A heat engine receives heat from a source at 1100 K at a rate of 400 kJ/s, and it rejects the waste heat to a medium at 320 K. The measured power output of the heat engine is 120 kW, and the environment temperature is 25∘C. Determine
(a) the reversible power,
(b) the rate of irreversibility and
(c) the second-law efficiency of this heat engine

part (a)   283.6 KW

part (b)  163.6 KW

part (c)   42.3%

Explanation:

The reversible power is the power produced by a reversible heat engine operating between the specified temperature limits.

part (a)  

η_th(max)=η_th(rev)=1-T_l/T_h

we replace the values in the equation

η_th(max)=1-(320 K/1100 K)

               = 0.7091

therefore

W_rev(out)=η_th(rev)Q_in

we replace the values in the equation

W_rev(out)=(0.7091)*(400 KJ/s)

W_rev(out)=283.6 KW

The irreversibly rate is the difference between the reversible power and the actual power output.

part (b)

I=W_rev(out)-W_u(out)

we replace the values in the equation

I=163.6 KW

The second law efficiency is determined from its definition.

part (c)

η_2=W_u(out)/W_rev(out)

we replace the values in the equation

η_2=42.3%

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explanation:

the answer would be   dignified - apex