CHAPTER 1
INTRODUCTION
1.1. A certain control volume of porous media received a discharge of 0.1 m3/s during a
period of 10.0 minutes. The volume of outflow was measured to be 58.0 m3 through the
same period of time. Determine the change in storage in the control volume using
equation 1.1.
Solution: Inflow to the control volume = 0.1 × 10 × 60 = 60.0 m3. Outflow from the
control volume = 58.0 m3. Using equation 1.1, change in storage = 60.0-58.0 = 2.0 m3.
1.2. In problem 1.1, if the change in storage in the control volume is zero, what is the
percentage of error encountered in measuring the outflow?
Solution: In this case the outflow must be equal to the inflow = 60.0 m3/s.
% of error in measuring the outflow = {(60.0-58.0)/60.0} × 100 = 3.33%.
1.3. A certain reach of an open channel 1.0 mile in length is excavated through permeable
soil. The rate of inflow to this reach is 100 ft3/s and the rate of outflow is 95 ft3/s. How
much of the water is being lost through this reach every year?
Solution: The difference between inflow and outflow is lost to the soil. Therefore, the
volume of water lost through this reach every year = (100-95) × 60 × 60 × 24 × 365 =
157,680,000 ft3/year.
1.4. A cylindrical tank is supplied by volume flow rate of 80 liter/min. Two outlet pipes
are connected to the base of the tank extracting water at the flow rate of 50 liter/min.
Calculate the rate of change of water volume inside the tank per min. If the diameter of
the tank is 60 cm, determine the change in the water level after one hour.
Solution:
Vin-Vout = Vstorage
Rate of change of volume = 80 – (50 + 35) = -5 l/min.
i.e. water volume will reduce at the rate of 5 l/min.
πd 2 h
2
volume of tank = πr h =
4
Height of water in the tank is reduced in one hour by
4
× 5000 cm 3 × 60 = 106.1 cm = 1.06 m.
=
π (60) 2
1.5. Water is flowing from tap 8 mm in diameter and is collected in a 10 liter bottle.
Determine the exit velocity of the water from the tap of diameter 8 mm, if the bottle is
filled in:
a) 30 sec.
b) 5.0 min.
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Elementary Hydraulics
Solution:
Cross section area of the tap = π/4d2 = π/4 × (8/1000 m)2 = 50.265 × 10-6 m2
Flow rate Q = volume/time.
a) Q = 10 litre/30 sec = 0.333 × 10-3 m3/s
Velocity = Q/area = (0.333 × 10-3 m3/s) / (50.265 × 10-6 m2) = 6.63 m/s.
0.01 / (5 × 60)
b) Velocity = Q/area =
= 0.663 m/s
50.265 × (10) −6
1.6. The dimensionless parameter Reynolds number is defined as Rn = V.d/ν, where V is
the water velocity, d is the diameter of the pipe, and n is the kinematic viscosity of water
(1.005 × 10-6 m2s). Determine Rn for the two cases (a and b) of Problem 1.5.
Solution:
a) Rn =
Vd 6.63× 0.008
=
= 52,776
V 1.005 ×10 −6
b) Rn =
Vd 0.663 × 0.008
=
= 5278
V
1.005 × 10 −6
1.7. The friction losses through a 10.0 km horizontal pipe of a certain diameter is given
by the relation
hf = 0.005 L
Where hf is the head loss due to friction, and L is the length of the pipe. The total energy
head at the end of the pipe is 21.0 m. Estimate the total energy head at the pipe inlet.
Solution:
The total loss through the pipe is
hf = 0.005 × 10 × 103 = 50.0 m
Energy head at pipe inlet = energy at pipe outlet + loss
= 21.0 + 50.0 = 71.0 m
1.8. Prove the following ratios are dimensionless:
Vy
v
ρVy
µ
V
p
ρV 2
gy
V2
p
γy
σ
γy 2
Q
2
gy
y gy
where V is the velocity, y is the flow depth, v is the kinematic velocity, ρ is the fluid
density, µ is the dynamic viscosity, p is the pressure, γ is the specific weight, σ is surface
tension, and Q is the discharge.
