tll(lrge of JO li/res/s. All e rror of
/.5 "'''' "'a5 ",ade "'bi/e ",,,aJ'uring Ill" ile"d Ol"er IIIe n"'cll. C,,/culale Ille percell/age eTTor in lile
disdwrge. Take Cd '" 0.61.
Soluti on. Give n :
Angle of V-no tc h.
e '" 900
Dis.:harg'"
Error in head.
Q '" 30 litis'" 30000 cm 1/s
dH:: l.5mm",0. 15cm
Cd:: 0.62
Let the he~d over the V- notch:: H
The d ischarge Q through a triangu lar notch is given by eq uation (8.2)
8
e
fi":"
Q'" - Cd ' tan - x ,,2g x H
15
2
30000 ::
Y.!
.! x 0.62 x tan ( _90_°) x , "lex"9","1 x HYl.
15
2
:: -8 X .62 x 1 x 44.29 X H Y2
15
30000 x 15
:: 2048.44
8 x .62 x 44.29
H '" (2048 .44):!I·~:: 2 1. 1 1 ern
I I
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1366
Fluid Mechanics
Usi ng equatio n (8 .7). we get
dQ '"
Q
~ dH
2 11
'" 2.5 x 0.15 '" 0 .0 1776 '" 1.77 % .
21.11
AD S.
Problem 8.11 Tire /!"lItl ojwMer Ol'e, II Irilmgu/ar II oleil of lingle 60' is 50 em ami co -effieien, of
discharge is 0.62. Ti,e flow measured by il is 10 be wililill all accuracy of 1.5% up or dowl!. Find Ihe
limiting )'a/ues of the Iwail.
So lution. Gi ve n :
Angle o f V-no tch ,
H~ad
o f waler,
9 = 60°
H "'SO cm
Cd'" 0 .62
dQ '"
±
1.5'll '" ± 0 .0 15
Q
The di sc harge Q over a triang ular no tc h is
8
Q'" -15
8
=-
15
c d .,f2i Carl
x 0.62 x
,
-::;- H Y2
_
Jz x 981
60·
x tan x (50)Y.!
2
'" 14.64 x 0.5773 x 17677.67 '" 149405.86 C111 3fs
Now appl yi ng eq uati on (8.7). we get
dH
dQ = 5 dH
ur :t.O I 5=2.5
Q
2 1/
if
.015
2.5
T h", lim iting va lues of th e he ad
" '
tlH
/I
-
.015
:± - -
2.5
.015
2.5
dH ::!: - - x H =::J::. - - x50=:!:0.3
" 1/ ± till " 50 ± 0.3 '" 50.3 e m. 49.7 elll
= 50.3 e m und 49.7 e m. AilS.
.. 8 .9. (a)
TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A
RECTANGULAR WEIR OR NOTCH
Consid er a rese rvo ir o r tank o f uniform cross·section al are a A. A rec ta ng ul ar weir or notch is
prov ided in one of its si des.
L = Length of crest of the wei r o r notc h
Cd = Co·efficien t of d i><: harge
H , = In it ial height o f liqu id ahove the crest of notc h
H2 = Fin al hei g ht of liqu id above th e crest o f no tc h
T = Ti me req ui red in seconds to lowe r the heig h! of liqui d from " , 10 H1 •
Le t at an y instant. the heig ht of liquid surface a bove the cre"t of wei r or notch be /, and in a small
time JT. let the li!.juid surfa\:e fall s by ·d''-. T hc n.
Let
- Adh=Q x dT
- ve s ig n is tak c n. as with th e im;rease of T. /, dec re ases.
I I
Ii
~ I
IL
367 1
Notches and Weirs
2
Q= - Cd x LX.fii
""'
3
X
h 3ll
,
""
~
__--=-~A~dl~,_ __
-Adh=iCJ XLx,,2g .11 xdTordT= Z
"3 Cd X LX/fiXhli!
Th" [01.11 lime Tis oblai",:,d by integrating Ih" :thovc equalion between the limits H, and H!"
lo'
,Fr-
fHl
-
If,
-Adh
~Cd XLx.fiixllm
3
0'
... (8.8)
(b)
TIME REQUIRED TO EMPTY A RESERVOIR OR A TANK WITH A TRIANGULAR
WEIR OR NOTCH
Consider II reservoir or tank of unifonn '-Toss-sectional area A.
one of ilS sides.
LeI a", Angle of the notch
havin~
11 triangular weir o r nOleh in
Cd == Co-emden! of discharge
HI'" Initial height of liquid above the apex of 1I00(h
H) '" Final heig ht of liquid above the apex of nOH;h
T = Time required in seconds. 10 lower the heighl frolll HI to II! above the apex o f the 1I0\cll.
LeI at any instant. Ihe height of liquid surface above the apex of weir or notc h be II and in a small
timc dT, let th~ liquid surface falls by 'dll', Then
- Adll=QxdT
- ve sign is taken, as wit h the increase of T. II decreases,
And Q for a triangular notch is
=
8
,
Q= - xCd xtan - .. 2g xlrV2
15
2
8
e"-::~,
,
- Adll= - xCd xtan -x .. 2g xII -xdT
15
2
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1368 Fluid Mechanics
tiT",
Adlt
8
e r;:>;: J/'
- xCd xtan - x,,2g xll 15
2
The total lime .,. is obtained by integrating the above equation between th e lim its III and 11 2,
T dT- J,
-Adh
r
1.
- JH, ~ C
15
Ian
d
~ vJ2; II SI !
2 "'-e;
f"'
"'
II
Si'2
dh
0,
'"
~15Ae
8xCd xtan
=
2
X
x..{fi
5A a
(-')[Ji"
, ]'"
3'"
4xCd xtan - x.j2i
[H Ill'
l
II,
1-
2
Problem 8 .12 "-ind the lime "~quired la low", the Waler Jerel from J III 10 2
dimension 80 In x80 m. by a reclangl jlar norch of len gIll 1.5 m. Take C'I '" 0.62.
Solution. Given:
Initial height of water.
HI=3 m
Final heigh! of wmcr.
" 1=2111
Dime nsion of reservo ir'" 80
or Are a.
Length of notCh.
In
...(8.9)
- Hl1l']'
-
III i'l
a
re,~erl'Oir
of
X 80 1ll
A=80x80=6400ml
L= 1.5m,C<I =O.62
Using Ih e relal ion given by the equ ation (8.8)
T
3A
[ ,
'" Cd X LX .j2i :.{Ji;
, ]
- J'H,
3x6400
[ '
']
= O.62xI.5xJ2 x9.81 J2 - Jj
'" 4661.35 [0.7071 - 0.5773J seconds
== 605.04 secunds == 10 min 5 sec. AilS.
Problem 8.13 If ill problem 8.12. illSI<'ad of a rectallgular lIo/ch. a righl·allgled V'lIotch is used.
filld /I'e lime requiTed. r"ke (11/ o/l,eT da/" same.
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Notches and Weirs
Solution. Gi ven:
Angle of nOlch.
Ini tial height of wate r.
Final he ight o f walcr.
Ar~a of rese rvoir.
369 1
H, =2 111
A= 80 x 80 = 6400 ml
Cd = 0.62
Using th e relation give n by equa ti o n (8.9)
5xMOO
[ "
'" 4 x .62 x Ian 90"x .)2 x 9.81
2
2" -]iT
J
'" 2913.34 x [ -'-
- - ' -]
2.8284 5.1961
'" 29 13.34 [0.3535 - 0. [924J seconds
= 469. 33 seco nds = 7 min 49.33 sec. Ail s.
Problem 8.1 4 A right-angled V-lIo/eh is illsert,'d ill Ihe side of a tallk of leng'" 4 m mId .....idth 2.5 m.
Initial heigh! o[ ...<lIer abo)'/! lile tlpex oflhe 1I0/eI, is 30 em. Find the heiglll of ll'<lIer abo)'/! IIII! apex if
Ihe lime fl!({lIired /0 {ali'a Ihe hend ill ("Ilk/rom 30 em ro jim,l heighl is 3 minlltes. Take Cd = 0.60.
Solution. Given:
Angl" of notch.
= 90°
Area o f lank .
A = Len gth x w id th", 4 x 25::: 10.0 m 2
Inilial he ig hl of waler.
1i ,,,,30 cm=O.3m
T:3 min =3 x60 = 180scconds
Time.
CJ = 0.60
leI lhe final he ighl of watc r above lh c ape x of nOld '" Hl
Using lh e re lat io n give n hy equation (8.9)
e
T=
SA 9
[~ - ~l
H
4x Cd xlan l x.[fi Hl "
180 =
4x.60xlan
I -
'l
5xlO
[ ,
(90' )
---y;y - ( )'"
2 xJ2x9.81 Hl
0 .3
'
]
= 4 x .60 x501 x 4.429 [ H/' '1 - (0.3)liZ
0'
I I
180 x 4 x 0.60 x 4.429 = 38.266.
50
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1370
Fluid Mechanics
I
"I'T - 6.0858 '" 38.266
or
H,
I
-,~
fI ;!
'" 38.266 + 6 .0858 '" 44.35 or H,
I.~
•
1
'" - - '" 0.0225
44.35
112 ", (0.0225)"1.5", (0.0 225 ),6667 = 0.0822
...
m = 8.22 COl, Ans .
VELOCITY OF APPROACH
8 . 10
Velocity of approach is defined as the veloci ty with which the wate r approaches or rc~chcs the
weir or nOld before it flows over it . Thus if Va is til.: velocity or 3pproach. then an additi onal head /10
cq uallo
V'
.....!L
2,
duc lU veloc ity of .tpproach. is acting on the water flowing over the notch. Then initial
heigh! of water over the nmelt bc\;OIllCS ( 11+ Il o) and I1nal hei ght becomes equal 10 11 0 , Then all the
formulae arc changed taking into co nsideration of ve loc ity of approach.
The ve loc it y of approach. Va is determined by finding the discharg e ove r the nOlch o r weir
neglecting veloci ty of appro ach. Then di viding th e discharg e by the cros.'H;•."Ctional area o f the channel
on the upstream sidc of the weir or notch. the ve locity of approach is obt ained. Mathe matically.
Q
Vo = CC---C'c----C
Arca of ch annel
This vel ocity o f approach is used to find an additional head
(II.
=
~.:). Again the discharge is
l:aJculatcd ,lI1d above process is repeated for l11or~ accurah: discharge.
Discharge over a rectan gular weir. wi th velocity of approach
2
X Cd X L x .,ffi [(HI + h o )Y2 - h,:n [
'" "3
.. .(K.10)
Problem 8.15 Wulu is flo,..illg ill II rec/Illlgu/w c/UWIlef of I III ,..ide "JIl/ 0.75 III deep. Filld 11,1'
disch"rge o,·u (I reCl,mgu/ar ,.."ir of aes//ellgll, 60c IIl. iflhe head "fwaler v,·er Ihe creSI of weir is
20 cm "",/ ,..(ller from dlm,"el flvws m ·er Ihe weir. Tilke Cd == 0.62. NeglecI elld COll/ractiOllS. Tilke
,·elOeily of approach illlO COllsidew/im, .
Solution. Givcn :
A", Width x d~pth '" 1.0 x 0.75 '" 0.75 m l
Are a o f channel.
L=60 cm=0.6m
Length of we ir.
~lcadofwmer.
H 1 =20cm =0.2 m
Cd = 0.62
Discharge over a rectangular weir without velocity of approach is given by
Q==
~
3
2
== -
3
I I
xCd xLx
J2i
x 0.62 x 0.6 x
Y2
xH 1
J2 x 9.81
x (0. 2)312 111]/5
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Notches and Weirs
371 1
'" 1.098 x 0.0894 '" 0.0982 1I1'/s
Q
.0982
Velocity of approach.
11. =-.:\ ",
:. Additional head.
"" '" - "- '" (. 1309)212 x 9.81'" .0008733 m
0.75 =0.1309l11/s
v'
2,
Then di scharge with velocity of approacl"t
Q '"
~
j~
X Cd X L
given by L-q ualion (8.10)
x
J2i
2
3
'" - x 0.62 x 0.6 x
[(If,
+ lI o)lIl_ 1I.3J2]
J2 x 9.8 1 ](0.2 + .OOO87)Jf2 -
(.OOO87) lI2 ]
= 1.098 [0.09002- .00002566]
= 1.098
Problem 8.16
x 0.09017 '" .09811 1 m3,s. An s.
Fil1d III" discharge o,"a u rec/(wgu/ur weir of /e"gll! 100 III. Th e head of "'{ller VI'er
Ille weir is 1. 5 m. The \'e1ocily ofaJlJlrOllcll is ghell {/,~ 0.5 mls. Take Cd '" 0.60.
Solution. Given:
L.:ngth of weir.
Head o f water.
Velocity of approach.
L = HlOm
1I ,= 1.5rn
V. '" 05 Illfs
Cd '" 0.60
11 2
If : - "-=
U5xQ5
=0.0 [27 111
2 x9.8 1
The diSl: hargc. Q over a rectangular weir due \0 velocity o f approac h is givc n by equat ion (R. I 0)
:. Additional head,
"
2g
Q=
3.. X Cd x Lx
3
J2i I(H I + h.):sn. _h. m ]
2
'" - x 0.6 x 100 x J2 x 9.111 ]( 1.5 + .0 I 27):sn. - .01 27 3/2 ]
3
'" 177. 16] 1.5127·V2 _ .0127m ]
m',s.
= 177. 16 ]1.8605 - .00 143 J = 3 29.35
Ans.
Problem 8.17 A rectangular weir of cre),tlengtll 50 cm is used to measure tile rate offlow ofll"ater
in a rectangular cliallnel of 80 elll ....ide and 70 em deep . Dererm ille rile discliarge in lire challnel ifrlle
warer lel'el i.f 80 111111 IlbOl'e rhe cresr of weir. Take I"elocity of approach illlo consideration and mlue of
Cd '" 0.61 .
Solution. Given:
1. =50cm =0.5 11)
Length of we ir.
Ar~a o f channel,
A = Wi d th x de pth = 80 CIII X 70 Cill == 0.80 x 0.70 == 0.56 m<
H~ad
over weir.
Il '" 80 111m '" 0.08 m
Cd == 0.62
The discharge ove r a rectangular weir without ve locity o f approach is g ive n by equation (8.1)
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1372 Fluid Mechanics
Q '"
~
3
x
cJ )( L)( Jii )( H lI!
2
3
= 0.9153)( .0226 = .0207 m 3fs
'" - x 0.62 x 0.5 x .}2 x 9.81 )( (D.08)3/"! m 3 1s
Velocity of approach.
Head due to Va>
Q ,0207
Vo '" - = - - =.0369mfs
A
056
Ir
"
2
(.0369)"
= V;t2g =
= .0000697 III
2 )(9.81
Discharge wilh velocity of approach is
,
=::. X 0.62 X 0.5 X .)2 )(9.81 [C08 + .0000(97)312 - .0000697~nl
3
'" 0.9 [53
X
[.0R00697I.S - .00006971.~J
3
'" .9153[.02265 - .0CKl00(582) '" 0.2073 111 /5. AIlS .
A suppre.ued reClimgu/ar weir I.! COIIJlrUCled acroJ.l a ellalilie/ of 0.77
Problem 8.18
II! widtll Willi
a head of 0.39 //I and Ihe crest 0.6 III (lbOl'1" the bed of the chaliliel. Eslimale the discharge oru ir.
Con sider \'e/ociry of approach and aSSU/II(' CJ '" 0.623.
Solution. Given;
Widlh of channel.
b = 0.77 111
Head over wdr.
H = 0.39 m
Hcigtll of ere,! from bed of channel", 0.6 m
Deplh of channel
'" 0.6 + 0.39 == 0.99
Value of
SuppresSI.""d weir me an s that the width of channel is equal to width of weir i.e .. there is no end
contraction.
Width of channel" Width of weir" 0.77 m
Now area of channel,
A" Width of chan nel x D<:pth of ch ann el
" 0.77 x 0.99
The diSCharge ove r a rectangular weir without velocity of approach is given by equation (8.1).
Q==~
3
xC" Xbx.j2iXH Y2
" "32 x 0.623 x 0.77 x J2 x 9.81
Now velocity of approach.
I I
v-
Q
• - Area of chan nel
0.345
~'"C'!c~
0.77 x 0.99
(,:
Here b " L)
x O.W Y2 '" 0.34 5 m'ls
" 0.4526 lIlls
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Notches and Weirs
373 1
Head due 10 velocity of approach.
2
1
V
0.4526
hu = - "- =
=0.0[04111
28
2 )( 9.81
Now th e discharge wil h velocity of approach is given by.
Q:
i
x
C" x b x ..fii
1(1/
+ 11~)·112 -
1l/ J
2
2
'" JX 0.623)( 0.77 x ./2 )(9.81 1(0.39 + 0.0 1(4)312 - (0.0 1(4)3n 1
2
'" 3" x 0.623 x 0.77 x 4.43 [0.2533 -
0.001061
'" 0.3573 mJ/s. An s.
Problem 8.19 A slwrp crested rectangular weir of I In heighl e.tfe/ld.i across a rec/(l/lgular
c/wrme/ of 3 m ",id,h. If lilt! head afwaler 0\'''' I/Ie weir is 0.45 Ill. co/cu/m(! the discharge. COllsida
I-e/oeily ojupproac/I (1111/ IlHume Cd '" 0.623.
Solu ti o n. Gi ven:
Width o f channel .
h=3 m
Heigh t of weir
H~ad of water over weir.
=I m
H '" 0.45 III
Depth of channe l
'" Heig ht of we ir + Head of water ove r we ir
'" I + 0.45= 1.45m
Value of
Cd '" 0.623
The discharge ove r a rectan gular we ir with out ve locity o f approac h is given by
2
Q'" 3
X Cd X b
x
(8. 1) as
,ffi X H lI!
'" 3. x 0.623 x 3 x ~2 x 9.81
3
Now ve loc il y of approach is given by
v-
~ qual io n
x O.45·V2
:=
1.665 1I1 3/s
Q
• - Area of c hanne l
= ~~~~~L~~S~~~~
Widlh of cha nn el x {kplh of channel
-"L66~SO
~
'" 0.382
3 x 1.45
1I1/s
Head due 10 ve locily of approac h is give n by.
V l
0382'
h = - "- '"
= 0.0074
"
2g
2x9.8 1
11\
Now the discharge Wilh ve locity of approach is give n by.
2
Q'" - X Cd X b x Jii I(H + 11)311 _ Ol).ln l
3
2
'" - x 0.623 x 3 x ~ 2 x 9.81 [(0.45 + 0.OO74)3ll _ (0.0074)311 1
3
= 1.703 m 3 /s. An~ .
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1374
Fluid Mechanics
.. 8. 11
EMPIRICAL FORMULAE FOR DISCHARGE OVER RECTANGULAR WEIR
The dis.chargc over a rectan gu lar wc ir is given by
Q ==
,
i Cd J2i X L x If/JIl l without velocity of approach
= 3: Cd Jfi X LX
3
[(f! + " a)3Il.
~ h/I':!t
... (il
wilh ve locit y of approach
...( ; i)
Equations (i) and (it) arc applicable to the wcir or notch for w hich the c rest length is equal lO Ihe
width of the channel. This type o f weir is ca ll ed Suppressed '...·;r. But if Ih" wdr is not suppressed, the
<'ffect of end contraction wi ll be tahn in to account.
";,,; .;.;.;.; .; .;.;.; .;.;,,;, ,
(a) Fr"ncis' s Fo nnuh., Francis on the basis of hi s experiments estab'" "" """"""""""",,
li shed thai end cOlllmcli on decre ases Ihe effective length of the crest of
weir and hence dcucascs the discharge . Each e nd contraction reduces Ihe
n cstlcngth by 0. 1 x H. where H is the he ad uver the weir. For a rectangu lar weir Ihere are two e nd cont ra<:1io ns onl y and he nce clfcctivc Icngth
L =(L - O.2H)
O. lH
..
'"",',',',', ', ',',',, ,', "
:::::::::::::::: ::::::::::::!::
~ ,~ :~'
2
Q = 3xCdX [L - 0.2XIl1x-!fi flYI
', ',','," ,', ','" " I
',', ',',','
"' ·I:'.:.'O."i~: ' ,
',' ,',', "',',', ','
r-O.l H
Fig. 8.8
Cd = 0.623 . g = 9 .8 1 m/s1. the n
If
2
Q = - x .623 x -J2 x 9.8 1 x [L - 0.2 x HI x 113f1
3
= 1.&4 IL - 0.2 x 1I1f1·>l2
•.. (8. 1 I)
If cnd contractions arc s upprcssed . tllc n
H = 1.&4 LIlY!.
... (8 . 12)
If vclocity of approacll is considered. tllen
Q = 1.84 L [(fI + 11~»)12 _ h}i21
... (8 . 13)
(b) Bazin 's Fo rmula . On tllc basis of rcsults of a series of exper im cnts. Bazin's proposed thc
fOllowing formula for til e discharge ovcr a rectangular wcir as
Q =m xLx-!fixH l12
whc r~ /Il
.. .(8 . 14 )
2
.003
= - X Cd = 0.405 + - 3
H
II = hei ght of water OVe r Ihe weir
If veloc ity of approach is considered. the n
Q =/Il , XLX,ffi [(H+holl12[
where 111 , = 0.405 +
.. .( 8. 15 )
.003
( H + II.l
Problem 8.20 The helld of "'(1ler ora (1 ~e("l{mg"'{lT ...e;r is 40 em. The lellglll of Ihe aes l of Ille
"'e;r \\";Ih end eo nlmelit", suppren'ed;s 1.5 III. F;lId Ihe d;seharge JM;lIg Ille Jollo ... ;lIg Jormuille:
(i) F mnei s' s F ormuli, Will (ii) BlIzin '.1 F orlllu/a.
II
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Notches and Weirs
375 1
Solution. Given:
~lcad of wate r.
H:40cnl=OAOm
L=I.5m
(I) Fr,mcis's Porm ula for e nd col11raclion suppressed is given by equation (8. 12),
u,ngth of weir.
Q'" 1.84 L x H3I! '" 1.84 x 1.5 x (.40)31:'
'" 0.6982 m .l/s
(Ii) Ba 7. in 's Formula is given by equat io n (8.1 4)
Q=mxLx,ffi xH J12
where
til '"
0.405 + .003 ", 0.405 + .003 '" 0.41 25
H
.40
Q'" .41 25 x 1.5 x
.J2 x9.8 1 x (.4),\12
'" 0.6932 ",.lIs. AIl-"
Problem 8.21
A weir
36 melfes /ollg is dil'ided inlO 11 eqlUll bays by !'<,nical pO,~I.f. eael! 60 em
wide. Determine Ille di,Kllarge o\'er Ille weir
approach is 2 metres per second.
Solution. Given:
Lengt h of wc ir.
L[ '" 36 10
if
Ihe head OI'er Ihe crest is 1.20
//I
allli \'e/OCify of
Numbcrofbays.
= 12
For 12 bays, no. of vertical post = I I
Width of c;l<:h POSt
Effective length.
Head On wdr,
Velocit y of approach.
=60cm=O.6m
L=
H
LI
- 11 xO.6'" 36 - 6.6 '" 29.4111
= 1.20
Vo :
III
2 mls
V2
Head due to Va'
Number of e nd con trac tio n,
11 := - "-
"
2g
2l
:=
:=
0.2038
III
2x9.81
= 2 x 12
{Each bay has twO e nd co ntrac tion s I
= 24
Di scharge by Francis Formula wi th end co ntracti o n and ve locit y of approach is
Q'" 1.84 [L - 0. 1 x n( H + Ilalll(H + /t)312 _ 1I/"2 J
= 1.84[29.4 - 0.1 x 24( 1.20 + .2038)J x [(1.2 + .2038/~ - .2OJ8L~ 1
II
: 1.841 29.4 - 3.369111.663 - .092 1
: 75.246 m J/s . Am .
A discharge of 2000 mJ/s is 10 pdS.1 OI'U a rectangular weir. Tile weir i.1 dil'ided imo
a III/Illbu of openings each of SpOIl /0 III. If IIII' I"eloeily of approach is 4 tills. find Ihe number of
0plmi,tgs needed in order l/te heml of ""diU ol"er lite cresl is ltol 10 exceed 2 m.
SolutIon. Given:
Problem 8.22
Total
disch~rge.
Leng th of each opening.
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1376 Fluid Mechanics
V" ", 4rnls
H=2m
Velocity of approach.
Head over weir.
:N
lei number of openings
Head duc to velocity of approach.
Ir: V} ", 4X4 '" 0.8155
" 2g 2x9.81
In
For each opcnillg. number of end contractions arc two. Hence discharge for each opening
consideri ng velocity of approad is given by Francis formul a
i.e..
Q'" I.MIL - 0. 1 x 2 x (I{ + I.a)][(/l + ha)3f1. - /'a3f1. )
'" 1.84110.0 - 0.2 x {l + .8155)1I2.81551~ _ .81SS u l
'" 17.3631 4 .7242 - 0.73641 '" 69.24 m]/s
2{J(J()
Total discharge
: =~='===
Discharge for 0"" opening
Number of opening
69.24
"" lS.88 (say 29) '" 29. Am .
.. 8. 12
CIPOLLETTI WEIR OR NOTCH
Cipollclli weir is a trapezoida l wcir. which has side slopes of
horizontal to 4 vertica l as ., hown in Fig. 8.9. Thus in 6ABC .
~ '1- r
' .- ..... -.., , f
f
e
AB H f 4
I
tan - '" - - : - - : 2
Be
H
4
r :=::::: -,
.
,
"
I
_ = tan _ = 14 ° 2'.
2
4
"
1
1
By giving this s lope to the sides. an increase in discharge through the
triangular portions ABC and DEF of the we ir is obtained. If this slope is
not provided the weir would be a rectang ular one. and duc to end Fig. 8.9
contraction. thc di scharge would decrease. T hu s in Ca.>C of cipolletti
weir. the factor of end contractio n is IIOt required which is shown be low.
The discha rge through a rectangular weir with twO end con tra ct ions is
Q=
~
1
X
Cd x (L-O.2
If).ffi xH
The cipollelli u·ejr.
Jl2
=~ xCd XLx.fii H3fJ. _ ~XCd X.fii xH sn
3
IS
Thus due to end cOnlr.lction. the discharge
discharge
Cilll
decre~scs
2
by 15 x Cd x
.ffii
x
,;noThis decrease in
be compe nsated by giving such a slope to the sides that the di scharge th rough twO
~
x Cd x..ffi x HV1. . Let the slope is given Ily 9/2. Tile discharg"
IS
through a V·notch of angle 9 is give" by
triangular portions is equal to
8
=-
15
I I
X
r:c
Cd X ,,2g
X
9
tan - Hm
2
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Notches and Weirs
Thus
-
8
X
IS
a2
e = -2 x15
[
- =-
Ian -
Thus
disch~rgc
2
15 8
through cipolieui weir is
'
Xlan _ H5I2= ~ xC x
CdX .fii
or
4
15
d
377 1
~xH'>r.
v~~
e/2 = t3n- 1 ~ = 14 ° 2'.
4
Q=~xCd)(Lx.fiiH3I2
... {8. 16)
3
If ve locity o f approach, Va is to be taken into consideration.
Q '"
Problem 8.23
~
3
X
Cd
X
Lx
J2i lUI + Ila)3I1 _ ha1l2
Find rhe diKilarge OI'er a cipo/klli wl'ir aile/18th 2.0
... (8 . 17)
1
III lI'ilell
Ihe he{jd orn tile weir
is /111. Take CJ = 0.62.
Solullon. Given:
li:ngth of we ir.
Head o\'er we ir.
L =20 m
1/ : 1.0111
CJ = 0.62
Using equation (8.16), the disc harge is given as
Q=~XCd XLX.ffi
3
x HJJ2
= -2 x 0.62 x 2.0 x
..j2 x 9.8 1 x ( 1)312 = 3 .66 1 m)/s. Ans.
3
Problem 8.24 A cipol/elli weir of cre.n lenglh 60 em diJClwrge.1 Wafa. The head afwaler ora IIII'
,,'eif is 360 mm. Find the disc/wrge ol'a I/Ie weir if Ihe ellmmel is 80 em ...ide and 50 em deep. Take
CJ = 0.60.
Solut ion. Given:
Cd '" 0.60
Length of weir.
L=6Ocm=0.6Om
Head o f water.
1f =36O mm =0.36m
Channel width
=oocm =0.80m
Channel depth
= 50cm= 0.50 m
A = uoss-scClion,ll area of dwnnel = 0.8 x 0.5 = 0.4 m1
To find velocity of appro ach. First ddermine discharge over the weir as
2
Q= - xCd xLx.fiixHln.
3
The velocity of approach, Vo = Q
A
,
Q '" ~ x 0.60 x 0.60 x
J2 x 9.8 1 x (0.36)ln. m /s '" 0.2296 m /s
3
3
.2296
0.40
V" = _ _ = 0.514 m/s
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1378
Fluid Mechanics
Head duc to ve loc it y o f approach,
h '" V "/2
"
Thus th e
d isc h~rgc
"
is give n by
g
'" (0.574 )2 '" 0.0 168 m
2x9.81
(8 ,17) as
cqu~tio ll
,
Q= i xCd xL x J2i I (H +hj -~ - h/~ l
=~XO .60 X .6X J2x9.8 1
[(.36 + .0 168)I.S _ ( .OI 68)I.S 1
'" 1.06296 x [.23 13 - .002 177 J '" 0. 2435 mJ/s. A" s.
to 8. 1J
DISCHARGE OVER A BROAD- CRESTED WEIR
A wei r ha vin g a w id ~ crest is kn ow n a~ broad-cres ted wei r.
Let If = height o f wate r aoove th e c r"st
L '" length o f th e creS!
ru=
7::,,:"::,,:~::~:~
: !r,
:"::":-".:?:~~-r~~
,,,,,,0 "'i''' O
~ ,
F;g . 8.10
----<
B~Qad<{;Tf'Jted wt:i~.
If 2L > H. tile we ir is ca ll ed broad ,uc~lcd weir
If 2L < H. the wei r is c alled a narrow-crested we ir
Fig. 8.10 s hows a broad-crested weir.
Let II '" head of water m the midd le of we ir w liich is CO nMan!
\' = ve locit y of flow over the we ir
Appl ying Bernoulli 's equatio n to th e still wate r surface o n the ups tre am side and runnin g water at
th e e nd of wei r.
1,2
-
28
Th~
= H - II
1. = J2 g ( H /r)
discharge Oller weir Q = Cd x Are a of n ow x Ve loc it y
=Cd x Lxll x J2g (H
II)
l
J
=C d XLXJ2 g( Hll - II )
I I
... (8. 18)
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Notches and Weirs
379 1
The di scharge will be maximum, if (1/111_ lll) is maximum
d (HII-, - II 3 )=Oor 2hxIJ - 3h-=Oor
'
2H= 311
dh
-
/,= -32 If
Q""" will be obtained by substitut in g this value of II in equation (IU8) as
'" Cd X L x
J2i J2~ H l
'" Cd X L x
.J2i
x 0.3849 X 11J.r.
"" .3849 x J2x9.8 1 x Cd x Lx HlI2 = 1.7047 x CJx t X 11lll.
'" 1.705xCJ xLxH
to 8. 14
JI2
. . . (8.19)
.
DISCHARGE OVER A NARROW-CRESTED WEIR
For a narrow -crested weir. 2L < H. It is simi lar lO a rectangular we ir or notc h hence. Q is given by
2
Q'" -
3
.. 8. 1S
X Cd X L
x
.fii
X
n ln
...(8.20)
DISCHARGE OVER AN aGEE WE IR
Fig. 8.1 1 shows an Ogee weir. in wh ich the crcS! of Ihe weir
rises upto maximum heigh! of 0. 11 5 x H (w here H is [he heigh! o f
Wale r above inlet of the weir) and then fall s as shown in f'ig. 8.11 .
The disc harge for an Ogee weir is the same as that of a rectangular
weir. and it is given by
2
Q= 3
... 8 . 16
X
Cd X f. x
.fii
x
If'!
,~e
CREST
.. ,(8 .21 )
Fig. S.1l
An Ogff 'Wt'ir .
DISCHARGE OVER SUB·MERGED OR DROWNED WEIR
When the water level on the downstream side of a weir is above the crest o f the weir. the n the wei r
is called to be a sub-mcrged or drowned weir. Fig. IU2 shows a sub-mc rged weir. T he tOial
discharge. ove r the we ir is obtained by di viding the we ir into two parts. T he port io n between upstream
and downstream water surface may be treated as free weir and ponion between dow nstream water
surface and erest of weir as a drowned wei r.
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1380 Fluid Mech a nics
Fig. 8.12
Slth-merged 'U:tir.
Lei H '" hciglil of water on the upstrea m side of the we ir
II '" height uf wate r on the dow nstream side of the we ir
Then
QI '" discharge ol'e r upper porti o n
,
"'~X Cd '
xL x fii IH- h l)f!
Q 2 '" di scharge thro ugh drowned porti on
'" Cd, x Are a o f flow x Ve locity of fl ow
'" Cd, xLxil x J2g( H - h)
.. TOI al di Sl' tJ argc .
Q'"
0 1 + Q2
2
'" J ed,
XLx,ffi 11f _ II I)I2 + Cd, x L xhx ~2 g( H - h) . ... (R.22)
Problem 6.25 (a) A broad-crested wei, of 50 m lengtll. /1<105 50 em heigh! O/lI'ater aborc i/$ eres l.
"-illd Ihe maximum discharge. T(lke Cd '" 0.60. Neg/eel I'C/Oeil)' of approach. (b) If Ihe I'e/oeil)' of
approach is /0 be taken ilifO cDllsidermioll.jilld the mlUimulII disclwrge when Ille challllel liaS a crosssecliOlw/ area of 50 "1 2 011 Ihe IIpstrcam side.
Solullon. Give n :
Le ngth of we ir.
L =50111
Head o f wata.
H ::50cm =0.5 m
Cd:: 0.60
(i) Nl'"glecting "elocily o f ap proach. Max imum di sc harge is givc n by equatio n (8 . 19) as
1.705 X Cd X L X H3f2
:: 1.705 X 0 .60 X 50 X (.5)'V2 = 18.084 mJ/s. AilS.
On",,::
(ii) T"kin g nloclly o r "ppn)!lch intn consldcnllion
Area of Channel .
A = 50 m 1
Velocity of appro ac h.
18.084
V:: - = - - = 0 .36 m/s
" A
50
: . Head du e toY".
h = V} = O.36x .36 =.0066 m
o
"
2g
2x9.8 1
Ma xi mum d isc harge , 0 ....., is give n hy
Qn",:: 1.705 X Cd X L x [(f/ + 11~) .\12 _ 11 • .\12 [
" 1.705 X 0 .6 X 50 x [(.50 + .(066)'~ - (.0066>'~ 1
= 5 1.1 5 [0 .3605 - .0005361 = 18.412 m l/s. AilS.
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Notches and Weirs
Problem 8.26
All Ogee weir 5 metres 101ig lIa,\" a Ilead of 40 em of wilier.
If Cd '"
381 1
0.6, filld Ille
disellarge ora Ille weir.
Solution. Given:
L=51O
Len gth of we ir.
H~ad
o f wala.
H= 4Qc rn '" DAD 111
Cd '" 0.6
Disdlargc over Ogee we ir is g iv en by equation (8.2 1) as
,
Q= ~
3
X Cd X L
x
.ffi X HY!.
'" ~ x 0.60 x 5.0 x
J2 x 9.81
x (O.4)3i'l '" 2..2409 nh~. A n s.
Problem 8.27 Tlte /JeiglllS o/ . . . aler 011 I/le IIps/reom (mil downs/rem" side of (l sllb·merged we ir of
J", /<'/lgl/' are 20 em ,md /0 em rnpeC/ire/y. If Cd /or free ,md dro,.."ed pOr/iolls (Ire 0.6 (lnd
0.8 r espec/il'ldy.
fill"
llie diJ ch(4rge on" Ihe weir.
Solution. Give n :
Height or waler on upslream s ide, H '" 20 em = 0.20 III
~I c ight of water 011 downstream side. II '" I0 e m '" 0.10
u,n g[h of weir .
L= 3 m
In
Cd, '" 0.6
0.8
Cd: '"
Total di scharge Q is the s um of discharge through free (>Onion and discharge through the drowned
[lonion. Th is is g ivcn by equation (8.22) as
,
Q==~X CJ, XLX,ffi
==
~
II2
IH _ II I· + Cd, xLxllx
xO.6 x3 x ,)2 x9.8 1
J2g (H II)
1.20 - .10(~+ 0.8 x3x . IOx
J2 x9.81 (.2 - .1 )
== 0.168 + 0.336 == 0.504 nI'/s. Ans .
HIGHLIGHTS
1. A nouh is a dcvice u.'iCd for measuring [hc rotc of flow of a liquid through a small channel. A weir is
a cOllcrete or masonary ,truewre placed ill the open channel over which Ihe flow occurs.
2. The discharge Ihrough 3 rectangular notch or weir i. gi,·en by
2
Q"'"3CJ XLXHJfl
whue C d '" Co-efticient of <iischmllc.
L ~ Length of notch or weir.
H ~ Hca<i of water OVer the notch or wcir.
J. The discharge over a triangular notch or weir is given by
Q3 15CJ
[an "2
8
' X ,,2g
"" xH
5fl
where e = \o[al angle of triangular notch.
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1382 Fluid Mechanics
4. The discharge through a trapezoidal notch or weir is equal to the sum of discharge through a
rectangular nOlCh and the discharge through a triangular notch. It is given as
2
Q""3
where
Cd ,
Cd, X Lx
Jfi
8
X H II2 + T5CJ, x tan
a ,,28
r::o::
"2)(
x It!'!
~ co-efficient (If discharge for rectangular notch.
Cd, ., co-efficient of discharge for triangular notch.
612 ., slope o f the side of lmpezoidal notch.
5. The error in discharge due \0 the errOr in the meaSurement of head OVer a rectangular and triangular notch
or weir is given by
dQ = 3dH
2 H
Q
...
r~r
5 tlH
.--2 H
where
a rC<:tnngular weir or notch
For a triangular weir or notch
Q" discharge through rectangular or trinngular notch or weir
H .. head over the notch or weir.
6. The lime required \0 empty a reservoir or a tank by a rectangular or a triangular notch is given by
... By a
r~ctangular
n()(ch
... By a triangular notch
A. cross·sc.:tional area of a lank or a reservoir
where
1/, _ initial height of liquid above the crest or apex of notch
111 - final height of liquid abo,·e. the. crest or apex of notch.
7. The wlocity with which the water approache, the weir or notch is called the velocity of approach. It is
denoted by V. and is gi"en by
V• •
Di seharge Over the notch or weir
Cross -sec tional area of channel
H. The head due to velocity of approach is gi"en by h.
~
v'
- '- .
2g
9. Discharge over a reClJngular weir. with velocity of approach.
Q10.
Fr~nci,s
%C,J- J2i Hil t + h)'" - h}I.'I.
Formula for a rectangular weir is given by
Q. 1.84[1. _ 0.2 II I IIlIl
.. 1.84 L llJi1.
.. 1.8..1 L [(II + h.,)lrl_ h}'2 J
where I. .. length of weir.
H ., height of water above the crest of the weir.
h • • head duc to "clocity of approach.
11. Ilazin's Fonnula for discharge over a rectangular weir.
~ I
... For tWO end conlraclions
... If end contractions are suppressed
... If "elocity of approach is considered
I~
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IL
Notches and Weirs
Q .. In L .j2;HJI2
... without velocity of approach
~ m L.f2i [(H + hdlrll
where
2
mK_
3
... with velocity of approach
cd"O. 405+ -~
.003
... without velocity of approach
H
~O.405+(
.003
H+ha
383 1
I
... wilh velocity of apP!U.lch.
12 . A trapezoidal weir. with side slope or I horizontal 10 4 vertical. is called Clpollelti weir. The dis.::harge
through Cipollctti
w~ir
is given by
o
Q = ~ Cd X L X
'"
J2i HJn.
.. , wilhout velocity of approach
~ CdxLx J2i [(H + h)lfl_ h}1:.'1
... with velocity of appro.1ch.
13 . The discharge over a broad-crestcd wcir is given by,
where
H = height of waler alx,,"e the
()r
cre~l
h ~ head
water at the middle of the wcir which is consl:m!
L .. length of the wcir.
,
14 . The condilion for maximum discharge ow. a broad-cres\cd weir is h. ~ H
and maximum diSCharge is given by Q_
K
1.705 Cd L H J".
2
IS. The discharge O\'er an Ogee weir i. gi"en by Q '" - C)..)(
3
.fii )( HlfI..
16. The discharge OYer sub-merged Or drowned weir is given by
Q "" discharge Over upper portion -+ discharge lhrough downed portion
.. %Cdl LX.[fi (H - h)lf! + Cd, UI
where
X
J2g(1f - h)
H _ heighl of waler On [he upstream side of lhe weir.
h: height of waler on lhe downstream side of [he weir.
EXERCISE
(A) THEORETICAL PROBLEMS
I. Deline lhe [cnns : notch. weir. nappe and crcs!.
2. How are the weirs and nolche< cla"ifjed 1
J. r ind an expression for the discharge oVer a rectangular weir in termS of head of waler Over the cre,l of
lhe weir.
4. Prove lha[ the discharge through a [riangular notch or weir is g;"en by
8
Q- - Cd xw.n
15
where 1/ .. he,[d of waler ovcr lhc n01Ch or
8 • angle of notch or weir.
I I
8
M- lll
-x,, 2g H
2
weir
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1384
Fluid Mechanics
5. What are the advantages of triangular notch or weir o,"cr ",,,[ang ular notch?
6. Pro"e that the error in discharge due 10 the error in the mea'DWlnen! of head over a rectangular notch is
given by
dQ
Q
:3 dJ/
."2"
where Q ~ discharge through rectangular notch
H ~ head O\'er the .cclangula. notch.
lind
7. Find an c~ pression for the time required to empty II tank of area of cross-section A. wilh a rcclnngular
notch.
II . What do you understand by 'Velocity of Approach " ? Find an e ~ prcssion for the discharge over a rectan gular weir with ~eloci ly of approa,h .
9. Define 'end conlrncl;on' of a weir . What is the effect of end contrnc(;on on the discharge through a weir ?
Ill . What is a Cipoliclti We ir 'I Prove thaI the discharge through CipolleU; we ir is given by
Q'"
%C.,I.. J2i H JIl.
wbere L '" length of weir. and H '" head of water o,'cr weir.
II . Differentiate betw""n Il roa,h,rested weir and Narrow-cre,ted weir. Fi nd the condition for maximum d is-
12.
13.
14.
15 .
16.
charge m'er a Ilroad-crested weir and hence deri,'e an expression for maximum di>eharge m'er a broad crested weir.
What do you mean by a drowned weir ? How will you detenninc the di>eharge for the downed weir?
Discuss 'cnd contraction' of a weir.
State the different devices that Can be used to meaSUre the discharge through a pipe also through an open
chan nel. Describe one of such devices with a ti Cat sketch and explain how one Can obtain the actual
discharge with its hclp ,
What is the difference between a notch and a weir?
Define vc!ocity of approach . How does the "elocity of approach aff(X:t the di><;harge over a weir?
(B) NUMERICAL PROBLEMS
I. Find the discharge of water flowing oVcr rcctangular notch of 3 m length when the conStant head of water
OVer tbe noteb is 40 ern. Take Cd'" 0.6.
[Ans . 1.344 ml/sl
2. Detenni"e the height of a rKtangular weir of Icngth 5 m to be built aCross a rectangular channe l. The
J.
4.
S.
6.
7.
I I
maximum depth of water on the upstream side of the weir is 1.5 m and dischargc is 1.5 m' per second.
Take C. ~ 0.6 and neglect end contractions,
[Ans. 1.194 ml
fi nd the d ischarge o\'er a triangular notch of angle 60° when the head over the triangular notch is 0.20 m.
Take C" '" 0.6.
[An s. 0,0164 111 l /s l
A rectangular channel 1,5 m wide has a di".:harge of 200 litres per second. which is measured by a right angled V-nolch weir. Find the position o f the apex of thc notch from the be<i o f lhe chann el if maximum
depth of water is not be exceed I tn, Ta~ e CJ • 0.62.
[Ans . .549 1111
Find lhe discharge lhrough a lrapezoidal notch which is 1.2 m wide at the lOp and 0.50 m al the bollom and
is 40 e111 in height , The head of walcr 011 the nolch is 30 cm. Assume Cd for rectang ular ponion as 0.62
while for triangular portion ~ 0.60.
[Ans. 0,22 m'/5[
A rectangula r notch 50 cm long is used for !)leasuring a discharge of 40 litres per second. An error of2 mill
was made in measuring the head over the nOlch. Calculate the percentage error in the discharge. Ta kc
Cd '" 0.6.
[A ns. 2.37%[
A right-angled V-notch is used for mea,uring a discharge of 30 litre."'s. An error of 2 I1l1n was made in
measuring the head o,'er the notch. Calculate the pereentage error in the discharge. Take CJ .. 0.62.
[An.• . 2.37%[
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Notches and Weirs
385 1
8 . fi nd the time required 10 lower the water level from 3 III to 1.5 m in a reservoi' of d imension 70 III x 70 m.
by a 'lX'wngular notch of length 2.0 m . Take Cd~ 0.60.
IAns. 11 min I 51
\I . If in the problem 8. instead of a rectangu lar notch. a right angled V-notch i~ used. find the lime required.
Take all other data -,arne.
IAns. I J min 31 s[
Ill . Water is flowing in a rectangular ~hannel of 1.2 rn wide and 0.8 III deep. Find the discharge over a feClangular weir of crest length 70 e1l1 if the head of water over the crest of weir is 25 em and Water from channel
flows o,'cr thc weir. Take CJ 0.60. Neglect end contractions bul con,ider vel<X'ily of approach.
IAn s. 0.1557 111 J /sl
II . Find Ihc discharge over a rectangu lar weir of length 80 m. The head of wuter over Ihc weir is 1.2 m. The
velocity of ~ppro~ch is gh·cn ~s 1.5 mls Take Cd ~ 3.6.
IAns. 208.11 m)lsl
11 . The head of water oyer a rectangular weir is 50 cm . The length of the crest of the weir with end contraction
suppres>ed is 1.4 m. Find the discharge using following fommlae : (i) Franciss Fonnula and (ii) Hazin·.
formula.
IAns. (i) 0.91 m'/s. (ii) .901 m' lsl
U . A di>charge of 1500 ml/s is to pass o,·cr a rectangu lar weir. The weir is dividcd into a number of opening.
each of span 7.5 m. If the velocity of approa~h is 3 m/s. fmd the number of openings nceded in order the
head nf water ovcr the crest is not to exceed 1.8.
IAIls. 37.5 s.ay J81
14. Find the discharge Over a cipolleui weir of length 1.8 m when the head oVCr the weir is 1.2 m. Take
Cd = 0.62
IAns. 4.331 m' lsl
15. (a) A broad ·nested weir of length 40 m. has 400 mm height of water aoo\"C its crest. Find the maximum
dischargc. Take Cd " 0.6. Neglect ,·elocity of approach
IAn •. 10.352 mllsl
(1)) If the ,·clocity o f approach is to be taken into consideration. find the ma~imum discharge when the
channel has a cross-sectional area of 40 m' On the upstrcam side.
IAns. 10.475 1111/sl
16. An Ogee weir 4 m long has a head of 500 mm of water. If Cd = 0.6. fi nd the discharge ovcr the wcir.
IAn s. 2.505 ml/sl
17. The heights of water On the upstrcam and downstream side of a sub-merged weir of length 3.5 mare
300 mm lind 150 mm rcspecti'·cly. If Cd for frcc and drowned portion arc 0.6 and 0. 8 rcspectively. fmd the
discharge o'·cr the weir.
IAns. 1.0807 ml/sl
E
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CR~EK
.. 9. 1 INTRODUCTION
This chapler deals with the flow of fluids whic h arc viM:ulIS and flowing at very low veloc ity. At
low ve locit y lh e fluid moves in laye rs. Each layer of fluid slidcsovert hc adjaccnllaYCf. Duc 10 relath'"
.
du
velocity betwee n two laye rs the velocity grad ient - exists and hence a shear stress 1 '"
<f)'
lhe layers.
The followi ng caso"s will be considered in this c hapler:
1111
j.l -
aCL~
On
<fy
I. Flow of vi scous flui d through ci rcular pipe.
2. Flow of vi<;C{}us fluid between 1WO parallel plates.
3. K inctic energy torrcction and momentum correctioll f;!ClOTS.
4 . Power absorbed in viscous flow through
(lI ) Journal bearings. (h ) Foot-step bearings. and (el Collar bearings.
to 9 .2
FLOW OF VISCOUS FLUID THROUGH CIRCULAR PIPE
ror the flow or viscous fluid through circular pipe, the velocity distribution across a secti on. the
ratio of maximum velocity to average veloci ty. the shear stre.'\S di~tribution and drop of prcs~ure for a
given length is to be dete nnined. The fluw throug h the circul~r pipe will be viscous ur I~min~r. if the
Reynolds number (R,·) is less thun 2000. The expression for Reynold number is given by
R '" p VD
. "
where p '" De nsity of fluid fl owing through pipe
V '" Aver~ge ve locit y of fluid
D", Diameter of pipe and
)1 '" ViM:Osity of fl uid .
DIRECTtON
OF FLOW
,
'"
VisWUI flow throllgh a pipf'.
(_)
Fig. 9.1
(b)
• For derivation, please refer 10 An . 12.8.1.
387
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1388
Fluid Mechanics
Consider 11 horizontal pipe of radius R, The "iSl."Qus fluid is flowing from len 10 right in the pipe as
shown in Pig. 9.1 (a). Cons ider a flui d clcmcllt of radi us r. sliding in a cylindrical fluid clemen! of
radius (r + dr). Let the length of fluid clemen! be Ax. If 'p ' is the intensity of pressure on the face All,
Ihen the intensi ty of pressure on face CD will be
(p + ~ a.x). Then the forces acting On the fluid
clcmcru ,m: :
L. The pressun: force, p X It,l on face AB.
2. The pressure fOrl:c.
(p +~ AX) rtr2 0n face CD.
3. The shear force. t X 2nrA. on the surface of fluid clement. As there
summation of all forces in til.: direction of flow must be zero i.e"
i~
no acceleration. hence lhe
' ra"J
'
ax
/lltr-
p+- lh
_ap
ax
1tr""-tX2ItrXal=O
dl7l,l - fx2n:rx6.t'=O
_ ilp .r - 21=0
ax
ap -,
... (9 . 1)
1: - -
(Ix 2
al' across a section is constant. Hence shear
ax
The shear stress t across a section varies with " as stress distribution across a section is linear
Ill;
shown in Fig. 9.2 (11).
SHEAR STRESS
DISTRIBU,)ON
VELOCITY
/ DISTRIBUTION
?'
)
Fig. 9.2
~
{.,
---, C-
'I
Sbt'ar urt'u and vl'iociry distribution across a st'Ction.
(i) Ve locity Distribution . To obtain the velocity distribution across a section. the value of shear
stress t =
~
du is substituted in equation (9.1).
dy
But in the relation t =
~
du ,y is m.::asured [rom the pipe wall. Hence
dy
y=R -r and
i/u
- i/r
Substituting this value in (9.1), we get
dy=-dr
i/u
dr
t=~ - = - ~ -
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Viscous Flow
dll
-, -
up ,
ax 2
= - --
dr
or
du
-
389 1
1 dp
"' - - ,
dr
2)1 a.r
Int egrati ng this above equation w,r.t ' , . we gel
,= _'_ dP,2+C
... (9 .2)
ax
4)1
where C is the constant of inwg ration and its va lue is obtained from the boundary conditio n that at
r= H. u=O.
o= - '_ QP n2+c
ax
4/1
c :- - '_ QP R2
4).1 ,h
Substituting this va lu e o f C in e quation (9.2) , we get
1 dP 2
l iJP 2
,= - , -4).1- R
4)1 fh
ax
I
op
2
2
= - - - [R - r l
41J
ax
In ' ljuation (9.3).
v~lucs of ).I. ~~
and R :Irc
... (9.3)
const~nt, which means the ve locit y, II va ries with the
squ are of T, T hus eq uati o n (9.3) is a equ ation o f parabola, This shows Ihm the vciocity di stribution
across Ih e sect io n of a pipe is parabolic. Th is ve loc ity distribution is sliow n in r ig. 9.2 (b).
(ii) Ra l lo uf !\Iud mum Veloc ll ,f lu AHrage Vel ocl t y. The velocity is maxi mu m. whe n f ' " 0 in
eq uation (9.3). Thus maximum veloci ty. Umo, is o btained as
u
.....
= __,_ op R2
4J.l ,ll-
...(9 .4)
The average ve loc ity. II. is ob tained by dividing the di sc harge of the nuid across th e seetion by the
area of the pipe (rrR\ The di...:hargc (Q) across the sectio n is Obtained by considerin g the now through
a circular ring ,dement of radius r and thickn ess dr a~ ~ho w n in Fig. 9 .1 (b ). Th e fiu id fiowing per
s<!(;ond Through Thi s c le memary ri ng
dQ = ve loci ty ~I a radiu s r x area of ring cleme nt
= II X 2rr r dr
1
ap
2
2
=- - - [R - r 1x2rrrdr
4J.l ax
Q=
I I
lOp
,
_ (R 2 _ r'")x2rrrdr
J,3' dQ= J,'3 _ _
4).lax
Ii
~ I
IL
1390
Fluid Mechanics
[-a")X2'
[!!'..C_~]'
: _1 [ - ''') x 2'['" _"']
ax
2
4 0 4 pfJx
2
4
: -1
4~
: _I [-ap)X2 1'tX~ : ~ [-aap)
R'
.•'
4)1,Jr
4
-,,-(-a
8j.I ax
p
-
,:
Average ",",ocity.
Q
--:
8!.1
) R'
,
rrR -
Area
u'" ~1 (-a~:) Rl
..,(9.5)
Dividing equation (9.4) by eq ua lion (9.5) ,
Ra1io of maximum velocity to :IVnagc velocity'" 2.0.
II pipe
(iii) Dro p of Prcs~ u ~ fo r a g j,'cn Len gth (L ) of
From equation (9.5). we have
(-,p)
"' [-a
):: 8R-1l:-; _
ax
ax
" : _I
R'
8p.
p
-
Integrating (he abov e equat ion w. r.t. x. we gel
-
J' dp: fl l
8)1 ,-1/
R"
l
d,
-
-
8~lU
8pu
- [PI- Pll = - ,- {XI - x, 1or (PI - p,) = -,- [x, - x d
R
-R--
8~lii
: -
"
(p _
I
Loss of pressure head
R'
L
I":
)."2 - ,( I '"
F ig . ';1.3
L from Fi g. 9.3)
8p.uL
(D /2)2
) _ 321.l11L
P! D' .
where PI - P! is Ihe drop o f pressure.
= P,-1'2
pg
:UpuL
'" Ilf = pg D '
...(9.6)
Equation (9.6) is called Jl a gen Po ise uiJI e Fo rmu h. ,
I I
Ii
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IL
Viscous Flow
391 1
Problem 9.1 A crude oil of \'i.~co.~iry 0.97 poi,le 01111 relalil'e delLlity 0.9 iJ flowing "uougll a
lIorizonra/ circular pipe of diameter tOO mm and of /eng/h 10 m. Calculate 1111' difference of pressure
alille 1"'0 "lIds a/ lire pipe. if 100 /(g of IIIe oil is collected in 1I lunk in 30 seconds.
