Engineering Fluid Mechanics Solution Manual
Prof. T.T. Al­Shemmeri
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Prof. T.T. Al-Shemmeri
Engineering Fluid Mechanics Solution
Manual
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2
Engineering Fluid Mechanics Solution Manual
© 2012 Prof. T.T. Al-Shemmeri & bookboon.com
ISBN 978-87-403-0263-9
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Engineering Fluid Mechanics Solution Manual
Contents
Contents
Book Description:
5
Author Details: 5
1
Chapter One Tutorial Problems
6
2
Chapter Two Tutorial Problems
13
3
Chapter Three Tutorial Problems
21
4
Chapter Four Tutorial Problems 28
5
Chapter Five Tutorial Problems
31
Sample Examination Paper
33
Class Test – Fluid Mechanics
34
Formulae Sheet
44
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Engineering Fluid Mechanics Solution Manual
Book Description
Book Description:
Title – Engineering Fluid Mechanics Solution Manual
Author – Prof. T.T. Al-Shemmeri
Fluid Mechanics is an essential subject in the study of the behaviour of fluids at rest and when in motion.
The book is complimentary follow up for the book “Engineering Fluid Mechanics” also published on
BOOKBOON, presenting the solutions to tutorial problems, to help students the option to see if they
got the correct answers, and if not, where they went wrong, and change it to get the correct answers.
Author Details:
Prof Tarik Al-Shemmeri – BSc, MSc, PhD, CEng
Professor of Renewable Energy Technology at Staffordshire University.
Current research interests in Renewable Energy and Environmental Technology.
Lecturing topics include: Energy management and Power generation.
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5
Engineering Fluid Mechanics Solution Manual
Chapter One Tutorial Problems
1 Chapter One Tutorial Problems
1.1
Show that the kinematic viscosity has the primary dimensions of L2T-1.
Solution:
The kinematic viscosity is defined as the ratio of the dynamic viscosity by the density of the fluid.
The density has units of mass (kg) divided by volume (m3); whereas the dynamic viscosity has the units
of mass (kg) per meter (m) per time (s).
Hence:
X
P
U
0/ 7 0/ /7 1.2In a fluid the velocity measured at a distance of 75mm from the boundary is 1.125m/s. The fluid
has absolute viscosity 0.048 Pa s and relative density 0.913. What is the velocity gradient and
shear stress at the boundary assuming a linear velocity distribution? Determine its kinematic
viscosity.
[Ans: 15 s-1, 0.72Pa.s; 5.257x10-5 m2/s]
Solution:
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Engineering Fluid Mechanics Solution Manual
*UDGLHQW
W
P
X
P
U
G9
G\
G9
G\
[
Chapter One Tutorial Problems
V 3DV
[ P V
1.3A dead-weight tester is used to calibrate a pressure transducer by the use of known weights
placed on a piston hence pressurizing the hydraulic oil contained. If the diameter of the piston
is 10 mm, determine the required weight to create a pressure of 2 bars.
[Ans: 1.6 kg]
Solution:
(a)
P = F/A
$
S
'
S
[ [ P Hence the weight =PxA/g
= 2 x 105 x7.854x10-5 / 9.81
= 1.60 kg
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Engineering Fluid Mechanics Solution Manual
Chapter One Tutorial Problems
1.4How deep can a diver descend in ocean water without damaging his watch, which will withstand
an absolute pressure of 5.5 bar?
Take the density of ocean water, r = 1025 kg/m3.
[Ans: 44.75 m]
Solution:
Use the static equation: S UJK
Hence the depth can be calculated as:
K
1.5
3
U J
[ [
P :DWHU
The U-tube manometer shown below, prove that the difference in pressure is given by:
  d 2 
P1 − P2 = ρ .g .z 2 1 +   
  D  
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8
Engineering Fluid Mechanics Solution Manual
Chapter One Tutorial Problems
Solution:
The relationship between liquid columns and the area of cross section is based on the conservation of
matter, ie continuity equation, hence:
volume 1 = volume 2
] [$
] [$
KHQFH o ]
] [S [G S [' G
] [ '
For the “U” tube manometer that the height different in the two columns gives the pressure difference,
therefore:
) ) 3 UJ]
3 UJ]
hence
3 3
UJ ] ] UJ] [> G ' @
Clearly if D is very much larger than d then (d/D)2 is very small so
3 3
UJ]
1.6A flat circular plate, 1.25 m diameter is immersed in sewage water (density 1200 kg/m3) such
that its greatest and least depths are 1.50 m and 0.60 m respectively. Determine the force exerted
on one face by the water pressure,
[Ans: 15180 N]
P
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9
P
D
Engineering Fluid Mechanics Solution Manual
Chapter One Tutorial Problems
Solution:
Area of laminar
$
S P
Depth to centroid
KF
P
Resultant Force
F
= r g A hc = 9.81 x 1200 x 1.228 x 1.05
= 15180 N
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Chapter One Tutorial Problems
1.7A rectangular block of wood, floats with one face horizontal in a fluid (RD = 0.9). The wood’s
density is 750 kg/m3. Determine the percentage of the wood, which is not submerged.
