ME 27 EF - FLUID MECHANICS/HYDRAULICS
By: Yuri G. Melliza
COURSE DESCRIPTION
This class provides students with an introduction to principal concepts and methods of fluid mechanics. Topics covered in the
course include pressure, hydrostatics, and buoyancy; open systems and control volume analysis; mass conservation and
momentum conservation for moving the models necessary to study, analyze, and design fluid systems through the application of
these concepts, and to develop the problem-solving skills essential to good engineering practice of fluid mechanics in practical
applications.
Students will have the opportunity to demonstrate a familiarity and ability to work on fluid mechanics. These outcomes will be
demonstrated through an assessment of quizzes, assignments and major examinations.
GENERAL OBJECTIVES
By the end of this course, you should expect to be able to calculate:
 fundamental fluid properties for different fluids and flows
 forces on objects submerged in both static and flowing fluids
 pressures in both static and flowing fluids, and the velocities associated with different flows
 forces in complicated momentum balance problems
 energy loss and the flow rates associated with different flow networks in channels and pipes
 dimensionless numbers important for design of experiments and practical engineering work
 numerical solutions for simple fluid flow problems
 properties of a boundary layer, both turbulent and laminar problems with the application of Bernoulli’s Principle
 Fluid Mechanics - it is a science that deals with the mechanics of fluids (liquid or gas) and is based on the same
fundamental principles employed in the mechanics of solids.
STUDENT COURSE OUTCOMES
This is an introductory course on fluid motion, the forces that fluids exert, and the forces that are exerted on them. The study of
fluid mechanics has numerous engineering applications. Fluids interact with structures such as high-rise buildings, dams, and
bridges and the static and dynamic loads imposed by the fluids must be considered in the design and construction of these
structures. Cars, aircraft, and ships all move through fluids, and frictional (fluid drag) forces represent a major energy sink. Water
is an important resource to California and fluid mechanical problems abound in the complex system of dams, aqueducts, treatment
plants, pipes, and valves used to deliver water to urban and agricultural consumers. Finally, the motions of contaminants in water
and air are governed by the mechanics of fluid flow. After this course, the students are expected to:
Branches of Fluid Mechanics
1. Fluid Static - deals with the forces exerted by or upon the fluid at rest.
2. Kinematics - deals with velocities and accelerations without considering forces and
energy.
3. Hydrodynamics - deals with the relation of velocities and accelerations and the forces exerted by or upon
the fluid in motion.
Fluid - is a substance capable of flowing and having particles which easily moved and changed
their relative position without the separation of mass.
Distinction between Liquid and Gas
Liquid
1. It has a free surface
2. A given quantity of liquid occupies a given volume of a container
3. It is incompressible
Gas
1. It has no free surface
2. A given quantity of gas occupies all portions of the container regardless of its shape and
size.
3. It is compressible
Free Surface - a surface in which all pressures can be removed except its own vapor
pressure.
Vapor - is a gas in which its pressure and temperature are such that very near to its liquid state.
PROPERTIES OF FLUIDS
1. Density (): It is the mass on a unit volume.
ρ m
kg
V m3
2. Specific Volume (): It is the volume per unit mass, or the reciprocal of its density.
m3
m kg
υ V
3. Specific Weight or Weight Density (): It is the weight on a unit volume.
γ W
KN
V m3
mg
ρg
γ

1000 V 1000
KN
m3
4. Specific Gravity or Relative Density (S):
For Liquids
It is the ratio of its density to density of water at standard temperature and pressure.
SL 
ρL
γ
 L
ρw γw
For Gases
It is the ratio of its density to the density of either air or hydrogen at some specified
temperature and pressure.
SG 
ρG
ρ AH
For Gases
ρ
P Kg
RT m 3
Where:
P - Absolute pressure in KPaa
R -Gas constant in KJ/kg-K
T -Absolute temperature in K
8.3143 KJ
M kg - K
M - molecular weight in kg/kg mole
R
where:
w = 1000 kg/m3 (Standard condition)
w = 9.81 KN/m3 (Standard condition)
AH = is the density of either air or hydrogen at some value of P and T.
5. Temperature: It is the measure of the intensity of heat of a fluid.
Celsius and Fahrenheit Scale
F  32
1.8
F  1.8(C)  32
C 
Absolute Scale
RANKINE:
KELVIN:
R  F  460
K  C  273
6. Pressure (P): It is the normal component of a force per unit area.
F KN
or KPa
A m2
KN
1 2  1 KPa (Kilo Pascal)
m
P
If a force dF acts on an infinitesimal area dA, the intensity of pressure is:
P
where:
F - force, KN
A - area , m2
dF KN
or KPa
A m2
PASCAL`S LAW
At any point in a homogeneous fluid at rest the pressures are the same in all directions.
Fy  0
Fx  0
P1 A 1  P3 A 3 sin  0
P1 A 1  P3 A 3 sin  eq. 1
P2 A 2  P3 A 3 cos 
P2 A 2  P3 A 3 cos   eq.3
From Figure
cos  
A1
;
A3
A 1  A 3 sin  eq. 2
Equation 2 to equation 1
sin 
P1  P3
A2
A3
A 2  A 3 cos   eq.4
Equation 4 to equation 3
P2  P3
therefore
P1  P2  P3
Atmospheric Pressure (Pa): It is the absolute pressure exerted by the atmosphere.
At sea level
Pa = 101.325 KPa
= 0.101325 MPa
= 1.01325 Bar
= 760 mm Hg
= 10.33 m of H2O
= 1.033 kg/cm2
= 14.7 psi
= 29.921 in Hg
= 33.878 ft of H2O
Absolute and Gage Pressure
Absolute Pressure - is the pressure measured referred to absolute zero and using absolute zero as the base.
Gage Pressure - is the pressure measured referred to the existing atmospheric pressure and using atmospheric pressure as the base.
Pgage - if above atmospheric
Pvac- vacuum or negative gage pressure, if below atmospheric
Pabs  Pa  PGage
Pabs  Pa  Pvacuum
7. Viscosity: It is the property of a fluid that determines the amount of its resistance to
shearing stress.
Assumptions:
1. Fluid particles in contact with the moving surface moves with the same velocity of that surface.
2. The rate of change of velocity dv/dx is constant in the direction perpendicular to the direction of motion.
3. The shearing is directly proportional to the rate of change of velocity.
let S - shearing stress in Pa or N/m2
dv
dx
dv
dv v
Sμ
but

dx
dx x
v
Sμ
x
x
N - sec
μ  S Pa - sec or
v
m2
S
Where
 - Absolute or dynamic viscosity in Pa-sec or N-sec/m2
S - shearing stress in Pascal (Pa)
x - distance apart in meters
v - velocity in m/sec
8. Kinematics Viscosity: It is the ratio of the absolute or dynamic viscosity to the mass
density.
ν
μ m2
ρ sec
Conversion
1 Pa-sec = 1 N-sec/m2
1 Pa-sec = 10 Poise
1 Pa-sec = 0.020885 lb-sec/ft2
1 Pa-sec = 0.10197 kg-sec/m2
1 Pa-sec = 1000 Centipoise
1 m2/sec = 10,000 stokes
1 m2/sec = 10.764 ft2/sec
1 m2/sec = 1,000,000 Centistokes
Stoke = 1 cm2/sec
1 kg/m3 = 0.062428 llb/ft3
1 lb/ft3 = 16.018 kg/m3
9. Elasticity: If a pressure is applied to a fluid, it contracts; if the pressure is released, it expands, the
elasticity of a fluid is related to the amount of deformation(expansion or contraction) for a given pressure
change. Quantitatively, the degree of elasticity is equal to:
P
Ev 
dP
KPa
dV

V
P
Where negative sign is used because dV/V is negative for a positive dP.
dP
KPa
dρ
ρ
dV dρ

V
ρ
Ev 
where:
Ev - bulk modulus of elasticity
dV - is the incremental volume change, m3
V - is the original volume, m3
dP - is the incremental pressure change in Kpa
10. Surface Tension
The attractive force exerted upon the surface molecules of a liquid by the molecules beneath that tends to draw the surface
molecules into the bulk of the liquid and makes the liquid assume the shape having the least surface area.
Surface tension is the elastic tendency of a fluid surface which makes it acquire the least surface area possible. Surface tension
allows insects (e.g. water spiders), usually denser than water, to float and stride on a water surface.
At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than
to the molecules in the air (due to adhesion). The net effect is an inward force at its surface that causes the liquid to behave as if its
surface were covered with a stretched elastic membrane. Thus, the surface becomes under tension from the imbalanced forces,
which is probably where the term "surface tension" came from.[1] Because of the relatively high attraction of water molecules for
each other through a web of hydrogen bonds, water has a higher surface tension (72.8 millinewtons per meter at 20 °C) compared
to that of most other liquids. Surface tension is an important factor in the phenomenon of capillarity.
Surface tension has the dimension of force per unit length, or of energy per unit area. The two are equivalent, but when referring to
energy per unit of area, it is common to use the term surface energy, which is a more general term in the sense that it applies also
to solids.
Floating objects
Cross-section of a needle floating on the surface of water. Fw is the weight and Fs are surface tension resultant forces.
When an object is placed on a liquid, its weight Fw depresses the surface, and if surface tension and downward force becomes
equal than is balanced by the surface tension forces on either side Fs, which are each parallel to the water's surface at the points
where it contacts the object. Notice that small movement in the body may cause the object to sink. As the angle of contact
decreases surface tension decreases the horizontal components of the two Fs arrows point in opposite directions, so they cancel
each other, but the vertical components point in the same direction and therefore add up [2] to balance Fw. The object's surface must
not be wettable for this to happen, and its weight must be low enough for the surface tension to support it.
Examples of Surface Tension
Drops of water. When using a water dropper, the water does not flow in a continuous stream, but rather in a series of drops.
The shape of the drops is caused by the surface tension of the water. The only reason the drop of water isn't completely spherical is
because of the force of gravity pulling down on it. In the absence of gravity, the drop would minimize the surface area in order to
minimize tension, which would result in a perfectly spherical shape.
Insects walking on water. Several insects are able to walk on water, such as the water strider. Their legs are formed to distribute
their weight, causing the surface of the liquid to become depressed, minimizing the potential energy to create a balance of forces
so that the strider can move across the surface of the water without breaking through the surface. This is similar in concept to
wearing snowshoes to walk across deep snowdrifts without your feet sinking.
Needle (or paper clip) floating on water. Even though the density of these objects is greater than water, the surface tension along
the depression is enough to counteract the force of gravity pulling down on the metal object. Click on the picture to the right, then
click "Next," to view a force diagram of this situation or try out the Floating Needle trick for yourself.
Capillarity
Capillary Action
Capillary action is the result of adhesion and surface tension.
Adhesion of water to the walls of a vessel will cause an upward
force on the liquid at the edges and result in a meniscus which
turns upward. The surface tension acts to hold the surface intact, so
instead of just the edges moving upward, the whole liquid surface
is dragged upward.




h
r
2σ cos θ
γr
h
Where:
 - surface tension, N/m
 - specific weight of liquid, N/m3
r – radius, m
h – capillary rise, m
Table 1.Surface Tension of Water
C

0
0.0756
10
0.0742
20
0.0728
30
0.0712
40
0.0696
60
0.0662
80
0.0626
100
0.0589
Sample Exercises
1. A mercury barometer at the ground floor of Kingston Tower in Chicago reads 735 mm Hg. At the same time another
barometer at the top of the tower reads 590 mmHg. Assuming the air to be constant at 1.21 kg/m 3, what is the
approximate height of the tower using g = 9.7 m/sec 2.
2
h
1
735 (101 .325 )
 98 KPa
760
590 (101 .325 )
P2 
 78 .7 KPa
760
ρg
1.21(9.7)
KN
γ

 11.64 x 10 -3 3
1000
1000
m
dP   γdh
P1 
P2  P1   γ(h 2  h1 )
h1  0; h 2  h
P2  P1   γ(h )
h
2.
P2  P1
 1658.08 meters
γ
A tank contains a mixture of 20 kg of nitrogen and 20 kg of carbon monoxide. The total tank volume is 20 m3. Determine
the density, specific volume and specific weight of the mixture if local g = 9.81 m/sec2
m  m1  m 2
m  20  20  40 kg
m 40
kg
ρ 
2 3
V 20
m
υ
1
m3
 0.5
ρ
kg
3.
A pressure gage registers 345 KPag in a region where the barometric pressure is 736.7 mm hg.
Find the absolute in Bars.
Pabs  Pa  Pgage
Bar
 736 .7(101 .325 ) 
Pabs  
 345  443.2 KPaa x

760
100 KPa


Pabs  4.432 Bars
4. The level of water in an enclosed water tank is 40 m above ground level. If the pressure of the
air space above the water is 120 Kpa, what is the pressure at ground level in KPa.
dP   γdh
By integratio n
Air
1

2

2
dP   γ dh
1
1
P2  P1   γ(h 2  h1 )
40 m
P2  P1  γ(h 2  h1 )
h1  0
2
h 2  40 m
P2  120  9.81(40 )  512.4 KPa
5.
A liquid has an absolute viscosity of 4.8 x 10 -4lb-sec/ft2. It weighs 54 lb/ft3. What are its absolute and kinematic
viscosities in SI.
μ  4.8 x 10 - 4
ρ  54
ν
lb
ft 3
lb  sec
ft 2
16 .018
x
1
lb
ft 3
1 Pa - sec
 0.023 Pa - sec
lb - sec
0.020885
ft 2
kg
m 3  864 .972 kg
m3
μ 0.023

