Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

Solving for \(Q_{v,milk}\), we have: \(Q_{v,milk} = \frac{Q}{\rho_{milk} \times C_{p,milk} \times (\Delta T)_{milk}}\) We are given the hot water flow rate of 5 kg/s and its initial temperature of 140°C. Considering a hot water exit temperature at least 10°C higher than the milk exit temperature, that would mean the hot water exit temperature is at least 70°C + 10°C = 80°C. So, the available temperature difference for the hot water is 140°C - 80°C = 60°C. Therefore, we can also calculate the heat transfer rate (Q) using the hot water side as follows: \(Q = m_{water} \times C_{p,water} \times (\Delta T)_{water}\) where: \(m_{water}\) = mass flow rate of hot water (given, 5 kg/s) \(C_{p,water}\) = specific heat capacity of water (\(4180\mathrm{~W/(kg\cdot K)}\)) \((\Delta T)_{water}\) = hot water temperature difference, which is \(60^{\circ} \mathrm{C}\) Thus, \(Q = 5 \times 4180 \times 60 = 1,254,000 \mathrm{~W}\) Now, using the formula for \(Q_{v,milk}\), we get: \(Q_{v,milk} = \frac{1,254,000}{1000 \times 4180 \times 50} = 0.006 \mathrm{~m^{3}/s}\) #tag_title# Step 2: Calculate the overall heat transfer coefficient#tag_content# Now we have the overall heat transfer coefficient: \(U = \frac{1}{\frac{1}{h_{milk}\times A_m} + \frac{1}{h_{water}\times A_w}}\) where: \(h_{milk}\) = heat transfer coefficient on the milk side (given, 450 W/m²·K) \(h_{water}\) = heat transfer coefficient on the water side (given, 1100 W/m²·K) \(A_m\) = heat transfer area on the milk side (to be determined) \(A_w\) = heat transfer area on the water side (to be determined) Since the heat transfer area is the same on both sides, we can rewrite this formula as: \(U = \frac{1}{\frac{1}{h_{milk}} + \frac{1}{h_{water}}} = \frac{1}{\frac{1}{450} + \frac{1}{1100}} = 299.2 \mathrm{~W/m^{2}\cdot K}\) #tag_title# Step 3: Use the LMTD method to determine the required area#tag_content# Using the logarithmic mean temperature difference (LMTD) method, we have: \(Q = U \times A \times \Delta T_{lm}\) where: \(\Delta T_{lm}\) = logarithmic mean temperature difference, which is given by the formula \(\frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1 / \Delta T_2)}\) In this case, \(\Delta T_1\) is the temperature difference at the inlet (140°C - 20°C = 120°C) and \(\Delta T_2\) is the temperature difference at the outlet (80°C - 70°C = 10°C). \(\Delta T_{lm} = \frac{(120 - 10)}{\ln(120 / 10)} = 40.7^{\circ} \mathrm{C}\) Now we can solve for the required heat transfer area (A): \(A = \frac{Q}{U \times \Delta T_{lm}} = \frac{1,254,000}{299.2 \times 40.7} = 102.29 \mathrm{~m^{2}}\) #tag_title# Step 4: Determine the shell width (tube length in each pass)#tag_content# The heat transfer area on the milk side can be calculated as follows: \(A = n \times L \times (D \times \pi)\) where: \(n\) = number of tubes (given, 30) \(L\) = length of each pass \(D\) = inside diameter of each tube (given, 20mm or 0.02m) We know that the milk makes 10 passes through the shell, so the total length of the tubes is \(L_{total} = 10 \times L\). Hence, \(A = 30 \times (10 \times L) \times (0.02 \times \pi)\) Now, solving for L: \(L = \frac{A}{30 \times 10 \times (0.02 \times \pi)} = \frac{102.29}{30 \times 10 \times (0.02 \times 3.14)} = 1.71 \mathrm{~m}\) So, the required shell width, or tube length in each pass, is approximately 1.71 meters.