[Ans: 17%]
Solution:
:HLJKW
:
)
For stable condition
Upthrust = weight force or F=W
The Upthrust due to Buoyancy = rseawater g Vx
The total weight of submersed wood = rwood g VL
Therefore the portion of block that is NOT submerged is
1- Vx/VL = (rwood / rwater ) = 1 - 750 / 900 = 17%
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Engineering Fluid Mechanics Solution Manual
Chapter One Tutorial Problems
1.8An empty balloon and its equipment weight 50 kg, is inflated to a diameter of 6m, with a gas
of density 0.6 kg/m3. What is the maximum weight of cargo that can be lifted on this balloon,
if air density is assumed constant at 1.2 kg/m3?
[Ans: 17.86 kg]
Solution:
:
)
Since the system is stable according to Newton’s second law of motion:
Upthrust = Weight force + Payload
The Upthrust is F
= r x g xVL = 1.2 x 9.81 x(px(4/3)x33) =1331.4 N
Weight of balloon gas = r xVL = 0.6 x 9.81 x(px(4/3)x33) =666 N
Weight of equipment
= 50 kg = 50x9.81 = 490.5 N
Therefore the payload that can be lifted
P
=F-W
= 1331.4 – (666 + 490.5)
= 175 N = 17.9 kg.
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2 Chapter Two Tutorial Problems
2.1A 20 mm dam pipe forks, one branch being 10 mm in diameter and the other 15 mm in diameter.
If the velocity in the 10 mm pipe is 0.3 m/s and that in the 15 mm pipe is 0.6 m/s, calculate the
rate of flow in cm3/s and velocity in m/s in the 20 mm diameter pipe.
(129.6 cm3/s, 0.413 m/s)
Solution:
Total mass flow into the junction = Total mass flow out of the junction
ρ1Q1 = ρ2Q2 + ρ3Q3
When the flow is incompressible (e.g. if it is water) ρ1 = ρ2 = ρ
Q
= A2 V2 + A3 V3
= (p/4) D22x V2 + (p/4) D32x V3
= (p/4) 0.012x 0.3+ (p/4) 0.0152x 0.6
= 2.356x10-5 + 10.6x10-5
= 12.96 x10-5 m3/s
since
4 $ [9
?9
4
S
[ P V Download free eBooks at bookboon.com
13
Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.2Water at 36 m above sea level has a velocity of 18 m/s and a pressure of 350 kN/m2. Determine
the potential, kinetic and pressure energy of the water in metres of head. Also determine the
total head.
Ans (35.68 m, 16.5 m, 36 m, 88.2 m)
Solution:
Take each term separately
S
UJ
9
J
[
[
P P [
z1 = 36m
The total head = 35.678 + 16.514 + 36 = 88.192 m
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.3The air supply to an engine on a test bed passes down a 180 mm diameter pipe fitted with an
orifice plate 90 mm diameter. The pressure drop across the orifice is 80 mm of paraffin. The
coefficient of discharge of the orifice is 0.62 and the densities of air and paraffin are 1.2 kg/m3
and 830 kg/m3 respectively. Calculate the mass flow rate of air to the engine.
Ans (0.16 kg/s)
Solution:
(i)p1 - p2 = rm x g x h
= 830 x 9.81 x 0.080 = 651 Pa
S S U > $ $ @
(ii)
Y
(iii) m = Cd V2 A2 x r
$
S
[
[
[> @
[ P
Q
= 0.62 x 34.03 x 0.00636 x 1.2
= 0.161 kg/s
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P V
Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.4Determine the pressure loss in a 100 m long, 10 mm diameter smooth pipe if the flow velocity
is 1 m/s for:
a) air whose density 1.0 kg/m3 and dynamic viscosity 1 x 10-5 Ns/m2.
b) water whose density 10003 kg/m3 and dynamic viscosity 1 x 10-3 Ns/m2.
