ME 363 - Fluid Mechanics
Homework #13
Due Friday, May 2, 2008
Spring Semester 2008
1] Water is to be withdrawn from a 3-m-high water reservoir by drilling a 1.5-cm-diameter hole in the bottom
surface. Disregarding the effect of the kinetic energy correction factor (α in equation 8-29), determine the flow
rate in [m3/s] of water through the hole if (a) the entrance of the hole is well-rounded and (b) the entrance is
sharp-edged.
2] Water at 20 degrees C is to be pumped from reservoir A (z A = 2 m) to reservoir B at a higher elevation (zB = 9
m) through two 25-m-long smooth pipes connected in parallel. The diameters of the two pipes are 3 cm and 5
cm. Water is to be pumped by a 68-percent-efficient motor-pump unit that draws 7 kW of electrical power
during operation. The minor losses and the head loss in the pipes that connect the parallel pipes to the two
reservoirs are considered to be negligible. Determine the flow rates through each of the parallel pipes in [m 3/s].
3] A clothes dryer discharges air at 1 atm and 120 degrees F at a rate of 1.2 ft3/s when its 5-in-diameter, wellrounded vent with negligible loss is not connected to any duct. Determine the flow rate when the vent is
connected to a 15-ft-long, 5-in-diameter duct made of galvanized iron (e = 0.0005 ft) with three 90-degree
flanged smooth bends (K = 0.3). Do not forget to include the exit loss. Take the friction factor f of the duct to
be 0.019, and assume the fan power input to remain constant.
no figure
Problem#1
Problem #2
Problem #3
Problem 1 Solution
Solution
Water is to be withdrawn from a water reservoir by drilling a hole at the bottom surface. The flow
rate of water through the hole is to be determined for the well-rounded and sharp-edged entrance cases.
Assumptions 1 The flow is steady and incompressible. 2 The reservoir is
open to the atmosphere so that the pressure is atmospheric pressure at the free
surface. 3 The effect of the kinetic energy correction factor is disregarded,
and thus  = 1.
Analysis
The loss coefficient is KL = 0.5 for the sharp-edged entrance,
and KL = 0.03 for the well-rounded entrance. We take point 1 at the free
surface of the reservoir and point 2 at the exit of the hole. We also take the
reference level at the exit of the hole (z2 = 0). Noting that the fluid at both
points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid
velocity at the free surface is zero (V1 = 0), the energy equation for a control
volume between these two points (in terms of heads) simplifies to
P1
V2
P
V2
  1 1  z1  hpump, u  2   2 2  z 2  hturbine,e  hL
g
2g
g
2g
where the head loss is expressed as h L  K L
z1   2
V22
V2
 KL 2
2g
2g

z1   2
V22
 hL
2g
V2
. Substituting and solving for V2 gives
2g
 2 gz1  V22 ( 2  K L )

V2 
2 gz1
2 gz1

2  KL
1 K L
since 2 = 1. Note that in the special case of KL = 0, it reduces to the Toricelli equation V2  2gz1 , as expected.
Then the volume flow rate becomes
V  AcV 2 
2
Dhole
4
2 gz1
1 K L
Substituting the numerical values, the flow rate for both cases are
determined to be
2
Dhole
2 gz1
 (0.015 m)2
Well-rounded entrance: V 

