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%