HEAT+MASS TRANSFER:FUND.+APPL.
6th Edition
ISBN: 9780073398198
Author: CENGEL
Publisher: RENT MCG
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Chapter 1, Problem 83P
To determine
The case for which energy required will be greater.
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Problem (14): A pump is being used to lift water from an underground
tank through a pipe of diameter (d) at discharge (Q). The total head
loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h
where (V) is the flow velocity in the pipe. The elevation difference
between the pump and tank surface is (h).
Given the values of h [cm], d [cm], and K [-], calculate the maximum
discharge Q [Lit/s] beyond which cavitation would take place at the
pump entrance. Assume Turbulent flow conditions.
Givens:
h = 120.31 cm
d = 14.455 cm
K = 8.976
Q
Answers:
(1) 94.917 lit/s
(2) 49.048 lit/s
( 3 ) 80.722 lit/s
68.588 lit/s
4
Problem (13): A pump is being used to lift water from the bottom
tank to the top tank in a galvanized iron pipe at a discharge (Q).
The length and diameter of the pipe section from the bottom tank
to the pump are (L₁) and (d₁), respectively. The length and
diameter of the pipe section from the pump to the top tank are
(L2) and (d2), respectively.
Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m],
calculate total head loss due to friction (i.e., major loss) in the
pipe (hmajor-loss) in [cm].
Givens:
L₁,d₁
Pump
L₂,d2
오
0.533 lit/s
L1 =
6920.729 m
d1 =
1.065 m
L2 =
70.946 m
d2
0.072 m
Answers:
(1)
3.069 cm
(2) 3.914 cm
( 3 ) 2.519 cm
( 4 ) 1.855 cm
TABLE 8.1
Equivalent Roughness for New Pipes
Pipe
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Equivalent Roughness, &
Feet
Millimeters
0.003-0.03 0.9-9.0
0.001-0.01 0.3-3.0
0.0006-0.003 0.18-0.9
0.00085
0.26
0.0005
0.15
0.045
0.000005
0.0015
0.0 (smooth) 0.0 (smooth)
Commercial steel or wrought iron 0.00015
Drawn…
The flow rate is 12.275 Liters/s and the diameter is 6.266 cm.
Chapter 1 Solutions
HEAT+MASS TRANSFER:FUND.+APPL.
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