Question

Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8 the compressor inlet temperature is \(300 \mathrm{K},\) and the turbine inlet temperature is \(1800 \mathrm{K}\). The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of \(60 \mathrm{MW},\) assuming both the compressor and the turbine have an isentropic efficiency of \((a) 100\) percent and \((b) 80\) percent.

Short answer

Question: Calculate the thermal efficiency and required mass flow rate of helium gas for a net power output of 60 MW, given a regenerative Brayton cycle with a pressure ratio of 8, compressor inlet temperature of 300 K, turbine inlet temperature of 1800 K, and regenerator effectiveness of 75%, for both the ideal and real cases with isentropic efficiencies of 100% and 80%, respectively. Answer: For the ideal case, the thermal efficiency is 2.62, and the required mass flow rate is 42.81 kg/s. For the 80% isentropic efficiency case, the thermal efficiency is 1.88 and the required mass flow rate is 29.52 kg/s.

Step-by-step solution

Calculate the ideal and real temperature at the exit of the compressor.

First, we need to find the ideal temperature at the exit of the compressor, using the isentropic relation: \(T_{2,ideal} = T_1 \cdot (PR)^{(\gamma-1)/\gamma}\) where \(T_1 = 300 \ \text{K}\), \(PR = 8\), and \(\gamma = 1.4\) for helium. This gives: \(T_{2,ideal} = 300 \cdot (8)^{(1.4-1)/1.4} = 673.4\ \text{K}\) For isentropic efficiency, we have the relation: \(\eta_{c,real} = \frac{T_{2,ideal}-T_1}{T_{2,real}-T_1}\) Solving for \(T_{2,real}\) gives: \(T_{2,real} = T_1 + \frac{T_{2,ideal}-T_1}{\eta_{c,real}}\) Substitute the values for the case of 80% isentropic efficiency: \(T_{2,real} = 300 + \frac{673.4-300}{0.8} = 768 \ \text{K}\)

Calculate the ideal and real temperature at the exit of the turbine.

For ideal case, use the isentropic relation again: \(T_{4,ideal} = T_{3} / (PR)^{(\gamma-1)/\gamma}\) where \(T_3 = 1800 \ \text{K}\). This gives: \(T_{4,ideal} = 1800 / (8)^{(1.4-1)/1.4} = 741.9 \ \text{K}\) For 80% isentropic efficiency, we have the relation: \(\eta_{t,real} = \frac{T_3-T_{4,real}}{T_3-T_{4,ideal}}\) Solving for \(T_{4,real}\) gives: \(T_{4,real} = T_3 - \eta_{t,real}(T_3-T_{4,ideal})\) Substitute the values for the case of 80% isentropic efficiency: \(T_{4,real} = 1800 - 0.8(1800-741.9) = 1057.3 \ \text{K}\)

Calculate the temperature at the exit of the regenerator for each case.

The regenerator effectiveness \(\epsilon\) is given as 75 percent and it is defined as: \(\epsilon = \frac{T_5 - T_2}{T_4 - T_2}\) Solving for \(T_5\) gives: \(T_5 = T_2 + \epsilon(T_4 - T_2)\) For the ideal case, \(T_5 = 673.4 + 0.75(741.9 - 673.4) = 723.6\ \text{K}\) And for the 80% isentropic efficiency case, \(T_5 = 768 + 0.75(1057.3 - 768) = 987\ \text{K}\)

Calculate the thermal efficiency for each case.

Thermal efficiency of the regenerative Brayton cycle can be calculated as: \(\eta_{th} = \frac{T_{3}-T_5}{T_5-T_1}\) For the ideal case, \(\eta_{th} = \frac{1800 - 723.6}{723.6 - 300} = 2.62\) And for the 80% isentropic efficiency case, \(\eta_{th} = \frac{1800 - 987}{987 - 300} = 1.88\)

Calculate the required mass flow rate for a net power output of 60 MW.

The net power output can be calculated as: \(W_{net} = \dot{m}C_p(T_{3} - T_{5}) - \eta_{th}C_p(T_{5} - T_{1})\Delta T_{21}\) Now, let \(\dot{m}\) be the mass flow rate of helium, and solve for it: \(\dot{m} = \frac{W_{net}}{C_p(T_{3} - T_{5}) - \eta_{th}C_p(T_{5} - T_{1})\Delta T_{21}}\) For the ideal case with \(C_p = 5.2\ \text{kJ/kgK}\), \(\dot{m} = \frac{60 \times 10^3}{5.2(1800 - 723.6) - 2.62(5.2)(723.6 - 300)} = 42.81 \ \text{kg/s}\) And for the 80% isentropic efficiency case, \(\dot{m} = \frac{60 \times 10^3}{5.2(1800 - 987) - 1.88(5.2)(987 - 300)} = 29.52 \ \text{kg/s}\) So, the required mass flow rates for cases (a) and (b) are 42.81 kg/s and 29.52 kg/s, respectively.