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Reheat cycle at 14 bar: ideal efficiency

Short answer

Using isentropic expansions and steam tables: With reheat, net work = 836.7 kJ/kg, ideal efficiency = 27.5%. Without reheat, net work = 747.8 kJ/kg, ideal efficiency = 26.8%.

Steam at a pressure of 14 bar and temperature 300°C is expanded through a HP turbine to a pressure of 5 bar, it is then reheated at constant pressure to a temperature of 300°C and then it completes expansion through the LP turbine to an exhaust pressure of 0.2 bar. Calculate the ideal efficiency of the plant and work done: a) Taking the reheating into account b) Without reheating

Assumptions: ideal cycle, isentropic expansion, feed pump work included (Wp=vfΔpW_p = v_f \Delta p). Values from steam tables (IAPWS-IF97).

States: 1 = 14 bar, 300 C; 2 = 5 bar after HP turbine; 3 = 5 bar, 300 C (after reheat); 4 = 0.2 bar exhaust; 5 = saturated water at 0.2 bar.

Steam table data

StateData
14 bar, 300 Ch1=3041.0h_1 = 3041.0, s1=6.9553s_1 = 6.9553
5 barsg=6.8206s_g = 6.8206 (so state 2 is superheated)
5 bar, 300 Ch3=3064.6h_3 = 3064.6, s3=7.4614s_3 = 7.4614
0.2 barhf=251.40h_f = 251.40, hfg=2357.55h_{fg} = 2357.55, sf=0.8320s_f = 0.8320, sfg=7.0753s_{fg} = 7.0753, vf=0.001017v_f = 0.001017

(h in kJ/kg, s in kJ/kg K)

State 2: s2=6.9553>sgs_2 = 6.9553 > s_g at 5 bar, so steam is slightly superheated (about 177.6 C). From the superheated table, h2=2807.1h_2 = 2807.1 kJ/kg.

State 4:

x4=7.4614−0.83207.0753=0.9370h4=251.40+0.9370×2357.55=2460.4 kJ/kg\begin{aligned} x_4 &= \frac{7.4614 - 0.8320}{7.0753} = 0.9370 \\ h_4 &= 251.40 + 0.9370 \times 2357.55 = 2460.4\ \text{kJ/kg} \end{aligned}

Pump work: Wp=0.001017(1400−20)=1.40W_p = 0.001017 (1400 - 20) = 1.40 kJ/kg

(a) With reheating

WT=(h1−h2)+(h3−h4)=(3041.0−2807.1)+(3064.6−2460.4)=233.9+604.2=838.1 kJ/kgWnet=838.1−1.4=836.7 kJ/kgQs=(h1−h5−Wp)+(h3−h2)=(3041.0−251.40−1.40)+(3064.6−2807.1)=2788.2+257.5=3045.7 kJ/kgη=836.73045.7=0.2747\begin{aligned} W_T &= (h_1 - h_2) + (h_3 - h_4) \\ &= (3041.0 - 2807.1) + (3064.6 - 2460.4) \\ &= 233.9 + 604.2 = 838.1\ \text{kJ/kg} \\ W_{net} &= 838.1 - 1.4 = 836.7\ \text{kJ/kg} \\ Q_s &= (h_1 - h_5 - W_p) + (h_3 - h_2) \\ &= (3041.0 - 251.40 - 1.40) + (3064.6 - 2807.1) \\ &= 2788.2 + 257.5 = 3045.7\ \text{kJ/kg} \\ \eta &= \frac{836.7}{3045.7} = 0.2747 \end{aligned}

Answer (a): work done ≈836.7\approx 836.7 kJ/kg (838.1 kJ/kg turbine work), ideal efficiency ≈27.5%\approx 27.5\%

(b) Without reheating

Steam expands directly from 14 bar, 300 C to 0.2 bar with s=6.9553s = 6.9553:

x=6.9553−0.83207.0753=0.8655h=251.40+0.8655×2357.55=2291.8 kJ/kgWT=3041.0−2291.8=749.2 kJ/kgWnet=749.2−1.4=747.8 kJ/kgQs=3041.0−251.40−1.40=2788.2 kJ/kgη=747.82788.2=0.2682\begin{aligned} x &= \frac{6.9553 - 0.8320}{7.0753} = 0.8655 \\ h &= 251.40 + 0.8655 \times 2357.55 = 2291.8\ \text{kJ/kg} \\ W_T &= 3041.0 - 2291.8 = 749.2\ \text{kJ/kg} \\ W_{net} &= 749.2 - 1.4 = 747.8\ \text{kJ/kg} \\ Q_s &= 3041.0 - 251.40 - 1.40 = 2788.2\ \text{kJ/kg} \\ \eta &= \frac{747.8}{2788.2} = 0.2682 \end{aligned}

Answer (b): work done ≈747.8\approx 747.8 kJ/kg (749.2 kJ/kg turbine work), ideal efficiency ≈26.8%\approx 26.8\%

CaseNet work (kJ/kg)EfficiencyExhaust dryness
With reheat836.727.5%0.937
Without reheat747.826.8%0.866

If pump work is neglected, the efficiencies are 27.5% and 26.9%. Reheat increases work by about 12%, efficiency slightly, and makes the exhaust much drier.

✓ Answer checked by Claude Opus 5.5 (high effort)

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