22-Mec-A1 Applied Thermodynamics and Heat Transfer · December 2016
Question 3 of 8: Regenerative Steam Power Plant with One Feedwater Heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — closed- and open-system energy balances, boundary work, reciprocating-compressor and vapour/gas power cycles, and gas-turbine refrigeration; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — radial composite-wall conduction, internal-flow decay, natural convection with radiation from a horizontal cylinder, and the ε–NTU cross-flow heat-exchanger method. Air and steam properties are evaluated from IAPWS-IF97 / ideal-air data; the gas-turbine cycle uses cold-air-standard constant specific heats.
Paper format: National Examination 07-Mec-A1, December 2016, 3 hours, open book. Part A — Thermodynamics (Q1–4); Part B — Heat Transfer (Q5–8). Each answer carries equal value; a complete paper is any five (three from one part and two from the other). All eight questions are solved in full below.
Question 3: Regenerative Steam Power Plant with One Feedwater Heater (equal value)
Figure 3 — $T$–$s$ diagram: 1→2 expansion to the 0.7 MPa bleed point (fraction $y$ drawn off to the open feedwater heater), 2→3 continued expansion into the wet region at the 25 kPa condenser, 3→4 condensation to saturated liquid, 4→6 condensate pumping and mixing to saturated liquid at 0.7 MPa in the heater, then feed pumping and 6 MPa heating back to state 1. States 1 and 2 lie to the right of the critical point and outside the dome (superheated); only state 3 is wet ($x_3=0.911$). Turbine legs are near-vertical because the 90 %-efficient expansion generates only ≈ 0.33 kJ/kg·K of entropy. Temperature axis compressed above the critical point.
Approach. Fix states 1–3 from the turbine inlet entropy and the 90 % stage efficiency, size the pump exits, close the open-feedwater-heater energy balance for the bleed fraction $y$, then sum the per-kilogram works and heat input; the plant power is $\dot m\,w_\text{net}$.
Turbine states. With $s_1=6.883$ kJ/kg·K, the isentropic targets give $h_{2s}=2842.5$ and $h_{3s}=2297.1$ kJ/kg; applying $\eta_t=0.9$,
$$h_2=h_1-0.9(h_1-h_{2s})=2900.6,\qquad h_3=h_1-0.9(h_1-h_{3s})=2409.7\ \text{kJ/kg}$$
The condenser exit quality $x_{3s}=0.863$ is acceptably high.
Bleed fraction (open FWH balance). Mixing to saturated liquid at 700 kPa,
$$y=\frac{h_6-h_5}{h_2-h_5}=\frac{697.0-272.7}{2900.6-272.7}$$
==**$y\approx0.161$ of the steam is extracted.**==
Net work per kilogram of boiler steam.
$$w_t=(h_1-h_2)+(1-y)(h_2-h_3)=934.2\ \text{kJ/kg},\quad w_p=(1-y)(0.77)+6.53=7.2\ \text{kJ/kg}$$
$$w_\text{net}=w_t-w_p\approx927\ \text{kJ/kg}$$
Efficiency and power. With $q_\text{in}=h_1-h_7=2719.6$ kJ/kg,
$$\eta_\text{th}=\frac{927}{2719.6}=34.1\%,\qquad \dot W_\text{net}=\dot m\,w_\text{net}=63.0\times927$$
==**$\eta_\text{th}\approx34\%$, $\dot W_\text{net}\approx58.4$ MW.**==
Check
Steam properties are from IAPWS-IF97; a ±0.1 % table sensitivity at the 6 MPa/500 °C inlet moves $\eta$ and $y$ by well under half a point. The 63.0 kg/s is taken as the boiler (turbine-inlet) flow, so $\dot W_\text{net}$ is the net plant output; the gross turbine power is $63.0\times934.2\approx58.8$ MW before pumping.
