Question

Heat is available at a rate of \( 2 \, \text{kW} \) from a thermal reservoir at \( 400 \, \text{K} \). A two-stage process harnesses this heat to produce power. Stages 1 and 2 reject heat at \( 360 \, \text{K} \) and \( 300 \, \text{K} \), respectively. Stage 2 is driven by the heat rejected by Stage 1. If the overall process efficiency is \( 50\% \) of the corresponding Carnot efficiency, the power delivered by the process, in \( \text{kW} \), rounded off to 2 decimal places, is:

Hint:

For multi-stage Carnot engines, calculate the effective Carnot efficiency and account for the actual efficiency percentage to find the power delivered.

Answer 1

Step 1: Carnot efficiency for each stage. The Carnot efficiency for a heat engine is given by: \[ \eta_\text{Carnot} = 1 - \frac{T_\text{cold}}{T_\text{hot}}, \] where \( T_\text{hot} \) and \( T_\text{cold} \) are the source and sink temperatures, respectively. 1. For Stage 1: \[ \eta_1 = 1 - \frac{360}{400} = 0.1 \, \text{(10\%)}. \] 2. For Stage 2: \[ \eta_2 = 1 - \frac{300}{360} = 0.167 \, \text{(16.7\%)}. \] Step 2: Effective Carnot efficiency of the process. The overall Carnot efficiency is: \[ \eta_\text{overall} = \eta_1 + (1 - \eta_1) \cdot \eta_2. \] Substitute values: \[ \eta_\text{overall} = 0.1 + (1 - 0.1) \cdot 0.167 = 0.1 + 0.9 \cdot 0.167 = 0.1 + 0.1503 = 0.2503 \, \text{(25.03\%)}. \] Step 3: Actual efficiency. The actual process efficiency is \( 50\% \) of the Carnot efficiency: \[ \eta_\text{actual} = 0.5 \cdot 0.2503 = 0.12515 \, \text{(12.52\%)}. \] Step 4: Power delivered by the process. The power delivered by the process is: \[ P = \eta_\text{actual} \cdot \text{Heat input}. \] Substitute \( \eta_\text{actual} = 0.12515 \) and \( \text{Heat input} = 2 \, \text{kW} \): \[ P = 0.12515 \cdot 2 = 0.25 \, \text{kW}. \] Step 5: Conclusion. The power delivered by the process is \( 0.25 \, \text{kW} \).