Question

A crossflow air-to-water heat exchanger with an effectiveness of \(0.65\) is used to heat water \(\left(c_{p}=4180\right.\) $\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})\( with hot air \)\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\(. Water enters the heat exchanger at \)20^{\circ} \mathrm{C}$ at a rate of \(4 \mathrm{~kg} / \mathrm{s}\), while air enters at $100^{\circ} \mathrm{C}\( at a rate of \)9 \mathrm{~kg} / \mathrm{s}$. If the overall heat transfer coefficient based on the water side is $260 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the heat transfer surface area of the heat exchanger on the water side. Assume both fluids are unmixed. Answer: \(52.4 \mathrm{~m}^{2}\)

Short answer

The heat transfer surface area on the water side of the heat exchanger is 52.4 m².

Step-by-step solution

Calculate the heat transfer rate

We'll use the effectiveness definition for a heat exchanger to calculate the heat transfer rate (\(\dot{Q}\)). Effectiveness (\(\varepsilon\)) is defined as: \(\varepsilon = \frac{\dot{Q}}{\dot{Q}_{\text{max}}}\) Where \(\dot{Q}_{\text{max}}\) is the maximum possible heat transfer rate and can be calculated as: \(\dot{Q}_{\text{max}} = \min(m_1c_{p1}, m_2c_{p2})(T_{1\text{in}} - T_{2\text{in}})\) Insert the given values: \(m_1 = 9\,\text{kg/s}\) \(c_{p1} = 1010\,\text{J/(kg}\cdot\text{K)}\) \(m_2 = 4\,\text{kg/s}\) \(c_{p2} = 4180\,\text{J/(kg}\cdot\text{K)}\) \(T_{1\text{in}} = 100 ^{\circ}\text{C}\) \(T_{2\text{in}} = 20 ^{\circ}\text{C}\) \(\varepsilon = 0.65\) Calculate \(\dot{Q}_{\text{max}}\): \(\dot{Q}_{\text{max}} = \min(9\times1010, 4\times4180)(100 - 20)\) Now we can calculate \(\dot{Q}\): \(\dot{Q} = \varepsilon \cdot \dot{Q}_{\text{max}}\)

Calculate the temperature difference

Calculate the temperature difference between the inlet and outlet of the heat exchanger: \(\Delta T = T_{1\text{in}} - T_{2\text{out}}\) As both fluids are unmixed, we can use the heat transfer rate to calculate the outlet temperature of the water: \(T_{2\text{out}} = T_{2\text{in}} + \frac{\dot{Q}}{m_2c_{p2}}\) Substitute the known values and calculate \(T_{2\text{out}}\). Then, find \(\Delta T\).

Calculate the heat transfer surface area

We're given the overall heat transfer coefficient based on the water side (\(U\)) as \(260\,\text{W/(m}^2\cdot\text{K)}\). We can use this to determine the heat transfer surface area (\(A\)) using the equation: \(A = \frac{\dot{Q}}{U\Delta T}\) Substitute the calculated values for \(\dot{Q}\) and \(\Delta T\), then calculate the heat transfer surface area (\(A\)). The heat transfer surface area on the water side of the heat exchanger is \(52.4\,\text{m}^2\).