Solution. Given:
I-l = 0.97 rois.c == 0.97 '" 0.097 Ns/m 1
10
RcI<llivc density
'" 0.9
'"
0.9 x 1000 ", 900 kg!nr'
Po. or dt'llsity.
Dia. of pipe.
D= l00mm=O. 1 111
L= 10m
Mass of oil collected.
M=l00kg
Time.
I'" 30 seconds
CaJculalC difference of pressure or (P, - PI)'
The differcm:c of press ure (P, - Il l) for viscous or laminar flow is given by
321-l uL
Q
PI - P2 '" - -,- , where II '" average ve locit y == - D'
Area
'00
= -
Now. mass of oil/sec
kg/s
30
=P o xQ:900xQ
(.:
Po= 9(0)
100 =900xQ
30
Q = 100 x-'30
ii == ~=
Area
==
0.0037 mJ{s
900
.0037 = .1)037 = 0.471 m/s.
~D1
~(.I) 1
4
4
For laminar or viscous n ow, the Reynolds nUlnocr (R, ) is less than 2000. Let us calcuJacc the
Reynolds numbe r for this problem.
Reynolds numbe r,
where
pVD
R, * = -
"
p=Po= 9oo. V= j; =0.471.D=0.1 m.~1=0.097
R = 900 x .471 xO.1 = 436.91
,
0.097
As Reynolds number is less than 2000. the flow is laminar.
32xO.097)(.47 1)(1 0 NI '
,
(.q-
m
= 1462.28 Nlm 1 = 1462.28 x 1O~ 4 N/cm 1 = 0.1462 Nlcm!. Ans .
• For derivation, plea"" refer to An . 12.8.1
I I
Ii
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IL
1392 Fluid Mechanics
Problem 9 .2 An oil of I'i.{cosily 0.1 Nsfm " 0111/ felMil'e detlJ'if)' 0.9 is fl o wing Ilrrougll a circular
pipe of diameter 50 mill and of /englll 300 m. The rale of jlQW of fluid [I"ough Ihe pipe is 3.5 li/reY! .
Find Ihe pressu,e drop in" lenglll of 300 m lind also Iile l'lle", SlreJ'$lIIlire pipe ...al/.
Solution. Given: Viscosity.).I '" 0.1 Nslm!
= 0.9
Relati ve dcr>sil y
Po or densi ty of oil
= 0.9 x 1000 = 900 kglm )
D = 50 nlln = .05 III
L =300 Ul
(,:
De nsity of waler = 1000 kg/ml)
Q = 3.5 Iitrc sJs = 3.5 = .0035 m1ts
1000
Find (0 Pressure d rop. P, ~ P l
(ii) SlIcar stress at piP<-' wall, t o
=
32).111L
---vr-
where
L782 m/s
,
The Rey nolds num ber (R, ) is given by, R, = pVD
,
where p = 900 kg/m". V ", average velocit y = II = 1.732 mls
R =900x 1.782:-: .05 =801.9
,
OJ
As Reynolds number is less than 2000, the flow is viscous or lam inar
32 xO. 1x 1.782 x 3000
PI - P2 =
(.05)'
= 6&4288 Nfm" = 6&428 x 10- 4 Nfcm ' = 68.43 Nfc m 2. AilS.
(ii) Shea r Stress a t tile pipe w:11I (1:0)
The shear stre ss at any rad ius r is give n by the equation (9. l)
dp'
dol 2
i.e ..
T'" - - -
Shear stress at pipe wall. where r '" R is gil'e n by
To =
Now
-ap
R
2
-~~
ax
~
-a~
p = --,
- (,-,
p,~--,-p,,,)
d.T
PI -Pl", f! I -P l
Xl - XI
Xl
XI
L
= 6&4288 Nfm l '" 2280.96 Nlm3
300
III
and
D
.05
R ="2=T= ·025m
t o = 2280 .9 6x
I I
.025 N
-~--,
Ill ·
2
,
= 28.512N/m ·. Ans.
Ii
~ I
IL
Viscous Flow
Problem 9.3
393 1
A lamillar flow is raking place in a pipe of diameter 200 mm. Ti,e maximum I'elociry
(If wl/ieb Illis OCCllrs. Also calculme Ih e I'e/Oeity at
is 1.5 mls. Find the mean \'elocil), olll/lile Tilt/ius
oJ em from Ihe wall of II,e pipe.
Solution. Given: Di a. of pipc. D ~ 200 mill = 0.20 In
U""" '" 15 m/s
find ( il Mean ve locity, u
( ii) Radius at w hi ch
U OCC lirs
(ii,) V~ locil y al 4 em frolll the wa ll.
(i) l\ lc3n n,' ]oc ily,
ii
u~, '"
Ratio of
or
2.0
.!J. = 2.0
II
( ii) Ra diu s a t wh ich
1.5
II '" = 0.75 m/s . Ails.
2.0
II
II (X;CU P.;
The ve loci ly. u, at an y radius 'r' is g ive n by (9 .3)
ap
ax
I - IR2 - r 2 1= u= - 4 ).1
- 1 -ap H1 [ 1- -" ,
4).1
ax
1
N-
BUI from equation (9.4) U""" is give n by
u
= __,_ aPR!
Now. [h e radiu s
r
al which
u
.•. ( 1 )
4 ).1 (J.r
"""
ii
==
= 0.75 m's
~: [",J0.5
~2
0.1
f =
0.1
x.J5 '" 0. 1 x .707 '" .0707 III
'" 70.7 mm. Ans.
(iii) Vclocil f HI 4 em rro m Ihe wull
r= R - 4.0= 10 - 4,0 = 6.0 em = 0.06
I I
n1
Ii
~ I
IL
1394
I
Fluid Mechanics
The velocity al a radius", 0.06 III
or
I
4 em from pipe wall is given by equation (I)
__
R~
r
.1__ . _ _
"U-,HJl="HO,6),] -
,.,,=
== 1.511.0 - .361 == 1.5 x .64 '" 0.96 m/s. Ans.
F ig . 9.4
Problem 9 .4 Crllde oil of ~ == /.5 poise and re/mil'l! densit), 0.9 flows Ihrougll a 20 mm diameter
I'crlical pipe. Tile pressure gaugesji.\ed 20 In apart read 58.86 Nkm : milt 19.62 Nkm ] <IS $110 ...11 in
Fig. 9.5. Filii! Ille direcliol1 ami mle of flow [llfOllgl! IIII! pipe.
I.S
1
10
== O. J 5 Ns.im
Solution. Given:
)l"
Relative de nsity
Density of oi l
Dia. of pipe.
=0.9
'" 0.9 x 1000 '" 900 k. g/m)
D = 20 mill = 0.02 111
1.5 poisc "
L" 20 III
PIl == 58.86 Nfcm-, = 58.86 x 10 Nfrn"
Ps = 19.62 Nfcm 2 = 19.62 x 104 N/ml.
"
Find (il Di rect ion of fluw
(ii) Rate of flow.
(I) Dirulion a rnow. To find th e diw:tion or flow, the IOwl energy ( L +
pg
~
+ z ) allhe lower end
2g
A :tnd Jt the uppe r end B is to be cakulmed. The direction uf fluw wil l be give n from th e higher ene rgy
to the lower ene rgy. As here the diJmeter of the pipe is same and hence kin etic energy at A Jnd B will
Ix: same. Hence to find Ihe d irec tion of flow , ca lcu la te
Taking Ihe leve l al A as dal uill. The value o f
A
(~ +
z)
at A and 8.
('.'g' + zJ "
= l!..4. + Z~
6x l0 4 x9.81
900x9.8 1
B- -
(:g+ Z)
<II
58.86 N/cm
,
I--!- 20mm
Fig . 9.5
=
As lh e value of
2
19.62 Nlcm
+0 1·: r=900kg/cm~1
= 66.67 III
The value of
-
>Om
.g
=
l-
2 xIO· x9.81
900x9.8 1
+ 20 = 22.22 + 20 = 42.22 m
(:~ + z Jis higher at A and Ilene" flow lakes place from A 10 B. An s.
(ii) Ral., o f flow. T he Joss of press ure IIcad for viscous flow lh rougll circular pipe is g iven by
I I
Ii
~ I
IL
Viscous Flow
For a vertical pi pe
395 1
/1/ '" Loss o f pc izo mclric head
'" ( : ; +ZA
24.45 "
u ==
"'
J-(:;
+28) '" 66.67 -42.22 '" 24.45 III
32 xO.l5xu x20.0
i
900 x 9.8 [ x (.02)"
24.45 x 900 x 9.81 x.()()()..I
32xO.l Sx20.0
== 0.889 ,. 0.9 mls.
The Rey no lds number stlOuld be calcul ated. If Reyno ld s number is less llian 2000. the flow w i ll be
laminar and th e above ex pression for loss of pressure head for laminar flow can be used.
Now Reynolds number
where
~
p" 900 k g/Ill) and V",
pVD
--
"
w
=900x O.9x.02 = 108
0.15
As Rey nolds number is less than 2000. th e flow is laminar.
" average velocity x area
Rate of flow
Rey nolds number
It
"II X "4
,
1t
D-=O.9X "4
X
'3
(.02)- m Is: 2.827 x 10-
43
10
Is
'" 0.2827 lil res/s. Ans.
Problem 9 .5 A fluid of "iscosily 0.7 Nsf",' (lnd specific grlll';/Y 1.3 is flo .... ing through" cireulM
pipe of diameter 100 "'III, The maximum silear Siress 01 Iile pipe wall is gi,'en OJ' 196.2 N/", 2, find
(i) Iile pre~'suf<' gmdie,lI, (ii) Iile ,,,'erage "'!/Ocily. alld (iii) Reyn old.f II I,,"ber of Ille flail".
N,
I.l = 0.7-,
Solution. Given:
Density
Dia. of pipe.
Shear Slres.~,
Find
(i) Pressure
m
Sp. gr. = 1.3
"1 .3 x 1000= 1300kglm 3
D = IOOll1m =O. llIl
t o = [96.2 NflT/
gr~dienl ,
dp
d.,
(ii) Ave ra ge ve locit y. II
(iii) Reynolds numbe r. R,
dp
(/) P ress ure g rad ie nt , r/x
The maximum shear stress (to) is gi ven by
dp f)
dp 0. 1
dp H
to=- - - O< [96.2= - - x - = - - x d.( 4
4
dX 2
,I,(
I I
Ii
~ I
IL
1396
Fluid Mechanics
op
196.2 x 4
,
'" -7848 Nlm' perm
0.1
Pressure Grndicnt:: - 7848 N/m l P"I' m. An s.
-
ax -- -
(ii) A \,.,rHge ""]<lcily,,,
~ _I ,(~aPlo'
- &!-t
d.f]'
1
,
:: 8xO.7 x (7848) x (.OSt
:: 3.50 Illfs
(iii) Rernold s numbe r , R,
N, = uxD ", ,,xD ",pxllx D
\'
III P
I.!
'" 1300 x 3.50xO.l '" 650.00. Ans .
07
Problem 9 .6 IV/uu power i,~ required per kilomdre of (I {rIle /0 OI'ercome (lie d.~co,,¥ re,~i.~/a'ice /0
the flow of glycerine lilrollgil II iloriZOIl/{11 pipe of diwlle1er 100 mm m (he rale of 10 Iilre,lIs ? Take
).l '" 8 puise fwd kinematic I'isn!Sily (I') '" 6.0 slakes.
Solution. Given:
Length of pipe.
L '" 1 km '" 1000
Dia. of pipe.
D= IOOlllm =O. l m
Di~hargc.
Q = 10 litrcsls '"
Viscosity.
.
8 Ns
1
1-l=8POISC=- - l ::O.8Ns/1ll
10 m
Kincr11mic Viscosity.
v= 6.0 Sluk.cS
III
~
1000
m3 /s:: .01 m 3/s
IpoiSC=-.!....NS/m l )
10
:: 6.0 en/Is = 6.0 x 10- 4 m"ls
Loss of pressure head is given by equ ation (9.6) as lit "
32'W~
pgD-
Power required == IV x I,/ watts
where IV = weight of oil flowing poer sec = pg x Q
SUbstituting the va lu es of Wand II/ in equation (i).
Power required
""
I I
== (pg x Q) x
(32 "L)
, wmts"
pg D-
...( i)
Q x 32 pUL
,
D-
(can(;elling pg)
u == ~= ~ = .Olx4 ==1.273 m!s
Area
!!..Dl
4
Itx (.l)'
Ii
~ I
IL
Viscous Flow
397 1
.0 1 x 32 x 0.8 x 1.273)( ]()OO
Power required
(.I )'
'" 32588.8 W" 32.5!SH kW. Ails.
~
9. 3
FLOW OF VISCOUS FLUID BETWEEN TWO PARALLEl PLATES
10 Illis ca se also, Ihe sllcar stress distribution, Ih c vdocily di stribut ion across a sec tion; Ihe ratio of
maximum ve locity \0 average ve locit y and diffcrcnc~ of pressure head for a give n length of parallel
plates, aTC to be calculated.
,
PARALLEL PlATE
-I-----~-"";...---~
DIRECTION
OF FLOW
-
1
,
Fig. 9.6
T,
I--<
PARALLEL PLATE
"
ViscOUJ flow bU'U'f't'n two paralld plall'S.
Consider two parallel fixed plates kepI at a distance '( apan as s ho wn in Fig. 9 .6. A viscous fluid is
flowing between these two plates from lcft to ri ght. Consider a fluid cJcmC(1{ of length AI" and thick ness Ily at a diMancc), from the lower fixed plale. If p is the intcnsily uf pressure on lhe f<lcc AD of lhe
fluid clemen! then ill1ensity of pressure on the face CD will be
(p + ~ I1r). Let 1: is the shear suess
acting on the face BC then the shear stress on the face AD will be
(1: + ~ Il.Y}
If lhe wid th of lhe
e lement in the d irection perpendic ular to the paper is unily then the forces aCling on the fluid element
arc:
1. The pressure force. P x Il.y x I 011 face AD.
2. The pressure force.
(p+ ~ Il....)
fly x 1 011 facc CD.
J. The shear force. 1: x Il.x x I on face Be.
4. The shear force. [r + dr Il.yj AI" X I 011 face AD.
a,
For steady and uniform flow. there is no acceleration and hence th e resultant force in the direc.tion
of flow is zero.
pll.yX I - (P + QP Ar)Il.YX l - tArx 1 + It+ dr ll.yjAxx I =0
(hay
_ dP Il.xll.,. +
dX
~ ll.yAr = 0
ax
Dividing by rull.y. we gel _ dP+at",Oor
a.1" ay
I I
... (9 .7)
Ii
~ I
IL
1398 Fluid Mechanics
(0
Ve lodt~'
""
stress T" fl -
"y
Integrating
Dist ri bution. To obtain the ve locity distribution across a section. th e value of s hear
from Newton's la w of viscosity for laminar flow is substi tuted in equation (9.7).
th~
above equation w. r.l. y, we gel
au I af>
- "' - - y + C
fJy
)l ax
I
f.,. ap isco nSlant)
1 ax
ap
I
)'~
u= --_·_ +C,)' + C2
).l ill 2
.
lmcgmting again
...(9.8)
where C, and C1 arc constants of integration. Their values ;lrc obtained from th e twO boundary condili ons lhJt is (0 at)' = O. 1/ = 0 (ii) at y '" I. II = O.
The substitu tion of
y = O. II = 0 in equ ation (9.8) gives
The substi tution of
0= 0 + C 1 X 0 + C2or C! " 0
Y = I. I. = 0 in equation (9.8) gives
I iJp [ 1
0= - - - + C, x 1+0
).l
ax
2
___'_ "P,
C _ _ ~ap ~ -
,-
f'Qx2XI
2)JJx
Substituting the va lu es of C, and C 2 in eq uat ion (9.8)
11= _'_ iJp
21-1 (Ix
I
II
l +y[ __'_'r,]
ax
2ft
ap
,
a( -
= - - - ltv - y-1
2)1
...(9.9)
'""'
In the aoove equation. )1.. -::;- and t are constant. It means Ij varies with the square of y. Hence
equation (9.9) is a equation of a paraoola. Hence velocity distribution across a section o f the parallcl
p late is paraoolic. This \'clocity distribution is show n in Fig. 9.1 (a).
1- to -I
-
- -..
Fig. 9.7
(&)
,0)
Vl'locity diuribution and Jbl'ar Jtrf'JJ diJtriblltion
a Jl'ction of parallel plates.
acrOIJ
~ I
I~
~ I
IL
Viscous Flow 399 1
(ii) Ratio of Maximum Vl'loc ily to An' rnge Ve locit y. The velocity is maximum. when y '" rIl.
Substituting this value in equation (9.9). wc gel
+- ap [I x+-('-]']
=_ 2~ ~~ [I; _t: ]=_ 2~ ~ r: =_ ~ ~/;
U""" '" -
d.(
. )1
_
2
,2
... (9. 10)
The average ve locity, ~. is obtained by dividing the d ischarge (Q) across th e section hy Ihe area of
the section (/ x I). And the d ischa rge Q is obtained by considering the ralC of flow of fluid through Ih~
strip of thickness Ify and inlcgrating it. The ralC of flow through strip is
dQ'" Velocity
al
a distance), x Area of strip
ap
J
2
= - - - [l), - y 1xdyx 1
2).1 ,h
_
-
Q
"=-~ = -
dP.1 J
12jl ih
Area
Ixl
I dp 1
=- ----,
[2)t ax
... (9. 1 I )
Dividing equation (9.10) by cqu;uion (9. 11 ). we get
I all 1
----,
U'!'." '"
II
_''1"'-' 'ap,
,',:- =8='2
12 3
1
... (9. 12)
- - -r
12j.l
ax
( iii) [)ro p uf Press ure h"Hd for" gh'e" Le";;t h . From equation (9. 1 I). we have
-
1
ap
u = - ---,
1211
Inte grating this equation w.r.1- x. we get
1
ax
I'
1
I I
ilp=
II
1
12)Ju
- , ~ dx
I
Ii
~ I
IL
1400
Fluid Mech a nics
(!J
12~IUL
PI - P2 '" - ,- ,-
.
!J '" PI - 1';
... (9 . 13 )
pg
(it·) S hear S tress Distribut io n . II is obtained by su bstituting
•
"
•
•
•
<
.
«
,
•
•
•
~<
••
Ihe va lu e of u from cljuat ion (9.9) int o
.
'. - -
d ~ l~
"
,
~-"
C_
'
I_
"J""
•
<
.
<
El.:
If ht is Ihe drop of pressure head. th e n
I
,
•. ' --
'
,
"
- -_ , ,
Fig. 9.8
1:11 -
t=11
dU
dy
I
ap
2).1
a.1
2).1
ax
.,.(9. 14)
,lIld, ,m: l"OnSlnnt. I'len<:e t
\'~rics Iincarly willi y. The shear S1ress diSlri bU1ion
2
~:
()y
11-2)'J
l=--~
In eq uation (9.1 4),
__
= fl~ [ l dP(ly _y1l]."[ __1 OP(/ _ 2Yl ]
ax
-
is show n in Fig. 9.7 (b). S hear stress is n1a!limul11. w hen y = 0 or I m the w all s uflhe plates, Shc~r stress
is zero , w hen y '" 1/2 Ihal is al the centre line between Ihe (WO plates. Max. s hear Slress (t o) is given by
l op
t o "' - - - t.
. .. (9 . 15)
ax
2
Calcu/me: (i) tile pressure grailielll along flo ..... (ii) l/ie a\'erage !'e{oeity. (wd
(iii) Ille discharge for Oil oil of !'iseasil), 0.02 Nslm ! j/O"'ing belween 1....0 SlatiQnary ptlralld plates I m
wide m(II' II"i»ed 10 mm tlfwrt. Tile !"docily mid ....ay bel,..enl l/ie plates is 2 mA
Problem 9.7
SolutIon. G ive n
IJ == .02 N sJm1
Viscosity.
Width.
b = I 111
Distance between plates.
1= 10 111111 = .0 I 111
Velocity midway be tween the plates. U mo , = 2 mrs.
(,) Press ure g rad;"'n!
(dd.,P]
or 2.0 = -
Usi ng eq uati on (9.10).
_'_(dP]
(.01)1
8x.02
<Ix
.2C·O:cx,8~XCC'.O..
2 == - 3200 N/ m , per m. An s .
.01 x.OI
tip ==
d.,
(ii) Ave rage "cloelt)" (;;:)
Usin g eq uati on (9.12),
U ... ,
--. -3
"
(iii) Disc harge (Q )
I I
2
-
2U",,,
3
2x2
3
u == - - - . - - == 1.3 m/s. An s.
-
3
-
,
= Area of n ow Xu = b XIX U == I X .01 X 1.33= .0133 m Is. An •.
Ii
1396 Fluid Mechanks
Proble~
Dt tennine (a) the prtssurt gradient, (b) the shear stress at the two horiumtal parallel
platts and (c) the discharge per metre width for the laminar flow of oil with a maximum Vt!locity of
2 mls betwu n two horiumtal porallel fued plates which are l oo mm apart. Given II = 2.4525 N s/m2•
Solution . Given :
U-.A = 2 mis, t = 100 mm = 0. 1 m. ~ = 2.4525 N/ml
Find
9.8
(i) Pressure gradient .. :
(il) Shear stres at the wall . to
(iii) Discharge per metre width. Q.
(i) Pressure gradient. dp
d..
Maximum ve locity. Um».' is given by equation (9. 10)
U
= __
1 OPt2
ox
mu
~
2.0= -
p
I
x aa x(. I )'
8)( 2.4525
x
Substituting the values
2 0 x8 x 2.4 525 = -ap
ax =- .I x .l
39"~ NI m• perm. Aos.
(ii) Shear stress at tbe wall, to
to is given by equation (9. 15) as to = - 1. op
2 ax
X
t=-
1. (- 3924) x 0.1 = 1.96.2 N/m1, Am.
2
(iii) Dlscbarge per metre wkllh . Q
= Mean velocity x Area
= ~ U_)( (I x 1) = ~ x 2.0 )( 0. 1 x 1 = 0.133 ml/s. ADS.
)
)
An oil of viscosity 10 poise flows betwu n two paraUe/fixed plales which are kept at
a distance of 50 mm apart. Find the rate of flow of oil betwu n the plates if the drop of pressure in a
length of / .2 m In 0.3 NIe,",. The width o/ the plates is 200 mm.
Solution. Give n :
~ = 10 poise
Problem 9.9
=..!..2. N s/m2 = I Nslm1
10
t=50mm=0.05m
PI - PI = 0.3 N/m2 = 0.3 x l et N/ml
L = 1.20 m
B = 200 mm = 0.20 m.
Width.
Find Q. rate of now
The difference of pressure is given by equation (9. 13)
IN')
.
( .: I IXHse= iO m 1
~ I
IL
1402 Fluid Mechanics
1 2 ~wL
PI - P2 == - ,- ,-
Substitutin g th e values,
w" gel
0.3 x 104
:
11 =
12x l.Ox
ux 1.20
.05 x .05
03 x 10 ' x 1.0 x .05 x .0 5 == 0.52 rn ls
12 x 1.20
= ii x Are a == 0.52 x (8 x /)
= 0.52 x 0.20 x .05 111 3/s = .0052
Rate of fl ow
1I1
3/s
1
== 0.0052 x 10 litreJs " 5.2 lit re's. A ns.
Problem 9.10 Wa/a at ISoC flows belweell two l(!fge parillie/ plales ill (l dis/arlce of 1.6 mm oparl.
D t!lermille (iJ ,lie mluimll'" !'e/ocir)' (ii) lile pressure drop per Illii/lenglil and (iii) Ihe shear slress (l/
Ihe "'(I /Is of Ihe p/lUes if Ihe Il\"emge veiocily is 0.2 rnA Tile "iscm-;Iy of water (II Ij OC is gi"en as
0.01 poiJe.
3
Solution. Gi\'cn :
t= 1.6 mm = 1.6 )(1 0 m
= 0.00 16 III
ii
.
== 0.2 m/sc\: , )l == .0 1 PO ISC
=
.OJ
ru
==
0.00 1
(i) Max imum ,·clocity. Un", is give n by cqu ,ni on (9. 12)
i.e.,
U" ....
3== "2"
= 1.5)( 0.2 = 0.3 m/s. Ans.
( ii) The prt."Ss ure dn.p . (P, - Pl) i ~ given by t'qu3t io n (9. 13)
12 ~I UL
PI - Pl= ~
or
prc~sllre
L211U
drop per un it le ngth = - ,-
r
' " - I 2 x .01 x -;-~0~.2~ = 937 .'N
44 Im ' pcrm .
-10 (.eX) 16i
ox -
01
(iii) Shea r s tress :.tthe walls is g ive n by equati on (9.I S)
I op
I
,
t o=- - - X I = - x 937.44 x .00 16 = 0.749 N/m ". Ans .
ox
2
2
Problem 9.11 TilerI' is a ilorilOllla{ c rack 40 111m ... ide and 2.5 mm deep ill a ...al/ of Illickllf'ss
100 mill. W<lIer leab' I/lrougil rile crack.. Find rile raIl' of lea kage of WlIIer Ihroug/l IIIe c rack. If Ihe
difference of pressu,e bel"'eell IiiI' 1"'0 endI of Ihe CTack iJ' 0.02943 Nkm 2. Take IiiI' "iscosil}' of ...ater
1',/11(1//00.01 poise.
Solution. G iven :
Width o f crac k.
D~plh of crac k.
Le nglh o f crack.
I I
b = 40 mm == 0.04 m
1= 2.5 nu n = .0025 111
l. = 100 111m = 0. 1 m
Ii
~ I
IL
Viscous Flow
403 1
/II - 1'2 '" 0.0294) Nlc m 2 '" 0.02943 x 104 N/m 2 '" 294.3 Nlm!
J..l
.0 1 Ns
10 Ill ·
= .0 1 poise = - ,
Find rate of kak agc (Q)
(PI - (2) is g iven by equa tion (9. 13) as
121l;;L or 294.3= 12x :Q!x
u xOJ
JO (.0025)(.0025)
PI - /l2= - , -, -
294.3 x 10 x .0025 )( .0025
II '"
Rate of lea kage'"
'"
12 ><.0 I x O.l
ii
1.5328
ill 'S
x area o f cross-section of crack
= 1.538x(bx/)
'" 1.538 x.04 x JXl25 rn J/s = 1.538 x [0- 4
111 3/s
= 1.538 x 10-4 x 10) litre's = 0.1 538 litre/s. Ans.
Problem 9 .12 The rIU1,,,{ c/ermmce be/wee" (I flyi/mulie plullger (md 'he cylinder walls iso.l mm:
the lenglh of lilt! plunger is 300 mm (lml diameter 100 111m. Fi"d the I'e/oeiry of leakage and mil.' of
iellkage plm the plunger at GIl ;IISWIlI wilell Ihe difference of Ihe press lire be/ween the IWO ellds of Ihe
plunger is 9 lIT o/lI'aler. Take /.I = 0.0127 poise.
Solution. G iven:
The flow through the clearance area will be the same as the flow between two parallel su rfaces.
I" 0. 1 mm ,,0.0001 m
L,,300mm:O.3m
Diameter.
D" IOOmm=O. lm
"
Differe nce of pressure
PI - P,
- " 9 mof waler
pg
PI - P2 : 9 x 1000 x 9.8 1 Nfm 2 " 88290 Nfm!
Viscosity.
Find
' . _ .0127 Ns
Il "'.0 1'7
_ po ise - - - ,
10 III
(i) Veloc ity of leakage. i,e .• mean ve locity
(ii) Rate of lea kage. Q
(i) Velocit y o f leaka ge
u
(i/). The average ve loc ity (i/) is given by equati o n (9. 11)
-
I()p ,
u: -----c
121.1 dol
I
PI-P,
x (.000 I ) x (.000 1)
= -~",
01
27
12x·
L
10
-
=
I
12 x.0127
88290
x - - x(.OOOI)x(.OOOI)
OJ
= .193 mfs == 19.3 cm/s. Ans.
I I
Ii
~ I
IL
1404
Fluid Mechanics
(ii) R:tte of leak;lge. Q
Q " ~ x area o f flow
,,0. 193 x nD x I m3/s= 0. 193 x II x.I x .0001 1I1 31s
=6.06x I O-6 m l fs = 6J)6 x 10-6)( IO] li1r(~ls
= 6.06 x 10-.1 lit,..,/s . A n s.
to 9.4 KINETIC ENERGY CORRECTION AND MOMENT UM CORRECTION
FACTORS
Kinetic energy correct ion factor is defined as the rat io of the kinetic energy of the flow pe r second
bascl! on acmal velocity across a sectiun to the kinetic energy of the flow per second based 011 avef'Jgc
velocity across the same sec tion. II is denOTed by 0.. HelKe nwthcnwtically,
"~
K.E.lscc oo
based on actual ve locity
:eO'C''C
'=.CCCCCCOOCC'7'C'K. E'/scc based o n average ve locity
... (9. 16)
Momen tum Cu rredlo n I' Hct Ot . It is defined as the rati o of momentum of th e flow per second
based 011 actual vcloc il y \0 the momentum of Ihe flow per second based 011 average veloc it y across u
section. It is denoted by ~. Henc~ mathematically.
~
= Momen tum per second based 011 actual vd ocity .
Momen tum pe r second based o n avera ge velocit y
. .. (9. 17)
Pro b le m 9.13 Show I/",//he momentum co rrection jador ""'{ energy co rrection judor jllr [llminar
flow /hrough" cirClll'" pipe lire -II) WId 2.0 respec/h·e/y.
Sol u llo n. (!) I\ lomc ntum Correction Factor or ~
The vciocity distribution through a circ ul ar pipe for laminar now at an y radius r is given by
equation (9.3)
"~ _' (_dP)
4~
(R2_?)
d.l
... ( I )
Consider an elementary area dA in the form of a ring at a radius r and of width dr. then
rlA = 21(r dr
-Fig. 9.9
Rate of fluid flow inll th rough Ih", ri ng
= dQ = velocity x area of ri ng clement
=ux2ltrdr
Momentum o f Ihe fluiu Ihrough ring per second
= mass x velocity
= p xdQ x I j = P x 2ltrdrx II x II = 2ltp u 2rdr
TOlal aClual momentum o f the fluid per second across the secti on
=
I I
Jr' 21(p 11 ,r tiT
o
Ii
~ I
IL
Viscous Flow
405 1
Substituting Ihe value of u from (I)
=
"',(")'[R'" +~_ 2R"']'
('I')!
+
<l.l
8W
2
a.T
'" IlP,
81-1"
6R
6
6
4
2R6 _ 6R
= PI',
0
('I' )'[~+~
_ 2R'
]
2
6
4
81l";h
6
12
- PI' ( ' , )"
- Sil l ax
x ~-....!2...(,p
)'
6 - 4&/1 1 ,h
R'
.. .(2)
Mome ntum of Ih e fluid per second based on ave rage veloc ity
=
mass of fluid
,
x ,we rag c ve locit y
~,
'" pAu
xii", pAUl
where A= Area uf cross-section = 7tR". Ii
'" average
velucity '" Umo,
2
--' (-"ax )R'
- RI-l
Momentum/sec based 0 11 average ve loc it y
'" p
x 7tHl x [ _ '
811
(al')
R, ] l '" P X 7tHl
ax
x 64' , (_
~1
~p)l R4
(IX
... (3)
p=
I I
MomCllIum I sec based on actual ve locity
Momentulll I sec based on average velocity
Ii
Viscous Flow
=
is?(:r,R'
(a
= --ax
!tp
p)
64~
R
= 64
, 48
401 1
=~.A"'.
3
(ii) Energy CorrtCtJon Factor. (L Kinetic energy of the fluid flowing through the elementary ring
of radius',' and of width · dr' per sec
=.!.. x mass x u2 =.! xpdQx';
2
2
=.1.. x p x (u x 27fr dr) x ,i = ..!..p x 21tru ] dr = 1tpru1 dr
2
..
2
Total actual kinetic energy of flow per second
=S: "p",'dr= S: "pr [4~ (-:)(R'- r')r dr
=!tp x
=
[4~ (- ::))' S: [R'- ?J' rdr
...!2...(_ ap )' [R' _R' _3R' + 3R']
64~'
ax 2 8 4 6
=...!2...(_ap )' R,[ 12-3-18+12]
64 .... 3
=
ax
...!2... (_ ap )' R'
64~l
ax
24
8
Kinetic energy of the flow based on average velocity
I
-z I
- -2 1
-J
=-xmassxu =-xpAuxu = -x pAu
2
2
2
A = (.if)
.
Substituting the value of
and
...(4)
1402
."-
Fluid Mechanics
Kinetic energy of the now/sec
=
.,
.
i x p x rul'X[8~ (- ~) R'r
I
= -x
p x ~x
2
I
64 x 1Ij1'
(il)'
_...1!... x H'
ilx
(ilp)'x
R'
ax
_
p.
- 128 x8~ 3 a=
K. EJsec based on actual velocity
Equation (4)
=
K.EJsec based on average velocity Equatiofl (5)
...!!£... (_
='
64~1
p
q8x8iJ l
~
9.S
... (5)
ilp )' x R'
ax
J8
P) II'
(d
- ax x
= 128 x8= 2.0.Aru.
64x8
POWER ABSORBED IN VISCOUS FLOW
For the lubrication of the machine pans. an oil is used. A ow of oil in bearings is an example of
viscous now. u a highly viscous ~i1 is used for lubrication of bearings. it will offer great resistance and
thus a grealCr power loss will take place. But if a light oil is used, a required film between the rotating
pan and stationary metal surface will not be possible. Hence. the wear of the two surface will take place.
Hence an oil of correct viscosity should be used for lubrication. The power required 10 overcome the
viscous resistance in the following cases wiU be detennincd :
I . Viscous resistance of Journal Bearings.
2. Viscous resistance ofFoot·step Bearings.
3. Viscous resistance of Collar Bearings.
9.5.1 Viscous Resistance of Journal Bearings. Consider a shaft of diameter D rolating in a
journal bearing. The clearance between the shaft andjoumal bearing is filled with a viscous oil. The oil
film in contact with the shaft rotates as the same speed as that of shaft while the oil film in contact with
journal bearing is stationary. Thus the viscous resistance will be offered by the oil to the rotating shaft.
Let
N = speed of shaft in r.p.m.
t = thickness of oil film
L = length of oil film
21CN
..
Angular speed of the shaft, ro = 00
..
Tangenllai speed of the sbaft= m x R or V = - - x - = - 60
2
60
,
' ••• 'I ' '
b
d.
The •••••
~ ....... stress In UI'C 0 1 IS given y;t = ~ dy
21CN
D
.DN
Viscous Flow
OIL
403 1
......-1 .
Fig. 9.10 JOllnlal btttrilig.
As the thickness of oil film is very small . the velocity distribution in the oil film can be assumed as
linear.
du
V -0 V 1fDN
Hence
dy
= - - = - = 60 X I
1tDN
t=IJ.
60 X1
Shear force o r viscous res istance = t
X
Area of surface of shaft
~1tDN
Illt' D' NL
= - - x rcDL = =~=
60t
601
Torque required to overcome the viscous resistance.
.
D
·
T = YISCOUS resIStance x 2
= JUt D l NL x D = !:~::",.;D:,J.:.:
N:::L
60t
2
t20,
1
Power absorbed in overcoming the viscous resistance
.p =
'E:!!!... = 2rcN X 1J.1[1 DJ NL
60
.
60
1201
JUtlDJN1L
= 60 60
watts. Ans.
x
xI
... (9. 18)
Problem 9.14 A shaft having D diam~l~r of 50 mm rotates centrally in ajoumol bearing having a
diamelt!r of 50. / 5 mm alld length tOO mm. The angular spaCt! ~lWun the shaft 011(1 the bearing is
filled with oil having viscosity 0/0.9 poiu. Dcruminc the powt!r absor~d in the inoring when the
spud of rotation is 60 T.p.m.
Solution . Given :
D =50 mmor .05 m
Dia. of shaft.
0 1 =50.15 mm o r 0.0501 5 m
Dia. of bearing.
L = 100 mm o rO. 1 m
Length.
2xN 2ttNT
21fNT
·Power P""T XOl "" T x - - - - - watlS: - - kW.
.
60
60
60,000
~ I
IL
Visco us Flow
409 1
09 poise= -O.9Ns
-,
10
~oo
f il =.
m-
N = 600 r.p.m.
Power= ?
Thickness of oi l film .
, ~ _D_, -_D
_ ~
2
~
Tange nti a l speed of sli afl.
Shear stress
Shear force (F)
"'0".,,":-2-,,',,0
0.15
- , - = 0.015 mm = 0.075 x 10
V = nON '" c'cXcO~.O~5,Xc600=
6IJ
6IJ
d,
t =Il - ~"
dy
Powe r
= 0.5 x IT
III
Illfs
;;-;;O~5~X::..;;'",
0.9
-; = -10- x 0.075 x 1O ~_1
= 1883.52 N/m2
= t X Ar~a = 1883.52 x itO X L
= 1883.52 x
Resistance torque
V
3
T= F x
7[
x .05 x 0.1 = 29.5&6 N
~ = 29586 x .~5
= 0.7387 Nrn
= 2 /tNT '" 21£ X 600 x 0.7387 = 46.4 1 W. An s.
6IJ
6IJ
Problem 9 .15 A silaft of /00 111m. diameter rQlaleI m60 r./I.m. in a 200 mm 10llg bearing. Taking
IIwl IIII' two surftlces are uniformly separated by a distance of 0.5 111m alld laking linear \'e/ocit)'
tfislribulioll ill I/le {'tbr icalillg oil IIm'illg dynamic ('; scosiry of -I cell(ipoiu s, find Ille power ab wrbed ill
tile beoril lg.
Solution. Giv~n :
Dia. of shaft.
Length of bearing,
D = I()() mm = O. l m
L = 200 unll = 0.2 11\
I=O.5mm=.5x 10-3 m
.
0.04 Ns
1.1 = 4 ce ntipoisc = .04 potse = - - --,
10
Ill '
N = 60 r. p.m.
Find power aowrbed
Using equation (9. 18),
.04
~ -x
It l X(.I») X (60)1 xO.2
10
6Ox60x0.5xlO ~
J
l
: 4 .% 1 )( 10- \V. ADS.
Problem 9.16 A ~'''afl of diameter 0.35 III rotales 1II 200 T.p."" ;"side a s/ee,'e 100 Ill'" [ollg. Th e
d)'llllmic ",'scosily of [llbricalitig oil ill Ihe 2 mm gap bellt'<'lm J'lee,'e ami J'/u'f/ is 8 poi,\'es. ell/cIl/are Ille
{lower [OJ'/ ill Ille beMing.
Solution. Gi ven:
Dia. of shaft.
I I
D = 0.35
111
Ii
~ I
IL
1410 Fluid Mechanics
N", 200 T.p.lll.
Speed of shafl,
L= l00l111n :O. 1 111
Length of s lct:\'c.
Distance between sleeve and shaft. I = 2 mill '" 2 x I 0- 3 III
Viscosity.
11 '" 8 poise '"
..!.
N~
10 m"
The power lost in [he bcaring is given by "qU31ion (9.18) as
P=
Illt l 0 ) N ~ L
60x60xl
8
to
walts
rr3 x (.35)) X(200)2 xO. 1
= -x
6Ox60x2xlO
= 5'X).!1 W = 0.59 kW. A ns.
1
"'II!. is n",n;ng {II !Zoo f.p.m., is
Problem 9 .17 A slee,'c, in which (I sh aft of di(llnder 75
radill/ clearance 0/0.1 mm. Calculate Ihe torque resi~'I"'lce
splJCe is filled wilh oil of d)'llllmic I'iscosily 0.96 poise.
Solution. Given:
Dia. of shaft.
0= 75 mm = 0.075 m
N = 1200 T.p.lll.
/=0.1 mm =O. l x [O-3 m
L = 100 Ilun = 0. 1111
Le ngth of sleeve.
11
nON
/t x .075x 1200
60
= 4.71 2 Illls
~
Shear stress.
!l - == -
Torque resis tan ce
iI(ll'ilig (I
length of ~'Ieel'e is 100 mm ami Ihe
'
0.96 Ns
=.096 poise
= - - --,
10 m "
Tangential ve locity of shaft. v=
Shear force,
if Ihe
t =
'"
V
.96
I
10
x
4.712
.lx lO J
,
4523.5 Nlm -
F =t xnDf.
= 4523.5xnx.075x.I '" 106.575N
= Fx 0
2
= 106.575 x .0;5 = 3.996 Nm. Ans .
Problem 9.18 A slrap of / 00 mm dim"eleT r""$ in a beMin!; of lenglll 200 mm willi II radial clear·
ance ofO.o25 mill al 30 r.p.lII. Filld Ille "elocily of IIIe oil, if ille power reqllired 10 overcome I/Ie l'iSCOliS
,e.\·i.~tallce is 183.94 'i'tlll.~.
Solullon. Gi ven:
D" loornm =O.1 111
L " 200 mm " 0,2 m
3
I " .025 111111 " 0.025 x 10 m
N" 30 r.p.rn . : H.P . = 0.25
Find viSl:osity ofoil.).t.
The h.p. is given by equat ion (9.18) as
I I
Ii
~ I
IL
Visco us Flow
"'
18 3.')4 ==
).In
, x( J ), x(30 )', xO.2
60 x60 xO.025x 10
411 1
J
133.94 x60x60x.025xlO- J Ns
It
=2.96
J
x .001 x 900 x 0.2
N~
m-
=2.96x 10= 29.6 po iw. Ans.
9 .S.2 Viscous Resistance of foot -Step Bearing. Fig. 9.11 shows Ihe foot-step hearing. in
which a vertical shaft is rotating. An o il film between the oonorn surface of the shaft and bearing is
provi ded, 10 reduce tile wear and lear. The viscous resistance is offered by lh" oil 10 [he shaft In this
case Ihe rad ius of the surface of the shaft in conlact willi oil is not conSlant as in the case of the journal
bearing. Hen!;c. viSl.,()us resistance in foot-step bearing is calculated by consid ering ~n eicmentary
circular ring of radius r and lhickne~s dr as shown in Fig. 9.1 I.
LeI
N", speed of the shaft
I = thickness ufuil film
R '" radius of the shaft
Are a of the ele mentary ring
'" 2rrrdr
,
Now shear stress is given by
d,
V
d)'
I
I = IJ-=).I~
t-*=1~;;;;~;;;;:;;;K
where V is Ihe tangential vclocily uf shan al radius r and is equallU
2~
Oil
WXr== - - H
6()
Shear force on the ring == dF == t
X
area of elementary ring
2 ttN
r
).l n l Nr l
= J.1 x - - x - X 2nr <lr '" - - - - d,
60
15,
I
Fig. '1.11
Torque required to ovacolllc the viscous resistance.
dT=dFxr
Foot-step beari llg .
ft n-N,"
)1' 3
" dr x r = n-Nr dr
151
151
Tot allUTque required to overcome the viscous resistance.
T=
fR1fT", j RL
Jo
0
n2N?dr
151
R
'" L
n2N r ? dr= L n2N
15/
Jo
15/
'" L
6(),
Power absorbed.
[~lR
= LnlN~
4
151
4
n2NR 4
0
...(9. 19A)
'«NT
p" - - - watts
60
2nN
2
~ = '"o
'
" - ' - J.1 nNR
- '"'"N",'R;:,.
60
601
60 X 301
I I
... (9. 19)
... (9.20)
Ii
~ I
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1412 Fluid Mechanics
Problem 9.19 find the torque required IQ rOlale II l'utieai J'lwft of diameler 100 mil! af 750 r.p.m.
TIlt' lower ''fId of IIII' shaft reSIS ill a F;ot-slep bearing. The end of 1/11' slwfl and sur/oce of the bearing
are holh flat al1</"re Jeporllled by WI oil film of IIlid:ness 0.5 mm. The \';swsily of IIII' o il is gil-en
1.5 poise.
Solution. Gi\'~n :
Dia. of Shaft.
D = l00 l111n =O. 1 111
R = D : Q:!. = O.05 m
2
2
N", 750 r.p.m.
Thickness of oil film ,
I = 05 mnl = 0.0005 m
1-1=
The to rqu e required isgivcn by
T=
1.5 Ns
c.
1 " P0l5C= - - ,
10 m "
cqu~lioJl
L
60,
(9.19) as
rc 1NH4 Nm
1.5
= - X
10
IT ' X750x (.05)4
= 0.2305 Nm . A ns.
60 x .0005
Problem 9.20 Find Ille po",er required /0 rO/ale a cireu/a, dis{' of diameler 200 mm at 1000 f.p.m,
Th e c ircIII", disc I/(Is " ckaran ce 0/0.01 mm from Ihe bollom f1" / plale Olllilile c/ea"mce co nlainJ' oil
of\,iscosily / .05 poise.
Solution. Given:
Dia. of di sc,
D = 200 mm = 0.2 m
Th ick ness of oil mm.
R = D '" 0.2 '" 0 .1 m
2
2
N ", 1000 r.p.m.
I '" 0.4 mm = 0.0Cl04 m
C'
1.05
_,2
I! '" I ..
O J poise = - - N ", m
10
The
pow~r
required to mtale the disc is given by equation (9.20) as
1'=
I!n 1N" R4
6Ox30 xI
1.05
10
= -- x
wa llS
n 1 x WOO" x (0.1 /
60 x 30 x .0004
= 452. 1 \V. A n s.
9. 5. 3 Viscous Resistance of Collar Bearing. Fig. 9. 12 shows th e coll ar bearing. where the
face of the collar is separated fro m bearing surface by an oil film of uniform thic kness.
N '" Specd of the s haft in r.p.m.
Let
/(1 '" Internal radius of the collar
I I
Ii
~ I
IL
Visco us Flow
413 1
R 2 '" Ex ternal radius of the co ll ar
I '" Th ickn ess o f o il film.
-<cFig. \1.\2 CQ/larinan"ng.
Consider an c lcn\cmary circular ring uf radius ' r' and wilhh dr of th e bearing surface. Then the
torque (dD required 10 overcome the viscous resistance on Ihe ekmc ntary circular ring is Ihe same as
give n by equa tion (9. 19A) or
dT"
L
",
N,3 df
It!
TOlall0rquc, required 10 overcome Ih e viscous reSiSl<lJlce, on Ih e whole collar is
T=
JR,Iff= JR
, J:..
H, 15r
It!
Nf)
H,
= - "-
151 x 4
tiT: L151411,
nlN [~l"
rr:!N IR,4 -R ~ 1 = L
-
I
601
rr: 2N rR ,~ - R"1
-
I
.. ,(9 .21)
Power absorbed in overcoming v isco us reSistance
p _ 2rr.NT _ 21(N
11
2N [R 4 R " [
- ( j ( ) - 6(j X
1l
2- I
601
=
)JI[ l N
2
60 x 301
•
•
IR2 - HI 1 wallS.
... (9 _22)
Problem 9 .21 A co ltor iJeMing 11I1I'ing e.lIemu/ (llIIl ;Ii(em(l/ dillmelers 150 /Il'" 'lnd 100 m",
respeclhdy iJ' uJ'ed 10 lake Ihe IlnuSI of II slillfl. An oil film of Ihicknen' 0 .25 """ is m(linlllined
beMet'1i IIII' col/ar surface and rhe bellrillg. Filld II,e po ...er 1051 in ol'ercomillg Ihe I'iscous resislallce
...hell the shaft rO/(ltes til 300 r,p.lII. Take jJ = 0.91 poise,
Solution. G iven:
EXlemal Dia , of co llar,
Dl = 150 111m = 0. 15 m
In lerna l Dia. of co llar,
Thickn ess of oil 1111\1,
D, .15
K,,,, - "' - ,,,0.0751ll
2
2
•
D , '" loo mm '" 0.1 III
D, 0 .1
K, ,,,-=-=0.05m
2
2
I '" 0.25 1\11\1 '" 0.00025 111
N = 3oo r.p.m.
0.9 1 Ns
I! = 0.91 poise = - - --,
10 III
I I
Ii
~ I
IL
1414
Fluid Mechanics
The power required is given by eq uation (9.22) or
p.
I-lIT1N
1
-'+::';;C-- IR
6O x 30xI
0.9 1
' -- x
4 _
41
R
1
I
I'l l )(300 1 )([.075" ~.054 l
10
60 x 30 x .00025
'" 56431 4 1.00003164 - .()()()()()625]
'" 564314 x .00002539 '" 14.327 W. Ans.
Problem 9.22 Th e a/ernul (md in(emu/ diameten of 1I CO/lOT heaTing (lTe 200 mm (lnd f 50 mm
respedi"d)'. Be/ween IIII' COIlM surface (llid Ihe hearing. an oil film oJ/hickness 0.25 mm alld of
I'isco:;;'), 0.9 poise. is main wined. Find Ihe IOrf/oe 01111 l/ie power /OJ/ in orereoming Ihe ,·ismus
resist,mc/' of Ihe oil when IiiI' shuft is "",,,ing 01 250 T.p.m.
Solution. Given:
D l '" 200 mm '" 0.2 m
N.
1
", .!2", 0.2
2
= 0.1
m
2
D ,,,, 150mm=O. 15m
o 0.15
RI
= - ' . - - = .075 III
2
2
1 = 0.25 111111
== .(0)25 III
I-l == 0.9 poisc = 0.9 N~
10 m-
Torque required is given by equmioll (9.21)
. ~(
,
4
4 0.9 1t~ X250[O. I ·-.075·]
, . /t"N [R , - H,I = x
Nm
601
•
10
60 x 0.00025
= 14804.4 [ .000 I - .000031641 = 1.0114 Nm. An s.
Power lost in visco us resistance
2nNT 21t x 250 x 1.01 14
• --~=
'" 26.48 W . An s.
6Q
... 9 .6
6Q
LOSS OF HEAD DUE TO FRICTION IN VISCOUS FLOW
The loss of pressure
h c ~d.
II, in a pipe of diameter D, in which a viscous Ouid of viscosity
).I
is
Ilowing with a velocity jj is given by Hagcn Poiscuille formula i.c .. by cq uation (9.6) as
II = 32).111L
,
pgO l
where L = length of pipe
The loss of head due to friction· is given by
,,_ 4 .j. L . V!:. 4 .j. L. ";/
,D x2g
Dx2g
... (ii)
[.: ve locity in pipe is always averag e I'elocity
- For derivmion.
I I
ple~sc
r<:fcr to Art. 10.3.1.
Ii
~ I
IL
Visco us Flow
415 1
where!'" co-cfri cicllt of friction between the pipe and fluid .
32~IUL
Equating (i) and (il). we gel
4./ . L.~ ~
D x2g
pg0 2 '"
16,
3211;;£ x D x 28
4 .L.u' . pg. V '
"
II .p. D
= 16X - ' - = 16 x....!....
pVD
where
fl
I
pVD
R,
--: -
R,
pVD
and N., == Rey nolds nurnb.:r '" - ).l
f=~.
. .. (9.23)
R,
Problem 9.23
Waler is flowing {lnOllS/' (I 200 111m diameter pipe ... ilh coefficient of friClion
2
(j/ a point 40 mm from Ihe pipe a.tis is O.()()981 Nkm . Calcuhlle II,e shear
stress allhe pipe wo/l.
Solution. G iven:
Dia. of pipe.
D '" 200 nlln '" 0.20 m
Co-efficient of Frict ion.
f'" 0.04
Shearslrcssat r= 40m m, t=0.00981 Nfcm ~
Let the shear stress at pipe wall '" t or irs! find w heth er the flow is viscousor not. The flow will be viscous if Rey no ld s number R, is less
(hall 2000.
/=
O.().I. The sileaf Siren'
Using equ ation (9.23). we get/=
R
,
~
R,
0'
""~=400
.04
Th is tneans flow is viscous. The s hear s tress in case of viscous flow tll ro ugh a pi"" is give n by tlw
equa ti on (9.1) as
ap ,
ax 2
t = - --
But
~~
is constant across 11 secti on. Across a sectio n. there is no
variation of p.
\'ari~tion of x
and there is no
,- ,
At the pi pe wa ll. radius", 100 nun and shear stress is to
0'(X1981 '" ~
100
40
100
]()() x 0.009&1
~ '" ..!.!L.
,
"'
40
I I
'" 0.0245 Nfcm
2
,
A n s.
Ii
~ I
IL
1416
Fluid Mechanics
Prob lem 9.24 A pipe of diameter 20 em Oll<llellglil 104 m i,~ laid at a slope of I ill 200. All oil of
sp. gr. 0.9 alii! \'iscosiry !.j {Jilise is pumped lip (1/ Ihe rale of 20 lilres per second. Find Ille f,ead lost
due ro frictioll. Also [(Ileulate the power reqllired to plmlp Ihe oil.
Solution. Given:
Dia. of pipe.
D=2Qcm=2.50rn
Length of pipe.
L = IOOOOm
Slope o f pipe.
i= I in 200:
Sp. gr. of oil.
S= 0.9
De nsity of oil.
2~
p" 0.9 x 1000 "" 900 kg/ml
1.5 Ns
-,
10 m'
Viscosity of oil.
)l" 1.5 poise == -
Di sc harge.
Q = 20 litre/s = 0.02 m )'.~
Velocity of flow.
ii
== ~ = 0.020 = 0.020 ,,0.6366 m/s
Are a ~ 02
~ ( .2 )2
4
4
R, = Reynolds number
= pVD = 900 x 0.6366 x .2
)l
1.5
10
=
900 x .6366 x.2 x 10
{": V:ii:O.6366 1
15
= 763.&9
As lhe Rey nold s number is less llian 2000. the fl ow is visco us. Th e co-efficien t of frict ion for
\'iscous fl ow is g ive n by equat io n (9.23) as
/==
~== _'6_ == 0.02094
R,
763.89
-,
Head lost due to frict ion,iI, ==
==
4 ./. L .u
D x 2g
4 X .02094 x I0000 x (.6366)'
0.2x2x9.8 1
m " 86.50 m. An s.
Due to slo pe of pipe I in 200. th e height throu gh wh ich oil is to be raised by pump
" S lope x Le ngth of pipe
=;x L= - ' - x 10000= 5010
200
Total head against which pUlnp is 10 work,
H"'lj +iXL"S6.50 +50 = 136.50 m
Power required to pump the oil
I I
Ii
~ I
IL
Visco us Flow
417 1
'" PI:. Q .11 '" 900 x 9.8 1 x 0.20 x 136.50
= 24. 1 kW. Ail S.
1000
1000
.. 9.7
MOVEMENT OF PISTON IN DASH·POT
Consider a piston moving in a vc nical dash-pot containing oil as
shown in f ig. 9. 13.
Let
D" Diame ter of pisto n,
L " Le ngth o f piMon.
IV
T
= Weight o f piston.
,
)..l :: Vi scosit y of o il.
V
= Velocity o f piston.
1
ii
= A I'c ragc veloci ty of oil in the clearance.
I = Clearance between the dash-pot and piscon.
to.p '"
Difference o f pressure imcilsitics between th e two end s
of the piston.
