Question

A reversible power cycle whose thermal efficiency is 43% receives 50 kJ by heat transfer from a hot reservoir at 300oC and rejects energy by heat transfer to a cold reservoir at temperature TC. Determine the energy rejected, in kJ, and TC, in oC. Determine the entropy production for the cycle, σcycle, in kJ/K.

Answer 1

To solve this problem it is necessary to take into account the Efficiency values in the energy cycles.

Efficiency can be defined as,

$\eta = 1 - \frac{Q_c}{Q_h} = 1- \frac{T_c}{T_h}$

Where,

Q = Heat Exchange

T = Temperature

The efficiency for this system is 0.43 then,

$0.43 = 1-\frac{Q_c}{50}$

Re-arrange to find $Q_c$

$Q_c = 50*(1-0.43)$

$Q_c = 28.5kJ$

And the temperature would be given as,

$\eta = 1- \frac{T_c}{T_h}$

$0.43 = 1 - \frac{T_c}{573.15K}$

$T_c = 513.15K(1-0.43)$

$T_c = 307.89K$

The entropy would be given as

$\Delta S = S_f - S_i$

$\Delta S = \frac{Q_c}{T_c}-\frac{Q_h}{T_h}$

$\Delta S = \frac{28.5}{ 307.89}-\frac{50}{573.15}$

$\Delta S = 5.32*10^{-3} kJ/K$