Combined Reheat–Regeneration Steam Power Cycle: 17-Phys-B6 Applied Thermodynamics and Heat Transfer, May 2016, Q2

Question 2: Combined Reheat–Regeneration Steam Power Cycle

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Given. HP turbine inlet 3.5 MPa/350°C; first extraction (open FWH1) at 0.8 MPa; reheat to 350°C at 0.8 MPa; second extraction (open FWH2) at 0.2 MPa; condenser at 10 kPa; both feedwater heaters open (mixing type); all expansions and pumps taken as ideal (isentropic) — no turbine/pump efficiency is given, so 100% is the only defensible reading.

Given data
Quantity	Symbol	Value
HP turbine inlet pressure / temperature	$P_1,\,T_1$	3.5 MPa / $350\,{}^{\circ}\text{C}$
FWH1 (extraction 1) pressure	$P_2$	0.8 MPa
Reheat temperature	$T_3$	$350\,{}^{\circ}\text{C}$
FWH2 (extraction 2) pressure	$P_4$	0.2 MPa
Condenser pressure	$P_5$	10 kPa

Fig. 2 — T-s diagram: HP turbine 1→2, reheat 2→3, LP turbine 3→4→5, condenser 5→6, FWH2 exit 8, FWH1 exit 10 (pump risers omitted for clarity).

Find. The net work output per kg of boiler flow and the cycle's thermal efficiency.

Approach. Fix all eight/nine state points with real steam properties (both turbine legs isentropic), find the extraction mass fractions $y$ (to FWH1) and $z$ (to FWH2) from energy balances on each open feedwater heater (exit at saturated liquid), then sum turbine work minus pump work over the correct mass fraction in each leg, and divide by the total heat added.

HP turbine, state 1→2 (isentropic to 0.8 MPa). $h_1=3104.8$ kJ/kg, $s_1=6.660\text{ kJ/kg}\cdot\text{K}$ ⇒ $h_2=2767.6$ kJ/kg ($T_2=170.4\,{}^{\circ}\text{C}$).
Reheat at 0.8 MPa to $350\,{}^{\circ}\text{C}$ (state 3). $h_3=3162.2$ kJ/kg, $s_3=7.411\text{ kJ/kg}\cdot\text{K}$.
LP turbine, state 3→4 (isentropic to 0.2 MPa) and 4→5 (isentropic to 10 kPa). $h_4=2825.7$ kJ/kg ($T_4=177.7\,{}^{\circ}\text{C}$); $h_5=2348.4$ kJ/kg at $x_5=0.902$ (wet at the condenser).
Condenser exit and pump 1 (10 kPa→0.2 MPa). Saturated liquid $h_6=191.81$ kJ/kg, $v_6=1.0103\times10^{-3}\text{ m}^3/\text{kg}$; $h_7=h_6+v_6(P_4-P_5)=191.81+0.19=191.99$ kJ/kg.
FWH2 exit and pump 2 (0.2 MPa→0.8 MPa). Saturated liquid $h_8=504.70$ kJ/kg, $v_8=1.0605\times10^{-3}\text{ m}^3/\text{kg}$; $h_9=h_8+v_8(P_2-P_4)=504.70+0.64=505.34$ kJ/kg.
FWH1 exit and pump 3 (0.8 MPa→3.5 MPa). Saturated liquid $h_{10}=720.86$ kJ/kg, $v_{10}=1.1148\times10^{-3}\text{ m}^3/\text{kg}$; $h_{11}=h_{10}+v_{10}(P_1-P_2)=720.86+3.01=723.87$ kJ/kg.
Extraction fractions from open-FWH energy balances (per kg of boiler flow). FWH1 mixes $y$ kg at $h_2$ with $(1-y)$ kg at $h_9$ to leave saturated liquid $h_{10}$: $$y=\frac{h_{10}-h_9}{h_2-h_9}=\frac{720.86-505.34}{2767.6-505.34}=0.0953$$ FWH2 mixes $z$ kg at $h_4$ with $(1-y-z)$ kg at $h_7$ (from the $(1-y)$ kg that passed through the LP turbine) to leave saturated liquid $h_8$: $$z=(1-y)\,\frac{h_8-h_7}{h_4-h_7}=(0.9047)\frac{504.70-191.99}{2825.7-191.99}=0.1074$$
Turbine and pump work. $$w_{\text{HP turb}}=h_1-h_2=337.23\text{ kJ/kg}$$ $$w_{\text{LP turb}}=(1-y)(h_3-h_4)+(1-y-z)(h_4-h_5)=304.53+286.44=684.97\text{ kJ/kg}$$ $$w_{\text{pump}}=(1-y-z)(h_7-h_6)+(1-y)(h_9-h_8)+1\cdot(h_{11}-h_{10})=0.153+0.576+3.010=3.74\text{ kJ/kg}$$ $$w_{\text{net}}=w_{\text{HP turb}}+w_{\text{LP turb}}-w_{\text{pump}}=337.23+684.97-3.74=\boxed{1018.5\text{ kJ/kg}}$$
Heat added and thermal efficiency. Heat is added in the boiler (state 11→1) and in the reheater (state 2→3, over the $(1-y)$ kg that is reheated): $$q_{\text{in}}=(h_1-h_{11})+(1-y)(h_3-h_2)=2380.98+357.0=2737.9\text{ kJ/kg}$$ $$\eta_{\text{th}}=\frac{w_{\text{net}}}{q_{\text{in}}}=\frac{1018.5}{2737.9}=\boxed{37.2\%}$$

Question 2 — results
Quantity	Value
Extraction fraction to FWH1, $y$	0.0953
Extraction fraction to FWH2, $z$	0.1074
Net work output, $w_{\text{net}}$	1018.5 kJ/kg
Heat added, $q_{\text{in}}$	2737.9 kJ/kg
Thermal efficiency, $\eta_{\text{th}}$	37.2%

Topic. Multi-stage reheat/regenerative Rankine cycle with open feedwater heaters.

Key relations. Isentropic turbine expansion ($s$ constant per leg); open FWH mass/energy balance to find the bleed fraction; pump work $w_p=v\Delta P$ for an (incompressible) liquid; $\eta_{\text{th}}=w_{\text{net}}/q_{\text{in}}$.

Why this works. Each open feedwater heater is a direct-contact mixing chamber that must leave the mixed stream at the SATURATED-LIQUID state of its own pressure — that single condition, applied twice (once per heater, propagating the correct upstream mass fraction), is what pins down both bleed fractions without any iteration. The reheat step is what lets the LP turbine's overall expansion avoid excessive moisture at the condenser, which is why $x_5$ is checked explicitly.

Common pitfall. Applying the FWH2 balance to the FULL 1 kg of boiler flow instead of the $(1-y)$ kg that actually survived the first extraction is the most common slip in a two-heater regenerative cycle; the reheat energy in $q_{\text{in}}$ must likewise be weighted by $(1-y)$, not the full flow, since the $y$ fraction bypasses the reheater entirely once extracted.

Source. Çengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 10 (Rankine cycle with reheat and regeneration, open vs. closed feedwater heaters).

Question 2 of 8: Combined Reheat–Regeneration Steam Power Cycle

Question 2: Combined Reheat–Regeneration Steam Power Cycle