Chapter 5, Problem 5.1P

To determine

The relation between the reversible and irreversible Q_L

Answer to Problem 5.1P

  $Q_{L,\text{reversible}}<Q_{L,\text{irreversible}}$

Explanation of Solution

Given information:

Two engines receive the same heat transfer in (Q_H). One engine is reversible and other engine is irreversible.

Concept used:

The efficiency of reversible heat engine is more than the efficiency of irreversible heat engine.

  ${\eta }_{\text{reversible}}>{\eta }_{\text{irreversible}}$

  $W=Q_H-Q_L$

Here, W is the work done, Q_H is the heat transfer in and Q_L is the heat transfer out.

Calculation:

We know that,

  $\begin{array}{l}{\eta }_{\text{reversible}}>{\eta }_{\text{irreversible}}\\ \frac{W_{\text{reversible}}}{Q_H}>\frac{W_{\text{irreversible}}}{Q_H}\end{array}$

Two engines have same Q_H, thus

  $\begin{array}{c}{W}_{\text{reversible}}>W_{\text{irreversible}}\\ Q_H-Q_{L,\text{reversible}}>Q_H-Q_{L,\text{irreversible}}\\ -Q_{L,\text{reversible}}>-Q_{L,\text{irreversible}}\\ Q_{L,\text{reversible}}<Q_{L,\text{irreversible}}\end{array}$

Conclusion:

The heat transfer out in reversible engine is less than the heat transfer out in irreversible engine.