864 .972
ρ
ν  2.66 x 10 -5 m
6.
x
2
sec
The absolute viscosity of water at 20C is 1.002 x 10-3 Pa-sec and the density is 998 kg/m3. What are its absolute and
kinematic viscosity in English units.
μ  1.002 x10 3 Pa - sec
ν
3
0.020885 lb - sec
Pa - sec
ft 2  2.09 x 10 -5 lb - sec
2
10.764
ft 2
ft 2
sec  1.08 x 10 -5 ft 2
μ 1.002 x10 Pa - sec
m

 1.004 x 10 -6
x
2
kg
ρ
sec
998
1m
3
sec
m
sec
IDEAL GAS PRINCIPLES
IDEAL OR PERFECT GAS
1. CHARACTERISTIC EQUATION
PV = mRT 1
P = RT 2
P
3
RT
RT
υ
4
P
PV
C  5
T
P1V1 P2 V2

 6
T1
T2
ρ
where:
P - absolute pressure in KPa
V - volume in m3
m -mass in kg
R -Gas constant in KJ/kg-K
T - absolute temperature in K
 - specific volumein m3/kg
 - density in kg/m3
2. GAS CONSTANT
R
KJ
 7
M kg - K
KJ
R  8.3143
(Universal Gas Constant)
kg mol - K
R
M  molecular weight of a gas in
kg
kg mol
3. BOYLE`S LAW ( T = C ): Robert Boyle (1627-1691)
If the temperature of a certain quantity of gas is held constant, the volume V is inversely proportional to
the pressure P, during a quasi-static change of state.
1
1
Vα or V  C
P
P
PV  C
P1V1  P2V2  8
4. CHARLE`S LAW ( P = C and V = C):Jacques Charles (1746-1823) and Joseph Louis Gay- Lussac
(1778-1850)
A) At constant pressure (P=C), the volume V of a certain quantity of gas is directly proportional to
the absolute temperature T, during a quasi static change of state.
V  T or V = CT
V
C
T
V1 V2

 9
T1 T2
B) At constant volume (V = C), the pressure P of a certain quantity of gas Is directly proportional to
the absolute temperature T, during a quasi-static change of state.
P  T or P = CT
P
C
T
P1 P2

 10
T1 T2
5. AVOGADRO`S LAW:Amadeo Avogadro (1776-1856)
All gases at the same temperature and pressure, under the action of a given value of g, have the
same number of molecules per unit of volume. From which it follows that the  M.
γ1 M1 R 2


γ 2 M 2 R1
6. SPECIFIC HEATS
Cp  Cv  R
Rk
k 1
R
Cv 
k 1
Cp
k
Cv
Cp 
7. ENTROPY CHANGE (S)
Entropy is that property of a substance that determines the amount of randomness and disorder of a substance.
If during a process, an amount of heat is taken and is by divided by the absolute temperature at which it is taken,
the result is called the ENTROPY CHANGE.
dQ
T
dQ
ΔS 
T
dS 

8. Isentropic Process: It is an internally reversible adiabatic process of an ideal or perfect gas in
which S = C or PVk = C.
P, V, & T relationship:
P1V1k  P2 V2 k
k 1
 k
T2  P2
 
T1  P1 
V 
  1 
 V2 
k 1
9. Polytropic Process: It is an internally reversible process of an ideal or perfect gas in
which PVn = C. Where n stands for any constant (but not equal to infinity).
P, V, & T relationship:
P1V1n  P2V2 n
T2  P2 
 
T1  P1 
n 1
n
V 
  1 
 V2 
n 1
1. Total moles of a mixture
n = ni
2. Mole Fraction
yi = ni/n
3. Total mass of a mixture
m = mi
4. Mass Fraction
xi = mi/m
5. Equation of State
A. Mass Basis
a. For the mixture
PV = mRT
b. For the components
PiVi = miRiTi
B. Mole Basis
a. For the mixture
PV = nRT
b. For the components
PiVi = niRTi
6. Amagat's Law: The total volume V of a mixture is equal to the sum of the volume occupied by each component at
The mixture pressure P, and temperature T.
V = Vi
n1
m1
V1
n2
m2
V2
P,T
n3
m3
V3
P = P1 = P2 = P3
T = T1 = T2 = T3
n = n1 + n2 + n3
V = V 1 + V2 + V3
 PV PV1 PV2 PV3 
 RT  RT  RT  RT 


V = V 1 + V 2 + V3
 RT 


 P 
V = V
Vi
V
yi 
7. Dalton's Law
The total pressure of a mixture P is equal to the sum of the partial pressure that each gas would exert at
the mixture volume V and temperature T.
P = Pi
T = T1 = T2 = T3
V = V1 = V 2 = V3
mixture
1
n1
P1
2
n2
P2
3
n3
P3
n
P
n = n1 + n2 + n3
 PV P1V P2 V P3 V   RT 
 RT  RT  RT  RT   P 



P = P1 + P2 + P3
yi 
Pi
P
8. Molecular Weight of a Mixture
M
y M
M
R 8.3143 kg

R
R kg mol
i
i
9. Gas Constant
R
x R
R
R 8.3143 KJ

M
R kg - K
i
i
10. Specific Heat
x C
Cv   x C
KJ
kg - K
KJ
i Vi
kg - K
KJ
CP  CV  R
kg - K
Rk
CP 
k 1
R
CV 
k 1
Cp 
i
Pi
11. Gravimetric and Volumetric Analysis
Gravimetric Analysis gives the mass fractions of the components in the mixture.
Volumetric Analysis gives the volumetric or molal fractions of the components in the mixture.
xi 
yi M i
y M
i
yi 
xi
Mi
xi
Mi


i
yi M i
M
Example 1
The specific weight of water at ordinary pressure and temperature is 9.81 KN/m3. The specific gravity of mercury is 13.55.
Compute the density of water, and the specific weight and density of mercury.
Example 2
A. Calculate the density, specific weight and specific volume of oxygen at 38C and 104 KPa.
B. What would be the temperature and pressure of the gas if it where compressed isentropically to 40 percent of its original
volume.
C. If the process described in (B) had been isothermal, what would be the temperature and pressure have been? (For O2: M = 32 ;
k = 1.395)
Example 3
A vessel contains 85L of water at 10C ( = 999.7 kg/m3) and atmospheric pressure. If it is heated to 70C ( = 977.8 kg/m3), what
will be the percentage change in its volume? What mas of water must be removed, to maintain the volume at the original volume?
VARIATION OF PRESSURE WITH ELEVATION
By applying equations for equilibrium
 Fy  0
(P  dP)A  PA  W  0
PA  dPA  PA  W  0
dPA  W
But
W  γV
V  A(dh)
dPA  γA(dh)
dP  γ(dh)
General Equation
dP  -γdh
If the specific weight γ or density ρ is constant
ΔP  -γΔh
Note: Negative sign is used because pressure decreases with increasing elevation and increases with decreasing elevation.
h is positive if measured upward
h is negative if measured downward
Where:
P – Pressure in KPa
 - specific weight in KN/m3
h – elevation in meters
a. For isothermal Condition (T = C)
P
RT
Pg
γ
1000 RT
dP   γdh
ρ
ln
h 2  h1  h
ln
Pg
dh
1000 RT
dP
g

dh
P
1000 RT
2
2
dP
g

dh
1000 RT 1
1 P
dP  

P2
g

(h 2  h1 )
P1
1000 RT

P1
gh

P2 1000 RT
1000 RT  P1 
ln   Elevation
g
 P2 
P1
P2 
 Final Pressure
gh
h
e1000RT
b. Isentropic Condition
2
 1 1 
P k 
1
 1    1 / k ( h 2  h1 )
C
1  
 k 1
k 1
(h )
k 1
k 1
P2 k  P1 k   k 1 / k
C
k

1
k 1 

 C1 / k P2 k  P1 k 

  Elevation
h
k 1
k
Pυ k  C
1
ρ
ργ
υ
P
C
γk
P
P
C  1k  2k
γ1
γ2
P1 / k
C1 / k
dP   γdh
γ
dP  
P1 / k



P2  P
dh
C1 / k
dP
1
  1 / k dh
1/ k
P
C
1
P 1 / k dP   1 / k dh
C
2
1 2
P 1 / k dP   1 / k dh
1
1
C

1
k 1
k

 
k 1
h
k
C1 / k


k
k 1
 Final Pressure

Pressure Variation in the Atmosphere
TROPOSPHERE: It is the layer in the atmosphere between sea level and 10.769 km and the temperature decreases linearly with
increasing elevation at a lapse rate of 6.5K/km.
P  Ts  0.0065h 


Ps 
Ts

6.5R




P g 

Ts 1   
  Ps 



h
0.0065
g
6.5R
meters
STRATOSPHERE: It is the layer that begins at the top of the troposphere and extends to an elevation of 32.3 km, and in this layer
the temperature is constant at -55C.
P

Ps
h
1
gh
e 1000RT
1000RT Ps
ln
meters
g
P
where:
P - pressure at elevation h, KPa
Ps - Pressure at sea level, KPa
h - elevation, meters
Ts - temperature at sea level in K
FROM NASA
1. Atmospheric pressure decreases at the rate of 83.312 mm Hg per 1000 meters rise
in elevation.
2. Atmospheric temperature decreases at the rate of 6.5K per 1000 meters rise in
elevation.
CONDITIONS OF PRESSURE VARIATION WITH ELEVATION
General Formula
dP  γdh
a.
Constant Density
2
2
 dP  γ dh
1
1
P2  P1  (h 2  h1 )
(h 2  h1 )  h
P2  P1  γh
P1  P2
γ
Note
h
 h  if measured upward
- h - if measured downward
b.
For Isothermal Condition (T = C)
P
RT
Pg
γ
1000 RT
dP   γdh
ρ
ln
h 2  h1  h
ln
Pg
dh
1000 RT
dP
g

dh
P
1000 RT
2 dP
2
g

dh
1000 RT 1
1 P
dP  

c.
P2
g

( h 2  h1 )
P1
1000 RT
P1
gh

P2 1000 RT
1000 RT  P1 
ln   Elevation
g
 P2 
P1
P2 
 Final Pressure
gh
h

e 1000RT
Isentropic Condition
Pυ k  C
1
υ
ρ
ργ
P
C
γk
P
P
C  1k  2k
γ1
γ2
γ
2
 1 1 
P k 
1
 1    1 / k ( h 2  h1 )
C
1  
 k 1
k 1
(h )
k 1
k 1
P2 k  P1 k   k 1 / k
C
k 1
k 1 

 C1 / k P2 k  P1 k 

  Elevation
h
k 1
k
P1 / k
1/ k
C
dP   γdh
dP  
P1 / k
dh
C1 / k
dP
1
  1 / k dh
1/ k
P
C
1
1 / k
P
dP   1 / k dh
C
2
2
1
P 1 / k dP   1 / k dh
1
1
C




P2  P

1
k 1
k
 
k 1
h
k

C1 / k


k
k 1
 Final Pressure
d.
Temperature decreases at a Standard Lapse Rate (6.5K per Kilometer)
T  (Ts  0.0065 h )
ρ
P
P

RT R (Ts  0.0065 h )
γ
ρg
Pg

1000 1000 R (Ts  0.0065 h )
dP  
Pg (dh)
1000 R (Ts  0.0065 h )

dP
g 
dh



P 1000 R  (Ts  0.0065 h ) 

2
1
dP
g

P 1000 R
2

  (T  0.0065 h) 
1
dh
s
Ts is constant

2
1
ln
dP
g

P (0.0065 )1000 R
2
(0.0065 )dh 
  (T  0.0065 h) 
1
s
 (T  0.0065 h ) 
P2
g

ln  s

P1 6.5R 
Ts

g
P2  (Ts  0.0065 h )  6.5R


P1 
Ts

g
 (T  0.0065 h )  6.5R
P2  P1  s

Ts


6. 5 R 

 P  g 
Ts 1   2 

P
  1



h
0.0065
MANOMETERS
Manometer is an instrument used in measuring gage pressure in length of some liquid column.
1. Open Type Manometer : It has an atmospheric surface and is capable in measuring gage pressure.
2. Differential Type Manometer : It has no atmospheric surface and is capable in measuring differences of pressure.
Determination of S using a U - tube
Specific gravity or relative density of an unknown liquid can be determine using a U - tube with both ends open to the atmosphere,
and applying the Variation of Pressure with Elevation principle, provided that one liquid of known specific gravity is available.
Problem No. 1
A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3 m above
A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3)
0
P1 = PB
1
1
2
P2 = PA
P0  133.4(1.2)  3(9.81)  PA
P0  0
PA  189.51 KPa
Problem no. 2
In the figure shown, fluid A is water ( = 9.81 KN/m3, fluid B is mercury ( = 133.42 KN/m3). Z = 450 mm and Y = 900 mm.
Compute the pressure difference (Pn – Pm) in KPa.
Pn = P1
1
Pm = P5
5
2
3
4
3
Pn  ( y  x )9.81  z(133.42)  z(9.81)  x(9.81)  Pm
Pn  Pm  ( y  x )9.81  z(133.42)  z(9.81)  x(9.81)
Pn  Pm   y(9.81)  x(9.81)  z(133.42)  z(9.81)  x(9.81)
Pn  Pm   y(9.81)  z(133.42)  z(9.81) KPa
Pn  Pm  64.5 KPa
Problem no. 3
A U tube with both ends open to the atmosphere contains mercury in the lower portion. In one leg, water stands 760 mm above the
surface of mercury, in the other leg , oil (S = 0.80) stands 450 mm above the surface of the mercury. What is the difference in
elevation between the surfaces of mercury in contact with oil and water columns?
HYDROSTATIC FORCES ACTING ON PLANE SURFACES
Fig. (a)
Fig. (b)
θ
θ
C.G. - center of gravity
C.P. - center of pressure
General Equation:
F  γhA
Therefore, the total hydrostatic force exerted by the fluid on any plane surfaces submerged in a homogeneous fluid at rest, is equal
to the product of the surface area A and the pressure at its centroids
h .
LOCATION OF THE CENTER OF PRESSURE
Ig  A y
yp 
Ay
2
yp  y  e
ye 
e
Ig
y
Ay
Ig
Ay
Ig
Ss
Ss  A y
e
where:
e - is the perpendicular distance between C.G. andC.P.
Ig - moment of inertia with respect to the axis at its centroids and lying on its plane
Ss- statical moment of inertia with respect to the axis SS not lying on its plane
HYDROSTATIC FORCE ACTING ON CURVED SURFACES
C'C = B'B = L
where L - length of the curved surface AB perpendicular to the paper
F  Fh
2

