A steam power plant operates on a simple ideal Rankine cycle between the pressur

Question

A steam power plant operates on a simple ideal Rankine cycle between the pressure limits of 3MPa and 50kPa. The temperature of the steam at the turbine inlet is 400'C and the mass flow rate of steam through the cycle is 60kg/s. Determine the thermal efficiency of the cycle and the net power output of the power plant.

Explanation / Answer

As the cycle is simple ideal rankine cycle, lets assume that steam at pump inlet (1) is saturated liquid at 50 kpa, steam at pump exit (2) is at 3 mpa and the compression process (1-2) is isentropic now turbine inlet (3) temperature is 600C and pressure at turbine is 3 mpa. Pressure at turbine exit (4) is 50 kpa and expansion process in turbine (3-4) is isentropic take down the values of enthalpy and entropy of steam at 50 kpa considering the steam tables (saturated water) and then water at 3mpa (sub cooled liquid) and entropy equal to the value of first state. Then again steam at 3 mpa and 600C (superheated) finally saturated steam value at 50kpa. The values from tables are as follows 1 => h = 340.613 kj/kg s = 1.09122 kj/kgk 2 => h = 343.613 kj/kg s = 1.09122 kj/kgk 3 => h = 3230.76 kj/kg s = 6.09211 kj/kgk 4 => h = 2407.175 kj/kg s = 6.09211 kj/kgk Now net work done is = 1 – (Q out / Q in) = 1 – ( h4 - h1 / h3 – h2 ) = 0.71578 = 71.578% Net power out put = turbine work - pump work = (h3 – h4) – (h2 – h1) = 820.55 kj/kg Given mass is 60 kg/s , so POWER = 820.55 * 60 = 49233 kwatts

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