Ans: (320 N/m2, 158 kN/m2)
Solution:
a) For air
Re = r V D/m = 1 x 1 x 0.01/ 1 x 10-5 = 1000 i.e. laminar
f = 16/Re=16/(1000)= 0.016
therefore hf = 4 xfx
L v2
D 2g
KI
[ [
[
3
U JK I
[[
P
1 P
b) for water
Re = r V D/m
= 1000 x 1 x 0.01/ 1 x 10-3 = 104 i.e. turbulent flow
f
= 0.079/Re0.25
= 0.079/(104)0.25 = 0.0079
therefore hf = 4 xfx
KI
3
L v2
D 2g
[ [
U JK I
[
P
[[ N1 P
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.5Determine the input power to an electric motor (hm = 90%) supplying a pump (hp = 90%)
delivering 50 l/s of water (r = 1000 kg/m3, m = 0.001 kg/ms) between two tanks with a difference
in elevation of 50m if the pipeline length is 100m long in total of 150 mm diameter, assume a
friction factor of 0.008 and neglect minor losses.
Ans: (33.6 kW)
Solution:
V = Q/A = 0.05/(p x 0.152/4) = 2.83 m/s
Re = rVD/m = 1000 x 2.83 x 0.15/0.001 = 4.244 x 105
k/D = 0.15/150 = 0.001;
KI
I / Y
' J
From Moody diagram f = 0.0051
[ [ [
[ P hsys = Hf + Hz + Ho = 8.708 + 50 + 0 = 58.808 m
Input power 3
UJ4+ V\V
K PK S
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.6A jet of water strikes a stationary flat plate “perpendicularly”, if the jet diameter is 7.5 cm and
its velocity upon impact is 30 m/s, determine the magnitude and direction of the resultant force
on the plate, neglect frictional effect and take water density as 1000 kg.m3.
Ans (3970 N)
Solution:
$
S
'
S
[ [ P )[
U$9 FRVT )[
[ [ [ [ )[
1
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.7A horizontally laid pipe carrying water has a sudden contraction in diameter from 0.4m to 0.2m
respectively. The pressure across the reducer reads 300 kPa and 200 kPa respectively when the
flow rate is 0.5 m3/s. Determine the force exerted on the section due to the flow, assuming that
friction losses are negligible.
Ans: (25.5 kN)
Solution:
$
$
Y
Y
S
S
[ P [ P 4
$
4
$
S
[ S
P V
[ P V
)[S$S$FRVT U 9 9FRVT9
)[[[±[[[ [[
)[ Hence
Fx = 25429 N
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Engineering Fluid Mechanics Solution Manual
Chapter Two Tutorial Problems
2.8A siphon has a uniform circular bore of 75 mm diameter and consists of a bent pipe with its crest
1.8 m above water level and a discharge to the atmosphere at a level 3.6 m below water level.
Find the velocity of flow, the discharge and the absolute pressure at crest level if the atmospheric
pressure is 98.1 kN/m2. Neglect losses due to friction.
Ans (0.0371 m3/s, 45.1 kN/m2)
Solution:
S Y
]
UJ J
S Y ]
UJ J
Y
J
[ [ KHQFH o 9
4
$[9
S
[
[ ?3
' [9
S
P V
[ [ S
UJ
[ [ 3D
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P V
Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
3 Chapter Three Tutorial Problems
3.1If the vertical component of the landing velocity of a parachute is equal to that acquired
during a free fall of 2m, find the diameter of the open parachute (hollow hemisphere) if
the total weight of parachute and the person is 950N. Assume for air at ambient conditions,
Density = 1.2 kg/m3 and Cd = 1.35
Ans (6.169m)
Solution:
678'<)25<2850$67(5©6'(*5((
&KDOPHUV8QLYHUVLW\RI7HFKQRORJ\FRQGXFWVUHVHDUFKDQGHGXFDWLRQLQHQJLQHHU
LQJDQGQDWXUDOVFLHQFHVDUFKLWHFWXUHWHFKQRORJ\UHODWHGPDWKHPDWLFDOVFLHQFHV
DQGQDXWLFDOVFLHQFHV%HKLQGDOOWKDW&KDOPHUVDFFRPSOLVKHVWKHDLPSHUVLVWV
IRUFRQWULEXWLQJWRDVXVWDLQDEOHIXWXUH¤ERWKQDWLRQDOO\DQGJOREDOO\
9LVLWXVRQ&KDOPHUVVHRU1H[W6WRS&KDOPHUVRQIDFHERRN
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9
Chapter Three Tutorial Problems
9 DV
[ [ 9
16/ P V
¦)
PJ U9 $I [&G
S
[ [ ' [
VROYH o '
P
3.2A buoy is attached to a weight resting on the seabed; the buoy is spherical with radius of 0.2m
and the density of sea water is 1020 kg/m3. Determine the minimum weight required to keep
the buoy afloat just above the water surface. Assume the buoy and the chain has a combined
weight of 1.2 kg.