The flo w of oil throug h c leara nce is s imilar 10 th e viscous fl ow
bet ween two paralle l plates. Th e differenc e o f pressure for parallel plates
for kngth .L' is giv"' ll by
.6.
p
~
12)tilL
Fig. 9.13
...( i)
- ,-,-
Also th e diffe rence of pressure a1 th e two e nds of piston is given by_
I'J.p =
Weight of pis ton
IV
4 1\1
-Area of pisto n - ~ D2 - rr.D ~
4
... (ii)
...
12)lilL 4 1V
Equaling (I) and (u). We get --,- ~ - ,
r
TtD "
_
Tt O-
1V/ 1
I~
41V
II '" - ,
x --.
,
...(iii)
12)1L 3n)1 L D "
V is lh e vdocily of piston or the veloc ity of o il in dash-poI in con tact wilh piston. The rate of fl ow
ofoi l in dash-pot
'" veloci ty x a rea o f dash-pot = V x ~ D !
4
Rat e of fl ow throu gh c learan ce = veloc ity throu gh cle aran ce x area of clearance = iI x nO x I
Due to continu ity equati on. ratc of flow through clearance must be equ al to rate of flow throu gh das h-pot.
Ii x nD xl=Vx ~02
4
_
II
Tt
",Vx - D
4
2
[
VD
TtDxl
41
x --- ~ -
... (1"1')
Equating the value o f Ii from <iii) and (il·). we get
VD
"
I I
Ii
~ I
IL
1418 Fluid Mechanics
... (9 .24 )
Problem 9 .25 All oil dash-flol COIIS;S/.! of II pislOn mOl'ing ill a cylinder IWI'iIiS oil. TI,;s arrangemelll is Iised 10 damp out the I'ibratiolls. The pislOn falls willi uniform speed (!lId co!'ers 5 em in
foo seconds. If on uddiliOlwll>'eighl of 1.36 N is placed 011 the top oflhe pis/oil. itfalls Ihrough 5 em
in 86 ~'ecollds wilh uniform speed, The diameter oflhe pisloll is 7,5 em lind its /ell gl/J is 10 em. Th e
clearance between IiiI' piS/Oil ,md Ihe cy/illder is 0./2 em which is uniform Iinoaghoul. Find the
\'iscosily of oil.
Solullon. Given:
Distance covered by piston due to self we ight, = 5 em
Time laken.
'" 100 sec
Additiunal weight.
'" 1.36 N
Time taken to cover 5 elll due to additiona l we ight. '" 86 sec
Dia. of pisto n.
D '" 7.5 em '" 0.075 111
Length of piston.
1_= 10cm=0.1 m
Clearance.
1=0.12cm:0.0012m
Let the viscos ity of oil
IV = Weight of piston.
V = Velocity of piston without addit ional weight.
¥" = Velocity of piston with addi tional weight.
Using equation (9.24). we have
=,
4 IVt J
_
3ltV J
].\ -
IV
"'
v
=
_
LV -
4[ W + 1.36] IJ
31tD ) LV *
IV + 136
V'
V
IV
V·
1V + 1.36
(
41V, ' )
Cancelling - -,-
31!D L
...(i)
V= Velocity of pis ton due to self weight of piston
""
Distance cove red
Time taken
5
- - em/s
100
¥" = Distalll'e cove red due to self weight + additional weight
Similarl y.
Time taken
5
= 86 em/s
V
5
86
- = - x -=0.86
V·
(00
5
Equating (i) and (ii). we get
"'
"'
I I
IV
W
+ 1.36
... (ii)
'" 0.86
1V =0.86W+.86xl .36
IV- 0.86 IV = 0. 14 W = .86 x 1.36
Ii
~ I
IL
Visco us Flow
419 1
IV", 0.86 x 1.36 "" 8.354 N
0.1 4
41VI
Using equation (9.24 ), we gel).l:=
J
'3C"~D'i'LCV~
4x8J54 x(.OO I2f
{
'" JIl x (O.07Sl x.1O x (-'- x _ '_ )
100
:=
.. 9.8
5
5')
": V ,. 100 emls = 100 x 100 mfs
100
0.29 N s/m 2 = 0.29 x 10 poise = 2.9 poi~.,. Ail s .
METHODS OF DETERMINATION OF CO - EfFICIENT OF VISCOSITY
The following arc the cx pcrirn cmal mctllods of dClcrmining llie co-efficient of viscosity of a liquid:
I, Capillary lube method.
2. Fa llin g sphere rcsis\JllcC met hod.
3. By rmating cylinder method. and
4. Orifice type viscometer.
T he apP;lralUS uscd for determining the viscosity of a liquid is called viscometer.
9 . S. 1 Capillary Tube Me thod . In capillary lube method. Ihe viscosity of a liquid is calculated
by measuring th e pressure difference for a given length of th e capillary lube. The Hagen I-'o isc uill e law
is used for "3lculatin g viscosity.
~~~~~-~~~ ~~ ~~ ~~
-
--
~
~ ~~
~.
-
-- -
~ ~ ~~ ~ ~
~_
T
1\
- :- : - :- : - :- j
)'
CONSTANT HEAD
,~,
Fig.9.14
jD
,' JI"\\J
ME~!~~ING --:=:::-'
Capillary tllbe Vi5CQmeter.
Fig. 9.14 shows lh~ capillary luhc viscomel,;,r. Th~ liquid whose viscosily is to be del,;,rmin,;,d is
fill,;,d in a consta nt head lank. The liquid is mainlained at constanl I,;,mpermure and is allowed 10 pass
through Ihe capilluy tube froU1the constant head lunk. Then, Ih~ liquid is collected in a measuring tank
for a given lime. Then Ihe rale of liquid collecled in Ih e tank IX'r second is determined. The pressure
head "II' is measured at a poim far away from the tank as shown in Fig. 9.14.
Then
1/ = Difference of pressure head for length L.
The pressure at oUl lel is atmospheric.
Lei
D = Diameter of (;api lla ry lUbe.
L = Length of tube for whic h difference o f pressure head is known.
p = Dt:nsity of flu id.
I I
Ii
~ I
IL
1420 Fluid Mechanics
~
and
'" Co-cfllciclU of viscosity.
Using Hagen Poiscuillc's Formula. II = 32 11 ii,L
pg D '
: -Q- = -QArea
I( D~
4
where Q is ralc of liquid flowing through lube.
So<
II
32f.l x
Q
~D2
xL
_
/, = _ _4" ,,_ - '"_. 128).l Q. I.
pgD2
).1=
1tpgD"
npghD'
128Q.L
.. .(9 ,25)
Measurement of D should be done very accurately.
9. 8 .2 Fall ing Sphe re Resis tance Method .
Theo r y. Th is method is based on Stoke's Jaw. according to which the drag force. F Oil a small
sphere moving with a l'OIiStam velocity_ V tli rough a viscous fluid of ViSl:osity. f.l for viscous conditions
is given by
... ( i)
F = 31t~IUd
d = diameter of sphere
U = velocity of sphere.
When the sphere attains a conSlant velocity U. Ihe drag
forc.: is Ihe difference between the weight of sphere and
where
and
~'~f
~ i.:o
0: ~
~
~
~
- -U - I - - -
-L- -
~
~I ~
SPHERE
buoyanl force acting 011 it.
Lei
CONSTANT
TEMPERATUR
BATH
FIXED
MARK
L '" dislalH:e Iravelled by sphere in visco us fluid.
I'" lime wken by sphere 10 cover distancc I.
p, '" dcnsity of sphere.
Pf '" density of fluid.
IY '" weight of sphere.
Fig. 9.15 Falling 5phaf: rf:5i5tam:f: method.
F(l '" buoyant force acting on sphere.
L
.
,
Then constant vclocny of sphere. U "'Weighl of spher~. IV", volume x dellsity of sphere x g
,
"'6d
;
Xp, Xg
f:
vol ume o f sphere ,. ~dJ}
and buoyant force. F 8'" weighl of fluid displaced
'" volumc of liquid displaced x dellsily of flui d x g
,
;
'" - d x Pf X g
6
~ I
I volume of liquid
dL~placed
'" volume of sphere J
I~
~ I
IL
Visco us Flow
421 1
For equilibrium.
Drag force'" Weigh! of sphe re - buoyam force
F=W - F8
or
Substituting the val ues of F. Wand P s . wc gel
IT
311~Ud= - d
)
1t)
It\
Xp, xg - - d xp, xg= - d- xgIP,- Pjl
6 6 6
11 dl x g[P, _ p, ] gd 1
~:6
3nUd
"'ISU ]P,- P/ ]
... {9 .16)
where PI '" Density of liquid
Hence in eq uation (9.26).lhc values of d. U. p, and PJ aTe ~nown and hence the viscosity of liquid
can be determined.
Mclhod. Thus this method consists of a tall vertical transparent cy lindrical wnk, which is filled
wilh. the liquid whose visco~ity is 10 be dctcrrllincd. This lank is surrounded by 31l011wr transparent
tank to keep the temperatu re of Ihc liquid in the cylindricallJnk lU be constant.
A spherical ball ofsrnall di~mclcr 'd' is r!;lccd un the surfa!;c of liquid. Provision is made to rele~sc
this ball. After a shon distance of travel. the ball attains a constant velocity. The time to travel a
known vertical distancc betwecn two fixed marks on the cylindrical tank is noted to calculate the
constant velocity U of the ball. Theil with the known values of d. p,. PI the I'iscosity tJ of the nuid is
calculated by using equation (9.26).
Rotating Cyli nder Method. This method
9.S.3
TORStONAL
.- SPRtNG
consists of two ,-,oncentri\: cylinders of r~dii R, and R2 as
show n in Fig. 9.16. The narrow space between the two
""~b,.,j§i,,,,;;'POINTER
cylinders is filled with the liquid whose viscosity is to be
determ incd. The inner cy linder is he Id stationary by means
of a tors ional spring while outer cylinder is rotaled at conVISCOUS
stant angular speed (I). The torque T acting on lhe in ner
LIQUID (>I)
,-,ylinder is measured by 'he torsional spring. The torque on
the inner cylinder must be equal and opposite to the torque
app lied on the outer cylinde r.
The to rque applied on the outer ,-,ylinder is due to
viscous resistance provided by liquid in the annular space
w
and at the bottom of the inner cylinder.
Let
OJ = angular speed of outer ,-,ylinder.
Fig. 9.16 Rotl1ting cylinder viscometer.
Tangential (peripheral) speed of outer cylinder
=OJxR 1
Tangential velocity of liquid layer in ,-,onwct with outer cylinder will be equ al to the tange ntial
velocity of outer cy linder.
Velocity of liquid layer with outer cylinder = OJ x R1
Velocity of liquid layer with inner cylinder = 0
!. Inner cylinder is stationary I
Velocity gradient over the radial distance (Rl - R I)
=du",wRl - O
Shear stress (t)
I I
Ii
~ I
IL
1422
Fluid Mechanics
Shear force (P)
'" she ar stress x are a of surface
'" t X l rr N I H
I' shea r stress is acting on surface area '" I n NIx N I
mil,
2
' ) XllR IH
( R2 - HI
The torqu e TI
o lllh ~
inn er c ylinder duc 10 shearin g ac ti on of the liqu id in th e annulM space is
'1'1'" shear fo l"('c x radiu s
=, ( ')
wR,
HI
_ 21fllwll
-
HI
x2n RI H xRI
H I2 R I
... ( i)
(R2 - Rd
appli~"d
If the gap betwee n th e bonom oflh e two cy linders is '/1 ' , th en Ihe torqu e
is give n by equatio n (9 .19A) as
on inn er cy li nde r
( T 2)
But he re
11
R =R 1· t = h thcn T1 ", 6011
It
2
N fl l
4
w'" 2/tN or N '" 6O{o
6il
2,
Il
'1'1= - - "
2
6000
4
1tJ.1 111
x - - xRI = - -
lit
T Ol al \orquc T ac ti ng on Ihe inn cr cy linder i s
fijI!
2//
HI
4
... (ii)
T : T 1 + T1
=
11 "
21'qlmHR,~ R2
(
R 1 - R,
)
IlJH])
4 _2
+ -- HI -
211
,[
IfI..lR I
2(R1 - RI)'!T
rrR,2w [41f1IR) + Hi
(Rl - HI)]
R
R1H + RI2]
1
R
I
4 11
xw
... (9.27)
where
T " IOrqu e meas ured by th e strain of the tors ional s prin g .
N I • N2 " radii of inner and o uter cy linder.
II " clearan ce at the botto m o f cy linders.
H " he ight o f liq uid in annul ar space.
~ := co-efficie nt of vi.<;cosity to be de te nnined .
Hence. th e va lue of ~ (;an be calc ul ~ltcd fro m equat io n (9.27).
9 .8.4 Orifice Type Viscometer. In thi s meth od. th e time taken by a ce na in quant ity o f the
liquid whose vis.;:osily is to be detc nnin cd , 10 fl ow thro ugh a s hort c ap illary lube is no ted down. The
co -efficie nt o f viscusit y is the n obwined by comparing with the co -efficient of viscosit y of a liqu id
whose viSl.'Os it y is kno wn or by th e usc convcrsio n fa ctors.
Visco melc rs s uch as Sa ybo lt. Red wood or En g ler arc usua ll y use d . Th e princ ipl e for ali lhe th ree
visco meter is same. In th e United Kin gdom. Red wood visco mete r is used whil e in U.S.A .. Sayboh
visco mda is co mmonl y used.
I I
Ii
~ I
IL
Visco us Flow
Pig. 9.17 shows thal
~aybo lt
viscometer. whi ch consists of a
tan~
at th e bonom o f which a sllon capillary tube is fitted. [n this tan k the
liquid w hose viscosity is to be determined is filled. This tan l.: is su r-
rounded by another tank. called constant te mperature bath. The liq uid ;, allowed
now through capillary tube
standard
'0
"
,
tcmpcraturc. The time laken by 60 c.c. of Ih" liqu id [0 now through
the ell pi liar y tube is noted down. The in itial hci g'ht of Ii uid in the tank
is previously adjusted to a standard height. From Ih" time measure·
ment. Ihe kinematic vis<'"Qsity of liquid is known from the relation.
V=
B
AI - -
,
Fcc
ree
· c '~&"
"
:-:
; 'LI aiD
-
::;
423 1
0 .
M
: 0-_
,~ ~9NSTAN T
. ;:-
,O ~; 0,
TEMPERATUR
,
BATH
-.~~
~.-_.
._. - --- - .
---:: .' .-=- -
III
l l J - .MEASURING
CYLINDER
Saybolt v i5Cometer.
where A = 0.24. B = 190. I = time noted in seconds. v = kinematic vis.;osily in slol; es.
Problem 9 .26 Tile \'is"osil), 0/ (In oil 0/ ~p. gr. 0.9 is mea~''' reJ by a capillary I"be 0/ diamela
50 mm. Th e differe1l ce o/press"re head belwe,," 111'0 poinlS Z m aparl is 0.5 m a/water. Tile mass 0/
oil collected ill tI metlsllri1lg t(/II1r, is 60 kg ill /00 secollds. Find Ihe I'ismsily 0/ oil.
Solu t ion. G iven
Sp. gr. of oil
=0.9
Dia. of capillary tube.
D = 50 mill = 5 cm = 0.05 III
Length of tube.
L = 2 III
Difference of pressure h~ad. 11 = 0.5 m
Mass o f oil.
At = 60 kg
Time.
1= 100 s
Mass of oi l per second
[knsity of oi l.
Di scharge.
Using
equ~tion
Fig . 9.17
60
100
= - - = 0.6 kg/s
I' = sp. gr. of oil x 1000 = 0.9 x 1000 = 900 kg/m 3
0.6
Q= Massofo i! Is = - m 3/s= 0000667
.
III ' Is
De nsity
900
(9.25). we get viscosity.
1tpgll D
4
[here II = 111 = 0.5 J
1'=128Q . L
It X 900 x9.81 x 0.5 x (.05)'
='Ci'C':~:;;;;;;C~~"'- = 0.5075 (SI Units) N shn!
128 x 0 .000667 x 2.0
= 0.5075 x 10 poise = 5.075 polsi.'". Ans.
Problem 9 .27 A ClIpillllr)' /lIbe 0/ iliameler Z mm amllellglh {OO mm is used/or mea.mrillg \'iscosit),
0/ tlliquid. nle difference o/pre.iSlIre ber ...een rile IlI"a enils o/rlle wbe is 0.6867 Nlcm ! ""'/Ihe \'iswsily
0/ liquid is 0 .25 poiJ'e. Find IIII! rale 0/ flow 0/ liquid Ihrougll Ihe lube.
Solullon. Give n :
Dia. of cap illary lube.
D=2m=2xIO l 1l1
Length of IUoc.
L= 100111111 = IOc1l1=0. 11I1
flp = 0.6867 Nkm2 = 0.6867 x 10" Nfrnl
Difference of pressure.
Difference of pressure head.
II
=_' _11 =~O~.6~86~7~x~I~0_·
1'8
I I
I'g
Ii
~ I
IL
1424
Fluid Mechanics
~
Viscosity.
'" 0.25 poise
= 0.25 Ns/ml
10
Ld the rmc of flow of liquid ", Q
0.6867 X 10 ' x (2 x IO -l)'
n p)ll1l)4
Using equation (9.25), we ge t 1-1 '"
128 . Q. L
= npg x
pg
--""~c-;;CCC;C--128)(Q)(O.l
,
0.25
10
It
=
Q=
"'
X 0.6867 x 104 X (2 X 10- 3)
128 x Q)( 0. 1
It
x 0 .6867 x [0' x 24 X 10- \1 x ]O
In ~
118 x 0.1x 0.25
'" 107.86 x 10 ~ m .lls '" 107.86 x [0 ~ x I06 crn 3/s
'" 107.86 x IQ-! C11I 3/S '" 1.078 cm l/s. AilS.
Problem 9.28 A sphere of diameter 2 mill falls 150 mm ill 20 J'eCOlll/S in a 1';5CO" 5 /iq"id. Til e
density of tile sphere iJ' 7500 kg!", J ' Illd of'iq"ill is 900 kgl", J. Find Ihe cQ-'1ficienl of I-isCQsily of Ih e
liql/id.
Solution. Given:
Dia.ofsphere.
<1= 2 mlll :2 x 10 .! m
Distance travelled by sphere
Time taken.
Veloc ity of s ph ere.
Density of s phere.
Densit y of liqu id.
'" 150 !TIm '" 0.15 III
I", 20 seconds
u '"
Dist ance
Time
0.1 5
'" .0075 mfs
20
p, '" 7500 kg/m ,l
PI = 900 kg/ m!
d1
Usin g re lation (9.26). we get!!= ~ Ip, - Pj l =
[8U
9.81 X[2 x 10
Jf
18 x 0 .0075
17500 - 9001
= 9.8 1x 4 xlO -li x6600 = 1.9 [ 7 Ns
18xO.0075
m1
'" 1.9 17 x 10 '" [9. [ 7 poise. AilS.
Problem 9.29 Find rhe I'iscosiry of" liquid of sp. gr. O.B. whe" " g"s bubble of diamerer 10 "''''
rise,~ UNUfi/y rhrougl, rhe lilfllid 111" I'e/odty of 1.2 cm!!l. Neglect rhe weight of the bubble.
Solution. Given:
Sp. gr. of liquid
= 0.8
De nsit y of liqu id,
PI = 0.8 x 1000 = 800 kgtlll 3
Dia. o f gas bubble.
D = 10 nlln = I elll = 0.01 III
Ve[ocily of bubble.
U = 1.2 emts = .01 2 I1ItS
As weight of bubble is neglected ,md density o f bubb le
I I
Ii
~ I
IL
Visco us Flow
425 1
d'
II '" ~ [p, - PI! which is for a falling sphere.
I8U
['or a rising bubble, the relalion will become ~s
Now using the rcl31ion.
j.I '" : : :
Substituting lh..,
valuc~.
[Pr p,1
9.8 I x.otx.O I
Ns
Ns
1800 - 01 ~ = 3.6]"2
18 x,0 12
III
111
we gel
'" 3.63 )( 10 '" 36,3 pil lse. Ail s.
Problem 9.30 Th e I'iscosily of II liq!lit/ is dc/ermined by rolaling cylinde r mel/wd. ill ...llieh c(lse
the i""er cylinde r of diameter 20 em is Sia/iunary. The OilIer cylinde r of diameter 20.5 em, C011/llillS
rhe liquid upro II heighl of 30 em. The clearance allile bOl/om of IIII! /h'O cylinders is 0.5 em. The ouler
cylinder is rO/(l/ed a/ 4()() r.p.lII. The /Orque registered 011 IIII! lorsioll meier (I//(Iched /0 rill! illller
cylinder is 5.886 NIII. Filld IIII! I'iseasily of fluid.
Solution. Given:
Dia. of inner I:ylindcr.
DI '" 20 ern
Radius o f inne r cylinder.NI = 10 em = 0.1 m
Dia. of outer cylinder,
Radius of oute r eylinder.N~ =
2~.5
= 10.25 e m = ,1025
In
Heigllt of liquid from bott om of outer cylinder = 30 em
Clearance at tile bonom of two cylinders. !J '" 0.5 CIlI '" .005 m
Height of inner cylinder immersed in liquid
= 30 -II = 30 - 0.5 '" 29.5 m
H",
0'
Speed of ou ter cy linde r.
29.5
CIlI '"
.295 m
N = 400 r.p.m.
= 2nN = 2)(/tx400 = 4 1.88
6<l
60
T", 5.886 Nm
III
Torque measured.
=
_-;-;-:,---=2"(1'.I0
:"2='"--;0".I)"X
".00
,,,,'::X:::'.:8116
-,,,:==--,.,;
It
x (.1 )1 x 4 1.88
[4 x .295 x .005 x .1025 + .1 1 (.1025 -. 1)]
2 x .0025 x .005 x 5.886
= -,-:;=~=~=::;
/tx.O I x4 1.88 [.0006047 .000025]
=0. 19286 Ns/rn 2 = 0.19286 x 10 = 1.9286 po ise. An s.
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IL
1426
Fluid Mechanics
Problem 9.31 A sphere of diamerer I mill /alh 11"014811 335 III ill 100 secQlIl1$ ill (j l'isCOIH fluid.
,e/alil'e den.!ilies of Ihe sphere (Illd Ihe liquid are 7.0 (lIId 0.96 respec/il'ely. de/ermine Ihe
If Ihe
dJllllmic l'isCQsily of Iile liquid.
Solution. Given:
Di3. of sphere.
d" 1 mm " 0.00 I m
Distance travelled by sphere = 335 mm == 0.335
Time taken.
I" 100 seconds
Distance 0335
- - = 0.00335 m/.<.cc
I ()()
Time
=7
p, == 7 x 1000 == 7000 kg/m l
= 0.96
P, = 0.96 x 1000 = 960 kglm 3
U"~==
Veloci ty of sphere.
Relative densi ty of sphere
Density of sphl're.
Relative density of liquid
Densi ty of liquid.
Using the relation (9.26).
t11
gd 1
We gC11J
= 18U
9.8 1 x 0.00 I'
Jr, - Pj l '" 18 X 0.00335 17{)()() - 9601
= 0.0000098 1 x 6040
[8xO.00335
= 0.98 1 Ns/rn 2
== 0.98 1 X 10", 9.81 pulse. Am .
Problem 9 .32 Delermille Ille fall I'e/ocily of 0.06 mm sand parlide (Ipecijic gradl)' '" 2.65) in
...mer at 20"C. /(Ike j1 '" 10 -J kg/mI.
Solu t Ion. Gi\'en :
d", 0.06 111111 '" 0.06 X [0- 3 m
Dia. of sand particle.
Specific gravity of sand
'" 2.65
Density of sand.
p, '" 2.65 x 1000 kglm l
(": P for wat<'r in S .I . uni t '" 1000 kg/rn l)
'" 2650 kg/Ill '
Viscosity of water.
)1. '"
10- 1 kg/ms '" 10-3 Nslrn 2
D.:nsity of water.
PI "" 1000 kg/Ill '
Sand pan il:1c isj uslli ke a sphere.
For equilibrium of sand p~rtic[e.
Drag force"" Weigl\! of sa[ld particle - buoyant force
F n ", W - F8
0'
F n '" Jltj.l x U x d. where U", Velocity of panil: lc
'" 31t x 10-' )( U x 0.06 X 10-1 N
IV"" Weight of sand particle
"m
""
~
6
Xtl'
...( i)
x P, Xg '" ~ x (0.06 X 10-,)1 x 2650 x9.81 N
6
P/J = Buoyant force = Weight of waler displaced
Hence vi""nsily '" (1 kg ~ 1 m ) )( --;- '" kglms. Hence kglms '"
S
I I
111
_N_~ .
m-
Ii
~ I
IL
Viscous Flow
427 1
"'~ xtFxp,xg= ~ x (0.06)( IO-')'x 1000)(9.81 N
6
6
Substituti ng the above va lu es in eq uat ion (i). we get
311 X 10- 3 X U x 0.06 X 10- 3 '"
~ x (0.06 X 10-")3 X 2650 x9.8 1 - ~ x (0.06 X 10-3)3 X 1000 x 9.8 1
6
6
Cancelling (/I x OJk) X 10-·)2 throughollt. we get
3x U
=~
X
6
'"
~
0.06 2 X 10 .l x 2650 )(9.81 -
~
6
X 0.06 2 X lO .l x IOOO x9.81
X 0.(6 2 )( 10-J X 9.81 (2650 - 10(0)
6
= ~ x 0.0036
X
10-J x 9.81 x 1650 = 0.0097 12
U '" 0.009712/3 '" 0.00323 nt/sec. An s.
HIGHLIGHTS
I . A flow is said to be viscous if the Reynolds number is less
th~n
2000. Or the fluid flows in layers.
2. For the viscous flow through circular pipes.
(I)
Op ,
af 2"
Shear Siress ...... t - -
(i/) Velocity ...... u _ - -'4jl
(iii) Ratio of velocities U : " .. 2.0
where
~
Loss of pressure head. Ill '"
"
32 IiL
~
pgD -
r", radiu.' al any poim.
'" pressure gradient.
u.... ~ maximum ,docity or velocity at,
H '" radius of the pipe.
Ii
~P [,r - ,zJ
OX
~ a,'crage ,'eloxily ~ ~.
•R
jJ ~
m
O.
co-cffident of viscoSity .
D '" diameter of the pipe.
J . For Ihe ,·iscou.' flow between two parallel plates.
I ilr
l
" " - - - ( 1 ) " - )") ... Vciocity distribution
2jJ
ar
Um>' .. 1.5
"
h
12j.luL
J .~
, 'p
l"-"2a.. 1t - 2y ]
where
I I
... Ratio of rna.imutn and avemge velocity
... Loss of pressure head
Shear stress dislributioll
I = thi~k"ess or distance between two plates.
y. distance in the "cMical direction from the lower plale.
1 '" sbear stress m any point in flow.
Ii
~ I
IL
1428 Fluid Mechanics
~.
The kinetic energy correction faclor a is gi.-cn as
o·
K.E.pcr second based on actual velocity
K. E. per second ba>Cd on average "clocity
~ 2.0 ... for a circular pipe.
5. Momentum correction factor. ~ is given by
Ii..
Momentulll per second based on actmt! velocity
Momentum pcr second based on average velocily
.. 3"4 ,-, for a circular pipe.
6 . For the viseou ••e,i.lance of Journal Hearing _
V",nDN dw=~=ItDN
60 'dy
I
601
Torque.
where L= lenglh of bearing, N .. speed of shaft
I" clearance between the shaft and bearing.
7. For the Foot -Step Bearing. the shem force. torque and h,p. absorbed are gj"cn as:
Shear force.
2
J
fl rr. N R
F. - - - 15 I
J
Torque.
T .. ~ l(lNK'
""
,. flIlJNl R'
6OX)OXI
where R_ radius of the shaft. N .. speed of the shaft
8. For the collar bearing the torque and power absorbed are given as
T _ _II
60,
,
..
leNIR,
.
" rrJNl
p", ..
- R,• I.
6Ox301
where H, .. internal radius oflhe collar.
I .. Ihickncss of oil film.
9. For the viscous flow the
IR,~ - R'l
•
,
Rl .. external radius of (he collar.
P = power in wallS.
co~fficicn( of friction
is given by.! ""
~
~
pVD VD
where R, . 1he Reynolds number .. - - . - .
"
'
10. The co-efficienl of viscosily is detennined by dash-pol
where IV "" weight of (he piston.
L", length of 'he piston.
\' "" "cioci,y of the piston .
I I
~rrungcl1lcn(
4IV/)
as)1
E
3/tLO JV
r = dcaran~c between dash -pot and piston.
0 .. diameter of the piston.
Ii
~I
IL
Visco us Flow
II. The co-efficienl of viscosity o f a liquid is also delcnnined
c~perjlnentally
429 1
hy the following method
/lpgI/O'
(i) Capillary wbe method. ~,. 128QL
(ii) Falling 'phc.c method. II '"
where
>I'
'" spe~ific
,,"1,,-,,1
lSU
L '" length of the lUbe,
Q '" rate of flow of fluid through capillary tube.
P, _ density of sphere.
weight of fluid.
D = diameter of the capillary lube.
d _ diameter of the sphere.
PI '" densi ty of fluid.
NI ~ rad ius of outer rotatin g cylinder.
T ~ torque.
U", "doci!y of sphere.
RI '" radius of inner stal;o"a,y cylinder.
EXERCISE
(A) THEORETICAL PROBLEMS
I . Define the terms: Vi scosity.
kincmati~
viscosity. '-ciOC;I}, gradient and pressure gradient
2. What do YOIl mcan by 'Vi>eou,; Flow"!
3. Dcrive an expression fOf the velocity distrib ution for "iscous flow through a tircular pipe. Also skc!ch th~
"docity distribution and shear stress distribution across a section of the pipe.
4. 1'<0"<" that the ma.~ imum vel","·ity in a ci rcular pipe for vis.:ou< flow is eq ual to two times the a'·erage
Velocity of th~ flow .
(Delili Unilusily. Deamber 1(01)
5. find an e ~pre"ion for the loss of head o f a "i",ous flu id flowing through a circular pipe.
6. What is Hagcn Poiseuillc· s fonnula ? Dcrivc an e ~ pression for Hagen Poiscuillc·s Fonnula.
7. Prove that the velocity distribution for viscous flow between two paral lel plates when bOlh plates arc fixed
across a .<;cClion i., parabolic in nature. AI"'.! prove that maximum ,·docity is equal to one and a half timcs
the a'·emgc vc locity.
8. Show that thc differencc of prcssurc hcad for a givcn length of the two para llel plates which are fi~cd and
through which viscous fluid is flowing is givcn by
l2fliiL
J
P81l
whcrell = Visco~ity of fluid.
Ii ~ Avcrage velocity.
1 ~ Distance between thc two pamllcl plates.
L ~ length of the plates.
9. Define the tenns , Kinetic energy correction factor and momentum correction factor.
HI . Prove that for viscous flow throuSh a circular pipe the kinetic enersy correction factor is equal to 2 while
4
momentum corrcction factor ~ J.
,. --
II . A shaft is rotal ins in a journal bearing. The clearance belween Ihe shaft and the bearing is filled wilh a
,·iscous oil. Find an cxpression for the power absorbed in overcoming vi>eous resistance.
12 . Prove Ihal power absorbed in o,·ereoming viscous resistance in foot-Slep bearing is given by
P.
""~60'~'xN~'301
~.r;-
where N ,. Radius of Ihe shaft,
t ~ Clearance between shafl and fool-slep bearing.
II
N ,. Speed of 1he shaft.
II ,.,. Visco,ity of fl uid.
Ii
~ I
IL
1430 Fluid Mecha nics
U . Show that the value of lhe
~o-efficien t
f .. ~,
where
of friction for v;><oous flow through a drculaf pipe is gil'en by,
R," Reynolds number.
14. I'rove that the co-efficient of viscosity by the da.;h-pOi arrange",,'"l is g;,'en hy.
4 (I'l l
where
).I '"' 3ltLD'V
IV ., Weight of Ihe piston.
L '" Length of piston.
I = Clearance between dash -pot and piston.
D '" Diamet.:, of piston.
V = Velocity of piston.
15. What arc the different methods o f dClcnnining the co-cFflcicnt of viscosity of a liquid? Describe any two
method in details.
16. Prow that the 10 •• of pre.sure head for lhe viscous flow through a circular pipe is giv<'n by
h
f =
32).1 u L
p:;;;iT""
where U ~ Avcrage velocity. w'" Sp<"ific weighl.
17. for a lami nar steady flow. prove thaI the pressure gradient in a direction of motion is equal to the shear
gradient normal to the di'e<;tion of motion.
18 . Describe Reynolds e~periments to demon>trate the two types of flow.
I~. For the laminar flow through a circular pipe. prove that'
(i) the shear stress variation acroSS the seclion of the pipe is linear and
(ii) thc vciocity variation is parabolic.
(B) NUMERICAL PROBLEMS
I . A crude oil of viscosity 0.9 poise and sp. gf. 0.8 is flowing through a horizontal circular pipe of diameter
80 111m and of length 15 m. Calculate Ihe difference of pre,sure at the two ends of the pipe. if 50 kg uf the
[Ans.0.559 N/cm' l
oil is collectcd in a tanK in 15 seconds.
2. A viscous flow is laking place in a pipe of diameter 100 mill. The maximum .-eiocity is 2 m/s. Find the
mean velocity and the radius at which this occurs. Al so calculate the "eiocity at 30 mm from the wall of
tbe pipe.
IAn s. I m/s. ,~ 35.35 mm. u ~ 1.68 mlsi
J . A fluid of viscosity 0.5 poise and specific gravity 1.20 is fluwing tbrougb :, circular pipe of diameter
100 mm. The maximum shcar stress at the pipe wall is given as 147.15 N/m ' . fmd (0) thc presslIre
gradient. (b) the average velocity. and (e) the Reynolds number of the flow.
I/\ n<. (a)- 64746 Nitn 1 per m. (b) 3.678m/s. (e) 882.721
4 . Detemline (II) the pressure gradiem. (b) the shear stress at the two horizontal parallel plates and (c) the
discharge per metre width for the laminar flow of oil with a maximum "elocity of 1.5 m/s between two
hor;~omal
•
parallel "xed
plates which arc 80 mm apart. T ake vi.cosity of oil as 1.961,Ns.
lIl'
IAn•• (a) - 3678. 7 N/m' per m. (b) 147.15 N/m'. (c) .08 m'lsl
5.. Water is flowing between two large parallel plates which arc 2.0 mm apart. o,,!enninc : {oj maximum
vciocity, (bj the pressure drop per unit lenglh and (c) Ihe shear stress a! walls of the piate if the awrage
velocity is 0.4 m/s. Take viscosity of water a.< O.QI poi~.
IAn s. (a) 0 .6 m/s. (b) 1199.7 N/ttt l per m. (e) 1.199 N/m! 1
6. There is a horizontal crac~ 50 tttm wide and 3 mm deep in a wall of thidness 150 mm. Water leaks
through the crack. find the rate of leakage of water through the crack if the difference of pressure between
the two ends of the crac~ is 245.25 N/ml. Take the viscosity of water 3> 0.01 poi.,e. [An.•. 183.9 em'lsl
I I
Ii
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IL
Visco us Flow
431 1
7. A shaft h""in£ ~ diameter of 10 em ro(ate<; centrally in ajoumal bearing ha ving a diameter of 10,02 em and
length 20 em. The annular space belw~n the shaft and the bearing is fillexJ with oil having viscosity of 0.8
poise. Dclcnninc the power absorbed in the t.earing when the speed of rotation is 500 r.p.m.
[An s. 343.6 WI
K. A shaft 150 mill diameter runs in a hearing of length 300 mill, with a radial clearance of 0.04 mm m
40 [ ,p_m. Find the viscosity of the oi l . if the power l"C'quired to o'-ercomc the viscous resistance is 220.725 W .
jAn s. 6.32 poise[
9. Find the torque required to rotalc a ...,"ical shaft of diameter 8 em at 800 r.p.m. The lower end of the shaft
rcs!s in a fool -.tep bearing. The end of the shaft and surface of the bearing are both fin! and are sCp.:1r:llcd
by an oil film of thickness 0.075 em . The viscosity of the oil is given as 1.2 poise.
[Ans . 0.0538 Nrnl
Ill . A wllar bearing having external ~nd intern:,l dimnclers 20 em and to ~'TTI respectively is used to take Ihc
thrust o f a shafl. An oil film of thickness 0.03 em is maintained between Ihe collar surface ~nd the bearing.
Find the power lost in overcoming the viscous resi.tance when the shaft rolates Ul 250 r.p.m. T~ke II s 0.9
poi,.,.
IAns. 30.165 WI
II . WateT is flowing through a ISO rmn diameler pipe with a co-efficienl of Frictionj ...05. The shear Slress al
a point 40 mm from the pipe wall is 0.01962 Nkm l. Calculate the shear Slress allhe pipe wall.
IAns. 0.0419 8 Nlem11
12. An oil dash-pot consisls of a piston moving in a cylinder having oil. The pislon falls wilh unifonn spe~'"d
and wve" 4.5 em in go seconds. If an addilional weight of 1.5 N is plated On the top of the piston . it failS
through 4.5 em in 70 seconds wilh uniform S]><.~d. The diameler of the piston is 1O em and ils length is
15 cm. The clearance between the pi,ton and the cylinder is 0.15 cm. which is unironn throughout. I'ind
the viscosity of oil.
IAns. 0.177 poisel
13 . The viscosity of oil of sp. gr. 08 is measured by a capillary lUbe of diameter 40 mm. The difference of
pressure head between two poinls 1.5 m apart is 0.3 m of water. The maSS of oil ~'Olleeled in a measuring
tank is40 kg in 120 seconds. Find the viscosity of the oi l.
[Ans. 2.36 poisel
14. A capillary tube of diameter 4 mm and length 150 mm is used for measuring viscosity of a liquid. The
difference of pressure between Ihe two ends of the lUbe is 0.7848 N/cm 1 and Ihe "iscosity of Ihe liquid is
0.2 poise. Find the rate of flow of liquid through the tube.
IA lls. 16.43 conJlsl
15. A sphere of diametu 3 mm falls 100 mm in 1.5 seconds in a viseous liquid. The density of the sphere is
7000 kg/m' and of liqyid is 800 k£hnJ. Find Ihe eo-efficient of viS<.'Osity of Ihe liquid. [Ans. 45 .61 poise]
16. The viscosity of a liquid is determined by rotming cylinder method. in which case the inner ,"ylinder of
diameter 25 ern is slalionary. The outer cylinder of diameter 25.5 em comains the liquid UplO a height of
40 em . The clearance at the bouom of the two cylinders is 0.6 cm. The ouler cylinder i, rotated at 300
r.p.m. The torque regislered on Ihe torsion melre attached to Ihe inner cyli nder is 4. 905 Nm. Find Ihe
IAII • . .77 poiscl
viscosity of liquid
17. Calculate: (a) the pres.ure gradient along the flow. (1)) the average ,·elocity. and (el the discharge for an
oil of viscosity 0.02 Nslm l flowing between IWO stalionary parallel piates I m wide mainlained 10 mm
apart. The "e\ocity midway between the plates is 2.5 ml,.
[,\ns. (a) - 4000 Nlrn2 per m. (b) 1.667 m/s. (el .01667 m'lsl
18 . Calculale:
(I) the pressure gradienl along the flow.
(ii) the average velocity. und
(iii) the discharge for an oil of viscosity 0.03 N slm~ /lowing between two stationary plates which arc par~lIc1
and arc at 10 mm ~p:1rt. Width of plates is 2 m. "The "elocity midway between the pl~tes is 2.0 mfs.
19. II cylinder of 100 mm diameter. 0.15 m length and weighing 10 N slides a~ially in a "enical pipe of
104 mm dia. If the space belween cylinder surface and pipe wall is filled with liquid of viscosily II and the
cylinder slides downwards at a velocity of 0.45 m/, . detennine II .
[Hlnt. /J ~ 100 111m _ 0. 1. L _ 0.15 m. LV _ 10 N. Dp _ 1.4 111m _ 0.104 m.
V = 0 .45 mi •. Hence t = (0.104 - 0.1)12 = 0.002 m.
I I
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1432 Fluid Mechanics
p20 . II liquid is pumped through a 15 em diameter and 300 111 long pipe al the rale o f 20 tonne5 pcr hour. The
density of liquid is 910 k glm) and "incm~lic viscosi ty ~ 0.002 m )!s . Dclennine the po" 'cr required and
show thaI the flow is viscous .
[Hi nt. D '" 15 em _0.15 Ill, L '" 300 m. IV", 20 tormeslhr
.. 20 x 1000 kgf/60 x 60 sec • 5.555 kgOsec • 5.555 x 9.81 Nts .
8C' '" 0.0061 111 , Is.
Q '" ~= ~5~.55~5~'~9~.
,
P8
910 )( 9 &1
v_ Q =
II
0.0061
~ (.l51)
~ O.l45 m/s. ,. _ O'()()2 m l!s.
R = pVD
Now
'
VxD
0 .345 x 0.15
,
0.002
P
= 25.87
which is less than 2000. Hence flow is viscous
I"" 32 I-ILVlpgD 1• where v ..
.e. :. !!" v)( p .. 0.002 x 910 .. 1.82
P
•
Hence.
f
32 x 1.82 x 300 x 0.345
•
(91O X 9.8 1 x 0.15 1)
•
30
I' '" pg.Q.hJ'IOOO '" 9 10 x 9 .81 x 0.0061 x 30!l{)()() '" I.6JJ kW.
21. An oil of specifIC gravity 0.9 and viscosity 10 poise is flowing through a pipe of diamdcr 110 mrn , The
velocity at the centre is 2 m/s. find: (I) pressure gradie nt in the direction of now. (il) shear mess at the pipe
wall: (iii) Reynolds number. and (il') vel ocily al a di,tance of 30 mm from the wall.
IUln t.p =900 kglmJ; IJ = 10 poise", IN s/m 2 ; 0= 110 mm =0.11 m,
U.... ~ 2 m/s; ii
(;) (-tlP)=
,Ix
(ii)
~ I mi• . V""" ~ - '-
41J
41J X V m ...
Rl
( iii ) R =
'
2
pxiixO
IJ
"'
tlx
R'
4xlx2
,
0.0551 = 2644.6 Niln
_ (-- dPx
) -Rz 2644 .x
6 -0.055
' otlx
( - 'p )
2
7' 7' NIm'.
=~.~
9OQ xlx O.ll
.,99 ' and
I
.
r)
" = _,_ (- dP ) (If _
= - ' - (2 644, 6) (0.055 2 _ 0.025 2) = 1.586 m/s.1
4)1
<Ix
4xl
12. Delennine (i) the pre"sure gradient . (ii) the ,hear stress al the two horil.Olltal plates. (ii;) the discharge per
metre width for laminar now of oil with a maximum ~elocily of 2 mls belween two plales which arc
150 mm apan. Given: IJ '" 2.5 N s/TIl'.
IHi" l. V""" = 2 m/s. 1 = 150 mm = 0.15 m.).I
(i) V
....
I tip,
" - - -I'
81J dx
I tip
(Ii) t o. - - 2 fix
X ,. -
dp
-
I
2
N shIll
-8).1 U"", -8x2.Sx2
= _1777.77NJml.
11
O.lS!
I
(- 1777.77) x 0.15 _ 133.33 N/m .
(iii ) Q = Mean velocity X Area.,
I I
-
dx -
(Delhi VII""",-ily. Drcember 20(1)
= 2.S
.... )
(~V
X (/ X I) '"
(~x
2)
X (0.15 X I)" 0.2 TIl lIs. I
Ii
CRM'J'RK
.. 10. 1
INTRODUCTION
The laminar now has been discussed in chapter 9. In laminar flow the fluid panicles move along
straight parallel palh in layers or lam in ae. such that the paths of individual fluid panicles do nO! cross
those of neighbouring panicles. Laminar flow is possible only at low velocities and when the fluid is
highly viscous. But when the velocity is increased or nuid is less viscous. the fluid panicles do nO!
move in straight paths. The fluid panicles move in random manner resuhing in genera l mixing of the
part icles. This type of flow is called IlImu Jcm flow.
1\ laminar flow changes!O turbulent flow when (I) veloc ity is increased Of UI) diameter of a pipe is
inc reased or (iii) the vi_'\Cosily of fluid is decrea_"",d. O. Reynold was fiut to demon~trate that the
transition from laminar to IlImuJcnt depends nOt on ly on the mean velocity but on the quantity pVD .
This quanlity p m is a
dimen~ionlessquantity
and is called
Reynold~ number (R, ).
"
In ca.<;£! of circular
"
pip'" if R, < 2000 the now is said to be laminar and if R, > 4000. the flow is said to be turbule nt. If
H, lies betwccn 2000 10 4000. the flow changes from la minar 10 tu rbulen t .
.. 10. 2
REYNOLDS EXPERIMENT
pV xd
The type of flow is ddermined from Ihe Reynolds number i.e" ~~~. This was demonstrated by
O. Reynold in 1883. His apparatus is
~hown
in Fig. 10.1.
"
Fig. 10.1 R ~nQ/d apparatus.
433
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1434
Fluid Mechanics
The apparatus consists of :
( i) A tank co ntaining water at constant head.
(il) A s mail tank containing so me dye.
(iii) A g lass tube ha ving a be ll -mouthed en tr~ncc at one end and a reg ulatin g va lu e at other ends.
The water from the tank was allowed 10 flow through the glass tube. The velocity of flow was varied
by [he regulating valve. A liq uid dye having sa llI e spo:ciflc weig ht as waler was ill lroduccd into the
g lass lubc as shown in Fig. 10.1.
D"
The following observati o ns were made by Re yno ld:
(i) Wh en tlie veloci ty o f flow was low. llie dye filamen t in the g lass tube was in Ihe (onn of a s trai ght line.
This straight li ne uf dye filame nt was parallel to the
~
WAVY
g lass tube. whkh was the case of laminar fl ow as show n
r~---:C--=--=
7"l FILAMENT
in Fig. 10.2 «(I).
(i/) With the increase of velocity of flow, the dyeTrans~ion
filament was no longer a strai ght-line but it became a
W3VY o ne as sho wn in Fig. 10.2 (b) . This sho ws that
flow is no longer laminar.
(e) Tu.wlent flow
(iil) With funh e r incre ase of ve loci ty o f flow. the
wavy dye- filam e m broke-up and finall y diffu sed in
Fig. 10.2 Diffe rent stages of filame nt.
wat er as Shown in Fig. 10.2 (c). Th is mea ns that the
[r;;=;;;=.;=====~;'
r ~'3IIlAMENT
(b)
•
mFDIIlFAFUSED
MENT
F ~ ~<~?,;*J
fluid panicles of the d ye at this hi ghe r veloc ity are mo vin g in random fashion. which shows th e case
of turbulent fl ow. Thus in c ase of turbulent fl ow th ~ mixing of dye- filame nt and water is intense and
flow is irregular, random and disorderly.
[n case o f laminar fl ow. th e loss of pressure head was found to b.: proponiona l to the ve locity but in
case of turbulent flow. Reynol d o bserved that loss of he ad is approximate ly proporti ona l to the square
of veloc ity. More exactly th e loss of head. 111 "" V · , where It varies from 1.75 to 2.0
to 10, l
FRICTIONAL LOSS IN PIPE flOW
Whc n a liquid is flowing throu gh a pipe, the veloc ity of th e liqu id layer adjaccm to th e pipe wall is
zero. The ve loc ity of liq uid goes on in creas in g from the wall and thu s ve[ocit y gradiem and hence
shea r Stre~s arc prodm;ed in th e whol e liquid due to vi><:osit y. This viscous action c au ses loss of
energy whkh is usually kn own as fric tionn[ loss.
On the bas is of his ex perime nts, Wi[lium Froude gave the following la ws of fluid fraction for
turbulent flow.
The frictional resistance for turhulent fl ow is :
(i) proportional to 11". whl,re " var ies from 1.5 to 2.0.
(ii) proportion al to thc de ns it y of fluid.
(iii) proport ion al to the area o f s urface in cOntact.
(i,') independe nt of pressure.
(v) depende nt on the nature of th e su rface in contact.
10.l . 1 Expression for Loss of Head Due to Friction in Pipes, Co nsider a uniform hori w mal pipe. havin g stead y fl ow as shown in Fig. 10 .3. Let I - I and 2-2 arc tWO sc<:lio ns of pi pe.
Let (It" pressure inte nsity 31 section I-I.
V, " ve locit y of flow at secti o n I - I,
~ I
I~
~ I
IL
Turbulent Flow
435 1
L = length of the pipe between sections I-I and 2-2.
d = di;ullclc r of pipe.
t' = frictional
resistance per Ulli! wetted area per unit ve loc ity.
Il, " loss of head due 10 friction.
and Pl' V! = are va lues of pressure intensity and veloci ty at section 2-2.
--
T,
"
-- "-1-, ------
Fig. 10.3
Uniform horizonr«l pipe.
Apply ing Bernoulli's eq uation s between sect io ll s 1- 1 and 2-2.
Total head 31 1- 1 '" Total head at 2-2
+ Loss of head due \0 friction between 1- 1 and 2-2
"'
~+ -'- +Zl = 1'1+V/+ Z2+I!
pg2g
!
pg 2g
Z, "" Z2 as pipe is horizontal
VI'" V 2 as dill. of pipe is same at I I and 2-2
~ '" Pl + II or " = ~ _ 112
pg
pg
f
f
...(i)
pgpg
But lit iS the head lost due \0 frictioll and hence inte ns it y of pressure wil l be reduced in the direc tion
of flow by frictional rcsi~tancc.
Now frictio nal resistance == fri cti on,1i rcs iswnce per unit wetted area per unit ve locit y x wetted area
x velocit/
or
F 1 ==l'xrr.dLxV 1 [-: wettedarea=rr.dxL, vc locit y= V== VI = V![
==!,XPXLXV2
[. ltd = Perimeter = PI ... (ii)
The forces acting on th e fluid between sections 1· 1 and 2·2 arc:
1. pressure force at section 1· 1 = PI x A
where A = Area o f pipe
2. pressure force at sect ion 2·2 = P! x A
]. frictional force FI as shown in Fig. 10.]_
Resolving all forces in the hori zontal direction. we have
PI A
- p0 -
... ( IO. J)
F I=O
(P I - Pl)A
= FI =1'
PI - P2 =
2
x Px L X V
r·
2
From (iil. FI =I'PLV [
I'XPXLXV 1
A
Btu from equat ion (il. PI - 112 = pgli,
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1436 Fluid Mechanics
Equating th e va lue of (PI - P2)' we ge l
,'X P XLXV
A
P8 11f =
Ilf '"
f'
p
- xpg A
X
P
Wetted perimeter
= ~=",c:==c
A
Area
2
,
L x V-
... (iii)
--= -
In equation (iii). -
f' 4
2 f'
4 L V!
" r = - X - XLXV = - x - pg d
pg
d
Puning
L= f, wherefis know n as co-effi cient of fricti o n.
p
2
Equation (ii') . becomes as
II = _4 ·_f
,
_LV_2 = c4,;
f Cc
·L
'o."V_'
28
d
... (1 0.2)
dx 2g
Equ ation (10.2) is know n as Darc y- W c isbad\ cquatio ll. This equation is commonly used for findin g
loss of head d ue to friction in pipes.
Someti mes equati on (10. 2) is wrintl! as
' f=
j .L.V
1
...(10 .2A)
dx2g
Then / is know n as fric tion facto r.
10.3.2
Expression for Co-efficient of friction in Terms of Shear Stress. The equation (10.1 )
g ives the forces acting on a fluid be tw een sectio ns 1- 1 and 2-2 o f r ig. 10.3 in ho ri zontal diredion as
PIA -
pz'l -
FI
(PI - pzli\
=0
= FI = Force due 10 shear siress t o
= shcar stress x surface arca
=tox ndx L
Cancelling nd frolll both s ides. we hav e
d
(PI -P 2l 4" = l O X L
(PI-Pi)=
Equation (10. 2) can be written
~ I
4to X L
d
a;;
II{ =
... (1 0.3)
PI - P2 __ 4/ . L. V
~cc:~
pg
1
dx2g
I~
~ I
IL
Turbulent Flow
(PI - P2l=
4j.L.V'"
xpg
d x2g
437 1
... (10 .4)
Equating [he value of (PI - fl 2l in equations (10.3) and (10.4),
41: o XL
4/.L. V 'd
'" dx2g xpg
t o :fV1 xpg
2g
pv!
0'
TO=[ -
jV 1 xpg
2g
... (10.5)
2
J= 2t o,.
. ..(1 0.6)
pV·
... 10.4
SHEAR STRESS IN TURBULENT flOW
The sllenr stress in viscous flow is given oy Newton ' s law of viscosity as
d,
where t " '" shear stress due to viscosity.
t,. '" j.I - ,
tly
Similar to the expression for viscous slieaT. J. Boussincsq expressed the tu rbulent shear ill math ematical fonn as
<Iii
t , "''1 dy
... (10.7)
t , = shea r stress due to turbulence
'1 = eddy viscosi ty
Ii = average veloc ity at a dislunce y from boundary.
The ratio of '1 (edd y viscosity) and p (ma ss den sity) is known as kinematic eddy viscosity :md is
denOlcd by {: (c psiloll). M alhemal icaliy it is wriucn as
where
..!l
... ( IO.S)
p
If the shear stress due 10 viscous now is al so considered. then the IOlal shear stress becomes as
£ '"
du
t"'t ,. +t,=Il - "
dy
du
dy
... (10.9)
The value of T] = 0 for laminar now . For other 1'1ISCS tbe value of T] may be scveral thousand times
the value of ).I. To find shear stress in turbulent flow , ~"<Iuation (10.7) given by Boussi nesq is used. But
as th e val ue of T] (eddy vL<;C()sity) cannot be predicted. th is equation is having limited U.'\C.
10.4.1 Reynolds Expression for Turbulent Sheilr Stress. Reynold s in 1886 developed
an expression for turb ulent shear stress be tween two layers of a n uid at a sllIali distance apart. which
is given as
... ( 10 . 10)
whefC u', \.' = nuctualing compon e nt of veloci ty in the direction of x and y due 10 turbulence.
As ,,' and ,.' are vary ing and herlCe twill al<;o vary. Hence to find the shear stres~, the lime average
on both Ihe sides of Ihe L'iJuation (10.10) is taken. Then equation ( I 0.10) h~com~s as
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IL
1438 Fluid Mechanics
...(1 0 . 11 )
The turbuicnt shear stress given by equation (10.11) is known as Reynold stress.
10.4 .2 Pr;lndtl Mixing Length Theory for Turbulent Shear Stress . In equation (10. 11).
the turbulent shear Slresscan only he calculated if the value of 11'1" is known. But it is very difficult to
measure ,i'v'. To overcome Ihis difficulty, L Prandtl in 1925. presented a millin~ length hYP01hcsis
which can be uscd 10 express turbulent shear stress in Icons of measurable quamilics.
According to PraHdtl, the mix ing length I. is lhal distance bo:lwccn two layers ill the transverse
direction sucli lhm the lumps of fluid particles from OnC layer could reach lhe other layer and the
particles are mixed in lhe other layer in such a way that lhe momentum o f the particles in the direction
of.( is same. Hc also assumed that thc velocity fluctuation in th e .(-direction ,,' is related (0 the mixing
lengt h I as
II'
=1 du
dy
and I"~ • the fluctuation componenl of velocity in y-dircction is of the same order of magnitude as ,,'
and hence
,,' =1 -d"
II)"
Now u'x,( becomes as
II',.'
