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Gas turbine with 78% effective regenerator

Short answer

Find T2 and T4 using compressor and turbine isentropic efficiencies, then regenerator exit Tx = T2 + 0.78(T4 - T2). Net work is 29.32 kJ/kg and heat input is 248.27 kJ/kg. Thermal efficiency is 11.81%.

In a gas turbine, the compressor takes in air at a temperature of 15°C and compresses it to four times the initial pressure with an isentropic efficiency of 82%. The temperature at the inlet of the turbine is 600°C and the isentropic efficiency of the turbine is 70%. A regenerator having an effectiveness of 78% is also incorporated in the cycle. Determine the thermal efficiency of the cycle. Take cp = 1.005 kJ/kg K and γ = 1.4.

Data: T1=15+273=288T_1 = 15 + 273 = 288 K, rp=4r_p = 4, ηc=0.82\eta_c = 0.82, T3=600+273=873T_3 = 600 + 273 = 873 K, ηt=0.70\eta_t = 0.70, ε=0.78\varepsilon = 0.78, cp=1.005c_p = 1.005 kJ/kg K, γ=1.4\gamma = 1.4; rp(γ−1)/γ=40.2857=1.4860r_p^{(\gamma-1)/\gamma} = 4^{0.2857} = 1.4860.

       +------ Regenerator ------+
       |  x                 y    |
 1     | 2                  4    |
 --> [C] --> [CC] --3--> [T] ----+

Compressor

T2s=288×1.4860=427.97 KT2=288+427.97−2880.82=458.69 KWc=1.005(458.69−288)=171.54 kJ/kg\begin{aligned} T_{2s} &= 288 \times 1.4860 = 427.97\ \text{K} \\ T_2 &= 288 + \frac{427.97 - 288}{0.82} = 458.69\ \text{K} \\ W_c &= 1.005(458.69 - 288) = 171.54\ \text{kJ/kg} \end{aligned}

Turbine

T4s=8731.4860=587.49 KT4=873−0.70(873−587.49)=673.14 KWt=1.005(873−673.14)=200.86 kJ/kg\begin{aligned} T_{4s} &= \frac{873}{1.4860} = 587.49\ \text{K} \\ T_4 &= 873 - 0.70(873 - 587.49) = 673.14\ \text{K} \\ W_t &= 1.005(873 - 673.14) = 200.86\ \text{kJ/kg} \end{aligned}

Regenerator

T4=673.14T_4 = 673.14 K >T2=458.69> T_2 = 458.69 K, so heat can be recovered:

Tx=T2+ε(T4−T2)=458.69+0.78(214.45)=625.96 KT_x = T_2 + \varepsilon(T_4 - T_2) = 458.69 + 0.78(214.45) = 625.96\ \text{K}

Thermal efficiency

Wnet=200.86−171.54=29.32 kJ/kgQs=cp(T3−Tx)=1.005(873−625.96)=248.27 kJ/kgηth=29.32248.27=0.1181=11.81%\begin{aligned} W_{net} &= 200.86 - 171.54 = 29.32\ \text{kJ/kg} \\ Q_s &= c_p (T_3 - T_x) = 1.005(873 - 625.96) = 248.27\ \text{kJ/kg} \\ \eta_{th} &= \frac{29.32}{248.27} = 0.1181 = 11.81\% \end{aligned}

(Without the regenerator: Qs=1.005(873−458.69)=416.38Q_s = 1.005(873 - 458.69) = 416.38 kJ/kg, η=7.04%\eta = 7.04\%.)

Answer: Thermal efficiency of the cycle = 11.81 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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