How is the pressure at which steam leaves the HP turbine found in this problem?

From s3 = s4 (LP expansion is isentropic) at 500 C, using the state-4 entropy from x = 0.896 at 5 kPa, then interpolating the superheated table

From the HP inlet entropy at 500 C

By assuming it is half of 20 MPa

From the condenser temperature only

2071 Shrawan · 12 marks

Reheat Rankine cycle: 20 MPa, 500°C

Short answer

By matching isentropic expansions: LP inlet entropy from exhaust dryness forces reheat pressure ≈ 1.5 MPa via interpolation. Net work = 1735.7 kJ/kg, heat added = 3906.7 kJ/kg, giving thermal efficiency ≈ 44.4%.

A steam power plant running on Rankine cycle has steam entering HP turbine at 20 MPa, 500°C and leaving LP turbine at 89.6 % dryness. Considering condenser pressure of 0.005 MPa and reheating occurring upto the temperature of 500°C determine: a) the pressure at which steam leaves HP turbine b) the thermal efficiency. [Refer the attached table for the properties of steam]

Assumptions: ideal reheat Rankine cycle, isentropic turbines and pump, pump work included. Values from steam tables (IAPWS-IF97).

States: 1 = 20 MPa, 500 C (HP inlet); 2 = HP exit at reheat pressure $p_r$ ; 3 = $p_r$ , 500 C (LP inlet); 4 = 0.005 MPa, $x = 0.896$ ; 5 = saturated water at 5 kPa; 6 = pump exit.

Steam table data

State	Data
5 kPa sat.	$t = 32.88$ C, $h_f = 137.77$ , $h_{fg} = 2423.0$ , $s_f = 0.4763$ , $s_{fg} = 7.9177$ , $v_f = 0.001005$
20 MPa, 500 C	$h_1 = 3241.2$ , $s_1 = 6.1445$

(a) Pressure at which steam leaves the HP turbine

State 4:

\begin{aligned} s_4 &= 0.4763 + 0.896 \times 7.9177 = 7.5705\ \text{kJ/kg K} \\ h_4 &= 137.77 + 0.896 \times 2423.0 = 2308.8\ \text{kJ/kg} \end{aligned}

LP expansion is isentropic, so $s_3 = s_4 = 7.5705$ at 500 C. From the superheated table at 500 C: $s = 7.6045$ at 1.4 MPa and $s = 7.5407$ at 1.6 MPa. Interpolating:

p_r = 1.4 + 0.2 \times \frac{7.6045 - 7.5705}{7.6045 - 7.5407} = 1.507 \approx 1.5\ \text{MPa}

Taking $p_r = 1.5$ MPa: at 1.5 MPa, 500 C, $h_3 = 3473.6$ kJ/kg.

Answer (a): steam leaves the HP turbine at about 1.5 MPa.

State 2 (1.5 MPa, $s_2 = s_1 = 6.1445$ )

At 1.5 MPa: $h_f = 844.7$ , $h_{fg} = 1946.3$ , $s_f = 2.3147$ , $s_{fg} = 4.1284$ .

\begin{aligned} x_2 &= \frac{6.1445 - 2.3147}{4.1284} = 0.9277 \\ h_2 &= 844.7 + 0.9277 \times 1946.3 = 2650.2\ \text{kJ/kg} \end{aligned}

Pump

\begin{aligned} W_p &= 0.001005 (20000 - 5) = 20.10\ \text{kJ/kg} \\ h_6 &= 137.77 + 20.10 = 157.87\ \text{kJ/kg} \end{aligned}

(b) Thermal efficiency

\begin{aligned} W_T &= (h_1 - h_2) + (h_3 - h_4) \\ &= (3241.2 - 2650.2) + (3473.6 - 2308.8) \\ &= 591.0 + 1164.8 = 1755.8\ \text{kJ/kg} \\ W_{net} &= 1755.8 - 20.1 = 1735.7\ \text{kJ/kg} \\ Q_s &= (h_1 - h_6) + (h_3 - h_2) \\ &= (3241.2 - 157.87) + (3473.6 - 2650.2) \\ &= 3083.3 + 823.4 = 3906.7\ \text{kJ/kg} \\ \eta &= \frac{1735.7}{3906.7} = 0.4443 \end{aligned}

Answer (b): thermal efficiency $\approx 44.4\%$ (net work 1735.7 kJ/kg).

  T
  ^                  1          3
  |                 /|         /|
  |                / |        / |
  |     f--------g'  |       /  |
  |    /             2------'   |
  |   /                         |
  |  6                          |
  |  5--------------------------4
  +--------------------------------> s
  5-6 pump, 6-f-g'-1 boiler (heat,
  boil, superheat), 1-2 HP turbine,
  2-3 reheater, 3-4 LP turbine,
  4-5 condenser

✓ Answer checked by Claude Opus 5.5 (high effort)

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