Solution:
Vy [ LT −1 ][ L]
=
=1
v
[ L2T −1 ]
ρVy [ ML−3 ][ LT −1 ][ L]
=
= [ MLT ]o
−1 −1
µ
[ ML T ]
V
gy
=
[ LT −1 ]
= [ MLT ]o
−2
1/ 2
([ LT ][ L])
2
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Chapter 1. Introduction
[ MLT −2 ][ L−2 ]
p
=
= [ M 1−1 L1− 2−( −3) − 2T − 2−( −2) ] = [ MLT ]o
2
−3
−1 2
[ ML ]([ LT ])
ρV
gy [ LT −2 ][ L]
=
= [ L1+1− 2T − 2−( −2 ) ] = [ LT ]o
2
−1 2
([ LT ])
V
p
[ MLT −2 ][ L−2 ]
=
= [ M 1−1 L1− 2 + 3−1−1T − 2+ 2 ] = [ MLT ] o
−
3
−
2
γy [ ML ][ LT ][ L]
σ
[ MLT −2 ][ L−1 ]
= [ M 1−1 L1−1+3−1− 2 T 2 + 2 ] = [ MLT ] o
=
[ ML−3 ][ LT − 2 ]([ L]) 2
γy 2
Q
y 2 gy
=
[ L3T −1 ]
[ L2 ]([ LT − 2 ][ L])1 / 2
= [ L3− 2 − (1+1) / 2 T −1+ ( 2 / 2) ] = [ LT ] o
1.9. A commonly used equation for calculating the velocity of uniform flow in open
channels, V, is
1
V = R 2 / 3 S 01 / 2
n
Where V is the velocity of uniform flow, n is the Manning coefficient, R is the hydraulic
radius, and So is the bed slope. Find the dimensions of the Manning coefficient to ensure
that the equation is homogeneous.
Solution:
1 2 / 3 1/ 2
R S0
n
The dimensions of V and R are LT-1 and L, respectively, and the bed slope, So, is a
dimensionless number. Therefore, substituting in the above equation:
V=
LT-1 =
1 2/3
L
n
,
1
= L1 / 3T −1
n
1.10. From example 1.1 and problem 1.9, find the relationship between the Chezy and
Manning coefficients. Check the homogeneity of this relationship.
Solution: From example 1.1
V = C R 1 / 2 S 01 / 2
and from problem 1.9,
1
V = R 2 / 3 S 01 / 2
n
Therefore,
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Elementary Hydraulics
C R 1 / 2 S 01 / 2 =
C=
1 2 / 3 1/ 2
R S0 ,
n
1 1/ 6
R
n
The dimensions of C, n, and R are L1/2T-1, L-1/3T, and L, respectively. For the above
equation, the dimensions of the left hand side are L1/2T-1, and the dimensions of the right
hand side are L1/3T-1 T1/6 = L1/2T-1. Hence, the homogeneity of the equation is satisfied.
1.11. The water power developed by a pump is given by the equation P = γQH, where P
is the power, γ is the specific weight, Q is the discharge, and H is the water head raised by
the pump. Check the condition of homogeneity of the equation.
Solution: The dimensions of P, γ, Q, and H are ML2T-3, ML-2T-2, L3T-1 and L,
respectively. For the given equation the dimensions of the left hand side are ≡ ML2T-3.
The dimensions of the right hand side are (ML-2T-2) (L3T-1) (L) = ML2T-3.
Therefore, the homogeneity of the dimensions of the given equation is satisfied.
1.12. The Darcy equation for groundwater flow is written as
q = -K
∂h
∂x
where q is the specific discharge (LT-1), K is the hydraulic conductivity, and ∂h/∂x is the
gradient of the piezometric head. Find the dimensions of K.
Solution: The gradient of the piezometric head ∂h/∂x is a dimensionless number. The
dimensions of K are the same as those of the specific discharge, q ≡ LT-1.
1.13. Determine the dimensions of the coefficients A and B in the following equation:
∂x
∂2x
+ Bx = 0
+A
∂t
∂t 2
where x is the length and t is the time.
Solution: The dimensions of the first term of the given equation LT-2. Therefore, other
terms must have the same dimensions. Thus the dimensions of A and B are thus T-1 and
T-2, respectively.
