14–93
14-110
Solution In a hydroelectric power plant, the flow rate of water, the available elevation head, and the combined
turbine-generator efficiency are given. The electric power output of the plant is to be determined.
Assumptions 1 The flow is steady and incompressible. 2 The entrance effects are negligible, and thus the flow is fully
developed and the friction factor is constant for the entire pipe. 3 The minor losses are given to be negligible. 4 The water
m/s 236.6
4/m) 35.0(
/sm 0.6
4/ 2
3
2
D
A
V
c
VV
6
3
3
skg/m 10002.1
m) m/s)(0.35 )(6.236kg/m (998
h
VD
which is greater than 4000. Therefore, the flow is turbulent. The
140 m
0.35 m
Water
1
14–94
14-111
Solution In a hydroelectric power plant, the flow rate of water, the available elevation head, and the combined
turbine-generator efficiency are given. The percent increase in the electric power output of the plant is to be determined
when the pipe diameter is tripled.
Assumptions 1 The flow is steady and incompressible. 2 Entrance effects are negligible, and thus the flow is fully
m/s 236.6
4/m) 35.0(
/sm 0.6
4/ 2
3
2
1
D
A
V
c
VV
,
/sm 0.6
3
VV
140 m
1
14–95
14-112E
Solution The drinking water needs of an office are met by siphoning water through a plastic hose inserted into a large
6.55610-4 lbm/fts. The plastic pipes are considered to be smooth, and thus their roughness is
= 0. The total minor loss
coefficient is given to be 2.8.
Analysis We take point 1 to be at the free surface of water in the bottle, and point 2 at the exit of the hose., which is
also taken to be the reference level (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2
2
22
g
g
g
g
g
where
2 = 1 and
)ft/s 2.32(2
8.2
ft 12/35.0
ft 6
22
2
2
2
2V
fh
g
V
K
D
L
fh LLL
(1)
The average velocity in the pipe and the Reynolds number are
(3)
4/)ft 12/35.0(
/sft
4/
2
3
2
2
2
V
D
A
V
c
VVV
0.35
1 ft
1
14–96
14-113E
mu=2.36/3600
nu=mu/rho
g=32.2
z1=4
KL=2.8
eps=0
rf=eps/D
14–97
14-114E
Solution The drinking water needs of an office are met by siphoning water through a plastic hose inserted into a large
6.55610-4 lbm/fts. The plastic pipes are considered to be smooth, and thus their roughness is
= 0. The total minor loss
coefficient is given to be 2.8 during filling.
Analysis We take point 1 to be at the free surface of water in the bottle, and point 2 at the exit of the hose, which is
also taken to be the reference level (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2
2
22
g
g
g
g
g
where
2 = 1 and
)ft/s 2.32(2
8.2
ft 12/35.0
ft 12
22
2
2
2
2V
fh
g
V
K
D
L
fh LLL
(1)
The average velocity in the pipe and the Reynolds number are
(3)
4/)ft 12/35.0(
/sft
4/
2
3
2
2
2
V
D
A
V
c
VVV
fff
D
f
h
Re
51.2
0log0.2
1
Re
51.2
7.3
/
log0.2
1
(5)
/sft
ft 00835.0
3
3
glass
VV
V
t
(6)
0.35 in
1 ft
1
2
14-115
Solution A water tank open to the atmosphere is initially filled with water. The tank is drained to the atmosphere
through a 90 horizontal bend of negligible length. The flow rate is to be determined for the cases of the bend being a
flanged smooth bend and a miter bend without vanes.