2
Fv KN
Fh  γhA KN
A = BC x L m2
where: A - area of the vertical projection of the curved surface AB
FV  γV KN
V = AABCDEA (L) m3
Fh - horizontal component of F in KN that passes through the center of pressure of the vertical
surface AB
Fv - vertical component of F, KN
F - total hydrostatic force acting on the curved surface AB
Note: If the liquid is underneath the surface the Force acts upward.
HOOP TENSION
D
H
1m
P = h
Considering a semi-circular segment of 1 m length
T
T
F
F = 0
F = 2T  1
T 
F
T
1m
t
F
22
On the vertical projection:
D
1m
P
F
A
where
A = 1D
F = P(1D)
let : S - tensile or hoop stress in KPa
T
S
At
where At = 1t (area subjected to tensile stress)
S
F
P(1D)

2(1t ) 2(1t )
S
PD
2t
T
projection of the curved
ARCHIMEDES PRINCIPLE
LAWS OF BUOYANCY:Any body partly or wholly submerged in a liquid is subjected to a Buoyant or Upward force, which is
equal to the weight of the liquid displaced. If the buoyant force is less than to the weight of the body, the body sinks, and if the
buoyant force is greater than the weight of the body the body floats.
A.
W = BF
W = BVBKN; W = BVB kg
BF = VsKN; BF = Vs kg
B.
W = BF - T
W = BVBKN; W = BVB kg
BF = VsKN; BF = Vs kg
C.
W = BF + T
W = BVBKN; W = BVB kg
BF = VsKN ; BF = Vs kg
D.
W = BF - T
W = BVBKN ; W = BVB kg
BF = VsKN ; BF = Vs kg
VB = V s
E.
W = BF + T
W = BVBKN ; W = BVB kg
BF = VsKN ; BF = Vs kg
VB = V s
where:
W - weight of the body; KN, kg
BF - buoyant force, KN, kg
VB - volume of body, m3
Vs - volume submerged, m3
B - specific gravity of the body, KN/m3
 - specific gravity of the liquid, KN/m3
B - density of the body, kg/m3
 - density of the liquid, kg/m3
SPECIFIC GRAVITY OF SOLIDS HEAVIER THAN WATER
W1
SB 
W1  W2
SPECIFIC GRAVITY OF AN UNKNOWN LIQUIDS
W  W3
SB  1
W1  W2
Where
W1 – weight of the body in air
W2 - weight of the body in water
W3 - weight of the body in an unknown liquid
SAMPLE PROBLEMS
Problem No. 1 (Manometer)
A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3 m above
A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa. ( of mercury = 133.4 KN/m3)
0
P1 = PB
1
1
2
P2 = PA
P0  133.4(1.2)  3(9.81)  PA
P0  0
PA  189.51 KPa
Problem no. 2 (Manometer)
In the figure shown, fluid A is water ( = 9.81 KN/m3, fluid B is mercury ( = 133.42 KN/m3). Z = 450 mm and Y = 900 mm.
Compute the pressure difference (Pn – Pm) in KPa.
Pn = P1
1
Pm = P5
5
2
4
3
3
Pn  ( y  x )9.81  z(133.42)  z(9.81)  x(9.81)  Pm
Pn  Pm  ( y  x )9.81  z(133.42)  z(9.81)  x(9.81)
Pn  Pm   y(9.81)  x(9.81)  z(133.42)  z(9.81)  x(9.81)
Pn  Pm   y(9.81)  z(133.42)  z(9.81) KPa
Pn  Pm  64.5 KPa
Problem No. 3 (Variation of Pressure)
An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at sea level is
15 C, what is the elevation assuming:
a. air at constant density
b. air under isothermal condition
c. air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer.
Given:
P1 = 762 mm Hg = 101.6 KPa
P2 = 736 mm Hg = 98.23 KPa
T1 = 15 + 273 = 288 K
a. Constant density
P
101.6
kg

 1.23 3
RT 0.287(288)
m
1.23(9.81)
KN

 0.01206 3
1000
m
P2  P1   (h2  h1 )
P P
h  1 2  279.4 meters


b. Constant Temperature
h
1000RT  P1 
ln 
g  P2 
h  284.2 meters
c. Air temperature decreases linearly with elevation at the rate of 6.5 K per kilometer
g
P2  T1  0.0065h  6.5R


P1 
T1

6.5R


g


P
2

T1 1    
  P1  

 meters
h 
0.0065
h = 283.04 meters
Problem No. 4 (Variation of Pressure)
The bottom of a river is 12 m below the water surface. Underneath which is a silt having a specific gravity of 1.75 and a thickness
t. The pressure at the bottom of silt is 450 KPa. Determine the thickness of the silt.
at 0 to 2
0  9.81(12)  1.75(9.81)t  450
t
450  (9.81)12
 19.355 M
1.75(9.81)
Problem No. 5 (Hydrostatic forces)
A rectangular gate 1.6 m wide and 2 m high has water on one side and is inclined at 45with the horizontal. Water is 1.5 m above
the top of the gate.
a) Compute the total force acting on the gate and its location
b) If the gate is hinged at the top(B)what force P is needed at the bottom(A) to open the gate.
h  1.5  1 sin 45
h  1.5  1 sin 45
h  2.207 m
M
A  1.6(2)  3.2 m2
2P  F(1.10)  0
F  9.81(2.207 )(3. 2)
P
69.3(1.10)
2
P  38.3 KN
F  69.3 KN
From Figure
sin 45 
h
y
2.207
 3.12 m
sin 45
For a rectangle
B
y
Ig 
yp 
yp’
bh3 1.6(2)3

 1.0666 m4
12
12
Ig  A y
2
Ay

2m
F
1.0666  3.2(3.12)2
3.2(3.12)
y p  3.23 m
P
1.5
1.5
 3.23 
sin 45
sin 45
 1.109 m
y p'  y p 
y p'
at A
A
Problem No. 6 (Forces of Curved Surfaces)
The curve surface represented by AB in the figure is the surface of a quadrant of a circular cylinder 3 m long. Determine the
horizontal and vertical component of the total hydrostatic force on the surface if the liquid is gasoline with S = 0.72.
Fv
Liquid surface
3m
L - length
L=3m
R = 2 m (radius)
L=3
m
2m
Fh
2m
h  4m
Fh  γ hA
A  2(3)  6 m 2
h  3 1  4 m
γ  0.72 (9.810 )  7.0632
KN
m3
Fh  169 .52 KN
Fv  γV
1

V   π 22  2(3) 3  27 .425 m3
4


Fv  7.0632 (27 .425 )  193 .71 KN
F
Fh 2  Fv 2
 257 .41 KN
Problem No. 7 (Buoyancy)
Two spheres each 1.2 m in diameter weighs 4 KN and 12 KN, respectively. They are connected with a short rope and placed in
water. What portion of the lighter sphere protrudes from the water.
W1  4 KN; W2  12 KN; R 
Volume of Sphere 
1.2
 0.60 m
2
4 3
πR
3
W1  W2  BF1  BF2
4

4  12  9.81VS1  9.81 π(0.60 )3 
3


VS1  0.73 m3
VF1  volume protrudes above liquid surface
4

VF1   π(0.60 )3   0.73  0.175 m3
3

Problem No. 8 (Buoyancy)
A block of material of an unknown volume is submerged in water and weighs 500 N. The same block weighs 650 N when weigh
in air. Determine the volume of the material in m3.
S
W1
650

 4.33
W1  W2 650  500
ρ  4.33(1000 )  4330
kg
m3
W  mg
650
 66 .26 kg
9.81
m
ρ
V
66 .26
V
 0.0153 m3
4330
m
Problem No. 9 (Forces on Plane surface)
A square gate (3 m x 3 m) is
= 0.80) stands on the left side of the gate to a height of 1.5 m above point A. Determine the amount of the hydrostatic force exerted
by the oil on the gate and its location of the center of pressure.
Ig for a square:
Ig =
bh 3
12
SOLUTION
h  1.5  1.5 sin 60  2.79 m
F  0.80(9.81) (2.79)(9)  197.06 KN
yp 
Ig  A y
2
Ay
3(3) 3
Ig 
 6.75 m
12
h
y
 3.22 m
sin 60
6.75  9(3.22) 2
yp 
 3.453 m
9(3.22)
Problem No. 10 (Hoop Tension)
A 1.2 m diameter steel pipe, 6 mm thick, carries oil with S = 0.822 under a head of 122 m of oil. Compute
a. The stress in the steel in KPa
b. The thickness of the steel pipe required to carry a pressure of 1724 KPa with an allowable stress of 124 MPa
PD
2t
P  0.822 (9.81)(122 )  983 .8 KPa
983 .8(1.2)
S
 98,379 KPa
2(0.006 )
1724(1.2)
t
 0.0083 m  8.3 mm
124 ,000 (2)
S
SAMPLE OF MANOMETER PROBLEMS
1.
The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine
a. The height h in mm in the open water column
b. The gage pressure acting on the bottom of the tank surface AB
c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute.
2.
The mercury manometer in the figure indicates a differential reading of 30 m when the pressure in pipe A is 30 mm Hg
vacuum. Determine the pressure in pipe B.
3.
In the figure pipe A contains carbon tetrachloride (S = 1.60) and the closed storage tank B contains a salt brine (S = 1.15).
Determine the air pressure in tank B in KPa if the gage pressure in pipe A is 1.75 kg/cm 2.
4.
A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine
the differential reading h. The specific weight of the air is negligible.
5.
A closed tank contains compressed air and oil (S = 0.90) as shown in the figure.a U – tube manometer using mercury (S =
13.6) is connected to the tank as shown. For column heights h1 = 90 cm; h2 = 15 cm and h3 = 22 cm, determine the
pressure reading of the gage.
6.
The pressure of gas in a pipeline is measured with a mercury manometer having one limb open to the atmosphere. If the
difference in the height of mercury in the limbs is 562 mm, calculate the absolute gas pressure. The barometer reads 761
mm Hg, the acceleration due to gravity is 9.79 m/sec2 and SHg = 13.64.
7.
A turbine is supplied with steam at a gauge pressure of 1.4 MPa, after expansion in the turbine the steam flows into a
condenser which is maintained at a vacuum of 710 mm Hg. The barometric pressure is 772mm Hg. Express the inlet and
exhaust pressure in kg/cm2 . Take the S of mercury is 13.6.
The pressure of steam flowing in a pipe line is measured with a mercury manometer. Some steam condenses in to water.
/estimate the steam pressure in KPa. Take the density of mercury as 13,600 kg/m3, the barometer reading as 76.1 cm Hg
and g = 9.806 m/sec2.
8.
9.
A manometer is attached to a tank containing three different fluids, as ashow in the figure below. What will be the
difference in elevation h of th mercury column in the manometer.
Quiz no. _______
1. The pressure on top of a mountain is 90 KPa. If the pressure and temperature ate sea level are 101 KPa and 288 K,
respectively, Determine the height of the mountain and the temperature on top, Assuming
a. Constant density
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
2. A manometer is attached to a tank containing different fluids, as shown in the figure below. What will be the
difference in elevation of the mercury column in the manometer.
1.
A differential manometer is shown in the figure below. Calculate the pressure difference (P A – PB) in kg/cm2.
(1 inch = 25.4 mm; 1 m = 1000 mm).
2.
On top of a mountain the pressure is 85 KPa. If the pressure and temperature ate sea level are 101.33 KPa and 21C
respectively, Determine the altitude and temperature on top of the mountain, assuming Constant density
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
FLUID MECHANICS/QUIZ NO. 1 (December 10, 2016)
Group 3
1 Lampago, Mikel N.
2 Leonor Jr., Eduardo E.
3 Nayon, Carl Christian J.
4 Ompoc, Horace B.
Opon, Rashiel Jan M.
6
Repollo, Ryan Z.
1.
For the configuration shown below, calculate the weight of the piston if the gage pressure reading is 70 KPa.
2.
On a certain day a PAL plane is flying at an altitude of 3000 m. If the atmospheric pressure and temperature at sea level
are 101 KPa and 288 K, respectively. Determine the pressure and temperature in the plane, assuming
a. Constant density from sea level to elevation 3000 m
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
FLUID MECHANICS/QUIZ NO. 1 (December 10, 2016)
Group 4
1 Somodlayon, Michael John T.
2 Tabamo, Brael Y.
3 Ubaub, Matt Vann Ernee V.
4 Vasallo Jr., Edward A.
5 Zalsos, Edzel A.
1.
A mountain is 2000 meters above sea level. If the pressure and temperature ate sea level are 101 KPa and 288 K,
respectively, Determine the temperature and pressure on top of the mountain, assuming
A .Constant density from sea level to the top of the mountain
b. Isothermal condition
c. Isentropic condition
d. Standard atmospheric lapse rate prevails
2. If the atmospheric pressure is 101.03 KPa and the absolute pressure at the bottom of the tank is 231.3 KPa.
What is the specific gravity of olive oil? (S of SAE oil = 0.89 ; S of Hg = 13.6)
101 .3  (0.89 )(9.81)1.5  2.5(9.81)   oil(2.9)  13 .6(9.81)(0.4)  231 .3
 oil  13 .34
KN
m3
Soil  1.36
SAMPLE PROBLEMS
1. A vertical piston cylinder device contains a gas at an unknown pressure. If the outside pressure is 100 kPa, determine (a)
the pressure of the gas if the piston has an area of 0.2 m2 and a mass of 20 kg. Assume g = 9.81 m/s2. (b) What-ifScenario: How would the answer in (a) changes if the orientation of the device is changed upside down. Answers: (a) 101
kPa, (b) 99 kPa.
2. A piston with a diameter of 50 cm and a thickness of 5 cm is made of a composite material with a density of 4000 kg/m 3.
(a) If the outside pressure is 101 KPa, determine the pressure inside the piston-cylinder assembly if the cylinder contains
air. (b) How would the answer change if the piston diameter was 100 cm instead? Answers: (a) 104.8 KPa
3.
Water flows through a variable-area pipe with a mass flow rate of 10,000 kg/min. Determine the minimum diameter of
the pipe if the flow velocity is not to exceed 5 m/s. Assume density of water to be 1000 kg/m3. Answers: 0.206 m
4. A bucket of concrete with a mass of 5000 kg is raised without any acceleration by a crane through a height of 20 m. (a)
Determine the work transferred into the bucket. (b) What happens to the energy as it is transferred to the bucket? (c) Also
determine the power delivered to the bucket if it is raised at a constant speed of 1 m/s.
Answers: (a) 981 kJ, (c) 49.05 kW
3. A building in Makati is 84.5 m high above the street level. The required static pressure of the water line
at the top of the building is 2.5 kg/cm2. What must be the pressure in KPa in the main water located 4.75
m below the street level.
P1 = 245.166 KPa
245.166  9.81(84.5)  4.75(9.81)  1120.71 KPa
14. A cylindrical tank 2 m diameter, 3 m high is full of oil. If the specific gravity of oil is 0.9, what is the
mass of oil in the tank?
  0.9(1000)  900
kg
m3

