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Minxue (Kevin) He
CEE151 W08
Homework #3 Solution Set
Page  1  of 11
1. Solution:
Assume that three pipes shown in Figure P4.5.2 are pipe A, pipe B, and pipe C from the left to
right, respectively.
Head loss in these pipes (
li
h
) can be expressed as:
2
2
2
2
2
2
A
A
lA
A
A
B
B
lB
B
B
C
C
lC
C
C
L V
h
f
D
g
L V
h
f
D
g
L V
h
f
D
g
=
=
=
……………………………………(11)
where
, ,
i
A B C
=
;
li
h
is
the head loss in pipe
i
;
i
f
is the friction coefficient in pipe
i
;
i
L
is the
length of pipe
i
;
i
D
is the diameter of pipe
i
;
i
V
is the flow velocity in pipe
i
.
It can be observed that there is a sudden expansion from pipe A to pipe B. According to Equation
(4.3.20) in the textbook, this head loss (
lAB
h
) can be expressed as:
2
(
)
2
A
B
lAB
V
V
h
g

=
……………………………….….
.(12)
It can also be observed that there is a sudden contraction from pipe B to pipe C. From Table 4.3.2
in the textbook, the head loss here (
lBC
h
) can be expressed as:
2
2
C
lBC
C
V
h
K
g
=
…………………………………….
.(13)
Using the center line of the pipes as the base line (elevation datum), then the elevation head at
both points 1 and 2 are zero. Applying energy equation between point 1 and point 2 with
Equations (11) to (13) considered:
2
2
2
2
2
2
2
1
2
(
)
0
0
2
2
2
2
2
2
2
C
C
C
C
A
A
A
B
B
A
B
A
B
C
C
A
B
C
V
L V
V
P
V
P
L V
L V
V
V
f
f
f
K
g
g
D
g
D
g
D
g
g
g
γ

+
+ =
+
+ +
+
+
+
+
….…(14)
where
1
P
and
2
P
represent the pressure at points 1 and 2, respectively.
From continuity equation:
3
2
2
2
3
2
2
3
2
2
4
4 0.05
/
2.831 /
(0.15)
4 0.05
/
0.874 /
(0.27)
4 0.05
/
1.316 /
(0.22)
A
A
A
B
C
Q
Q
m
s
V
m s
A
D
m
m
s
V
m s
m
m
s
V
m s
m
π
×
=
=
=
=
×
=
=
×
=
=
……….….
.(15)
Assume
3
9810
/
N m
=
, substituting the value of
1
P
and Equation (15) above into Equation
(14) gives:
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View Full Document Minxue (Kevin) He
CEE151 W08
Homework #3 Solution Set
Page  2  of 11
2
2
2
3
2
2
2
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
2
260000
2.831
/
9810
/
2(9.81) /
1.316
/
300
2.831
/
250
0.874
/
9810
/
2(9.81)
/
0.15
2(9.81)
/
0.27
2(9.81)
/
400
1.316
/
(2.831 0.874)
/
0.22
2(9.81) /
2(9.8
A
B
C
Pa
m
s
N m
m s
P
m
s
m
m
s
m
m
s
f
f
N m
m s
m
m s
m
m s
m
m
s
m
s
f
m
m s
+
=
+
+
+

+
+
2
2
2
2
2
1.316
/
0.05
1)
/
2(9.81)
/
m
s
m s
m s
+
……(16)
Equation (16) above can be simplified as:
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This note was uploaded on 03/11/2008 for the course CEE 151 taught by Professor Mugg during the Spring '08 term at UCLA.
 Spring '08
 Mugg

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