--- title: "Reheat cycle at 14 bar: ideal efficiency" subject: "Power Plant Equipment" chapter: 7 key: "semester_7/power_plant_equipment:reheat-14-bar-efficiency" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reheat-14-bar-efficiency site: BCT Engineering --- # Reheat cycle at 14 bar: ideal efficiency Power Plant Equipment · Chapter 7: Thermal (Steam) Power Plant - Asked in: 2079 Bhadra, 8 marks - Video: [Reheat Cycle Reheat Rankine Cycle](https://www.youtube.com/watch?v=sZlchIqsvjY) — BEST MECHANICAL ENGINEERING - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reheat-14-bar-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-7 ## Question Steam at a pressure of 14 bar and temperature 300°C is expanded through a HP turbine to a pressure of 5 bar, it is then reheated at constant pressure to a temperature of 300°C and then it completes expansion through the LP turbine to an exhaust pressure of 0.2 bar. Calculate the ideal efficiency of the plant and work done: a) Taking the reheating into account b) Without reheating ## Short answer Using isentropic expansions and steam tables: With reheat, net work = 836.7 kJ/kg, ideal efficiency = 27.5%. Without reheat, net work = 747.8 kJ/kg, ideal efficiency = 26.8%. ## Answer **Assumptions:** ideal cycle, isentropic expansion, feed pump work included ($W_p = v_f \Delta p$). Values from steam tables (IAPWS-IF97). States: 1 = 14 bar, 300 C; 2 = 5 bar after HP turbine; 3 = 5 bar, 300 C (after reheat); 4 = 0.2 bar exhaust; 5 = saturated water at 0.2 bar. #### Steam table data | State | Data | |---|---| | 14 bar, 300 C | $h_1 = 3041.0$, $s_1 = 6.9553$ | | 5 bar | $s_g = 6.8206$ (so state 2 is superheated) | | 5 bar, 300 C | $h_3 = 3064.6$, $s_3 = 7.4614$ | | 0.2 bar | $h_f = 251.40$, $h_{fg} = 2357.55$, $s_f = 0.8320$, $s_{fg} = 7.0753$, $v_f = 0.001017$ | (h in kJ/kg, s in kJ/kg K) **State 2:** $s_2 = 6.9553 > s_g$ at 5 bar, so steam is slightly superheated (about 177.6 C). From the superheated table, $h_2 = 2807.1$ kJ/kg. **State 4:** $$ \begin{aligned} x_4 &= \frac{7.4614 - 0.8320}{7.0753} = 0.9370 \\ h_4 &= 251.40 + 0.9370 \times 2357.55 = 2460.4\ \text{kJ/kg} \end{aligned} $$ **Pump work:** $W_p = 0.001017 (1400 - 20) = 1.40$ kJ/kg #### (a) With reheating $$ \begin{aligned} W_T &= (h_1 - h_2) + (h_3 - h_4) \\ &= (3041.0 - 2807.1) + (3064.6 - 2460.4) \\ &= 233.9 + 604.2 = 838.1\ \text{kJ/kg} \\ W_{net} &= 838.1 - 1.4 = 836.7\ \text{kJ/kg} \\ Q_s &= (h_1 - h_5 - W_p) + (h_3 - h_2) \\ &= (3041.0 - 251.40 - 1.40) + (3064.6 - 2807.1) \\ &= 2788.2 + 257.5 = 3045.7\ \text{kJ/kg} \\ \eta &= \frac{836.7}{3045.7} = 0.2747 \end{aligned} $$ **Answer (a): work done $\approx 836.7$ kJ/kg (838.1 kJ/kg turbine work), ideal efficiency $\approx 27.5\%$** #### (b) Without reheating Steam expands directly from 14 bar, 300 C to 0.2 bar with $s = 6.9553$: $$ \begin{aligned} x &= \frac{6.9553 - 0.8320}{7.0753} = 0.8655 \\ h &= 251.40 + 0.8655 \times 2357.55 = 2291.8\ \text{kJ/kg} \\ W_T &= 3041.0 - 2291.8 = 749.2\ \text{kJ/kg} \\ W_{net} &= 749.2 - 1.4 = 747.8\ \text{kJ/kg} \\ Q_s &= 3041.0 - 251.40 - 1.40 = 2788.2\ \text{kJ/kg} \\ \eta &= \frac{747.8}{2788.2} = 0.2682 \end{aligned} $$ **Answer (b): work done $\approx 747.8$ kJ/kg (749.2 kJ/kg turbine work), ideal efficiency $\approx 26.8\%$** | Case | Net work (kJ/kg) | Efficiency | Exhaust dryness | |---|---|---|---| | With reheat | 836.7 | 27.5% | 0.937 | | Without reheat | 747.8 | 26.8% | 0.866 | If pump work is neglected, the efficiencies are 27.5% and 26.9%. Reheat increases work by about 12%, efficiency slightly, and makes the exhaust much drier. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is state 2 (5 bar after HP expansion) superheated? A. Because s2 = 6.9553 is greater than s_g = 6.8206 at 5 bar B. Because pressure is below 1 bar C. Because the turbine is not isentropic D. Because it is reheated Answer: A. With s2 = s1 above s_g at 5 bar, the steam is still superheated, so h2 comes from the superheated table. (Key point: State 2 condition) 2. What is the total turbine work with reheat? A. (h1 - h2) only B. (h1 - h2) + (h3 - h4) = 838.1 kJ/kg C. (h3 - h4) - (h1 - h2) D. h1 - h5 Answer: B. HP work 233.9 plus LP work 604.2 gives 838.1 kJ/kg. (Key point: Turbine work with reheat) 3. Heat supplied in the reheat cycle includes: A. Only the boiler heat h1 - h5 B. Boiler heat (h1 - h5 - Wp) plus reheat heat (h3 - h2) C. Only reheat heat D. Condenser heat rejected Answer: B. Q_s = 2788.2 + 257.5 = 3045.7 kJ/kg. (Key point: Heat supplied with reheat) 4. Which gives the correct efficiencies with and without reheat? A. 27.5% with, 26.8% without B. 26.8% with, 27.5% without C. 35% with, 27% without D. Both exactly 27.5% Answer: A. 836.7/3045.7 = 27.5% with reheat; 747.8/2788.2 = 26.8% without. (Key point: Efficiency results) 5. How does reheat change exhaust dryness here? A. 0.866 to 0.937 (drier) B. 0.937 to 0.866 (wetter) C. No change D. Both become superheated Answer: A. Without reheat x = 0.866; with reheat x4 = 0.937, so the exhaust is much drier. 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