V  (2)2 3  9.425 m3
4
m

V
m  900(9.425)  8482.5 kg
5. A cubical tank 1 m on a side, contains a mixture of 1.8 kg of nitrogen (M = 28; k = 1.399) and 2.8 kg of an
unknown gas. The mixture pressure and temperature are 290 KPa and 340 K. Determine
a) Molecular weight and gas constant of the unknown gas
b) the volumetric analysis
Given:
8.3143
Mx 
 72.93 kg/kg m
mN2 = 1.8 kg ; mx = 2.8 kg
0.114
P = 290 KPa ; T = 340K
xi 0.391 0.609
 Mi  28  72.93  0.014  0.0084
m  1.8  2.8  4.6 kg
1.8
xi
xN 2 
 0.391  39.1%
 Mi  0.0223
4.6
x x  0.609  60.9%
y N2  0.6278  62.78%
PV  mRT  mixture
y x  0.3722  37.22%
V  1(1)(1)  1 m3
290(1)
KJ
R 
 0.1854
4.6(340)
kg - K
R  xN2R N2  X xR x
R N2 
8.4143
KJ
 0.297
28
kg  K
0.1854  0.391(0.297)  0.609(Rx)
KJ
R x  0.114
kg
6. A closed vessel of 0.7 m-3Kinternal volume contains a gas at 58 KPa and 18C and with R = 0.27 KJ/kgK.If now 0.35 kg of another gas at 18C and R = 0.29 KJ/kg-K is also admitted into the vessel. Calculate
the final pressure of the mixture.
Given:
P1V  m1R1 T1
V = 0.7 m3
58(0.7)
m1 
 0.517 kg
P1 = 58 KPa; T1 = 18 + 273 = 291 K
0.27(291)
R1 = 0.27 KJ/kg-K
m  m1  m2  0.517  0.35
m2 = 0.35 kg
m  0.867 kg
T2 = 291 K
R2 = 0.29 KJ/kg-K
0.517
x1 
 0.5963
0.867
0.35
x2 
 0.4037
0.867
R  x1R1  x2R 2
R  0.5963(0.27)  0.4037(0.29)
KJ
R  0.278
kg - K
mRT 0.867(0.278)(291)
P

V
0.7
 100 .2are
KPa
7. In the figure shown the diameters of the twoPcylinders
75 mm and 600 mm, the face of the piston is 6 m above the
face of the weight “W” and the intervening passages is filled with oil (S = 0.80). What force F is required if W = 35 KN.
F
W = 35 KN
8.
If 1 100 N force F1 is applied to the piston with 5 cm diameter, what is the magnitude of the force F2 that can be resisted
by the piston with the 10 cm diameter? Neglect the weights of the piston.
F2
10 cm
oil
2m
F1
5 cm
2.2 m
9.
A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down into a tank
of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa, and neglecting vapor
pressure, how high will water rise in the tube. (Assume isothermal conditions for air)
1.5 m
1.2 - x
1.2 m
x
MANOMETER
1. A mercury manometer shown is connected to a pipeline carrying water ( = 9.81 KN/m3). If the elevation at point B is 3
m above A and the mercury reading y = 1200 mm, what is the pressure in the pipe in KPa.( of mercury = 133.4 KN/m3)
Open
y = 1200 mm
B
3m
A
2. A is water, Fluid B is oil (S = 0.80), z = 350 mm. Compute the pressure difference between m and n in KPa.
Fluid B
z
m
n
Fluid A
Example 2
What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has no
bottom.
80 cm
ID = 20 cm
400 cm
R = 160
cm
160 cm
water
h  1.6  4  5.6 m
V - Volume of imaginary volume of liquid
V  Vc ylinder - Vdome
V  (1.6)2 5.6  
1 4
3
 1.6 
2 3
2

2
V  1.6  5.6  1.6   36.46 m3
3


Tension in bolts  9.81(36.46 ) - 6  351.7 KN
Example 3
A hemispherical dome surrounds a closed tank as shown. If the tank and dome are filled with gasoline (S = 0.72) and the gage
indicates a pressure of 60 KPa. What is the total tension in the bolts holding the dome in place.
5.5 m
R=2m
Gasoline
S = 0.72
3.0 m
P =60 KPa
By extending the curved surface to the imaginary free surface of the liquid
with 0 gage
0  0.72(9.81) h  60
h  8.5 m
H - height of imaginary cylindrica l prism
H  8.5 - 3  5.5 m
the weight of imaginary fluid prism
V  Vcy linder - Vdome
14
2
3
V  2 5.5    2  52.4 m3
23
W  0.72(9.81)(52.4)  370 KN
Example 4
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm below the
interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What is the mass and
density of the block?
FLUID MECHANICS (PRE – FINAL S4)
March 15, 2017
Name ____________________________________
1. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
kg
m3
V1  0.08(0.08)(1.5)  0.96 m3
ρ w ood  0.651(1000)  651
W1  0.96(651)  6.25 kg
W2  11,200V2
BF1  (1.5  0.305)(0.08)21000  7.65 kg
BF2  1000V2
W1  W2  BF1  BF2
6.25  11,200V2  7.65  1000V2
7.65 - 6.25
 1.4 x 10- 4 m3
10,200
V2 
2.
W2  1.4 x 10 - 4 (11200)  1.54 kg
A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m
wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
W  BF
W1  W2  Vs
(150  35)(1000)  1000(5.5)(16)h
h  2.1m
3.
A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh
50 N. What is its weight in air and its specific gravity?
W1 
W2  50 Newton
V1  (0.20)(0.20)(0.4)  0.016 m3
1 
m1
V1
m1
W1
V1
S

1000 W1  W2
W1  m1g Newton
W1
W1
gV1

1000 W1  W2
W1
W1

9810 V1 W1  W2
1
1

9810 V1 W1  W2
W1  W2  9,810(0.016)
W1  156.96  50
W1  206.96 Newton
S
4.
206.96
 1.312
206.96 - 50
A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025).
Discharge of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh
water (S = 1)
W1  BF1
3630(1000)  A 1L(6.7)(1.025)(1000)
3,630
 Eq.1
6.7(1.025)
W2  BF2
A 1L 
(3,630  181)(1000)  A 2L(6.4)(1.025)(1000)
A 2L 
3449
 Eq.2
(6.4)(1.025)
FINAL EXAM (March 25, 2017)
SET 2
NAME _____________________________________
1.
An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S =
0.68). Determine the specific weight of the object.
2.
Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has
a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980
kg/m3;  = 9.6 KN/m3). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The
Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at
the same elevation.
3.
A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3
must be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
4.
At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second
point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first.
FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016)
Group No. 1
1
ABRERA, JEHOYAH CHRISTI E.
2
ANDAM, ROBERTO JR. M.
3
Arañas, Kieth d.
4
BONAYOG, STEVEN ANGEL M.
5
Cabral, Christian Mark o.
6
COLIPANO, HUBERT C.
7
DOFELIZ, MEAGAN ROSE G.
8
LLERA, KRISTIAN RAY D.
9
Loking, James Ralph v.
10 LOPEZ, NORMELYN G.
11 MENORO, ALCRIS JOYCE B.
12 Munalem, Paul John B
1. A circular gate 1.5 m in diameter is inclined at an angle of 45. Sea water (S = 1.02) stands on one side of the gate to a
height of 10 m above the center of the gate. Determine the total hydrostatic force F and the location of C.P.
F  hA  1.02(9.81)(10)
sin45 
F  hA
A
e
 2
D
4
Ig
h
y
10
y
 14.142 m
sin 45
A  1.767 m 2
Ig 
Ay
h
y
sin 45
D 4
Ig 
64

(1.5) 2  176.82 KN
4
yp 
D 4
 0.2485 m 4
64
Ig  A y
Ay
2
 14.15 m
2. An airplane barometer reads 762 mm Hg at sea level and 736 mm Hg at some unknown elevation. If the temperature at
sea level is 15 C, what is the elevation assuming:
a. air at constant density (h = 287.8 m)
b. air under isothermal condition(h = 292.8 m)
c. air under isentropic conditions (h = 291.35 m)
d. air condition follows standard atmosphere condition (h = 291.83 m)
SOLUTION:
762
101.325  101.6 KPa
Ps 
760
Ts  15  273  288 K
736
101.325  98.13 KPa
P
760
FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016)
Group no. 2
1
Abacahin, Ruvie Grace s.
2
Acosta, Faradiban O.
3
BUARON, EMMAN REY T.
4
CALIXTRO, JUNIEL F.
5
Ellevera, Glenn Eric P.
6
Mabanta, Chris Dilon F.
7
Pactores, Garret Oliver r.
8
Tapitan, Romel y.
9
Virtudazo, Erick Jan b.
10 TALADO, JOHN KENNETH E.
11 REYES, REYVENCER T.
12 Onda, Ruffy q.
1.
What force must be exerted through the bolts to hold the dome in place? The metal dome and pipe weighs 6 KN and has
no bottom.
h  1.6  4  5.6 m
V - Volume of imaginary volume of liquid
V  Vcy linder - V dome
1 4
3
V  (1.6) 2 5.6    1.6 
23
2

