Chapter 11
166
CHAPTER ELEVEN
SERIES PIPE LINE SYSTEMS
SERIES PIPE LINE SYSTEMS
169
SERIES PIPE LINE SYSTEMS
171
C
l
a
ss
II
s
y
s
t
e
m
s
11.8 Class II hL = 30.0 ft = f
2
2
L
D g
Method IIC Iteration
Try f = 0.02; then
= 24.64 / f = 35.1 ft/s
Class II [Repeated using computational approach – Sec. 11.4]
hL = 30.0 ft; D/ = 2120; Use Eq. 11-3.
11.9 Class II
22
1 1 2 2
21
2 2
L
p p
z h z
g g
: 1 2
1 2
z z
Method IIC Iteration
hL =
22
1 2
3
268 kN/ m = 7.70 m = ; =
((0.90)(9.81 kN/ m ) 2g
LDhgpp L
f = D fL
Spreads
heet solutions to Problems 11.8, 11.9, and 11.10 are on next page.
Method IIA
SERIES PIPE LINE SYSTEMS
173
SERIES PIPE LINE SYSTEMS
175
Chapter 11
178
11.1
4
C
l
ass
I
I
P
t
.
1
a
t
s
u
r
f
ace
of
tan
k
A
;
Pt.
2
in
s
t
rea
m
out
s
i
de
p
i
pe.
1
= 0,
p
2
= 0
222
212211
2121
: ( )
2 2 2 LL
p p p
z h z z z h
g g g
In Eq. I:
13.59 m =
222
222 (2.36 638 )
2 2 2
f f
g g g
11.15 Class II with 2 pipes: Pts. A and B at tank surfaces. Method IIC
22
A A B B
BA
2 2
L
p p
z h z
g g
: zA zB = hL = 10 m: A B
A B
0
0
p p
= 0.0203
Entrance Elbows Friction Enlarge. 2
0.1593 1.504
D
Friction Elbow Valve Exit
(3.36 638 )
Copper Tube:
OD = 50 mm ; 1.5 mm wall
Chapter 11
180
[11.15 solution continued]
Solve for 6
L
gh
Iterate for both f4 and f6:
4
5
3444
4
74
(0.1059)
0.1059 m 883: 1.614 10 ( )
1.2 10 m 6.56 10
R
DD N
4 = 2.263
6 = 3.76 m/s
6 = 196.2
SERIES PIPE LINE SYSTEMS
181
11.16
C
l
ass
II
w
i
th
t
w
o
p
i
p
e
s
P
t
. B in
s
t
r
ea
m
outsi
d
e
p
i
pe.
p
B
= 0
Metho
d
II
C
2222
A A B B A B A
BABA
: + +
2 2 2 LL
ooo
p p p
z h z z z h
g g g
hL = (0.52 + 638f)
2
2
BA
8
(0.53 1075 )
2f
g
In Eq. I
14.68 =
2222
BBBB
BABA
15.3
+ [8.49 + 9761 + 1075 ] = [ 5.81 + 9761 + 1075 ]
2g
+
= 0.3 mm
Chapter 11
184
SERIES PIPE LINE SYSTEMS
185