Question

The radiator in an automobile is a crossflow heat exchanger $\left(U A_{s}=10 \mathrm{~kW} / \mathrm{K}\right)\( that uses air \)\left(c_{p}=1.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ to cool the engine-coolant fluid \(\left(c_{p}=4.00 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\). The engine fan draws \(22^{\circ} \mathrm{C}\) air through this radiator at a rate of \(8 \mathrm{~kg} / \mathrm{s}\) while the coolant pump circulates the engine coolant at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The coolant enters this radiator at \(80^{\circ} \mathrm{C}\). Under these conditions, the effectiveness of the radiator is \(0.4\). Determine \((a)\) the outlet temperature of the air and \((b)\) the rate of heat transfer between the two fluids.

Short answer

Answer: The outlet temperature of the air is 45.2°C, and the rate of heat transfer between the two fluids is 185.6 kW.

Step-by-step solution

Calculate the heat capacity rates of the air and the coolant fluid

To do this, we will use the formula \(\dot{C} = m \cdot c_{p}\), where \(\dot{C}\) is the heat capacity rate, \(m\) is the mass flow rate, and \(c_{p}\) is the specific heat at constant pressure. For the air: \(\dot{C}_{air} = m_{air} \cdot c_{p_{air}} = 8 \, kg/s \cdot 1.00 \, kJ/(kg \cdot K) = 8 \, kW/K\) For the coolant fluid: \(\dot{C}_{coolant} = m_{coolant} \cdot c_{p_{coolant}} = 5 \, kg/s \cdot 4.00 \, kJ/(kg \cdot K) = 20 \, kW/K\)

Determine the minimum heat capacity rate

We need to find the minimum heat capacity rate (\(\dot{C}_{min}\)) to calculate the maximum possible heat transfer. In this case, the minimum heat capacity rate is associated with the air because \(\dot{C}_{air} < \dot{C}_{coolant}\). \(\dot{C}_{min} = \dot{C}_{air} = 8 \, kW/K\)

Calculate the actual heat transfer between the two fluids

Using the effectiveness of the radiator (\(\epsilon = 0.4\)) and the maximum possible heat transfer, we can calculate the actual heat transfer (\(q\)) as follows: \(q = \epsilon \cdot \dot{C}_{min} \cdot (T_{inlet \, coolant} - T_{inlet \, air})\) \(q = 0.4 \cdot 8 \, kW/K \cdot (80^{\circ}C - 22^{\circ}C) = 0.4 \cdot 8 \, kW/K \cdot 58^{\circ}C = 185.6 \, kW\)

Determine the outlet temperature of the air

Now that we have the actual heat transfer between the two fluids, we can calculate the outlet temperature of the air using the formula: \(T_{outlet \, air} = T_{inlet \, air} + \frac{q}{\dot{C}_{air}}\) \(T_{outlet \, air} = 22^{\circ}C + \frac{185.6 \, kW}{8 \, kW/K} = 22^{\circ}C + 23.2^{\circ}C = 45.2^{\circ}C\)

Calculate the rate of heat transfer between the two fluids

We have already calculated the rate of heat transfer (actual) between the two fluids in Step 3. The rate of heat transfer between the two fluids is \(q = 185.6 \, kW\). In conclusion, the outlet temperature of the air is \(45.2^{\circ}C\), and the rate of heat transfer between the two fluids is \(185.6 \, kW\).