2
V  1.6  5.6  1.6   36.46 m 3
3


Tension in bolts  9.81(36.46 ) - 6  351.7 KN
2.
A U tube mercury manometer is connected to a closed pressurized tank as shown. If the air pressure is 14 KPa, determine
the differential reading h. The specific weight of the air is negligible.
14  9.81(0.6)  9.81(0.6)  9.81h  13.6(9.81)h  14
19.81(0.6)  9.81(0.6)  9.81h  13.6(9.81)h  0
h
(9.81)(1.2)
 0.095 m  9.5 cm
12.6(9.81)
FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016)
Group no. 3
1
VENCER, KYLE IAN B.
2
Salac, James Rhode E.
3
Nuñez, Insan John L.
4
ESTOSO, LADY LEE I.
5
Golosino Jr., Virgilio G.
6
Levi, Daniel L.
7
Jimenez, James V.
8
Enanod, Lourence A.
9
CARIGA, SAMS M.
10 BACUÑATA, ERROLJOHN B.
11 Agbu, Jovan o.
12 ABERO, SWEETSEL MAE M.
1. A BOEING 747 flies at an altitude of 2500 m above sea level where Ps = 101.33 KPa and Ts = 288 K. Determine the
pressure and temperature at this altitude considering
a. air under isothermal condition (P2 = 75.31 KPa ; T = 288 K)
b. air under isentropic conditions (P2 = 74.31 KPa ; T = 263.6 K)
c. air condition follows standard atmosphere condition (P2 = 74.66 KPa ; T = 271.75 K)
h  2500 m
Ps  101.33 KPa
Ts  288 K
FLUID MECHANICS (ACTIVITY NO. 2/January 14, 2016)
Group no. 4
1
ABUG, BON JASON A.
2
Amarille, Freniel Ian A.
3
BUNTAG, GILBERT G.
4
Dedumo, Francis Dominique C.
5
Estoya, Mark Louie s.
6
Joloyohoy, Ful Yuri a.
7
Mamucay, Jirah o.
8
Paran, Marc Joseph L.
9
Sobrepeña, Ferden s.
10 Tuble, Ian Stephen d.
11 Igot, Neeco Adrienn r.
12
1. The closed tank in the figure is filled with water. The pressure gage on the tank reads 48 KPa. Determine
a. The height h in mm in the open water column
b. The gage pressure acting on the bottom of the tank surface AB
c. The absolute pressure of the air in the top of the tank if the local atmospheric pressure is 101 KPa absolute.
48  0.60(9.81)  h(9.81)  0
h  5.5 m
48  0.6(9.81)  (0.6)(9.81 )  PAB
PAB  59.772 KPa
Pabs  48  101  149 KPa
2. The air above the liquid is under a pressure of 40 KPa gage, and the specific gravity of the liquid in the tank is 0.80. If the
rectangular gate is 1 m wide and if y1 = 1 m and y2 = 3 m , What force P is required to hold the gate in place?
SOLUTION
F  40  0.8(9.81)( 2.5)(1)(3)  178.86 KN
Ig
e
Ay
1(3) 3
 2.25 m 4
12
A y  3(2.5)  7.5 m 3
Ig 
e  0 .3 m
M
@ Hinge
0
F(1.5  0.3)  3P
p
178.86(1.8)
 107.32 KN
3
FLUID MECHANICS QUIZ NO. 3
1.
A block of wood has a vertical projection of 15.24 cm when placed in water and 10.2 cm when placed in alcohol. If the
specific gravity of alcohol is 0.82, find the specific gravity of wood gravity of wood.
In water
W  BF  1000A(h - 0.1524)  1
In Alcohol
W  BF  0.82(1000) A(h - 0.102)  2
Equating eq. 1 and eq. 2
(h - 0.1524)  0.82(h - 0.102)
h - 0.82h  0.1524  0.82(0.102 )
h  0.382 m
W  1000(0.2296)A  3
W   B A(0.382)  4
eq.3  eq. 4
1000(0.229 6)  0.382 B
SB 
B
0.2296

 0.601
1000 0.382
2.
A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface 1.50 cm
below the interface. The density of the oil is 790 kg/m3 . What is the gage pressure at the lower face of the block? What
is the mass and density of the block?
3.
Compartments A and B of the tank shown in the figure below are closed and filled with air and a liquid with S = 0.6. If
the atmospheric pressure is 101 KPa (abs) and the pressure gage reads 3.5 KPa (gage), determine the manometer reading
h in cm.
3.5  9.81h  0.6(9.81)h  0.6(9.81)0.02  13.6(9.81)(0.03)  0
h
The gate shown is hinged at A and rests on a smooth floor at B. The gate is 3 m square. Oil (S = 0.80) stands on the left
side of the gate to a height of 1.5 m above A. Above the oil surface is a gas under a gage pressure of -6.9 KPa. Determine
the amount of the vertical force P applied at B that would be required to open the gate.
P = -6.9 KPag
Free Surface
1.5m
h
A
F
 CG
 CP
45
m
P
3
4.
3.5  0.6(9.81)0.02  13.6(9.81)(0.03)
 1.9 m
9.81(1  0.6)
h  1.5  1.5 sin 45  2.56 m
F   6.9  0.80(9.81)2.56(3)(3)  118.8 KN
5.
A spherical buoy 2 m in diameter floats half submerge in a liquid with S = 1.5. What is the weight of the lead anchor weighing
7000 kg/m3 will completely submerged the buoy in the liquid.
 L  1.5(1000)  1500
kg
m3
4
R 3  4.18 m 3
3
1 4

Vs   (1)3   2.09 m 3  volume submerge
23

W1  BF1   L ( Vs )
Vsphere 
W1  1500(2.09)  3135 kg
from figure 2 :
W1  W2  BF1  BF2
BF1  1500(Vsphere )  1500(4.18)  6270 kg
W1  3135 kg
W2  7000 V2
BF2  1500 V2
3135  7000 V2  6270  1500 V2
V2  0.57 m 3
W2  3990 kg
6.
A glass tube 1.5 m long and 25 mm diameter with one end closed is inserted vertically with the open end down
into a tank of water until the open end is submerged to a depth of 1.2 m. If the barometric pressure is 98 KPa,
and neglecting vapor pressure, how high will water rise in the tube. (Assume isothermal conditions for air)
fOR isothermal :
P1V1  P2 V2

(0.025)2 (1.5)  0.00074 m3
4
P2  98  9.81(1.2  x )  1
V1 

(0.025)2 (1.5  x )  2
4




98  (0.025)2 (1.5)  98  9.81(1.2  x ) (0.025)2 (1.5  x )
4
4




98(1.5)  98  9.81(1.2  x )(1.5  x )
V2 
147  (98  11.772  9.81x )(1.5  x )
147  (109.772  9.81x )(1.5  x )
147  164.658  109.772x  14.715 x  9.81x 2
9.81x 2  124.487 x  17.658  0
x 2  12.7 x  1.8  0
 12.7  12.98
2
x  0.14 m  14 cm
x
1.
A brass cube 152.4 mm on a side weighs 298.2 N. We want to hold this cube in equilibrium under water by attaching a
light foam buoy to it. If the foam weighs 707.3 N/m3, what is the minimum required volume of the buoy?
W1  298.2 N
BF1  (0.1524) 3 (9810)  34.72 N
W2  707.3 V2  1
BF2  V2 (9810)  2
298.2  707.3 V2  34.72  V2 (9810)
V2 
2.
298.2  34.72
 0.029 m 3
9810  707.3
A man dives into a lake and tries to lift a large rock weighing 170 kg. If the density of the granite rock is 2700 kg/m3,
find the force that the man needs to apply to lift it from the bottom of the lake. Assume density of lake water to be 1000
kg/m3.
W  2700 V  170 kg
170
 0.063 m 3
2700
BF  1000(0.063)  63 kg
W  BF  T
V
T  170 - 63  107 kg
Example 1
A large pipe called a penstock in hydraulic work is 1.5 m in diameter. Here it is composed of wooden staves bound together by
steel hoops each 3.23 cm2 in area, and is used to conduct water from a reservoir to a powerhouse. If the maximum tensile stress
permitted in the hoops is 130 MPa, what is the maximum spacing between hoops under a head of 30.5 m.
1.5 m
L
L
P  9.81(30.5)  299 .205 KPa
D  1.5 m
On the vertical projection
F
; A  LD
A
F  299.205(1. 5)L  448 .81L KN
P
F
 224 .4L KN
2
Tensile Stress on the hoop
T 
T
; A  3.23 cm2
A
A  0.000323 m2
S
S  130000 KPa
224.4L
0.000323
L  0.187 m  18.7 cm
130000 
Example 2
Compute the wall stress in a 1200 mm diameter steel pipe 6 mm thick under a pressure of 970 KPa.
Given:
D = 1.2 m
t = 0.006 m
P = 970 KPa
S
PD 970(1.2)

 97,000 KPa
2t
2(0.006)
Example 3
What is the minimum allowable thickness of 600 mm diameter steel pipe under an internal pressure of 860
KPa with a working stress in the steel of 70,000KPa.
S
PD
2t
860(0.600)
2t
t  0.004 m  4 mm
70,000 
Example 4
A wood stave pipe is bound by steel rods which take the entire bursting stress. Find the proper spacing for
25 mm steel rods for a 1800 mm diameter wood stave pipe under a pressure of 590 KPa if the working
stress in the steel is 105,000 KPa.
Given
Dr = 0.025 m
D = 1.8 m
P = 590 KPa
S = 105,000 KPa
F
2T

A DL
For the rods
P
S
T

0.0252
4
105,000 
4T
0.025 
2
T  51.54 KN
L 
2(51.54)
 0.10 m
590(1.8)
L  10 cm
Example 5
A vertical cylindrical tank, 2 m in diameter and 4 m high, is held together by means of two steel hoops, one at the top and one at
the bottom. When molasses (S = 1.50) stands to a depth of 3 m in the tank, what is the tensile force in each hoop?
A. NOZZLE
2
2
P1 v 1
P v

 Z 1  2  2  Z 2  HL
 2g

2g
From continuity equation: Q = Av ;
for 1 = 2 = 
Q  A1v1  A2 v2
For a nozzle the head loss HL is equal to:
 1
v
 2  1 2
C
 2g
 v

2
where: Cv - velocity coefficient
JET POWER
2
PJet  Q
v2
2g
B. VENTURI METER
P1 v12
P
v 2

 z1  2  2  z 2
γ 2g
γ
2g
A1v1  A 2 v 2
2
d 
v1   2  v 2
 d1 
4
d 
v1   2  v 2 2
 d1 
In the meter
2
P1  γ w (h) - γ Hg (h)  P2
P1 - P2  h(γ Hg  γ w )
P1  P2
v 2  v12 v 2 2
 (z1  z 2 )  2

γ
2g
2g
v2 
  d 4 
1   2  
  d1  


P  P

2g  1 2  (z1  z 2 )
γ


  d 4 
1   2  
  d1  


C. ORIFICE
An orifice is an opening with a closed perimeter in which fluid flows.
By applying Bernoulli's Energy theorem:
2
2
P1 v 1
P v

 Z1  2  2  Z2
 2g
 2g
But P1 = P2 = Pa and v1 is negligible, then
2
v2
 Z1  Z 2
2g
and from figure: Z1 - Z2 = h, therefore
2
v2
h
2g
v 2  2gh
let v2 = vt
v t  2gh
where:
vt - theoretical velocity, m/sec
h - head producing the flow, meters
g - gravitational acceleration, m/sec2
COEFFICIENT OF VELOCITY (Cv)
actual velocity
theoretica l velocity
v'
Cv 
vt
Cv 
COEFFICIENT OF CONTRACTION
area of jet @ vena contracta
area of the orifice
a
Cc 
A
Cc 
where: a - area of jet at vena contracta, m2
A - area of the orifice, m2
COEFFICIENT OF DISCHARGE
actual flow
theoretica l flow
Q'
Cd 
Q
Cd  C v Cc
Cd 
where:
v' - actual velocity
vt - theoretical velocity
a - area of jet at vena contracta
A - area of orifice
Q' - actual flow
Q - theoretical flow
Cv - coefficient of velocity
Cc - coefficient of contraction
Cd - coefficient of discharge
JET TRAJECTORY:
v  velocity of water at tip of nozzle
at 1 to 0
v o  v 1 - 2gd
2
2
0  v 1 - 2gd
2
(vsin )2
1
2g
d
v 0  v 1  gt
v sin 
2
g
t
at 0 to 2
d  v 1t 
d0
d
1
2
1
1
2
2
gt 2
gt 2  3
(vsin )2

2g
t2 
t
gt 2
1
2
gt 2
(vsin )2
g2
(vsin )
4
g
R  v cos (2t )
R
2(vsin )( v cos ) v 2 (2sin cos )

g
g
(2sin cos )  sin( 2)
R
v 2 (sin2)
5
g
If the jet is flowing from a vertical orifice: If the jet is initially horizontal v x = v.
D. PUMP
Pump is a steady - state, steady - flow machine in which mechanical energy is added to the liquid from one point to another point
of higher pressure.
2
B
TANK
Points 1 & 2 are the reference points at suction and discharge, respectively.
a) Total Dynamic Head (Ht)
v 2  v 2 
 P2  P1 
1   Z  Z  H meters
   2
h  

