or
1
2
Also, from Eq. (1),
312
V
VV=−
(11)
Solution method:
(a) Guess values of f1, f2, and f3 (A good starting value is the large Re value for
Problem 8.119
The three water-filled tanks shown in the figure below are connected by pipes as indicated.
If minor losses are neglected, determine the flowrate in each pipe.
Solution 8.119
or
−= +
22
12
22
200 m 200 m
60 m 20 m (0.015) (0.020)
mm
0.1 m 0.08 m
29.81 29.81
ss
VV
(2)
Hence,
Elevation = 20 m
Elevation = 60 m
Elevation = 0
= 0.10 m
= 200 m
= 0.015
D
𝓵
f
= 0.08 m
= 400 m
= 0.020
D
𝓵
f
= 0.08 m
= 200 m
= 0.020
D
𝓵
f
Elevation = 20 m
Elevation = 60 m
Elevation = 0
= 0.10 m
= 200 m
D
𝓵
(B)
•
(A)
•
(C)
•
or
=+
2
23
1
22
200 m 400 m
60 m (0.015) (0.020)
mm
0.1 m 0.08 m
2 9.81 2 9.81
ss
V
V (3)
Hence,
22
60 1.529 5.10VV=+
Thus, from Eqs. (4) and (5),
()
2
22
33 3
8
.14 2 784 95.8 0VV V+−+−=
This can be simplified to
Thus,
π
== =
3
2
333
mm
(0.08 m) 2.80 0.0141
4ss
QAV
Also, from Eq. (3):
22
1
60 1.529 5.10(2.80)V=+ or =
1
m
3.62 s
V
Problem 8.120
Five oil fields, each producing an output of Q barrels per day, are connected to the 28-in.-
diameter “mainline pipe” (A – B – C) by 16-in.-diameter “lateral pipes” as shown in the
figure below. The friction factor is the same for each of the pipes and elevation effects are
negligible. (a) For section A – B, determine the ratio of the pressure drop per mile in the
mainline pipe to that in the lateral pipes. (b) Repeat the calculations for section B – C.
Solution 8.120
For any of the pipe sections,
2
pV
fDg
Δ= or 2
1
p
Vf
ρ
Δ
=
Also,
QAB = 3Q so that 22
3
44
AB AB lat lat
DV DV
ππ
= or
2
3lat
AB
lat AB
D
V
VD
=
Thus, Eq. (1) becomes
p
(b) Similarly, for section BC:
p
Lateral
Main line
Q
Q
Q
Q
Q
ABC
Δ
Δ
Also,
Problem 8.122
Water flows through the orifice meter shown in the figure below at a rate of 0.10 cfs. If d =
0.1 ft, determine the value of h and the pressure difference associated with this h.
Solution 8.122
12
00 1 2
4
2( ) 0.1 ft
where 0.6,
2
(1 ) ft
12
pp d
QCA p p h gh
D
βγρ
ρβ
−
====−==
− (1)
Also,
h
Q
d
2 in.
0.66
0.64
2
D
D
DV
d
Therefore, from Eq. (1):
Problem 8.123
Water flows through the orifice meter shown in the figure below such that h = 1.6 ft with d
= 1.5 in. Determine the flowrate.
Solution 8.123
12
00 1 2
4
2( ) 1.5 in.
where 0.75, and
2in.
(1 )
pp d
QCA p p h gh
D
βγρ
ρβ
−
====−==
− (1)
Thus,
h
Q
d
2 in.
From the figure above, with this Re and
β
, C0 = 0.62 ≠ 0.6 (the assumed value)
Assume C0 = 0.62 or ==
3
ft
0.151(0.62) 0.936 s
Q. Thus V = 45.8(0.0936) or =ft
4.29 s
V
and
0.66
0.64
2
D
D
DV
d
Problem 8.124
Water flows through the orifice meter shown in the figure below at a rate of 0.10 cfs. If h =
3.8 ft, determine the value of d.
Solution 8.124
12
00 1 2
4
2( ) ,where ,
2
(1 ) ft
12
pp dd
QCA p p h gh
D
βγρ
ρβ
−
===−==
− (1)
and
h
Q
d
2 in.
Trial and error solution: From Eq. (1)
Assume
β
= 0.6, or 22
0.6 0.10 ft
12 12
d
β
== = . Thus from Eq. (2), C0 = 0.759. However, from
d
d
Problem 8.125
Water flows through a 40-mm-diameter nozzle meter in a
7
5-mm-diameter pipe at a rate of
3
m
0.15 s. Determine the pressure difference across the nozzle if the temperature is (a) 10 °C,
or (b) 80 °C.
Solution 8.125
or
1
2
12 12
32
kg N
8.09 , where ~ , ~
mm
n
C
pp pp
ρρ
−= − (1)
Also, (0.075 m)
Re VD V
vv
== , with
ππ
== =
3
22
m
0.015 m
s3.40 s
(0.075 m)
44
Q
V
D
(a) Assume T = 10 °C, or from Table B.2 Physical Properties of Water (SI Units):
Vd
From Eq. (1):
1
2
12
0.986 8.09(999.7)pp−= or 4
12 2
N
6.73 10 m
p
p−= ×
Thus, p1 – p2 = 67.3 kPa
(b) Assume T = 80 °C, or from the table above: 3
kg
971.8 m
ρ
=, −
=×
2
7m
3.65 10 s
v,
p
1.00
0.98
0.6
0.4
0.2
Problem 8.126
Gasoline flows through a 35-mm-diameter pipe at a rate of
3
m
0.0032 s. Determine the
pressure drop across a flow nozzle placed in the line if the nozzle diameter is 20 mm.
Solution 8.126
2
12
4
2( ) 20 mm
, where 0.571,
35 mm 4
(1 )
nn n
pp d
QCA A d
D
β
β
π
ρ
−
=====
− (1)
From Table 1.6 Approximate Physical Properties of Some Common Liquids (SI Units)
Hence, from the figure above, Cn = 0.986
From Eq. (1),
1.00
0.98
0.6
0.4
0.2