Question

An air handler is a large unmixed heat exchanger used for comfort control in large buildings. In one such application, chilled water $\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ enters an air handler at \(5^{\circ} \mathrm{C}\) and leaves at \(12^{\circ} \mathrm{C}\) with a flow rate of \(1000 \mathrm{~kg} / \mathrm{h}\). This cold water cools air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\). The rate of heat transfer between the two streams is (a) \(8.2 \mathrm{~kW}\) (b) \(23.7 \mathrm{~kW}\) (c) \(33.8 \mathrm{~kW}\) (d) \(44.8 \mathrm{~kW}\) (e) \(52.8 \mathrm{~kW}\)

Short answer

Answer: (a) 8.2 kW

Step-by-step solution

Convert flow rate of chilled water to kg/s

We are given the flow rate of the chilled water in kg/h. To make the calculations easier, let's convert this flow rate to kg/s. \(Flow\,rate\,in\,kg/s = \frac{1000 \, kg/h}{3600 \, s}\)

Calculate heat gained by the chilled water

To find the heat gained by the chilled water, we can use the following equation: \(Q_{water} = m_{water} \cdot c_{p_{water}} \cdot \Delta T_{water}\) where \(Q_{water}\) is the heat gained by the chilled water, \(m_{water}\) is the mass flow rate of the chilled water in kg/s, \(c_{p_{water}}\) is the specific heat capacity of the water in kJ/(kg·K), and \(\Delta T_{water}\) is the temperature change of the water in K. We have values for all the variables, so we can put them into the equation: \(Q_{water} = \frac{1000}{3600} \cdot 4.2 \cdot (12 - 5)\)

Calculate heat lost by the air

Since the heat transferred between the two streams must be equal, the heat lost by the air can be found by setting the heat gained by the chilled water equal to the heat loss by the air: \(Q_{air} = Q_{water}\) We must now find the mass flow rate of the air using the following equation: \(Q_{air} = m_{air} \cdot c_{p_{air}} \cdot \Delta T_{air}\) where \(Q_{air}\) is the heat transmitted from the air, \(m_{air}\) is the mass flow rate of the air in kg/s, \(c_{p_{air}}\) is the specific heat capacity of the air in kJ/(kg·K), and \(\Delta T_{air}\) is the temperature change of the air in K. We know the specific heat capacity and temperature change of the air. Let's put everything into the equation: \(\frac{1000}{3600} \cdot 4.2 \cdot (12 - 5) = m_{air} \cdot 1.0 \cdot (25 - 15)\)

Solve for the rate of heat transfer

Now that we have the equations set up, we can solve for the rate of heat transfer between the two streams. The heat gained by the chilled water is equal to the heat lost by the air: \(Q_{water} = Q_{air}\) \(\frac{1000}{3600} \cdot 4.2 \cdot (12 - 5) = m_{air} \cdot 1.0 \cdot (25 - 15)\) We can now solve for \(Q_{water}\), which is also the rate of heat transfer between the two streams: \(Q_{water} = 0.278 \times 29.4 = 8.1732 \, kW\) This value is closest to option (a), so the answer is: (a) \(8.2 \mathrm{~kW}\)