Question

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine inlet temperature is \(800^{\circ} \mathrm{F}\). The turbine isentropic efficiency is 90 percent, the pump losses are negligible, and the cycle is sized to produce \(2500 \mathrm{kW}\) of power. Calculate the mass flow rate through the boiler, the power produced by the turbine, the rate of heat supply in the boiler, and the thermal efficiency.

Short answer

Based on the given information and calculations, the mass flow rate through the boiler is 5.49 kg/s, the power produced by the turbine is 2500 kW, the rate of heat supply in the boiler is 17944.38 kW, and the thermal efficiency of the steam Rankine cycle is 13.93%.

Step-by-step solution

Convert the given temperatures and pressures to SI units.

Since we are given the pressures and temperatures in different units, we need to convert them to SI units. 1 psia = 6.89476 kPa, so 1500 psia = 1500 * 6.89476 kPa = 10342.14 kPa (boiler pressure) 2 psia = 2 * 6.89476 kPa = 13.78952 kPa (condenser pressure) Temperature conversion: °F to °C = (°F - 32) * 5/9 800°F to °C: (800 - 32) * 5/9 = 426.67°C Add 273.15 to convert to Kelvin: 426.67°C + 273.15 = 699.82 K (turbine inlet temperature)

Determine the enthalpy at different stages of the cycle.

Using steam tables, we can find the specific enthalpies (\(h_1, h_2, h_3, h_4\)) and specific entropy (\(s_1, s_2, s_3\)) corresponding to different pressures and temperatures. 1. At the turbine inlet (Point 1): pressure \(P_1\) = 10342.14 kPa and temperature \(T_1\) = 699.82 K Using steam tables, find the enthalpy and entropy: \(h_1\) = 3494.33 kJ/kg \(s_1\) = 6.8355 kJ/kg·K 2. At the turbine outlet (Point 2): pressure \(P_2\) = 13.78952 kPa, and isentropic (i.e., \(s_2\) = \(s_1\)) Find the enthalpy for an isentropic process, \(h_{2s}\): \(s_{2}\) = 6.8355 kJ/kg·K \(h_{2s}\) = 2323.17 kJ/kg Now, find the actual enthalpy at point 2 using the turbine isentropic efficiency: \(\frac{h_1 - h_2}{h_1 - h_{2s}}\) = 0.9 \(h_2\) = \(h_1\) - 0.9(\(h_1 - h_{2s}\)) = 2563.93 kJ/kg 3. At the condenser outlet (Point 3): pressure \(P_3\) = 13.78952 kPa, saturated liquid Find the enthalpy: \(h_3\) = 206 kJ/kg 4. At the boiler inlet (Point 4): pressure \(P_4\) = 10342.14 kPa Enthalpy remains the same as in point 3 due to negligible pump loss: \(h_4\) = \(h_3\) = 206 kJ/kg

Calculate the mass flow rate through the boiler.

Using the given power production and the enthalpy difference in the turbine, we can find the mass flow rate. The rate of power production is equal to the product of mass flow rate (\(\dot{m}\)) and enthalpy difference in the turbine: \(2500 kW\) = \(\dot{m}\)(\(h_1 - h_2\)) Rearranging the equation, we can find mass flow rate: \(\dot{m}\) = \(\frac{2500 kW}{h_1 - h_2}\) = \(\frac{2500}{3494.33 - 2563.93}\) kg/s = 5.49 kg/s

Calculate the rate of heat supply in the boiler.

We can find the rate of heat supply (\(\dot{Q}_b\)) in the boiler using the mass flow rate and the enthalpy difference in the boiler: \(\dot{Q}_b\) = \(\dot{m}\)(\(h_1 - h_4\)) = 5.49(\(3494.33 - 206\)) kW = 17944.38 kW

Calculate the thermal efficiency of the cycle.

The thermal efficiency (\(\eta\)) of the Rankine cycle can be calculated as the ratio of the net power output to the heat supplied by the boiler: \(\eta\) = \(\frac{2500 kW}{17944.38 kW}\) = 0.1393 = 13.93% So, the mass flow rate through the boiler is 5.49 kg/s, the power produced by the turbine is 2500 kW, the rate of heat supply in the boiler is 17944.38 kW, and the thermal efficiency is 13.93%.