=
(I IIU]X
(I dU]=/l
(d"dy ]'
d)'
tly
Substituting the value of II' ,,' in equation (10.11). we get the ex pression for shear stress in turbulent
flow due 10 Prandtl as
( ]'
_T = pl-,d"
-
... { 10.12)
dy
Thus the total shcar stress at any point in turbulent flow is the sum o f shear stress due to viscous
shear and turbulent shear and can be written as
-T=].I -+p
<1" /,(<1"]'
'd)'
... ( 10. 13)
d)'
But the vis<:uus shear stress is negligible except ncar the boundary. Eq uatiun (10.13) is used fur
must of turbulent fluid flow problems for dctennining shc~r stress in turbulent flow.
to 10 .S VELOCITY DISTRIBUTION IN TURBULENT FLOW IN PIPES
In case of turbulent flow, the (Otal shear stress at .my point is the sum uf viscuus shear suess and
turbulent shear stress. Also the viscous shear stress is negligible cxcept ncar the boundary. Hence it
can be assumed that the shear stress in turbulent flow is given by equa lion ( 10.12). From this equation.
the ve locit y distribution can be obtained if the relation between I, the mixing length and ), is ~nown.
Prandtl assumed that the mixing length. I is a linear functiun of the di stance y from the pipe wall i.e..
1= ky. where k is a t"Onstalll. knuwn as Kannan consta nt and = 0.4.
Substituting the value of I in equntion (10.12), we get
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Ii
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IL
Turbulent Flow
439 1
d" '
( dy J
_
,
t o rt :p x (ky),x
...( 10. 14)
For small values of y that is very close to th e boundary of Ihe pipe, Pral1 (.hl assum ed shear stress T
to ~ COnstant and app rox imakl y equal to t o wlli ch prese nts Ihe turbulen t shear Slress at the pip.:
boundary. Substitu ti ng t "" t o in equ ati on ( 10 . 14), we get
dd"Y
In c"uation
( 10.1 5),
'i
fi
=
,'yVp
~
p"
...( 10. 15)
fi
!IIL lr l
ML '
pO ha s th e dim ensio ns
~Ll _ _L
L
r ) - T " BUI -T is ve locit y and he nce
has th e d i mension of ve lo ci ty. whi c h i s kno wn as shear ve loc ity and is deno ted by" •.
fi
n
I
dll
'
= u•. th en cqumlOn ( 10. 15) !J;.,comcs - = - II •.
P
dy ky
For a give n case of lurhul cnl now. u. is conSlant. Hence integ rating ahiwc equ ati on. we gel
Thu s
-
",
... (10. 16)
u =T1og,.y + C
where C = co nstant o f integra tio n.
Eq uati o n ( 10 .1 6) shows [hat in lurbule nt flow. lhe vel oc it y varies di reclly with lhe log arilhm of lhe
di slance fro m lhe boundary o r in other words the veloc ity distributio n in turbulent fl ow is log arithmic
in natu re . To determin e the co nstant of integ ration. C the boundary condition that at)' '" R (rad ius o f
pipe ). U '" II """ is s ubstituted in equat io n ([ 0. [6).
lin,", ""
k",
",
,
log, R + C
C=U m,,-, - - log R
'
Substitutin g the va[u e o f C in equ ati on ( 10 . [6). we get
II.
II '"
k
[og, ), +
"" "mlU + -0",4
'" limo,
+ 2.5
u""" -
II .
k
log, R "" II,,,,,, +
log, (yI R)
II .
log, (y IR)
II.
k
(log, ), - log, R)
[ .: t. = 0.4 "" Karm an con, tan t]
... (10. 17)
Equ ati o n ( 10. 17) is ca ll ed ' Prand tr s un il'cT.'ia l vcloc ity di stri but ion equati o n for turbul ent flow in
p ipes. Thi s equati on is appli cabl e to smooth as we ll as roug h pipe boundari es. Equation ( 10. 17) is al so
wrine n as
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Ii
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IL
1440
Fluid Mechanics
"m., -
Dividing by
u., we gel
U '" -
2.5 u. log, (yIR) '" 2.5 u. log, (Rly)
'" 2.5 log.. (Rly) '" 2.5 x 2.3 Iog lo (Rly) (.: log.. (RIy) '" 2.3 Iog lo (IVy)]
" .."" - u '" 5.75 10gIO (Rly)
.. .( 10. 18)
,.
In equation (10. 18), the difference b<:lwccn the mal<jnwill velocity" ..." and
point i.e .. (urna, - ul is known as " "c locity deleeI'.
IOCil1
velocity
1/
al any
10.5 . 1 Hydrodynamiully Smooth and Rough Boundaries.
Let k is Ihe average height of
the irregularities projec ting from the surface o f a boundary as shown in Fig. lOA. If Ih<: value o f k is
IM!;c for a boundary then the boundary is called roug h ooulldary and if the value of k is less, Ihen
boundary is known as smooth boundary. in general. This is the classilkalioll of rough and smoot h
boundary based on boundary \.Characteristics. BUI for proper ciassifkation, the flow and fluid daracteristics arc also to be co nsidered.
LAM INAR SUBLAYER
T ---------- - - -- -
'L ~
'
(a) Smooth bounda<y
Fig. 10.4
LAMI NAR SUBLAYER
II
J
_ " i~
t (b) Rough
___ _
bounda<y
Smootb and rougb boundariel.
For turbulent flow analysis along a boundary, the flow is divided in two runions. The first rurtion
consists of a thin layer of fluid in th e itnm",diate ndghbourhood of the boundary. where viscous shear
stress predominates whi l", the shear st ress due to turbulence is negligihle. This run ion is known as
laminar sub- la yer. The height upto which the effect of viscosity predominates in this zone is denot~d
by /)'. T he second portion of flow. where shear stress due 10 turbulence arc large as compared to
viscous stress is ~nown as turbulent zone.
If the average height k of the irregularities. projecting from the surface of a boundary is much less
than 0'. the thic kness of laminar sub -layer as shown in Fig. 10.4 (a). the boundary is "ailed smoot h
boundary. This is because. outside the laminar sub-layer the flow is turbulent a nd eddies of various
size present in turbulent flow try to penetrate the lamimlr sub-layer and reach the su rfac e of Ihe
boundary. BUI due to greal thickn ess o f lamin ar sub-laye r Ihe eddies arc unable to reach the s urfa"e
irregularities and hence the boundary behaves as a smoot h boundary. This type of boundary is called
hydrodyn:llni"ally smoot h boundary.
Now, if the Reynolds number of the flow is increased the n Ihe Ihidnes~ of laminar sub-la yer will
d"'creasc. If Iht: thiCkn ess of laminar suh-Iayer bcco ln"'s much sl)1all er Ihan the avcragt: height k of
irregularities of Ih e surface as show n in Fig. 10.4 (h). the boundary wi ll aCI as rough boundary. This is
because the irr"'gu larities of Ihe surface arc above th '" laminar sub- laye r and the eddies prt'scnt in
lurbulent zone will come in contact with th'" irreg ularities of the surface and lot of cnergy will be 1051.
Such a boundary is ca ll ed hydrodynamically rough bou ndary .
From Nikuradsc's experim ent :
k
k
I. If /)' is less than 0.25 or S' < 0.25. the boundary iscallcd smooth bounda ry.
I I
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IL
Turbulent Flow
4411
2. If :' is greater (han 6.0, the bounda ry is rough.
3. If 0.25 < ( ;, ) < 6.0. the boundary is in transition.
In terms of roughness Reynolds number -"k ,
,
h
I. If -lI,k,
< . boun d
ary"IS l'Onsldcrcd smoot.
"
Ilk
,
2. If - '- lies betwee n 4 and 100. boundary is in transition stage. and
3. If lI.k > 100. the boundary is rough.
10.5.2
"
Velocity Distribution for Turbulent flow in Smooth Pipes . The velocity distribution for turbulem now in smoOl h urruugh pipe is given by cqumion (10.16) as
11=
"
- Iog,y + C
k
It may be sccnlhal at )' '' 0, the velocity /I at wall is - ... This means that velocity II is positive at
some distalll:e far away from the wall and-"" (min us infinity);n the wall. Hc",:e at >01Il1' finite distance
from wall, the ve locity w ill be equal lO zero. Lei (his distance from pipe wall is y'. Now the eonstanl
C is determined from the boundary condition i.e .• at y '" y' . II '" O. Hence above equation becomes as
.. ,
O= T log,
f
11"
+corc=-Tlog,y
Substituting the value of C in the above equation. we get
II
=
II .
T
log, y -
II.
T
, I I.
log, Y =
T
,
log, (yl)')
Substituting the val ue of" '" 0.4, we get
II
= ~ log, (yly') = 2.5
0.4
.!!.... '" 2.5 x 2.3 Jog 1o (yly')
,
.
"'
.!!.... = 5.75 logl o (yly')
",
fl.
log, (yly')
I': log, (yly') '" 2.3 loglo (yly'»)
... ( 10 . 19)
For thc smooth boundary. there cxists a laminar sub·layer as shown in Fig. 10.4 (a). The velocity
distribution in the laminar sub·layer is parabolic in nature. Thus in the laminar sub-layer. logarithmic
velocily d iMribulion does nOI hold good. Thus il can be a~sumcd (hal y' is proportional 10 0', whe re 0'
is the lhickness of lam inar sub· layer. From Ni kurad sc's e)(P<"rimenl lhe va lue of / is give n a~
I I
Ii
~ I
IL
1442
Fluid Mechanics
where S' = I L6v . where
".
It
= kinemmic viscosity of fluid.
,
11.6v
J
u,
107
)' = - - x -
O.108v
= --u.
SUbsti tut ing this value of )" io equation ( 10.19), we obtain
I~' '" 5.75 loglo [.]({sv]
".
y
'" 5.7510&10 (~) = 5.7510gJO (u. X9.259 )
.I OS\'
\'
II ·Y
'" 5.75 10&1 0 -
,
!l.y
"" 5.75 loglo -
10.S.3
,
+ 5.75 10&10 9.259
[.: 0.i08 =9.259]
+ 5.55
.. .( 10.20)
Velocity Distribution for Tu rbulent Flow in Rough Pipes .
In case of rough
bound aries. lite thic kness of laminar sub- layer is I'cry sma ll as shown in Fig. 10.4 (b). The surface
irregularit ies arc abol'c the lam in ar sub- layer and hence the laminar s ub-layer is com p letely destroyed.
Thus y' ca ll be considered proport io nal 10 th e Ilcighl of protru sions k. Nikuradse 's e~pcrimc l11 ~hows
the va lu e o f y' for p ipes COaled wit h unifonn sand (rough p ipes) as y' '" -
k
30
.
Substituting this va lu e o f y' in equation ( 10. 19). we gct
~ :: 5.7510g lo
14.
[-y-)::
I.:I?IJ
5.75 [loglO (yll.:) x 30]
'" 5.75 logl o ()"II.:) + 5.75 101:10 (30.0) '" 5.75 log 10 0-11.:) + 8.5 ... (10.21)
Problem 10 .1 A pipe-lille carryillg water has orerage height of irregularilies projecling from Ihe
sllrface of Ihe bOlilidory of Ille pipe as 0.15 mm. W/lal Iype of bOlllldary is il ? The slteM Slress
dln'eloped is 4.9 Nlm'. Tile kill ematic riscosit)' of Wilier is .01 stol.:es.
Solution. Givcn :
Average height of irreg u larities. 1.::: 0. 15 111m '" 0.15 x 10- 3 m
Sh~a r stress d~veloped.
to'" 4.9 Nlm2
v'" 0.01 stok~s '" .01 cm1ls", .01 x 10-1 m 21s
Kin~m:nic viscosity.
[knsity of water.
p '" WOO kg/J11 3
S h ~ar
I I
vdocity,
".:: Jr: o I p '" J[000
4.9 "" JO.0049 :: 0.07 tnls
Ii
~ I
IL
Turbulent Flow
u,k
= -
Roughness Reynold number
"
=
0.07 x 0.15 x IO-J
443 1
'" 10.5.
,
Since - '- lies between 4 and 100 and hence pipe s urface behaves as in transilio rl .
"
Problem 10.2
A rOI/gl! pip" is of diomeler 8.0 em. Tile \'('IQcily <II a point 3.0 em from Willi is 30%
more Ilwn 1/11' !'elocily til a point I em frolll pipe wall. Delermille IIII' at'erage IJeiglll of Ihe roug/me.f.!.
Solution. Given:
Dia. of rough pipe.
D " 8 elll '" .08 tl1
LeI ve locit y of flow al I crn from pipe wall
'" II
Then velocity of flow al :I em from pipe wall
"1.3 II
The ve locity distributio rl for rouglt pipe is giwn by equation (10.21) as
~ '" 5.75 loglo (ylk) + 8.5, where k'" height of rough ness.
".
For a poilU. 1 elll from pipe wall. we have
,
-
".
= 5.75 log 10 (I.OIk) + 8.5
... ( i)
For a point. 3 ern from pipe wall. velocity is I,) " and hence
1.311
'" 5.75 log,o (3.0/k) + 8.5
... (ii)
".
3
Di viding (ii) by (i). we ge l 1, =
c',,"'~5c'O:"~'""(i3'cOc'-i'C)+",,"C
'
5.75 log ,o(J 1 k) + 8.5
or
1.3[5.75Iog lO (Ilk) + 8.5[ = 5.75Iog I0 0.0Ik) + 8.5
or
7.475 logl o ( I/k) + 11.05", 5.75 lug lO (3.0/k) + 8.5
7.475 log lO ( l /k) - 5.75 log 10 O/k) '" 8.5 - 11.05 = - 2.55
Of
or
or
7.475 [log lo 1.0 -Ioglo k[ - 5.75 [Iogl o 3.0 - log lo k[ = - 2.55
7.475 [0 - loglo kI - 5.75 [.4 771 - 10); 10 kJ '" - 2.55
or
- 7.475 log lo k - 2.7433 + 5.75 log 10 k '" - 2.55
or
- 1.725 logl o k '" 2.7433 - 2.55 '" 0.1933
"'
0. 1933
log w k = - 1.725 = - 0.1120= 1.888
k", .772fi
COl.
An s.
Problem 10.3 A .\1/100111 pip" of diameter 80 mill atlll 800 //I Io/Ig carri,'.1 Wafer af II,e rale of
0,480 m J/lllinule. Calculate Ille 10.1$ of head. wall .Ihearillg .lIre.lS, celltre lill<' 1·e/OCily. ["elocil), alld
she!" SlreSJ (1/ 30 mm from pipe wa/l. Also ca/Culmt' II,e II,icklless of lamillar .Illb-layer. Take kill emU/ic ["iscosit)' of water (jJ 0,015 stokes, Take the mlile of co-f'jficiellt offriclioll ,from Ihe re/alioll
girt''' (IS
.0791
f= ( R,)'I~' where
I I
R, =
ReYliolds IIllmber.
Ii
~ I
IL
1444
Fluid Mechanics
Solut ion. Given:
Dia. of smooth pipe.
Leng th of p ipe.
L=800m
Discharge.
0048
Q =.
Kinematic vi!\Cosily.
Density of wata.
v :: .015 sto kes", .0 15 X 10-4 ml{s
P'" 1000 kg/Ill )
Mean veloc ity.
V -_
d =RO nun =.ORm
In
3Iminute '" 0.48
60 = .0081031s
-.iL '" _°
=8
"=.00
7
Area
It
(.08) 2
= 1.59 1 mls
4
R '" Vxd = 1.591 xO.08 :8.485 x 104
Rey nolds numbe r.
,
V
.015 x I0.....
As the Rey nolds number is more IhJIl 4000. th e flow is turbulent.
Now th e va lu e of l' is give n by /=
.079 1
.079 1
------;{.I=
(8.485 x 10
R,
4
1/4
= ,004636
)
(i) Head lost is given by equ atio n (10.2) as
4. f. L. Vl
4 x .tXl4636 x 800 x 1.591 l
II, '"
d x 2g
'"
.08x2x9.8 1
: 23.42 m. Ans.
(ii) Wall s hearing stress. to is give n by eq uat ion (10.5) as
to=
(i ii ) Ce ntre- line ve locity.
U""'-,
7 '"
Jj V !
lI. y
1/
,. = 5.75 login - ,
II.
=
is s henr veloci ty and
x 1.591 1 = 5.866 N/m l, An s.
for smoo th pipe is given by eq uation ( 10.20) as
-
where
21000
.004636 x
+ 5.55
~ = J~:
... ( i)
= 0.0765 ntis
The veloci ty will be maximum whe n ), = !!.- = .08 = .04 nt ,
2
Hence
31 )'
= .04
2
Substituting these values in (0, we get
In . II '" II ...."
1/ "",
.0765
'0;
= 5.75 logl a 0.0765)(
,0 15)(10
+ 5.55
= 5.75 logw 2040 + 5.55
(i\') The shear stress, !
"" 5.75)( 3 .309 + 5.55 = 19.03 + 5.55 = 24.58
"""" = .0765 )( 24.58 = 1.88 mls. An s,
at an y point is give n by
a"
ax 2
~
1=- - -
I I
... (A)
Ii
~ I
IL
Turbulent Flow
445 1
where r = distance from ce mre of pipe
and hence slica r stress at pipe wall where r = R is
up R
to" - - 2
Dividing equation (A) by equa tion (IJ). we gct
... (8)
ax
, ,
'.
Slicar stress
R
,. -"R '
A poin t 30 mm fmlll pipe wa ll is having r
= .0 1 Ill) =
t o x .01
=4 -
3 '" I ern
= .01
5.866
III
z
- - = 1.4665 Nfm . A ns •
.'"
4
Velodly at a point ) em from pipe wa ll meanS y = 3 ern :: .03 III
! a1 (r
U
and is given by equat ion ( 10.20) as -
u y
'" 5.75 loglo - ' - + 5.55, where u. = .0765. y:: .03
II,
V
u
.0765 x .03
.0765 '" 5.75 log w .015 x 10 ..... + 5.55
:: 5.75 JOS 10 1530 + 5.55 = 23.86
0.0765 x 23.86:: 1.825 OIls. ADS.
(v) Thickness of lamin ar s ub-layer is given by
1/::
1)'= 1J.6xv
11.6)(.0 15xlO--I =2.274xIO-lm
.0765
1
:: 2.274 X 10- em = .02274 em. A ns .
Prob lem 10 .4 De/ermine Ihe wall siJe(lTing J'ln:ss in (l pipe of diameter 100 111m which ~'lIrries ",,,Ier.
The ,-e/OCilie;- al Ihe pipe cenlre and 30 mmfrom Ihe pipe ,'enlre are 2 ",Is ""'/ /.5 mls respeCI;'·e/y.
Th e flow in pipe is gi"ell (IS lurbulent,
Soluti o n. Given
D", 100 mm '" 0. 10 m
Dia. of pipe.
Radius,
R=
o~o
= 0.05 m
Veloc ity at ce ntre.
" ..... '" 2 mls
Velocity at 30 111m or 0.03 m from centre'" I.S m/s
Velocity (at r = 0.03 m). II = 1.5 mls
Let the wall shearin g stress
=lu
['or tu rbulent now. th e ve locit y distrihution in terms of centre tine ve tocity ( un",,) is given by equa tion (to. t8) as
-::.' =.515
-"'".~u_
,. loglQ ,.
y
(K)
where
I I
II '"
[ .S IIl/s at y = (R - r) = 0.05 - 0.03 = .02 III
Ii
~ I
IL
1446
Fluid Mechanics
2.0 -1 5_
_5751
.
og, & .05_")288
__ .
ur 0.5_':>288
__ .
.02
u.
U . '"
Using the relation
u. '"
0.2185=
2~~8
II.
= 0.2185
mfs
FoTP. where p fur water'" 1000 kglm l
Jl~or 1~ =O.2185~=O.0477
t o == 0.0477 x 1000", 47.676 Nfml, An s.
or
10.5.4
Velocity Distribution for Turbulent Flow in Terms of Average Velocity. The
average veloci ty U . through the pipe is obtained by first finding the 100ai discharge
Q and then dividing
the total disdJ.arge by the area of the pipe.
ELEMENTARY C IRCULAR RING
/
- - I---FLOW
.
- r --
'
~
,,
A vtTagl' velocity for IUrb"lt'nf flow.
Fig. 10.5
Consider an elementary circular ring of rad ius 'r' and thickn ess dr as shown in Fig. 10.5. The
distance ofille ring from pipe wal l is y '" (N - r), where R '" radius o f pipe.
Then the discharge. dQ. through the ring is given by
dQ = area of ring X veloci ty
= 2rrrdr X II = II X 2rrrdr
H
TOlal discharge.
Q= f dQ= f o
..
... ( I0.22)
x2rrrdr
(a) for sm oo th pip es. For smooth pipes. the velocity diSlribulion is given by cqu;uion ( 10.20) as
u
-
II.
"'
11=
Bm
u.),
= 5.75 Jog lo + 5.5
v
,.y 1xu.
[5.75Jog lO --;-+5.5
)' =(R - r)
Substituting the va lu e o f " in equation ( 10.22). we get
Q=
I I
r'[
Ja
5.75 1og lo , .( Hv ,)
1
+ 5.5 II. X 2tt rdr
Ii
~ I
IL
Turbulent Flow
447 1
Average veloc ity.
'"
~ ( It [5.75 Jog lo11, ( R rrR
Ja
r) +
v
5.5]11. 2Tt rdr
Integ ration of the above eq uation and subscqucllt simp lification g ives th e ave rage veloci ty for
turbu Jcm flow in smoo th pipes as
U
II . R
- '" 5.75 10lll(l + 1.75
II.
v
...( 10 .23)
(b ) For ro u gh pip"s. For rou gh pipes. t he vdoc ily at any poinl in lurbulenl fl ow is gi ve n by
equation (10.2 1) as
" '" 5.75 101110 (y /t) + 8.5
".
-
"m
y=(R - r)
" '" 5.75 loglo (H
-o,)
- + 8.5
-
k
u,
"'
Average vcioc it y,
u- = - Q =
1tR ~
'R'
Int eg rati on o f the abo ve eq ualion and sub.""qu ell[ simplifi cat io n wi ll g ive Ih e following relation for
'lVcragc velocity.
fJ
for turbul ent flow in rough pipe as
fJ
-
".
R
'" 5.75 10gIO - + 4.75
,
... ( 10.24 )
(el Dirrc~nce of the \' ~Iodty at any point and an' rage ve locity for s mooth and rough Ilipes.
The veloc it y at any poin t for turbul ent n ow for s mooth pipes is g ive n by equat io n ( 10.20) as
u
u.(R - r)
+5.5
II.
v
and Ihe average vdocity is given by equation (10.23) as
- ~5.75Jogl o
U
II.
~
u. R
5.75 Jog lo + 1 75
I'
Differe nce of ve loc ily u and U for
I I
[·: y=R - r [
~1ll00111
pip.: is oblained
a~
Ii
~ I
IL
1448
Fluid Mechanics
, -iJ :
-
[ 5.75 10g lO
,.IN-')+ 5.5] - [5.75 101,: .0-,.R+ 1.751
v
II.U,
II-V
[
U.(R-r)
- - '" 5.75 10g.o:::CC--'"
0'
,
'.
:5 .75 10g lO
[
v
,.N]+
log lO~
5.5 - 1.75
,.R]
,.IRv- ,) + ~
+3.75
') + 3.75
R ,- '" 5.75 login ( '" 5.75 101:'0 (yIR) + 3.75
...( 10.25) 1'; R-r= y ]
S imi larl y the veloc ity. II at any poillt for rough pipe is g ive n by equati on ( 10.2 1) as
,
-
,. '" 5.75 log 10 {ylk ) + 8.5
and ave rage ve locity is g ive n by equat io n (10.24) as
iJ
-
,. '" 5.75 logwCRfk) + 4.75
Diffcrc llcc of ve loc ity II and U for ro ugh pipe is given by
-
,
iJ
- -
II.
'" [5.75 101:'0 (ylk) + 8.5] - [5.75 log 10 (Rlk) + 4.75 1
II.
'" 5.75 101:)0 [(ylk ) + (Rlk)] + 8.5 - 4.75
,-u
- - '" 5.75 log 10 CyfR) + 3.75
,.
...( I 0.26)
Equ ati ons (10.25) and (10.26) arc the Same. This shows that Ih e difference of ve locit y al any point
and th e average velocity will be the sam" in case of smoo th as we ll as ro ugh pipes.
Problem 10.5 Detert/lint' IIIe diJ'/(/lice frolll Ille pipe ....all al .... hicll Ille local ("dodl)' is equal to Ihe
al"erage I"e/adl)' for turbulent fla .... in pipes.
Solution. Given:
Local ve locity a l a point = ave rage velocilY
u= U
For a smooth or rough pipe. Ihe di fference o f ve locit y at any point a nd average I'elocity is gil'e n hy
equa li on ( 10.25) o r eq uation (10.26) as
II-U
- - = 5.75 loglo (yIR) + 3.75
,.
Substit utin g th e given co nditi on i.e .. u =
fJ -
fJ . we
gel
U
- - = 0 = 5.75 logl o (yIR) + 3.75 or 5.75 IOIlIO (yIR) = 3.75
,.
~ I
I~
~ I
IL
Turbulent Flow
logl o (yfR)= - 3.75 = - 0.6521
5.75
"'
449 1
=_ 1.3479
ylR", 0.22279 "" 0.2228 or y '" .2228 R. Ans.
Problem 10.6 For /Il,hu/,,,,/ flo", ill II pipe of dium eter 300 III"'. Jim/ Ille di sellarge "'/1(:11 Ihe CIOn/reline ,-e/OCily iJ' ].0 ",Is 'lIId Ihe ,-e/Oeily 01 a poi,,/ 100 mm from Ihe centre <IS mem,.",ed liy pilOHube
i$ 1.6 mls.
Solution. Given:
Dia. of pi pe.
D=300mm=0.3m
Velocity at cenlre,
03
Ro - :O.15m
2
u nWl '" 2.0 m/s
Velocity
(atr= IOOI111l1 =O. lm).u= 1.6mfs
Now
y '" R - r= 0.15 - 0.10 '" 0.05 m
(a l r=O.1 10 or aty = 0.05 In)." = 1.6 OIls
Rad ius.
Velocity
The veloci ty in terms of cent re-li ne veloc ily is given by equation (10. 18) as
-"'.·,.,c-:.::' = 5.75 10&10 (Rly )
,.
~
Substilulin g Ihe va lu es. we gel 2.0 - 1.6 - 5751
.
og lO
II .
.05
", YO ,05",]
[ R =O.15rn
'" 5.75 logl o 3.0 = 2.7434
0.4
-
'.
u.
= 2,7434
OA
= - - - '" 0.1 458 m/s
...(i)
2.74 34
Using equalion (10.26) whicll gives relat ion between veloci ty al any point and average vdocily, we
have
II - U
- - '" 5.75 log 10 (yIR) + 3.75
,.
y'" H, ve locity II becomes'" lin""
"'
-"'._••~---"Uc '" 5.75 loglo (RIR) + 3.75 '" 5.75 X 0 + 3.75 '" 3.75
,.
"0<
II ....... : 2.0 and u. from (I) '" 0.14511
2.0
- U '" 3.75
0.1 458
fJ :
Di scharge.
I I
2.0 - .1458 x 3.75 '" 2.0 - 0.5467 '" 1.4533 m/s
Q'" Area )( average veloc ity
Ii
~ I
IL
1450
Fluid Mechanics
==
10.5.5
'41t
~
_
D )( U
"'41t
J
1
(0.3) x 1.4533" 0.1027 m Is . '\n s.
Velocity Distribution for Turbulent Flow in Smooth Pipes by Power Law. The
ve loc ity distribution for tur bulent flow as give n by equations ([0. 18). (10.20) and ( 10.21) arc
logarithmic in nature. T hese equatio ns arc nO! cOIlVcniclll 10 usc. Nikuradsc carried OUl experiments
for difFerent Reynolds number 10 determine the vd oc it y dis tribu tion law i n smoot h pipes. H e cxprcsscd
the ve loc ity distribution in exponenti al form as
-"- = (yIR) lin
•.,( 10 .27)
" ~,
I
where cx ponelll - depends on Reynol d s number
"
T he val ue of (~) decreases. with incrca..;;ing Reynolds number.
- =-
Por
"
6
I
- =-
I
I
"
7
"
7
/I
10
Thus if - = - . the vel oci ty distribution law becomes as
u:.
=
( ~yn
...(10.28)
Equiltion ( 10.28) is kflown as I n ih power law of velocity disnibution for s mooth pip.:s.
to 10. 6
RES ISTANCE OF SMOOTH AND ROUGH PIPES
The loss of h.-:ad. due to fri cti on in pipes is given by equation (10.2) as
II =
I
~4~.!~.~L~.~V_'
rlx2g
In this equation. the val ue of co-"fficie nt of friction. f should h,:, known accurately for predi cting th.-:
loss of head due to friction in pipes. On the basis of dimensional analysis. it can h,:, shown that the
pressure Joss in a stra ight pipe of d iarne ta D , length L. roughness k. average velocity o f now fJ.
viscosity and density o f nuid J.l and p is
L\P = p~2$ [R,,~.~]
or
p&l=Q [R,.~.~]
2
E~perirnenla lJy it was found that
~ I
pressure drop is a fUIll:tion of
~
to the first power and hence
I~
~ I
IL
Turbulent Flow
pi;l
=
~ 9 [n,.~J
451 1
or
2
The te rm of lh e rig ht hand side is called co"c fficic nl offri cli on f. Thus/ = 1\1
[R" ~]
Th is equati on shows that friction co-e m de n! is a functio n of Rey no lds num ber a nd kID rati o . where
k is the ave rage hei ght of pi pe wall roughn ess protrusio ns.
(a) Variation of 'I' for Laminar Flow. In vi sco us flow chapte r. it is show n that co-effic ie nt of
fric tion "f fo r la min ar fl ow in pipes is g ive n by
f= ~
... (10.29)
R,
Thus friction co-e ffi c ient is olll y a funcli OI) o f Rey no lds number in case of lam inar n ow. It is indepe ndent of (kiD ) rat io.
(b) Variation uf l' fu r Turbul~nl Fl ow. For turhule nt flow . the co-dfi c ient of fri ct io n is a funct ion
of R, and kID ratio. For rel at i ve rou ghness (kiD), in th e Itlrbulc nt !lo w Ihe boundary may be smooth or
rough and hence the va lue of J ' wi ll be diffcrCIII for thcse bound ari es.
(,) '/' rur s muuth pip es. For turbul ent n ow in smooth pipes. co -e Ffici~nt o f frictio n is a func tion
of Rey no lds num ber o nl y. The value of laminar sub- laye r in case of s mooth pipe is large as compared
to the a vc ra ge hc ight of surface roughn ess k. The va lu c of 't' fo r s mooth pi pe for Rcy no ld s num be r
varying from 4000 to [00000 is give n by the re lati on
/=
.0791
(R,t
...(10.30 )
4
The equati on (1 0 .30) is give n by Blasiu s.
The valucof J ' for H, :> JO~ is obta ined from equati on ( [0. 23) whi c h gives til<' vel ocit y di stribu tion
for smooth pipe in te nn s o f avc ra~e ve locity
fJ
-
!t.
From equat iOll (10. 6), we
(U)
as
("oR)
= 5.75 10g lo - -+ 1.75
v
have/= ~ , where
pV'
...( [0 .3 1)
V= ave ra!!:", velocit y
... ( 10 .3 IA )
"'
Su bstitutin g the valu e o f u. in equatio n ( 10.3 1). we get
fj
U J/ 12
I I
=5.75JOg 1o [ J /1 2 )R +t.75
v
Ii
~ I
IL
Fluid Mechanics
~
vf
Taking R
'" 5.75 10g( D
12
[fIRv ..JTT2) + 1.75
= on and simplifying, tile above equation is wriucn as
I :2.03 Jog 1o
r<-.:
v4!
But U D = R, and hence above
"
~
l~uillion
[UD
- ,ffJ ) - 0.9 1
v
is writ1cn as
= 2.03 Jog lo (R,..[4j) - 0.91
, 4/
... (10.32)
Equ ation (10.32) is valid UplO R, = 4 x 1O~
Nikuradse's experimental resuit for t UTbtlk"! now in SlllOOlh pipe for 'ris
I
fA7 '"
,4 /
2.0 ]ogl o (R,
J41) - 0.8
... (10.33)
7
This is applicahk uPIO R, = 4 X 10 . But the equation ( 10.33) is solved by hit and tria l method. The
val ue of'j ' (i.e .. co -c fficicnl of friction) can alternate ly be obtained as
000
/=.8+
.05525
... (10.34)
(N,) 0'\1
.•.
The valueuf '1'Ii.e., fril:tion factor which is used in cqualion (ID.2A)] is given by
0.22 1
... ( I O.3·M )
/= 0.0032 + (R,t2l7
(ii) Va lu e of 1 ' fo r rough !liIJes. ror turbulent flow in rough pipes. the co-efficielll of frict ion is
,\ fu nction of rclmive roughness (kiD) and it is independent o f Reynolds number. This is because the
val ue of laminar sub-layer for rough pipes is very s mall as compared to the heig ht of sutface roug hness. The average velocity for rough pipes is give n by (10.24) as
U
-
= 5.75 log lO (Rlk) + 4.75
",
8m
Substituting the val ue o f
Ij.
in the above equa tion, we get
U
~
= 5.75 Jog 1o (Rlk) + 4.75
U ,, /12
which is simplified
But Nik uT<ldsc's
10
the fontt as
e~perintental
I
r;;-;
, 4/
I
~
, 4/
= 2.03
10g lO
(Rlk)
+ 1.68
result gave for rough pipe the following
=2 tog lO (Rlk) + 1.74
... ( I 0.35)
rel~t ion
for '1 '
~s
... (10.36)
(e) Va lu e of 'f ' fo r co mm ercial pi lles. The \'~Iue of '/' for commercial pipes such as pipes made
of metal. connc tc and wood is obtained from NikuT<ldsc's e~pe rimen tal data fo r s mooth and roug h
I I
Ii
~ I
IL
Turbulent Flow
453 1
pipes. Accordi ng to Colebrook. by subtract in g 2 log,o (Rlk) fro m both sides of eq uati o ns (1 0.33) and
(10.36). th e value of -I' is o btain ed fo r cO l11l11 ercia l s moo th a nd ro ugh pipes as :
1. Smooth pi lles
~
, 4f
- 2 laglO (Rlk)
= 2 log 10 (R,.j4j) - 0.8 - 2 log,o (Rlk)
Fl]
= 2 10gIO ( R,
R ik
... ( I 0.37)
- 0 .8
2. KOIIRh pi llt'll
I
~
,4 f
- 2 10&'0 (Rlk) = 2 log 10 (Rlk) + 1.74 - 2 10gIO (Rlk )
= 1.74 .
... (1 0.38)
Problem 10.7 For tile problem /O.6,jind tile co-effie;ell! offriC/ion and Ihe orange 11t>;glll ofrol,gllness projections.
Solution. From th c sol mio n of pro bl em 10.6. we have
R=0. 15111
u. = 0. 1458 lOIs
iJ = 1.4533 m's
r or co-e ffi cien t o f fric tio n. we know Ih at
II,
=
ff .J112
0'
0. 1458 = 1.4533.JTi2
0'
JT12 = 0 .1458 '" 0. 1
1.4533
f= 2.0 x (. l)~ = .02. Aus.
Hdg tll o f roughness project ion is o btained fro m equa tion ( 10.36) as
I
r:=-:
,4 f
~
2 loglO ( Rlk) + 1.74
Su bstitut ing the va lu es o r R andf. we get
I
.)4 x 0.2
0'
.15 )
logl o ( k
~
(0.")
2 loglo - - + 1.74 or 3 .5355 = 2 loglo
k
(k.")
+ 1.74
3.~
53~5~5~-~I~
.74C = 0.8977 == logl o 7.90
= -~
2
0 .1 5 == 7.90
k
k == 0.15 == 0.0 1898
7.90
In
== 18.98 111111. AilS.
Problem 10.8 IYater is flowitlg thmllgll a rOl jgll pipe of diometer 5()() mm atld Ie/lgtll 4(X){) m ot tile
rate of 0.5 m 3Is. Fitld the powe r required 10 maillwin this flow . Take ti,e m'erage heig/II of mug/mess
as k == 0.40 mm.
~ I
I~
~ I
IL
1454
Fluid Mechanics
Solution. Given:
Dia. of rough pipe.
D", 500
:, Radius,
R"'2=O.25m
111111 '"
0.50
III
D
Length of pipe.
L '" 4000 III
DiM: hargc.
Q'" 0.5 111 3fs
3
Average height of roughness. k '" 0.40 II1Ill '" 0.4 X 10- III
First fi nd the value of ,",o-efficient of friction. Then cakulmc the head lost duc 10 friction ilnd then
power required.
ror a rough pipe. Ihe value of l' is given by the equation (10.36) as
'2
I":"":" '"
.;4 J
, ( .25 )
]oglo (Rlk) + 1.74",. Ing lC
= 2 1og lo (625.0) + 1.74 = 5.591
.4 xlO
J
+ 1.74
+ 1.74 = 7.331
I 1 '" 0.1364 or /= (0. 1364t14
, = .00465
,,'4/" = 733
"'
Also the average velocity.
Head
lo~t
fJ
= Discharge =~=~ '" 2.546
If 0 2
It (.5)!
Area
4
4
duc to friction. "1=
4. f . L . V'
d x2g
'"
4 x .00465 x 4000 x 2.546!
O.Sx2x9.81
[": V=U =2.546.d=J)=Q.5J
=49 .1 6m
Power required.
p =
=
IV x 11/
w.Q.iI!
pxgxQxll,
IlXlO
IlXlO
11100
kW
1000 x 9.8 1 x 0.5 x 49.16
1000
"" 241.1 3 kW. Ans.
Problem 10.9 A JmOOlil pipe of diameler 400 mm and /ellglll 800 m carries ...aler al Ille mIl' of
0.04 m)/s. D elermine Ille head IOJ'I due 10/riCliOIl ....all shear Slres~·. ce"lre-/ille \'e!ocily (md Ihickness
o/imlli/llir sub-layer. Take Ihe killematic l"iSfMily of \I'll/a as 0.018 Siokes.
Solution. Givcn :
Dia. of pipe.
Radiu s.
Length of pipe.
DiSl:hargc.
Kincmatic viSl:osity.
Avcragc velocity.
D "" 400 mm "" 0.40 m
D
R""2,,0.20m
L"
800 m
Q" 0.04 mlls
v " 0.018 stokcs " 0.018 l:m 2ls " 0.0 18 x 10.... m 21s
fJ "" ....Q..... '"
Arl:a
0.04
"" 0.3183 m/s
It (0.4 )1
4
Rey nolds number.
I I
R, '" VxD" U xD '" 0.3J83 x 0.4 ,,7.073 x 104
V
V
.0 18 x 10-4
Ii
~ I
IL
Turbulent Flow
455 1
The flow is turbul ent.
J'
The co-effi cie nt of fric tion
f
is obtained from equation ( 10 .30) as
= .0791 =
0.0791
( H, ) '"
(7.073x 10 ' )'"
f".__L:c.~V_l
,4 ~
.
(,) Head lo~t du e to friction. "! "'-
'"
= .0791 '" .00485
16'0
..,
4. f . L. fJ;:'
Dx2g
Dx 2g
4 x .00485 x 800 x (.3183)l
0.40x2x9.81
'" 0.20 m. An s.
( ii) Wall she ar stress (tol is give n by equation (10.5) as
f
_
(I -
f· p· Vl _
2
j. p. fP
[-o'
2
v= U I
,
= 0 .00485 x 1000 x (.3 I84 )l N/m = 0.245 Nlm',' AilS.
2.0
(iii) The ce ntre- Jin e ve locit y (II"",) for smooth pipe is g ive n by equatio n ( 10 .20) as in which
j j : II ....
at y "' R
II
....!!:O!. '"
II .
whe re th e she ar ve locit y
II , =
II . R
5.75 Jog 1o -
v
J¥ = J~:
+ 5.55 jPut in equation ( 10 .20),
II '"
II".., at y '" HI
= JO.OOO245 = 0.01 56 Ill/s
Substitutin g the valu es o f u •• R and v in th e abo ve equation, we gel
~: 5. 7 5Jogl o 0.0 156xO,;0 + 5 .55 :24.1 73
0.0156
.0 18 xlO
II ....... : 24. 173 x .0 156 '" 0.377 mls. A DS.
(i ,') Th e lh ickn ess o f laminar s utl·la yer (0') is give n by
0': 11.6xv '" 11.6x.0 18xlO-4 '" .001338 m '"
1.33H 111m. An s.
.0156
Problem 10.10 A rOl jgh pipe of i/iaml!ler 400 mm wid leng/II /000 II! carries »"ater (If /he rUle of
0.4 ", J/s. The 1\"(11/ rOllghness is 0.012 mm. Determine the co-efficie'll of f riction. \\"(11/ s/JelJr ~·Iress.
centre-line "d ocity and "e/ocity (It a distan ce of 150 mm f rom the pipe 11'(11/.
Solut ion. Given:
D ", 400mm ",0.4 m
Dia. of rough pipe.
II,
Radiu s.
2
2
u:n glh of pipe.
L", 1000 m
DiSl:hargc.
Q : 0.4 ml/s
Wall TOu ghness.
I I
R ", 0 = 0.4 = 0 .20 m
I: = 0.0 12 rnm '" 0 .0 12 x IO- J III
Ii
~ I
IL
1456
Fluid Mechanics
'f'
(i) T he va lue o f co-cfricicllt of fricti on
1.0 = 2 log
~
, 4[
1.0
ri"7 = 2
,,4[
=2
4/ =
10
]ogl o (
IO~IO
for ro ugh pipe is given by the eq uat ion (10.36) as
(RII<;) + L74
0.20 1 ) + 1.74
.012x l0
(16666.67) + 1.74 = 10. 183
(_,_)2 '"
.00964
10.1&1
_ .00964 _ 00'
[ _ - - - _ . _41. Ans.
4.0
(ii) Cen t,..., -li"" velocity (II ..... ) for rough pipe is give n by equation ( 10.21 ) in which u is made
:=
"m.., at )' '" R and hence
"nu,
".
= 5.75 log lO (Rlk) + 8.5
...(il
where shear velocity.
and t o'" wall shear stress '"
where V",
~[ PC'.V_'
'C.
2
Di sc harge
Q
Q
"'7'''''''"
--"
--= 3 . 183 m /s . A ns .
Area
~D2
~ ( .4 )l
4
4
= .24
00 I x l 000
11.=
f:=Jlt~
=0.11
3.1831
x -- "
2.0
Illis
Substituti ng the value of II .. R, k in equation (I). we get
II"",. '" 5.75 log lo (
0.1 1
0.2
.0 12 x 10
J) +
8.5 '" 32.77
II ...., = 32.77 x 0.11 = 3.60 m/s. A ns.
(i l') Velocity (u)
~t
a distance y= 150 111m
= 0.15m
The velocity (u) at any poinT for rough pipe is giv en by equation ( 10.21) as
- " = 5 ,75 Jog 1o (ylk) + 8.5
,,'
whae
I I
It." 0.11 m/s and y" 0.15 111. k = 0.0 12 X 10-J m
Ii
~ I
IL
Turbulent Flow
0.15
-, - '" 5.75 10);10 (
0.11
.01 2 x 10
J
457 1
) + &.5 '" 32.05
/I = 32.05 x 0.11'" 3.52 m /s . An s.
Problem 10.11 A smooth pipe line of 100 mm diameter c(!rries 2.27 m3 per millule ai,,'mer (If 20°C
wilh kinem(llic "iscosily of 0.()098 Ilokes. Calculate III I' friction Judor. m(l.limllIM refoei/)' (lJ we ll (l~'
x/wur SlrellS
(l/
IiiI' bo undary.
SOlution. G ive n :
Dia. of pi pe.
D = lOO rnm =O. 1 111
R '" 0.05 In
Radiu s of pipe.
Di ~ h arg".
Kine/lHll k viscos ity, v '" O.{l098 stokes'" 0.0098 e m'ls = 0.0098 x 10--1 m~ls
. g .ive n ..vy
Now average vI' IOC .lt y IS
u '" -Q-
" c,O~.O~3~78'C
Area
U xD
Rey nolds nu mbe r is give n by. R,= -
v
,
~( O .l )'
=
0.0378x4 -_4.8 17 m/,
If
x om
4.817xO. l
.... =4.9 154 x lOs,
0.0098 x 10
The fl ow is lurbul clll and R, is mo re (hall lOS, Hence for smooth pipe. th e co·crfkicnl of fric ti on
'f ' is o bla incd from equ ation (10.33) as
I
J4i == 2.0 logl o (R, J41) - 0.8
I
fA"i
"'
,,4/
" - O.S
== 2.0 logl o (4.9 154 x 10, x ..."4[)
== 2.0 Il og lO 4.9 154 x lOs + loglo ,f4i 1- 0.8
= 2.0 15.69 15 + loglo J4l J - 0.8" 2 x 5.6915 + 2 loglo
" 11.3830 + 10g\ 0 (,f4i)1
I
,f4i -log 1o (4j)"
"'
- 0.8 =
,f4i - 0.8
11.383 + loglO (4j) - 0.8
11.383 - 0.8 " 10.583
...(i)
(i) FriCli01I[aClo r
Now. friction fac tor (J* ) = 4 x co·cfficic nt of friction " 4[
Substitu ti ng the va lu e of '4F in equa ti o n (i), we gel
I
~
, ['
- log 10
r"
10.583
.•. (i i)
The above eq uati o n is so lved by hit and trial method.
Lei
= o. t, th e n L.H.S. o f equation (ii), bn'()mes as
r
L. H.S."
I I
I
f'i<:O - log 1o 0. 1 ,,3. 16 - (- 1.0) ,, 4.16
... 0.1
Ii
~ I
IL
1458
Let
Fluid Mechanics
r
== 0.01. then L.H.S. of equation (ii). becomes as
I
=
",0.01
L.H.S.=
log1 oO.Ol '" [0 - (- 2)= 12
BUI (orexael solution. L.H.S. sliould be 10.583. Hence value ofr lies betwccn 0.1 and 0.01.
Le!.t" '" 0.013 the n L. H.S. of equation (ii), becomes as
=I
L. H.S. '"
"O.Ot3
- 101: )0 0.013", 8.77 - (- 1.886) '" 8.77 + 1.886 == 10.656
which is approximately .-:qualto 10.583.
Hence the va lue of is equal \00.013.
Friction factor.
= 0.0 13. An s.
(ii) M" ximum "e/ueil), (u"",,)
Now we kROW that
= 4/
r
r
r
f· '" 0.0 13 = 0.00325
Co-efficient of friction. ! =
4
4
Now the shear ve locity (If.) in krrns of co-eFfIcient of friction and average velocity is given by
cqu31ion (IO.3IA) as
u.==u-
fi
f7 =4.817x
r·
OO325
2
=4.817xO.0403=O.J94
For smooth pipe. the velocity al any point is g iven by equation ( 10.20)
,,=
Il.
[S.75 10g lO
II. : ) '
+ 5.55]
The velocity will be maximum ~t the ce ntre of the pipe,
where y = R = 0.05
i.e.. radius of pipe. Hence the above equation becomes as
1
U"",-,='" [ 5.75 10g 1o -,.xR
,- + 555
= 0.194 [ 5.75 1og lO
(iii ) Sllear
1
0. 194 x 0 .05
• + 5.55
0.0098 x 10
'" 0.194 [22.974 + 5.55[
(t o)
J'lres,~ allhe bOllllliary
".=
We k now that
= 5.528
m ls. Ans.
Pi or,,~= ~
t o = P ,, 2. = 1000 X 0.194 2 = 37.fi3 N/m ~ . An s.
Problem 10.12
(p = J()()() kg/m
J
fly<frod}'lwl1Iically smoot!! pipe carries waler aI the rale of 300 lis at 20°C
,
v = I~ m2/s) wilh a head loss of
diamerer. Use f = 0.0031 +
I I
0.221
(R,)
0.2J7
eqlwlion fo r f.
/englh of pipe. Determine Ihe pipe
/xLxV z
pVD
wltere /If = ~""~~ and R< = - .
D X2g
~l
3 III ill
100
111
Ii
~ I
IL
459 1
Turbulent Flow
Solul io n. Gi ven:
Dis.: hargc.
Q = 300 lis ", 0.3 mlfs
Density.
Kine mat ic viscosit y.
p = 1000 kg/m l
v '" 1O-~ ml/s
~!cad
Leng th o f pipe.
I,, "" 3m
L '" 100 m
Va lue of fri ctiun fa(;lor.
f '" 0.0032 +
loss.
Rey no lds numbe r.
pVD
0.221
n ",
(R.)""
V)( D
R, : - , - : -,-,-
vx D
6
: - - , - = VX D XIO
10-
Find: Diamcler of pi pe.
Let D = Di ameter o f pipe
I'!cad loss in ten us of friction fad or is g iven as
'
!J: -,J'cX<oL'cX-,-v_
I
Dx 2g
3=
LJ~X~I~OO~X""V~'
C:
D:><2 :><9 .81
3)(D)( 2 )( 9.8 1
100 V ~
/=
No w
or / =
/', = 3, L = 100 Ill)
0.58860
V2
...( i)
Q = AxV
O.3 = ~ D2 xVor dXV ", 4x 0.3 = 0.382
,
4
v = 0 .382
D'
A lso
"'
_
f-
... Ui)
'}
O.!X)3_ +
022 1
• '"
(R, )-
0.588,6 D = 0.0032 + _ _O~,2~2~1",,,,
VxDxlO '
V' I, , )Oll1
C
From equation (i) .f '"
"'
0.5886 D '" 0 0032
,
( O~~2 r
.
+
0.22 1
(O~~2 x Dx
10
05~S: 0
IO~ )
0 '37
6
) -'
( ":
I I
and R, '" V x D x
. ..
0.382 )
D'
From CqU3UOU (U ), V = - -'-
Ii
~ I
IL
1460
Fluid Mechanics
0.5886 X, D~ '" 0.0032 +
0.382'
-;-_~O.~22~',",=
(0382 x IO ~
t!J1
D U.!J1- -
4.0.33 rY '" 0.0032 + 0.0 105 x Do.m
4.033 D~ - 0.0105 DD.m - 0.0032 '" 0
0'
0'
... (iii)
T he 3OO\'C equat ion (iii) will be solved by hit and trial method .
(il Assume D " I m. then L. H.S. of equation (iii). hecomes as
L. H. S. '" 4.033 x 13 _ 0.0105 x 1°. 137 - 0.0032
= 4.033 - 0 .0105 - 0.0032 = 4.0 193
By inc re asing the val ue of f) more than 1 111. the L.H.S. will go on in creasing. He nce decrease the
val ue of O.
( ii) Assume D = 0.3 Ill. then L.H.S. of eq uati on (iii).
becomes as L.H.S. '" 4. 033 x O.3~ - 0.0 105 x 0.]°·231 - 0.0032
= 0.0098 - 0.00789 - 0.0032 = - 0.00129
As this va lue is negative. th e val ue of D wi ll be s li gh tl y more th an 0.],
(iii) Assume D '" 0.306 m. Ihen L. H.S. of equati on (iii). becomes as
L. H.S. = 4.033 x O.306~ _ 0.0 I 05 x 0.306 o.m - 0.0032
= 0.0108 - 0.00793 - 0.0032 = - 0.00033
This value of L. H.S. is approximatel y equa l to zero. AClllally the va lu e of f) wi ll he sligh tl y more
than 0.306 m say 0 .308 m . Ail S.
Problem 10.13 Waler is /lowing IllrOl18h a rough pipe of diameler 600 mill (jf file rme of
600 Ii/res/second. The 'mil roug/lne.ls is J 111111. Find Ille power /OSI for I kill leng/h of pipe.
Solution. Gi\"en :
Dia. of pipe.
Radius uf pipe.
Discharge.
Wall roughness.
f) '"
600 mm == 0.6 m
8= °·6 :0.3m
2
Q = 600 litrels = 0.6 m 3 /s
k == 3 mm '" 3 x 10- 3 m '" 0.003 m
Length of pipe.
L = 1 km '" 1000 1Il
Fur rough pipes, the co-efficienl of fri<.:tiun in lenns of wall roughness, k is gi\"en by equalion (10.36)
as
,
( 0.3)
fA7 =2 10g lo (Rlk)+ 1.74=2Io8 1O - - +1 .74= 5.74
-.J4 f
0.003
0'
Fl '" -5.76
'- == 0. 174 2 or 4f == (0. 174 2)2 '" 0 .03035
1
4 fXLXV
The head loss due 10 friction is given tly. hf = -"--;;-";"-D x2 8
I I
Ii
~ I
IL
Turbulent Flow
w here
v:
Q ",
A
0.6
~ (O.62)
'" 2. 122
461 1
111/5
", ,,,
0.03035 x HlOO x 2. 122 '
0.6x2:><9.81
: 11.6 m
T he pow er* 10SI is give n by, P '" pg x Q x h! '" 1000 x9.8 1 x 0.6 x 11.6
lOOO
~w '" fiS.27 kW. An s.
1000
HIGHLIGHTS
I . If the Reynold number is less Ih:m 2000 in a pipe. the flow is laminar while if the Reynold number is
more than 4000. the flow is turbule!!! in pipes.
2. Loss of pressure head in a laminar flow is proporlionallo the mean velocity of flow. while in case of
turbulent now il is approximately proportional to the square of wlocity.
3. Ex['ression for head loss due to friction in pipes is gh"en by Darcy -Weisbach equation.
Z
h '" 4 x f x Lx V . where! '" co-eff,cicll t of friction
J
dx 21(
'" J x Lx V' . where! '" friction factor
dx 2g
4. Co-<:fficicnl of friction is expressed in tenns of shear Slress as ..
3
'v
where V = mean velocity of now. p .. mass densi ty of nuid.
5. Shear Stress in turbulent now is sum of shear StreSs due to vis.cosity and shear Stress due to turbulence.
i.t .•
r '" t ,. -+ r,. where
t, '" shear stress due to viscosity
t,. shear stress due to turbulence
dii
(fii
., - +, -
d)'
d}'
6. Turbu lcnt shcar Siress by Reynolds is given as
where
7.
t '"
P w'y'
u' and ,r. nuctuating component of velocity.
TIle expression for shear
stres.~ in turbulent now due tQ I>r:mdtl is t ~ pI" [d")l.
where 1_ mixing length.
,Iy
8. The velocity distrib uti on in the turbulent now for pipes is given by the expression
u ~ u_ -+ 2.5 u· log, (yfR)
where
11_ .. is the centre-line velocity.
}' ~ distance from the pipe wall,
R .. radius of the pipe.
and
· Pow~r
I I
11. '" shear velocity wbich is equal 10
.. pg x Qxhf wa1t ..
Iff.
1000
Ii
~ I
IL
1462 Fluid Mechanics
9 . Velocity defect is the difference between the ma~ jmurn vdocity (u_,l and local velocity (u) at any
point and is gi"cn by (u ..... - u) _ 5.75 X w. logIn ( Nlv).
10 . The ooundary is k.nown as hydrooynamically smooth if t. Ihe average h<'ighl of Ihe irregularities project ing from thc surface of Ihe Doundary is small compared \0 Ihc thickness of Ihc laminar sub-layer (0') and
boundary is rough if k is large in comparison with Ihe thickness of Ihc sub-layer.
