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Answered Same Day Dec 23, 2021

Solution

David answered on Dec 23 2021

114 Votes

Problem 2-29
Answer
Depth of the well = 75m , that means pump have to raise water to a height of 75m
g = 9.8m/s2
hence increase in potential energy of water = mgh
where m is mass of water to be raised with the help of pump.
Now , for this work to be done, pump has to provide mechanical energy must be equal to the potential
energy of water, so
Energy supplied by the pump = mgh
Energy supplied per kilogram of water(per unit mass) = gh (mgh/m)
= 9.8*75
= 735 Joules (Answer)
Problem 3-23
Answer
From the given relation we have
PVn = Constant
So for a two set of conditions we can write
P1 (V1)n = P2 (V2)n
Now as per question we have
P1 =14.6psia
P2 = 120psia
V1 = 0.33ft3/lbm
V2 = 0.57ft3/lbm
14.6*0.33n = 120*0.56n
On rea
anging it will give
(0.33/0.56)n = 120/14.6
On taking log on both sides
n log0.5892 = log8.219
n = log8.219/ log0.5892
n = 0.9148/0.2297
n = 3.98 = 4 (Answer) since n is an integer
Now, wok done in a polytropic process can be calculated as
W = (P2V2 – P1V1)/(1-n)
W = (120psia*0.57ft3/lbm - 14.6psia*0.33ft3/lbm)/(1-4)
W = (68.4-4.818)/(3)
W = -21.194 psia*ft3/lbm (answer)
Problem 4-11
Answer
(a) Flow rate = 60lbm/s
T = 680F
Flow rate at throat, station B and at exit station C can be determined using the concept of ratio
Flownozzle/Flowthroat = Athroat/Anozzle
60/Flowthroat = 3.14*12/3.14*2.52
Flow throat = 135lbm/s (Answer)
Flownozzle/Flowexit = Aexit/Anozzle...

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