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Allushta [10]
4 years ago
6

A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses

heat to atmosphere at 300 K. Assuming the block to be at steady state, determine (a) the net rate of heat transfer in kW; (b) the rate of entropy generation in the system's universe
Engineering
1 answer:
LUCKY_DIMON [66]4 years ago
6 0

Answer:

a) Zero

b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K

Explanation:

a) In steady state  

Net rate of Heat transfer = net rate of heat gain -  net rate of heat lost  

Hence, the rate of heat transfer = 0

b) In steady state, entropy generated  

ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]

Substituting the given values, we get –  

ds/dt = -[5/1500 + 3/1000 – (5+3)/300]

ds/dt = - [0.0033 + 0.003 -0.2666]

ds/dt = 0.2603 KW/K

 

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Given:

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% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

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