Or if
,<
,
6'
~
and if
,
0 .25. the boundary is smoolh : if 0' > 6.0, the boundary is ro ugh
lie Dctwcen 0.25 \0 6.0 , thc boundary is in transition.
II. Velocity distribution for lurbulent flow is
•
'.
_ 5.75 [og lO u.y + 5.55 for smooth pipes
"
~
5.75 10SI0 0'11::)
where " " velocity at 3ny point in Ihe turbulent flow.
u.
~ shear vcb::ity nnd ~
J*.
\I
*
+ 8.5 for rough pipes
~inematic viS<.'()sity of fluid.
y .. disunce from pipe wall. and k .. roughness factor.
11. Veloc ity distribution in tenns of avernge velocity is
U
-
'.
w.N
.. 5.7510g tO- - + 1.75 for smooth pipes.
"
'" 5.7510)\' 0 NI t + 4.75 for rough pipes.
Il. Difference uf local velocity and ave"'~l!e velocity for smooth and ro ugh pipes is
u
-fJ
- - '" 5.75 10gt O(yIR) + 3.75.
'.
14. The co-efficient of friction is given by
f.,
•
"
R; ..... for lami nar flow •
0.07~~
for turbulent flow in smooth pipes for R,
(R,)
.05525
n'
•.0008 + -----.... for R, S hE but
(H.)"'
~
'"
<!;
4 x 10
<!;
4000 by S
1O~
,
• 2 10l! to (Rlk) + 1.74 for rough pipes whcre R, . Reynolds number.
EXERCISE
(A) THEORETICAL PROBLEMS
1. What do you understand by turbulem fluw ? What factor decides the tyPe of flow in pipes "
2. (tI) De.i,·c ~n exprcssion for the loss of head due to friction in pipes.
<I» Derh'c Darcy -Wcisbach C<jll.:t.tion.
<J.NT.U.. Hyderal>ad. S 20(2)
J . E ~plain the tcnn c'O-efficicnt of friclion. On what factors docs this co-cfticient depend ?
I I
Ii
~ I
IL
Turbulent Flow
463 1
4 . Obtain an expression for the co-.,fficienl of friction in the lenns of shear ,tress.
5. What do )'OU mea" by Pmndll mixing u,ngth Theory? Find an expression for shear slress duc to l>randt!.
6. Dcriw un cxpre3>ion for Prundl!'. universal ,-"I<>city distribution for lurbulent now in pipes. Why this
vciocity distribution is called universal ?
7. What is a velocity derect ? Dcrive an expression for velocity defect in pipc •.
g, How wou ld you distinguish between hydrodynamically smooth and rough boundarie s?
9. Obta in an expression for the "c!ocity distrib ution for turbulent flow in smooth pipes.
III . Show that "eloc.ily distribution for turbu lent flow through rough pipc is gi "en by
-"
~
5.75
log l~
(,-Ik) + 8.5
".
where u.", shear velocity.)," distance from pipe wall. k., roughness factor.
II . Obtain an expression for wlodty distribution in temts of ""crage velocity for
(<I) smooth pipes ~nd (b) rough pipes,
I l . Prove tl1.al the difference of loc~1 velocily and avcrage "clocity for lurbulent now through rough or 511looth
pipes is given by
" -u
- '" 5.75 10g lO 6lR) + 3,75,
'.
13. Obtain an
e~pression
for velocity distribution in turbulent now for (I) smooth pipes and (iiI roug h pipes.
(Delhi Uni,'er.,ily. I)eceml>er. ZOO])
(B) NUMERICAL PROBLEMS
J. A pipe-line carrying water lias average height o f irregUlarities projecting from the surface of the
boundary of tile pipe as 0.20 rum. What type of the boundary is it ? The shear stress development is
1
7.848 N/m . Take value of kinematic viscosity for water as 0.01 stokes. IAn s. Boundary is in transition I
2. Detenninc the average height of the roughness for a roullh pipe of diameter 10,0 em when the velocity at
a I"'int 4 em from wall is 40% more than the "elocity at a point I em from pipe wall.
[Ans. 0_94 eml
J . A smooth pipe of diameter 10 em and 1000 m long carries water at the mte 0[0.70 m1/minute, Calcu late
the loss of head. wall _,hearing stress. centre line velocity. velocity and shear stress at 3 em from pipe wall.
Al so calcu late the thickness of the laminar sub-layer. Take kinematic viscosity of water as 0.015 stokes
and value of co-efficient of friction J as
f·
4.
S.
6.
7.
I I
.0791
--,,-4
. where R, '" Reynolds number.
(I<)
IAns. 20.05 m. 4,9 Nlml ; 1.774 mls ; 1.65 mls ; 19.62 N/m1 : 0.248 mml
The velocities of water thro ugh a pipe of diameter 10 cm. are 4 mls and 3.5 mls at the centre of tMe pipe and
2 cm from the pipe centre respectiYely . Delennine the wall shearing stress in the pipe for turbulent now.
[Ans. 15.66 kgflnt' l
f or turbulent flow in a pipe of diameter 200 mm. find the discharge when the centre· line velocity is
30 m/s and veloc ity at a point 80 mm from the centre as measured by pilOt·tube is 2,0 m/s.
[Ans. 64,9 litres/51
For problem 5. find the co-efficient of friction and the average height of roughness projections.
IAns. 0.029. 25 ,2 mml
Water is nowing through a rough pipe of diameter 40 em and length 3000 m at the rate of 0.4 ml , •. (:ind
the power required to maintain this flow . T"kc the average height of roughness as K _ 0.3 mm.
[Ans. 278.5 kNI
Ii
~ I
IL
1464
Fluid Mechanics
K. A ,mooth pipe of diameter 300 111m and length 600 m carrie. waler at rate of 0.04 ",2/s. OClcnninc thc
head losl due [0 friction. wall shear stress. cenlre-line ~elOCilY and thickness of laminar sub·luycr.
Take the kinematic viscosity of water as 0,018 stokes. lAos. 0.588 m, 0.72 NiemI, 0.665 m/s. 0.779 mml
9. A rough pipe of diameter 300 mm and length 800 111 carries water at the rate of 0.4 mIls. The wall
roughness is 0 ,015 mm. DClcnninc thc co-dflcicn! of friction. wall shear stress. centre line "elocil), and
velocity at a distance of I ()() min fmm the pipe wall.
IAn s. f .. .flO263. TO", 42. 08 N/cm'. w.... '" 6.457 mIs, U = 6.2 49 m/s]
I ll . Dctcmlinc the d;,lance from the
~-en(re
velocity for turbulent now in pipes.
I I
of Ihe pipe. a( which the local
~elocity
is equal to Ihc :l"cragc
[Ans. 0.7772 RI
Ii
CHAptER
to
I 1. 1 INTRODUCTION
In chapters 9 and 10. la minar flow and turbulent flow have ~"n discussed. We have ..;;cen lilat when
th" Reynolds number is less than 2000 for pipe flow. the flnw is known as laminar flow whereas when
the Reyno lds number is more than 4000. the flow is known ;IS turbulent flow. In this chapter. 1he
lUrbulent flow of fluids through pipes running full will be considered. If 1he pipes are panially full as in
the case of sewer lines. the pressure inside the pipe is same and equ:1110 atmospheric pressure. Then the
flow of fluid in the pipe is not under pressure. This case will be taken in the chapter of flow o f water
through open channels. Here we will consider flow of fluids 1hrough pipes under pressure only.
to I 1.2
LOSS OF ENERGY IN PIPES
When a fluid is flowing through a pipe. the fluid expe riences some resistance due 10 which some of
the energy of fl uid is losl. This loS1S of energy is dassified as :
Energy LolSes
I
I
1. Major Energy Losses
2. Mioo< Energy Losses
I
I
This is d~ to friction and it is
cak:ulated by the f,"lowing
formula .. ,
This is du.. to
(s) Sudden expansion of pipe
(b) Sudd<!n eoolractioo of pipe
(e) Bend in pipe
(0) Pipe fittir.gs ..Ie.
(s) Darcy·Weisbach Formula
(b) Ch .. zy's Formula
(e) An
to 11.3
obslruc~oo
in pipe .
LOSS OF ENERGY (OR HEAD) DUE TO FRICTION
(a) Darcy· Wels bach formula. The luss of head (or energy) in pipes due IU fri<.:liun is calcul;ncd
from Darcy-Weisbach equ~lion which has been derived in chapter 10 and is given by
,
Ie'."L".V,-
,,4 ".
11=,
d x 2g
whe re
... ( 11.1}
II, : loss of head due to frict ion
'65
I I
Ii
~ I
IL
1466
Fluid Mechanics
f
== co-efficient of friction whic h is a fUllction of Reynolds number
~
16
for R, <
.
2000 (VISCOUS
R,
now)
0.079,or R, varymg
. f w m 4000
== ----;!4
R,
10'
!O
L == length of pipe.
V == mean velocity of flow.
d ==
diamc l~r
of pipe.
(b) Ch el ~" s I' UI-mulll for loss uf IIl."lI d du .. to fr ldkm in p ili...". Rdcr \0 chapler 10 ankle 10.3.1
in which expression for loss of head duc \0 friction in pipes is derived. Equation (iii) o f anide 10.3.1. is
f'
P
pg
A
Ilf == - x~XLXV2
whe re II ", loss of head duc to fril:tion.
f
A == ~rca of noss-scdiun uf pipe.
and
V = mean ve locity o f flow.
A [ ==
Now the ratio of ~
P
... ( I 1.2)
P == welled perimeter of pipe.
L = length of pipe.
Area of flow
Perimeter (wetted )
1
is I:allcd hydraulil:
UlCilll
depth or hydmuli<.: radius and
is denoted hy m.
Hydraulic mean depth,
SubstilUting
In
A
P
_
==
:0
A
~d 2
4
d
P
'"'
4
- = -- ~ -
In
l In
. equntlOn
. (II .2). we get
or -P
:0 A
III
pg
II
xmx _I ", _xlnx
--.L
L
f'
L
... (11.3)
Let
rei = C, whe,"" C is a constant known
~~
per unit length of pipe.
Substituting the valucsuf
ff
and
JEt
3.'i
Che zy's constant and!!.L '" i. where i is loss of head
L
in equation (11.3), we get
..r;;;;
V=C
... ( I 1.4)
Equation (11.4) is known as ChCl.y'S formula. Thus thc loss o f head due to friction in pipe from
Chczy's formula can be obtained if thc veloc ity ofllow through pipe and also the value of C is ~nown.
The value of III for pipe is always equal to d14.
I I
Ii
~ I
IL
Flow Through Pipes
467 1
Problem 11.1 Find tile head la,I'/ due 10 fricliol! ill a pipe of diameler 300 mm alld lellg/II 50 III,
Ihrough which Wafer is flowing a/ a re/ociry of 3 mls /Ising (i) Darc), [orlllll/o, (ii) Che;.y'sformula for
.... hich C '" 60.
Take y jar water'" 0.01 sioke.
Solution. Given:
Dia. of pipe.
d", 300 mill '" 0.30 m
L",
50 In
lA:ngth of pipe.
V:3 rnls
Veloc ity of flow.
Clw zy's con,tan!.
C= 60
Kinematic viM'oshy,
1'= 0.0 1 stoke = 0.01 cm!/s
= 0.01 x lO--4m Z/s.
(i) Durey ~'ormul" is given by equation (Il.l) as
where
II =
~4~.f'c"
. L~.~V_'
r
d x2g
J' = co-efficient of friction is a function o f Reynolds number. R,
= 3.0 x 0.30 =9 x lOS
R = Vxd
But R, is given by
,
.01 x 10"'"
v
Value of
.00256 x 50 x 3!
"1=4 x03x2.0x9.8J
'" .78211 rn. Ail S.
Head los!.
(il) C helY's Formul a. Using equation (1 1.4)
V= C
where C '" 60.
d
4
III '" -
0.30
4
'" - - "
.r;;r
0.075
1ll
3", 60 ,).075 x I or i"
,
h,
hi
L
50
(3)' I
60
x .075 '" 0.0333
,= - = -
/,
Equating the two valu es of i. we ha ve 5~ '" .0333
II! '" 50 x .0333 '" 1.665 m . t\ns.
Problem 11 .2 Find the dlallieter of a pipe of lengrh 2IXXJ m .... Ilen rhe rate off/ow of\\'arer rhrougll
the pipe is 200 litreYs (md the hemf losr due to friction is 4 III. Take rile \'alue of C '" 50 ill Chezy's
fo rmulae.
I I
Ii
~ I
IL
1468 Fluid Mech a nics
Solution. Given:
L : 2000 111
Length o f pipe.
Di sc harge.
Q == 200 litre/s == 0.2 m3/s
Head lost due to frict ion.
I,! '" 4 m
Value of Chczy's constant. C = 50
lei the diamete r of pipe '" II
Velocity of flow.
V'" Disdlarg c
Area
H ydmulic mean depth.
m= -
d
4
4
/If
Loss of ltead pcr unit leng th . i = -
'" - - = .002
2000
L
Chczy's formu la is given by eq uation ( 11.4) as
v=
c,;;;;
Substituting the values o f V. m. i and C. we ge l
0.2 )(• 4 _
_.,cO
'ltd -
. bo t h SI·d CS, -d
Squarmg
4
X .
~~ x .002 ur J~ x. 002 __
4
4
.OO509
~
•
002 _
.~259 or
d
d
d= >lJO.0518 =( .0518) " ~ ==
,r '"
0.2, x 4 __ .00509
,
nd- )(50
11-
4 x .0000259 '" 0.0518
.002
O.553111 == 553 111m. Ans.
Problem 11 .3 A crude oil of kinematic I'iscosiry 0.4 sto lce is flowing Iitmugil a pipe of diameter
300 mm at 111l' mil' of 300 lilres pc, sec. Find (he head 1051 due ro friction/or a {'!IIg/h of 50 In oflhe pipe.
Solution. Given:
Kine m;l1ic viscosi ty,
v = 0.4 stoke = 0.4 em'/s =.4 x 10--1 It/Is
Dia. of pipe.
Discharge.
Lenglh of pipe.
d = 300 Illm = 0.30 nI
Q = 300 lilrcs/s = 0.3 ur' /s
Velocity of fl ow.
V= -
L = 50
III
~O~.3~
= -;;
= 4.24 m/s
Q
~ ( 0.3 )1
Area
4
.". Rey nolds numbe r.
R = Vxd = 4.24x OJO =3.18 x 10~
,
v
0.4 X 10--1
As R, li es t>ctween 4000 and 100000.
th~
.079
v;tlue of f is g ive n by
.079
f = - -,,-, =
"'
iN,)
(118x 10' )
I I
= .00591
Ii
~ I
IL
Flow Through Pipes
He ad lost du e to fr ictio n.
"J ~
4. / .L.V'dx2g
'"
4691
4 x.0059 I x50x 4.24 '
'" 3.61 m . An s.
03 x 2 x9.81
Problem 11 .4 An oil of Sf!. gr. 0.7 is flowing through" pipe of diameter 300 m", a/ the mil' of
500 li/re:i/s. Find Ille head /OJ'l dill' /0 friction amI power req"ired 10 mail/luin Ihe flow for a lenglh of
1000 m. r ake v '" .29 stokes.
Solution. G iven:
Sp. gr. of oil.
s= 0.7
Dia. of pi pe,
d= 300 m in '" 0.3 III
Di sc h il rg~.
Q '" 500 lilresls = 0.5 1I1 )/s
Le ngth of pipe.
L = IOOOm
v= ~ = ~_
Veloci1y,
Area
1'1:
d' -
0.5 )( 4 = 7.073 m/s
X 0.3 '
If
4
7.073 )(0.3 = 7.3 16)((10)4
0.29 X 10 -4
Rey no ld s numbe r.
.
. ,
.079
0.79
Co "cffic lcnl o f fnctlO n. /= ---w- =
~ = .0048
R,
(7.3 16 )(104)"
He ad lost d ue to fri cti on. IIj =
Power required
'"
4x
f x L X V· 4 x .0 048 x 1000 x 7.0 73'd x2g
pg .Q. ll j
I ()()()
0.3 x 2 x9.81
'" 163.1 R In
kW
where p = density of oil = 0.7 x 1000 = 700 kg/ml
Powe r requircd = 7()() x 9.8 1 x 0.5 x 163.18 = 560.28 kW . Ans.
I ()()()
Problem 11 .S Ca/cu/(1Ie Ihe t/isc/wrge Ihrougl, II pipe of dhm'e/er 200 """ Irhen IIII.' difference of
pressure helld be/wee" Ihe IIro ends of (l pipe 500 iii oporl is 4 /Ii of 1I·Il/er. Tilke llie 1·llli,e of
"/. == 0.009 ill Ihe formula h ==
f
c4~,f'c'"L~,~V_'
d x 2g
Solution. Givc n :
Dia. of pi pe.
d=200 mm =0.20 m
Lc nglh of pipe.
t=500 m
Differe nce of pressure head.
hI '" 4 m of water
f== JI09
Usi ng equation ( 11.1). we ha vc
0'
I I
Ilf '"
4.0 '"
4x f xLxV l
d x 2g
4x.OO9x500x Vl or vl = 4.0'1'.0.2'1'. 2 '1'. 9.8 1 ",0.872
0.2 '1'.2 '1'.9.8 1
4.0 x.{)()9 x 500
Ii
~ I
IL
1470 Fluid Mechanics
V'" J0.872 '" 0.9338 '" 0.934 Illis
Dischar£c,
Q'" veloci ty x area
11
It
,
,
= 0 .934 x - d- = 0.934 x - (0.2)4
'" 0.0293
111 3{S
4
= 29.3 Jilres/s. ,\"s.
Problem 11.6 Water i~' flolt'i"g through II pipe 0/ diameter 200 mm lI"ilh Ii nducil), of J m/J'. Filii! the
head lost due /0 [ric/ion [o r a /ength of 5 II! if the co ·«fficiell/ of f riction i~' gil''''' by f = 0.02
+
~O:j'
,
where R, ;$ ReYllolds number. Th e kinemaric I'iscosiry a/waler '" .0 1 .Hoke.
Solution. Given:
= 0.20
Dia. of pipe.
Ve locity,
d = 200
Length,
Kincrn:J.Iic vL'iCosil y.
L =5 m
v = 0.01 stoke = .01 x lQ---I m1fs
:. Reynolds number .
R = Vxd = 3)(0.20 :6x 10'
Value of
111111
III
V = 3mls
,
\'
f~
.0 1 x 10"'"
09
02 + -0-)=02+
R,
0.09
.09
(6XIO' )"
1
=.02+ - 54.1 3
'" .02 + .00166 '" 0.02166
.. Head lost due 10 frict io n. Ilr
4 x! x 1- x V~ =24".0~X~.~
02cIC66
::;;:Xc5~.O~X7"-3'
dx2g
0.20x2.0x9.8 1
== 0.993 III of Wilie r . Ail S.
Problem 11 .7 A" oil of sp. gr. 0.9 and \·isw sity 0.06 poise is /loII'i"g IllrOllg/l a pipe of diameter
200 mm at Ihe rale of6O liIles/s. Fimf Ihe /,,'ad losl dlle to friction for {/ 500 m "'''Slir of pipe. Find
Ihe po"·er required to ",aintoi" this/low.
Solution. Gi ve n :
Sp. gr. of o il
== 0.9
Viscosity.
~=
0.06 poise = 0.06 Ns/m 2
10
Dia. of pipe.
Disdarge .
Q = 60 litresls = 0.06 m 31s
Leng th.
[knsity
L = 500m
p = 0.9 x 1000 == 900 kg/m3
.". Reynolds number.
d = 200 mm == 0.2 m
pVd
R, = - - =900x
"
whe re
I I
V xO.2
006
10
1.909 lOIs '" 1.9 1 nils
Ii
~ I
IL
Flow Through Pipes
471 1
R '" 9(0)( 1.91 )(0. 2 )( 10 "" 57300
,
0.06
- As R, lies be tween 4CHXl and 1O~. the va lu e of en-effic ient of fricti on./ is given by
f= 0.079 '"
R,o. !)
Head lost d ue to friction. "1 =
4x j xLxV l
4x.OOSl x SOOxL91 l
'" "-C"~"-,",""~o"'''-dx2g
0.2x2x9.8 1
'" 9 .411 III of wa te r. A il s.
Power required
... 11.4
pg. Q. h!
'"
1000
'"
900 x 9 .81
x 0.06 x 9.48
1000
= 5.02 kW . Am _
MINOR ENERGY (HEAD) LOSSES
The loss of h ~ad or energy due to frictioll in a pip.: is !;nown as major loss while the loss of energy
due to change of vdocil y of the following nuid in magnitude or directio n is c all ed minor loss of
energy. The minor loss of energy (or head) includes th e following C3.'>eS:
I . L oss of head d ue 10 sudden l-nlargcrncn t.
2. Loss of he ad due to sudden contraction .
3 . Loss of head at the clltrancc of a pipe.
4. Loss of head at the exit of a pipe.
5. Loss of head due to an obstruetiun in a pipe.
6. Loss of he~d duc to bend in th e pipe.
7. Loss of he ad in various pipe fittings.
In (;asc of long pipe the above losses are small as compared with the loss of head due to fri(;t ion and
hen(;e they are called minor losses and even may be neglected with out seriuus e rror. But in casc of a
short pipe. these losses arc comparable with the loss of head due to friction.
I 1.4. 1 lou of Head Due to Sudden Enlargement. Consider ~ liquid nowing through a pipe
which has sudden enlargemenl a~ shown in Fig. 11.1. Consider two sec lions (1) -(1) and (2) -(2) before
and after the en largemen l.
o
Sudden eniargemelll.
PI : pressure intensi ly at seclion I- I.
Vt : ve locity of flow at sectio n I- I.
At : area of pipe at sect ion I- I.
Fig. 11.1
I I
Ii
~ I
IL
1472 Fluid Mechanics
P2' V 1 and A 2 '" correspond in g values at section 2·2.
Due to sudden change of diameter of tlie pipe from VI 10 D2_tile liquid l10w in g from the sma ller
pipe is not able to follow the abrupt cha nge of the boundary. Thus the now separates from the boundary
and turbulcm eddies are fomK-Q as shown in Fig. II. J. The loss of hc'1I.1 (or energy) takes pla",e due to
the fonn mion of these eddies.
Let
p' = pressure inlclI~ity of the liquid eddies on the area (Al - AI)
Ii , = lo.'\S of IIcad due to sudde n ~nlargernenl
Apply ing Bernoulli's equation at sections I-I and 2·2.
v,'
fil
VI!
fI,
p,l;
2g
pg 2 g -
- + - + I I = - ' + - '- + l, + loss of head due to sudden enlargement
BUl
ZI
= l, as pipe is horizontal
.
b.. + ~=
{J,
,
+ V, +11
pg2gpg2g
0'
'
h, =U~ - ~;)+(~~ - ~~)
... ( i)
Consider the contml vo lume of liqllid he twecn sections 1-1 and 2 ·2. Then the force acting on the
liquid in Ihe comrol volume in the dire"tion of flow is given by
,.-~ '" PIAl + p'(A 2 - AI) - P02
Bllt experi mentally it is fOllnd that p' '" PI
F, '" /lIAI + PI(A 1 - AI) - P?"t-2 '" /lIA, - /102
... (1 i)
'" (PI - P,JA 1
Momentum of liquid/sec al sec tion 1-1 = mass x velocily
,
'" pAl VI X VI '" pAIV I
Momentum of liquid!sec al seclion 2·2 '" pA,V2 x V 2 = pA 2V/
Change of mome ntum/sec", pA 2 Vl 2 _ pA l VI2
Bllt from continuity equation. we have
>
Change of momcnt ulnfscc '" PA2V~- - P x
A;Vj
Ie
,
>
X
,
VI- '" pA,V1- - pA1VIV1
'" pA 2 [V/ - VI V,I
... (iii)
Now ncl force acting on the control volume in the direction of now must be eq ual co Ihe rate of
change of mom~ntum or change of momentum iX'r secoJld. Hence equaling (Ii) and (iii)
(PI - p z)A j '" pA1[Vl' - VI Vl l
Dividing by g on bolh sides. we have PI - Pl.
pg
I I
1', '" ,V",'~~
--:V""VL'
= 'V./_~CV""VLl. or -P I - -'--'g
pgpg
g
Ii
~ I
IL
Flow Through Pipes
Substituting the va lue of
(li -Ji2)
pg
pg
473 1
in equation (0. we get
v,--
2 ' 2 V, V, +V,'"- V,'
V,, - V,V,
V," V;
,
' + - , -'- = '
.
.
g
2g2g
28
V , V' 'V V
= ' + ' - - 1;
2,
II ,
=
(ltj _ V, )'
2,
.
. .. ( I 1.5)
I 1.4 .2 Lou of Head due to Sudden Contraction . Consider a liquid flowing in II pipe which
has a sudden COll trn clion in area as shown in Fig. 11.2. Co nsider two sect io ns I- I and 2-2 before and
af!cr (;ontraction. As Ihe liquid flows from large pipe 10 small er pipe. Ihe area of flow goes on
decreasing and beco mes minimum ,11 a sect ion C-C as shown in Fig. 11.2. This sect ion C-C is called
Vcna-comracta. After section e·c, 11 sudden cnl~rgcmc n l of the area takes place. The loss of head duc
10 slldd~ n contraction is actually due to sudden ClllaT)! Cmcnl from Vcna -cnntracla to smaller pipe.
Lei
Ac '" Area o f now at section C-C
V,. '" Velocily of flow .11 se<: lion C-C
A! '" Area of flow al section 2-2
V 2 ", Velocity of flow at seclion 2-2
~,~ lIe '" Los~ of head due 10 sudden contraction.
Now iI,. '" actu al loss of he ad due to en largement from
section C-C to section 2-2 and is given by eq uat ion (1 1.5) as
'"
(~. ;gV.)l ~; l ~
-Ir
. .(il
_ ._ '- '....,.-
Fig. 11.2
~
,
o
Sudden con/radion.
From continuity equation. we have
Suh~lituling
the valu e of
V
....£..
V,
;n (I). we gel
... ( 11.6)
'"
k~l ,wherck'" l ~c -Ir
If lhe value of Ce is assumed 10 be cquallo 0.62. lhclI
k
= [ _ 1_ ,
0.62
I I
Ill ", 0.375
Ii
~ I
IL
1474 Fluid Mechanics
Th eil /1< beco mes as
t V,'
V,"
2g
28
h< '" -"- '" 0.175 - "-
If the value of Co is not give n th en the head loss duc 10 co ntractio n is take n as
V}
: 0.5 -
2g
V'
o r ll<",0.5 - 1 .
. .. (11.7)
2g
Find the Ion" of head ",hen (' pipe of dilllneta 200 m m is ~'udde"ly en/Mgt!tI to
diameter of 400 mm . Tile raIl" of flow of '\"lIler {hroug" the pipe iJ' 250 lilres/s.
Solution. Gi ve n :
Dia. of small er pipe. VI'" 200 nun ", 0.20 III
Problem 11.8
1t
2
1t
,-
'
A I'" - D I = - (.2) = 0 .031 4 1 111-
Area.
4
4
Dia. of large pipe.
D!", 400mm =0.4m
:. Area.
A 2 '"
~
DiSl: hargc.
X (0 .4)' = 0. 12564 m'
4
Q = 250 litrcsls '" 0 .25 1I1 )/s
Ve loci ty,
Q
0.25
YI=-: - - =7.96 m!s
AI
.03 14 1
v,,,, iL ,,,
Ve locity.
u:>ss of head
(l
du ~
A,
0.25 = 1.99 mls
.12564
10 e nlarge me nt is give n by equat ion (1 1.5) as
/', =
(VI - Vll'
=
(7.96 _ 1.99)'
= 1.816m o f wut cr. An s.
2g
28
Problem 11 .9 AI (I sudden etl/argemelll of II W<ller mid" from 1-10 mm 10 480 mm diame ler. Ille
I!ydwulic grlUlielll rise,~ by 10 mm . EJ'limtlle rile rtlle of flow.
O .N.T.U .• S 2002)
Solution. Give n :
Dia. of small er pipe.
D , "" 240 nlln "" 0 .24
:. Area.
A, ""
Dia. of large pip".
D2 "" 480 mm "" 0.48 m
Area.
Al ""
Ri se o f hydrauli c gradi ent·. ;.1'.•
It
It
2
11\
4D, "" 4 (.24)
1
~ (0.48)2
4
(Zl + ~;) - ( ;~ + l,) =
Let th e rate o f now = Q
Appl ying Berno ulli 's eq uati on 10 b01h se ctio ns. ;,1' .. small e r pi p.: sec ti on. and large pi pe sec ti o n.
~2
Ji + _'_ + l , =
pg
2g
V"
Pl
pg
+ _,_ + I I + He ad loss du e 10 e nlarge me nt
2g
...( i)
• Pica"" refer An. 11.5, I.
I I
Ii
~ I
IL
Flow Through Pipes
475 1
BUl head toss due 10 en largement,
... (ii)
Substituting this va lu e in (ii) , we get
Now s uhSlitulin ll th e va lu e of II, alld VI in equati on (il.
PI
-
(4V2 )'
P2
V"
9V/'
+ - 2 + Z, + -pg 2g
' 2g
+ - - - +Z, == -
pg2g
I~~} _ ~~ _ 9~" =(~; + Zl) - (~~+ZI)
0'
Btu hydraulic gradient ri sc
==
16V,'
V,"
9 V,"
2g
28
2g
- - - - - . -. - - . - :
(~; + Zl ) - (;~ +ZI) I ~
==
-
I
100
Vl ",
6V,1
I
28
[00
0< -
-. : -
=O. 1808=O. 181mfs
.I~26x~'~9~.8!'
100
Disctlargc.
It
2
= 4 D, X V,-=
"4J'[
1
'-
(.48) x. 181 =O.OJ275m Is
= 32.75 lilres/s. Ails.
Problem 11 .10 HIe rafe offlow a/li'a1er IllrQuglt a horizontal pipe is 0.25 mJIJ. Ti,e diameter of IIII'
pipe whic/I is 200 111111 is suddenly enlarged 10 400 111m. The pressure intensifY in Ille mwl/a pipe is
11.772 N/cIl1". Delallline :
(il IOJJ of/lead dill' 10 SI4ddell enlargement. (ii) preSSllre ill/ensif)'
(iii) !,ower 1051 dlle /0 elliargemelli.
Solution. Given:
Disch~rge,
Dia. of small er pipe.
I I
ill
Ihe large pipe.
Q = 0.25 11I 31s
D I = 200 111111 = 0.20 111
Ii
~ I
IL
1476 Fluid Mechanics
A,'"
Area.
'4" {.2)-=O.OJI 41
m
,
Dia. of large pipe. D2 = 400 mm = 0.40 III
"
:. Area,
Al"' "4 (O.4)- '" 0.12566 111 ,-
Pressure in smaller pipe,
PI = 11.772 Nfc11l 1 = 11.772 X 104 Nlm"
Now ve loc ity.
VI'"
Veloc ity .
Q
0.25
V,=-= - - : 1.99 m/s
•
A2 .12566
Q
0.25
AI
.03141
- =- - '" 7.96
mfs
(i) Luss of head duc 10 sudden c llJargcmc nl,
(7.96 _ 1.99 )2
2x9.81
= 1.816 m. Ans .
(ii) Let the pressure in te ns ity in large pipe '" Pl"
Then applying Bernoulli"s cq untio ll before and aft er the sudden c nlarg cme nl ,
""'
,
,
.!!J....+ V,P8
_ fll
2g - pg
+ V1 +/1
28
11.772
X
10"
lOOOx9.81
h
PI
' or pg = P8
+
,
V,
+"'2g 1.99 "
2x9.81
2x9.8 1
,
V,
(G ive n horizontal pipe)
2~ -II,
- 1.816
= 12.0 + 3.229 - 0.2018 - 1.81 60
= 15.229 - 2.0178 = 13.2 1 m of water
P2 '" 13.21 x pg '" 13.21 x [000 x 9.8 1 N/m l
4
'" 13.2 1 x 1000 x 9.81 X 10- Nlcm 2 = 12.96 Nfl,m l , Ans.
(iii) Power loS! due to suddell en largemem.
p=
pg .Q.h,
I()OO
x 0.25 x 1.81 6 _ 4 453 kW
= 1000 x9.811000
- .
. AilS.
Problem 11 . 11 A horizoilial pipe of diameter 500 mm i$ .lIIddl'lll), cOlltracled to a dilllneler
of 250 mm. 1'111' pressure illlelisities ill 1/11' large allil smaller pipe is gil'ell as 13.734 Nkm< alld
J /. 772 Nkm" respectil·ely. Find the 10.iJ of head due 10 con traction if C~ = 0.62. Also determille Ihe
r ail' of flow ofwlllu.
Solution. G iven:
Dia. of la rge pipe.
DI = 500 U1m = 0.5 11\
Area.
I I
AI =
~ (0.5)"
= 0.1963
m~
Ii
~ I
IL
FlowThrough PipeS
Dia. of s maller pipe.
°
:. Area.
A 1 ",
Pressure in large pipe.
PI'" 13.734 NIemi. '" 13.734 x 104 Nfm!
PI = 11.772 NIemi. = 11.772 x 104 Nlml
Cc = 0.62
2 ",
Pressure in smaller pipe,
Head 10M due to comraction
4771
250 mill '" 0.25 m
"4" (.25)- '" 0.04908 Ill',
V' [ = ....L.
2g
1 - 1.0]1 = _V,___
2 [1 _ 1.0 ]" = 0.375 ---.L
VI
Co
0.62
2g
28
From continuity equation, we ha "e A ,V ,:: AlV!
,
,
4 D1X Yl
V,
[ V,
~ Dl
J' xV,-~0(0.'']'
V, ~ V,
.50-4
4 '
Apply ing Bernoulli's equation before
~nd
~1
VI
pg2g
pg2s
after contral:tion.
.fL+ -'- +z '" 1:2+ _,_ + z, + II,.
8m
'
(pipe is horizontal)
l, '" <':2
~+
VI! '" P2 + V} + II
C
pg2gpg2g
v,l
VI
2,
4
" 0=0.375 - '- and VI '"
Substituting the se
valu~s
13.734 x 10' +
9.81 x 1000
14.0 +
"'
in
th~
above equation. we gel
(V! 14)2 =
2g
11.772 x 10' + V} + 0.375 VI'
lOOOx9.81
2g
2g
VI!
V,·
cc'''-o-=
12.0 + 1.375-'
16x2g
28
1 V,l
V,'
V,2
14- 12 = 1.375 -'- - - -'- '" 1.3125 - '
2g 16 2g
2g
2.0 '" 1.3125 x -v':-' or V 2 ",
"'
2g
(i)
LO!iS
.I~2~82x~2[x;;9.~8Il
1.3 125
'" 5.467 mfs.
V,2
0.375 X (5.467)2
of Ilc ad duc to contraction.il e '" 0.375 - '- '"
'" 0.571 m .
2g
2x9.8 1
(ii) Rate of flow of watcr.
An.~ .
Q '" A 2V 2 '" 0. 0490& x 5.467 '" 0.26&3 ml/s '" 26!U liUs. AilS.
Problem 11 .12 If in Ille problem II.ff. Ihe rare of flow of lI"arer is J()() lilres/s. alher (Imil
remO;1Ijllg Ihe some. fi/ld the I"(llue of CQ·efficiell/ of call/rocliall. Ce.
I I
Ii
~ I
IL
1478 Fluid Mechanics
Solution. Gi ven:
VI == 0.5 m. l) 2 " 0.25 m. PI" 13 .734 X 104 Nfml.
4
P2"" 11.772 X 10 N/m 2, Q '" 300 li tis", 0.3 1I1 3/s
Also from rmbi cill 11. 1 I . VI
= VI. WhH~ VI = .Q. = -i0~."30,,-:: = 1.528
4
AI
~ ( O.5) 1
mfs
4
V 2 =4 X VI = 4 x 1.528= 6.112 mfs
From Bernoulli 's c!.juntiu ll. we
P
\C
1
h~ vc
P
VI
- ' + -'-=~ + -'- +" c'
pg 28
pg 28
"'
"'
,13~.7~3=4~,,,I~Oc· (1.528)2 11.772 x (10)' (6.1 12f
-:
+
=
+
+ /l c
9.8 1)(1000
2)(9.8 1
9.8 1x1 000
2x9.81
14.0 + 0.1 19 = 12.0 + 1.904 + he
[4.11 9= 13.904+11..
li e'" 14. 11 9 - 13.904:0.215
[I ]'
v'
Bm fro m cqum ion ( 11.6). 11=-'----1
c
28 Co
Hence eq uali ng th e twO va lu es of he. we get
V
,' [I ]'
2~
Co - I
== 0.2 15
1
6 .11 2 [_I
2x9.81 Co
_1]" =0.2[.'1
]1 == =;~=--~"
0.215)(2.0)(9.8 1
:0. 11 29
I
[ -C,. - 1
6.(12)(6.112
'rJ'=9
LO = 1.0 + 0.336 = 1.336
-LO - 1.0 == "v.!
[2'1 = 0.336 or Co
Co
LO
Cc = - - = 0.748. An s.
1.336
Problem 11 .13 A 150 mill Jiamerer pipe reduces ill diame/er abrllplly 10 100 mm diameter. If IIII'
pipe carries 'Hiler m 30 lilres per second. caJculme IIII' preSSllre loss across IIII' cOil/rae/ion. Take the
co·effieielll of con/rac/ion as 0.6.
I I
Ii
~ I
IL
Flow Through Pipes
479 1
Solution. Gi ven:
Dia. of large pipe.
v, '"
150
of liorgc pi pe,
A,'"
'4" (.15)
Dia. of smalle r pipe.
D1
100 111111 = LI Om
Area of smalle r pi pe.
A , '"
Ar~a
",
-
I11Ill '"
0. 15
III
=0.0 1767111"'
~ (. 1 0)~ = 0.007854 Illl
4
Q = 30 litrcsls = .03 m1/s
Dis.;hargc.
Co-cffi<:i c l11 o f contract ion .Cc '" 0.6
QO
VI'" - = - .G3
- - '" 1.6 9 7 Ill is
A,
.0 1767
v, '" .Q. = -:;;.~03;;:-
and
•
Al
= 3.82 Ill/s
J)()7854
Apply ing Bernoulli's equation before and after co ntrac ti on,
,
1i+l+ Z _
pg
'- pg
2g
o.
.
fI, + V,"
Z, '"
2g
+2:1+
/1
... ( i)
C
Z:!
and h c' the head loss due [0 contraction is gi ve n by eq uat ion (11.6) as
he
v' [= -;_8
1-I ]' =-,!'-""o
3.82 ' [-1
- - I ]' =0.33
2 x9.8 ! 0.6
Co
S ubMitulin g these va lues in eq uat io n (I). we gel
P,
1.697'
PI
3.82 ~
- +
=- +
+ 0.33
pg 2 x9.8 1 pg 2 x9.8 1
~ +O. 1467= P1 +.7438+.33
pg
pg
li _
pg
p, '" .7438
+ .33 _ .1467 '" 0.927 1 m of wate r
pg
(PI - P2)
= pg x 0.9271'" lOoo x 9.81 x 0.927 1 Nfm 2
'" 0.909 x 10 Nfm 2 = 0.909 N/cm 2
4
Pressure loss across con trac tio n
= PI - II! '" 0.909 Nfem 2. AilS.
~ I
I~
~ I
IL
1480
Fluid Mechanics
Problem 11 .14 In Fig. 11.3 below, 11'/11'11 a suddell cOlltraction is ,illlmdllced in a, horizontal pipe,
line from 50 em ro 25 em. tile pressure c/umge.! from fO,500 /(g/l11- (103005 NinO fa 6900 kg/m(67689 Nlm!). Ca/cuiare Ihl' rale a/flow, Alsllme co·.jjicielll of cOlI/rac/ioll of jet ro be 0.65.
FollolI'ing Ihis if Illere is a sudden enlargemellf from 25 em to 50 em alld if Ihe pressure (1/ Ihe
25 em seclio/l is 6900 kg/",l (67689 Nlm') wlul! is the pressure allhe 50 em elilarged sec/ion?
Solu tion. Given:
Dia. of large pipe.
DI '" 50cm '" 0.5 III
Arc~.
AI = ~ (.5)'''' 0.1963
4
D, '" 25 em '" 0.25
Dia. of smaller pipe.
Arca.
A 2 ""
Pressure in large pipe.
Pressure in small er pipe.
'4" (.25)
m'
III
'" 0.04908
III
,
PI'" 10500 kgfm' ur 103005 Nlm'
Pl '" 6900 kg/m' OJ 67689 N/m'
CD
' 0
-
pJE6900 kglm'
D,= 25 em
D , ~50cm
p, zlOSOO
kglm l
@
f-----'
Fig. l1.J
Hc~d
los! duc to l'omraction is given by equation (I 1,6).
h~ =
l
Vl [ _I _
2g
Cc
Loll = 2gv/ (_'0.65_ _,)1 =0.2899 v/2g
... ( i)
From continui ty equation. we Ital'c
" ( 0.
°l'
5
0.25
x V, '" Y1
•
4
... (i i)
Apply ing Bcmoulli's equation at sections I·] and 2·2.
B",
I I
2 1 ", 21 (as pipe is horizontal)
Ii
~ I
IL
Flow Through Pipes
\C ~
~ + _,_
pg
4811
1
V
_ Pl + ---..L
+ II
2g-pg
2g
C
Substi tutin g th e val ues of PI_ fi 2. lic and VI _ we get
103005
+ (V,f4)2 =
IOOOx9.81
28
10.5 +
,.,;6~7~68~9c;-;-+ _V,_" + .2899 v/
1000 )(9.81
2~
1.2899
V,"
V,'
_0- _I , _0'"
28
3.6 '" 1.2274 x
"'
V1 ",
28
v'
v/
,..c"-o-=
6.9 +
16 x 2 8
10.5 - 6.9 "" 1.2899
"'
28
16
V
1.2274 _,_'
2g
28
vo'
_02,
I '" 7.586 m/s
.I~3~6~'~2~'~9~8I
1.2274
(r) Rate of flow o f wa ter. Q'" A1V1 '" 0.04908)( 7.586
'" 0.3723 m J/s or 372.3 liUs. A ns .
(ii) Applying Be rno ulli' s equati on to sections 3-3 and 4-4.
,
V,
_
P.
v.;°
+ -"- + ZJ '" - + - + Z4 + head loss d ue to sudden enlar£crncnl (II,)
pg
28
pg
2g
Pl
-
Pl '" 6900 kg/Ill ' , or 67689 N/m 1
""'
VJ = V 1 ", 7.586 Ill/s
V,
7.586
Y4 = V, '"
= - - '" 1.8965
_0
4
4
Z}= Z4
And head Joss due 10 sudden enlargement is given by equa tion ( I 1.5) as
(VJ - VJ
II, '" '--'-:;-"-'- '"
28
S ub~tituling
(7.586 - 1.8965/
2x9.8 \
= 1.65
III
the se va lues in Berno ulli' s equation. we ge l
~6,,
76:c89~
7.586 ' =
1000 x9.81 + 2 x 9.81
:;;;~I~"=~ +
1000 x9.81
1.8965' + 1.65
2 x 9.81
6.9 + 2.933 = -;:;;;!-I'o,"",,,"
-;+ 0.183 + 1.65
1000 x 9.81
I I
Ii
~ I
IL
1482 Fluid Mechanics
-;c=PC''-c= "" 6.9 + 2.'0133 - 0.1 &3 - 1.65 '" 9.833 - 1.833 '" 8.00
1000 x 9.8 1
p~ '" 8 x 1000 x 9.8 1 '" 711480 N/m l. An s .
I 104. 3
Lon of Head at the Entrance of a Pipe.
This is the loss of energy which occurs
when a liquid enters a pipe which is connected to a large tank or reservoir. This Joss is simila r to Ihe
loss of head duc (0 sudden contraction. This loss depends on Ihe fonn of entrance. For a sharp edge
cntram:e. this loss is sligtnly more ttJan a rounded or be ll mouthed entrance. In practice the v,duc of
V'
loss of head at the cntram;c (or ililct) of a pipe wit h sharp cornered Cnlrance is taken", 05 - . where
V", velocity of liquid in pipe.
2g
This loss is denoted by h i
v'
/I j = 0.5 -
...( I 1.8)
2,
I 1.4.4
Loss of Head at the Exit of Pipe .
velocity uf liquiu
~t
This is the 105s of head (or energy ) duc \0 the
outlet o f the pipe which is uissipawd either in the fonn
of~
free jet (i f oullcl
ofth~
pipe is free) or it is lost in the tank or reservoir (if the outlet of the pipe is connected to the tank or
V'
reservoir). This loss is equal to - . where V is the velocity o f liquid 3t the outlel of pipe. This loss is
2g
denoted
"0.
. ..(11. 9)
where V'" veloc ily al oUi let of pipe.
I 1.4 .S loss of Head Due to an Obs truction in a Pipe. Whenever there is 3n obstruction
in a pipe. tbe loss of enagy takes place due to reduction of tbc area of tbc cross-section of the pipe at
the place where obstruction is present. There is a sudden enlargement of the are~ of flow beyond the
obstruction due to which loss of heau takes place as shown in Fig. 11.3 (a)
Consider a pipe of area of cross-section A hav ing an
obstruction as shown in Fi g. 11.3 (a).
u,t
(I = Maximum area o f obstruction
A = Area of pipe
V= Velocity of liquid in pipe
Thcn (A - a) = Area of flow o f liquid at section I-I.
As the liqu id flows and passes through section
I-I. a vcna-l·ontracta is formeu beyond $Cction I-I.
after which tbe stream of liq uid widens again anu
ve locity o f flow at section 2-2 becomes un iform and
equal to velocity. V in the pipe. This situa tion is si milarto
the flow of liquiu throu gh sudden enlargement.
Let
®,
:
:
-~-
•
I
-
Fig . 11.3 (a)
- -
~-
.- - -
An obstruction in a pipt .
Vc '" Veloci ty of liqu id 3t I'cn a-contrac!a.
Then loss of heau due to obstruction = loss o f heau due 10 enlargement from l'ena·l"Unlracta to
sect ion 2-2.
I I
Ii
~ I
IL
Flow Through Pipes
(v, - V)'
.. .( i)
2,
From co ntinuit y. we ha ve
a e x V( '" A x V
whe re a e " area o f cross-scctio n at vc na-contracta
If Co '" l'O -c ffi cicnt o f contracti o n
...(i i)
c _ area al vena - CO nlracta
Th en
'"
( A - d)
c -
483 1
Q<
(A
a)
" c =Cc X( A-a)
Substitutin g lhi s value in (ii). we get
C~ X(A - a)x
Su bstitutin g Ihi s value of
V~
Vr =A x V
in eq uat ion (i). we get
, ( AxV - v)'
Head loss due 10 ohstrl,.clion '"
(V.. -V )28
C« A - a)
'" "-'''''''7''---"2g
V
' ( A -,)1...
28 Ce (A - a)
( 11.1 0)
I 1.4.6 lo5S of Head due to Bend in Pipe. Wh e n there is an y bend in a pipe. the veloc ity of
flow c hanges. du e 10 whic h Ih e se parat io n o f th e fl ow from th e bound ary and al so fo rmation of eddi es
takes place. Thu s th e ene rgy is lost. Loss o f Itcad in pi pe due 10 be nd is ex pressed as
kV ~
lib
= --
2,
where " b ::: loss of he ad due [0 he nd . V= ve loci!y of flow . k "" co·cffi cie nt of be nd
The valu e of k dqlC nds o n
(i) Ang le o f be nd.
I 1.4.7
(iil Radius of curva ture o f bend.
(iiI) Dia meter of pipe .
loss of Head in Various Pipe Fittings . Th e loss of he ad in th e vario us pipe fillin gs
such as va lves. couplin gs etc .. is expressed as
...( I 1. 1 I)
where V = ve loci ty of tl ow. k = co-efficie nt o f pi pe fil li ng.
Problem 11 .15 IViller is flowillg Ihrollgll a horizolllal pip/' of diometer 200 mm at a "e/ocity of
3 mls. A circuli" solid plute of diameter' 50 mm is placed ill IIII' pipe /0 obstruct IIII' flow. Filld Ihe loss
of Ilead due 10 oDslrue/ioll ill Ihe flipe if Co = 0.62.
Solullon. G iven:
Di a. of pi pe.
Ve loci ty.
D = 200 mill = 0.20 m
V= 3.0 m's
Arca o f pi pe.
A=
Dia. of obstru cti o n.
d= 150m lll =0. 15 m
Area of o bstruc tio n.
I I
~ 0' = ~
"
(0.2) ' = 0.03 141 m'
u= "4 (. 15) =0.0 1767 m
,
Ii
~ I
IL
1484
Fluid Mechanics
The head los! due [0 obstruction is give n by equati o n ( II 10) as
1.0
J
3x3 [
.03 141
2x9.81 0.62 [.03 141- .01767]
LOr
9
""':;-;;713.687 - 1.011 = 3 .3 11 m. An s.
2 x 9.81
Problem 11 .16 Determine the rale of flail' of Iruler Ihrougll (l pipe of diameter 20 em and length
50 III when olle end of Ihe pipe is connected 10 (I limk (Ind other end of Ihe pipe is open /0 111ft
atmosphere. The pipe j~. horiZo,,'a/ {lIId Ihe height of wo ler in the lank is 4 m aboI''' Ihe centre of Ihe
pipe. Consider all minor losses and /(Ike f == .009 ill Ihe formllio h! " ./. f . L. V I
d x2g
Solution. Dia. of pipe.
d= 20 cm == 0.20 m
Lengt h of pipe,
L = 50
In
","fER SUflf"'"
of waler.
H=4 m
Co-efficient of friction.
/= .009
H ~igh l
Let the ve loc ity o f water in pipe
_ ~~
__'-4
" V mls.
Appl ying Bernoulli 's eq uation al the top of th e wate r
su rface in the tank and at the o utl et of pipe, we have
ITakin g point I on the top and point 2 at the oU llet of
p ipe l·
Fig. 11.4
~,
E.!... + _,_ +<, '" ~ + -v''- + l, + all
pg
2g
pg
2g
•
losses
Considerin g datum line pass in g through th e centre of pipe
0+ 0 + 4.0= 0+
v;
4.0= - '- + h;+ ",
2,
But the ve locity ill pipe = V,
:. V = V1
V'
4.0 = -
2,
V'
From equatio n (11.8). h; '" O.S -
2,
~ I
+11; +",
...(i)
and h, fro m eq uat ion ( 11.1) is g ive n as
I~
~ I
IL
Flow Through Pipes
4851
Substituting these val ues. we have
2
4.0= V" + O.5V + 4 .!.L.V"
2g
=
2g
dx2g
~ [ 1.0+0.5 + 4 X.OO9X50 l= ~
28
0.2
= 10.5 x
V'
2,
4 x 2 x 9.81
V=
[1.0 + 0.5 + 9.0]
2g
10.5
= 2.734 m/sec
Q = A x V = ~ X (0.2)2 x 2.734 = 0.08589 111 3,s
Rate of now,
4
= 85.89 lil res/s. AilS.
Problem 11 .17 A iloriw,,/a/ p ipe /i",' 40 m /Oll8;S cOllllec/cd 10 II ...Mer /<wk al 011e end and
diSc/larges freely inlO the armosphere at Ihe Ollwr end. Por I/le firs t 25 III of its /e,lgl/l/rom file /allk.
Ihe pipe is 150 mm diameter and il5 diameter is sIIddenly e'''llrged /0 300 mm. Th e ',eight of WilIer
/er el ill Iile I"nk is 8 m abo ..e Ihe centre of Ihe p ipe. Considering all /o~'ses of Ilead ... I,ie" at""',
delermilie (lie rule of flo ..... Take f = .01 for boll! seC/iOtIS of Ihe pipe.
Solution. Given:
Totallcng1l1 of pipe.
L
= 40 m
lA:ngth of 1M pipe.
LI=25m
DiJ. of lSI pipe.
lA:ngth of 2nd pipe.
L 2 =40-25= 15m
ill = 150 111111 = 0. 15 m
=
=
Dia. of 2nd pipe.
d 2 300 mm
0.3 m
Hcight of wate r.
fI = 8 m
Co-e flici ent of fric tio n.
/= 0.01
Applying Bernoulli' s th corem 10 the free surface of water in the tank and oUllet of pipe as shown in
Fig. I 1.5 and t<[king refe rence line passing through the centre of pipe.
'm
j L__
_
"V~, __r---~::;,~-
L, " 25 m
Fig . 11 .5
v'
0+0+8= p~ +-'- + O+all iosscs
pg 2g
I I
Ii
~ I
IL
1486 Fluid Mechanics
v,!
8.0:0+ 2~ +/J i +l!f, +11, +11"
0'
... (i)
~'
whe re h j = loss of head al enlfance = 0.5 -'-
2,
/'f,
= head
InS! due \0 friction i ll piP<" I
4X[ X L,XV;l
= :::'-'f:~:'-'"­
d 1 xl,
(V, _ ¥1)1
2,
I"~ = loss head duc \0 sudden enlargement '" ,",-;;---"'-
hf = Head lost duc to friction in pipe 2 =
,
4xfxL xv,l
2_
dz X2g
BUI from continui ty equation. we have
A ,V, '" A 2V 2
I, . = _0_' _"_,1 '" "0.,,
' "x ,,
(4"V"J_'
•
2,
2,
28
4 xO.O I x 25 X(4Vlr
II" =
=
11, '"
0.15x2x8
4x.0 ] x25xI6
0.1 5
v/
x~
2,
V'
2,
= 106.67 -'-
(V, - V2)2
(4Vl - Vj
1
9V1
28
2,
2,
_ 4x.O l x l 5xYll _ 4x.0 I x [5
I' f -
'
Suh~liluling
0.3x2,
Ihe values of tllese
lo~scs
0.3
V}
x ~
2,
-20
- . x
V}
~
2g
in equation (i ), we gel
V,l
8V,'
V,"
9V,"
1
8.0= -'- + - - +106.67 -'-+-'- + 2x V:
2,
v'
2,
= -'- [I
2,
-+-
8
2,
-+-
106.67
8.0x2xg '"
126.67
..
~ I
-+-
2,
2,
v',
9 + 2J = 126.67Tg
8.0x2x9.8 1 ", JL2391 = J 113 IIl/s
126.67
, ,
,
Ralc of flow, Q == A, x V, == - (0.3) ' x I 113 " 0.01861 III Is" 18.61111,..,s/s, ADS.
,
4
I~
~ I
IL
Flow Through Pipes
4871
Problem 11 .18 D etermine Ihe difference in Ille <'iel'lliioll ,~ bet ..... een the ..... lIIer sllrfaces in Ille 110'0
tallks which are cOlineCled by a horizontal pipe of diameter 300 mm (//11/ Jeng lll 400 m. The rate 0/11QI<'
ofwmef through Ihe pipe is 300 li/res/s. Consider all/asses (lnd take Ihe ra/lle of! == .008.
Solution. Given :
Dia. of pipe,
Length.
Disc h.arge .
Co-cffjcicnl of fric tion.
