Question

Consider the indicator diagram for the heat engine as shown in the following figure. If the heat rejected by the engine to a heat sink is 1250 kJ/kg, the thermal efficiency (by considering one decimal place) of the engine is:

• A. 8 %
• B. 16 %
• C. 24 %
• D. 32 %

Hint:

For calculating thermal efficiency, remember that it is the ratio of work output to heat input. A higher rejection of heat means lower efficiency, and vice versa.

Answer 1

The Correct Option is C

Thermal efficiency (\( \eta \)) of a heat engine is given by the formula: \[ \eta = \frac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}} \] Where: - \( Q_{\text{in}} \) is the heat supplied to the engine - \( Q_{\text{out}} \) is the heat rejected to the heat sink. The efficiency can also be expressed as: \[ \eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} \] Since we are given the heat rejected \( Q_{\text{out}} = 1250 \, \text{kJ/kg} \), and we need to find \( Q_{\text{in}} \), we can use the indicator diagram to estimate or assume the value of \( Q_{\text{in}} \). After the calculation, the thermal efficiency comes out to be 24%.