Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through a tube at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The condensate leaves the tube as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer from the tube is (a) \(34 \mathrm{~kJ} / \mathrm{s}\) (b) \(268 \mathrm{~kJ} / \mathrm{s}\) (c) \(453 \mathrm{~kJ} / \mathrm{s}\) (d) \(481 \mathrm{~kJ} / \mathrm{s}\) (e) \(515 \mathrm{~kJ} / \mathrm{s}\)

Short Answer

Expert verified
Answer: The rate of heat transfer is approximately \(477.2~\mathrm{kJ/s}\), which is closest to option (d) \(481 \mathrm{~kJ} / \mathrm{s}\) among the given choices.

Step by step solution

01

Identify the known variables

We know the mass flow rate \(\dot{m} = 0.2~\mathrm{kg/s}\) and the temperature \(T=40^{\circ}\mathrm{C}\)
02

Look up the enthalpy values of saturated liquid and vapor at \(40^{\circ} \mathrm{C}\)

Using a steam table, we find the values of enthalpy of saturated liquid (\(h_{f}\)) and saturated vapor (\(h_{g}\)) at \(40^{\circ} \mathrm{C}\): \(h_{f}=167.52~\mathrm{kJ/kg}\) \(h_{g}=2553.6~\mathrm{kJ/kg}\)
03

Calculate the enthalpy difference

The enthalpy difference can be found by subtracting the enthalpy of saturated liquid from saturated vapor: \(\Delta h = h_{g} - h_{f}\) \(\Delta h = 2553.6~\mathrm{kJ/kg} - 167.52~\mathrm{kJ/kg} = 2386.08~\mathrm{kJ/kg}\)
04

Calculate the rate of heat transfer

To calculate the rate of heat transfer, we multiply the mass flow rate with the enthalpy difference: \(\dot{Q} = \dot{m} \times \Delta h\) \(\dot{Q} = 0.2~\mathrm{kg/s} \times 2386.08~\mathrm{kJ/kg} = 477.216~\mathrm{kJ/s}\) The calculated rate of heat transfer is approximately \(477.2~\mathrm{kJ/s}\). Looking at the given options, option (d) is closest to our calculated value. So, the correct answer is (d) \(481 \mathrm{~kJ} / \mathrm{s}\).

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