11= 300 l111n == 0.30 III
L = 400111
Q = 300 li tis = 0.3 m 3,~
/= 0.008
v
Vc locily,
~ _ Q_ ~ -;;-~O.~3--:-
= 4.244 IlIfs
~ X {.3 }l
Are a
4
Le t the Iwo tank s im: conne<:tcd by a piP<' as show n in rig. 11 .6.
-0-1
!"'
H, -
l: 400 m
d- O.30m
H,
l
t
- v
Fi g . 11.6
Let HI = he ig h1 of wmer in lSI lank above th e cent re o f pipe
Hl = he igh1 of wme r ill 2nd lank :Itxwc th e ",,"He of pipe
The n difference in elev ati ons belwee n wat er surfaces = H I - Hl
Appl ying Be rno ulli 's equaliOIl to lh" fr"" surface of wa leT in the lw o tallks , we hav"
H , '" Hz + losses
", f1 1 +
...( i)
h i + H f , + 110
v"
0.5 )( 4.244 "
where I. i ", Loss of head at e ntran ce'" 0.5 ~
28
2 x9.8 [
"':;c",,,,'-- '" 0.459 III
hJ, '" Loss of head due to frict ioll '"
4 x [ x L x Vl
4 x .00&x 400x 4.244 '
'" ""~';;';'::",";;';;'C''--dx2g
0.3x2 x 9.81
v" = 4.244 '
h" '" Loss o f head at out le t ", -
2g
2 x 9.81
'" 0.9 18
'" 39 . 16 III
III
Substitutin g lhese values in (i ). we get
H , '" Hl + 0.459 + 39.[6 + 0 .9 18 '" Hz + 40.537
Problem 11 .19
H , - Ill '" Differe nce in e lev ati o ns
'" 40 .537 rn. Ans.
The Irk/ioll jtKlor [or /Ilrblliell/ flow l" rOllgh rOllgh pipes CUll be delermined by
Kamw,,· Pramlll elfllalin",
h
= 2 loglo (Rolk) + I .N
...llere ['" [ric/iOI' factor, Ro '" pipe radiuI, k = aI'erage rouglmess.
I I
Ii
~ I
IL
1488
Fluid Mechanics
Two resen'oirs Willi a ,I'ur/ace lel'eI difference 0[20 melre.1 are to be (onne(fed by f melre diam eter
pipe 6 kill long. What will be Ille tli.leilarg" when a caSI iron pipe of rOllglllle$s k '" 0.3 mill is IUi'd ?
Whm will be Ihe percel1lt1ge increase in rhe discharge iflhe caS I iron pipe is replaced by a Sleet pipe of
roughness Ii. = 0. 1 mm ? Neg /eel aI/locH/losses.
Solution. Given:
1,:20m
<1= I.Om
Di ffere nce in le vel s.
Dia. of pipe.
:. Radius. Ro = 0.5
11\
= 500
111111
Length of pipe.
L = 6 kill = 6 x [000 '" 6000 In
Roughn ess of c ast iron pipe. Ii. = 0.3 mm
Roughness of st~cI pipe,
Ii. = 0.1 mm
lst Case. Ca~llro" Pipe. First fi nd the v alue of friction factor usin g
,
"Jj=2 10 g lO (R J k)+ 1. 74
= 210g IO (500/0.3) + 1.74
f~
(_,
_)1
8.1837
...(1)
= 8.1837
= 0. 0 149
Local losscs arc to be negl ected . Th.is means onl y head loss due to frict ion is 10 be considered.
loss due to friction is
~lcad
20",! XLXV '
d x2g
[Here f is the friction factor and not co-efficie nt o f friction
.: Friction factor", 4 x co-efficient o f friction]
20 '"
v'"
.0149 x 6000 x V ' '" 4.556
LOx2x9.81
J4.~~6
.
v2
'" 2.095 rn ls
11
2
II
,
3
DlschHrge_ Q 1 = Vx Area = 2.()95 x - x If '" 2.()95 x - x 1-= 1.645 rn Is
4
4
2nd Case. Sleell'ipe.
k",O.1 mm.R o = 500mm
Substitutin g th ese va lues in equatio n (I). we get
,
J1 = 2 loglo (50010. 1) + 1.74 '" 9. 1379
f'"
(9J~79
r'"
0. 0 11 9
f xLxV 1
.0119x6000xV 2
Hcad loss duc 10 fricti on. 20 '" L",~,,_ or 20 '" "C"c"'-;"'~"-_ " 3.639
Ifx2g
1.0x 2x9.81
I I
v2
Ii
~ I
IL
Flow Through Pipes
:. Discharge,
v=
J3.639
20
'" 2.344 mls
01 =
VxAn:~
'" 2.344 x
pcn;cllwgc inncasc in the di scha rge ==
- 0I
Q
1
'4" X
X 100:=
Q,
I '" 1.841
( 1.84 1- 1.645 )
1.645
111
4891
,Is
x 100 '" 11 .9 1<,t,. An s.
Desig" IIII' dill"'eler of (l sleel pipe 10 ( .Try IIlIl/er II",';ng kinematic 1';KO~ily
IO-l! ",lis willi II mean I"e/Ocily of I m/s. The /,eat! loss iJ" /0 be limited /05 m per I()() m /ength of
pipe. Con sider the ",/,,;m"'"( sa",/ rollg/mess Iwig/II of pipe k, := 45 x fO • em. AISllme I/Wlille Darcy
Problem 11.20
1' =
Weisbach friction jaClor oreT Iile whole rmlge of lurbulent jlOII" ellll be e.l'(Jressed <I S
w/lere D = Diameter of pipe and R, = ReYliolds lIumber.
Solution. Gi ven:
y:= IO-~ 11111s
Kinematic viscosi ty.
M~an velocity.
Head loss.
V=lmfs
h[ =5 m in a knglh L= 100m
k, := 45 X 10---1 em = 45 x 10...... m
Value of
Value of
f = 0.0055 [ I + ( 20 X tO
Usin g Darc y Weisbach equa ti on. hf =
i=
J
~ + IZ,~ )"l]
... ( i)
4X f XLXV 1
Dx2 g
h, xDx2g
, = 5xDx 2x9.81 :0.2452D
4xlOOxi l
4x LxV -
Now the Reynolds number is given hy.
= IXD =I0 6 D
IO -~
Substitutin g the va lu es oFf. R, and k, in
cqu~t io l1
(i), we gel
l
10'D
0.2452D=0.0055 1+ ( 20X IO x 45 xD10-' + I O~
[
0.2452
0.9+ I
- - 0 =[1+(0.0055
I I
0
0
J'' ]
J'' ]
Ii
~ I
IL
1490
Fluid Mechanics
44.581J :
0'
[ (i19)''']
1+
Of
44.58 1J - 1 '"
(-t-19)'"
(44.58 D - IY'= .!.:2. or D(44.58D-ll= 1.9
D
... ( i i)
Equation (ii) is so lved by h.i! and trial method.
(il Let D = 0. 1 111. then LH .S. of equation (ii) oc(:omcs as
L.H.S.=O.I (44.58xO. I - l »)=O.1 X 3.45S3 = 4. 135
This is mo re than the R. Il .S.
( ii) LeI D '" 0.08 m. tllen L. H.S . of equation (iJ)
bcCQIll CS
as
L. H.S. '" 0.08 (44.58 x 0.08 - 1)3", 0.08 (2.5664)3 '" l.J52
This is less than the R.H.S .
He nce va lu e of D Ik s between 0.1 ~[l(l 0.08
(iii) LeI D "" 0 .085. th e n L.H.S. of equation (il) bcl'Omcs as
L.II.S. == 0.085 (44.58 x 0.085 _ 1)3 == 1.S44
This val ue is sli ght ly less than R. H.S. He nce increase the va lue of 0 slightl y.
(i,') LeI D '" 0.0854 m. then L. H.S. o f equati o n (iil hecorncs a~
L.H.S. '" 0 .0854 (44.58 x 0.0854 _ 1)3 '" 1.889
Thi s va lue is nea rl y equa l to R.H.S.
:. Com~et va lu~ o r D = 0.0854 m . Ans.
Problem 11 .21 A pipe line AR of diameter 300 mm and of lenglll -100 m ,'arries waler al Ilw HilI' of
50Iilres/s. Tlwflow lakes place from A 10 B wi,erI' poinl B iJ' 30 melres aho\"(! A. Fi",l lile press"re al
A if Ihe pUJs" re al B is 19.61 Nkm !. Take f = .008.
S o lution. G ive n:
Dia. or pipe.
d = 300 mm = 0.30 m
Len gth or pipe.
L =400 m
Di sc h~rge.
Q = 50 litresls = 0.05 m 3fs
:. Velocity.
V=
--.SL =
Area
0.05 =
0.05
~d 1
~ x (.3)1
4
4
= 0.7074 mls
/Is = 19.62 Nfcm 1 = 19.62 x 104 Nfm l
Pres,ure at n .
/= .008
Apply ing Bernoulli's eljuatio ns at points A and lJ and taking datum line passing throu gh A. we have
vi
PA V;
P8
- + - + lA = - + pg
Bo'
2g
pg
VA
2g
+ ZII+ h!
= VB
[.: Dia. is same [
ZA=O.lf/ = JO
"od
I I
4 2X~f~X~L:;-::X ~V_'
11= -~
f
dx2g
Ii
~ I
IL
Flow Through Pipes
[9.62x I0
4
491 1
+30+ 4 x.008x 400x.7014 '
1000:><9.8 1
0.3 x 2x9.8 1
:20+30+ 1.088:51.088
f!A '"
=
51.08& x 1000 x 9.81
5 1.088 x 1000 x 9.81
'" 50. 12 N/cm 1, An s.
... 11 .5
\
Fig. 11 .7
HYDRAULIC GRADIENT AND TOTAL ENERGY LINE
The concept of hydraulic gradien! line and total energy line is very us.cful in the siudy of flow of
fluids through pipes. They 3fC defined as:
[1 .5 . 1
Hydr;Julic Gradient Line.
h is defined as the line which gives the sum uf pressure head
( ~ ) and datum head (2) of a flowing fluid in a pipe with respect to some reference line or it is the line
which is obtained by joining the lOp of all venical ordinates. showing the pressure head (plw) of a
flowing fluid in a pip<! from the centre of the pipe. It is briefly wriucn as H.G.L. (Hydraulic Gradient
Line).
I 1.5 .2 Total Energy Line . II is defined as the line which gives the sum of pressure head. datum
head and kinetic head o f a flowing fluid in a pipe with respect to some referen~ line. II is also dernwd as
the tine which is obtained by joining the tops of aU I'ertical ordinates showing the sum ofpressurc head
and kinetic head from the centre of the pipe. It is briefly written as TEL. (Total Energy Line).
Problem 11.22
For tile problem 11./6. draw Ih;: ffydrm,/ic G",di;:,1/ Line (H.G.L.) and Total
Ellergy Lille (T.EL).
Solution. Given:
L" 50 m. d" 200 mm" 0.2 m
H"
4 m./==.OO9
Velocity, Y through pip.: is calculated in problem 11.16 and its value is y" 2.7J4 mfs
Now.
== Head Inst at entrance of pipe
y 1 0.sx2.734 1
== 0.5 - +
==O.19m
2g
2x9.8 1
"j
,
1:e_L
4m
1
- - hJ;,j.
----~ 2 ___..g,
B.... - - - - --- - -
38
L:50 m
d:20=
Fig. 11.8
and h, == Head loss due to friction
I I
Ii
~ I
IL
1492 Fluid Mechanics
== 4 X j XLXV ':: 4xO.009x50x (2.734 )' = 3.428 n1.
dx2g
O.2x2x9.81
(a) To tal E ne rgy Lin e (T .E.L.). Conside r three points. A, Band C on Ihe free s urface of wat er in
th e tank. at the inle t of the pipe and at th e o utlet of the pi pe respecti ve ly as show n in r ig. 11 .8. Lei us
find total e ne rgy at these points. taking the ce ntre o f pipe as rcfc~ncc lin e.
I. Total e nergy alA
II
=:
-
pg
V"
+-
2g
+ l = 0 + 0+ 4 .0
2 . Total e nergy a1 B :: Total e nergy at A - Il j
p
3. Total e nergy at C = - '
pg
V
:=
=:
4111
4.0 - 0 .1 9 == 3.81 III
V1
l
+ 0:: 2.734' '" 0.38 m.
2 x 9.8 1
2g
Hence t Ol ~1 ene rgy line will coincide wi lh free surface uf water in th e tank. At the inlet of the pipe.
it w ill decrease by II; (= O. [9 111 ) from free s urface ami at outle t of pipe IO ta l ene rgy is 0 .38 m. He nce
in Fig. [1 .8,
(i) Puin! D re prese nts total e ne rgy at A
(;1) Po int E. where DE " I" , represc nts total e nergy at inl et o f th e p ipe
(iii) Po int F. where CF ", 0 .38 represe nts total e nergy at outl et o r pi pe.
Jo in D to E and E to P. T he n DEF represc nts th e to tal cncrgy lin c.
(b) Hydra ulic Gra di ent Line ( H.G.L.). H.G. L. g ives the sum o r (plw + z) with refercnce to the
+-'- + Z,. :: 0 + 2g
dalUm · lin e. Hence hydrauli c gradient line is
V'
by subtractin g -
o htai n ~d
2g
from to tal ene rgy lin e. At
Vl
V'
outlet of th e pipe. tot al e ll crgy " - . By subtracting fro m total e nergy at thi s point. we sha ll ge t
2g
2g
poin t C, whic h li es on the centre lin e o f pipe. Fro m C, draw a lill e CO p arall c l to EF. The n CO
represc llts the hydraulic gradi cnt line .
Problem 11 .23
For Ihe problem 11. 17. draw Ihe II),dralllic gradiem am/lOwl ellerg), lille.
Solution. Refer to problem 11. 17.
Gi ve n :
L ",25m. d l ",0.1 5 m
I
L2 ", 15m. d 2 : O. 3 m. /" .0 1.1I :8 m
The ve locity V2 as
e al c u l at~d
in probl e m! 1.1 7 is
V2 ", 1.1 13 mfs
VI '" 4 V2 '" 4 x l. ln '" 4.452 m/s
vi
The vario us he ad losses arc iI; '" O.5 x " -"O".5cx,,40.4,5"2~' " 0 .50 m
2g
2 x9.8(
"Ji
I I
=
~
4 f,-,;X~~'7
X ~V,,'
d l x2g
=
4 x .OI X25x (4.452)'
3
= 6 .7 III
0.15x2x9.8 1
Ii
~ I
IL
Flow Through Pipes
hi,
_
~
4 ~'2f~'~L,
~_~X~V
~,'C '" 4 x .oJ x 15 x (1.1 13f
-
V,1
Ll l3'
II == - -=
o
0.3x2x 9.8 1
d"x2g
2g
2 x9.81
493 1
",D. 126m
== 0.06] III
Alw
T otal E nergy Line
(I) Point A lies on free surface of water.
(ii) Take All "" h; == 0.5 m.
(iii) From R. draw a hori zo nta l line. Tak e BL
~qual
[Ollie leng th of pipe. i.e .. 1- 1" From L draw a
vt' rtical line downward.
(il') CUllhc li ne LC" h" == 6.73 m.
(.,) Join the point lJ \0 C. From C. take a line CD verticall y do wnward equa l 10 11, := 0.568 m.
(d) From D. draw DM hori zontal and fro m point F whi c h is lying on the cenlre o f Ihe pipe. d ra w a
vCT1ic~ l linc
in the upward direc tion, meeling at M. From M. tak e a dislam::e ME = h" = 0.126.
10il1 DE.
The n lin e ABCDE rc prcsc llIs th e total e nergy li ne.
I
Fig . 11.9
Hydrpulic Gradicnt Lin e (H.G.L.)
V'
0) From IJ. take BG '" - '-
2g
= 1.0 rn.
(ii) Draw th e line GH parallel to the line BC
(iii) From F. draw a line Fl parallel to the lin e ED .
(il') Join the poi nt H and I.
Then th e line GHIF represen ts the hydraul ic gradicntline ( H.G .L.).
Pro blem 11 .24 For Problem 11./8. draw Ihe hydrrw/ic gradienl "",{Io l,,/ e"ergy /i"e.
So lution . Refer to Problem II.IS.
Given:
d= :\00 mm "" 0.3 m
L = 400 m. Q "" JOO liucs/s "" 0.3 11I 3/s
f=
I I
.OOS
Ii
~ I
IL
1494
Fluid Mechanics
II, '" 50
11 2 ",
BUIU 1 - 11 2 '" 40.53 7
50 - 40.537 '" 9.463 Ill.
Ill.
In
(Calculated in Problem 11.18)
The calcu lated losses arc:
(!)
"i'" 0.459 m
(iii) 11,,:0.918
(a)
= 39.16 m
(ii) ""
III
T.E.L.
(i) Point A is on [h e fr<)c surface of W3Icr in 1st tank.. From A. take All '" h i '" 0.459 m.
(ii) Draw a horizont al line Bf', Take HF equal 10 the length o f pipe. From F. dmw a vertical line in
the downward di rection. CUI Fe", h/, '" 39.16 m.
lie. From C take CD '" 110 = 0.918 m. T he point D should coinci de with free surface of water
in 2nd tank. Then line ABeD is the IOtal e nergy li ne.
(iii) Join
-.L h,=0 .459 m
A
l"'4wm
d=O.3m
Fig. 11 .10
(b) H. G.L. From D, draw a li ne DE parJlIcI to line Be. The n DE is the H.G.L.
0,
v'
From IJ, take BE '" -
2,
'" 0.9 18
III
and from Edraw a line ED parallel
10
BC. The poim D should
coi llcide with free ~urface of wate r in the 2nd tan k. Th~n line ED represcnls the H.G.L.
Problem 11 .25 Ti,e rale ofjlow ofwllter pllmped illlo a pipe ABC. which is 200 mlong. is 20 litres/s. The
pipe i~· laid on an upward slope of I in 40. TIl/' length of the porlion AB i5 100 m Imd it~· di",neler
il· /00 mill. while Ihe length of the portion BC is also 100 m but il~· diameter is 200 mm. The change of
di",nt!/er (It B i5 l·udden. Th e jlOII' is laking pia,·" from A 10 C. wlwre Ihe pressure al A is 19.62 Nkm 2
and end C i5 connecled to (I /imk. Find Ihe pressu'e al C (Illd draw IIII! /lyil,,""ic gradient lind lo/al
energy line. Take f = .008.
Solu t ion. Given:
Leng[h of pipe.
ABC = 200 m
Discharge.
Slope of pipe.
Leng[h of pipe.
Lenglh of pipe,
Pressure a[ A.
Co·efficicnt of friction.
Q = 20 litrcsls = 0.02 m 31s
.
I
= 111140 = -
40
AB = 100 Ill. Dia. of pipe AB = 100 mm = 0.1 m
Dia. of pipe BC = 200 mm = 0.2
PI! = 19.62 Nfcm" = 19.62 x 10· Nlm"
BC = 100
111.
111
f = .008
Velocity of water in pipe AB. VI =
Disch~rge
Area of AS
0.02
= - - = 2.54 lilts
~ ( .l )2
4
I I
Ii
~ I
IL
Flow Through Pipes
Velocity of water in pipe
1Jc.
Q
V2 '"
Area o f
Be
495 1
'" 0.02 '" 0.63 111/s
'::' (.2)2
4
Apply ing Bernoulli's cqumion \0 points A and C.
V1
~+~+
pg
28
P
z,\ '"
_C
pg
V!
+ _<_ + Zc + total
2g
loss from A to C
... (i)
TOlal loss from A to C", Loss duc to friction in pipe AB + loss of head duc to en largement at B +
loss of head duc to friction in pipe Be.
... (ii)
Now loss of head duc 10 friction in pipe AB.
/1 , =
4 J1-V2
4 x.OO8 X 100 X (2 54)2
" d x2g
Loss of head due 10 fric tion in pipe
II
/,
==
O.lx2x9.8 1
=10.52m
Be.
4 X.008 X100 X(0.63)1
'" 0.323
0.2 x2 x9.8 1
III
Loss of Itcad due to en largeme nt at B,
TOl allos.~froIllA
IOC
= h/l +h, + hI,
= 10.52+ .186+ .323= 11.029'= 11 .03111
SubstitUTing this v:1luc in (I). we gel
V'
JI
V1
P
--=l..+.....i..
+ z - ....!:. + _c_ +
pg 28
'\- P8 28
l
+ 11 .03
. ,.( iii)
c
Taking datum line passing through. A. we have
Z,\ == 0
<;- '"
Also
4~
x lotal length of pipe '"
4~
x 200:0 5 m
PA'" 19.62 X 104 N/m"
VA'" V I = 2.54 mls. Vc = V1 = 0.63 mls
Substituting these va lues ill (iit). we get
19.62 x (10)" + (2.54)" + 0 =.&+ (0.63)'
+ 5.0 + 11.03
lOOOx9.81
2 x 9.81
pg 2x9.81
20 + 0.328:: .& + 0.D2 + 5.0 + 11.03
pg
20.328:: .& + 16.05
pg
I I
Ii
~ I
IL
1496
Fluid Mechanics
Pc == 20.328 _ 16.05 == 4.278 m
pg
Pc = 4.278 x I OOOx 9. 8 1 N/m 2
4.278 X 1000 x 9.8 1
= -'~"""'~";'''''""'
Nlc(11" == 4. 196 N/c l1l ~. ADS.
(0'
H ydrllullc Gradient and Total E""rg," Un.,
p
, 1 ~''1052
,
f"
'~
."
"'G,. , " 1
,
tt(
20 328
~
~
Lhf ·n3 ,
'll '
c
....
---~
-
__ -
,
'",
Fig. 11.11
PIp.: AU . Assumin g tlie datum lin e pass in g th ro ugh A. th en IOlal e nergy 3l A
V;
p~
pg 2g
== 20.328 III
=
Total e nergy at B
4
19.62 X 10 + (2.54), + 0" 20 + 0.328
l 000x9.81
'" Tot al e nergy at A !
A lso
~ + ~ +"
Vc 11g
=
(0.63)2
1 x9.RI
" ft
=:
2x9.8 1
20.328 - 10.52", 9.&08
III
== 0.02.
Tota l En"rgy l.ine . Draw <I ho ri m nlal line AX as ~h ow n in Fig. I J. I I. Th e co) nlfC- lill e o f Ih" pipe is
dra wn in s uc h a way [h al s lope of pipe is I in 40. Thus the po int C will he at a hei ght of
,
40 x 200 '" 5
111
fro m th e lin e AX. Now draw a vc ni ca l line A D cq ua lto IOta l ene rgy at A . i.e .. A D
= 20.)28 m. f'ro m poi nt D. draw a ho ri zont al line and fro m poi nt /J. a I'c n ica l line. meetin g at Q. From
Q. lake I'e ni cal di stan ce QE == h" = 10.52 1Il. Join DE. f'rorn E. tak e EF = h, = 0. 186 m. Fro m F,
draw a hori zo nt al line and from C. a ve ni ca l line meet in g at R. From R ta ke RG = h" == 0.323 Ill. Join
F to G. Then DEFG re prese nts th e lota l energy line.
1I~' d raul ic Gr.ldl enl Line. Draw the line 1M parallel to the line DE at a d istan ce in the dow nward
V'
direc tion eq ua l to 0.328 Ill . Also draw the line PN para llel to the line GF at a d islan ce of - '- = 0.02.
2,
Jo in po int M to N. Then lin e UlINP re prese nts the hydrau lic g rad ie nt lin e.
Problem 11.26 A pipe lille. 300 111111 ill diameter (md 3200 III 10llg is IIsed to plllllp lip 50 kg per
second of (Ill oil ...llOse dellsity is 950 kg/m J (md ... llOse kinematic viscosiry is 2.1 stokt'S. Tile centre of
Ihe pipe line oll/u: IIpper elld is -10 m abore th(l1l Illal al rile lo·....er end. The disclwrge (II Ihe IIpper end
is armospheric. Fi/ld Ihe pressllre at IIII' lower elld WId draw the Il)'dralllir gradient alld Ihe total
energy line.
I I
Ii
~ I
IL
Flow Through Pipes
Solution. Given:
Dia. of pipe.
497 1
d = 300 IlUll '" 0.3 III
L=32oom
u,ngth of pip!:,
Mass.
M =50kgfs=p.Q
50 50
3
Q= - = '" 0.0526 m Is
p
950
Discharge,
Density.
p '" 950 kg/ml
Kin ematic viscos it y. \' == 2.1 stokes" 2.1 cmlls
= 2.1 x 10-4 m1fs
Hcigtn of upper end
" 40
Pressure at upper end
" atmospheric = 0
In
vxd
Discharge
R, == - - . where V ==
Reynolds number.
"'7"''''
Area
v
R, '"
0.744 x 0.30
2.1 x 10
=I
~ (O.3)1
== 0.744 mrs
06'
~.S
f=~= _I6_ =O.QI5
:. Co..efficicnl of friction.
R,
Head losl due \0 friction.
•
0.0526
11 ==
f
=
1062.s
~4~X~f~X~L",X~V_'
d x 19
4 x 0.015 )( 3200 x (0.7 44 )1
0.3)(2x9.8 1
= 18.05 m ofoil
Apply ing the Bern oulli 's equation ;1( the lowe r and upper cnd of the pipe and tak in g datum line
through the lower end. we have
p~ssing
But ZI '" O. Zl '" 40 m. VI = V1 as diameter is M ill e
Pl=O.llf = 18.05m
:. SubSlituting lhese values. we have
l!.!. '" 40 + 18.05" 58.05
pg
In
of oil
PI'" 58.05 x pg '" 58.05 x 950 x 9.8 1
I': p for o il '" 950)
H.G.L. and T.E.I ,.
VI
2g
~ I
( .744)2
"
2x9 .81
'" 0.0282 m
I~
~ I
IL
1498 Fluid Mechanics
J!J... = 58.05 111 of oil
pg
,
0.0282 m
r
fH: G./.-- ,
i, l;, (
/
58.05 m
T
<Om
Draw a horizontal line AX as shown in Fig. 11.12. From
/
/
A. draw the CClllrc line of llie pipe in such <I way that point C
/
/
is a disI 31lCC of 40 III aooyc the horizontal line. Draw a ven iA
A
cal line AB through A such that AIJ " 58.05 111. Join IJ with
Fig. 1l.12
C. Then BC is the hydraulic gr,l(licllI line.
Draw a line DE parallel (0 BC a1 a hdghl of 0.0282 III abo ve the hydraulic gradient line. Thl' n DE is
the total energy line .
.. I 1.6
flOW THROUGH SYPHON
Syphon is a long ben! pipe which is llscd to transfer liquid from a reservoir al a higher elevation to
another reservoir a1 a lowe r level wilen the (WO reservoirs arc separated by a hill or high level ground
as show n in Fig. I [.1 J.
:;;:C;:;:;~ SUMMIT
A
- ...,,--- -- --
,
-----
Fig. 11.13
The point C which is at the highest of the syphon is called the summit. As the point C is above the free
surface of the water in the tan~ A, the pressure 3t C will be less than atmospheric pressure. Theoretically.
the pressure at C may be r~-duccd to - 10.3 III of water but in actual pra~1ice this pressure is unly - 7.6 III
of water or 10.3 - 7.6 = 2.7 10 of water absolutc. If the prcssure at C be~'Qllles less tl1:ln 2.7 III of water
absolute , the dissolved ai r and other gases would eomc out from water and ~"OlIcct at the summi t. The
flow uf W3ler will be obstructed. Syphon is used in the fullowing cases:
!. To carry water from one reservoir tu another reservoi r sepa rmed by a hill or ridge.
2. To t:lke out the liquid from ~ tank which is not having nny outlet.
3. To cmpty n channel nOl provided with nny outlet s luice.
zoo
P ro blem 11 .2 7 A J)"p ilon of diomelU
mm co"nects 11"0 reserl"oirs Iwdng a diffuence in efem·
lioll of 10 m. The lellglh of I/Ie sypholl is 500 m amllhe Siumnil is 3.0 111 abOl'e Ille Waler {ere/ ill Ille
upper reserl"oir. The le"glh of Ihe pipe from upper resu}"oir 10 Ihe su mmil is 100 m. Delermi"e the
di~'charge Illrough Ille $)"1'/10" alld "Iso press"re <II Ihe summil. Neg/e"1 milior IOSJe~·. The co·efficielll
of friClioll. f = .005.
Solution. Give n:
Dia. of syphon.
Difference in level uf twu reservoirs.
Length of syphon,
I I
d", 200 mm '" 0.20 III
H=201ll
L=500m
Ii
~ I
IL
Flow Through Pipes
Hciglll of s Ulllmit frolll upper reservo ir.
Le ngth of sypho n UplO su mmit.
Co-efficient of friction.
4991
h '" J.O III
L1 ", 100m
f == ,OOS
c
,
Fig. 11 .14
If minor losse s arc neglected thcI11hc luss of head lakes place only duc 10 friction.
Apply ing Bernoulli' s eq uati on to poi nt s A and B.
V1
J!.ii + ~ + ZA
pg
2g
VI
'"
~ + ~ + z~ + Loss o f head duc
pg 2g
to friction from A 10 8
0+ 0 + LA '" 0 + 0 + l /l+ IIJ I'; PA '" I'D '" atmospheric pressure. VA '" VB'" OJ
"'
1... -Z8= /IJ =
B"
l ,\ -18'"
20
4xjxLxV 1
dx2g
III
20= 4X.OO5X500XV
0.20)(2)(9.8 1
v=~
l
=2.548V1
20 =2.80mls
2.548
Q'" Veloci ty x Area
:. Disc harge ,
'" 2.80 x : (.2)1", 0.0879 mJ/s '" 87.9 Iilresls. An s.
Press ure at S ummit. Applyi ng Bernoulli' s equ ation \0 poin ts A and C,
P V1
--<l.. + ....L + L ..
pg
28
P
VI
V'
"'
, .
= - ' + -'- + l c + Loss of head due 10 fneltOH b<:twccn A and C
pg 2g
O+O+O",&+ - + 3.0+hj,
pg
2g
,
ITaking datum line passing through AI
p, + ;;;-~"~'=
4
00~5cX:cc'00
""X+.(2=8"--)'
O" _pg
+ 3.0
+ - .,X~.,,
2x9.8 1
0.2x2x9.81
=
&
pg
I I
lVc =V= 2 .801
& + 0.399 + 3.0 + 4.00 == & + 7.399
pg
pg
== - 7.399 m or wute- r. AilS.
Ii
~ I
IL
1500
Fluid Mechanics
Problem 11 .28 A SypllOIl of diameter 200 mill C()IweCf$ two re~·er\'oi'.1 hal'ing II differene<' in
eiel'arion of 15 m. Ti,e 10iailenglh oflhe syphon is 600 In oll<llhe sUlllmil is 4 III abore Ihe water level
in II,e tipper resumir.lfrlle separariO/I lakes place at 2.8 III a/wafer absolure,find Ihe maximum length
of syphon from upper f<'SenGi, 10 rhe summit. Take f = .004 and atmospheric pressure", 10.3 III
of ",a/a.
Solution. Gi ven:
Dia. of sypllon,
d"" 200 nUll
== 15 III
Differe nce o f level in two rC!i.ervoirs
TOlat ]englli of pipe
Hei glll of summit from uppe r
= 0.2
III
== 600 III
== 4 III
rc~rvoir
.E.::.. = 2.8 III of water absolute
Press ure head al su mmit.
P8
& = 10.3
Atmospheric prcssure licad .
III
of water absolute
P8
,'" .004
Co-efficient of frictio n. f
A
,
Fig. 11.15 (a)
Apply in g Bernoulli's eq uation 10 points A and C and taking the datum line passing throu gh. A ,
v~
Vl
l!..i. + ..:..d... + ZA '" .&. + _c_ + z.- + Loss of head due to frict ion between A and C
pg 2g
pg 2g
Substitutin g the va lu es o f pressures in terms of absolute, we have
v'
10.3+0 +0 =2.8+ -
2g
+4.0+ II ,
vo!
h~
,
[.: V< '" velocity in pipe '" V]
"
VI
'" 10.3 - 2.8 - 4.0 - - = 3.5 - -
2g
Apply ing Bernoulli's equation to points A and 8 and taking datum line
=
Bm
~ + vi
pg
2g
pa ~sin g
th rough 8.
+ Zs + Loss of head due to friction from A to B
l!..i. '"
pg
...(i)
2g
Ps "at mospheric pressure
pg
VA" O. VB=O.z,,= IS. zB"'O
0+0+ 15 =0+0+0+h,
I I
Ii
~ I
IL
Flow Through Pipes
11 =150r
J
4 x.()(}4 X 600 X VI
0.2)(2)(9.& 1
501 1
4x /X LxV 1
=15
dx2g
=150rV=
15)(0.2)(2)(9.81
4 x.OO4 X 600
2.47 m/s
Substituting Ihis value of V in equation (il. we get
B.
It!,
2.471
=3.5- 2)(9.81 = 3.5 -0.311 = 3.189
II
"' -
/,
... (; i)
III
o4~'~f~',-,:!"~'~V,-'
... (iii)
<lxlg
where L] = inlet leg of syphon or length of syphon fronl upper reservo ir 10 the
It
~uJ1\mit.
_ 4)(.004)(L, x (2.47)1
0.2)(2)<9.81
=0.0248)(L 1
I, -
Substituting this value in cquntioll (iil ,
0.0248 LI = 3.189
3.189
L, '" .0 248 '" 128.511 Ill. ADs.
Problem 11 .29 A syphon of di"meter 200 mm COlineCIS 111'0 rcsen'o;,s whoJ'c water J'llff<lcC {CI"e!
differ by 40 m. The rOlllllengll1 oJlhe pipe is 8000 m. The pipe crosses a ridge. Tile .mmmil a/ridge j ,Y
8 III (Ibm'" lire /el'el 0/",(l/e1 in Ihe upper f.-Jerl'ojr. V elermille /lJe minimum deplll oflile pipe be/ow Ille
sUfflmil of II,e ridge, if Ille absolule pre.ullre lIead a/ IIII' .mmlllil of s)'pllon is nOl 10 fall below 3.0 m of
•.-ater. Take f "" 0.006 alld atmospheric pressure head", 10.3 m of ....ater. Th e lellglll of syphollfrom Ille
IIpper resen·oir to tile slmlmil is 500 m. Filld the discharge a/so.
Solution. Gil'en :
d",200mm =0.20 m
Dia. of sypllOn.
Difference in lel' els of two reservoirs. fI " 40 m
Total le ngth of pipe,
I." 8000 m
Heig ht of ridge summit from wa ter level in upper reservoir = 8 III
Let th e depth of the pipe below Ihe summit of rid ge = x
III
Hei ght of syphon from water surface in Ihe upper reservoir " (8 - .r)
Pressure head a( C,
III
!!..:...." 3.0 m of water absolute
pg
. pressure head . ....!L
I'
Atmospheric
pg
"
10.3 m of water
Co-erfkient of friction
f = .006
Length of syphon from up per rese rvo ir to the s ummit. 1.( = 500 m
I I
Ii
~ I
IL
1502
Fluid Mechanics
Fig . 11.15 (b)
Apply ing Bernoulli's ~qumion to points A nnd B and taking d:llul11 line passing through B. we have
p
VI
P
VI
.....d.. + .....d... + z~ '" ....!!.. +....!!.... + ll/ + head loss due 10 friction A to B
pg
2g
pg
28
0+ 0 +40 = 0+ 0+ 0+
4XjXLXV l
~"o::C,'~~
d x2g
40= 4xQ.OO6x8000x V
1
0.2x2x9.81
v: ,~40~x3o~.2gx'122x~9~.8i!:1
= 0.904 Ill/s
4 x .006 x 8000
Now applyi ng Bernoulli 's cquatiol110 points A and C and assuming datum line passing through A.
we have
~
pg
Substituting
l!..!.. and i!L
pg
pg
+ V; + z~ = .f!..::... + V} + 'c+ head
28
pg
2g
loss due to fric tion from
A lu
C
in (erms of absolute p .........wre
Vl
4xfx~ xV l
10.3 + 0 + 0 = 3.0 + - + (8 - x) + :..:":;~C:'-'-28
d x2g
•
•
10.3 = 3.0 + (0.904 ) + (8 _ .r) + 4 :dK16 x 500 x (0.904)2x9.8 1
0.2x2x9.81
0'
=3.0+0.041 + (8- .r)+2.499= 13.54-x
x= 13.54-10.3=3.24m. An5.
Discharge.
.. 1l.7
Q = Area x Velocity = ..::. x (.2) 2 X 0.904 = 0.0283 m J/s. Ans.
4
fLOW THROUGH PIPES IN SERIES OR FLOW THROUGH COMPOUND PIPES
Pipes in series nr compound pipes are defined a~ the pip es of different lengths and different diam·
etas connected end to end (in scries) to form a pipe line as show n in Fig. 11.16.
I I
Ii
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IL
Flow Through Pipes
Let,
Lt_~.
503 1
L)" length of pipes ],2 and 3 respectively
d l _ d!. d)" diameter o f pipes 1. 2. 3 respectively
VI_ V2• V)" velocity of flow through pipes 1,2,3
f l'!!' f J " co-efficient of frictions for pipes 1.2.3
H = difference of water leve l in the two tan~S.
,
,
F ig . 11.16
The discharge passing through e ach pipe is same.
Q=AIVj"AzVt=AlVl
The d ifFt're nce in liquid surface levels is equal 10
H=O.5 Vl
l
th~
sum of tile lOla] head loss in the pipes.
+ 4Ji~V/ +05V/ + 4f!~V}
2g
2g
d 1 x2g
d 2 x 2g
+ (V2 ,
vS + 4fl~VJ~
+ v/
d'
2
.tg
J
x.tg
g
...( 11. 12)
If minor losses arc ncgLcclL-d. then above equation becomes as
,
,
'
H= 4fl L, VI + 4f,L,V~ + 4fJ~V;
d 1 x2g
d 2 x2g
d J x2g
... ( 11.13)
If the co.efficknt of friction is smne for all pipes
II '" h '" fl = f. then equation ( 1 1.1 3) beco mes as
i.e..
H= 4 fL,V/
d,x2g
== 4f
2g
+ 4 JL'lV1l + C4~JL,
,,,,V,c'
d 2 X2g
d ) x 2g
[ ' , '1
L,V,· + L 2V1 + L)V,
d,
dl
d)
.. _(11. 14)
Problem 11 .30 The difference in ...aler !J'urface lel'els in 1"'0 lallks ....hich are conneCled by Ihree
pipes ill series of {engl/,s 300 m. 170 III alld 210 III alld of dlamelerI 300 mill. 200 ,,"" allli 400 111m
respeclil'ely, i,f 12 III. Delermine Ihe rale of flo ... of \l"(/(cr if co-efficielll of frietiOIl are .005, .0052 IIl1d
.1J048 respeeli"ely, cOllsiderillg : (i) millor losses also (iO IIcglee/illg millor 10S,Ie.!.
Solution. Given:
Difference o f water le"eI.
lA:ngth of pipe I.
lA:ngth of pipe 2.
~ I
H
= 12
In
L , == 300 m and dia .. tI, == 300 mm= 0.3 m
L1 == 170 m and dia., 1f1 = 200 111111 :0 0.2 11\
I~
~ I
IL
1504
Fluid Mechanics
Len gth of pi pe 3.
L J '" 210 m and dia .. d.l '" 400 n1ln '" 0 .4 m
Also.
I, '" .005./, '" .0052 and!} '" .OM S
(i) Co ns id e ring Minor Losses. Lei VI' V, and V3 are th e veloc iti es in the l SI. 2nd and 3 rd pi pe
respecti vel y.
From co ntinui ty. we have AI VI '" A1 V1 '" A JVJ
"d,'
,
()' X VI '" 2.25 VI
A,V,
4
d,
0.3
V, : - - . - - V , = - , V, '" Al
d
O
2
- d;
1
.
' 0
o
4
V == A, V, '"
and
Aj
J
0
d,l
d;
=(03 )'
\I:
0.4
I
VI '" 0 .5625 VI
No w us in g eq uati on (II 12). we li avc
Substitut ing V1,md Y1•
O.5IC '
4x ,005 x 300 X \1: 1 O,SX (2.25\11lt
12.0", - - ' - +
' + ---''0--'-'28
0.3 x 28
28
12.0 : VI' [0.5+20.0+ 2.53+ 89.505+2.&4 7 +3. 189 + 0.3 16[
"'
2,
"'2,
• - '- 111 8.8871
12 x 2 x 9.81
v,-R ale of now,
11&887
= 1.407 mls
Q = A rea x Ve loc it y = AI x VI
It
,
It
,
1
'" "4 (dlt X VI '" "4 (.3)- x 1.407 = 0.09945 111 Is
= 99.45 litrcs/s . Ans.
(ii ) Neg lectin g !\linor Losses. Us in g
Cq U3Ii Oll
( 11. 13), we have
1
H = 4.fiI4 V1 + 4f.'-!. V} + 4fl ~Vll
tf1 x2 g
d 1 x 2g
d J x2 g
or
[ 2.0= ~
o [ 4x.OO5x300
2g
I I
0.3
+ 4x.OO52xI70x (2.25) , + 4x.()()48x2 10 x (.5625j"oJ
0.2
0.4
Ii
~ I
IL
Flow Through Pipes
vl
= _,_ [20.0 + 89.505 + 3. 189]
2g
2x9.81x 120
11 2.694
V'
5051
X 112.694
'=' _ L
2g
== 1.445 m/s
Di.o;chargc. Q'" VI X AI '" 1.445 x ~ (.3)'" 0.1021 1113/S == 102. 1 lilrcsls. Ans.
4
Problem 11 .30 ( A). Tllret' pi{!1'S of.Joo mm. 200 111m mill 300 mm dialllelers Iw,'c (engll,s of 400 m.
200 III. ,md 300 III re~peclil'l.<l)', They liTe COllnected ill series 10 make (l compOlmd pipe. Tile ends of
Ihis compound pipe UTe connected wilh Iwo tanks whose differen ce of w(ller iel'e/J' is 16 m. If
co-efficient of/rictioll for Iliesc pipes is same and equililo OJJ05, de/ermine the disc/wrge 111rougll the
compOlmd pipe neg/llCIifl8 firSI the minor losses and Ihell including II,em.
Solution. Given:
Difference of water levels, H= 16m
Length and d ia. of pip.:: I. LI ",400 m and d l '" 400 mm '" 0.4 m
Length and di a. of pip.: 2. L 2 '" 200 m and d 2 '" 200 mm '" 0.2 m
Length and di a. of pipe J. L 1 ", 300 m and d) '" 300 mm '" 0.3 m
Also
11"'12"'13=0.005
(i ) Disc harge thro ugh th e comllo und pipe fi rst neglec ting minor losses.
Let VI' V2 and V3 are the velocities in the lst. 2nd and 3rd pipe respectively.
From continuity. we have AIV I '" A 2 V2 '" AJVJ
, ,
AV = 4 dl
dl1, VI = (00
4)'
- X VI '" -
V1 ", - ' -'
d;
4 .
Al
d;
7r
0.2
VI'" 4V I
and
Now using equ ati on ( ] 1.13). we have
H",
4JiI4 Vt' + 4fl ~Vl1 + 4 fJ L JV/
d l x2g
"'
16=
4xO.OO5x400xVI'
0.4x2x9.8 1
1
l
'" 2 x9.81
V
16=
~'
I
2x9.81
+
d 1 x2g
4 x 0.005 x 200 x (4Vd1
( 4 x 0.005 x 400
0.4
d l x2g
0.2x2x9.81
+
4 x 0.005 x
(20+320 + 63.14)=
200 x 16
0.2
V,'
I
2x 9.81
+
4 xO.005x300
03x2x9.8 1
,
x (1.77 V)"
I
c4CXCO
coOO
"""cX
"3"OO
,,,-,X
C3coO"c' .)
+-
OJ
x403.14
16 x2 x9.81
'" 0.882 mfs
403.1 4
I I
Ii
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IL
1506
Fluid Mechanics
"
,
Q=A , X VI '" - (0.4) x 0.882 ", O. '108m Is. Ails.
4
( ri) Disc ha rge through th e compound ]Iipe co ns idering minor losses also.
Discharge.
Minor IOHes (lrt':
0.5 VI~
II : - -
(a) At inl et.
,
2,
(b) Between 1st pipe ;lml2nd pipe. due to contraction.
h: _
05_V_,1 =,-O'.':..,;-(4-,
V,C')L
2,
c
=
-
2g
'
°
'
5x 16xV, =8x V,2g
2g
(e) Between 2nd pipe and 3rd pipe. du e 10 sudden enl argeme nt.
/, '"
(V, -
'
vS
2,
(4 VI _ 1.77V )l
I
= "'"--~'-".L
28
V1
\.\ '
= (2.23)1)( _ ,_ '" 4.973 - '2g
v1
(t.77Vd"
2g
28
(d) At lh~ o lltl el o f 3rd pipe. 11" = - '- =
The major losses arc
=
=
28
4Ji X 1'1 XV/
,~1
\1: '
= 1.77' x - '- = 3. 1329 -'2g
+ 4 fl x ~
d1 x2g
4 x 0.005 x 400 x
0.4x2x9.81
X V1'
d1 x2g
2,
+ 4 f J x~
XV)'
d J x2g
v/ + 4 x Q.()()5 x 200 x (4 VI) 1 + -,-4,-XCOC,OO:;;::5CX"300
:;:::OX,,(~,,-77CV,, ,-)'
'" 4{)3.14
0.2x2x9.81
x
0.3 )(2x9.81
v'I
2 x 9.81
:. Sum of minor losses and major losses
2
:
2
05 V/ +8x-+
VI
~'
VI
., V12]
4.973-+3.13_9
- + 403.14[ 2g
2g
2g
2g
2g
v'
'" 4 19.746 - '-
2,
But tOlalloss IIll1st b<: equal 10 H (or 16 m)
4[9. 746)( -V''- :
2g
:. Discharge.
I I
16
I = O.864m/s
VI:,I~16~X~2~X~9~,8I
4[9.746
Q:A I V 1 : -It (0 .4) 1 x 0.864 = O. 10HS m J Is. An s.
4
Ii
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Flow Through Pipes
5071
EQUIVALENT PIPE
.. 11 .8
This is d~rlncd as the pipe of uniform diameter having loss of head and discharge equa l 10 [he loss
of head and discharge of a compound pire consisting of several pipes of different lengthS and diam Ckrs. The uniform diameter of the equi va lent pipe is called equivalent size of lh e pipe. The length of
cqu;valcm pipe is equal to sum of lengths of the compound pir<: consisting uf different pipes.
u.l L [ '" le ngth of pipe 1 and ill'" diameter o f pipe I
L 2 ", le ngth of pipe 2 and ill" diameter of pipe 2
L3 = knglh of pipe 3 and d 3 = diameter of pipe 3
II '" 100ai head loss
L = length of equ ival ent pipe
d", diameter oflhe equivalent pipe
Then L=LI+L2+L3
Total head Joss in the compound pipe. neglecting minor losses
,
,
,
11= 4JiL,V[- + 4fl. ~V; + 4 /l~ V)
d l x2g
d l x2g
dJ x2g
... (11. 14A)
II '"'Il'"'h==1
Assuming
Discharge,
4Q
4Q
4Q
red,
red;
redj
VI'" - , . V 2 ", - , 311d V 3 ", - ,
Substituting these values in equmioll (11.14A), we have
... ( 11.1 5)
Head loss in the equivalent pipe.
where V '"
Q
Q
"A'" ~d!
H",
4/.L.V
d x2g
l
IT aking sallie
value
0
ff
as in compound pipe]
4Q
'" rtd l
4
4Q
H= 4/.L. ( ;;if
dx2g
J' ,", 4 x,16Ql/[~1
re - x2g
d~
... ( 11. 16)
Head loss in compound pipe and in equivalenl pip.: is same hellce equating equations (I 1.15) and
(11. 16),wchavc
~ I
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IL
1508
Fluid Mechanics
... ( 11.1 7)
0<
Equati o n ( I I 17) is kn ow n as Dupuit' scq u3t ion. In thi s equat ion L ", L ] + L2 + Ll and dl' 112 and d J
are kn ow n. He nce the equi val ent 5i].e of Ihe pipe. i.e .. value of II ca n be o bta in ed .
Pro blem 11 .31 Tilfee pipo of ":ngrl!.~ 800 m, 500 In ami 400 In allli of dialneter.! 500 min. 400 mm
and 300 mill respeCfil'ely are connecled ill serres. TilesI' pipes are 10 be replaced by a single pipe of
length 1700 m. Find Ihe diameter of Ihe single pipe.
Solullon. Given :
Len gth o f pipe I.
LI '" 800!ll and dia.• III = 500 111m '" 0.5 III
L" nglh o f pipc 2.
~ '" 500 III and o ia .. d~ '" 4 00 mm '" 0.4 In
Le ngth o f pipe 3,
L 3 ", 400 III and « ia .. d] '" 300 mm '" 0.3 In
Le ngth of singic pi pe .
L = 1700 m
Le t the diame ter of eq ui va icnt sin g le pirc = d
L
1.,
Appl ying c quuli on (I J 17), - , = - s
d
ii,
L,
L"
<11
ill
+---t + - ,
1700 800 500 400
-,- = - , + - , + - s = 25600 + 48828.1 25 + 164609 == 239037
d
.5
.4
0.3
1700
.007 118
239037
d == (.007 188)°·2 == 0 .3 71 8 == 37 1.8 nlnI. A ns •
tiS==
.. 1 1.9
FLOW THROUGH PARALLEL PIPES
Consider a ma in pi pe whic h di vides inlo two or more bran ches as s how n in Fi g. 11.1 7 and aga in join
toge ther dow nstream to form a s in g le pipe. the n Ihe bran ch pipes arc sa id 10 be conncclcd in paralle l.
The di scharge Ih ro ugh th e ma in is increa sed by co nn ecting pipes in parallel.
B R A N CH PtP E :2
B R A N C H PIPE 1
Fig. 11 .1 7
The rate of fl ow in Ih e main pipe is equ a l 10 th e s um of rale of flo w Ih roug h bra nch pipes. Hence
from l"' ig . 11. 17. we have
...( 11.18)
Q=Q I +Q 2
In th is. arTang emcl11. the loss of head for each branch pipe is sa me.
:. Loss o f h" ad for bran c h pipe I = Loss of head for hrauc h pipe 2
I I
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IL
Flow Through Pipes
4f,L,V,l == 4 f2~V/
tI, x2g
til x2g
I, ==/"
If
,hell
•
5091
... ( 1 1.19)
L,V,'
L"V}
... ( 11.20)
d, xlg d 1 x2g
Problem 11.32 A main pipe tfi\'ides illll) /WI) parallel pipes which again formJ QlIe pipe (/$ .fholi'll ill
Fig. f 1.17. The lellglh alld diameter for the first parallel pipe are 2000 m and 1.0 m respecril·e!y ....hile
l/ie lengll. and dim/leler of ]"'/ parallel pipe <Ire 2000 In wId 0.8 m. Find II,e mil' of flow ill ellcl.
paralld pipe. if TOla' flow ill 11,1' lIIain is 3.0 m)k Ti,e co-efficient ofIric/IQlI for l'(lell paf<lllei pipe is
.mme atilt equal rn .005.
Solution. Give n :
lA:ngt h of pipe I.
L,=200011l
d,,,,, 1.0rn
Dia. of pipe 1.
Length of pipe 2.
L2 == 2{)()() 111
Dia. of pipe 2.
d,=O.8rn
To tal flow.
3.0 lOlls
Q'"
I, == 12 =/= .005
0, '" disctlarge in
pipe 1
Q1 == discharge;n pipe 2
Q "" Q, + Q2 == 3.0
... ( i)
From equ ation ( II 18).
Using eq uation (1 1.19). we It avc
4j, ~V," '" 4j , L..,V,l
d , x 2g
d l x 2g
4 x .005 x 2000 x Vll.
4 x .005 x 2000 x V,
=
0.8x2x9.81
1.0 x 2 x9.81
" = -Vi.· or V, 1 ", -V,',
V,
-
0'
1.0
0.8
0.8
Now
and
Substituling the va lu e o f Q, and Ql in equa tion (I). we get
1(
VI.
It
"4 xO.894 +"4 x.64 V1 '" 3.0 or 0.8785 V1 +O.5026 V1 =3.0
0'
I I
or V =
...lQ... = 2.17
Ill/S.
1.3811
Ii
~ I
IL
1510 Fluid Mechanics
Substituting this va lue in equation (ii) ,
V,
2.1 7
VI = - -- = - -
.894
Hence
11:
0.894
= 2.427 m/s
2
It
0 1 ", - d, )( VI '" -
4
4
l
,
x I x 2 .427 = 1.906 nOs. Ans .
Q1 =Q-Q 1 '" 3.0-1.906= I.094m J /s. AilS.
Problem 11 .33 A pipe line 0/0.6 III dillmeTer is /.5 km lOll/!.. To ;/lCT/'(Ise Ihe disc/"" ge. (Jnolher line
of JIl" J'ome diameter iJ' in/rot/fleed pnralld 10 Ille first iii Ihe second half of Ihe /ength. Neglecting
minor IOSJ'cs, find Ihe increase in tlischarg(' if 4[ = 0.04. The head lI/ inlet iJ' 300 mm.
Solution. Given:
Dia. of pipe line.
Length of pipe line.
Head a1 inlet.
Head at outlet.
:. Head loss.
D =O.6 rn
L = 1.5 kill = 1.5 x 1000", 1500
III
4f =0.04or/:.0 1
II = 300 rmn = 0.3 m
= atmospheric head = 0
h, ,,, 0.3 III
Lenglh of another parallel pipe. LI '"
lS~
= 750 m
Dill. of anolher parallel pillt'. d l = 0.6 10
Fig. I 1.18 shows the arrangement of pipe system.
L :750m.d :O.6m
' l m<---='''''''---='==
T~-_cQi;;ii'i~-~'
~::-...;.
0,
C
O:O.6m
..........
l___
Q,
,
L,"'750m.d,:O.6m
Fig. 11.18
1s t Ca se. Discharge for a sing le pipe of length 1500 III and dia. '" 0.6 m.
2
This head lost due to friction in singic pipe is
4JLy·
"r=-'<~Cdx2g
whereV" = veloc it y of flow for s ingle pipe
.,
4x.OlxI500xY·
"'
0.3 =
V"=
0.6 x2g
0.3xO.6x2x9.81 =0.2426m/s
4x .01x 15oo
Discharge. Q. = V" x Area = 0.2426 x ~ (.6) 2 "" 0.06&5 m3/s
... ( i)
4
2nd Case. When an additional pipe o f length 750 111 and diamete r 0.6 m is connected in parallel with
the last half length of th e pipe.
I I
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Flow Through Pipes
Let
511 1
Q 1 '" di scharge in I SI paral lel pi pe
Q2 '" d ischarge in 2nd pa rall el pipe
Q '" Q, + Q!
where Q= d isc harge in main pi pe when pipes J rc pa rall el.