1.14. Determine whether the following equation is dimensionally homogenous or not.
F = 16.83 µVd + 29.78 ρQ2d-2
Where F is force, ρ is the density, µ is dynamic viscosity, Q is the discharge, V is velocity
and d is diameter.
Solution: Assuming that the constants have no dimension, following dimensional
analysis is done:
[M.L.T-2] = [ML-1T-1].[LT-1][L] + [ML-3][L3T-2]2.[L-2]
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Chapter 1. Introduction
= [MLT-2] + [ML-3][L6T-2][L-2]
= [MLT-2] + [MLT-2]
= [MLT-2].
Therefore, the equation is dimensionally homogenous.
1.15. The head loss for flow inside a pipe is often expressed as
LV 2
hf = f
d 2g
where f is the friction coefficient, L and d are length and diameter of the pipe,
respectively, V is the average water velocity and g is the acceleration of gravity.
Determine the units of friction coefficient f. If the system of units is changed from SI to
BG, how this will affect the value of ‘f’?
Solution: using dimensional analysis.
[L] [L2. T -2 ]
[L] = [f] .
= [f] [L]
.
[L] [LT -2 ]
f is dimensionless, therefore, the value of f will not change by changing the system of
units.
1.16. A person weighs 150 lb at the earth’s surface. Determine the person’s mass in slugs
and kilograms.
Solution: Weight (lb) = mass (slugs) x gravity (ft/s2)
150 = m × 32.2
The person’s mass = 150/32.2 = 4.658 slug
= 4.658 × 14.594 = 67.98 kg ≅ 68.0 kg
1.17. The discharge of water through an open channel is 10.0 m3/s. What is the discharge
in liters/min, and gallons/min? What is the volume of water (in gallons) delivered by this
channel per year?
Solution:
Q = 10.0 m3/s
= 10.0 × 103 × 60 = 6.0 × 105 lit/min
= 1.3216 × 105 gallons/min
Volume of water per year = 1.3216 × 105 × 60 × 24 × 365
= 6.946 × 1010 gallons
1.18. A certain object weighs 150.0 Newton at the earth’s surface. Determine the mass of
the object (in kilograms) and its weight (in Newton) when located on a planet with an
acceleration of gravity equal to 4.0 m/s2.
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Solution: 150 (N) = m (kg) × 9.81 (m/s2). Therefore, m = 15.29 kg. The mass of the
object is not affected by any variation in the acceleration due to gravity. The object will
have the same mass irrespective of where it is placed. Weight of the object on the other
planet = 15.29 × 4.0 = 61.16 N.
1.19. A person having a total mass of 80.0 kg rides an elevator. Determine the force (in
Newton) exerted on the floor of the elevator when it is accelerating upward at 5.0 m/s2.
Solution:
W (N) = m (kg) × g(m/s2)
W = 80.0 × (9.81 + 5.0) = 1184.8 N
1.20. In problem 1.19, if the elevator is accelerating downwards at 5.0 m/s2, determine
the force (in Newton) exerted on the floor.
Solution: If the elevator is accelerating downwards at 5.0 m/s2, g will be equal to (9.815.0) = 4.81 m/s2. Therefore, W = 80.0 × 4.81 = 384.8 N.
1.21. If the force exerted on the floor of the elevator in problem 1.19 is zero, what will be
the direction and acceleration of the elevator?
Solution: The elevator must be accelerating downward at 9.81 m/s2.
1.22. A tank of oil has a mass of 30 slug. Determine its weight in pound and in Newton at
the earth’s surface.
Solution:
W (lb) = mass (slug) x gravity (ft/s2)
W = 30 × 32.2 = 966 lb = 966 × 4.4482 = 4296.96 N
1.23. If the tank in problem 1.22 is placed on the surface of the moon where the
gravitational attraction is one-sixth of that at the earth’s surface, what would be its weight
in Newton?
Solution: the weight on the moon will be one-sixth of that at the earth’s surface
W = 4296.96/6 = 716.16 N
1.24. How many hectares and acres in an area of 1 km2?
Solution: 1 km2 = 100 hectare = 247.1 acre
1.25. Convert the following:
a) Discharge of 60 ft3/min to lit/s.
b) Force of 20 pounds to dynes.
c) Pressure intensity of 30 lb/in2 to N/cm2.
d) Specific weight of 62.4 lb/ft3 to N/m3.
e) Dynamic viscosity of 255 dyne.s/cm2 to lb.s/ft2.
f) Dynamic viscosity of 100 gm/cm.s to slug/ft.s.
