Question

An ideal reheat Rankine cycle operates between the pressure limits of \(10 \mathrm{kPa}\) and \(8 \mathrm{MPa}\), with reheat occurring at 4 MPa. The temperature of steam at the inlets of both turbines is \(500^{\circ} \mathrm{C}\), and the enthalpy of steam is \(3185 \mathrm{kJ} / \mathrm{kg}\) at the exit of the high-pressure turbine, and \(2247 \mathrm{kJ} / \mathrm{kg}\) at the exit of the low-pressure turbine. Disregarding the pump work, the cycle efficiency is \((a) 29\) percent \((b) 32\) percent \((c) 36\) percent \((d) 41\) percent \((e) 49\) percent

Short answer

Disregard the pump work. Solution: 1. Calculate the total heat added during the cycle: \(Q_{\text{in}} = (h_{2} - h_{1}) + (h_{4} - h_{3})\) 2. Determine the work done during the cycle: \(W_{\text{cycle}} = Q_{\text{in}} - (h_{4} - h_{5})\) 3. Calculate the cycle efficiency: \(\eta = \frac{W_{\text{cycle}}}{Q_{\text{in}}} \times 100\%\) 4. Find the correct answer choice based on the calculated efficiency.

Step-by-step solution

Determine the heat added during the cycle

To find the heat added during the cycle, we need to calculate the heat added in both the boiler and the reheater. Let's denote the specific enthalpy and specific heat for each state as \(h_{1}, h_{2}, h_{3}, h_{4}\). We already know \(h_{1} = 3185 \text{ kJ/kg}\) and \(h_{4} = 2247 \text{ kJ/kg}\). From the given information, we know that reheating occurs at \(4 \text{ MPa}\) and \(500^{\circ}C\). Using steam tables, we can find the specific enthalpy at these conditions: \(h_{2} = h_{\text{500^{\circ}C, 4 MPa}} = 3428 \text{ kJ/kg}\) Similarly, the specific enthalpy at the inlet of the low-pressure turbine (state 3) will be the same as the specific enthalpy at the exit of the high-pressure turbine: \(h_{3} = h_{1} = 3185 \text{ kJ/kg}\) Now, we can calculate the total heat added during the cycle: \(Q_{\text{in}} = (h_{2} - h_{1}) + (h_{4} - h_{3})\)

Determine the work done during the cycle

Since we are disregarding the pump work, the work done during the cycle, \(W_{\text{cycle}}\), can be calculated by subtracting the heat rejected in the condenser from the total heat added: \(W_{\text{cycle}} = Q_{\text{in}} - Q_{\text{out}}\) We know \(Q_{\text{in}}\) from Step 1, and we can find \(Q_{\text{out}}\) by calculating the enthalpy difference in the condenser: \(Q_{\text{out}} = h_{4} - h_{5}\) Using the given condition of \(10 \text{ kPa}\) at the condenser exit and assuming saturated liquid, we can find the specific enthalpy from the steam tables: \(h_{5} = h_{\text{10 kPa, sat liquid}} = 191.8 \text{ kJ/kg}\) Now, we can calculate \(W_{\text{cycle}}\): \(W_{\text{cycle}} = Q_{\text{in}} - (h_{4} - h_{5})\)

Calculate the cycle efficiency

With \(W_{\text{cycle}}\) and \(Q_{\text{in}}\) values calculated in the previous steps, we can now find the cycle efficiency: \(\eta = \frac{W_{\text{cycle}}}{Q_{\text{in}}} \times 100\%\)

Find the correct answer choice

After calculating the cycle efficiency in Step 3, look for the answer choice that matches the result and select it as the correct answer.