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The steam of the Rankine cycle expands in the turbine heat zone, its enthalpy decreases by 150 kJ/kg, after which the condensation of saturated steam at constant pressure in the heat exchanger releases 280 kJ/kg of heat. Determine the thermal efficiency of the cycle.

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Answer:

The thermal efficiency of the Rankine cycle can be calculated using the formula:

$\eta_{th}=\frac{(h_1-h_2)-(h_4-h_3)}{(h_1-h_4)}$

where:

h1 = enthalpy at turbine inlet

h2 = enthalpy at turbine outlet

h3 = enthalpy at condenser inlet

h4 = enthalpy at condenser outlet

Given:

- The enthalpy decrease in the turbine is 150 kJ/kg (h1 - h2 = 150 kJ/kg)

- The heat released during condensation is 280 kJ/kg (h1 - h4 = 280 kJ/kg)

Assuming the pump work is negligible, we can simplify the efficiency equation:

$\eta_{th}=\frac{(h_1-h_2)}{(h_1-h_4)}=\frac{150}{280}=0.536$

Therefore, the thermal efficiency of the Rankine cycle is 53.6%.

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