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Chapter 1. Introduction
Solution:
a)
b)
c)
d)
e)
Discharge of 60 ft3/min = 60 (0.3048)3 × 60/1000 = 0.1019 lit/s.
Force of 20 pounds = 20 × 444,822 dynes = 8.896 × 106 dynes
Pressure intensity of 30 lb/in2 = 30 × 4.448/(2.54)2 = 20.68 N/cm2.
Specific weight of 62.4 lb/ft3 = 62.4 × 4.448/(0.02832) = 9800 N/m3.
Dynamic viscosity of 255 dyne.s/cm2 = 255 × (30.48)2/(444,822)
= 0.5326 lb.s/ft2.
f) Dynamic viscosity of 100 gm/cm.s = 100 × 30.48/14,594 = 0.209 slug/ft.s.
1.26. Making use of the conversion factors given in Appendix A, express the following
quantities in BG units: (a) 16.7 km, (b) 9.30 N/m3, (c) 1.95 kg/m2, (d) 0.045 N.m/s, (e)
6.35 mm/hr.
Solution:
a)
b)
c)
d)
e)
16.7 km = 16.7/1.6093 miles = 10.377 miles.
9.30 N/m3 = 9.3 × 0.2248/(3.28)2 = 0.059 lb/ft3.
1.95 kg/m2 = 1.95 × 0.06852/(3.28)2 = 0.0124 slug/ft2.
0.045 N.m/s = 0.045 × 0.2248 × 3.28 = 0.0332 ft-lb/s.
6.35 mm/hr = 6.35/25.4 = 0.25 in/hr.
1.27. Express the following quantities in SI units using the conversion factors in
Appendix A: (a) 10.25 in/min, (b) 4.20 slugs, (c) 4.0 lb, (d) 82.2 ft/s2, and (e) 0.0315
lb.s/ft2.
Solution:
a)
b)
c)
d)
e)
10.25 in/min = 10.25 × 2.54 × 60/100 = 15.621 m/s.
4.20 slugs = 4.2 × 14.6 kg = 61.32 kg.
40.0 lb = 4 × 0.4536 = 1.814 kg.
82.2 ft/s2 = 82.2 × 0.3048 = 25/05 m/s2.
0.0315 lb.s/ft2 = 0.0315 × 0.4536/(0.3048)2 = 0.1538 kg.s/m2.
1.28. Verify the conversion factors for (a) acceleration, (b) density, (c) specific weight
and (d) dynamic and kinematic viscosity. Use the basic conversion relations, 1 ft =
0.3048 m, 1 lb = 4.4482 N, and 1 slug = 14.594 kg.
Solution:
a) acceleration: 1 ft/s2 = 1 × (0.3048 m)/s2 = 0.3048 m/s2.
b) Density: 1 slug/ft3 = (14.594 kg)/(0.3048 m)3 = 515.38 kg/m3.
c) Specific weight: 1 lb/ft3 = (4.4482 N)/(0.3048 m)3 = 515.38 N/m3.
d) Dynamic viscosity: 1 lb.s/ft2 = (4.4482 N.s)/(0.3048 m)2 = 47.88 N.s/m2.
Kinematic viscosity: 1 ft2/s = (0.3048 m)2/s = 0.0929 m2/s.
1.29. Water flows from a large drainage pipe at a rate of 1350 gal/min. What is the flow
rate in m3/s and lit/min.
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Solution:
1350 gal/min = (1350 × 3.7854 × 10-3 m3)/(60 s) = 0.0852 m3/s =
(0.0852 × 1000 l)/(1/60 min) = 5,110 l/min
1.30. The viscosity of a certain fluid is 3.5 × 104 poise. Determine its viscosity in both SI
and BG units.
Solution:
3.5 × 10-4 poise = 3.5 × 10-5 kg/(m.s) = (3.5 × 10-5)/47.88 =
7.31 × 10-7 slug/(ft.s).
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