2
1
L


t
γ
2g






b) Fluid or Water Power (FP)
FP  Qh t KW
c) Capacity or Discharge (Q)
Q  A1 v1  A 2 v 2 m3/sec
d) Brake or Shaft Power (BP)
2TN
KW
60,000
BP 
e) Motor Power (MP) (For motor driven pump)
1. For single - phase motors
MP 
EI (cos )
KW
1000
2. For 3 - phase motors
3 EI (cos )
KW
1000
MP 
f) Pump Efficiency
p 
FP
x 100%
BP
g) Motor Efficiency
m 
BP
x 100%
MP
h) Combined Pump-Motor Efficiency (Overall Efficiency)
c 
MP
x 100 %
IP
where:
ht - total dynamic head, m
FP - Fluid or Water Power, Kw
BP - Brake or Shaft Power, KW
MP - Motor Power. KW
IP - Power input to motor, KW
Q - Capacity or Discharge, m3/sec
 - Specific weight of the liquid pumped, KN/m3
v - velocity, m/sec
P - pressure, KPa
g - gravitational acceleration, m/sec2
Z - elevation, m (Positive if measured above datum, and negative if measured below datum)
HL - head loss (due to fluid friction, fittings and turbulence in pipes)
T - Brake torque, N-m
N - no. of RPM (revolutions per minute)
E - Volts
I - current drawn by the motor, Amperes
cos - Power Factor
E. HYDRAULIC TURBINE
Impulse Type (Pelton Type)
Reaction Type( Francis Turbine)
Dam
1
Head Water
Head Gates
(Fully Open
Penstock
Y – Gross Head
PA
A
Draft Tube
zA
B
2
Tail Water
Fundamental Equations
1. TOTAL DYNAMIC HEAD (h):
a. For an Impulse Type
h = (Z1 - Z2) - HL meters
h = Y - HL meters
Y = Z1 - Z2 meters
b. For a Reaction Type
h = (Z1 - Z2) - HL meters
h = Y - HL meters
Y = Z1 - Z2 meters
Y - gross head at plant, m
at A to 2
h
PA v 2A

 z A meters
 2g
2. WATER POWER (WP):
FP = Qh KW
3. DISCHARGE OR RATE OF FLOW (Q):
Q = Av m3/sec
4. BRAKE POWER (BP):
BP = 2TN
KW
60 000
T - brake torque, N-m
N - rotative speed, RPM
5. TURBINE EFFICIENCY (e)
e = BP x 100%
FP
e = evehem
where:
ev - volumetric efficiency
eh - hydraulic efficiency
em - mechanical efficiency
6. ROTATIVE SPEED (N):
N
120 f
n
where: f - frequency, cps or Hertz
n - no. of generator poles (usually divisible by 4)
QUIZ NO. 4
Problem No. 1
The velocity of water in a 10 cm diameter pipe is 3 m/sec. At the end of the pipe is a nozzle whose velocity coefficient is 0.98. If
the pressure in the pipe is 55 KPa, what is the velocity in the jet? What is the diameter of the jet? What is the rate of discharge?
What is the head loss?
d1  0.10 m
m
v1  3
sec
C v  0.98
P1  55 KPa
Applying Bernoulli' s equation
2
2
Problem No. 2
A centrifugal pump draws water from a well at the rate of 142 L/sec of water through a 203 mm ID suction line and a 152 mm ID
discharge line. The suction gauge located on the pump centerline reads 254 mm Hg vacuum, while the discharge gauge is 6 m
above the pump centerline. If the power input to the water is 75 KW, find the reading of the discharge gauge in KPa.
m3
sec
d1  0.203 m ; A 1  0.032 m2
Q  0.142
d2  0.152 m; A 2  0.018 m2
 2
d
4
Q
v
A
A
m
m
; v 2  7 .8
sec
sec
z1  0; z 2  6 m
v 1  4 .4
 101.325 
P1  -254
  33.86 KPa
 760 
WP  Qh t
ht 
75
 53.84 m
(0.142)(9.81)
ht 
P2  P1 v 2  v 1

 z 2  z1   HL

2g
2
2
2
2

 P
v  v1
P2   h t  2
 z 2  z1   HL   1
2g

 
P2  414.45 KPa
Problem No. 3
A 15 KW suction pump draws water from a suction line whose diameter is 200 mm and discharges through a line whose diameter
is 150 mm. The velocity in 150 mm line is 3.6 m/sec. If the pressure at point A in the suction line is 34.5 KPa below the
atmosphere where A is 1.8 m below that of B on the 150 mm line, Determine the maximum elevation above B to which water can
be raised assuming a head loss of 3 m due to friction.
FP
d1
d2
15
0.2
0.15
KW
m
m
v2
P1
P2
HL
A1
A2
Q
v1
SW
P1/SW
P2/SW
v1^2/2g
v2^2/2g
z1
z2
ht
D(Phead)
D(Vhead)
m/sec
KPa
KPa
m
m^2
m^2
m^3/sec
m/sec
KN/m^3
m
m
m
m
m
D(Ehead)
3.6
-34.5
0
3
0.03
0.02
0.064
2.025
9.81
-3.52
0.000
0.209
0.66
0.0
1.8 + h
24.04
3.52
0.45
(1.8 +
h)
h
15.27
m
m
Problem No. 4
A power nozzle throws a jet of water that is 50 mm in diameter. The diameter of the base of the nozzle and of the approach pipe
is150 mm. If the power of the nozzle jet is 42 HP and the pressure head at the base of the nozzle is 54 m, compute the head lost in
the nozzle.
Problem No. 5
A fire pump delivers water through a 150 mm main to a hydrant to which is connected a 75 mm hose, terminating in a 25 mm
nozzle. The nozzle is 1.5 m above the hydrant and 10 m above the pump. Assuming frictional losses of 3 m from the pump to the
hydrant, 2 m in the hydrant and, and 12 m from the hydrant to the base of the nozzle, and a loss in the nozzle of 6% of the velocity
head in the jet, to what vertical height can the jet be thrown if the gage pressure at the pump is 550KPa.
Problem No. 6
Water issues from a circular orifice under a head of 12 m. The diameter of the orifice is 10 cm. If the discharge is found to be 75
L/sec, what is the coefficient of discharge? If the diameter at the vena cotracta is measured to be 8 cm, what is the coefficient of
contraction and what is the coefficient of velocity.
v  2gh  theoretica l velocity
v  2(9.81)(12 )  15.34
Q'  0.075
m
sec
m3
sec

m3
(0.10)2 (15.34)  0.12
4
sec
Q'
Cd 
 0.625
Q
a d' 8
Cc   
 0.67
A D 12
Cd  Cc(Cv )
Q
Cv 
0.625
 0.93
0.67
Problem No. 7
A jet discharges from an orifice in a vertical plane under a head of 3.65 m. The diameter of the orifice is 3.75 cm and the measured
discharge is 6 L/sec. The coordinates of the centerline of the jet are 3.46 m horizontally from the vena contracta and 0.9 m below
the center of the orifice. Find the coefficient of discharge, velocity and contraction.
Problem No. 8
The inside diameters of the suction and discharge pipes of a pump are 20 cm and 15 cm, respectively. The discharge pressure is
read by a gage at a point 2 m above the centerline of the pump, and the suction pressure is read by a gage
1 m below the pump
centerline. If the pressure gage reads 145 KPa and the suction gage reads a vacuum of 250 mm Hg when diesel fuel (S = 0.82) is
pumped at the rate of 30 L/sec, Find the KW power of the driving motor if overall pump efficiency is 75%.
d1
d2
z1
z2
P1
P2
S
SW(water)
SW
Q
A1
A2
v1
v2
P1/SW
P2/SW
v1^2/2g
v2^2/2g
D(Phead)
D(Vhead)
D(ElHead)
HL
ht
WP
e
BP
0.20
0.15
-1
2
-33.33
145
0.82
9.81
8.0442
0.03
0.031
0.018
0.955
1.70
-4.14
18.03
0.05
0.15
22.17
0.10
3.00
0.00
25.27
6.10
0.75
8.13
m
m
m
m
Kpa
KPa
KN/m^3
KN/m^3
m^3/sec
m^2
m^2
m/sec
m/sec
m
m
m
m
m
m
m
m
m
KW
KW
Problem No. 9
A jet of water 7.6 cm in diameter discharges through a nozzle whose velocity coefficient is 0.96. If the pressure in the pipe is 82.7
KPa and the pipe diameter is 20 cm and if it is assumed that there is no contraction of the jet, what is the velocity at the tip of the
nozzle? What is the rate of discharge?
Q  A 1v 1  A 2 v 2
4
v
2
1
d  2
  2  v 2
 d1 
4
2
d  v 2
v1
  2  2
2g  d1  2g
2
2
P1 v 1
P v

 z1  2  2  z 2  HL
 2g

2g
2
2
 1
v 2
P1 v 1
P v

 z1  2  2  z 2   2  1 2
 2g

2g
 Cv
 2g
4
2
2
 1
v 2
P1  d2  v 2
P v
  
 z1  2  2  z 2   2  1 2
  d1  2g

2g
 Cv
 2g
2
v2
2g
d1
d2
Cv
P1
P2
v1
v2
SW
g
v2
A1
A2
Q
0.20
0.076
0.96
82.7
0
1.80
12.467
9.81
9.81
12.47
0.031
0.005
0.057
m
m
Kpa
KPa
m/sec
m/sec
KN/m^3
m/sec^2
m/sec
m^2
m^2
m^3/sec
4
 1

 d2   P1  P2



 1  1     
 ( z1  z 2 )
2

 C v
 d1  

4
2
v 2  1  d2   P1  P2
 2     
 ( z1  z 2 )
2g  C v

 d1  

 P1  P2

 ( z1  z 2 )


v2


2g
 1  d 4 
 2   2  
 C v
 d1  
2
SAMPLE PROBLEMS
APPLICATION OF BERNOULLI’S EQUATION
ExampleNo. 1
The water in a 10 m diameter, 2 m high aboveground swimming pool is to be emptied by unplugging a 3 cm diameter, 25 m long
horizontal pipe attached to the bottom of the pool. Determine the maximum discharge rate of water through the pipe.
1
h=2
m
2
P1 v12
P
v 2

 Z1  2  2  Z2
γ 2g
γ
2g
P1  0; v 1  0; P2  0; Z1  0; Z 2  -2 m
0000
2
Q  Av
π
Q  (0.03)2 (6.23)  0.0044 m3/sec
4
v22
 (2)
2g
v22
2g
v 2  2(9.81)( 2 )  6.23 m/sec
Example No. 2
A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is
now opened and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet.
1
P1 v12
P
v 2

 Z1  2  2  Z2
γ 2g
γ
2g
P1  0; v 1  0; P2  0; Z1  0; Z 2  - 5 m
0000
5
v22
 (5)
2g
v22
2g
v 2  2(9.81)5  9.623 m/sec
Example No. 3
The water level of a tank on a building roof is 20 m above the ground. A hose leads from the tank bottom to the ground. The end of
the hose has a nozzle, which is pointed straight up. What is the maximum height to which the water could rise.
P1 v12
P
v 2

 Z1  2  2  Z2
γ 2g
γ
2g
P1  0; P2  0; Z1  0; Z 2  h; v 2  0
0
v12
000h
2g
v12
2g
h  max imum height
h
v1  velocity at the tip of the nozzle
Example no. 4
Water flows through a horizontal pipe at the rate of 1 Gal./sec. The pipe consist of two sections of diameter 4 in. and 2 in with a
smooth reducing section. The pressure difference between the two pipe sections is measured by mercury manometer . Neglecting
frictional effects, determine the differential height of mercury between the two pipe sections.
Q = 1 gal/sec = 0.0038 m3/sec
D1 = 4 in. = 0.1016 m
D2 = 2 in = 0.0508 m
2
1
x
h
P1  9.81x  9.81h  13.6(9.81
)h  9.81x  P
Mercury S = 13.6 2
P2  P1  9.81h  13.6(9.81)h
P2  P1  123 .606  1
from 1 to 2
P1 v 1 2
P v 2

 Z1  2  2  Z2
γ 2g
γ 2g
Q
0.0038
v1 

 0.47 m/sec
A1 π
(0.1016 )2
4
Q
0.0038
v2 

 1.875 m/sec
π
A2
(0.0508 )2
4
2
v1
(0.47)2

 0.0113 m
2g 2(9.81)
v 2 2 (1.875)2

 0.18 m
2g
2(9.81)
P1
P
 0.0113  0  2  0.18  0
γ
γ
P2  P1
 0.0113  0.18  0.1687
γ
P2  P1  9.81(0.1687 )  1.655 KPa  2
Equating eq.1 and eq. 2
- 123.606h  -1.655
- 1.655
- 123.606
h  0.0134 meters
h
h  0.53 inches
Example No. 5
A 300 mm pipe is connected by a reducer to a 100 mm pipe. Points 1 and 2 are at the same elevation. The pressure at point 1 is
200 KPa. Q = 30 L/sec flowing from 1 to 2, and the energy lost between 1 and 2 is equivalent to 20 KPa. Compute the pressure at
2 if the liquid is oil with S = 0.80. (174.2 KPa)
P1 v12
P
v 2

 z1  2  2  z 2  H L
γ 2g
γ
2g
Q  A1v1  A 2 v 2
v
Q
π
; A  D2
A
4
v1 
0.030
m v12 (0.424 ) 2
 0.424
;

 0.009 m
π
sec 2g
2(9.81)
(0.3) 2
4
0.030
m v 2 2 (3.82 ) 2
 3.82
;

 0.744 m
π
sec 2g
2(9.81)
(0.1) 2
4
KN
γ  0.8(9.81)  7.848 3
m
20
HL 
 2.55 m
7.848
v2 
P1 v12
P
v 2

 Z1  2  2  Z 2  H L
γ 2g
γ
2g
Z1  Z2 : (Z 2 - Z1 )  0
P2 P1 v12 v 2 2
200
 

 HL 
 0.009  0.744  2.55
γ
γ 2g 2g
7.848
P2  174 .22 KPa
Example No. 6
A venturi meter having a diameter of 150 mm at the throat is installed in a 300 mm water main. In a differential gage partly filled
with mercury (the remainder of the tube being filled with water) and connected with the meter at the inlet and at the throat, what
would be the difference in level of the mercury columns if the discharge is 150 L/sec? Neglect loss of head. (h=273 mm)
L
1m 3
m3
x
 0.150
sec 1000 L
sec
S of mercury  13.6
Q  150
γ  13.6(9.81)  133.416
KN
m3
d1  0.30 m
d 2  0.15 m
Q  0.150 m 3 / sec
HL  0
v
Q
A
v1 
v2 
0.150
π
(0.30 ) 2
4
0.150
π
(0.15 ) 2
4
 2.122 m / sec;
v12
 0.230 m
2g
 8.488 m / sec;
v22
 3.672 m
2g
P1 v12
P v 2

