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A simple Rankine cycle uses water as the working fluid. The boiler operates at 6000 kPa and the condenser at 50 kPa. At the entrance of the turbine the temperature is 450 deg C. The isentropic efficiency of the turbine is 94 percent, pressure and pump losses are negligible, and the water in the condenser is subcooled by 6.3 degC. The boiler is sized for a mass flow rate of 20 kg/s. Determine the rate at which heat is added in the boiler, the power required to operate the pumps, the net power produced by the cycle, and the thermal efficiency.

User Kmansoor
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1 Answer

22 votes
22 votes

Answer:

the rate at which heat is added in the boiler = 59597.4 kW

the power required to operate the pumps = 122.57 kW

The net power produced by the cycle = 17925 kW.

The thermal efficiency = 30%.

Step-by-step explanation:

The specific enthalpy of saturated liquid is equal to the enthalpy of the first point which is equal to 314 kJ/ kg.

The second enthalpy is calculated from the pump work. Therefore, the second enthalpy = first enthalpy point + specific volume of water [ the pressure of the boiler - the pressure of the condenser].

The second enthalpy = 314 + 0.00103 [ 6000 - 50 ] = 320.13 kJ/kg.

The specific enthalpy for the third point = 3300 kJ/kg.

Therefore, the rate at which heat is added in the boiler = 20 × [3300 - 320.13] = 59597.4 kW.

The rate at which heat is added in the boiler = 59597.4 kW.

Also, the power required to operate the pumps = 20 × 0.00103 [6000 - 50] = 122.57 kW.

The power produced by the turbine = 20 [ 300 - ( the fourth enthalpy value)].

The fourth enthalpy value = 3300 - 0.94 [ 3300 - 2340] = 2397.6 kJ/kg

Thus, the power produced by the turbine = 20 [ 300 - 2397.6] = 18048 kW.

The power produced by the turbine = 18048 kW.

The net power produced = 18048 + 122.57 = 17925 kW.

The thermal efficiency = [net power produced] / [the rate at which heat is added in the boiler].

The thermal efficiency = 17925/ 59597.4 = 30%.

User Sascha Kolberg
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