# Step-by-Step Solution: Regenerative Brayton Cycle Power Output ## **Given Data** - **Regeneration effectiveness:** \( \varepsilon = 100\% \) - **Compressor inlet:** \( P_1 = 95 \) kPa, \( T_1 = 290 \) K - **Turbine inlet:** \( P_3 = 880 \) kPa, \( T_3 = 110 \) K - **Heat supplied:** \( \dot{Q}_{in} = 34,000 \) kJ/s - **Air properties at room temp:** \( c_p = 1.005 \) kJ/kg⋅K, \( k = 1.4 \) - **Working fluid:** Air (ideal gas) - **Assume:** Steady-state, negligible kinetic/potential energy changes --- ## **1. Ideal Brayton Cycle with Perfect Regeneration** ### **a. Sketch and State Points** ``` 1 — Compressor inlet (P1, T1) 2 — Compressor exit (P2, T2) 3 — Turbine inlet (P3, T3) 4 — Turbine exit (P4, T4) Regenerator: 4→5 (pre-heats air before combustion) ``` Since effectiveness is 100%, air entering the combustor (point 2') is heated by the turbine exhaust to the max possible temp. --- ### **b. Find Pressure Ratio** \[ r_p = \frac{P_2}{P_1} = \frac{880}{95} \approx 9.263 \] --- ### **c. Find Compressor Outlet Temperature (\(T_2\))** For isentropic compression (ideal Brayton, constant \(k\)): \[ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} \] \[ T_2 = T_1 \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}} \] \[ T_2 = 290 \times (9.263)^{.286} \] \[ \ln(9.263) = 2.227;\quad .286 \times 2.227 \approx .637 \] \[ e^{.637} \approx 1.891 \] \[ T_2 = 290 \times 1.891 \approx 549.4~\text{K} \] --- ### **d. Find Turbine Exit Temperature (\(T_4\))** Isentropic expansion from \(P_3\) to \(P_4\) (which equals \(P_1\)): \[ \frac{T_4}{T_3} = \left(\frac{P_4}{P_3}\right)^{\frac{k-1}{k}} \] \[ T_4 = T_3 \left(\frac{P_1}{P_3}\right)^{.286} \] \[ T_4 = 110 \times (95/880)^{.286} \] \[ 95/880 = .108 \] \[ \ln(.108) = -2.224;\quad -2.224 \times .286 = -.636 \] \[ e^{-.636} \approx .529 \] \[ T_4 = 110 \times .529 \approx 581.9~\text{K} \] --- ### **e. Regenerator (Perfect, ε = 100%)** With perfect regeneration, the air leaving the regenerator (point 2') is heated to \(T_4\). So, \( T_{2'} = T_4 \). --- ### **f. Find Mass Flow Rate (\(\dot{m}\))** Net heat supplied in the combustor is: \[ \dot{Q}_{in} = \dot{m} c_p (T_3 - T_{2'}) \] \[ T_{2'} = T_4 \implies \dot{Q}_{in} = \dot{m} c_p (T_3 - T_4) \] \[ \dot{m} = \frac{\dot{Q}_{in}}{c_p (T_3 - T_4)} \] \[ T_3 - T_4 = 110 - 581.9 = 518.1~\text{K} \] \[ \dot{m} = \frac{34,000}{1.005 \times 518.1} \] \[ 1.005 \times 518.1 = 520.7 \] \[ \dot{m} = \frac{34,000}{520.7} \approx 65.3~\text{kg/s} \] --- ### **g. Find Net Power Delivered (\(\dot{W}_{\text{net}}\))** \[ \dot{W}_{\text{net}} = \dot{m} \left[ (T_3 - T_4) - (T_2 - T_1) \right] c_p \] \[ (T_3 - T_4) = 518.1~\text{K} \] \[ (T_2 - T_1) = 549.4 - 290 = 259.4~\text{K} \] \[ \dot{W}_{\text{net}} = 65.3 \times 1.005 \times (518.1 - 259.4) \] \[ = 65.3 \times 1.005 \times 258.7 \] \[ 1.005 \times 258.7 = 259.9935 \] \[ 65.3 \times 259.9935 \approx 16,980~\text{kW} \] --- ### **h. Final Answer (Constant Specific Heats)** \[ \boxed{16,980~\text{kW}} \] is the net power delivered by the plant, **assuming constant specific heats**. --- ## **2. Accounting for Variation of Specific Heats** ### **a. Use Table or Software Data** At higher temperatures, \( c_p \) increases. **Typical average values:** - For air, average \( c_p \) between 290 K and 110 K ≈ 1.13 kJ/kg⋅K. ### **b. Recalculate \( \dot{m} \) with Average \( c_p \) (\( \bar{c_p} \))** \[ \dot{m} = \frac{34,000}{1.13 \times 518.1} = \frac{34,000}{585.45} \approx 58.1~\text{kg/s} \] ### **c. Recalculate Net Power (\( \dot{W}_{\text{net}} \)), Using Enthalpy Differences** For variable \( c_p \), use enthalpy values \( h(T) \) from property tables: \[ \dot{W}_{\text{net}} = \dot{m} \left[ (h_3 - h_4) - (h_2 - h_1) \right] \] From air tables (\(h\) in kJ/kg): - \( h_1 = h(290~K) \approx 290.2 \) - \( h_2 = h(549.4~K) \approx 560.6 \) - \( h_3 = h(110~K) \approx 1164.6 \) - \( h_4 = h(581.9~K) \approx 597.5 \) So: - \( h_3 - h_4 = 1164.6 - 597.5 = 567.1 \) - \( h_2 - h_1 = 560.6 - 290.2 = 270.4 \) \[ \dot{W}_{\text{net}} = 58.1 \times (567.1 - 270.4) = 58.1 \times 296.7 \approx 17,240~\text{kW} \] --- ### **d. Final Answer (Variable Specific Heats)** \[ \boxed{17,240~\text{kW}} \] is the net power delivered by the plant, **accounting for variable specific heats**. --- ## **Summary Table** | Case | Net Power Output (kW) | |--------------------------|----------------------| | Constant \(c_p\) | 16,980 | | Variable \(c_p\) | 17,240 | --- ## **Key Points** - **More accurate:** Accounting for variable \(c_p\) gives a slightly higher net power output. - **Regeneration:** Perfect regeneration (100%) means the regenerator preheats air as much as possible, reducing heat supplied in the combustor. --- ## **Image: Ideal Regenerative Brayton Cycle**  *Alt text: Temperature-entropy diagram for ideal regenerative Brayton cycle, showing compression, preheating, heat addition, and expansion steps.* --- If you need **step details for specific states or calculations**, let me know!