Ideal Reheat Rankine Cycle — Turbine Work and Exergy Destruction: 04-BS-10, May 2013, Q3

Question 3: Ideal Reheat Rankine Cycle — Turbine Work and Exergy Destruction (20 marks)

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Given. Ideal (internally reversible turbines and pump) reheat Rankine cycle. HP turbine inlet $P_3=12.5$ MPa, $T_3=600\ ^\circ\text{C}$; HP turbine exhausts to the reheat pressure $P_4=6$ MPa; reheat returns steam to $T_5=600\ ^\circ\text{C}$ at 6 MPa; LP turbine expands to the condenser pressure $P_6=25$ kPa. Heat-source temperature $T_H=1500$ K; sink/dead-state temperature $T_0=300$ K.

State	Description	P	T
1	Condenser exit, sat. liquid	25 kPa	$T_{sat}$
2	Pump exit (isentropic)	12.5 MPa	—
3	HP turbine inlet	12.5 MPa	600°C
4	HP turbine exit / reheat inlet	6 MPa	—
5	Reheat exit / LP turbine inlet	6 MPa	600°C
6	LP turbine exit / condenser inlet	25 kPa	—

Find. (a) total turbine work; (b) $\eta_{th}$; (c) exergy destruction of each process and the total; (d) $\eta_{II}$.

Fig. Q3 — T–s diagram of the ideal reheat Rankine cycle against the saturation dome (1→2 pump, 2→3 boiler, 3→4 HP turbine, 4→5 reheater, 5→6 LP turbine, 6→1 condenser). States 1 and 2 nearly coincide because the isentropic liquid-temperature rise across the pump is only about 1–2°C at this scale.

Approach

Fix all six states from the given pressures/temperatures using steam properties (pump and both turbines isentropic, since the cycle is ideal), sum the two turbine work terms for part (a), take thermal efficiency from the net work over the boiler-plus-reheat heat input, and evaluate exergy destruction process-by-process — noting that the ideal turbines and pump contribute none, so all of the cycle's exergy destruction is concentrated in the finite-$\Delta T$ heat-transfer processes (boiler, reheater, condenser).

Condenser exit and pump (1→2). State 1 is saturated liquid at 25 kPa: $h_1=271.96$ kJ/kg, $s_1=0.8932$ kJ/kg·K. The ideal pump is isentropic ($s_2=s_1$) to 12.5 MPa: $$w_p=h_2-h_1=284.65-271.96=12.69\ \text{kJ/kg}\quad(T_2=65.53\ ^\circ\text{C}).$$
HP turbine (3→4), isentropic. $h_3=3604.64$ kJ/kg, $s_3=6.7828$ kJ/kg·K at 12.5 MPa/600°C; expanding isentropically to 6 MPa gives $h_4=3347.54$ kJ/kg ($T_4=468.4\ ^\circ\text{C}$, superheated): $$w_{t,HP}=h_3-h_4=257.10\ \text{kJ/kg}.$$
Reheat and LP turbine (5→6), isentropic. Reheating at 6 MPa to 600°C: $h_5=3658.75$ kJ/kg, $s_5=7.1693$ kJ/kg·K. Isentropic expansion to 25 kPa gives $h_6=2393.98$ kJ/kg at quality $x_6=0.9047$: $$w_{t,LP}=h_5-h_6=1264.77\ \text{kJ/kg}.$$
Total turbine work (part a). $$w_{turbine}=w_{t,HP}+w_{t,LP}=257.10+1264.77=\boxed{1521.86\ \text{kJ/kg}}.$$
Heat input and thermal efficiency (part b). Boiler duty $q_{boiler}=h_3-h_2=3319.99$ kJ/kg; reheat duty $q_{reheat}=h_5-h_4=311.21$ kJ/kg; $q_{in}=3631.20$ kJ/kg. Net work $w_{net}=w_{turbine}-w_p=1509.17$ kJ/kg, so $$\eta_{th}=\frac{w_{net}}{q_{in}}=\frac{1509.17}{3631.20}=\boxed{41.56\%}.$$
Exergy destruction per process (part c). The pump and both turbines are internally reversible (ideal cycle) — zero destruction. All destruction occurs where heat crosses a finite temperature difference: boiler/reheat draw from the source at $T_H=1500$ K, and the condenser rejects to the sink at $T_{sink}=300$ K while the steam itself condenses near 65°C ($\approx338$ K): $$X_{dest,boiler}=T_0\!\left[(s_3-s_2)-\frac{q_{boiler}}{T_H}\right]=1102.89\ \text{kJ/kg},\qquad X_{dest,reheat}=T_0\!\left[(s_5-s_4)-\frac{q_{reheat}}{T_H}\right]=53.69\ \text{kJ/kg}.$$ $$X_{dest,cond}=T_0\!\left[(s_1-s_6)+\frac{q_{out}}{T_{sink}}\right]=239.20\ \text{kJ/kg}.$$ $$X_{dest,total}=\boxed{1395.78\ \text{kJ/kg}}.$$
Second-law efficiency (part d). Exergy supplied by the source, $X_{in}=q_{in}(1-T_0/T_H)=2904.96$ kJ/kg: $$\eta_{II}=\frac{w_{net}}{X_{in}}=\frac{1509.17}{2904.96}=\boxed{51.95\%}.$$

Quantity	Result
(a) $w_{turbine}$	1521.86 kJ/kg
(b) $\eta_{th}$	41.56%
(c) $X_{dest}$: boiler / reheat / condenser / total	1102.89 / 53.69 / 239.20 / 1395.78 kJ/kg
(d) $\eta_{II}$	51.95%

Concept & Theory

Topic. Ideal reheat Rankine cycle performance and second-law (exergy) accounting of a steam power plant.

Key relations. Isentropic turbine/pump ($s_{out}=s_{in}$) for the ideal cycle; $\eta_{th}=w_{net}/q_{in}$; exergy destruction of heat transfer from a reservoir at fixed $T_{res}$: $X_{dest}=T_0[\Delta s_{steam}\mp q/T_{res}]$ (minus for heat received from a source, plus for heat rejected to a sink); second-law efficiency $\eta_{II}=w_{net}/X_{in}$ with $X_{in}=q_{in}(1-T_0/T_H)$.

Why this works. Reheating raises the average temperature at which the second half of the heat is added and keeps the LP-turbine exhaust quality high (here $x_6\approx0.90$, above the ∼90% wet-steam blade-erosion limit), both of which improve $\eta_{th}$ relative to a single-expansion cycle at the same boiler pressure — but even an "ideal" cycle with reversible turbomachinery still destroys most of its exergy in the boiler, because 1500 K combustion-gas heat is being transferred into water/steam that never exceeds ∼875 K, a very large temperature gap.

Common pitfall. Assuming an "ideal" cycle has zero total exergy destruction — internal reversibility of the turbines/pump only means those components destroy no exergy; heat addition/rejection across a finite $\Delta T$ to the external source/sink remains irreversible regardless of how the working-fluid processes themselves are idealized.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 10 (Vapor and Combined Power Cycles) & Ch. 8 (Exergy Analysis).

Question 3 of 9: Ideal Reheat Rankine Cycle — Turbine Work and Exergy Destruction

Question 3: Ideal Reheat Rankine Cycle — Turbine Work and Exergy Destruction (20 marks)

Approach