 Z1  2  2  Z2
γ 2g
γ
2g
Z1  Z2  0
P1  P2 v 2 2 v12


 3.672  0.230
γ
2 g 2g
P1  P2
 3.442 m
γ
P1  P2  3.442 (9.81)  33 .766 KPa
P1  9.81( x )  9.81h  133 .416 h  9.81x  P2
P1  9.81h  133 .416 h  P2
P1  P2  133 .416 h  9.81h
P1  P2  h (133 .416  9.81)
P1  P2
33 .766

(133 .416  9.81) (133 .416  9.81)
h  0.273 m
h  273 mm
h
Example no. 7
A mechanical engineer of an industrial plant is required to install a centrifugal pump to lift 15 L/sec of water from a sump to a
storage tank on a tower. The water is to be delivered into a 105 KPag tank and the water level in the tank is 20 m above the water
level in the sump. Pump centerline is 4 m above the water level in the sump. The suction pipe is 100 mm diameter and discharge
pipe is 65 mm diameter. Head loss at suction is 3 times the velocity head in the suction line and head loss at discharge is 20 times
the velocity head in the discharge pipeline. Other data are as follows:
p = 75 %
m = 80 %
E = 220 Volts
Motor – 3-Phase
Power Factor = 0.92
Requirements:
a. Sketch of the problem
b. Total dynamic head in m
c. Water Power in KW
d. Motor power in KW
e. Line current drawn by the motor in amperes
f. Total power cost per day for 10 hours a day continuous operation and a power costs of P 5.00/KW-hr
105 KPa
2
65 mm
100 mm
4m
1
Q  Av
Q
A
Q
0.015
m
vs 

 1.91
A s π (0.10 ) 2
sec
4
Q
0.015
m
vd 

 4.52
As π (0.065 ) 2
sec
4
v
vs 2 (1.91)2

 0.186 m
2g 2(9.81)
vd 2 (4.52 )2

 1.041 m
2g 2(9.81)
v 2
h Ls  3 s   0.56 m
 2g 


v 2
h Ld  20  d   20 .83 m
 2g 


20
m
H L  0.56  20.83  21 .39 m
P2  P1 v 2 2  v12

 Z2  Z1  H L
γ
2g
at 1 to 2, datum line through point 1
ht 
105 - 0
 0  (20  0)  20 .83
9.81
h t  52 .091 m
ht 
WP  0.015(9.81 )(52.091)  7.67 KW
7.67
 10 .22 KW
0.75
10.22
MP 
 12 .8 KW
0.75
BP 
MP 
3 EI(P.F.)
1000
3 (220)I(0.9 2)
1000
I  36 .44 amperes
12.8 
Cost  12 .8(10 )(5)  P639 .00
Example
No.
8
Figure below shows a siphon discharging oil (sp gr 0.90). The siphon is composed of 8 cm. pipe from A to B followed by 10 cm.
pipe from B to the open discharge at C. The head losses are from 1 to 2, 0.34 m; from 2 to 3, 0.2 m; from 3 to 4;0.8 m. Compute
the
discharge
in
L/sec
4.5 m
3m
0
Example No. 9
The liquid in the figure has a specific gravity of 1.5. The gas pressure P A is 35 KPa and PBis -15 KPa. The orifice is 100 mm in
diameter with Cd = Cv = 0.95. Determine the velocity in the jet and the discharge when h = 1.2. (9.025 m/sec; 0.071 m 3/sec)
PA
Without considering head lost
1
P1 v12
P v 2

 Z1  2  2  Z 2
γ 2g
γ
2g
Datum at point 1 :
h
PB
2
Z1  0 ; Z 2  - h; v1  0
v 2 2 P1  P2

 Z1  Z 2
2g
γ
P  P

v 2  2g  1 2  h   theoretical velocity
γ


KN
γ  1.5(9.81)  14 .715
m3
 35  15

v 2  v t  2(9.81) 
 1.2  9.5 m/sec
14
.
715


Example no. 10
A pump draws water from reservoir A and lifts it to reservoir B as shown in the figure. The loss of head from A to 1 is 3 times the
velocity head in the 150 mm pipe and the loss of head from 2 to B is 20 times the velocity head in the 100 mm pipe. Compute the
horsepower output of the pump and the pressure heads at 1 and 2 when the discharge is 20 L/sec. (FP= 20.73 HP; 5.74 m ; 84.3 m)
2
72 m
100 mm
1
A
6m
B
I-1
150
mm
20
m3
 0.020
1000
sec
Q
v
A
0.020
vs 
 1.13 m/sec
π
(0.150 ) 2
4
Q
vd 
0.020
I-2
V-1
V-2
pump
Water Power  0.020(9.81 )(78.63)
Water Power  15 .43 KW  20 .7 HP
At 1 to A (Point 1 on datum)
P1 v12
P
v2

 Z1  A  S  ZB  h Ls
γ 2g
γ
2g
 2.55 m / sec
π
(0.100 ) 2
4
v 2
h Ls  3 s   0.20 m
 2g 


000
v 2
h Ld  20  d   6.63 m
 2g 


PA (1.13) 2

 6  0.20
γ 2(9.81)
PA
 5.73 meters ; PA  56 .2 KPa
γ
At B to 2 (Point B on datum)
PB vd 2
P v 2

 ZB  2  2  Z2  h Ld
γ
2g
γ 2g
H L  h Ls  h Ld  6.83 m
At point 1 to 2
PB 2.55 ) 2

 0  0  0  78  6.63
γ 2(9.81)
P1 v12
P v 2

 Z1  ht  2  2  Z2  H L
γ 2g
γ 2g
v1 and v 2 are negligible
P1  P2  0 gage (Atmospheric)
h t  Z2  Z1  H L
h t  72  6.83  78 .63 m
PB
 84 .3 meters; PB  827 KPa
γ
Example No. 11
The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100 mm,
respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction pressure gauge
which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction, turbulence and fittings
amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec, find
a. The total dynamic head developed by the pump
b. The fluid power in KW
c. The brake or shaft power delivered to the fluid for a pump efficiency of 75%
d. The brake torque if the pump speed is 1200 RPM
e. The electrical power input to the pump motor for a motor efficiency of 92%
f. The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor
g. The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q
S
SW(water)
SW
g
d1
d2
P1
P2
z1
z2
HL
FP
A1
A2
v1
v2
v1^2/2g
v2^2/2g
P1/sw
P2/sw
D(Phead)
D(vhead)
D(Zhead)
ht
Water Power
Pump
Efficiency
Brake Power
Motor
Efficiency
Motor Power
E (Volts)
Power Factor
Phase
N(RPM)
Torque(N-m)
Line Current
PowerCost
Cost
0.035
0.75
9.81
7.3575
9.81
0.150
0.100
-60.00
320.00
-4
10
15
75
0.018
0.008
2.0
4.5
0.20
1.01
-8.15
43.49
51.65
0.81
14
81.46
20.98
m^3/sec
75.00
%
27.97
KW
92.00
%
30.40
220.00
0.90
3.00
1200.00
222.57
88.65
1.50
228.01
KW
Volts
KN/m^3
KN/m^3
m/sec^2
m
m
KPa
KPa
m
m
m
KW
m^2
m^2
m/sec
m/sec
m
m
m
m
m
m
m
m
KW
Phase
RPM
N-m
Amperes
Pesos/KW-hr
Pesos
Example No. 12
A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec. Neglecting air
friction, determine its diameter at a point 4 m above A.
d1
v1
P1
P2
g
A1
Q
SW
P1/SW
P2/SW
v1^2/2g
z1
z2
HL
v2^2/2g
v2
A2
d2
d2
0.075
10
0
0
9.81
0.0044
0.0442
9.81
0
0
5.097
0
4
0
9.097
13.36
0.0033
0.065
6.5
m
m/sec
Kpa
KPa
m/sec^2
m^2
m^3/sec
KN/m^3
m
m
m
m
m
m
m
m/sec
m^2
m
cm
Example No. 13
A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829 N/cm 2 above
atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow of water from orifice
if Cd = 0.6.
Example No. 14
The 600 mm pipe shown in the figure conducts water from a reservoir A to a pressure turbine, which discharges through another
600 mm pipe into tailrace B. The loss of head from A to 1 is 5 times the velocity head in the pipe and the loss of head from 2 to B
is 0.2 times the velocity head in the pipe. If the discharge is 700 L/sec , what horsepower is being given up by the water to the
turbine and what are the pressure heads at 1 and 2.(FP = 537.4 HP; 53.628 m; -4.75 m)
Q  700 L/sec  0.70 m 3 /sec
v
Q 4(0.70)

 2.5 m/sec
A (0.6)2
v2
( 2 .5 ) 2

 0 .3 2 m
2g 2(9.81)
hLA -1  5(0.320  1.6 m
hL2-B  0.20(0.32)  0.064 m
HL  1.6  0.064  1.664 meters
At A to B
2
2
PA v A
P
v

 Z A  B  B  Z B  HL  h

2g

2g
With datum line through poin B
Z B  0; Z A  60 m
PA  PB  0 gage
v A  VB  0
h  ( Z A  Z B )  HL
h  60  1.664  58.336 meters
WP  Qh  0.70(9.81) (58.336)  400.6 KW  537 HP
h
P1
g
SW
P1/SW
d(orifice)
A(Orifice)
Theoretical
Velocity
Cd
Q(m^3/sec)
Q(L/sec)
2
88.29
9.81
9.81
9.0
0.15
0.018
m
Kpa
m/sec^2
KN/m^3
m
m
m^2
14.69
m/sec
0.60
0.1558
155.8
m^3/sec
L/sec
At A to 1
PA v A 2
P v2

 ZA  1  1  Z1  h LA1
γ
2g
γ 2g
with datum through point A
ZA  0 ; Z1  -55.5 m
000 
P1
 0.32  55 .5
γ
P1
 55 .5  0.32  55 .18 meters
γ
P1  55.18(9.81 )  541.3 KPa
At 2 to B, with datum through point 2
Z2  0 ; Z B  -4.5 m
P2 v 2 2
P
v 2

 Z2  B  B  ZB  h L 2  B
γ
2g
γ
2g
P2
 0.32  0  0  0  4.5  0.064
γ
P2
 4.5  0.32  0.064  4.756 meters
γ
P2  -4.756(9.8 1)  -46.7 KPa  46.7 KPa vacuum
FLUID MECHANICS (PRE – FINAL S4)
March 15, 2017
Name ____________________________________
5. A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
ρ wood  0.651(1000 )  651
kg
m3
V1  0.08 (0.08 )(1.5)  0.96 m 3
W1  0.96 (651)  624.96 kg
W2  11,200V 2
BF1  (1.5  0.305 )1000  1195 kg
BF2  1000V 2
W1  W2  BF1  BF2
624.96  11,200V 2  1195  1000V 2
1195 - 624.96
 0.056 m 3
10,200
W2  625.93 kg
V2 
6.
A barge is loaded with 150 Metric tons of coal. The weight of the empty barge in air is 35 Metric ton. If the barge is 5.5 m
wide, 16 m long and 3 m high, what is its draft. (Depth below the water surface)
W  BF
W1  W2  ρVs
(150  35)(1000 )  1000 (5.5)(16 )h
h  2.1 m
7.
A prismatic object 20 cm thick by 20 cm wide by 40 cm long is weighed in water at a depth of 50 cm and found to weigh
50 N. What is its weight in air and its specific gravity?
W1 
W2  50 Newton
V1  (0.20 )( 0.20 )( 0.4)  0.016 m 3
ρ1 
m1
V1
m1
V
W1
S 1 
1000 W1  W2
W1  m1g Newton
W1
gV1
W1