BUI as the ic ngth and dimnctcrs of eac h pa r,tlie ) pi pe is same
Q, =Q2'" Q!2
Consider the now thro ugh pipe ABC o r ABO
Head loss tll ro ugh A BC", H ead lost th rough A n + head losl l hro ugh BC
...(ii)
BUl lic ad l ost du e to fri ct io n th rough ABC", 0.3 m gi ve n
Head loss d ue to fric tion th ro ugh All '"
4 x/ x 750xV l
• w here V = ve locity o f fl ow t hrough A ll
0.6x 2 /9.81
Q
40
H ead loss d ue to f ric tion th rough. A lJ
= 4x .Olx 750
0.6 x 2 x9.81
X [~l2 =3 1.87 Q2
n:x .36
Head loss du e to frictio n tli rough. BC
= 4x / xL. XV,l
d, x 2g
=
~; ~o;: ;~~ x[ xt(61'1
2
4x.oJx 750
. . V _ Distance
[
•
I -
~(.6f
Q
1
4
16
'" 0.6 x 2 x 9.81 x 4xl1:'" x .36" Q2:7.969Q2
Substitut ing these va lues in cqu atioll (ii). we ge l
0.3 = 31.87 Q1 + 7.969
Q"
01 = 39.839 0 2
¥'153
39.839 "" 0 .0867 m,Is
Incrc aSoe in discharge", Q - Q. = 0.0867 - 0.0685 = 0.0 182 m 3 /s.
AD S.
Problem 11 .34 A I'umpili g pfali/ fvrces ,mler Ih rough a 600 mm di(/meler main. Ihe friClivn head
being 27 III. III order IV retlllce Ihe po,..", COII~'umplion . il is proposed 10 lay {mother main of "ppropri·
ale tli",ne/'" afOiIS Ihe "ide of Ille <'xiS/ins one. so Ilwl 1"'0 pipes may ,..ork ill parallel fo r Ihe enlire
Jeng/II (/lid reduce 111<' frie/ioll 11<'(ld 10 9.6 m ollly. Filld Ihe diameter of Ille lIeli' maill if. ,..ilh 111<'
e.(cepli011 of diameter, il i,~ simi/M 10 Ihe exiJlillg Ol1e in el'''')' respect.
Solution. G ivc n :
Dia. of si ng lc main pipe.
d = 600 mm = 0.6 m
hf = 27 m
Frictio n head.
Frictio n he ad for two parall el pipes
"" 9.6 m
1s t Case.
For a sin g lc mai n [Fig. 11. I 9 (a) ]
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1512 Fluid Mechanics
/1 '" 4 . f
I
2
. L. v or 27. 0 == c~4~X~f~X:;-::L~X~V.,-'
dxlg
0.6x2x9.8 1
f LV2 = 27.0 x O.6x2x9.8 1 ",3 17.844
4
5
=79.46 1. whcrcV= Q
A
Q'
n(i)
fL A' = 79.461
2nd Case. Two pipes arc in parallel [Fig. 11.19 (b) ]
Loss of head in anyone pipe == 9.6 m
:. For 1st pipe.
111,=
,
4 .!. L. V,"
-1
d, x2g
9
=.6
0 ---
n
-
- - -
n
n
-I
(a) Single main
-
----
-~'- - - - --- - -- - - - h,- "'9 .6m
,
" - _. _ _ _
_Q_'I. ____ _ • __ • _ _ __ _ _ _ _ _ _ _ • __ _
(b) Two para llel pipes
Fig. 11.19
Om
.fl=~
L, =L . V, '"
A,
A
d,,,,d=O.6
4c·~fc'"L= Q,'
,c-:c
Al '" 0.6
0.6)(2)(9.81
f
"'
For the 2nd pipe,
Q,! - 9.6 XO.6x 2 x9.8 1_
'8
xX
L -:-r - 282
.5_
A
4
I,f, -
... (ii)
4 jxL, XV,'
• • == 9.6,
dl x 2g
i,
4 / X LX Q =9.6
d,x2gx A;
/ xLxQ; '" 9.6 x 2 x 9.81 '" 47.088
ill X A~
Di vidi ng
I I
~qualion
4
... (iii)
(i) by equal io n (iiI, we gel
Ii
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Flow Through Pipes
Q2
79.461
Q I! '"
28.2528 '" 2.8 125
..£ ",./2.8125
Q,
513 1
'" 1.667
Q
Q 1 '" 1.667 '" .596 Q
QI+Q 2 :Q
Ql= Q- Q 1 = Q- .596 Q=O.404 Q
Dividing eqU3lion (ii) by cqumion (iii),
ai' xd, xA;
A l XQ;
= 28.2528 '" 0.6
47.088
Bw
.596 Q )' x di, =0.6
( A04 Q
.)6"
d/ '" 0.6
1
X
.36 x
(::~r
= 0.03537
<I, '" (.03537)"~ = 0.5125 In '" 5 12.5 mill. Ans.
A pipe of (Iiame/a 20 em and /el1gl11 2000 III collnects IWO reserroirs, IIm'ing
lew'ls (IS 20 III. De/ermine Ihe discharge II"OI'81i Ihe pipe.
adell/iV/wi pipe of diameter 20 em amllnlglll 1200 III is allllciled IQ Ilu~ lasl 1200 '" 1"'115111 of
Problem 11 .35
difference of
"'at'"
If wi
Ille <'.1"1.11;118 pipe, Jim/rile illerea,51! ill Ihe discharge. Take f
'"
.015 lind lIegleC/mitlQr 10$,ro.
Solution. Given:
Dia. of pipe.
Len glh of pipe,
d=20cm =O.20m
L=2000m
Difference of waler levels. H = 20 In
Co-cfflcicn! of frictio n.
/=0.Ql5
1s t Ca se. When a si n g l ~ pipe connects the two reservoirs
11= 4./ . L .V
dx2}:
1
4[. L [ Q ]'
= dx2g : d1
32f·L.Q1
= rr. " xgxd '
I I
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1514 Fluid Mechanics
20 = "32""xc·Oc'c'o'c'cOOOccoxCiQ,-) '
" 30985.01 Q-,
I( ' X9.8 I X(0.2)
Q ==
~
o
309X5.07
,
,, 0.0254 m Is. An s.
2nd Case.
Lei Q 1 "diSl: hargc through pipe CD.
Q1 == di sch arge through pipe DE.
QJ = d isc harge through pi pe DF.
Le ngth of pipe CD.
LI == 800 111 and its dia .• d l " 0.20 111
L2 ", 1200 III and its dia., (/2 " 0.20 III
L) = 1200 11\ and its dia .. d1 == 0.20 m.
Length of pipe DE.
Length of pipe DF.
Since the diam eters and leng th s o f th e pipes DE and DF arc equal. Hell ec Q2 will be equal 10 Ql'
Also for parallcl pipes. we have
Q 1 == Q! + QJ" Q1 + Q! = 2Q!
Q,=~
,
2
A
c
T
0,
;:-..;.
°
,
20m
2
•
800"" 20ciii ~ -----
'200",03
. 20c",
FL----'
Fig. 11.20
Apply in g Bernoulli' s equati on (0 points A and B and tak i ng th e fl ow throu gh CDE. we ha l'e
20 =
4/ .t.. v/ + 4/ .L,. V/
d t x2g
d 2 x2g
,
,
- 4x.015x800 , ( -4Q,
- - )" + 4 x.015xl200
0.2x2 x 9 .81
Ifx.04
0.2x2x9.81
lQ, )'
x( --/tx.04
'" 12394 01 1 + 46470/ '" 1704 1 Qi l
0
1 '"
~ 1 7~1
'" 0.0342 ml/s
Increase in di scharge'" Q I - Q '" 0.0342 - 0.0254 = .00811 mJ/s. AilS.
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Flow Through Pipes
515 1
Problem 11 .36 Two I'ipes /rare a /ellglh L each . aile of I/Ielll lias {/ diameter D. and Ille otller a
diolllefer d. If Ihe pipes (Ir,. arwnged ill powllef. Ihe loss of head. ",lleli a IOral 'II/anllty of Willer Q
flows II/rougll (llem is h. bU!. iflhe pipe.! are arrtmged ill series and the s(lme qlwlllif), Q flows throll8'1
loss of head ;.1 H. If II = 0 , find Ille ralio of H
2
assuming 1110' pipe co-efficient f Iws a COIUlalll \'al'le.
them.
Jill'
10
II, n ..glee/ing
.~econd(lfy
IOHeJ' {md
Solution. Given
u:ngth of pipe I.
L, '" L and its dia. il, '" D
Length of piP<' 2.
L2 '" L and its dia .• d 2 '" d
Total discharge
'" Q
~1c'1(.l loss wlien pipes are arranged in parallel'" II
Head loss when pipes arc arranged in se ries", H
d
= 0 and f is conSlant
2
ht Case. Whell pipes are connected to parallel
Q'" Q , + Q2
Loss of head in each pipe = II
... ( i)
For pipe AB,
d, =0
co
Fig.
4JL x [ nO'
4Q~)'
]1.2]
32fLQ,!
---,"'-";'--'-= " or n 10 ) xg '" "
Dx2g
['or pipe AC.
32jf·Qi
n ! d ) x8
= II
... (ii)
...(iii)
1
32jLQ , '" 32fLQi
n 1 D )g
n l d )g
[':0=2d[
'" 2) '" 32
t '" ./Yi '"
I I
5.657 or Q , '" 5.657 Q1
Ii
~ I
IL
1516
Fluid Mechanics
Substituting Ihe valu es of Q, in eq uation 0). we get
Q '" 5.657 Q: + 0 1 '" 6.657 Q1
Ql : ~ :::O. 1 5Q
... (il·)
6.657
Q,=Q - Q1=Q - O. 15Q=0.85Q
From (i)
2nd Case.
Wh ~ n
... (v)
th e pipes arc ron ncclcd in se ri es.
To tal
los_~
= Sum of he ad losses in lIie two pip.:s
1/ : 4 j .L.V,2 + 4 f ·L.V~!
d,x2g
u 1 x2g
WhcreV, :~= 4Q"Vl = ~= 4~
~D l
rrD -
7Uf
!!.- dl
4
4
C.'
C. D
-
_ v,
0
-
0
Fig. 11.22
II =
11 =
"'
From eq uation (ii).
4J. LX (-,:-;-ZY 4fL (~~r
---;;-cC2'-"+ -;=:-:"Dx2 g
dxlg
32fl_Q !
S'
Dn "
32JLQ!
+ S '
x g d n·
xJ\
32jL
-,,'?f"-- -"
n 1 S x g - Q,!
and from equ ation (iii).
D
h
;
"
2f
:::
Q
£
tr' d x g
SubstituTin g th ese valu es in equation ("i), we have
Q1
[Q
-1
,+ -Q,l]
1
Il
!
II
Q'
H = Q x - , +Q X- , =-,11 +-,11 ="
Q,Qi
Q,Qi
0,
11
-
II
Q2
= - , +- ,
0, Qi
But from equa tio ns (il') a nd (v).
Ii
-=
h
Q2
Ol
0, = .85 Q and 0, = 0. 15 Q
1.384 + 44.444 = 4S.H2H. Ans.
Problem 11 .36 (A). Tlm·e piJll·s of the same /ength L, diameter D. WId friction faCiO r f Me
connec/ed in p"",IIe1. Determine l/ie diwneter of Ille pipe of length L GIld friction filClor f ...lIich ...iII
carry ti,e SlIIlIe discharge for the .tame Ilead loss. Use ti,e form ulllilf '" f x Lx V"'/2g D.
I I
Ii
~ I
IL
517 1
Flow Through Pipes
Solution. Given:
Length of cacti pipe
'" L
Diamete r of c 31:h pipe
== 0
Friction factor of cadI pipe
=f
1
Head loss,
Ilr '" fX L x V f2gD
When th e th ree pipes arc connected in parallel. then head loss in each pi P<' will be same. Allu IOtal
head loss will be equa l to Ihe head loss in each pipe.
Let
h ", Total he ad loss.
f
hi, '" Head toss in lSI pipe.
hi, '" Head loss in 2nd pipe. and IIf, '" Head loss ill3 rl1 pipe.
f xLx
Then
Let
hI "" IIf, == Il f , == 1If, or II! '"
V~
'-'';,O',"D
'-'--
...(il
Q1 '" Discharge through IS! pipe. Q2 '" Discharge Th rough 2nd pipe.
Q3 = Discharge throug h 3rd pipe. and Q = Total discharge.
When th e three pipes arc con nected in parallel. Ihen
Q = 0, + Q1 + QJ = 3 X Q,
=)xA , xV t
:3)(
'4" 0 2 )( V (wlicrcA, = '4D
" ', and V, = V)
... (ii)
For a si ng le pipe (or leng th L: friction facior j) wtJ ich will carry same discharge as lhe lh ree pipes in
parallel
LeI d = dia. of lhe sing le pipe
,. = ve loc ily through si ngle piP<'
Then dischurge. Q = Area x Velocity
= (:
d')
...(iii)
x v
Equatillg the two values of discharge. given by equations (ii) and (iii). we get
If
!
If
D' ,·
,
3x - D xV= - <l" XI' or 3x - .! = ... (i\')
4
4
If
V
The head loss for the si llg le pipe is also equal to lhe lotal head loss for three pipes when lhey arc in
l)ara ll ~ 1.
But head loss for lhe sing le pipe of lengt h L, dia. <I, friclion factor f and I'e loc il y "is give n hy
IJ =
,
~f_'cL~'~"_'
... ( 1')
d x 2g
Equaling Ihe IWO valu es o f '"give n by equations (i) and (v). we gel
,
f x L X V' = ~f~';--"L~'C'c"_ oc
Dx2 g
dx2g
If
-
D
,
_V_'
,
= _"_
D
d
y.!
= -,
V'
Substituting the va lu e o f "IV in eq uation (i,'). we gel
I I
Ii
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IL
1518 Fluid Mechanics
(dJ'"
3 , -D',= dD
:!... '" 3;'5 "
0'
(d J'" (dJ' (<f J'"
3° 0
D
x D
ur 3 = D
'"
,, D
1.55
d" 1.55 D. Ans .
He nce d ia. of singl e pipe sho uld be 1.55 ti mes tlie dia . o f the th ree pipes connected in pa rall c l.
Pro blem ' 1,37 For a /O ll'n wU/«r supply, II main pipe /i" e ojdiomele r 0 ../ '" is required. As piPH
more than 0.35 m drame/a are 1101 readily oWlilable. flVO parallel pipeJ of I/Ie same diameter were
usedfar Waler supply. Iflhe IOral discharge ill Ihe parallel pipes is same as in the single moill pipe.Jind
the (1;(IIlIeler of the parallel pipe. Assllml:' IIII.' co- efficienT of[rielion S(lme for (11/ pipes.
So lution. Gi ve n :
Dia. of sing le ma ill pip" line.
d" 0 .4 m
LeIth" length of single pipe lin e
"L
Co-cfticicnt of fric tio n
=f
._ .
.
.
.
4JLVl
d x 2g
4fLV '
Loss o f head duc to t nctlOn 111 sin g le P'PC " - - - • ~",,:o--
... ( i)
0.4 )(2)( g
where V:: Ve loc it y of n o w in the sin gle pi()C.
In case o f parallel pilX'. as the diameters aud k llg ills o f th e two pipes are sam e. He!)ec di sc harge in
each pi pe will he half the di scharge of s in g le main pipe. As di.<;ehargc in each paralie l pipe is same.
hence ve loc ity will also be same .
V. = Ve loc it y in eac h p~rallcl pipe
Let
d. = Dia. o f eac h paralle l pire
The n loss of head du e 10 fri<:ti on in parallel pipes =
Equatillg Ihe two losses g ive n by equil ti o ns (i)
4 /. L.V l
0.4x 2g
v'
Cance ll ing 4 f L .
2,
0.4
~Ild
4/ X L XV,!
J. x 2g
...(ii)
(ii). we ha ve
4/ xL x V.!
d. x 2g
. -,v.! "' - ,' d.
V"
0.4
V.-
d.
... (iii)
From continuit y
TOl al no w i n sin gle mai n :: SUIll of fl ow in two parallel pi pes
Of Ve loc ity of ma in x Are a :: 2 X Veloc ity in each paralle l pi pe x Area
II
,
1(
,
V
V x - (0.4)-:: 2 x V.)( - d-.or - ::
4
4
V.
2x!!.-d:
4
.~-;-
~ (O .4 ) l
4
u}
0. 16
Squarin g both sides.
Comp ar ing equati ons (iii) and (il'). we get
I I
Ii
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IL
Flow Through Pipes
519 1
0.4 '" 4d;
or ~ '" 0.4 x .0256 '" .00256
d.
.0256
4
d. '" (J)0256)1/~ '" 0.303 III '" 30.3 cm. ADS.
USC two pipes of 30.3 CIll diameter.
All old waIn supp!.,' diSlribUli01l pipe of 150 "'Ill di(alleler of (, cil)' is 10 be replaced
by /"'0 p"ralld pipes of smfll/er eq""{ diameter having equal /""gllis "",/ idenlical friction jaclor
I·u/aes. Find 011' tile new di",,,eler "-''1"ired.
Problem 11 .38
Solution. Given:
Dia. of old pipe. D '" 250 mm '" 0.25 rn
Lei
= Di a. of each o f para lid pipes
Q = Discharge in old pipe
Q 1 = Discharge in first paralld pipe
Q,- = Discharge in second paralic I pire
d
f = Fri clion factor.
When a single pipe is rcpla~'Cd by two parallcl pipes. the head loss will be same in the s in gle pipe
and in eac h ufl he p3 rallcl pipes. Also the discharge in single pip e will be equal I() the total d ischarge ill
twO p;lT<llIcl pipes i. e"
... (i)
and
Q" Q 1 + Q1
As the dia. of each parallel pipe is same and also length of each parallel pipe is equal. hence
QI = Q1 or QI=Q1=Qf2
Now
h, ,, Head luss ill single pipe
=f X LXV '
Dx2g
JX LX[
=
It
wheref= Friction factor
Q
X 0.25 '
J'
4
[.: V =
0.25 x 2 x 9.81
=
••• (ii)
f x tX (4Q )1
0.25 x 2 x 9.81 x( It x 0.25')"
A~a = : ~l )
... (iii}
IlJi = Head loss in IS! parallel pipe
"
=
JX Lx (~i
II x2 g
(": Dia. of parallel pipe" II and VI is the velocity
2X~d'
I x LX [
r
in lSI parallel pipe)
dx2g
I I
Ii
~ I
IL
1520 Fluid Mechanics
=
! XLX
C (4Q)'
-d-,--:-,_,L9".8 ", X:-;(2"'X""CX-d~'~)"
[X Lx(4Ql
f X L X (4Q)!
2 '
0.25x2x9.81 x(n x 0.25
r
==
d
x 2 ><9.8 1>«2x Itxd l
,
d x {2 Ittf~)2 '" 0.25 X ( It x 0.25 2)2
4
d X 4 x It '" 0.25 x 0.25
0'
"'
"'
5
O.2 SS
0.25
d '" - - ord= - ,,-,
0.25
=-- =
0.1894m
-= 0.19m. An s.
4
(4 )
1.31 95
Problem 11.39 A pipe of diam eter 0.4 In and of 1<'11gll1 2000 //I i,~ conllected 10 II reserl'Qir at 0111'
end. '/'l,e oiller end oflile pip" is connected to a june/ioll from ..... hich IWO pipe.! of /ellglhs /000 1/1 alld
diwneler 300 mill n Ul ill parallel. Tllese parallel pipes are connected /0 mlO/iler reurl'Oir. which is
IUll·ing lew/ a/waler /0
III
below Ille water lel'eI of Ihe (lbm"e re.~efl'ojr. Determine Iile lawl discharge
Iff'" 0.0 15. Neglecl millor losse.!.
Solution. G i ven:
Dia. of pi pe.
d = 0.4 111
L ength of pipe.
L '" 2000 1lI
Dia. of parallcl pipes.
d l '" d 2 '" 300 mm '" 0.30 111
Le ngth of parallcl pipes.
LI '" ~ '" 1000 m
Difference o f water leve l in two reservo ir. H = 10 m. ! = .015
Appl ying Bernoulli's eq uati on to po ints E and f -. Tak.ing fl ow throu gh AllC.
4 fL V1 -4,'
10= - + - f"XC~"'c:''-"'V'_
dx2g
<II )(2g
'" 4 x.O 15 x 2000 x V
0.4)(2x9.81
2
+ :4C',-;;.O~,,,5"X:;-'"OOO
::;C;X;-:VL
"
0.3)(2)(9.8 1
...(i)
,
----.---
T
A
,
.om
-'-
("2000
0"04 m L -1 000-:n.<l
. Ill. \I
<0j"
~
----,,--------C
°
1
;>,,0.3
Fig. 11.23
From continuit y equat ion
Discharge through All '" disc harge through Be + d isc harge throu gh llD
I I
Ii
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IL
Flow Through Pipes
521 1
"'
13m il, '" il2 and also the len gth s of pi pes Be and BD arc equ al and he nce discharge throug h Be and
lJD wi ll be sallie. Tll is mea ns VI '" til also
(0.4/ x V", 2 X (O.3)!V, or .16V= O. HI V,
"'
V '" 0.16 V= 0.888 V
I
0. 18
Substitutin g thi s va lu e o f VI in equ ati on (I). we get
[0", 15.29 V! + ( I O. 1 9)(.888)~V2 = 15.29 V' + 8.035 V 1 = 23.325 VI
v=~ 21325
10
Discharge
= 0.654 I11ls
= Vx Area
= 0.654 x ~ d! = 0.6 54 x ~ (0.4)1 = .0822 ..h •. Am.
4
4
Problem 1 1.40 Two J'/w,p ended pipes of iliame/us 50 mm lim/ {OO mm respeClil'e/y. cach 0/
/eng/h 100 III are connected in p"ral/l'l betwN:1l IWO resen'oirs which lw,"e II difference of ie"ei of
10 m. If the co- efficienr of fricrion for each pipe iJ' (-If) 0.32, calcuh'lt' Ille rate of flow for eilcil pipe
and (1/.10 Ihe dilllneler of II single pipe 100 //I long which would gil'e the same discharge. if i/ were
subs/illll"d for rhe original 1\\'0 pipe.t.
Solution . Gi l'~ n :
d l '" 50 mm '" 0.051ll
Dia. of 1st pipe.
Le ngth of lst pipe.
LI = 1{lO 1ll
Dia. of 2nd pipe.
d 2 '" 100 mm '" 0. 10 1ll
L.cngth of 2nd pipe.
L,. '" 100 111
Differe nce in level in reservoirs. H '" 10111
Co-e ffici ent of frict io n
4f'" 0.32
i
.om
L
Fig. 11 .24
VI = ve locity of flow in pipe I. and
V! = ve loci ty of flow in pipe 2.
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1522
Fluid Mechanics
When the pipes arc connected in parallel. llie loss of head will be same in both the pipes.
For tlie fir.>i pipe. loss of head is given as
-4fL"X"~~Xc"",'_ . 0.32 x 100 x V/
/f: -
dl x2g
10" 32.619
VI'"
VI
~ 32.61019
("; 4/" .32)
0.05x2x9.81
1
'" 0.5535 m/s
:. Rate of flow in 1st pire. Q, '" VI X AI"" 0.5536 x
~ (dill
4
'" .5536 x ~ (O.OS)! '" '(MJ]087 11I ]/s '" 1.087 litresls. A ilS.
4
For the 2nd pipe. loss of head is given by.
IO=H= 4f X~XV/
til. x 2g
'" 0.32xlOOxV}
0.lOx2x9.8 1
V2 ", " ' CIO=X~.I,OCX;;2~X;;-c9·c'c'
.32 x 100
,
,
'" "4 (. 1)- x .783 "
'"
0.783 111/5
0.006 15
111
,Is'" 6. 15 lilres/s. Ans.
D" diameter of a single pipe which is substituted for the twO original pipes
L" length ursinglc pipe '" ]00
V = velocity through pipe
III
The discharge ItJrough single pipe.
Q = Q 1 + Q1 = 1.087 + 6. 15 = 7.237 ]itrests = .007237 11I 3/s
v=
-.JL =
Area
.007237
11: 0 "
4 x.007237
11:0 "
.009214 Illfs
0:
4
Loss of head through sing le pi pe is
O.32XIOOX(·()()D9~_ [4 )'
4 /X LXV 1
H"
0'
0'
"
0 x 2 x 9.8 [
10.0= 32x IOOx .0092 [4 ' '" .0001384
2 x9.81 x f) l
D~
OS", .0001384 = .0000[384
f) '"
I I
-'-~DCxC20g-
10
(.00001384)"} = 0. 1067
III '"
106.7 mm. Ans.
Ii
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Flow Through Pipes
5231
Pro blem 11.41 Two reJeFI'oir life COII/leeled by a pipe lille of diameter 600 mill and /englll 4000 m.
7'11,' di/fere/tce of water lel'e! in Ihe res(:'fmirs is 20 m. AI a diswnce of I (){)() III from the upper reserl'Oi r.
a small pipe is connected /0 Ifle pip/' line. T ill! warer ntt1 be lak ell from {he small p ipe. Find Ihe
discharge /0 II,e lower resen'oir. if
(0 No wmer is {I' ken f rom the small pipe. and
(ii) 100 /ilrl's/S of waler is /(I ke" f rom smol/ pipe.
Take! '" .005 and "eg{ec/ mill or losses.
Solution. Gi ven:
Dia. of pi pe.
d " 600 mm == 0.60 III
Le ngth o f pi r<'.
L = 400m
Diff,,~n ce o f water le ve l.
H == 20 rn .f= .005
(i) Nu w"t.,r Is taken from small pipe
E
- =o.::~:i":---
t
A:-
"'m
~
C
l."4Qoo
1
OOOrn
~
rn.
<1'<0.6""
>
,
. ---
-
Fi g. ]1.25
4/ xLxV 1
4x .OO5 x 4000 x V "
The head loss du e to fri cti on in pipe A B = -'--;cc-;;- - or 20 =
0.6)(2x9.8 1
d x2g
v=
:. Disc harg<).
20xO.6x2x9.8 1 _ h .943 " 1.715rnls
4 x .005 x 4000
"
4
( ii) 100 litrl'S o f wa ter is takc n from s mall pillC
Le t
Q I '" discharge through pipe AC
Ttwn for pnral kl pi""s
,
Q '" Ar<)a x V " - (0.6)" x 1.7 15" 0.485 rn Is. AilS.
Q1 '" di sc harge throug h pi p" CB
Q I '" Q1 + 100 litresls '" Q, + 0. 1 mJls
-
,
Ql"(QI - O. I) m ls
...(1)
L I ", 1000 III
Len gth of pi"" AC,
Len glh of pipe Cll.
L 1 '" 4000 - 1000 = 3000 III
Appl ying Berno ulli' s equation 10 points E and F alld lak ing now th roug h ABC, we have
1
' "'-.11 + 4 jL,- Vl
20= -,,4,fL=,
d l x2 g d 1 x2g
where VI '" ve locity th roug h pi pe AC '"
dl
I I
'"
QI
~(0.6)'
... (ii)
=~
It
x .36
dia. of pipe A C = 0.6
Ii
~ I
IL
1524 Fluid Mechanics
-:- __
V2 ", velocity through pipe CB '" -'00,-,
..~(O.6) 1
4
4 Ql
1T x.36
d 1 == dia. of pipe CB '" 0.6
Substituting these values in equation (ii). we get
[4Q' )" + 4 x .005 x 3()()() x [4Q'
- - ' -)'
"0- 4 x .005 x 1000 x - - - - O.6x2x9.81
1tx.36
1tx.36
20 = 21.25 QI2 + 63.75 Q/
Ql= Q1 - 0.1 or OJ '" Ql + 0.1
Bm from (i),
Substituting the value of
or
0.6x2x9.8 1
0 , in
... (i ii)
equation (iii). we get
20", 21.25 (Q, + 0.1)' + 63.75 Q/
= 21.55 10/ +.01 + 0.2 OJI + 63.75 Q/
= 21.25 Q/ + 0.2125 + 4.250 Q 2 + 63.75
= 85 Q2" + 4.25 0 2 + .2125
19.7875'" 0
Q/
Q/
85
+ 4. 25 0 2This is a quadrnlil: equaTion in Q 1
- 4.25± J4.2S l + 4 x 85 x 19.7875
QJ=
2x85
=
"" 0.458 rn l ls
- 44.25 ± 82.13
=
82.1 3- 4.25
170
170
(Neg lecting negative TOOl)
Discharge 10 lower reservoir'" Q, = 0.458 rn J/s. Am.
to I 1. 10
flOW THROUGH BRANCHED PIPES
When th ree or more reservoi rs are connec ted by meanS of pipes. baving one or more junctions. tbe
syste m is called a branclling pipt) system. Fig. 11.26 sbows tllrcc reservoirs at different levels connected to a single junctiOll. by means of pipes whicll arc called brallc bcd pipt)s. Tile lengtlls. diameters
and co-efficient of friction of eacb pipes is given. It is required to find the disellarge alld direction of
now in eacll pipe. Tile basic equations used for solving sucll problems arc:
!. Conti nuity equatlol1 whicb means tbe il1f1ow of fluid at tile jUl1ctiol1 should be equal to tbe
outflow of fluid.
2. Uerno lllll's .,quutJon. " nd
3. Darcy-Welsbach equ;.\lo l1
Also it is assumed tbat reservoirs are very large and tile water surface levels ill the rese rvoirs arc
constant so that steady conditions exist in tile pipes. Also minor losses arc assumed very smull. The
flow from reservoir A takes place to junction D. Tile flow from junction D is lowards reservo irs C.
Now the now from jUllction D lowards reservoir B will wke place only when piezometric head al D
( WhiCh is equ al to
~; + Zn ) is mOre than the piezometric lIead at B (i.e .• ZIJ). Let us consi der that flow
is from D to reservoir B.
~ I
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IL
Flow Through Pipes
r
525 1
---:-: -.-~ ~~
,
--=::
::::::::- ,
~
L"
'.
~ '>-
a,. 't,
".
"
'" ,
"
,~
,
.-:.-
c
~
I
1
'.
'"
Fig. 11 .26
For flow from A to D from Bernoulli's equation
Z.t =Zo +ft+I,J,
pg
,
For now from D to B from BcnlOulli's equation
Po
ZB+ /If I
ZO+ pg=
... (r i)
For now from D to C from Bernoulli's equation
Z,, + -Po : Zc + /'/
OK
...(iii)
•
From continuity equation,
Discllarge through AD '" Disc harge through DR + Discharge through. DC
n,
It,
It,
4" til-V, '" '4 d 2- X V 2 + '4 d,·V.!
'"
d/V) '" iI/Vl + If/V,
There arc four unk nowns i.e., VI' V1 • Vj and
... ( iI')
.!!..IL and there arc four equations (i), (ii), (iii) and (11').
pg
H.:ncc unknown can be calculated.
Problem 11.42 Three resen'uiTS A, Band C (lie connee/cd by" pipe S)'.~lem ,y/'oll'1I in ""ig . 11.27.
Find ll1e discharge illlD or from lite rcurI'vir,l' Blind C if Ihe rafe of flow from rt'Jerl'oirs A i.~
60 Ii(res/s. Find (lie !leig/II of water lel'el ill lilt' ",serl'oir C. Take f '" .006 for al/ pipes.
Solution. Given:
lA:ngth uf pipe AD.
L, '" 1200 III
Dia. uf pipe AD.
d , '" 30 elll '" 0.30 III
Discharge through AD.
0 , '" 60 litres/s = 0.06 ml/s
Height uf water level in A from reference line. Z... = 40 m
For pipe DB. length
L2 '" 600 m. dia .. d2 ", 20 cm = 0.20 m. ZI! '" 38.0
For pirc Dc' length
L3 = 800 Ill. dia .. 113 = 30 Clll = 0.30 m
I I
Ii
~ I
IL
1526
Fluid Mechanics
Fig. 11 .27
Apply ing Bcmoulli's cqualio ll s to poinls E and D. Z... '" Zv +
where
Ii" '"
~
+ h~,
pg
l
4. f· ~ . Vi . where VI '" ~ '" 0.06 '" 0.M8 mlsec
til x 2g
Area
IT (.3)2
4
"Ji '"
4 x.OO6 x 1200 x.848 2
= 3.518 m
0.3x2)(9.81
0 0 =Zo+ -flo + 3.518
Z... =Zv+ -p" + 3.5180r4.
pg
pg
[
ZD
+.E.L] = 40.0 -
3.518 = 36.482
J1\
P8
~!cncc picwlllclric head at D = 36.482. But ZB'" 38 m. ]'[enee water Oow s from B w D.
Apply ing Bernoulli's equmion to points lJ and D
Z8'"
(20+ :; )+hfl or 38= 36.482 + 1,1,
h!, =38-36.482= 1.518m
Om
1.:'i18 =
V2 =
Dischilrgc,
4 x .cXJ6 x 600 x V22
0.2x2)(9.8 1
1 = 0.643 m/s .
,I~1.~'~I8~'~O~.2~'~2~X~9~.8~
4 x.OO6 x 600
1(
= 0.0202
I I
,
Q2 = V 2 X "4 (d 2t = 0.643 x
111
It
,
"2 x (.2)"
3
1s '" 20.2 lil resls. A ilS.
Ii
~ I
IL
Flow Through Pipes
527 1
Appl ying Bcm o ul!i' s equati on 10 poillts D and C
-
I'D
ZD '~=Zc
+hl,
pg
"'
Bu! from continuity
Q 1 + Q2
'"
QJ
Q3 = Q 1 + Q!" 0.06 + 0.0202 = 0.0802 m3 /s
V =~ = 0.0802 = 1.I34m/s
3
/t
,
/t
4 (·3t
"4(.09)
'
82
4 x.OIX> X !lOOX 1.I]4 !
9
_, 6.4 = Zc +
= Zc + 4.1 4
03x2x9.81
Zc " 36.482 - 4. [94 = 32.288 m . Ans.
Problem 11 .43 Three THal'oirs, A. LJ and C are cOIll!ec led by a pipe .fys/em ,~ I/O\Vn in Pig . 11.2S.
The leng/lls and dillmelers of pipes I. 2 alld 3 are 800 m. WOO III. 800 m. (md 300 mm. 200 mill all d
150 1II1n respeClil'eiy. Determ ine IIIe piezometric head m jUllc /ioll D. Take f " JXJ5.
Solution. Gi\'en :
The length o f pipe I.
LI = 800 III and its dia ., til = 300 mill = 0 .3 m
The le ngth o f pire 2.
L2 " 1000 In and it s dia .. d 2 " 200 mm " 0.2 111
The le ngth o f pipe 3.
L} = 800 III and its dia .. til = I SO mill = O. IS 111
Hcig lll of rese rvoir, A from d<llu11l lin e. Z... = 60 m
Similarly.
Z8 = 40 III alld Zc = 30 1)1 .
The di rectio n o f n ow in pipes a re ~hown (give n) in Fig . 11.28. Appl ying B~fn oulli 's equation 10
pni nt s A and D
Z...
[
z. . -
"'
=
(Zo+ ~; )+ IIf'
f x ~ xV/ -_ 4 x.OO5x800xV
(Zo + fpgll Jl-- IIf , _4x
d x2g
0.3x2x9.&1
2
l
~
l
60 -
(Zo+ ~; ) = 2.7 1& V11
...( i)
Appl ying Bcrno ulli· s equali on to pnints D and B
( "oJ Z,, + ~
pg
= 28 + Il f = 40+
,
= ..-v.,.
.,,, ,
d 2 x2 g
c4cX~.OCOC;::X~IOC;COOCXCV~
!! = 40.0 + 5.09
(z/> +~;) -40.0 = 5.09 V/
I I
4f xL.,xV,~
-
0.2x2x9.81
V11
...(ii)
Ii
~ I
IL
1528 Fluid Mechanics
I·":··,
l
'
'o;<>o~ @ ~,,~~
80m
T
':0>:' ...".,..
T'10m
_____ ~~~~N.: ___ '.:a~~ ____ L
-vr" a
C
30m
Fig. 11.28
Apply ing BcmooJ!i's equation to poillts D and C
[
'C·
ZD+POJ=4: +h
pg
( Zv +
J,
OO
",,',,,,V,C'
=30 + 4jxL.,xV/ :: 30+ -,4C ""5CX"8COO
ilJ x2g
O.l5x2x9.81
~; ) = 30.0 + 5.436 V/
... (iii)
Adding (il :md (ii), we hav e 60 - 40 = 2.711\ V12 + 5.09
V/ + 5.09 V/
20:: 2.718
or
VJ
... ( i v)
Adding (i) and (iii). we have 60 :: 2.71 R V I1 + 30.0 + 5.436 Vi
or
60 - 30:: 30 = 2.7 18 1'1 2 + 5.436
Also from cOll1inuily cqu~lion. we have
Q,:: Ql + Q,l
"41t ill 2 X VI :: "41t
"'
"'
2
2
iI! V! +
1
"4J!
D.3 V 1 '" O.2 V1 + 0.15
And from (v) ,
.. .{v)
222
d j V1 ur ill VI :: til V1 + tl3 V1
2
X
VJ or .09\', :: .04 V2 + .0225 V]
2.7 18 v/
5.09
... ( vii)
30 - 2.7 18Vll
5.4 36
... (I'iii)
20
Now from (il'),
1
V/
Substituting the value of V" and VJ in (I'i), we gel
0.09 VI'" .04
20 - 2.7 18V/
+ .0225
5.09
30
2.7 18 ~2
5.436
Squari ng bolh sides. we gel
(0.09 V ) 2 '" (.04)2 X
I
(20 -5.09
2.718 VI?] + (0.0225)2 X 30 - 2.718 V/ + 2 x.04
5.436
2~7~18~"CLll "v
x .0225 X \1,~20,--~
5.09
I I
Ii
~ I
IL
Flow Through Pipes
529 1
.0081 VI! '" JlO628 - .000854 VI! + .00279 - .000253 VI! + .0018
.0081 VI! + .000854 VI! + .000253 V i l '" .00628 + .00279 + .0018", .01087
.009207 V12 ", .01087
or
or
or
VI ==
' g08~7~ '" 1.0&6 mls
J~.O~
.009207
Substituting this value o f VI in (1·;1) and (I'iii)
V =
2
,lo20,--~c2".7~'~8cXC~"-ll
=
5.09
30
2.718x 1.086
Jo2<l,--~c2c.70'~8~Xc'c.Oc86,--1
= 1.816m/s
5.09
2
-222
_.
InI S
5.436
Pi~zornClric
head 31
D = Zo + ~ = 30.0 + 5.436 x V/
pg
'" 30.0 + 5.436 X (2.22)2 '" 56.79 m. AilS.
Pro blem 11 .44 A pipe line 60 em dil"neler bifurcales (II a Y-jU1IClion into Iwo branches 40 em (lnd
JO em ill diameter. If tile rale of flow ill the mai" pipe is /.5 ",J/J' and mean "eloeily of flo .... in JO em
iliameler pipe is 7.5 mis, determine tile rule of jlOK' ill Ihe"'O em diameter pipe.
Solution. Given:
Dia. of main pip;.:.
Dia. of branch pipe I.
Dia. of brancb pipe 2.
D
= 60 em '" 0.6 10
DI = 40 cm = 0.4 m
D1 = 30 cm == 0.3 10
Velocity in brand piP<' 2. Vl = 7.5 m/s
Rate of flow in main pipe. Q = 1.5 1I1 3fs
Fig. 11.29
Q I = Rate o f now in bwncb pipe J.
Q1 = Rat~ of now in branch pipe 2.
Q = Rate of flow in rnain pipe.
Now rate of flow in main pipe is equal to the surn o f rate of flow in branch piP<'s.
Let
Q=QI+Q 1
But Q2 = Area of branch pipe 2 x Velocity in branch pipe 2
I I
...( i)
Ii
~ I
IL
1530
Fluid Mechanics
,
,
",A,x V,: - D ,'x V,=
•
•
4 '
•
Substituting the va lues Df
to I 1. 11
'4~ (0.3) ! x
3
7.5 "" 0.53 m Is
0
and Q! in equation (i) , we ge t
1.5 :Q 1 +0.53
0 1 = 1.5 - 0.53 = 0.97 Ill/s. AilS.
POWER TRANSM ISSION THROUGH PIPES
PlOwer is tran smill cd through pipes by flowing waler or mhcr liq uid s /lowing through Ilwm. The
power Ira nsmillcu depends upon (i) the weight of liquid flowing through the pipe and (ii) the total head
available at the end of the p ipe. Consider a pipe A B conncclcd to a tank as s hu wn in Fi g. 11.30. The
]lOwer ava ilable at the end B of th e pipe and t he condition for max i mullllransmi ssio n o f power w ill be
obtained as lllc rl tioncd belDW :
....-.-.-.-.
Fig. 11.30
Pow('r trammi!!;on through pip('.
L = len gt h o f th e pipe.
d == diamdcr of the pipe.
H == to tal h~ad a vailahlc at the inle t o f pipe.
V", ve loc it y of flow in pipe.
II/ '" loss of head due to friction. and! '" co·efflcie nt of friction.
The he ad available at th e out let of the pipe. if minor losses an: neglected
'" Total he ad at inlet - loss o f head due to friction
Let
J.:
1
", H - II =H- 4! X I. xV "
/
dx2g
h '" 4! XLXV "1
j
dx2g
Weigh t of wa te r n owi ng th rough pipe per sec.
II' '" pg x vo lum e of wmer pe r sec = pg x Area x Velocity
,
,
=pgx~d-xV
4
The powe r tr3nSmilled at the outlet of the piP<'
= weight of water per see x head
~t
o Ull et
Olpg X!!..d 1 XV ) X ( H - 4!XLXV 1)
4
dx2g
Wa".~
Power tran smitted at ou tlet of the pipe ,
2
1'=...ELX !!.. d 2 XV ( H _ 4 fL V
1000 4
dx2 g
I I
),w
...{ II.2I )
Ii
~ I
IL
Flow Through Pipes 531 1
Efficie ncy of power tran smission.
Powe r available at ou tlet of the pi pe
" ~ ~==~~~C'C=;"C:"'-'Power suppli ed al lIie inle t of the pipe
== Weight of wat~r per seC)( Head avai lable al oullet
We ight of water pe r sec x Head at inle t
IVX ( H-h/ ) '" If - h!
IVxH
If
...( 11.22)
I 1. 11 . 1 Condition for Maximum Trilnsminion of Power. The condition for maximum
trnnsmission of power is obtained by differentiati ng cquatio ll ( 11.2 1) with respect \0 V an d cquJling
the sain e to zero.
..!!.....(P)=O
Thus
<IV
or
.d! .V. . [...£LX!!..d~(IfV_
4JLVlll~o
1000 4
d x2g
or
~X!!.-d!( H - 4X3X f XLXV11~o
1000
"'
d x2 g
4
4jL V l
11 - 3)( - - - =0 or fI - 3xll/ =O
d x2 g
...( 1 1.23)
Equaling ( 11.23) is the cond ition for maximulIl tran smi ssio n or power. It states tlla! power transmi tted throug h a pipe is maximum wile n the loss of head due to friction is one-third of the total head at
inlet.
I 1. 11 . 2 Maximum Efficiency of Transmission of Power.
sion lhrough piJ}t' is given by eq uation ( 11.22) as
Effilciency of power transmis-
H - hI
fI
I] :
For maximum power transmission through piJ}t' the cond ition is given by eq uati on (11.23) as
H
", : 3
Subst ilUl ing the va lue of hf in emciency. we get maximum "1].
1]" ... :
H - H I3
H
I
2
=' - 3"="3 0r66.7%.
...( 1 1.24)
Problem 11.45 A pipe of diameter 300 ""Il aml/ength 3500 '" is u~·ed for tile trallsmissioll of power
by waler. The Iota/ "eat! at Ihe illiel of the pip'· i~· 500 Ill. Fi",llhe m(lxim,,," po,,·er a!"ai/ah/e at lile
ouliet of Ihe pipe. if lile ,",,/lle off .006.
Solution. Given:
Diame ter of the pipe.
t! = 300 mm = 0.30 III
=
I I
Ii
~ I
IL
1532 Fluid Mechanics
3500
Len gth of th e pipe.
Tot~1 h e ~d m inl et.
H =500 m
Co-efficient of frict ion.
/= .006
For ma xi mum power
L ""
Iran~rnission.
III
usin g eq uatio n (1 1.23)
H
500
3
3
h, ,,, - '" Now
1,=
r
'"
166.7 III
4X!X L XV 1
dx2g
=
4x.OO6x3500XV l
0.3x2)(9.8 1
= 14.27
v2
Equating the two values o f ",. we get
166.7= 14.27 V1 0r
V= t66.7
14.27
=3.41 7 mfs
Q= VXArea
Discharge.
= 3.4 11
It
,
1t
>
J
x - (dt = 3.417 x - (.3t = 0.24 15 m Is
4
4
Head available alth e end of th e pipe
= H- h =
j
M a~ imum
power ava ilable =
I/ - !!... '"
3
2 11 "" 2x500 '" 333.33 111
3
3
pg x Q x head a1 the end of pipe
kW
1000
= 1000x9.8 1 x .241 5 x 333.33 kW = 689.7 kW . Ah S.
1000
Problem 11 .46 A pipe /i1le 0/ /enstll 2000 m is used for power /rtlllsm iiiJ'iOI1. If 110.3625 kW power
is 10 be {"",smilIe" through ti,e pipc iii wilie" ,w,'er hlll'ing a pressure of 490.5 N/cm l al ;lllel is
flowing. Find Ille "iameler 0/ Ille pipe and e/ficie'lcy O/lfllrlsmil·siOIl If the pressure drop orer tile
Jength a/pipe is 98.1 Nkm ". Take/: .0065.
Solution. Given:
Len gth o f pipe.
Powe r lransmineo
Pressure al in let.
:. Pressure head at inl el.
Pressure drop
:. Loss of head.
L=
2000 m
: 11 0.3625 kW
p = 490.5 Nfcm-, = 490.5 x 10 N/m"
"
II ='!!"": 490.5 x 10 ' : 500 m
pg
1(00)(9.8 1 .
[.: P'" 1000J
= 98. 1 Nlcm 2 : 98. 1 )( 10· N/m 2
98.1 x 1O~
pg
1000)(9.8 1
: 100m
Co"efficien t of fric tion.
/ '" .0065
He <ld avail<lblc Jt th e end of the pipe: H - h, : 500 - 100 '" 400 m
Let the diameler of th e pipe '" d
I I
Ii
~ I
IL
Flow Through Pipes
Now power transll1iucd is g i ve n by. P "
110.3625 "
pgXQX ( H- hj
1000
533 1
)
kW
IOOOx9.8 1 xQx 400
1()()()
1103625 x lOOl = 0.02812
lOOOx9.8lx 400
8m
d ischarge. Q" Area x Velocity ""
~ If x V
4
~ d' x V: .028 12
4
V"" .28 12)( 4 " 0.0358
rrd '
d"
4/XLxV
-"-';:";C:-'-'
The IIc ad lost do~ to fricti on, "1"
Bm
...(i)
d x2 g
Ill "" 100 m
J00
= 4 x f x L x V1
""
,4"X,·,llOc605~'c2cllOO~cX,V,--'
")(2x9.8 1
dx2g
2.65 X V ' = 2.65 x (.0358 )1 = .003396
d
d"
d~
v=
': Prom equat io n (i).
.0358
d'
100= JXJ3396
d'
d=
[
.003396)'"
[00
""
8
'
.1_77
III
=
127.7 mm . An s.
Efficiency of power tran sm i.<;sioll is give n by equali on (1 1.22).
T]
~HC-i'h'L = 500 - 100 = 0.80 '" 80%. An s.
= -
H
500
Problem 11 .47 For Problem /1.46. find : (i) rhe diameter of the pipe correIponding 10 maximum
efficiency of lrilll5l11issiOll . (ii) diameter of the pipe corresponding /0 90% efficiency of
Irallwni.~,~ioll.
Solution. 0) Diamctcr of pipe co rrespondin g to max imum e fficie ncy.
Le t th e dia. of pipe for 1]m., "" d
But fro m e qumi o n ( 11.24), 1]"",,:0 66.67% ""
H - h
2
II
3
- - -j = - or
I I
~
3
500 - Ir j
2
- ""
500
3
Ii
~ I
IL
1534
Fluid Mechanics
1500 - 1000
500
= - - '" 166.7111
3
3
2
3
1,/ ",SOO - SOOx -",
0'
The other data given from Problem 11.46.
Power transmincd
'" 110.3625
Length of piP<'.
L = 2000 111
Co-crticicn t of friction.
f = .0065
Powe r transmiucd is give n by Ihe relation.
pg XQX ( H - h/)
p==-~~-",
1()()()
110.3625= lOOOX9.8IXQx (500-166.7)
"'
I ()()()
Q=
1103625 x 1000
l000x9.81 x (500
'" 0.03375 n/fs
166.7)
Q = area of pipe x veloc ity of fl ow
"0<
=
~
4
Ir x V {where V = velocity o f no w }
, ,
0.03375= "4 d x V
V = 0.03375 ,X 4 = 0.04297
/tx d "
... ( i)
dl
4jLV l
Now Ihe head lost duc \0 fricti on. h/ = - - -
d x2g
"1=166.7 III
O6 CC'c20()()()~C'CV~'
166.7= -.4C'C·~Ooe5
IIx2x9.8 1
= 2.65 V' = 2.65 X ( .04297 )' = .00489
d
d
(/1
~
dS= .00489
(.: V =
.~2197)
= .()(X)(}2933
[66.7
<1= (.0CKl02933)'~ = 0. 1240
111
= 124
111111.
Ans.
(il) Lei th e diame ter o f pi pe. wilen e fficiency of tran smi ssio n is 90% = d
11=90%=0,9
But
BOI
I I
I]
is given by equation ( I 1.22) as,
I] "
" 0.9
H,,500 m
Ii
~ I
IL
Flow Through Pipes
535 1
500 - h,
500
'" 0.9 or 500 - 500 x 0.9 = h, or 500 - 450 = hI
/'1 = 500 - 450 = 50
The
olh~r
JIl
P = 110.3625. L '" 2000. f = .0065
given data is.
Using relation fo r power trans mi ssion. P =
pgXQX (H - h,)
'-'--'c;;;';;-~'-'
1000
'~1~'~Q~'_(~5~
OO~-~
50,-,)
110.3625 = ~IOOO~_'~9~
1000
0'
62,,5,-,7c:'OOO
"",c"c =. 0'_S m-fs
'
Q= :-CCICI~O=.3"
[000 x9.8 1 x (500 - SO)
rr ,
Q= - d' xV
Bo<
4
.!:.
tf~ x
v =.025
or V : .025x4 = 0.03 183
nd 1
4
... ( i)
tfl
, ,
4 fLV 1
Now the head 10SI duc to fnC\lon. hi = - - d x2g
0'
50= 4 x.OO6S x 2000 x
II x 28
dS = .002685 = .00Cl0537
50
If = (.OOOO537)l ll = . 1399 III
... I 1. 12
= 140 mm.
AIlS .
FLOW THROUGH NOZZlES
Fig. 11.31 shows a nozz le fitted a1 th e e nd o f a long pipe. The 100al e nergy at [h e e nd o flhe pipe
consists of pressure energy and kin etic ene rgy. By rUling the nozz le at the end of the pipe. 111e \olal
energy is conve ned into kin etic energy. T hu s nozzles are use d. where higher ve lociti es of now arc
required. Th e exam pl es are :
---" 'c{}f
H
--rr-'
OIA=O
,
__ l __ -~ - -1+ --
BASE OF
~E
! PIPE
1•
f ig. lUI
I I
L
NOZZLE
·1
Nozzle [imd to a pipe.
Ii
~ I
IL
1536
Fluid Mechanics
L In case of Pe lton turbin e. the nozz ic is fmed at the end of the pi pe (callcd penstock) to increase
ve loc ity.
2. In case oflhe ex tinguishing nrc. a nozzle is fined at th e end oflhe hose pipe to increase velocity.
Lei
D", diameter of tile pipe, L = le ngth of the pipe.
A ", area of the pipe = ~ 0 2,
4
v= ve locit y of flow in pi pe.
H", to tal he ad at the inlet o f Ihe pipe.
d == diam eter of nozz le al ou l]e1,
\' = ve locit y of flow at oU ile! of noale.
a'" area of Ihe noz7.le at n Ulle! '"
~
Ifl,
4
f = co·cfficic nt of friction for pipe.
2
Loss of head duc to frictio n in pipe. h, = 4fLV
28 X D
Head available al lh" e nd of th e pipe or al th e ba..,., of nozz le
'" Head al in ld of pipe - head lost due to fricti on
=
H- hf = (H_4JLV2]
2g xD
Neglecting minor losses a nd also assumi ng losses in th e nozzle neglig ibl e. we have
Tota l Itead at inkt of pipe == totalltead (energy) at tlte ou tk t of nozzle + losses
1,2
But total Itead at o utlet of noa l" == kinetic
h ~ad
vl
== -
2,
4JLV l
V
11= - + hf = - , - - 2g
2g
2gD
(
0: h = 4JLVl] ... U)
I
2gD
From continuit y eq uat io n in the pi pe and o utl et of nou k.
A V==m'
a l'
V= -
A
Substituting this va lu e in equati on (I). we ge t
,.'
4 fL
2gD
H= - , - - ,
2g
I'
=
(,n.)'
-
A
2gH
(I
4fL
D
fU
"( 1< 4DA'
.']
-
-
2g
... ( I 1.25)
" ]
A'
+ -- , ~
Discharge th ro ugh nozz le = I' X v.
I I
Ii
~ I
IL
5371
Flow Through Pipes
I 1. 12. 1
nozzle"
Power Transmitted Through Nozzle.
1
The kin etic energy of Ihe jet a1th" outlet of
,
"2 mv-
Now mass o f liquid aT Ihe outlet of nuzzle per second" pm'
Kinetic c llcrgy of the jd allhc oUlle t p.:r SIX.
= -I
2
P(lV
2
1
X ,. = - pm'
2
1
Power ill kW at the oUllet of nozzle = (K. E.lscc) x
3
I~
=
,
pm·
.'c]])=OO-
Effi ciency o f power transmission through nozz ie,
1
- pm'
Powe r at outlet of nozzele
11 = Power at Ihe in let of pipe =
,
. ~'~]])~(](]~
1'g, Q. H
1000
~ p(n •. \,l
~p(l\ •. \ , l
pg.Q.H
pg.av.H
: ~'c-;;-,,: ,
= 2gH
"
= [ 1 4 fL1 III
+, ()
1
Q=al')
... ( 11.26)
Al
[
I 1. 12.2
1':
F,"m,q"",;O" (ll '25)'~:[
...
2g11
Condition for Maximum Power Transmitted Through Nozzle.
' 11
1
1+ _4 /_1. _" _.
D A'
We know Ihal,
Ihe lo[al head at inlet of pipe = total head at the olllle! of Ihe noale + losses
tota l head at outlel of nozzle '"
i.e ..
/1, '"
4
f L
V
Dx2g
1
~ and
"
j
'" Joss o f liq UId III pI pe
: _v
1 + C4~.f~.L,,'''V_'
2g
D x2g
;; =(11 _4.~~~~Vl)
I
- pa v [ (
But power transmiucd through nozzle = - ' - - : -'-- X 1, 1=_'__ 2g H
I I
!
3
- pav
I
- pm'
1000
1000
1000
-
4
.
f •L.
V'l]
lJ x 28
-
Ii
~ I
IL
1538 Fluid Mechanics
l
LV1
'" Plla)' [H _ 4/
1000
D x2g
...( 1 1.27)
Now from cOnTinuity cqumion. AV '" (IV
v=~
A
Substituting tile value o f V in equation (11.27). we get
1
1
pgm'
4 fLa
' -vPower transmitted through nozzle" - [ II - -,
lOOO
D x 2g A
.
