--- title: "Reheat Rankine cycle: 20 MPa, 500°C" subject: "Power Plant Equipment" chapter: 7 key: "semester_7/power_plant_equipment:rankine-reheat-20mpa" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/rankine-reheat-20mpa site: BCT Engineering --- # Reheat Rankine cycle: 20 MPa, 500°C Power Plant Equipment · Chapter 7: Thermal (Steam) Power Plant - Asked in: 2071 Shrawan, 12 marks - Video: [Thermodynamics - IDEAL REHEAT RANKINE CYCLE](https://www.youtube.com/watch?v=WTDW_AqqY-I) — JG ENGINEERING TUTORIALS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/rankine-reheat-20mpa - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-7 ## Question A steam power plant running on Rankine cycle has steam entering HP turbine at 20 MPa, 500°C and leaving LP turbine at 89.6 % dryness. Considering condenser pressure of 0.005 MPa and reheating occurring upto the temperature of 500°C determine: a) the pressure at which steam leaves HP turbine b) the thermal efficiency. [Refer the attached table for the properties of steam] ## Short answer By matching isentropic expansions: LP inlet entropy from exhaust dryness forces reheat pressure ≈ 1.5 MPa via interpolation. Net work = 1735.7 kJ/kg, heat added = 3906.7 kJ/kg, giving thermal efficiency ≈ 44.4%. ## Answer **Assumptions:** ideal reheat Rankine cycle, isentropic turbines and pump, pump work included. Values from steam tables (IAPWS-IF97). States: 1 = 20 MPa, 500 C (HP inlet); 2 = HP exit at reheat pressure $p_r$; 3 = $p_r$, 500 C (LP inlet); 4 = 0.005 MPa, $x = 0.896$; 5 = saturated water at 5 kPa; 6 = pump exit. #### Steam table data | State | Data | |---|---| | 5 kPa sat. | $t = 32.88$ C, $h_f = 137.77$, $h_{fg} = 2423.0$, $s_f = 0.4763$, $s_{fg} = 7.9177$, $v_f = 0.001005$ | | 20 MPa, 500 C | $h_1 = 3241.2$, $s_1 = 6.1445$ | #### (a) Pressure at which steam leaves the HP turbine State 4: $$ \begin{aligned} s_4 &= 0.4763 + 0.896 \times 7.9177 = 7.5705\ \text{kJ/kg K} \\ h_4 &= 137.77 + 0.896 \times 2423.0 = 2308.8\ \text{kJ/kg} \end{aligned} $$ LP expansion is isentropic, so $s_3 = s_4 = 7.5705$ at 500 C. From the superheated table at 500 C: $s = 7.6045$ at 1.4 MPa and $s = 7.5407$ at 1.6 MPa. Interpolating: $$ p_r = 1.4 + 0.2 \times \frac{7.6045 - 7.5705}{7.6045 - 7.5407} = 1.507 \approx 1.5\ \text{MPa} $$ Taking $p_r = 1.5$ MPa: at 1.5 MPa, 500 C, $h_3 = 3473.6$ kJ/kg. **Answer (a): steam leaves the HP turbine at about 1.5 MPa.** #### State 2 (1.5 MPa, $s_2 = s_1 = 6.1445$) At 1.5 MPa: $h_f = 844.7$, $h_{fg} = 1946.3$, $s_f = 2.3147$, $s_{fg} = 4.1284$. $$ \begin{aligned} x_2 &= \frac{6.1445 - 2.3147}{4.1284} = 0.9277 \\ h_2 &= 844.7 + 0.9277 \times 1946.3 = 2650.2\ \text{kJ/kg} \end{aligned} $$ #### Pump $$ \begin{aligned} W_p &= 0.001005 (20000 - 5) = 20.10\ \text{kJ/kg} \\ h_6 &= 137.77 + 20.10 = 157.87\ \text{kJ/kg} \end{aligned} $$ #### (b) Thermal efficiency $$ \begin{aligned} W_T &= (h_1 - h_2) + (h_3 - h_4) \\ &= (3241.2 - 2650.2) + (3473.6 - 2308.8) \\ &= 591.0 + 1164.8 = 1755.8\ \text{kJ/kg} \\ W_{net} &= 1755.8 - 20.1 = 1735.7\ \text{kJ/kg} \\ Q_s &= (h_1 - h_6) + (h_3 - h_2) \\ &= (3241.2 - 157.87) + (3473.6 - 2650.2) \\ &= 3083.3 + 823.4 = 3906.7\ \text{kJ/kg} \\ \eta &= \frac{1735.7}{3906.7} = 0.4443 \end{aligned} $$ **Answer (b): thermal efficiency $\approx 44.4\%$** (net work 1735.7 kJ/kg). ```text T ^ 1 3 | /| /| | / | / | | f--------g' | / | | / 2------' | | / | | 6 | | 5--------------------------4 +--------------------------------> s 5-6 pump, 6-f-g'-1 boiler (heat, boil, superheat), 1-2 HP turbine, 2-3 reheater, 3-4 LP turbine, 4-5 condenser ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the pressure at which steam leaves the HP turbine found in this problem? A. From the HP inlet entropy at 500 C B. From s3 = s4 (LP expansion is isentropic) at 500 C, using the state-4 entropy from x = 0.896 at 5 kPa, then interpolating the superheated table C. By assuming it is half of 20 MPa D. From the condenser temperature only Answer: B. State 4 is known from dryness and pressure, so its entropy equals s3 at 500 C; the superheated table then gives the pressure (about 1.5 MPa). (Key point: Finding reheat pressure) 2. What is the HP turbine exit (reheat) pressure found? A. About 15 MPa B. About 5 MPa C. About 1.5 MPa D. About 0.15 MPa Answer: C. Interpolating between 1.4 MPa (s = 7.6045) and 1.6 MPa (s = 7.5407) for s = 7.5705 gives about 1.507, i.e. 1.5 MPa. (Key point: Result of part (a)) 3. Which expression gives the total heat supplied in this reheat cycle? A. (h1 - h6) only B. (h1 - h6) + (h3 - h2) C. (h1 - h2) + (h3 - h4) D. (h3 - h4) - (h6 - h5) Answer: B. Heat is added in the boiler (h1 - h6) and again in the reheater (h3 - h2). (Key point: Heat supplied with reheat) 4. With W_T = 1755.8 kJ/kg, pump work 20.1 kJ/kg and heat supplied 3906.7 kJ/kg, what is the thermal efficiency? A. About 44.4 % B. About 45.0 % C. About 50.2 % D. About 38.0 % Answer: A. Net work 1735.7 divided by 3906.7 gives 0.444, i.e. about 44.4 %. 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