Question

Saturated steam coming off the turbine of a steam power plant at 40oC condenses on the outside of a 7.421-cm-outer-diameter, 2.916-m-long tube at a rate of 16.154 kg/h. Calculate the rate of heat transfer in kW (give your answer with 3 decimals, do NOT enter units!!!).

Answer 1

Answer:

Rate of heat transfer will be 10.69 kw

Explanation:

We have given temperature $T=40^{\circ}C$

From standard table enthalpy of vaporization 2382.75 KJ/kg

Steam is coming out from turbine at rate of 16.154 kg/hour

We have to calculate the rate of heat transfer

Heat transfer is equal to $Q=mass\times enthalpy\ of \ vaporization$

So rate of heat transfer $=16.154\times 2382.75=38490.94KJ/hr$

= $=16.154\times 2382.75=38490.94\times \frac{1}{3600}watt=10.691kwatt$

So rate of heat transfer will be 10.69 kw