Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity at inlet
velocity at outlet
steady flow energy equation



solving wc = 1830.64 kJ/kg
wc in KWH
we know that



Answer:
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Explanation:
yww <33
Answer:
Gravitational force (pulled downward by the Earth)
Normal force (pushed upward by the ground)
Applied force (pushed by the person)
Friction force (pulled opposite the direction of motion by the roughness of the ground)
Answer:
Mechanical power of pump is 74.07%.
Explanation:
Power of motor = 15 KW
Efficiency of motor= 90%
So the actual power(P) supplied by motor = 0.9 x 15 KW
P=13.5 KW
Water flow rate = 50 L/s
Volume flow rate = 50 L/s
We know that

So

We know that pump is an open system and work input for open system can be calculated as
W=VΔP
ΔP is the pressure difference
V is the volume flow rate
So by putting the values
W=0.05 (300-100) (here ΔP=300 - 100=200 KPa)
W=10 KW
So mechanical power of pump
η =0.7407
Mechanical power of pump is 74.07%.
Answer:
See explaination and attachment
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Have the solution as an attachment.