.
d(P)
The power (P) will be maXllllum, when - - " 0
ill'
~
"- [ pg",, ( II 0':'
dv 1000
D x2g A -
"'
II
"0
"[pga ("'dv 1000
"'
P<O [H - 3 4 fL <l1~1
[ 1000
D x2g A+
1l=oorfl _ 3x
4 fL
D x2g
xv!=o( . . V"-""l
A
4 /L
· '" II == licad loss in pi pe
-V-
"'
D x2g
!
1
... { 11 .28)
"'
Equatioll (11.28) gives the condition for maximum power transrnillcd through nozzle. II states 1hat
power transmitted through nozzle is maximum when tlie head lost due to friction in pipe is one-third
the total head supplied at the inle t or pipe.
11. 12.l
Diameter of Nozzle for Maximum Transmission of Power Through Nozzle . For
/I
rna.<imum lransmi.'l sion of power. the condition i~ gi wn by equation (I 1.28) as, h! = -
3
Bo<
/,/
1
4fLV
Dx2g
= 4! .L .V'
D x2g
1
H
4!t V
orH=3x - - 3
D x2g
---= -
Bu( H is also = (Olal head at outlet of nozzle + losses
I I
Ii
IS34
Fluid Mechanics
Equaling the two values o f H, we get
3x
4jLV 1
v 2 4fLV!
12jLY'
= -+
or
O x 2g
2g O x 2g
D x2g
8jLV 2
or
O x2g
4jLY'
D x 2g
v2
2g
=-
=2g
... (i )
But from continuity. AV = av or V = ~.
A
Substituting this value of V in equation (I), we get
8jL
0'
o
A
_
-x-:-z -
or
_8 jL
_ x
-:(~",,-d_'-7)" = 1 or
o (~o' r
8jL x
0
d= ( -0' ) '"
8jL
From equation (ii).
8jL
1
( Divide by ;: ) ... (ii)
d: = lord' = 0'
0
~
... ( 11.29)
A'
- o =-,
0
~o =rjL
0
... ( 11.30)
Equation ( 11 .30) gives the ratio of the area of the supply pipe to the area of the nozzle and hence
from this equation. the diameter of the nozzle can be obtained.
Problem 11 .48 A nozzie is jilted 01 the end of a pi~ of lenglh 300 m and of diameter 100 mm. For
the ma;cimllm transmission of poKIer through the noWe, find the diameter a/nozzle. Take f = .009.
Solution. Given :
Length of pipe.
L=300m
D= I00mm=O. lm
Diameter of pipe.
Co-efficient of friction.
/= .009
Let the diameter of nozzle
=d
For rnuimum trolnsmission of power. the diameter of nozzle is given by relation ( 11 .29) as
d = ( 0'
8jL
)1'.=(8 x.OO9
OJ!
)1'. = 0.02608 m = 26.08 mm. ADS.
x 300
Problem 11.49
The hl!ad of watl!r al thl! inlet of a pip#! Z(J(X) m long and 500 mm dial1ll!rl!r is
60 m. A nOWI! of diaml!tl!r 100 mm at iu outll!t is fitted to thl! pi~. Find (hI! l'l!locity ofwall!r at thl!
ourll!t of rhl! nOWI! iff = .01 for thl! pipl!.
SOlution. Given :
Head of water at inlet of pipe. H = 60 m
~ I
IL
1540
Fluid Mechanics
Length of pipe.
Dia. of pipe.
Dia. of n01:zle at oUllet.
L '" 2000
D '" 500
III
111111 '"
0.50
III
<1=100111111=0.1111
Co-cflkicm of frict ion.
/= .01
The velocity at outkt of nozzk is given by eqUal ion ( 11.25) as
2x9.8 1 x6(}
2gfl
(I
AIL
"' )
+- ' CT
D
A
1-r:====C2C'~9~>~'~'~60~==~~, '" 30.61 m/s. Ans.
X.I)"
1 + 4 x.O lx2000 , (O.I
-0.5
0.5 x.5
Problem 11.50 Filld Ille mllXinwm power IWI/smilled by a jel of waler disc/!{jfgi,lg freely ow of
noule jilTed 10 a pipe'" 300 III long ami /(}() mill diameter willi co·efficienl of [riujOlI as O.OJ. Ti,e
(H'ailllhie head (lIllie "ou le is 90 m.
Solution. Given:
Length of pipe.
L=300m
Dia. of pipe.
D", HlO Illlll '" 0.1 III
Co-efficient of frict ion.
f= .01
Head available at nozzle.
'" 90 III
For maximum pow er transmission through the nozzle. the diameter at the outlel of nozzle is given
by cqu~lIiun (11.29) as
,,_(D')IN_[
8fL
Area allh e nozz le.
/t,/t
(I" -
4
(O.l)S
8x .0 I x300
]114".0254 m
,
,
d-" - (.0254,-" .0005067 m-.
A
The nozz le allhe oU IIet. discharges waler into almosphere and hence the 10lal head availabl e allhe
nou le is converted inlO kinetic he ad.
Head available at outlet " 1,1'2g ur 90" 1,212g
v" .)2 x9.8 1 x90 ,,42.02 mls
DiSl:hargc through nozzle.
Q" a x v" .0005067 x 42.02" 0.02129 m Jls
HC'~'"dC';'oOc'e'ele"'"Oerc""""'='""'"
Max imum power transrniued " jP"g",'"Q,-"",
1000
_ IOOOx9.8 IxO.02 129x90 _ 8796kW
1000
_ I .
. Am.
Problem 11 .51 The rate of flow of ",(lfer throllgll (I pipe of length 2000 m and diallieter I III is
2 IIIJh. At the end of the pipe (I nozzle of olltIhle diallieter 300 mill is filled. Find 'he power trimsmilled
I I
Ii
~ I
IL
Flow Through Pipes
Illrough Ihe Iioule
If Ihe
head of !I'll/a at inler of Ille pipe is 200
//I
541 1
(lnd co-efficient of frielioll for
pipe 1$ 0.0 1,
Solution. Given
u:ngth uf pipe.
Dia. of pipe,
Discharge,
Dia. of noule,
H~ad a1 iold of pipe.
L =2000m
D = 1m
Q=2m 3/s
,,== 300 111m '" 0.3 111
H",
200
Co·dOcicn! of friction.
f= .01
Now area of pipe.
A
==.::.
4
rn
D 2 ='::' x 12 == O.7S54 m l
4
~ = 2546 m/s
A 0.7854
Power lransmiltcd thro ugh nozz le is given by equation ( 11.27) as
Velocity of water throu gh pipe. V", Q ==
p=
=
4jl.V' ]
pg .a, l'[ H- - 1000
Dx2 g
WOO x9.81 x 2.0 [
4 x .01 x 2000 x (2.546)2]
200 (.: av= Q)
]()OO
lx2x9.81
= 3405.43 kW . A n s .
.. 11. 13
WATER HAMMER IN PIPES
Consider a long pipe All as shown in Fig. 11.32 conncclcd at one end lO a lank cOlUaining water at
a he ight of H from the CCnlrc of the pipe. At the other end of the pipe. a valve 10 rcgulalc thc n ow of
water is providcd. When the valvc is complctcly open. the water is nowing wit h a veloci ty. V in the
pipe. If now the vJlve is suddenly dosed. the momentum of th c flowing wJler will be destroycd ~nd
("()nsequcntly ~ wave of high pressure will be sc t up. This wave of high pressure will be translllillcd
along the pipe with a velocity eq ual to the ve locity of sound wav e and may creale noise called knock ing. Also this wa,'e of high pressure has lhe effect of hamme ring action on the wa ll s of the pipe and
hence il is also known as water hammer.
,
FI g. 11 .32
VALVE
Waltr ba mmt!r.
Thc prcss urc ri se duc to water hammer depends upon: (i) the velocity of flo w of water in pipe.
(ii) Ihe length of pipe. (iii) tim c taken 10 c lose the va lve. (il'l claslic properties of Ihc malerial of Ihc
pipe. The following cases of water hammer in pipes will be considcred :
I. Gradual closure of valve.
2. Sudden closure of valve :md I:onsidering pipe rigid, and
I I
Ii
~ I
IL
1542
Fluid Mechanics
3 . Sudden cl osure of va lve and considerin g pipe clastic.
I 1. 1J. I Gradual Closure of Valve. Let the wate r is flowing through the pipe AB s hoWIl in
Fig. 11.32, and the valve provided ill the end of the pipe is dosed gradually.
Let
A = area of cross· scClion of the piP<' AB,
L = length o f pipe.
V = ve locit y of flow of water through pipe,
T = lime in second required \0 cl ose th e va lve. a nd
p = intensity o f pressure wave produced.
Mass of water in pipe All '" P x vo lum e of water = p x A x L
The valve is closed g radu all y in tim e 'r seconds ami hellce the wate r is brought from init ial
veloci ty V to ze ro velocit y in time sc~"oilds.
:. Rem rdat ion of Waler
Change of ve locity
v- 0
V
Time
T
T
= Ma'is x
Retarding force
Retardation
= pAl.
x
!.
...( i)
T
If P is the intens it y of pressure wave produced due \0 closure of the va lve. the force due to
pressure wave.
... (ii)
"'P x area of pipe '" p x A
Equating the twO fo rces. givcn by equat ions (i) and (ii).
V
pALx - "'pxA
T
pLV
...( 11.31)
/!"' - T
Head of pressure.
P
pg
pLV
pgx T
H ~ - ~--
~
pLV
oc
pxgxT
(i) The val ve c losure is said to be gradual if T> 21.
e
LV
gT
H~ ­
... ( I 1.32)
.. .( 1 1.33)
where I " time in sec. C" veloc ity o f pressure wave
2L
(ii) The valve c los ure is said to be s udden if T < -
e
.. .{ 11.34 )
where C", ve locit y of pressure wave.
11.13.2 Sudden Closure of Valve and Pipe is Rigid. Equation ( 11.31 ) gives the relation
betw.:.:n iner.:ase of pressur.: duc to wa ter hammcr in pipe and th.: time r.:quircd to close th.: valve. If
1= O. tit.: increase in pressu re will be infinit e. Bul from cxpcri m.:nt s. il is ob""rved thaI thc increase in
pressure due to wa ler hamlll~r is finite. eve ll for a very rapid closure of \·alve. Thu.~ eq uati on ( I 1.31) is
valid on ly for (i) incompressible nuid s and (ii) whe n pipe is rig id. But when a WaVe of high pressure is
crea ted. the liqui ds ge l co mpressed 10 some ext en t a nd also pipe material ge lS stretched . Por a sudden
closure of valve [lhe valve of I is small and hence a wave o f high pressure is created ] the following lwo
cases wi ll he conside red:
(i) Sudden closure of va lve and p ipe is rigid, and
(ii) Sudd~n closure of va lve and pipe is c lastic.
I I
Ii
~ I
IL
Flow Through Pipes
543 1
Consider a pipe An in which wate r is flowing as s hown in Fig. 11.32. LeI the pi pe is rigid and va ll'c
fitted at Ihe end IJ is dosed suddenl y.
LeI
A" Area of cross-sect ion of pipe AB.
L", Length of pipe.
V= Velocity of flow of water through pipe,
P'" InlcIISi er of pressure wave produced.
K = Bulk modulu s of water.
When Ih e valve is closed suddenl y. Ihe kinetic energy of [he flow ing water is converted into strain
ene rgy of wa ter if the effect of friction is neglected and pipe wa ll is assumed perfect ly rigid.
I
,
Loss of kineti c energy'"
x mass of wa ter in pipe x V'
'2
I
,
= - xpALxV'
2
Gain of strain ene rgy '"
(': mass" p x vo lume" p x A xL)
i
~ ( J; J x volume", I; x AL
Equati ng loss of kinetic energy 10 gain of stra in energy
I
,
I 1/
- pALxV' = - xAL
2
2 K
,
I
1
2K
p'= -pALxV
x = pK v'
2
p=
AL
,
~PKV- =
JKp!
P
... ( I 1.35)
V.,JKP=V
( .. ,JKlp =
=pVxC
cj ...(1116)
where C" vclocily* of pressure wave.
I 1. 13. 3 Sudden Closure of Valve and Pipe is Elastic. Consider the pipe AB in wh ich water
is flowing as shown in Fig. 11.32. LeI the thickness .r" o r th e pipc wall is s ma ll compared 10 the
diameter 0 of the pipc and also let the pipe is clastic.
Let
E = Modulus of Elasticity of the pipe material.
I
- " Pois;;on·s ratio for pipe matcrial.
m
p" Increase of pressure due 10 water hammer,
I" Thkkness of the pipe wa ll.
0" Diameter of the pipe.
When the valve is closed suddenly, a wave of high pressure of in knsity p will be produced in the
water. Due to this high pres~urc p. circumferen tial and lo ngitudinal stresses in Ihe pipe wall wi ll be
produced.
li '" Lo ngillld inal stress in pipe
Let
Ie" Circu mferent ial stress in pipe,
pO
The magnitude ofthesc stresses arc given aS li= -
pD
and / c " 41
21
Now from the knowl~dge of strength of material we know , strain cn~rgy slored in pipe material per
unit volume
• For derivation of velocity of pressure wave. please refer 10 chapler 15.
I I
Ii
~ I
IL
1544
Fluid Mechanics
=
_'2£!
_ [t:1 + f}, _ 2fi X fc]
m
2, pD , PO ]
, [( PD)' [PD)'
'" 2£ 4r + 2r
-
Taking
m
41
21
m
(i.e.,
= 4
Poisson ratio '" ~)
:. Strain energy stored in pipe material pe r unit vo lume
,
1
[ p1 lJ
'" 2£
pl /) l
/
1)1
+"""'4T - 4/ " x4
16/ "
1
1
pl D l
pl D "
= 2£ x----:1T'" 8EI l
Total volume of pipe material", rt f) x I X L.
TOl al strain energy stored in pipe material
'" Strain energy per unil vulume x total volume
plnD'L
P"01
= , x itO X IX L '" '-c~-'
8Er
8£1
c- n;~l '"
Now loss of kmetlc energy of wa ter
Gain of strai n energy in water '"
= -I
2
1
mV
= -I
2
1
pAL x V
~ ( ~ ) x volume = +~
Area o f pip:: =
A)
C: '" = pAL)
x AI.
Then. loss of kinetic energy o f waler = Gain of strain energy in wale r + Suain energy stored in pipe
111<11crial.
..!...PAl. x
2
Divide by AL.
Vl= .2!. . (L]
xAL+ p l AXDL
K
2EI
"'..!... p! + p" O = p! [~ + .!::]
22 K2EI2KEI
pV"
,
p =
pv'
~+.!::
K
I I
EI
Of P=
/f.'iV'
~+.!:: '"
K £1
Vx
JH
~+~
K
... ( 11 .37)
B
Ii
~ I
IL
Flow Through Pipes
I 1. 13.4
LeI
5451
Time Tilken by Pressure WOlve to Trilvel from the Villve to the Tank i1nd
from Tank to the Valve
T", The required time taken by pressure walle
L = Length of the pipe
C = Velocity of pressure wave
1'11<,," lotal distance = L + L = 21.
:, Time,
2L
Distance
T:
=='7'=,--,----=
: -C .
Velocity of pressure walle
. ..( l l ,}8)
Problem 11 .52 Tile w(l/er isf/owing wilh (l "e/oCil)' of 1.5 ",Is ill (1 pipe oJ/eng/h 2500 m (wd of
diameter 500 mm. AI/he end of Ihe pipe. " \'GI,''! i~' prorided. Find Iile rise i"pressure if the ,"u/,"e is
closed ill 25 seCOllds. Take Ille ..a/"e olC = 1-160 mls.
Solution. Gil'i.:n :
V=l.5m/s
Ve locity of water.
L ", 2S00 III
Length of pipe.
D = 500 111111 = O.S III
Diameter of pipe.
T", 25 seconds
Time to close the \'alve.
Value of.
C = 1460 m/s
:p
Let the rise in pressure
The ratio .
2L = 2 x 2500 =3.42
C
1460
From equation ( 11.33). we have if T> 2L. the closure of valve is said to be gradual.
C
Here
T ", 25 sec and
'L
~
C
'" 3.42
T> 2L and hcnce valve is dosed gr:ldually.
C
For gradually closure of va ll·c. the rise in press ure is given by equation (I [.31 ) as
p
=
P~L
'" 1000 x 2500 x
~~
'" 150000 N/m
1
150000 N
N
'" 15.0 - ,' An s .
[0'
em 1
ern
Problem 11 .53 If itl Problem 11.52. lit e mire is c1o ,~ed in 2 ICC. Jiml lite rise itl pre.l$ure behind
the I'oll'e. Assume lite pipe 10 be rigid one (lnd /(Ike Bulk modulus of ",mer.
i.e" K = 19.62 x 1(1 N/cm~.
Solullon. Given:
v = l.5mfs. L =2500 m
f) = 500 mm '" 0.5 1ll
T",2sec
Time to close the valve.
K '" 19.62 X [0' N/cml
Bulk modu lu s of water.
'" 19.62 x [0 4 )( [0 4 N/m ' "" 19.62)( [0 8 N/m'
'"
Velocity of press ure wave is g iven by.
I I
Ii
~ I
IL
1546
Fluid Mechanics
c= fK=
19.62x 108 '" 1400 mfs
1000
Vp
The ratio.
(": p
C '"
1400
1000)
n
2x25oo
lL
=
'
7 < -
'" 3.57
C
.
:. From cq u:lli on ( I 1.34). ifT<. 2 L. va lve iscJoscd sudde nl y. For s udde n closure o f va lve . when
C
pipe is rigid. the risc in pressure is g iven by equatio n ( 11 .35) o r (I [.36) as
P'" V.[KP '" 1.5
~19.62 x 109 x
("; P'"
WOO
[(00)
'" 210. 1 X [04 N/m 1 '" 210. 1 N/c m ~. Ans.
Problem 11 .54 Ifin Problem 11.52. Ille lhicknes,f ofrhe pipe is IV mm and Ihe mll'e is suddenly
closed (1/ Ihe end of Ihe pipe. find IIII' rise in p re.fsure if Ille pipe is considered 10 be eias/le. Take
E '" 19.62 X {OiO Nlm 1 for pipe lila/erial a llli K '" /9.62 X 104 Nkm! for water. Cu/cu/me Ihe
circumferential srress and longiluriinal stress tll'I'e/oped ill IIII' pipe Will/.
Solution. Given:
V'" 1.5 m/s. L", 2500 m. D", 0.5 m
r:lO mm : .OI m
Thic kn ess of pipe.
E: 19.62 x 10'0 N/m'
Modu lus of elastici ty.
Bulk modulu s,
K : 19.62 x 10~ Nlcrn 1 = 19.62 x 10 8 N/rn'
For sudden c losu re of the va lve for an c1aSlic pipe. the rise in pre:;sure is g ivc n by equat ion ( 11 37) as
p = Vx
( ~:~)
K
'" 1.5 x
= 1.5 x
Et
[ 1x lOR+
19.62
1000
05)
19.62 X 1010 x .0 1
1000
(S.Cl9XIO 1O +2.54xlO 10 )
: 17 15510 Nlm 1
Circu mferential stress ife) is given by
'"
171.55 N/cm !. AilS.
'" pxD = 17 1.55 x 0.5 = 4286.9 Nlm l
2,
2 x .OI
..
. .
p xD
171.55xO.5
2
wngltudlnal stress tS given by,/, = - - =
= 2 143.45 N/m • AilS.
4r
4 x.Ol
Problem 11 .55 A ,·,d..e is provided al rile end of a casr iron pipe of dialllerer 150 111111 m,d of
rllid:lless 10 mm. Tile warer is [lowing rllrougll rile pipe, which is suddenly sropped by closing Ille
\"a/l·t'. Pind rile maximum I"t'locify of Wafer. wilen rile rise of pre.~surc due 10 slIddeli clO.lllre of
\"all"e is 196 .2 Nlcm J• Take K for wmcr as 19.62 x 104 Nkm ' allil E for casr irOiI pipe as
11. 772 x ul Nkml.
Solution. Givcn :
Diameter of pipe.
Thic kn ess of pipe.
I I
D", 150 111111 '" 0. 15 m
I '" 10 111111 '" .0 1 III
Ii
~ I
IL
FlowThroughPipeS
Bulk modulu s.
p= 196.2 Nlc rn 2 : 196.2 x 104 Nfm ~
K = 19.62 x t o'! N/cm 2 '" \9.62 x 10 8 N/m 2
Modulus of elastic ity.
E" 11.772 X 106 Nlcm 2 " 11.772 x 10 10 N/m 2
Risc of pressure .
5471
For sudden c losure of va lve and when pipe is clastic. the pressure risc is given by equation (11.37) as
p
P" Vx
196.2 10
X
"'
4
",
Vx
=Vx
v=
(-'K- +,,)
"
1000
Vx
I
0.15
( 19.62xlO! + 1 1.772 X 10 10 x.OI
1
J:;::=:;::~I
~OOO
~;;::=:;::~
5,(19 x 10 10 + 1.274 x 10 10
~;:;;:
w =VxI2S.27xI04
.Ir;:;;I~OOO
6.364 x to
196.2
X
10
4
125.27 x 1 0~
= I 566 mls
= 1.566 m/s. An s .
Ma!<imuill vdoc hy
... 11 . 14
'"
PIPE NETWORK
A pipe network: is an imc rcon nCl1cd system o f pipes funning several loops or cin: ui[s. T he pipe
netwurk is s hown in Fig. 11.33. The examp les of suell networks of pipes arc the municipal wa ter
distribution systems in cities and labormury suppl y syste m. In such syste m, it is requin-d to determine
the distribution of flow through the various pipes of th e network. The follow in g are the necessary
cond itions for any network of pipes:
(,) The flow into eac h junction must be equa l to the flow o ut of th e junction. This is due to
con tinui ty eq uati on.
(ii) The a lgeb ra ic sum of head losses round each loop must be zero. This mea ns that in each loop.
the loss of head due to flow in cloc kwise direction must be equa l to the loss of he ad due 10 flow in
anticlockwisc direction.
(iii) T he head loss in each pipe is expressed as hJ ==rQ". The value of r d~pends upon th~ leng th of
pipe. diam eter of pipe and co-cfficient of fri ction o f pipe. The va lue of II for turbulent fl ow is 2. We
know that.
II ==
J
==
~4~X~f~X~L:ccX~V_' :
Dx2 g
4fLXQl
DX 2g
==
4JLX ( QA' )'
X(: D l y
Dx2 g
==
4/LXQ '
DX2gX(:)l
XD4
4f xLxQ'
2gX(~r xD l
I I
Ii
• if.'
•
Row Through Pipes
5431
...( 11.39)
This head loss will be positive. when the pipe is a part of loop and the now in the pipe is clockwise.
Generally. the pipe network problems arc: difficult to solve analytically. Hence the methods of
successive approximations are used. ' Hardy Cross Method' is one such method which is common ly
used.
I 1. 14. 1 Hardy Cross Method.
The procedure for Hardy Cross Method is as follows :
I . In this method a trial diSiribution of disc harges is made arbitrarily but in such a way that
conti nuity equation is satisfi ed at each function (or node).
2. With the assumed values orQ, me head loss in each pipe is calculated according to equation ( 11 .39).
3. Now consider any loop (or circuits). The algebraic sum of head losses round each loop must be
zero. Th is means that in each loop, the loss of head due to flow in clockwise direction must be equal
10 the loss of head due to fl ow in anliclockwise direction .
4 . Now calculate the nel head loss around each loop considering the head loss to be positive in
clockwise fl ow Ilnd to be negative in anticlockwise flow .
If the net head loss due to a.'isumed values of Q round the loop is zero, then the assumed values
or Q in that loop is correct. But if the nel head loss due to assu med "alues of Q is not zero. then the
assumed values of Q are corrected by introducing It correction 6Q for the flows. till thc circuit is
balanced.
The correction factor 6Q· is obtained by
dQ- -
~> a;
... ( 11.40)
- L,rn Q; -I
For turbulent flow. lhe value of n = 2 and hence above correction factor becomes as
- L ,QJ
dQ= ~
... ( 11.41 )
.... 2'0.
5. If the value of 6Q comes out to be positive. then it should be added to the flows in the clock·
wise direction C'; the flows in clockwise direction in a loops nt"e considered positive) and subtracted
from the flows in the anticlockwise direction.
6. Some pipes may be common to two circuits (or two loops). then the two corrections are
applied to these pi pes .
• L.e:t for any pipe Qo = assumw discharge and Q = correct discharge. then
Q=Qo+dQ
: . Head loSJ forthr pipe.
h,- nr=r(Qo+tJ.Q)2.
For complete circui!. the net head loss. IiI, = t (~) = II' (Qo + dQ)l = &(~ + 2Qo.1.Q + dQl)
= !r (Q02 + 2Qo dQ) As dQ is small compared wilh Qo and hence dQl can be neilect~.
..
I ref '"' I rQ02 +!r x 2QoAQ
For the conttl distributton. the net head loss for Il circuit should be uro (i.r .•
I (rtf) '" 0)
:.
IrQJ +!r X 2QoAQ=O
+ 4Q II' x 2Qo " 0 (As AQ is same for one circuit. hence it can be taken 001 of the summationJ
f.h,-
or
IrQJ
AQ=
-}; , ot .
); 2,
a.
~ I
IL
Flow Through Pipes
5491
7. Afte r tlie correctiuns have been applied to each pipe in a loop and \0 all loops. a second trial
(;akulalion is made for all loops. The procedure is repeated Till t..Q becomes negligible.
~-~--I "
Fig.I1.33 Pipellei1J}Qrk.
[Loops arc, ABCGFA- FEGf'. GEHG. GHDG
"III) GCDG]
Problem 11 .56 C('/cll/ale Ihe diIchllrge ill each 1';1'1' of IIII' Ildwork SilO"''' ill Fig. '/.3-1. Th e pipe
network f Oll sis/S of Spires. The head loss "f ill a pipe is g;,'en by hI = ..Q2. The l'alw:s of r fOT "Mious
pipes ami {,Iso IiiI' inflow or UifljIOWS at "odes are silo,.." ill Ihe figure.
,.,
..
•
,. ,
, '"
,. ,
,. ,
Fi g . 11.34
Solution. Given:
Inflow at node A '" 90. outflow at B "" 30. at C '" 40 and at D '" 20.
Values of rfor AB = 2. forBC = 1. for CD", 2. forAD = 4 and for liD = 1.
Por the first trial. the discharges arc a.%umc(/ as sliown in Fig. 11.34 (a) so that continuity is satisfied
at each node (i.e .. flow into a node == flow out of the node). rOT this distribution of diSl:hargc. the
corrediuns IlQ for the loops ABD and BCD are calculated.
'~----~----~~
90
60
8
30
Fig . 11.34(a)
I I
Ii
~ I
IL
1550
Fluid Mechanics
fi rs t Tri:1I
LoopADB
Pipe
,
Q,
,
, '"
"
'"
'"
'<I.
h," rQo'
Pil'"
AD
4 X 30' = 3600
08
_lxIO' ~ _ IOO
2xlxlO=20
C8
-2x60' . - 7200
2><2><60 .. 2-10
80
2
2x4x30=140
r." d
= - 3700.
AQ~
-I,o,;
:l: 2r
o.o
DC
l:2rQo = .500,
..
-(-3700)
un(iciockwisc.
As toQ is positive for loop ADIl. hence it soould be lId<kd
10 the now in (he dockwise di"'Clioo and sublracle</from
the flow in {he lli11idockwise din."Clioo. lIenee the correcled
flow for second trial for loop ADn will be as follows:
Pipe AD = 30 + 7.4 = 37.4 (flow is doc~ wi..,)
Pipe All = 60 - 7.4 = 52.6 (flow is amic10dwisel
Pipe nD .. 10 - 7.4 .2.6 (flow is an1icloc~wisc)
,
Loop OCB
Q.
,
, "
2rQ.
hJ" rQo'
20
2x20" =800
2x2"20~80
20
_ l x21i-~ _ 400
2x I
1><10'., 100
t2
= .500
2><1><10 ., 20.
!2rQo= 140
f(!;
x20~40
aQ .. - I rQi' -500 __ 3.57
1: 2r<4;. 140
. 7A
500
[n the loop AIlR. the head loss h, is Jl/:8~li\'e in pipes DR
nnd All as the direction of dischargu in lh~<c pipes is
,
= - 3.6.
The head loss in pipe Be for loop OCR;s negat;ve
,", d irection
pipe BC
di.< d'a'g"
nnlidockwise .
As lIQ is negative for loop VCR. hence ;1 should be
subtracted from the flow ;n {he dock"'i"" direchon
a"d added 10 Ihe flow in lhe illllic1""~wise dirrelion.
li enee correeled flow for second lrial for loop ocn
will be as follows:
Pi pe DC = 20 - 3.6 = 16.4
,,.
"
'"
'"
Pi pe Ile = 20 + 3.6 = 23.6
Pipe BO· .. 26 - 3.6. - I
N"I ~. 11lc pipe IlD is (."Qmm<>n 10 (wo loops (i.e .. loop ADn and loop (>e8). Hence Ihis pipe will gel Iwo corm:li'ms.
Afler Ihe Iwo corm:li{ms. lhe r"sul!;.",( flow in pipe 80 is ncgali"e in I(JOp OC8. Ik nce Ihe di ....i:I;on of flnw will be
anlidockwisc in pipe nD for loop DCn.
37.4
,. ,
The di<lribulion of discbarges in ,'arlo", pipe' for second lrial is shown in Fi g. 11.34 (bl. For second lrial Ihe
correclion t.Q for loop. AD8 and DC8 are cakulalcd as follow.:
~ I
I~
~ I
IL
Flow Through Pipes
Pire
AD
DB
'"
,
,
,
w.,
ADB
hr'" rQ/
Q,
,
37,4
.j
'''',
x 37.4' ~ 5595
2 x.t X 37..1
I'ip.e
~
299.2
DC
CB
2 52.6 _ 2 x 52_6' .. - 5533.5 2x2x52.6 .. 210A RD
2><1><1 .. 2
I X I ' .. 1
rrQ o' " 62.54.
",I.'IQ "
&'0, DCB
Q,
h,. 1001
,
,
L 2r<?o
2rQ o
16.4 2" 16.4' ~ 537.9 2><2>< 16.4~65,6
~ I x 23.6' - 5.56,9 2 x I x 2... 6_ 47.2
_ 1><1' .. _ 1
2><1><1 .. 2
,
, ,
23.6
"irQo' .. - 2il.
UrQo" 511.6
- I, rQJ
551 1
-~> a;
62j~
,,- - -
,'.<.\Q" L/ 2rQ
-5 \ 1.6
o
nrQo'" II.t S
- (_20)
.~
=....!£.... =0,174
=-0.122"'-0,1
11 4.8
::: 0.2
A. AQ is ""gati"~, hence;1 should be subtracted from lhe
flow in (he doclwis<- dirtttion and added to the flow in the
As dQ is I"'siti\·e. heoce i1 should be added 10 1he
flow in the clock",i ... direction and subtracted from
""tidockw;", di=tion
the flow in the antidockwise direction.
As the correction \l!.Q) i. small (i. ~ .. "'Q = - 0 . 1). this
As the correction ("Q ) is small (i.e .• t.Q = 0.2). this
rorrcclion is applied and further trial, are discontinued.
lIeoce colTCCled I10w for
loop ADB will be as follows:
For pipe AD. 0 0 = 37,4 - 0. 1 = 37.3 (as flow is clockwise)
For pipe /)B. Qo = I - 0.1 ~ 0,9 (as flow is clockwise)
""""'tion is applied and further trials"", discontinued.
li enee COlTCCtoo flow
for loop DCB ,,"'ill be as follows:
For pipe DC. Qo = 16,4 + 0.2 = 16.6 (clockwi ... flow)
For pipe eB. 0" = 23.6 - 0,2 = 23,4 (amiclockwise
flow)
For pipe 1m. QQ'" 09 - 0.2 .. 0.7 (antidockwise
flow)
For pipe All. QQ'" 526 + 0.1 '"' 52 ,7 (as flow is anticlockwisc )
Th" final distribution o f dis.charges in each pipe is as follows:
Di s.c harge in pipe AD = 37.3 from A to D
AB = 52.7 from A to B
DB = 0.7 fro m D to B
DC = 16.6. from D 10 C
BC = 23. 4 from B to C
T he final discharge in each pi pe is sho wn in Fig. 11.34 (c)
ok-"'____",;.,___-..:;c, ,,
37.3
23.4
A~
90
____
~
____
52,7
~~_
B 30
Fig. 11 .34 (el
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IL
1552 Fluid Mechanics
No te. The pipe DB is CommOn to two loop (i,t' .• loops ADB and loop DBC). Hence this pipe will get tWO
corrections. For loop AOB, lhe correction ""Q" - 0. I ami hence lhe rorTe<:ted flow in pipe DB is I - 0. I .. 0.9.
Now again. the corrc~lion is applied to pipe DB when we consider loop DBe. For loop DBC. the ~'Orrcdion
IlQ .. 0.2 but flow is antic1oc~wisc and hence the final correct 110'" in pipe DB will be 0.9 - 0.2 .. 0,7 ,
HIGHLIGHTS
1. The energy loss in pipe is classified as major energy loss and minor energy losses. Major energy loss
is due \0 friction whi le minor energy losses are duc to sudden expansion of pipe. sudden C01\lr~Cl;on
of pipe. bend in pipe and an obstruction in pipe.
_ _ . _
4/ LV'
2. Energy loss duc to fflcllOn IS gtven by Darcy Formula.it _ - - - .
l
<I x 28
J. The head loss due to friction in pipe can also be calculated by Chczy"s fonnula.
wh~rc
C.
Ch~lis
V ~ c..r;;li Che~y's fonnula
Constant
m _ Hydruali, mcan depth
V ", Velocity of flow
~~
(for pipe running full)
4
h
i_Loss of he'ld per unit length.. ;
hi '" L )( i. where i is oblai ned from Chezy's fonnula.
(V, _ V,j2
4 . Loss of head due to sudden expansion of pipe. he "
2,
-
where V, ,. Velocity in small pipe. VI = Velocity in large pipe.
S. Loss of head due 10 sudden eo"tmetion of pipe.h, '"
[..!...
- llYlC
_II
c
V'
where Cc _ co·efficient of contraction • 0.375 ~
... (For Co _ 0.62)
2,
~0.5 :i.
"
V'
6. Loss of head allhe ell lrJnCe of a pipe. hi .. 0.5 7. Loss of head atlhe exit of pipe. ho
V'
2,
" -.
"
" .(if value of Co is not gi ven)
.
8. The linc representing Ihe sum of pressure head and datum head wilh respttl 10 some reference line is
c~tlC<.l hydraulic gradient line (H.GL) while the linc representing the sum of pressure head. datum head
and velocity head wilh resp<."t to SOme reference line is known as toml energy line (T. E.L).
9. Syphon is a long bent pipe used 10 transfer liquids fron' a reservoir at a higher level \0 another reservoir
~t a lower level. whcn the tWO reservoirs are separated by a high level ground.
10. The maxim um vacuum created allhe summit of syphon is only 7.4 m of waler.
II. When pipes of different length. and different diameters are eonneeled end to end. pipes are called in
series or compound pipes. The ratc of now through each pipe connected in series is same.
12. i\ single pipe of unifonn diameter. having same dischnrge and same loss of head as compound pipe
consi,ting of .>cwml pipes of different lengthS und diameters. is known as equivalent pipe. The diamet~r
of equivalent pi pe is called equivalent site of the pipe.
I I
Ii
1548
Fluid Mechanics
13. The equivalent sile of the pipe is
obtal~
from
L _I, +L,+.!:l.
-.J"
d - ,
d'J
,
7 d!
where
L = equn'alent length of pipe = L, + ~ + ~
d l• d z• d J = art. diameters of pipes connected in series
d = equivalent size of the pipes.
14. ~n the pipes are connected in parallel. the Joss of Mad in each pipe is same. The rate of now in
mam pipe is equal 10 sum of the rate of now in each pipe. connected in parallel.
15. For solving problems for bnmchcd pipes, the three basic, equations I. ~.. continuity. Bemoulli"s and
Darty's equations are used _
pgxQX(H - h, )
.6. Po~er transmitted in kW through pipe is given by P = ::::..-~=-=
1000
where Q = discharge through pipe :: area )( "elocity '"
~ JJ x Y
4
H • total head at Inlet of pipe
hi = head lost due to friction
4/LV'
• - - - . when: L = Length of pipe
dx2g
pgXQ X(H - h, )
In SI. unJlS. po",er transmitted is ai,en by. Power'"
11. Efficiency of
po~er
transmISsion through pipes. 11 '"
kW.
1000
H-h,
H
.
18. Condition for maximum transmission of power through Pipe. hl :ll'
H
3
and maximum efficiency
",66.61Q
19. The , 'clocIIY of water at the outlet of the noule is v =
2gH
4JL . '
[ 1+ o-)( • 2
where H "" head at the inlet of the pipe. L = length of the pipe.
a = area of the nOlzle outlet.
D • dmmetcr of the pipe.
A = arra of the pipe.
po~er trunsmitted through nozzle. P= pg)(
Q [H _ 4jLy
1(0)
D)(2g
1
20. The
j
and the efficiency of po~er transmission through nonle. J'j '" ,..---'-....,.,
1+
11 . Condition for m.u.imum
4jL
)(
a~
o •1
po~er transmission through noule. hI = ~ .
j
~ I
IL
1554
Fluid Mechanics
22 . Diameter of !lou;]e for maximuill power transmission through nOlzle is. II ..
(;~ )".
where" z diameter of (he nozzle at oUlict, D .. diameter of Ihe pipe.
L .. IcnJ!lh of (lie pipe., ~ co-efficicn! of friclion for pip<:.
23 . When a liquid js flowing through <I long pipe fitled Wilh a valve ~l the end of the pip" and the va lve is
closed suddenly, a pressure waVe of high imcnsily is produced behind the valve. This pressure wave of
high intensity is having Ihe effect of hammering action on (he walls of Ihe pipe. This phenomenon is
known as waler hammer.
24. The inlcnsity of pressure rise due to watcr hammer is given by
P •
pLV
T
'" when valve is closed gradually .
'" v.JKP
... when '-alve is closed suddenly and pipe is assumed rigi"
cv xJ<1J
K
... when ,'al,·c is closcd suddenly and pipe is ciaslic.
£1
L = L<:ngth of pipe,
T " Time required to close the '·al"c.
D • Diameter of the pipe.
I '" Thickness of the pipe wall.
25. If the time required to close the VJh'c:
where
T>
3.!:
... the val"e closure is said 10 be
c
T< 2L
C
where
V,. Velocity of flow.
K .. Bulk modu lus of water.
E. Mod ulus of ciasticity for pipe material.
gr~dual.
... the valve closure is S:lid to be sudden
L .. length of pipe.
e m "clocilY of pressure wa,'e produced due to wala hammer
~ ~.
EXERCISE
(A) THEORETICAL PROBLEMS
J . How will you detem,inc the loss of head due to friction in pipes by using (i) Darey Formula and
(ii) Chczy's formula?
2. (a) What do you understand by the tenlls : Major eTlcrgy loss and minor energy losses in pipes?
(b) What do you understand by tolal energy line. hydraulic gmdicnt line, pipes in scric~. pipes in p.~r.. ncl
and equivalent pipe
J.
?
(0 : (I) Sudden enlargement and (ii) Sudden contraction
of a pipe.
(b) Obtain expression for head loss in a sudden expansion in the pipe. List all lhe a!isumptions made in
the derh·ation.
4. Define and explain tile tenTls: (I) Hydraulic gradient line and (iI) Total energy line.
I I
(a) Derive an expression for the loss of head due
Ii
~ I
IL
Flow Through Pipes
5. Show that Ihe loss of head due to sudden
6 . What is a syphon? On
e~pansion
555 1
in pipe line is a fun ction of veloci ly head.
wh~l
principle it works?
7. What is a compound pipe? What will be loss ()f head when pipes are connected in series ?
8. E ~plajn the lcnns: (i) Pipes in pnrallcl Ui) Equivalent pipe and (iii) Equivalent size of the pipe .
'.I. Find an expression for Ihe pow"r transmission through pipes. Wh at is Ihc cond it ion for maximum lruns-
mission of power and corresponding efficiency of transmission?
Ill. Prove Ihm the head loss duc 10 friction is equal to one -third of the lo\al head a\ inlet for max imum power
transmission through pipes Or nouJes.
II . Prove that Ihc velocity through nozzle is gj"en by \'
~
2gH
,i
,.--"-,
D ,.
4fL
where a" Area of nozzle a1 outlet. A '" Area of the pipe.
12. Show that the diameter ofthc noule for maximum tran,mission of power is gi"en by d '"
(~:)"~
where /) Diameter of pipe. L Leng th of pipe.
13. Find an expression for the rat;o of the oUllet area of the nO'-1.1e 10 the area of the pipe for maximum
transmission of power.
14. Explain thc phenomenon of Watcr Hammer. Obtain an expression for the risc of press ure when the flowing
water in a pipe is bmught to rest by closing the valve gradually.
IS. Show that the pressure risc due to suddcn closure of a "al\,c at the cnd of a pipe. through which water is
E
flowing is given by I'
E
Z
V
'
JH
~
+!2
K
E<
where V. Velocity of flow, 0 .. Diameter of pipe. E z Young's Modulus. K .. Bulk Modulus and
I .. Thickness of pipe.
16. Three pipes of different diameters and di ffcrent lengths are connected in series to make a compound
pipe. The ends of this compound pipe arc eonn«led with IWO tan ks whose difference of waler le"cl is
H. If co--efficicnt of friction for these pipes is S:lme. then deri>'c the formula for the total head loss.
ncgl«ting first the minor losses and then including them.
17, For the two c"ses of flow in a sudden contruction in a pipeline and flow in a sudden cxp.1nsion in a pipe
linc. draw the flow p.1ltCm. piclomctric gr~dc line and total energy line.
III, What do yo u mean by "l"<juivalent pipc" and "flow through paral lcl pipcs"?
19. (a) Define and explain the tcnns: (I) Hydraulic gradicnt line and Ui) tOlal energy line.
(b) What do yo u mcan by equivalent pipe? Obtain an expression for cquivalent pipe
(Delhi U" iwrsi/),. December 2(02)
(8) NUMERICAL PROBLEMS
1. fi nd the head loss due to friction in a pipe of diameter 250 mm and length 60 tn. Ihrough which watcr is
llowing at a "~Iocity of 3.0 mls using (i) Darcy formula and (ii) Chny"s I'onnuia for which
C. 55. Take v for wafer" .01 sfokc.
IAn s. (I) 1.182. (ii) 2.8561
2. Find fhe diameter of a pipe of lenllth 2500 m when the rute of flow ofwafcr through the pipe is 0.25 mJls and
head loss duc to fricfion is 5 m. Take C z 50 in Chezy's fomlU la.
I,\ os. 605 mml
I I
Ii
~ I
IL
1556
Fluid Mechanics
J . An oil of Kinematic ViSCQsily 0.5 ~loke is Oowing through a pipe of diameter 300 nlln at the rate of
320 1;lres per sc<;. Find thc head lost duc to friction for a length of 60 111 of thc pipe.
[Ans. 5.14 m]
4 . Calculate the rale of flow of waler through a pip" of diameter 300 nnn, when the diff""'n,,e of pressure
head belween the two ends of a pipe 400 m apart is 5 m of wat .. ,. Take the value off ~.009 in Ihe fonnula
h • 4 flyl
f
d x 2g
IAn •. 0101 m'/sl
5. The discharge through a pipe is 200 lilres/s. Find Ihc loss of head when Ihe pipe is suddenly enlarged from
150 mm to 300 mm diameter.
[A ilS. 3.672 Inl
6. The ralc of flow of waler through a horizontal pipe is 0 .3 mJ/s. The diameter of thc pipe is suddenly
enlarged from 250 mm to 500 mm. The pressure intensity in the smaller pipe is 13.734 N/Clnl. Determine:
(.) loss of head due to sudden enlargement. (ii) pressure intensity in the large pipe and (iii) power lost due
10 enlargement.
[Ans. (i) 1.07 m, (ii) 14.43 Nfcml. (iii) 3.15 kWI
7. A horizonl~1 pipe of <liameter 400 mm is su<l<lenly eonlraele<ltu a diamder of 200 mm . The pres;wre
in1ensities in the large and smaller pipe is given as 14.715 Nlcm" and 12.753 Nfcm '! re'ipeclivcly. If
C, " 0.62. find Ihe loss of head due 10 conlraclion. Also dctenninc Ihc rale of flow of waler.
[Ans. {Il 0 571 m, (;0 171. 71ilreslsl
8. Waler is flowing Ihrough a horizontal pipe of diameler 300 mm al a w locily of 4 m/s. A circular solid plale
of diameter 200 mm is placed in the pipe to obSlructthe flow. If C< '" 0.62. find Ihc loss of head d ue 10
obmuction in Ihc pipe.
[Ans. 2.953 ml
9. Detemline Ihc rate of flow of water Ihruugh a pipe of di~meter 10 em an<llength 60 em when one end of
Ihe pipe is ~'Onnccted to a tank and olher end of the pipe is open to the atmosphere. Thc height of water in
Ihe lanl.: from Ihc centre of the pipe i, 5 em. Pipe i, gi"en as horizontal and value of! .01. Conside r minor
los>cs.
IAns.15.41ilre>lsl
Ill. A horizontal pipe -line 50 m long is connected to a water tank at one end and diseharges freely into the
atmosphere at the othcr cnd f or the first 30 m of its length from the tank. the pipe is 200 mm diameter and
its diameter is su<l<lenly enlarged to 400 mm. The height of water le"el in Ihe tanl.: is 10 m above the centre
of Ihc pipe. Considering all minor losses_ delcnnine Ihe r~te of flow. Take! ~ .01 for bolh -"eetions of the
pipe.
[Ans_ 164.l.1litre>lsl
II . !)etenninc the difference in the elevations between the waler .urface, in the two tanks which are connecled by a h"riwnlal pipe of diameter 400 Illm and lenglh 500 Ill. The rale of flow of waler through the
pipe is 200 litres/s. Cons ider alllo'i"'s and takc Ihc value of!= .009.
[Ans_ 11.79 ml
12. For the problems 9. 10 and II draw the hydraulic gradient lines (H.GL) and tO(a1 energy lines (TEL )
13. A syphon of <liameter 150 n1ln connects two reservoirs h,wing a differencc in elevation of 15 nt. The
lenglh of Ihc syphon is 400 m and summil is 4.0 m above the water levcl in lhe upper reseTvoir. The length
of Ihe pipe fro111 up!"'r rc>crvoir to Ihc .um",it i. SO ",. !)clenlline the discharge Ihrough Ihe syphon and
also pre"ure at the ~Ulnl11i1. Neglect minor losses. The co-efficient of friction_! '" .005.
[Ans. 41 .52 lim:>I•. - 7.281 m of walerl
14. A syphon of diamNer 200 mm connects two re<;ervoirs having a difference in elevalion a., 20 m. The lotal
length of Ihe syphon i, ROO m and Ihc sUlmnil is 5 m above Ihc waler le,'cl in Ihe upper reservoir. If the
scparation takes place at 2.8 m of water absolulC find thc maximum length of syphon from upper reservoir
to Ihc summit. Ta~e f",.OO4 and almospheric pressure = 10.3 m of water.
[,\ns. 87.52 IllI
IS. Three pipes of lengths ROO Tn. 600 m and 300 III an<l of dintneters 400 mm. 300 nun and 200 mm respectively are connected in <;eries. The ends of the compound pipe is con necled to two tanks, whose water
surface Ic"cls are mainlained al a difference of 15 m. DClennine thc Me offlow of waler Ihrough the pipes
iff" .OOS . What will be diameter of a singlc pipe of length 1700 Tn and! _ .OOS, which replaces the three
pipes?
IAns. 0.0848 m!/s. 266.5 mml
E
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I~
~ I
IL
Flow Through Pipes
557 1
16 . Two pipes of lengths 2500 In each and diameters 80 em and 60 em respectively. are connected in
""ralle!. The co -efficient of frktion for each pipe is 0.006. The total now is C<jual to 250 [ilres/s. Find the
rate of now in euch pipe.
IAns. 0.1683 ml ls. 0.0817 m'/sl
17 . A pipe of diameter 300 mm and Ic" glh 1000 m connects two reservoirs. having difference of water
I,,\'els as 15
18 .
19 .
20.
21 .
22 .
23 .
24 .
25 .
26 .
27.
211 .
29.
~ I
In.
Dc\cnninc thc discharge thmugh thc pipe. If an additional pipe of dian\e!", 300 mm and
length 600 In is auachcd 10 Ihc last 600 m Icn~th of the cxisling pipe. find the increase in the discharge.
Take / _ .02 and neglcct minor losses.
IAns. 0.0742 m'ls. 0.0258 m'/sl
Two sharp ended pipes of diameters 60 mm and 100 nun respectively. each of length 150 In arc
CQnncrted in parallel betw",," two reservoi" which have a diff~r~nc~ of lev~1 of 15 m. If w-eff,ci~nt of
friction for each pipe is 0.08. culculute the rate of flow for each pipe and also the diameter of a singlc
pipe 150 m long Which would give the same discharge if it were subst it uted for Ihe original IWO pipe •.
(Ans.0.0017 .. 00615, 110 mml
Three reservoirs A. Band Care rormccled by a pipe syslem having length 700 m. 1200 m and 500 m and
diameters 400 mm. 300 mm and 200 mm respectively . The water Ic"cis in reservoir A and B from a datum
line arc 50 111 and 45 m respeeli,·ely. The level of waler in reservoir C is tlelow thc level of waler in
rescH'oir 8. Find the dischar~e inlo or from the rescH'oirs lJ and C if the rale of flow from reservoir A is
150 lilres per '!Cc. Find Ihe height of waler lewl in the r6crvoir C. Take! ~ .005 for all pipes.
[Ans . .005 mJls •. 095 m)ls. 24.16 ml
A pipe of diameler 300 mm and kn~lh 3000 m is used for Ihe lTansmission of power by water. The lolal
head al Ihe inlet of Ihe pipe is 400 m. Find the maximum power available al Ihe oUllel of Ihc pipe , Take
!~.005.
(An o.667.07l.:WI
A pipe li ne of length 2100 m is used for Iransmining 103 kW. The pressure at the inlet of the pipe is
392.4 Nlcm l. If Ihe efficiency of transmi.sion is 80%. find the diameter of the pipe. Take f .. ,005.
(Ans. l36mml
A noule is filled at the end of a pipe of len ~lh 400 m and of diameter 150 mm . For the maximum
lransmi,sion of power Ihrough Ihe noale. find Ihe diameter of Ihc noale . Take f ~ ,008. [Ans. 41.5 mml
The head of waler at the inlet of a pipe of length 1500 m and of dialneter 400 mm is 50 m. A nOllle of
diameter 80 nl111 at the outlet. is fitted to the pipe. Find the velocil y of water at the outlet of the n01.zle
iff" .01 for the pipe.
[Ans. 28.12 mlsl
The mte of flow of water throu~h a pipe of length 1500 m and diameler I!OO mIll is 2 m)ls. At the end of
the pipe a nozzle of outside diameler 200 nl1n is filled , Find the power tmnsmilled through Ihe nonle if
the head of water at the inlet of the pipe is 180 m andf: .01 for pipe.
(Ans. 2344.7I: WI
The waler is flowing with ,) "eloci ty of 2 mls in a pipe of length 2000 m and of diameter 600 mm. At thc
end of lhe pipe. a valve is provided, Find thc rise in pressure if thc val"e is closed in 20 s«onds, Take the
value of C ~ 1420 mis,
IA ns. 20 N/cml l
If the \'al,·c in the problem 25 is closed in 1.5 sec. find the rise in pressure. Take bulk modulus of water
= 19.62 x 10' Nk,n' and consider pipe as rigid one.
(Ans.I86,75 Nlcm1 1
If in the prohlem 25. the thickne ... of the pipe i.' 10 mm and the valve is closed sudden ly. Find the ri'iC
in pressure if th~ pipe is wnsidcred to be clastic. Take value of E ~ 19,62 X 106 Nkm l for pipe material
and K ~ 19.62 X 10' Nlcm 2 for water. Calculate the cireumferentia l stress and longitudinal stress
(An•. " = 221.47 Nkm1.f," 6644.1 Nicm './I" 3322 Nlem1 1
deve loped in the pipe wnll.
The difference in water surface levels in two tanks. which are connected by two pipes in -"cries of
lengths 600 m and 400 m and of diameters 30 em and 20 em respecti,'ely. is 15 III . Detennine Ihe mte of
flow of water if Ih~ co-efficient of friction is 0.005 for both the pipes. Neglect mi nor losses.
Water is flowing ,·erti.al1y downwards through a 10 Cm diameter pipe al the rate of 50 I.p.s. At a
particular location the pipe suddenly enlarges 10 20 em diameler. A point P is located 50 cm above the
section of enlar~emcnl and another point Q is located 50 em below it in Ihe enlarg~"<I portion. A
pressure gauge connected at P gives a reading of 19.62 N/cm l . Calculate the pressure at location Q
neglecting friction loss between I' and Q but con.,idering the los .. due to sudden ~n largemcnt. What
I~
•
Flow Through Pipes
5531
neglecting friction loss between P and Q but (:oRsidering the lou due to sudden enlargement. What
will be the pressure al Q if the same discharge flows upwards assuming that the pressure P remains the
same ? Consider the lou due 10 contraction with C~ _ 0.60 but neglect friction loss bet.....een P and Q.
lAos. 21 .36 N/cm . 23.4 N/cm J (A.M.I .E.. Summer, 1985)
30. Two tanks are connected with the help of two pipes in series. The lengths of the: pipes are 1000 m and
800 m whereu the diameters are 400 mm and 200 rom respectively. The c~flicicnt of friction for both
the pipe5 i. 0.008. 1be difference of water level in the two tanks il 15 m. Find the rJte of flow of water
througb the pipes. considering aU losses. Also draw the total energy line and hydraulic gradient lines
for the system.
(Delhi University. May /998) (AM. 0.0464 ml /s1
[Hint. L, = 1000 m : Lz = 800 m :d1:: 400 mm = 0.4 m ;d1 = 200 mm =0.2 m,f= 0.008 : H - 15 m.
Now
H =h/ +hfl+hc+hJ2+ho.
,
.,
H _ O.SV,2 + 4/xL,X \.jl + O.5Vl + 4/ 'Xl:;zxVl
2g
d)(2g
28
d z x2g
or
1
1
IS = 0.5l'J + 4xO.OO8XI(XlO.xV. +
2x9.81
OAx2x9.81
o.svl
2x9.81
+::L
2g
+ 4 xO.OO8x800xvl +Y1..
O.2x2x9.SI
28
AI",
31 . A pipe of diameter 25 ern and length 2000 rn conneclS two reservoirs, having difference of water level
2!i m. Det~nnine the di5Charg~ throuSh the pipe. If an additional pipe of diamct~r 2S em and length
1000 m is auached lO the last 1000 m length of the existing pipe, find the increase in discharge. Tale!
= O.OI!i. Neglect minor losscs. (IHlhi Univu.fity. lkcem~r 2(02) lAD&. (1149.62 I/s, (m 13.14I1sl