Ans (33 kg)
Solution:
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Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
Since the system is stable and motionless, Newton’s second law of motion reduces to:
Weight = Upthrust (Buoyancy)
Fb = Fg
Volume of sphere = (4.pi/3) x R3 = 0.0335 m3
The Upthrust is FB= rfluid x VL x g =1020 x 0.0335x9.81= 335.3 N
Weight = 9.81 x(1.2 + Load )
Hence the payload = (335.3 / 9.81) – 1.2 = 32.98 kg
3.3An aeroplane weighing 65 kN, has a wing area of 27.5 m2 and a drag coefficient (based on wing
area) CD=0.02+0.061 xCL2. Assume for air at ambient conditions, Density = 0.96 kg/m3. Determine
the following when the craft is cruising at 700 km/h:
1. the lift coefficient
2. the drag coefficient, and
3. the power to propel the craft.
Ans (0.13, 0.021, 2040 kW)
Solution:
&/
)/
U$9 [ [ [ [9 &'
[& /
)'
U9 $I [& '
ª
º
[ [ «
»
¬ [ [ [9 ¼
hence
[ [9 [ [ [ [9 Download free eBooks at bookboon.com
23
Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
Power
9 [ 9 3
)' [9
9
NP K
[
P V
hence
&/
&'
3
[ [ [ [ [ N:
3.4 A racing car shown below is fitted with an inverted NACA2415 aerofoil with lift to drag given
as: Cd=0.01 + 0.008 x Cl2
The aerofoil surface area is 1 m2 and the car weight is 1 kN; the car maintains a constant speed
of 40 m/s, determine at this speed:
1. The aerodynamic drag force on the aerofoil
2. The power required to overcome this drag force
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I joined MITAS because
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Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
Assume for air at ambient conditions, take Density = 1.2 kg/m3
Ans (18 N, 0.7 kW)
Solution:
)/
U$9 &'
[&/
)'
U9 $I [&G
[ [[ [ 1
3RZHU
3.5
[ [ [[ &/
)' [9
[ [ N:
Air flows over a sharp edged flat plate, 3m long and 3m wide at a velocity of 2 m/s.
1. Determine the drag force
2. Determine drag force if the plate was mounted perpendicular to the flow direction
assume Cd = 1.4.
For air, take density as 1.23 kg/m3, and kinematic viscosity as 1.46x10-5 m/s2.
Ans (0.05N, 31N)
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Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
Solution:
1.
5H O
9f [/
X
[
[ [ [ o /D PLQDU
&'
5H
)'
U$9f [& '
[[[ [ [ U$9f [& '
[[[ [ [
1
2.
&G
)'
1
3.6(a) An airplane wing has a 7.62m span and 2.13m chord. Estimate the drag on the wing (two
sides) treating it as a flat plate and the flight speed of 89.4 m/s to be turbulent from the
leading edge onward.
(b) Determine the reduction in power that can be saved if the boundary layer control device is
installed on the wing to ensure laminar flow over the entire wing’s surface.
For air, take density as 1.01 kg/m3, and kinematic viscosity as 1.3x10-5 m/s2.
Ans (358 N, 86%)
Solution:
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26
Engineering Fluid Mechanics Solution Manual
Chapter Three Tutorial Problems
a)
5H O
&'
9f [/
X
5H
[ [ ! [ o 7XUE XOHQW
[
[[ [ [ [ [ 1
U$9f [&'
3RZHU )' [9 [ N:
)'
b) laminar.
