Question

Cold water $\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( enters a heat exchanger at \)15^{\circ} \mathrm{C}$ at a rate of \(0.5 \mathrm{~kg} / \mathrm{s}\), where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(50^{\circ} \mathrm{C}\) at a rate of $1.8 \mathrm{~kg} / \mathrm{s}$. The maximum possible heat transfer rate in this heat exchanger is (a) \(51.1 \mathrm{~kW}\) (b) \(63.0 \mathrm{~kW}\) (c) \(66.8 \mathrm{~kW}\) (d) \(73.2 \mathrm{~kW}\) (e) \(80.0 \mathrm{~kW}\)

Short answer

Based on the solution provided, the maximum heat transfer rate in the heat exchanger is \(32.16 \mathrm{~kW}\).

Step-by-step solution

Identify known and unknown variables

We are given the following information: \(c_{p_{water}} = 4.18 \mathrm{~kJ/ (kg \cdot K)}\) - specific heat of water \(T_{water,initial} = 15^\circ \mathrm{C}\) - initial water temperature \(m_{water} = 0.5 \mathrm{~kg/s}\) - mass flow rate of water \(c_{p_{air}} = 1.0 \mathrm{~kJ/ (kg \cdot K)}\) - specific heat of air \(T_{air,initial} = 50^\circ \mathrm{C}\) - initial air temperature \(m_{air} = 1.8 \mathrm{~kg/s}\) - mass flow rate of air Unknown: \(Q\) - maximum heat transfer rate

Calculate the maximum temperature change

The maximum possible heat transfer occurs when the temperature of the air and water is equalized. Therefore, we need to calculate the maximum temperature change for both air and water: \(\Delta T_{max} = T_{mixed} - T_{initial}\) We also need to determine the mixed temperature \(T_{mixed}\). Since the heat exchange happens within a closed system, we have the relationship: \(m_{air} \cdot c_{p_{air}} \cdot (T_{air,initial} - T_{mixed}) = m_{water} \cdot c_{p_{water}} \cdot (T_{mixed} - T_{water,initial})\)

Solve for the mixed temperature

Let's solve the equation above to find \(T_{mixed}\): \(1.8 \cdot 1.0 \cdot (50 - T_{mixed}) = 0.5 \cdot 4.18 \cdot (T_{mixed} - 15)\) Simplify and solve for \(T_{mixed}\): \(T_{mixed} = \frac{1.8 \cdot 1.0 \cdot 50 + 0.5 \cdot 4.18 \cdot 15}{1.8 \cdot 1.0 + 0.5 \cdot 4.18} = 30.32^\circ \mathrm{C}\)

Calculate the heat transfer rate

Now, we can calculate the heat transfer rate for both air and water. For water: \(Q_{water} = m_{water} \cdot c_{p_{water}} \cdot (T_{mixed} - T_{water,initial})\) \(Q_{water} = 0.5 \cdot 4.18 \cdot (30.32 - 15) = 32.16 \mathrm{~kW}\) For air: \(Q_{air} = m_{air} \cdot c_{p_{air}} \cdot (T_{air,initial} - T_{mixed})\) \(Q_{air} = 1.8 \cdot 1.0 \cdot (50 - 30.32) = 35.41 \mathrm{~kW}\) Since in a heat exchanger, the heat transfer rate must be the same for air and water: \(Q = Q_{water} = Q_{air} = 32.16 \mathrm{~kW}\) The closest answer choice is: (a) \(51.1 \mathrm{~kW}\)