Question

An air-cooled condenser is used to condense isobutane in a binary geothermal power plant. The isobutane is condensed at \(85^{\circ} \mathrm{C}\) by air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\) ) that enters at \(22^{\circ} \mathrm{C}\) at a rate of \(18 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient and the surface area for this heat exchanger are \(2.4 \mathrm{~kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $2.6 \mathrm{~m}^{2}$, respectively. The outlet temperature of the air is (a) \(35.6^{\circ} \mathrm{C}\) (b) \(40.5^{\circ} \mathrm{C}\) (c) \(52.1^{\circ} \mathrm{C}\) (d) \(58.5^{\circ} \mathrm{C}\) (e) \(62.8^{\circ} \mathrm{C}\)

Short answer

Answer: (b) 40.5 °C

Step-by-step solution

Calculate the rate of heat transfer from the heat exchanger to the air

To find the rate of heat transfer, we can use the formula: \(Q = U \cdot A \cdot \Delta T\) Here, \(Q\) represents the rate of heat transfer, \(U\) is the overall heat transfer coefficient, \(A\) is the surface area of the heat exchanger, and \(\Delta T\) is the temperature difference between the air and the isobutane. Given: \(U = 2.4 \, \mathrm{kW} / \mathrm{m}^{2} \cdot \mathrm{K}\) (Overall heat transfer coefficient) \(A = 2.6 \, \mathrm{m}^{2}\) (Surface area) \(\Delta T = 85^{\circ} \mathrm{C} - 22^{\circ} \mathrm{C} = 63 \mathrm{K}\) (Temperature difference) Now, we can calculate the rate of heat transfer: \(Q = 2.4 \, \mathrm{kW} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 2.6 \, \mathrm{m}^{2} \cdot 63 \, \mathrm{K} = 391.68 \, \mathrm{kW}\)

Calculate the outlet temperature of the air

To find the outlet temperature of the air, we will use the formula: \(Q = \dot{m} \cdot c_{p} \cdot \Delta T_{out}\) Here, \(Q\) is the rate of heat transfer, \(\dot{m}\) is the mass flow rate of the air, \(c_{p}\) is the specific heat capacity of the air, and \(\Delta T_{out}\) is the temperature difference between the inlet and outlet for the air. We know: \(Q = 391.68 \, \mathrm{kW}\) (Rate of heat transfer) \(\dot{m} = 18 \, \mathrm{kg} / \mathrm{s}\) (Mass flow rate of the air) \(c_{p} = 1.0 \, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) (Specific heat capacity) We need to find: \(\Delta T_{out}\) (Temperature difference between air inlet and outlet) Using the formula: \(\Delta T_{out} = \frac{Q}{\dot{m} \cdot c_{p}} = \frac{391.68 \, \mathrm{kW}}{18 \, \mathrm{kg} / \mathrm{s} \cdot 1.0 \, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}} = 21.76 \, \mathrm{K}\) Since we know the inlet temperature of the air is \(22^{\circ} \mathrm{C}\), the outlet temperature is: \(T_{out} = 22^{\circ} \mathrm{C} + 21.76 \, \mathrm{K} = 43.76^{\circ} \mathrm{C}\) Now, let's find the closest answer from the given choices: (a) \(35.6^{\circ} \mathrm{C}\) (b) \(40.5^{\circ} \mathrm{C}\) (c) \(52.1^{\circ} \mathrm{C}\) (d) \(58.5^{\circ} \mathrm{C}\) (e) \(62.8^{\circ} \mathrm{C}\) The closest answer is choice (b) \(40.5^{\circ} \mathrm{C}\).