--- title: "Regenerative cycle: 4 MPa, 400°C, open heater" subject: "Power Plant Equipment" chapter: 7 key: "semester_7/power_plant_equipment:regenerative-cycle-4mpa" source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/regenerative-cycle-4mpa site: BCT Engineering --- # Regenerative cycle: 4 MPa, 400°C, open heater Power Plant Equipment · Chapter 7: Thermal (Steam) Power Plant - Asked in: 2074 Asoj, 12 marks; 2070 Asar, 12 marks (asked 2 times) - Video: [Regenerative Rankine Cycle | Problem Solving | Thermodynamics](https://www.youtube.com/watch?v=68ySu1_uCd4) — Learn in 30 Seconds - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/regenerative-cycle-4mpa - Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-7 ## Question On a regenerative cycle, steam leaves the boiler and enters the turbine at 4 MPa, 400°C. After expansion to 400 kPa, some of the steam is extracted from the turbine to heat the feed water in an open feed water heater. The pressure in the feed water heater is 400 kPa, and the water leaving it is saturated liquid at 400 kPa. The steam not extracted expands to 10kPa. Determine the cycle efficiency. [Refer the attached table for the properties of steam] ## Short answer Cycle efficiency ≈ 37.5%. Using steam table data: h1=3214.4, h2=2686.3, h3=2144.5 kJ/kg; pump work 4.23 kJ/kg; bleed fraction m=0.1654 from open FWH energy balance. Net work 976.1 kJ/kg, heat input 2605.7 kJ/kg. ## Answer **Assumptions:** ideal regenerative Rankine cycle with one **open feed water heater**; turbine and pumps isentropic; exit of heater is saturated liquid at 400 kPa; pump work included. Property values are taken from the steam tables (IAPWS-IF97 values, as in Cengel's tables). ```text 1 +--------+------>+---------------+ +-----+ | Boiler | | Turbine |===| Gen | | + S/H | +---------------+ +-----+ +--------+ | m kg (2) | (1-m) kg (3) ^ 7 v v | +-----------+ +-----------+ +--------+ 6 | Open FWH | | Condenser | | Pump 2 |<---| (mixing) | +-----------+ +--------+ +-----------+ | 4 ^ 5 | +--------+ | | Pump 1 |<--------+ +--------+ ``` #### Steam table data | Pressure | $t_{sat}$ (°C) | $h_f$ | $h_{fg}$ | $s_f$ | $s_{fg}$ | $v_f$ (m³/kg) | |---|---|---|---|---|---|---| | 10 kPa | 45.81 | 191.81 | 2392.07 | 0.6492 | 7.4997 | 0.001010 | | 400 kPa | 143.61 | 604.72 | 2133.33 | 1.7766 | 5.1188 | 0.001084 | At 4 MPa, 400 °C (superheated): $h_1 = 3214.4$ kJ/kg, $s_1 = 6.7712$ kJ/kg K. #### State 2 (bled steam, 400 kPa, $s_2 = s_1$) $$ \begin{aligned} x_2 &= \frac{6.7712 - 1.7766}{5.1188} = 0.9757 \\ h_2 &= 604.72 + 0.9757 \times 2133.33 = 2686.3\ \text{kJ/kg} \end{aligned} $$ #### State 3 (exhaust, 10 kPa, $s_3 = s_1$) $$ \begin{aligned} x_3 &= \frac{6.7712 - 0.6492}{7.4997} = 0.8163 \\ h_3 &= 191.81 + 0.8163 \times 2392.07 = 2144.5\ \text{kJ/kg} \end{aligned} $$ #### Pumps $$ \begin{aligned} W_{p1} &= v_f (p_5 - p_4) = 0.001010(400 - 10) = 0.39\ \text{kJ/kg} \\ h_5 &= 191.81 + 0.39 = 192.20\ \text{kJ/kg} \\ W_{p2} &= 0.001084(4000 - 400) = 3.90\ \text{kJ/kg} \\ h_7 &= h_6 + W_{p2} = 604.72 + 3.90 = 608.62\ \text{kJ/kg} \end{aligned} $$ #### Bled steam fraction (energy balance on open FWH) $m h_2 + (1-m) h_5 = h_6$: $$ m = \frac{h_6 - h_5}{h_2 - h_5} = \frac{604.72 - 192.20}{2686.3 - 192.20} = 0.1654\ \text{kg/kg} $$ #### Work and heat (per kg of steam at boiler exit) $$ \begin{aligned} W_T &= (h_1 - h_2) + (1-m)(h_2 - h_3) \\ &= 528.1 + 0.8346 \times 541.8 = 980.3\ \text{kJ/kg} \\ W_P &= (1-m) W_{p1} + W_{p2} = 0.33 + 3.90 = 4.23\ \text{kJ/kg} \\ W_{net} &= 980.3 - 4.23 = 976.1\ \text{kJ/kg} \\ Q_s &= h_1 - h_7 = 3214.4 - 608.62 = 2605.7\ \text{kJ/kg} \end{aligned} $$ #### Cycle efficiency $$ \eta = \frac{W_{net}}{Q_s} = \frac{976.1}{2605.7} = 0.3746 $$ **Answer: cycle efficiency $\approx 37.5\%$** (bled fraction $m = 0.165$ kg per kg of steam). ```text T ^ 1 | /| | f----------g | | / | | 7 | | 6 . . . . . . . 2 bleed (m kg) | / | | 5 | | 4-----------------3 (1-m) kg +-------------------------> s 4-5 pump 1, 5 and 2 mix in FWH -> 6, 6-7 pump 2, 7-f-g-1 boiler, 1-2-3 turbine, 3-4 condenser ``` For comparison, a simple Rankine cycle between the same limits gives about 35.3%, so one open heater raises the efficiency by about 2 percentage points. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the quality at the 400 kPa bleed point found? A. Using s2 = s1 (isentropic expansion) to get x2 = 0.9757 B. Using h2 = h1 C. Using constant temperature D. Using v2 = v1 Answer: A. The turbine is isentropic, so entropy at 400 kPa equals s1 and x2 = (s1 - sf)/sfg. (Key point: Bled steam state) 2. The bled fraction m is found from: A. m h2 + (1-m) h5 = h6 B. m h2 = h6 C. m = h1/h2 D. m h5 + (1-m) h2 = h1 Answer: A. Mixing bled steam (h2) with pump-1 water (h5) must give saturated liquid h6 at 400 kPa, so m = 0.1654. (Key point: Open FWH energy balance) 3. Turbine work per kg of boiler steam is: A. (h1 - h2) + (1-m)(h2 - h3) B. (h1 - h3) C. m(h1 - h2) + (h2 - h3) D. (h1 - h2)(1-m) Answer: A. Full flow expands from 1 to 2; only (1-m) continues from 2 to 3. (Key point: Turbine and heat terms per kg) 4. Cycle efficiency and the heat supplied are: A. 976.1 / 2605.7 = 37.5 %, with Qs = h1 - h7 B. 976.1 / 3214.4, with Qs = h1 C. 980.3 / 2605.7, with Qs = h1 - h6 D. 4.23 / 976.1 Answer: A. Net work 976.1 kJ/kg divided by Qs = 3214.4 - 608.62 = 2605.7 kJ/kg gives about 37.5 %, about 2 points better than simple Rankine. 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