[ )'
[[ [ [ [ [ U$9f [&'
3RZHU )' [9 [ N:
&'
1
UHGXFWLRQ GOT-THE-ENERGY-TO-LEAD.COM
Biz, bir enerji şirketinin de yenilenebilir olması gerektiğine inanıyoruz. Bu nedenle bizimle birlikte
dünyayı değiştirmek isteyen, yaratıcı fikirlere sahip, meraklı iş arkadaşları arıyoruz. Size sunduğumuz
fırsatların neler olduğunu ve hep beraber yarının enerji ihtiyacını nasıl karşılayacağımızı, internet
sitemizden öğrenebilirsiniz.
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Engineering Fluid Mechanics Solution Manual
Chapter Four Tutorial Problems
4 Chapter Four Tutorial Problems
4.1Assuming the ideal gas model holds, determine the velocity of sound in
a) air (mwt 28.96) at 25°C, with g = 1.4,
b) argon (mwt 39.95) at 25°C, with g = 1.667.
Ans: [346 m/s, 321.5 m/s]
Solution:
For air at 25 oC
R = 8314.4/ 28.96 = 287.096 J/kgK
a= [γ R T]1/2 = (1.4 (287.096 J/K kg) (298K))1/2 = 346.029 (m/s)
For Argon
R = 8314.3 / 39.95 = 208.117 J/kgK
a= [γ R T]1/2 = (1.667 (208.117 J/K kg) (298K))1/2 = 321.536 (m/s)
4.2An airplane can fly at a speed of 800km/h at sea-level where the temperature is 15°C. lf the
airplane flies at the same Mach number at an altitude where the temperature is -44°C, find the
speed at which the airplane is flying at this altitude.
Ans: [195m/s]
Solution:
a= [γ R T]1/2
= (1.4 x287x(273+15))1/2 = 346.17 (m/s)
V1= 800 km/h = 800 x 1000/3600 = 222.222 m/s
M = V/a = 222.222 / 346.17 = 0.642
a2 = [γ R T]1/2
= (1.4 x287x(273-44))1/2 = 303.33 (m/s)
Hence
V2 = M x a2 = 303.33 x 0.642 = 195 m/s
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Engineering Fluid Mechanics Solution Manual
Chapter Four Tutorial Problems
4.3 A low flying missile develops a nose temperature of 2500K when the ambient temperature and
pressure are 250K and 0.01 bar respectively. Determine the missile velocity and its stagnation
pressure. Assume for air: g = 1.4. Cp = 1005 J/kgK
Ans: [2126 m/s, 31.6 bar]
Solution:
&S
&Y 5
&S
&S J 5
hence
J 5
[&S
J
7R 7
9
[
9 &S
[&S[ 7R 7 [ [ P V
J
3R
3
9 J > @
&S7
3R
ª
9 º J 3[ « »
¬ &S7 ¼
- NJ.
J
ª
º [ « »
¬ [ [ ¼
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EDU
Engineering Fluid Mechanics Solution Manual
Chapter Four Tutorial Problems
4.4An airplane is flying at a relative speed of 200 m/s when the ambient air condition is 1.013 bar,
288 K. Determine the temperature, pressure and density at the nose of the airplane. Assume for
air: g = 1.4, density at ambient condition = 1.2 kg/m3 and Cp =1005 J/kgK.
Ans: [To=307.9K, Po=1.28 bar, r =1.42 kg/m3]
Solution:
7R
9
7
[&S
[
3R
3
9 J > @
&S7
3R
ª
9 º J 3[ « »
¬ &S7 ¼
UR
3R
5[7R
U R
3
U [ > R @J
3
.
J
J
[ [ ª
º [ « »
¬ [[ ¼
E DU
NJ P ª º
[ «
»
¬ ¼
NJ P Download free eBooks at bookboon.com
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Engineering Fluid Mechanics Solution Manual
Chapter Five Tutorial Problems
5 Chapter Five Tutorial Problems
5.1
A small-scale hydraulic power system has an elevation difference between the reservoir water
surface and the pond water surface downstream of the turbine is 10 m. The flow rate through
the turbine is 1 m3/s. The turbine/generator efficiency is 83%. Determine the power produced if:
a) Flow losses are neglected.
b) Assume friction loss equivalent to 1 m head.
