Question 10.4: Problem: A regenerator with an effectiveness of 80% is added...
Problem: A regenerator with an effectiveness of 80% is added to the Brayton cycle of Example 10.3. Find the efficiency of the cycle with the regenerator.
Find: The efficiency η_{th, \text{Brayton}} of the cycle in Example 10.3 with a regenerator.
Known: Regenerator effectiveness ε = 0.8, Example 10.3 Brayton cycle with regeneration.
Diagram: Figure E10.4.
Assumptions: Air is an ideal gas with temperature dependent properties.
Governing Equation:
Effectiveness of Brayton cycle ε = \frac{ h_5 – h_2}{h_4 – h_2}
Efficiency of Brayton cycle η_{th, \text{Brayton}} = \frac{w_{net}}{q_H} = 1 – \frac{q_C}{q_H} = 1- \frac{ h_4 – h_1}{h_3 – h_2}
From the solution to Example 10.3, specific enthalpies h_1 = 300.19 \ kJ / kg, \ h_2 = 579.8650 \ kJ / kg, \ h_3 = 1348.55 \ kJ / kg, \text{ and } h_4 = 715.1786 \ kJ / kg.
Enthalpy of the air at the exit of the regenerator is
h_5 = h_2 + ε (h_4 – h_2),
h_5 = 579.8650 \ kJ/ kg + 0.8 \times (715.1786 \ kJ / kg – 579.8650 \ kJ / kg) = 688.12 \ kJ/kg.
The new Brayton cycle efficiency is then
η_{th, \text{Brayton}} = \frac{w_{net}}{q_{H}} = \frac{w_{t} – w_{c}}{q_{H}} = \frac{(h_{3} – h_{4}) – (h_{2} – h_{1})}{h_{3} – h_{5}}
η_{th, \text{Brayton}} = \frac{ (1348.55 \ kJ/ kg – 715.1786 \ kJ / kg ) – ( 579.8650 \ kJ / kg – 300.19 \ kJ / kg ) }{1348.55 \ kJ / kg – 688.12 \ kJ / kg } = 0.535 \ 55 .
Answer: The new cycle efficiency is 53.6%.