1000 W1  W2
W1
W1

9810 V1 W1  W2
1
1

9810 V1 W1  W2
W1  W2  9,810 (0.016 )
W1  156.96  50
W1  206.96 Newton
S
8.
206.96
 1.312
206.96 - 50
A ship, with vertical sides near the water line, weighs 3630 Metric ton and draws 6.7 m in saltwater (S = 1.025).
Discharge of 181 Metric ton of water ballast decreases the draft to 6.4 m. What would be the draft d of the ship in fresh
water (S = 1)
W1  BF1
3630 (1000 )  A1L(6.7)(1.025 )(1000 )
3,630
 Eq .1
6.7(1.025 )
W2  BF2
A1L 
(3,630  181)(1000 )  A 2 L(6.4)(1.025 )(1000 )
A 2L 
3449
 Eq .2
(6.4)(1.025 )
In a hydroelectric power plant, the water surface on the crest of the dam is at elevation 75.3 m while the water surface just at the
outlet of the head gate is at elevation 70.4 m. The head gate has 5 gates of 0.91 m x 0.91 m leading to the penstock and are fully
opened. Assume 61% as coefficient of discharge, determine
a. The quantity of water that enters the hydraulic turbine in m3/sec
b. The KW power that the turbine will developed, assuming e turbine = 90% efficiency and the turbine is 122 m below the
entrance of the penstock
c. The brake torque for a speed N = 1800 rpm
d. The number of generator poles if f = 60 Hertz
e. The electrical power developed by the generator if electrical and windage loses amounts to 18%
h  75.3  70.4  4.9 m  head producing the flow
v  2gh  theoretica l velocity
Q  5(Av)  theoretica l flow for 5 gates
Q'  C d Q  Actual flow
Q'  0.61(5)(0. 91x0.91) 2(9.81)4.9  24.76
m3
sec
h t  total dynamic head
h t  122  4.9  126.9 m
BP  e Turbine Q' h t  24.76(9.81)(126.9)(0.90)  27,746.2 KW
2TN
60,000
T  147,198 N  m
BP 
GP  22, 751KW
FINAL EXAM (March 25, 2017)
SET 2
NAME _____________________________________
1.
An object weighs 25.95 N when submerged in kerosene (S = 0.81) and weighs 26.6 N when submerged in gasoline (S =
0.68). Determine the specific weight of the object.
  6.23(9.81)  61.116
KN
m3
2.
Determine the water power and mechanical efficiency of a centrifugal pump which has an input of 3 KW. If the pump has
a 203 mm diameter suction line and a 152 mm diameter discharge line and handles 10 L/sec of water at 66C ( =980
kg/m3;  = 9.6 KN/m3). The suction line gauge shows 102 mm Hg vacuum and the discharge gauge shows 180 KPa. The
Discharge gauge is located 61 cm above the center of the discharge pipeline and the pump inlet and discharge lines are at
the same elevation.
BP
3
KW
D1
0.203
m
D2
0.152
m
Q
0.01
m^3/sec
A1
0.032
m^2
A2
0.018
m^2
v1
0.309
m/sec
v2
0.551
m/sec
g
9.810
m/sec^2
v1^2/2g
0.005
m
v2^2/2g
0.015
m
SW
9.6
KN/m^3
P1
-13.60
Kpa
P2
180
KPa
z2
0.61
m
z1
0
m
(P2-P1)/sw 20.17
m
(v2^20.01
m
v1^2)/2g
(z2-z1)
0.61
m
HL
0.00
m
Ht
20.79
m
WP
1.996
KW
em
66.52
%
3.
A piece of wood of S = 0.651 is 8 cm square and 150 cm long. How many kilograms of lead weighing 11,200 kg/m 3 must
be fastened at one end of the stick so that it will float upright with 30.5 cm out of water?
S  0.651
ρ wood  0.651(1000 )  651
kg
m3
V1  0.08 (0.08 )(1.5)  0.96 m 3
W1  0.96 (651)  624.96 kg
W2  11,200V 2
BF1  (1.5  0.305 )1000  1195 kg
BF2  1000V 2
W1  W2  BF1  BF2
624.96  11,200V 2  1195  1000V 2
1195 - 624.96
 0.056 m 3
10,200
W2  625.93 kg
V2 
4.
At one point in a pipeline the water speed is 3 m/sec and the gage pressure is 50 KPa. Find the gage pressure at a second
point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice at the first.
FLUID MECHANICS (PRE – FINAL S2)
March 04, 2017
Name ____________________________________
1.
The diameters of the suction and discharge pipe of a fuel pump for a day tank of a gasoline engine are 150 mm and 100
mm, respectively. The discharge pressure gauge located 10 m above the pump centerline reads 320 KPa and the suction
pressure gauge which is 4 m below the pump centerline reads a vacuum of 60 KPa. Head losses due to pipe friction,
turbulence and fittings amounts to 15 m. If gasoline with a relative density S = 0.75 is pumped at the rate of 35 L/sec,
find
a.
b.
c.
d.
e.
f.
g.
The total dynamic head developed by the pump
The fluid power in KW
The brake or shaft power delivered to the fluid for a pump efficiency of 75%
The brake torque if the pump speed is 1200 RPM
The electrical power input to the pump motor for a motor efficiency of 92%
The line current drawn by the motor if the motor is 3 – phase, 240 volts, and 0.9 Power Factor
The cost of power for 5 hours operation, if electricity costs P 1.50/KW-hr
Q
S
SW(water)
SW
g
d1
d2
P1
P2
z1
z2
HL
FP
A1
A2
v1
v2
v1^2/2g
v2^2/2g
P1/sw
P2/sw
D(Phead)
D(vhead)
D(Zhead)
ht
Water Power
Pump
Efficiency
Brake Power
Motor
Efficiency
Motor Power
E (Volts)
Power Factor
Phase
N(RPM)
Torque(N-m)
Line Current
PowerCost
Cost
0.035
0.75
9.81
7.3575
9.81
0.150
0.100
-60.00
320.00
-4
10
15
75
0.018
0.008
2.0
4.5
0.20
1.01
-8.15
43.49
51.65
0.81
14
81.46
20.98
m^3/sec
75.00
%
27.97
KW
92.00
%
30.40
220.00
0.90
3.00
1200.00
222.57
88.65
1.50
228.01
KW
Volts
KN/m^3
KN/m^3
m/sec^2
m
m
KPa
KPa
m
m
m
KW
m^2
m^2
m/sec
m/sec
m
m
m
m
m
m
m
m
KW
Phase
RPM
N-m
Amperes
Pesos/KW-hr
Pesos
2.
A jet of liquid is directly vertically upward. At A (Nozzle tip) its diameter is 75 mm and its velocity is 10 m/sec.
Neglecting air friction, determine its diameter at a point 4 m above A.
d1
v1
P1
P2
g
A1
Q
SW
P1/SW
P2/SW
v1^2/2g
z1
z2
HL
v2^2/2g
v2
A2
d2
d2
3.
0.075
10
0
0
9.81
0.0044
0.0442
9.81
0
0
5.097
0
4
0
9.097
13.36
0.0033
0.065
6.5
m
m/sec
Kpa
KPa
m/sec^2
m^2
m^3/sec
KN/m^3
m
m
m
m
m
m
m
m/sec
m^2
m
cm
A closed vessel contains water up to a height of 2 m and over the water surface there is air having a pressure of 8.829
N/cm2 above atmospheric pressure. At the bottom of the vessel there is an orifice of diameter 15 cm. Find the rate of flow
of water from orifice if Cd = 0.6.
h
P1
g
SW
P1/SW
d(orifice)
A(Orifice)
Theoretical
Velocity
Cd
Q(m^3/sec)
Q(L/sec)
2
88.29
9.81
9.81
9.0
0.15
0.018
m
Kpa
m/sec^2
KN/m^3
m
m
m^2
14.69
m/sec
0.60
0.1558
155.8
m^3/sec
L/sec
Example No. 15
A Francis turbine is installed with a vertical draft tube. The pressure gauge located at the penstock leading to the turbine casing
reads 372.6 KPa and velocity of water at inlet is 6 m/sec. The discharge is 2.5 m 3/sec. The hydraulic efficiency is 85%, and the
overall efficiency is 82%. The top of the draft tube is 1.5 m below the centerline of the spiral casing, while the tailrace level is 2.5
m from the top of the draft tube. There is no velocity of whirl at the top or bottom of the draft tube and leakage losses are
negligible. Calculate,
a) the net effective head in meters (43.817 m)
b) the brake power in kw. (881.2 kw)
c) the plant output for a generator efficiency of 92%. (810.7 kw)
d) the mechanical efficiency (96.550)
GIVEN:
P1 = 372.6 KPa
v = 6 m/sec
Q = 2.5 m3/sec
eh = 0.85
e = 0.82
ZB = (1.5 + 2.5) = 4 m
at point b to 2 (datum line through point 2)
PA v A 2
P
v 2

 ZA  2  2  Z2  h L A  2  h
γ
2g
γ
2g
P2  0
v2  0
Z2  0
h L A  2  0 (negligibl e)
h
PA v A 2

 ZA
γ
2g
372 .6
(6) 2

4
9.81 2(9.81)
h  43 .817 m
WP  Qγh
h
WP  2.5(9.81)( 43 .817 )  1074 .6 kw
BP  1074.6(0.8 2)  881 .2 KW
PLANT OUTPUT  881.2(0.92 )  810 .7 KW
e  emeh e v
0.82  e m (0.85 )(1)
em 
0.82
 0.965  96 .5%
0.85
Example no. 16
A 4 m3/hr pump delivers water to a pressure tank. At the start, the gauge reads 138 KPa until it reads 276 KPa and then the pump
was shut off. The volume of the tank is 160 Liters. At 276 KPa, the water occupied 2/3 of the tank volume.
a) Determine the volume of water that can be taken out until the gauge reads 138 KPa.
b) If 1 m3/hr of water is constantly used, in how many minutes from 138 KPa will the pump run until the
gauge reads 276 KPa.
P1  138  101 .325  239 .325 KPa
P2  276  101.325  377.325 KPa
1
V2  (0.160 )  0.0533 m3
3
P1V1  P2 V2
V1  0.084 m3
Vw@138  (0.160 - 0.084)  0.076  76 Liters
2
(0.16 )  0.1067  106 .7 Liters
3
Vtaken out  0.1067  0.076  0.0307  30 .7 Liters
Vw @ 276 
4 t  1t  0.0307
t  0.01023 hrs  0.614 min  36 .84 sec
HEAD LOSSES
HL = Major loss + Minor losses
Major Loss: Head loss due to friction and turbulence in pipes
Minor Losses: Minor losses include, losses due to valves and fittings, enlargement, contraction, pipe entrance and pipe exit. Minor
losses are most easily obtained in terms of equivalent length of pipe "Le".the advantage of this approach is that both pipe and
fittings are expressed in terms of "Equivalent Length" of pipe of the same relative roughness.
Considering Major loss only
Darcy-Weisbach Equation
hf 
fLv 2
meters
2gD
Considering Major and Minor losses
f(L  L )v 2
e
hf 
meters
2gD
where; f - friction factor from Moody's Chart
L - length of pipe, m
Le - equivalent length in straight pipe of valves and fittings, m
V - velocity, m/sec
D - pipe inside diameter, m
g - gravitational acceleration, m/sec20
REYNOLD'S NUMBER: Reynold's Number is a non dimensional one which combines the physical quantities which describes
the flow either Laminar or Turbulent flow.
The friction loss in a pipeline is also dependent upon this dimensionless factor.
NR 
ρvD vD

μ
ν
where;  - absolute or dynamic viscosity, Pa-sec
 - kinematic viscosity, m2/sec
For a Reynold's Number of less 2100 flow is said to Laminar
For a Reynold's Number of greater than 3000 the flow is Turbulent
FRICTION FACTOR: Moody's Chart
f
/D
 - absolute roughness
D - inside diameter
/D - relative roughness
NR
VALUES OF ABSOLUTE ROUGHNESS  FOR NEW PIPES
Type of Material
Feet
Drawn tubing, brass, lead, glass
centrifugally spun cement, bituminous
lining, transite
0.000005
Commercial Steel, Wrought iron
0.00015
Welded steel pipe
0.00015
Asphalt-dipped cast iron
0.0004
Galvanized iron
0.0005
Cast iron, average
0.00085
Wood stave
0.0006 to
0.003
Millimeter
0.0015
0.046
0.046
0.12
0.15
0.25
0.18 to
0.9
Concrete
Riveted steel
0.001 to 0.01
0.003 to 0.03
0.3 to 3.0
0.9 to 9
For Laminar Flow
f 
64
NR
PIPE, TUBING AND FITTINGS
Nominal Pipe Diameter: Pipe sizes are based on the approximate diameter and are reported as nominal pipe sizes. Regardless of
wall thickness, pipes of the same nominal diameter have the same outside diameter. This permits interchange of fittings. Pipe may
be manufactured with different and various wall thickness, so some standardization is necessary. A method of identifying pipe
sizes has been established by ANSI (American National Standard Institute). By convention, pipe size and fittings are characterized
in terms of Nominal Diameter and wall thickness.
For steel pipes, nominal diameter is approximately the same as the inside diameter for 12" and smaller. For sizes of 14" and larger,
the nominal diameter is exactly the outside diameter.
SCHEDULE NUMBER: The wall thickness of pipe is indicated by a schedule number, which is a function of internal pressure
and allowable stress.
Schedule Number  1000P/S
where P - internal working pressure, KPa
S - allowable stress, KPa
Schedule number in use: 10,20,30, 40,60, 80, 100, 120, 140, and 160.
Schedule 40 "Standard Pipe"
Schedule 80 " Extra Strong Pipe"
TUBINGS: Tubing specifications are based on the actual outside diameter with a designated wall thickness. Conventional systems
such as the Birmingham WireGauge "BWG" are used to indicate the wall thickness.
FITTING: The term fitting refers to a piece of pipe that can:
1. Join two pieces of pipe
ex. couplings and unions
2. Change pipeline directions
ex. elbows and tees
3. Change pipeline diameters
ex. reducers
4. Terminate a pipeline
ex. plugs and valves
5. Join two streams to form a third
ex. tees, wyes, and crosses
6. Control the flow
ex. valves
VALVES: A valve is also a fitting, but it has more important uses than simply to connect pipe. Valves are used either to control
the flow rate or to shut off the flow of fluid.
DESIGN OF A PIPING SYSTEM
The engineer should consider the following items when he is developing the design of a piping system.
1. Choice of material and sizes
2. Effects of temperature level and temperature changes.
a. insulation
b. thermal expansion
c. freezing
3. Flexibility of the system for physical and thermal shocks.
4. Adequate support and anchorage
5. Alteration in the system and the service.
6. Maintenance and inspection.
7. Ease of installation
8. Auxiliary and standby pumps and lines
9. Safety
a. Design factors
b. Relief valves and flare systems