Ans: (81 kW, 73 kW)
Solution:
a)
ORVVHV
3$
b)
P
U J 4 + K [ [[ [ N:
ORVVHV P
3%
[ [[ [ 0:
5.2 A hydro-electric power plant based on the Loch Sloy in Scotland has an effective head of 250
metres. If the flow rate of 16 m3/s can be maintained, determine:
a) the total power input to the turbine assuming a hydraulic efficiency of 98%; and
b) the pressure difference across the turbine.
Ans: (38 MW, 2.4 MPa)
Solution:
The power available from water is
P
= ρ Q g h x h = 1000 x 16 x9.81 x 250 x0.98
= 38455.2 kW which is 38 MW.
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Engineering Fluid Mechanics Solution Manual
Chapter Five Tutorial Problems
The power can also be expressed as = P = Q x Δp
Hence
Dp =P/Q
= 38455200 / 16
= 2.403 MPa 5.3A proposed hydropower plant to be built using a reservoir with a typical head of 18m and
estimated power of 15 MW. You are given the task to select an appropriate type of turbine for
this site if the generator requires the turbine to run at a fixed speed of 120 rpm.
Ans: (Ns=396, Francis or Kaplan)
Solution
Type of Turbine
1V
1
3
K Specific speed range 1V
Francis
70–500
Propeller
600–900
Kaplan
350–1000
Cross-flow
20–90
Turgo
20–80
Pelton, 1-jet
10–35
Pelton, 2-jet
10–45
[
hence – Francis or Kaplan
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1
3
K Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Solution of Mock Exam
Sample Examination Paper
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Sample Examination Paper
Class Test – Fluid Mechanics
Module Tutor
T. Al-Shemmeri
This Paper contains TEN questions. Attempt all questions.
A formulae sheet is provided.
Place your Answers in the space provided. No detailed solution required.
Print your name on every page. Submit all together for marking.
Marking Grid Leave Blank Please
question
1
2
3
4
5
6
7
8
9
10
total
1st Marker
2nd marker
Agreed percentage
Recommended grade
Question One
List TWO types of instrument used to measure the pressure of a toxic fluid contained in a sealed tank.
Complete the table below:
Type
Principle
Marks
Standard Bourdon gauge
Expansion of a helical element when stressed by the action
of pressure/force
/ 4 marks
Piezometer – closed
A piezometer is a device used to measure static liquid
pressure in a system by measuring the height to which a
column of the liquid rises against gravity
/ 4 marks
Total (8 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Two
a) Draw (not to scale) the pressure distribution of the water on the dam shown below:
(6 marks)
b) Indicate on the sketch, the direction of the resultant force on the dam?
(2 marks)
c) Approximately, indicate the position of the centre of pressure on both sides.
(2 marks)
Total (10 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Three
List Three methods used to improve the resolution of detecting a small pressure reading in a manometer.
Complete the table below:
Method
Principle
Marks
Incline one limb
Trigonometry will mean the sensitivity can be improved
by reading the longest side of a triangle.
Read (Lx Sin(theta)) instead of reading (h)
/ 3 marks
Use different size (cross-sectional
area) for either side of the
manometer.
Help is derived from the continuity law, or displaced
volume on either side to remain the same, A. z = constant,
so z2 can be much larger if A2 is smaller than A1.
/ 3 marks
Use different fluids with appreciably
different density values.
This uses the conservation of mass, for the displaced
column on either side of the manometer.
/ 3 marks
Total (9 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Four
N3D
NJ
NJ
FP
Complete the table below:
/ 4 marks
Theoretical reading of the pressure
P = F / A = (2.5+0.5)x9.81 /(3.14x0.032/4)
41635
% error
/ 3 marks
=[(41.635-40.000)/41.635]x100%
3.93%
The maximum load if the gauge limit is 100 kPa
/ 3 marks
F = P x A = 100x103 x(3.14x0.032/4)
70.686 N
= 7.2 kg
Total (10 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Five
If the fan, below, circulates air at the rate of 0.30 m3/s, determine the velocity in each section. Complete
the table below.
