Air at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle, where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s, the rate of heat transfer to the air in the heat exchanger, by taking the enthalpy of air as h = Cpt, where Cp is the specific heat equal to 1.005 KJ/Kg and t being the temperature, is

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Air at a temperature of 15°C passes through a heat exchanger at a velocity of 30 m/s where its temperature is raised to 800°C. It then enters a turbine with the same velocity of 30 m/s and expands until the temperature falls to 650°C. On leaving the turbine, the air is taken at a velocity of 60 m/s to a nozzle, where it expands until the temperature has fallen to 500°C. If the air flow rate is 2 kg/s, the rate of heat transfer to the air in the heat exchanger, by taking the enthalpy of air as h = Cpt, where Cp is the specific heat equal to 1.005 KJ/Kg and t being the temperature, is

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BPSC AE Paper IV General Engineering 19 Dec 2024 Official Paper

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1050 KJ/sec
4500 KJ/sec
1300 KJ/sec
1580 KJ/sec

Answer 1

Concept:

The rate of heat transfer in a heat exchanger is calculated using enthalpy change when there is no change in kinetic or potential energy. Enthalpy for air is given by:

\( h = C_p \cdot T \)

Calculation:

Initial temperature: \( T_1 = 15^\circ C = 288~K \)
Final temperature after heat exchanger: \( T_2 = 800^\circ C = 1073~K \)
Specific heat: \( C_p = 1.005~kJ/kg·K \)
Mass flow rate: \( \dot{m} = 2~kg/s \)

Heat added in exchanger:

\( \dot{Q} = \dot{m} \times C_p \times (T_2 - T_1) \)

\( \dot{Q} = 2 \times 1.005 \times (1073 - 288) = 2 \times 1.005 \times 785 = 1580~kJ/s \)

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