--- title: "Reheat cycle: 4 MPa, 400°C" subject: "Power Plant Equipment" chapter: 7 key: "semester_7/power_plant_equipment:reheat-cycle-4mpa" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/reheat-cycle-4mpa site: BCT Engineering --- # Reheat cycle: 4 MPa, 400°C Power Plant Equipment · Chapter 7: Thermal (Steam) Power Plant - Asked in: 2072 Kartik, 10 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/reheat-cycle-4mpa - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-7 ## Question On a reheat cycle, steam leaves the boiler and enters the turbine at 4 MPa, 400°C. After expansion in the turbine to 400 kPa, the steam is reheated to 400°C and then expanded in the low-pressure turbine to 10 kPa. Determine the cycle efficiency. [Refer the attached table for the properties of steam]. ## Short answer Using ideal reheat cycle with isentropic expansion and pump work, net work is 1293.4 kJ/kg and heat added is 3606.1 kJ/kg. Cycle efficiency is approximately 35.9%. ## Answer **Assumptions:** ideal reheat Rankine cycle; isentropic turbines and pump; pump work included. Values from steam tables (IAPWS-IF97). States: 1 = 4 MPa, 400 C; 2 = 400 kPa (HP exit); 3 = 400 kPa, 400 C; 4 = 10 kPa; 5 = saturated water at 10 kPa; 6 = pump exit. ```text 1 +----------+----->+------+ +------+ | Boiler | | HP |=====| LP |===> Generator | + S/H | 2 | turb | | turb | | Reheater |<-----+------+ +->+------+ | |------------3---+ | 4 +----------+ v ^ 6 +-----------+ | +------+ 5 | Condenser | +------| Pump |<-------+-----------+ +------+ ``` #### Steam table data | State | Data | |---|---| | 4 MPa, 400 C | $h_1 = 3214.4$, $s_1 = 6.7712$ | | 400 kPa sat. | $h_f = 604.72$, $h_{fg} = 2133.33$, $s_f = 1.7766$, $s_{fg} = 5.1188$ | | 400 kPa, 400 C | $h_3 = 3273.9$, $s_3 = 7.9001$ | | 10 kPa sat. | $h_f = 191.81$, $h_{fg} = 2392.07$, $s_f = 0.6492$, $s_{fg} = 7.4997$, $v_f = 0.001010$ | #### State 2 ($s_2 = s_1$) $$ \begin{aligned} x_2 &= \frac{6.7712 - 1.7766}{5.1188} = 0.9757 \\ h_2 &= 604.72 + 0.9757 \times 2133.33 = 2686.3\ \text{kJ/kg} \end{aligned} $$ #### State 4 ($s_4 = s_3$) $$ \begin{aligned} x_4 &= \frac{7.9001 - 0.6492}{7.4997} = 0.9668 \\ h_4 &= 191.81 + 0.9668 \times 2392.07 = 2504.5\ \text{kJ/kg} \end{aligned} $$ #### Pump $$ \begin{aligned} W_p &= v_f (p_6 - p_5) = 0.001010 (4000 - 10) = 4.03\ \text{kJ/kg} \\ h_6 &= 191.81 + 4.03 = 195.84\ \text{kJ/kg} \end{aligned} $$ #### Work, heat and efficiency $$ \begin{aligned} W_T &= (h_1 - h_2) + (h_3 - h_4) \\ &= (3214.4 - 2686.3) + (3273.9 - 2504.5) \\ &= 528.1 + 769.3 = 1297.4\ \text{kJ/kg} \\ W_{net} &= 1297.4 - 4.03 = 1293.4\ \text{kJ/kg} \\ Q_s &= (h_1 - h_6) + (h_3 - h_2) \\ &= (3214.4 - 195.84) + (3273.9 - 2686.3) \\ &= 3018.5 + 587.6 = 3606.1\ \text{kJ/kg} \\ \eta &= \frac{1293.4}{3606.1} = 0.3587 \end{aligned} $$ **Answer: cycle efficiency $\approx 35.9\%$** (net work 1293.4 kJ/kg, exhaust dryness 0.967). ```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 ``` Compared with a simple Rankine cycle between 4 MPa, 400 C and 10 kPa (about 35.3%, exhaust dryness 0.816), reheat increases efficiency slightly, raises the net work by about 21% and makes the exhaust steam much drier. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is state 2 found? A. Isentropic: x2 = (s1 - sf)/sfg at 400 kPa = 0.9757 B. Constant enthalpy C. Constant volume D. From s3 Answer: A. Isentropic HP expansion gives s2 = s1, so x2 = (6.7712-1.7766)/5.1188 = 0.9757. (Key point: State 2 dryness) 2. State 4 is found using: A. s4 = s3 at 10 kPa, giving x4 = 0.9668 and h4 = 2504.5 kJ/kg B. s4 = s1 C. h4 = h3 D. x4 = 1 Answer: A. LP expansion is isentropic from state 3. (Key point: State 4) 3. Pump work is calculated as: A. v_f (p6 - p5) = 4.03 kJ/kg B. h1 - h2 C. h4 - h5 D. zero Answer: A. 0.001010 x (4000 - 10) = 4.03 kJ/kg. (Key point: Pump work) 4. Cycle efficiency is about: A. 35.9% B. 25% C. 45% D. 29.3% Answer: A. Net work 1293.4 over heat 3606.1 gives 0.3587. 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