--- title: "Gas turbine with reheat and 90% regenerator" subject: "Power Plant Equipment" chapter: 6 key: "semester_7/power_plant_equipment:gt-reheat-regenerator-1400k" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-reheat-regenerator-1400k site: BCT Engineering --- # Gas turbine with reheat and 90% regenerator Power Plant Equipment · Chapter 6: Gas Turbine Power Plant - Asked in: 2070 Chaitra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/gt-reheat-regenerator-1400k - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-6 ## Question 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. ## 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%. ## Answer **Assumptions:** air-standard ideal cycle; compression in one stage (100 to 1000 kPa) and two-stage expansion are isentropic; no pressure losses; $c_p = 1.005$ kJ/kg K, $\gamma = 1.4$, so $\frac{\gamma-1}{\gamma} = 0.2857$. ```text 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 $$ \begin{aligned} T_2 &= T_1 \left(\frac{p_2}{p_1}\right)^{0.2857} = 300 \times 10^{0.2857} \\ &= 300 \times 1.9307 = 579.21\ \text{K} \end{aligned} $$ #### 2. Turbine exit temperatures First stage, 1000 kPa to 300 kPa: $$ T_4 = \frac{T_3}{(1000/300)^{0.2857}} = \frac{1400}{1.4106} = 992.51\ \text{K} $$ Second stage, 300 kPa to 100 kPa, after reheat to $T_5 = 1400$ K: $$ T_6 = \frac{T_5}{(300/100)^{0.2857}} = \frac{1400}{1.3687} = 1022.84\ \text{K} $$ #### 3. Regenerator Effectiveness $\varepsilon = \dfrac{T_x - T_2}{T_6 - T_2}$, so $$ \begin{aligned} T_x &= T_2 + 0.9(T_6 - T_2) \\ &= 579.21 + 0.9(1022.84 - 579.21) = 978.48\ \text{K} \end{aligned} $$ #### 4. Work and heat (per kg of air) $$ \begin{aligned} W_C &= c_p (T_2 - T_1) = 1.005(579.21 - 300) = 280.61\ \text{kJ/kg} \\ W_{T1} &= c_p (T_3 - T_4) = 1.005(1400 - 992.51) = 409.53\ \text{kJ/kg} \\ W_{T2} &= c_p (T_5 - T_6) = 1.005(1400 - 1022.84) = 379.05\ \text{kJ/kg} \\ W_{net} &= 409.53 + 379.05 - 280.61 = 507.97\ \text{kJ/kg} \end{aligned} $$ Heat supplied in the combustor (x to 3) and reheater (4 to 5): $$ \begin{aligned} Q_s &= c_p (T_3 - T_x) + c_p (T_5 - T_4) \\ &= 1.005(1400 - 978.48) + 1.005(1400 - 992.51) \\ &= 423.63 + 409.53 = 833.16\ \text{kJ/kg} \end{aligned} $$ #### 5. Thermal efficiency $$ \eta_{th} = \frac{W_{net}}{Q_s} = \frac{507.97}{833.16} = 0.6097 $$ | Quantity | Value | |---|---| | $T_2$ | 579.21 K | | $T_4$ / $T_6$ | 992.51 K / 1022.84 K | | $T_x$ (after regenerator) | 978.48 K | | Net work | 507.97 kJ/kg | | Heat supplied | 833.16 kJ/kg | **Answer: thermal efficiency $\eta_{th} \approx 61.0\%$** (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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With isentropic turbines, how is T4 found for the first stage (1000 to 300 kPa)? A. T3 x (1000/300)^0.2857 B. T3 / (1000/300)^0.2857 = 992.51 K C. T3 / 10^0.2857 D. T3 - 300 Answer: B. Isentropic expansion gives T4 = T3 / (pressure ratio)^0.2857 = 1400/1.4106 = 992.51 K. (Key point: Turbine exit temperatures) 2. The regenerator effectiveness uses which exhaust temperature? A. T4 (HP turbine exit) B. T6 (LP turbine exit, 1022.84 K) C. T3 D. T1 Answer: B. Hot gas entering the regenerator is the LP turbine exhaust T6, so Tx = T2 + 0.9(T6 - T2) = 978.48 K. (Key point: Regenerator hot-side temperature) 3. Total heat supplied includes: A. Only the main combustor, x to 3 B. Combustor (x to 3) and reheater (4 to 5) C. Only reheater D. Regenerator heat Answer: B. Qs = cp(T3 - Tx) + cp(T5 - T4) = 423.63 + 409.53 = 833.16 kJ/kg. (Key point: Heat supplied) 4. What is the thermal efficiency? A. 507.97 / 833.16 = 61.0 % B. 280.61 / 833.16 = 33.7 % C. 833.16 / 507.97 = 164 % D. 409.53 / 833.16 = 49.2 % Answer: A. Wnet = 409.53 + 379.05 - 280.61 = 507.97 kJ/kg; divide by Qs to get about 61 %. 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