Assumptions 1 The flow is steady and incompressible. 2 The flow is turbulent so that the tabulated value of the loss
2
22
2
2
21e turbine,2
2
2
2
2
upump,1
2
1
1
1
LL h
g
V
zhhz
g
V
g
P
hz
g
V
g
P
where the head loss is expressed as
g
V
Kh LL 2
2
. Substituting and solving for V2 gives
22
2
2 2 1
1 2 1 2 2 2
2
2
2
22
LL
L
V V gz
z K gz V ( K ) V
g g K
Then the flow rate becomes
L
K
gz
D
VA
2
1
2
2pipe
2
4
V
L/s 10.0/sm 0.0100 3
1.11.05
m) 7)(m/s 81.9(2
4
)m 04.0(
2
4
22
2
1
2
2c
L
K
gz
D
VA
V
7 m
4 cm
1
2
14–99
14-116
Solution A swimming pool is initially filled with water. A pipe with a well-rounded entrance at the bottom drains the
pool to the atmosphere. The initial rate of discharge from the pool and the time required to empty the pool completely are to
= 0), and take the positive direction of z to be upwards. 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 very low (V1 0), the energy equation for a control
volume between these two points (in terms of heads) simplifies to
2
22
2
2
21e turbine,2
2
2
2
2
upump,1
2
1
1
1
LL h
g
V
zhhz
g
V
g
P
hz
g
V
g
P
where
g
V
D
L
fhL2
2
since the minor losses are negligible. Substituting and solving for V2 gives
DfL
gz
V
g
V
D
L
f
g
V
z/
2
22 2
1
2
2
2
2
2
21
m/s 808.1
m) m)/(0.05 0.022(251
m) 2)(m/s 81.9(2
/1
22
1
,2
DfL
gz
Vi
L/s 3.55 /sm 1055.3/4]m) (0.05m/s)[ 808.1()4/( 3322
,2,2
DVAV iciinitial
V
DfL
gz
D
VAc/1
2
4
2
2
V
Then the amount of water that flows through the pipe during a differential time interval dt is
dt
DfL
gz
D
dtd/1
2
4
2
VV
(1)
which, from conservation of mass, must be equal to the decrease in the volume of water in the pool,
dz
D
dzAd kc 4
)(
2
0
tan,
V
(2)
10 m
2 m
25 m
5 cm
Swimming
pool
2
1
14–100
where dz is the change in the water level in the pool during dt. (Note that dz is a negative quantity since the positive
direction of z is upwards. Therefore, we used –dz to get a positive quantity for the amount of water discharged). Setting
Eqs. (1) and (2) equal to each other and rearranging,
dzz
g
DfL
D
D
dz
gz
DfL
D
D
dtdz
D
dt
DfL
gz
D2
1
2
/1
2
/1
4/1
2
42
2
0
2
2
0
2
0
2
The last relation can be integrated easily since the variables are separated. Letting tf be the discharge time and integrating it
from t = 0 when z = z1 to t = tf when z = 0 (completely drained pool) gives
2
1
1
2
1
1
1
2
2
0
0
2
12
2
0
0
2/1
2
2
0
0 2
/1
2
2
/1
–
2
/1
z
g
DfL
D
D
z
g
DfL
D
D
tdzz
g
DfL
D
D
dt
z
f
zz
t
t
f
Simplifying and substituting the values given, the draining time is determined to be
h 24.6
s 480,88
m/s 81.9
m)] m)/(0.05 25)(022.0(m)[1 2(2
m) 05.0(
m) 10(
)/1(2
22
2
1
2
2
0
g
DfLz
D
D
tf
Checking: For plastic pipes, the surface roughness and thus the roughness factor is zero. The Reynolds number at the
beginning of draining process is
040,90
skg/m 10002.1
m) m/s)(0.05 808.1)(kg/m 998(
Re 3
3
2
DV
which is greater than 4000. The friction factor can be determined from the Moody chart, but to avoid the reading error, we
determine it from the Colebrook equation using an equation solver (or an iterative scheme),
D
51.2
1
51.2
/
1
14–101
14-117
Solution In the previous problem, the effect of the discharge pipe diameter on the time required to empty the pool
g=9.81
f=0.022
z1=2
14–102
14-118
Solution A swimming pool is initially filled with water. A pipe with a sharp-edged entrance at the bottom drains the
pool to the atmosphere. The initial rate of discharge from the pool and the time required to empty the pool completely are to
= 0), and take the positive direction of z to be upwards. 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 very low (V1 0), the energy equation for a control
volume between these two points (in terms of heads) simplifies to
2
22
2
2
21e turbine,2
2
2
2
2
upump,1
2
1
1
1
LL h
g
V
zhhz
g
V
g
P
hz
g
V
g
P
where
2 = 1 and
g
V
K
D
L
f
g
V
K
D
L
fhhhh LLLLLL 22
2
2
2
minor,major,total,
since the diameter of the piping system is constant. Substituting and solving for V2 gives
L
LKDfL
gz
V
g
V
K
D
L
f
g
V
z
/1
2
22
1
2
2
2
2
2
1
Noting that initially z1 = 2 m, the initial velocity and flow rate are determined to be
m/s 407.1
0.5m) m)/(0.03 0.022(251
m) 2)(m/s 81.9(2
/1
22
1
,2
L
iKDfL
gz
V
L/s 0.994/sm 1094.9/4]m) (0.03m/s)[ 407.1()4/( 3422
,2,2
DVAVV iciinitial
L
cKDfL
gz
D
VA
/1
2
4
2
2
V
2 m
2
10 m
1
14–103
dz
D
dzAd kc 4
)(
2
0
tan,
V
(2)
14–104
Design and Essay Problems
14-119 to 14-120