4
1 KL
4
Sharp-edged entrance:
V 
2
Dhole
4
2 gz1
 (0.015m ) 2

1 K L
4
2(9.81 m/s 2 )(3 m)
 1.34  10 3 m3 /s
1  0.03
2(9.81 m/s 2 )(3 m)
 1.11 10 3 m 3 /s
1  0.5
Discussion
The flow rate in the case of frictionless flow (KL = 0) is 1.3610-3 m3/s. Note that the frictional losses
cause the flow rate to decrease by 1.5% for well-rounded entrance, and 18.5% for the sharp-edged entrance.
Problem 2
Solution
The pumping power input to a piping system with two parallel pipes between two reservoirs is
given. The flow rates are to be determined.
Res
Assumptions 1 The flow is
steady and
incompressible. 2 The entrance effects are negligible,
and thus the flow is fully developed. 3 The elevations
of the reservoirs remain constant. 4 The minor losses
and the head loss in pipes other than the parallel pipes
are said to be negligible. 5 The flows through both
pipes are turbulent (to be verified).
ervoir B
zB=
Res
2
ervoir A
zA=
2m
35 m
cm
2m
1
2
Properties
The density and dynamic viscosity of
water at 20C are  = 998 kg/m3 and  = 1.00210-3
kg/ms. Plastic pipes are smooth, and their roughness is
zero,  = 0.
5
P
cm
ump
Analysis
This problem cannot be solved directly since the velocities (or flow rates) in the pipes are not known.
Therefore, we would normally use a trial-and-error approach here. However, nowadays the equation solvers such as
EES are widely available, and thus below we will simply set up the equations to be solved by an equation solver. The
head supplied by the pump to the fluid is determined from
Vghpump, u
(998 kg/m 3 )V (9.81 m/s 2 )hpump, u
W elect,in 
 7000 W 
(1)
 pump -motor
0.68
We choose points A and B at the free surfaces of the two reservoirs. Noting that the fluid at both points is open to the
atmosphere (and thus PA = PB = Patm) and that the fluid velocities at both points are zero (VA = VB =0), the energy
equation for a control volume between these two points simplifies to
PA
V2
P
V2
  A A  z A  hpump, u  B   B B  z B  hturbine,e  hL
 hpump, u  ( z B  z A )  hL
g
2g
g
2g
or
hpump, u  (9  2)  hL (2)
where
h L  h L,1  h L, 2
(3) (4)
We designate the 3-cm diameter pipe by 1 and the 5-cm diameter pipe by 2. The average velocity, Reynolds
number, friction factor, and the head loss in each pipe are expressed as
V1 
V2 
V1
Ac,1
V
2
Ac,2


V1
D12
/4
V2
D22
Re 1 
V1 D1

Re 2 
V2 D2

/4
 V1 
V1
 (0.03m) 2 / 4
 V2 
 Re 1 
V2
 (0.05 m) 2 / 4
(6)
(998 kg/m 3 )V1 (0.03 m)
 Re 2 
1.002  10 3 kg/m  s
(998 kg/m 3 )V 2 (0.05 m)
1.002  10 3 kg/m  s
 / D
2.51
1
 2.0 log

 3.7
f1
Re 1 f 1

1
(5)

 


(7)
(8)

2.51
 2.0 log 0 

f1
Re 1 f 1

1

 (9)


1
f2
 / D
2.51
2
 2.0 log

 3.7
Re 2 f 2

hL,1  f 1
L1 V12
D1 2 g
h L,2  f 2

L2 V22
D2 2 g

hL,1  f 1


  1  2.0 log 0  2.51


f2
Re 2 f 2


V12
25 m
0.03 m 2(9.81 m/s 2 )
h L,2  f 2
V  V1 V2

 (10


(11)
V22
25 m
0.05 m 2(9.81 m/s 2 )
(12)
(13)
This is a system of 13 equations in 13 unknowns, and their simultaneous solution by an equation solver gives
V  0.0183m 3 /s, V1  0.0037m 3 /s,
V2  0.0146m 3 /s ,
V1 = 5.30 m/s, V2 = 7.42 m/s, h L  h L,1  h L, 2  19.5 m , hpump,u = 26.5 m
Re1 = 158,300,
Re2 = 369,700, f1 = 0.0164,
f2 = 0.0139
Note that Re > 4000 for both pipes, and thus the assumption of turbulent flow is verified.
Discussion
This problem can also be solved by using an iterative approach, but it will be very time consuming.
Equation solvers such as EES are invaluable for this kind of problems.
Problem 3