Question

In a steam power plant based on Rankine cycle, steam is initially expanded in a high-pressure turbine. The steam is then reheated in a reheater and finally expanded in a low-pressure turbine. The expansion work in the high-pressure turbine is 400 kJ/kg and in the low-pressure turbine is 850 kJ/kg, whereas the pump work is 15 kJ/kg. If the cycle efficiency is 32%, the heat rejected in the condenser is ________________ kJ/kg (round off to 2 decimal places).

Hint:

In Rankine cycle problems, the heat rejected can be found by subtracting the net work output from the total heat input.

Answer 1

Correct Answer: 2620

The efficiency of the Rankine cycle is given by: \[ \eta = \frac{\text{Net Work Output}}{\text{Total Heat Input}} = \frac{W_{\text{HP}} + W_{\text{LP}} - W_{\text{pump}}}{Q_{\text{in}}} \] Where:
- \( W_{\text{HP}} = 400 \, \text{kJ/kg} \) is the work output in the high-pressure turbine,
- \( W_{\text{LP}} = 850 \, \text{kJ/kg} \) is the work output in the low-pressure turbine,
- \( W_{\text{pump}} = 15 \, \text{kJ/kg} \) is the pump work.
The total work output is: \[ W_{\text{total}} = W_{\text{HP}} + W_{\text{LP}} - W_{\text{pump}} = 400 + 850 - 15 = 1235 \, \text{kJ/kg}. \] Now, using the given cycle efficiency (\( \eta = 32% = 0.32 \)): \[ 0.32 = \frac{1235}{Q_{\text{in}}} \] Solving for \( Q_{\text{in}} \): \[ Q_{\text{in}} = \frac{1235}{0.32} = 3859.375 \, \text{kJ/kg}. \] The heat rejected in the condenser \( Q_{\text{out}} \) is: \[ Q_{\text{out}} = Q_{\text{in}} - W_{\text{total}} = 3859.375 - 1235 = 2624.375 \, \text{kJ/kg}. \] Thus, the heat rejected in the condenser is \( \boxed{2620.00} \, \text{kJ/kg} \).