Question

A single-pass crossflow heat exchanger uses hot air (mixed) to heat water (unmixed), flowing with a mass flow rate of \(3 \mathrm{~kg} / \mathrm{s}\), from \(30^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). The hot air enters and exits the heat exchanger at \(220^{\circ} \mathrm{C}\) and $100^{\circ} \mathrm{C}\(, respectively. If the overall heat transfer coefficient is \)200 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}$, determine the required surface area.

Short answer

Answer: The required surface area of the single-pass crossflow heat exchanger is approximately 30.69 m².

Step-by-step solution

Calculate the heat capacity rate of the water

First, let's calculate the heat capacity rate of the water. The specific heat capacity of water is approximately \(4.18 \mathrm{~kJ/kg·K}\). The heat capacity rate can be calculated using the formula \(\dot{C} = m\dot{c}_p\), where \(\dot{C}\) is the heat capacity rate, \(m\) is the mass flow rate, and \(\dot{c}_p\) is the specific heat capacity. The heat capacity rate for the water is: \(\dot{C}_w = m\dot{c}_{p,w} = (3 \mathrm{~kg/s}) (4.18 \mathrm{~kJ/kg·K}) = 12.54 \mathrm{~kJ/s·K}\)

Calculate the heat transfer rate

The heat transfer rate can be calculated using the heat capacity rate and the temperature difference of the water before and after the heat exchanger: \(\dot{Q} = \dot{C}_w (\Delta T_w) = 12.54 \mathrm{~kJ/s·K} (80^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C}) = 12.54 \times 50 \mathrm{~kJ/s} = 627 \mathrm{~kW}\)

Calculate the required temperature difference

To calculate the required surface area, we need to know the Log Mean Temperature Difference (LMTD) for the heat exchanger. The LMTD can be calculated as follows: \(\mathrm{LMTD} = \dfrac{\Delta T_1 - \Delta T_2}{\ln \frac{\Delta T_1}{\Delta T_2}}\) Where: \(\Delta T_1 = T_{h,i} - T_{c,o} = 220^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C} = 140^{\circ} \mathrm{C}\) \(\Delta T_2 = T_{h,o} - T_{c,i} = 100^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 70^{\circ} \mathrm{C}\) \(\mathrm{LMTD} = \dfrac{140^{\circ} \mathrm{C} -70^{\circ} \mathrm{C}}{\ln \frac {140^{\circ} \mathrm{C}}{70^{\circ} \mathrm{C}}} = 102.2^{\circ} \mathrm{C}\)

Calculate the required surface area

Now, we will use the overall heat transfer coefficient and the LMTD to calculate the required surface area for the heat exchanger using the formula: \(A = \dfrac{\dot{Q}}{U \times \mathrm{LMTD}}\) Where \(A\) is the required surface area, \(\dot{Q}\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, and LMTD is the log mean temperature difference. \(A = \dfrac{627 \times 10^{3} \mathrm{~W}}{200 \mathrm{~W/m^2·K} \times 102.2^{\circ} \mathrm{C}} = 30.69 \mathrm{~m}^2\) So, the required surface area of the single-pass crossflow heat exchanger is approximately \(30.69 \mathrm{~m}^2\).