Question

The condenser of a room air conditioner is designed to reject heat at a rate of \(22,500 \mathrm{~kJ} / \mathrm{h}\) from refrigerant- \(134 \mathrm{a}\) as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(32^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the heat transfer area on the refrigerant side.

Short answer

Answer: The heat transfer area on the refrigerant side of the finned condenser coils is approximately 3.6232 square meters.

Step-by-step solution

Convert the units of heat rejection rate to the same units as the heat transfer coefficient

The heat rejection rate is given in kJ/h, while the heat transfer coefficient is given in W/m²·K. To make them consistent, we need to convert the heat rejection rate to W (1 kW = 1,000 W and 1 h = 3600 s). $$ Q = 22,500 \frac{kJ}{h}\cdot\frac{1000 W}{kJ}\cdot\frac{1h}{3,600s} = 6,250 W $$ Now that we have \(Q = 6,250 W\), we can move to the next step.

Determine the temperature difference between the refrigerant and air

To find the temperature difference, we should first calculate the average air temperature as it flows across the condenser coils. Avg air temperature = \(\frac{T_{inlet} + T_{outlet}}{2}\) Avg air temperature = \(\frac{25 + 32}{2} = 28.5 ^\circ C\) The refrigerant is condensed at 40 degrees Celsius. So, the temperature difference is: $$\Delta T = T_{refrigerant} - T_{avg\,air} = 40 - 28.5 = 11.5 K$$ Now that we have the temperature difference, we can move to the next step.

Calculate the mass flow rate of air

Given \(Q = mc_{p}\Delta T\), where \(m\) is the mass flow rate of air, \(c_{p}\) is the specific heat capacity of air (\(1005 J/kg·K\)), and \(\Delta T\) is the temperature difference between the air entering and leaving the condenser coils. We can rearrange the equation to find \(m\): $$ m = \frac{Q}{c_{p}\Delta T} = \frac{6,250}{1005 \times(32 - 25)} = \frac{6,250}{1005 \times 7} = 0.9009\, kg/s $$ Now, we know the mass flow rate of air. We can move to the next step.

Determine the heat transfer area on the refrigerant side

We can now use the overall heat transfer coefficient (\(U = 150 W/m²·K\)) to find the heat transfer area (\(A\)). The heat transfer equation is: $$ Q = UA\Delta T $$ We can rearrange this equation to solve for the heat transfer area \(A\): $$ A = \frac{Q}{U\Delta T} = \frac{6,250}{150 \times 11.5} = 3.6232\, m^{2} $$ Therefore, the heat transfer area on the refrigerant side is approximately 3.6232 square meters.