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In a steam power plant based on ideal Rankine cycle, superheated steam enters the turbine at 1.4 MPa and $500^{\circ} C$ where it expands to a condensor pressure of 0.01 MPa. Superheated steam properties are; At 1.4 MPa and $500^{\circ} C, h = 3474.1 k J / k g a n d s = 7.6027 k J / k g K$ Saturated water properties are given in below table:

The cycle efficiency is ____% (Correct upto two decimal places).
Take specific volume of saturated liquid at 0.01.MPa as $0.00101 m^{3} / k g$ .
32.36

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Solution

The correct option is A 32.36

Given data:

$h_{1} = 3474.1 k J / k g$

$s_{1} = 7.6027 k J / k g K$

$h_{3} = (h_{f}) a t 0.01 M P a$

$= 191.81 k J / k g$

Process 1-2 is isentropic

$∴ s_{1} = s_{2}$

$7.6027 = (s_{f} + x s_{f g}) a t 0.01 M P a$

(where x = Dryness fraction)

$7.6027 = 0.6492 + x (8.1454 - 0.6492)$

$x = 0.9276$

$h_{2} = (h_{f} + x h_{f g}) a t 0.01 M P a$

$= 191.81 + 0.927 \times (2583.9 - 191.81)$

$h_{2} = 2410.72 k J / k g$

$W_{p} = h_{4} - h_{3} = v_{3} d P = v_{3} (P_{4} - P_{3}) (P_{4} = P_{1} = 1.4 M P a, P_{3} = 0.01 M P a$

$W_{p} = h_{4} - h_{3} = 0.00101 (1.4 - 0.01) \times 1000 k J / k g$

$W_{p} = h_{4} - h_{3} = 1.404 k J / k g$

$h_{4} = h_{3} + W_{p} = h_{3} + 1.404 = 191.81 + 1.404$

$h_{4} = 193.214 k J / k g$

$η_{c y c l e} = \frac{W_{n e t}}{Q_{i n p u t}} = \frac{W_{T} - W_{p}}{Q_{b o i l e r}}$

$= \frac{(h_{1} - h_{2}) - (h_{4} - h_{3})}{(h_{1} - h_{4})} = \frac{(3474.1 - 2410.72) - (1.404)}{(3474.1 - 193.214)}$

$= \frac{1061.976}{3280.886} = 0.3236 o r 32.36 %$

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