Question

Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through the tubes of an air-cooled condenser at a rate of $0.2 \mathrm{~kg} / \mathrm{s}$. The condensate leaves the tubes as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer to air 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

a) 350 kJ/s b) 400 kJ/s c) 453 kJ/s d) 500 kJ/s Answer: c) 453 kJ/s

Step-by-step solution

Determine the enthalpy of saturated vapor and saturated liquid

Using a steam table or other thermodynamic source, look up the enthalpy of saturated vapor \(h_{fg}\) and saturated liquid \(h_{f}\) for water at \(40^{\circ} \mathrm{C}\). In this case, we find: $$h_{fg} = 2406.7 \mathrm{~kJ} / \mathrm{kg}$$ $$h_{f} = 167.57 \mathrm{~kJ} / \mathrm{kg}$$

Calculate the enthalpy difference

Subtract the enthalpy of the saturated liquid \(h_{f}\) from the enthalpy of the saturated vapor \(h_{fg}\) to find the enthalpy difference \(\Delta h\): $$\Delta h = h_{fg} - h_{f}$$ $$\Delta h = 2406.7 \mathrm{~kJ} / \mathrm{kg} - 167.57 \mathrm{~kJ} / \mathrm{kg}$$ $$\Delta h = 2239.13 \mathrm{~kJ} / \mathrm{kg}$$

Calculate the rate of heat transfer

Use the mass flow rate \(m = 0.2 \mathrm{~kg} / \mathrm{s}\) and the enthalpy difference \(\Delta h\) to calculate the rate of heat transfer \(Q\) using the heat transfer equation: $$Q = m \times \Delta h$$ $$Q = 0.2 \mathrm{~kg} / \mathrm{s} \times 2239.13 \mathrm{~kJ} / \mathrm{kg}$$ $$Q = 447.826 \mathrm{~kJ} / \mathrm{s}$$ Comparing this value to the given choices, we find that option (c) \(453 \mathrm{~kJ} / \mathrm{s}\) is the closest to our calculated value. Rounding errors and approximations in the steam table may account for the slight difference. The rate of heat transfer to air is approximately \(453 \mathrm{~kJ} / \mathrm{s}\). So, the correct answer is (c).