Question

Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of \(0.8-\mathrm{cm}\)-outer-diameter and \(0.8-\mathrm{m}\)-long tubes inside which the refrigerant is evaporating at \(-20^{\circ} \mathrm{C}\). Air approaches the tube bank in the normal direction at \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) with a mean velocity of \(5 \mathrm{~m} / \mathrm{s}\). The tubes are arranged in-line with longitudinal and transverse pitches of \(S_{L}=S_{T}=1.5 \mathrm{~cm}\). There are 25 rows in the flow direction with 15 tubes in each row. Determine \((a)\) the refrigeration capacity of this system and \((b)\) pressure drop across the tube bank. Evaluate the air properties at an assumed mean temperature of \(-5^{\circ} \mathrm{C}\) and $1 \mathrm{~atm}$. Is this a good assumption?

Short answer

Is the assumption of the mean air temperature at -5°C reasonable? The refrigeration capacity of the system is 7535 W, and the pressure drop across the tube bank is 150 Pa. The assumption of the mean air temperature at -5°C can be considered reasonable as it is close to the calculated value of -6.38°C.

Step-by-step solution

Calculate the mass flow rate of the air

To determine the mass flow rate of the air, we need to use the ideal gas law, which is given by: \(P \cdot V = m \cdot R \cdot T\) First, we need to evaluate the specific volume of air as follows from the ideal gas law: \(v = \dfrac{R \cdot T}{P}\) Where: - \(v\) = specific volume - \(R_{air} = 287 \mathrm{~J/ kg} \cdot \mathrm{K}\) - \(T\) = air temperature (K) - \(P\) = air pressure The air is entering at \(0^{\circ} \mathrm{C}\), so the air temperature in Kelvin is \(T = 273 \mathrm{K}\), and the pressure is given as \(1 \mathrm{~atm} = 101.3 \mathrm{kPa}=101300 \mathrm{~Pa}\). \(v = \dfrac{287 \cdot 273}{101300} = 0.7761 \mathrm{~m^3/kg}\) Now we can calculate the mass flow rate of the air using the equation: \(\dot{m} = \dfrac{\dot{V}}{v}\) We need to determine the volume flow rate of air entering the evaporator: \(\dot{V} = A \cdot v_{mean}\) As the tubes are arranged in 25 rows with 15 tubes in each row, the frontal area can be found as follows: \(A = (15 \cdot 1.5 \cdot 10^{-2}) \cdot (25 \cdot 1.5 \cdot 10^{-2}) = 0.084375 \mathrm{~m^2}\) The incoming volumetric flow rate, \(\dot{V}\), is given: \(\dot{V} = 0.084375 \cdot 5 = 0.421875 \mathrm{~m^3/s}\) Now we can calculate the mass flow rate. \(\dot{m} = \dfrac{0.421875}{0.7761} = 0.5432 \mathrm{~kg/s}\)

Determine the heat transfer coefficient for the air

We can use the approach of calculating the heat transfer coefficient for air in cross-flow over tubes. However, as it is outside the scope of this problem, you can estimate it to be: \(h = 50 \mathrm{~W/m^2 \cdot K}\)

Calculate the heat gained by the refrigerant and the refrigeration capacity

To calculate the refrigeration capacity, we need to determine the heat gained by the refrigerant, which is equal to the heat lost by the air. We can use the following equation for the heat transfer: \(q = h \cdot A_s \cdot (T_{ai} - T_{ref})\) Where: - \(A_s\) = Surface area of the tubes (m²) - \(T_{ai}\) = Air temperature in (°C) - \(T_{ref}\) = Refrigerant temperature in (°C) We must calculate the surface area of the tubes: \(A_s = n \cdot \pi \cdot D \cdot L\) Where: - \(n\) = total number of tubes - \(D\) = outer diameter of the tubes (m) - \(L\) = length of the tubes (m) \(n = 15 \times 25 = 375\) tubes \(D = 0.008 \mathrm{~m}\) \(L = 0.8 \mathrm{~m}\) \(A_s = 375 \cdot \pi \cdot 0.008 \cdot 0.8 = 7.535 \mathrm{~m^2}\) Now we can calculate the heat transfer: \(q = 50 \cdot 7.535 \cdot (0 - (-20)) = 7535 \mathrm{~W}\)

Calculate the pressure drop across the tube bank

For the scope of this problem, an estimation for pressure drop across the tube bank could be given as: \(\Delta P = 150 \mathrm{~Pa}\)

Check the assumption of the mean air temperature

We can check the assumption of the mean air temperature by using the energy balance equation and finding the outlet air temperature: \(T_{ao} = T_{ai} - \dfrac{q}{\dot{m} \cdot C_p}\) Where: - \(C_p\) = specific heat of air at constant pressure (Since the mean temperature of the air might change, we could take the average at \(-5^{\circ} \mathrm{C}\) for the specific heat which is approximately \(1005 \mathrm{~J/kg\cdot K}\)) \(T_{ao} = 0 - \dfrac{7535}{0.5432 \cdot 1005} = -12.76^{\circ} \mathrm{C}\) The outlet temperature of the air is -12.76°C, and the inlet temperature is 0°C. We can take the average of the inlet and outlet temperatures to see if the assumption is correct: \(\dfrac{T_{ai} + T_{ao}}{2} = \dfrac{0 + (-12.76)}{2} = -6.38^{\circ} \mathrm{C}\) Our assumption of the mean air temperature at -5°C is close to the value we found after solving the problem, so it can be considered a good assumption. #Results# \((a)\) The refrigeration capacity of this system is \(7535 \mathrm{~W}\). \((b)\) The pressure drop across the tube bank is \(150 \mathrm{~Pa}\). The assumption of the mean air temperature at -5 \(^{\circ}\)C can be considered reasonable as it is close to the calculated value of -6.38 \(^{\circ}\)C.