A heat engine receives heat from a source of 1200 K at a rate of 400 kW and rejects the waste heat to a medium at 300 K. The power output of the heat engine is 160 kW. What will be the irreversibility rate for this process?

Concept:

The irreversibility rate in a heat engine is calculated using the formula:

\( I = T_0 \Delta S_{\text{gen}} \)

Where:

• \( I \) = Irreversibility rate (kW)
• \( T_0 \) = Surrounding temperature (300 K)
• \( \Delta S_{\text{gen}} \) = Entropy generation rate

The entropy generation rate is given by:

\( \Delta S_{\text{gen}} = \frac{Q_{\text{in}}}{T_{\text{source}}} - \frac{Q_{\text{out}}}{T_{\text{sink}}} \)

Calculation:

Given:

• Heat supplied: \( Q_{\text{in}} = 400~kW\)
• Heat rejected: \( Q_{\text{out}} \) (to be determined)
• Power output: \( W_{\text{out}} = 160~kW\)
• Source temperature: \( T_{\text{source}} = 1200~K\)
• Sink temperature: \( T_{\text{sink}} = 300~K\)

Applying the First Law of Thermodynamics:

\( Q_{\text{in}} = Q_{\text{out}} + W_{\text{out}} \)

\( Q_{\text{out}} = 400 - 160 = 240~kW\)

Step 2: Compute Entropy Generation Rate

\( \Delta S_{\text{gen}} = \frac{400}{1200} - \frac{240}{300} \)

\( = 0.333 - 0.8 \)

\( = -0.467~kW/K\)

Taking absolute value,

\( \Delta S_{\text{gen}} = 0.467~kW/K\)

Step 3: Compute Irreversibility Rate

\( I = T_0 \times \Delta S_{\text{gen}} \)

\( I = 300 \times 0.467 \)

\( I = 140~kW\)