Question

Writing the first- and second-law relations and \(\operatorname{sim}-\) plifying, obtain the reversible work relation for a steady-flow system that exchanges heat with the surrounding medium at \(T_{0}\) a rate of \(Q_{0}\) as well as a thermal reservoir at \(T_{R}\) at a rate of \(Q_{R} .\) (Hint: Eliminate \(\dot{Q}_{0}\) between the two equations.)

Short answer

The reversible work relation for the system is given by the following expression: \(\dot{W}_{rev} = \dot{Q}_R \left( \frac{T_0}{T_R} - 1\right)\)

Step-by-step solution

Apply the first law of thermodynamics for a steady-flow system

The first law of thermodynamics for a steady-flow system can be written as: \(\dot{W}_{rev} = \dot{Q} - \dot{Q}_{R}\) where \(\dot{W}_{rev}\) is the reversible work, \(\dot{Q}\) is the heat exchange with the surrounding medium, and \(\dot{Q}_{R}\) is the heat exchange with the thermal reservoir.

Apply the second law of thermodynamics for a steady-flow system

The second law of thermodynamics for a steady-flow system can be expressed as: \(\dot{S}_{gen} = \frac{\dot{Q}_0}{T_0} - \frac{\dot{Q}_R}{T_R}\) where \(\dot{S}_{gen}\) is the entropy generation for the system.

Use the entropy generation definition

In any reversible process, the entropy generated is zero. Therefore, for a reversible work relation, we set \(\dot{S}_{gen} = 0\). Consequently, we obtain: \(0 = \frac{\dot{Q}_0}{T_0} - \frac{\dot{Q}_R}{T_R}\)

Eliminate \(\dot{Q}_{0}\) using both equations

To eliminate \(\dot{Q}_{0}\), we first express it in terms of \(\dot{Q}_R\) from the second law equation: \(\dot{Q}_{0} = T_{0} \left( \frac{\dot{Q}_R}{T_R} \right)\) Now, substitute this expression for \(\dot{Q}_{0}\) in the first law equation: \(\dot{W}_{rev} = \left( T_{0} \left( \frac{\dot{Q}_{R}}{T_R} \right) \right) - \dot{Q}_{R}\)

Obtain the reversible work relation

By simplifying the above equation, we can obtain the reversible work relation for the steady-flow system: \(\dot{W}_{rev} = \dot{Q}_R \left( \frac{T_0}{T_R} - 1\right)\) This is the reversible work relation for the steady-flow system that exchanges heat with the surrounding medium at \(T_{0}\) and a rate of \(Q_{0}\) as well as a thermal reservoir at \(T_{R}\) at a rate of \(Q_{R}\).