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 reheated 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).

The correct answer is 2624.37.

Step-by-Step Explanation:

To find the heat rejected in the condenser, we can analyze the energy distribution in the reheating Rankine cycle. Let's first list the given parameters:
- Expansion work in the high-pressure turbine (HPT), W_T1 = 400 kJ/kg
- Expansion work in the low-pressure turbine (LPT), W_T2 = 850 kJ/kg
- Pump work input, W_P = 15 kJ/kg
- Thermal efficiency of the cycle, η_cycle = 32% = 0.32

Step 1: Calculate the net work output (W_net) of the cycle
The net work output of a steam power plant is the total work produced by the turbines minus the work consumed by the pump:
$$W_{net}=W_{T1}+W_{T2}-W_P$$
Substituting the given values:
$$W_{net}=400+850-15=1235\text{ kJ/kg}$$

Step 2: Calculate the total heat input (Q_in) to the cycle
The thermal efficiency of the cycle is defined as the ratio of net work output to the total heat input:
$${\eta }_{cycl}=\frac{W_{net}}{Q_{in}}$$
Rearranging the equation to solve for Q_in:
$$Q_{in}=\frac{W_{net}}{{\eta }_{cycl}}$$
Substituting the values:
$$Q_{in}=\frac{1235}{0.32}=3859.375\text{ kJ/kg}$$

Step 3: Calculate the heat rejected in the condenser (Q_out)
Applying the first law of thermodynamics to the cycle, the net work output is equal to the net heat input (total heat input minus heat rejected):
$$W_{net}=Q_{in}-Q_{out}$$
Rearranging to find Q_out:
$$Q_{out}=Q_{in}-W_{net}$$
Substituting the calculated values:
$$Q_{out}=3859.375-1235=2624.375\text{ kJ/kg}$$
Rounding off to 2 decimal places as per the question requirements:
$$Q_{out}\approx 2624.37\text{ kJ/kg}$$