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Gas turbine with reheat and 90% regenerator

Short answer

Isentropic compression to 1000 kPa gives T2=579.21 K. Two-stage expansion with reheat yields T4=992.51 K, T6=1022.84 K. Regenerator effectiveness 0.9 preheats air to 978.48 K. Net work 507.97 kJ/kg, heat input 833.16 kJ/kg, thermal efficiency 61.0%.

Air enters the compressor at 100 kPa, 300 K and is compressed to 1000 kPa. The temperature at the inlet to the first turbine stage is 1400 K. The expansion takes place isentropically in two stages, with reheat to 1400 K between the stages at a constant pressure of 300 kPa. A regenerator having an effectiveness of 90% is also incorporated in the cycle. Determine the thermal efficiency. Take cp = 1.005 kJ/kg K and γ = 1.4.

Assumptions: air-standard ideal cycle; compression in one stage (100 to 1000 kPa) and two-stage expansion are isentropic; no pressure losses; cp=1.005c_p = 1.005 kJ/kg K, γ=1.4\gamma = 1.4, so γ−1γ=0.2857\frac{\gamma-1}{\gamma} = 0.2857.

 1 ->[ C ]-2->[ Regen ]-x->[ CC ]-3->[ T1 ]-4
                 ^   |
                 |   v y (to exhaust)
 4 ->[ Reheater ]-5->[ T2 ]-6 -> hot side of Regen

States: 1 (100 kPa, 300 K) - compressor - 2 (1000 kPa) - regenerator - x - combustor - 3 (1400 K) - HP turbine - 4 (300 kPa) - reheater - 5 (1400 K) - LP turbine - 6 (100 kPa) - regenerator - y (exhaust).

1. Compressor exit temperature

T2=T1(p2p1)0.2857=300×100.2857=300×1.9307=579.21 K\begin{aligned} T_2 &= T_1 \left(\frac{p_2}{p_1}\right)^{0.2857} = 300 \times 10^{0.2857} \\ &= 300 \times 1.9307 = 579.21\ \text{K} \end{aligned}

2. Turbine exit temperatures

First stage, 1000 kPa to 300 kPa:

T4=T3(1000/300)0.2857=14001.4106=992.51 KT_4 = \frac{T_3}{(1000/300)^{0.2857}} = \frac{1400}{1.4106} = 992.51\ \text{K}

Second stage, 300 kPa to 100 kPa, after reheat to T5=1400T_5 = 1400 K:

T6=T5(300/100)0.2857=14001.3687=1022.84 KT_6 = \frac{T_5}{(300/100)^{0.2857}} = \frac{1400}{1.3687} = 1022.84\ \text{K}

3. Regenerator

Effectiveness ε=Tx−T2T6−T2\varepsilon = \dfrac{T_x - T_2}{T_6 - T_2}, so

Tx=T2+0.9(T6−T2)=579.21+0.9(1022.84−579.21)=978.48 K\begin{aligned} T_x &= T_2 + 0.9(T_6 - T_2) \\ &= 579.21 + 0.9(1022.84 - 579.21) = 978.48\ \text{K} \end{aligned}

4. Work and heat (per kg of air)

WC=cp(T2−T1)=1.005(579.21−300)=280.61 kJ/kgWT1=cp(T3−T4)=1.005(1400−992.51)=409.53 kJ/kgWT2=cp(T5−T6)=1.005(1400−1022.84)=379.05 kJ/kgWnet=409.53+379.05−280.61=507.97 kJ/kg\begin{aligned} W_C &= c_p (T_2 - T_1) = 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\ W_{T1} &= c_p (T_3 - T_4) = 1.005(1400 - 992.51) = 409.53\ \text{kJ/kg} \\ W_{T2} &= c_p (T_5 - T_6) = 1.005(1400 - 1022.84) = 379.05\ \text{kJ/kg} \\ W_{net} &= 409.53 + 379.05 - 280.61 = 507.97\ \text{kJ/kg} \end{aligned}

Heat supplied in the combustor (x to 3) and reheater (4 to 5):

Qs=cp(T3−Tx)+cp(T5−T4)=1.005(1400−978.48)+1.005(1400−992.51)=423.63+409.53=833.16 kJ/kg\begin{aligned} Q_s &= c_p (T_3 - T_x) + c_p (T_5 - T_4) \\ &= 1.005(1400 - 978.48) + 1.005(1400 - 992.51) \\ &= 423.63 + 409.53 = 833.16\ \text{kJ/kg} \end{aligned}

5. Thermal efficiency

ηth=WnetQs=507.97833.16=0.6097\eta_{th} = \frac{W_{net}}{Q_s} = \frac{507.97}{833.16} = 0.6097
QuantityValue
T2T_2579.21 K
T4T_4 / T6T_6992.51 K / 1022.84 K
TxT_x (after regenerator)978.48 K
Net work507.97 kJ/kg
Heat supplied833.16 kJ/kg

Answer: thermal efficiency ηth≈61.0%\eta_{th} \approx 61.0\% (0.6097).

The high value comes from the 90% regenerator: the hot turbine exhaust (1022.84 K) preheats the compressed air from 579.21 K to 978.48 K, so the fuel only has to supply the remaining heat. Reheat raises the work output, and the regenerator recovers most of the extra exhaust heat that reheat produces.

✓ Answer checked by Claude Opus 5.5 (high effort)

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