Section
Dimensions
m
Area
m2
1
0.25 square
0.25x0.25 = 0.0625 m2
2
0.20 diameter
Velocity
m/s
(3.14/4)x0.22 = 0.0314 m2
V1 =Q/A
= 0.3/0.0625
= 4.8 m/s
V2 =Q/A
= 0.3/0.0314
= 9.5 m/s
Marks
/ 5 marks
/ 5 marks
Total (10 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Six
Oil of relative density 0.90 flows at the rate of 100 kg/s in a horizontal pipe of 200 mm diameter, 1 km
long. If the friction factor for the pipe is 0.006, complete the following table:
Quantity
Value
Units
Marks
flow velocity
m = density x A x V
hence
V = m / ( den x A)
3.537
m/s
/ 3 marks
frictional head loss
h = (4f L/D) x V2/2g
= (4x0.006x1000/0.2)x (3.5372/19.62
= 76.5
m
/ 3 marks
frictional pressure loss
P = den x g x h
= 900x9.81x76.5
= 675
kPa
/ 2 marks
energy to overcome friction
E = m.g.h
= 100x9.81x76.5
= 75
kW
/ 2 marks
Total (10 marks)
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Sample Examination Paper
Question Seven
Show that Bernoulli’s equation is dimensionally homogeneous
3 marks for the p-term,
3 marks for the v-term, and
2 marks for the z-term and
2 for stating that all dimensions have/have not the same dimensions
Total (10 marks)
Solution:
Take each term separately
S
0/7 D
œ
œ / œ P
UJ
0/ /7 / 7 9
E
œ
œ / œ P
J
/7 F = œ / œ P
All terms have the same units of meters. The equation is therefore dimensionally balanced.
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Question Eight
Oil (relative density 0.85, kinematic viscosity 80cs) flows at the rate of 90 tonne per hour along a 100
mm bore smooth pipe. Determine for the flow:
Quantity
Value
Marks
flow velocity
/ 5 marks
0.00785 m2
A= (pi/4)xD2
m = 90 t/h
25 kg/s
= 90 x 1000/3600
3.745 m/s
V = m / A/ density
/ 5 marks
Nature of the flow
New = 80 cs
= 80x10-6 m2/s
Re = den x V x D/ mu
= V x D / New
Re =3.745 x 0.1 / 80x10-6
Re = 4682
Hence flow turbulent
/ 2 marks
frictional factor
f = 0.0095
f = 0.079 / Re0.25
Total (12 marks)
Question Nine
List two instruments for measuring the flow rate of air through a rectangular duct.
Method
Principle
marks
Hot wire Anemometer
use a very fine wire (on the order of several
micrometres) electrically heated up to some
temperature above the ambient. Air flowing past
the wire has a cooling effect on the wire. Uses the
relationship between the electrical resistance of the
wire and the flow speed.
/ 4 marks
Pitot-static tube
Relies on the measured static and dynamic pressures
to determine the fluid velocity using Bernoulli’s
equation.
/ 4 marks
Total (8 marks)
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
QUESTION TEN
Draw the body force diagram for a parachute jumper.
If the vertical component of the landing velocity of a parachute is 6 m/s, find the total weight of the
parachutist and the parachute (hollow hemisphere Diameter 5m).
Assume for air at ambient conditions, Density = 1.2 kg/m3 and Cd = 2.3
For correct body force diagram
/2 marks
For correct use of formula
/ 6 marks
For correct answer
/ 2 marks
Total (10 marks)
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Sample Examination Paper
Solution
The free body diagram shows that the body is affected by 3 forces.
Newton’s second law of motion states that when the object is moving at steady state then the net force
on the object must be zero
Hence F g = Fb + Fd
Since the parachute is open, the buoyancy force will be negligible and only due to the body of the person,
which is small in comparison to the parachute size. Fb = 0
16/ ¦)
)Z )Z
o )Z )G
U9 $I [&G S
[ [ [ [ 1
NJ
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Engineering Fluid Mechanics Solution Manual
Sample Examination Paper
Formulae Sheet
FLUID STATICS:
P=rgh
CONTINUITY EQUATION:
mass flow rate m=rAV
volume flow rate Q=AV
Energy Equation
(P/r g) + ( V2/ 2g ) + Z = constant
DARCY’S EQUATION
Hf = (4 f L /D) ( V2/ 2g )
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FRICTION FACTOR FOR A SMOOTH PIPE
f = 16/ Re if Re < 2000
f = 0.079 / Re0.25 if Re > 4000
MOMENTUM EQUATION
F = m ( V2 cosq - V1 )
Drag FORCE = Cd x (1/2) x r. A.V2
FLUID POWER
E=rghQ
45