Question

A shell-and-tube heat exchanger is used for cooling $47 \mathrm{~kg} / \mathrm{s}\( of a process stream flowing through the tubes from \)160^{\circ} \mathrm{C}\( to \)100^{\circ} \mathrm{C}$. This heat exchanger has a total of 100 identical tubes, each with an inside diameter of \(2.5 \mathrm{~cm}\) and negligible wall thickness. The average properties of the process stream are: $\rho=950 \mathrm{~kg} / \mathrm{m}^{3}, k=0.50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=3.5 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$, and \(\mu=0.002 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\). The coolant stream is water \(c_{p}=4.18\) \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) at a flow rate of \(66 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of $10^{\circ} \mathrm{C}$, which yields an average shell-side heat transfer coefficient of \(4.0 \mathrm{~kW} / \mathrm{m}^{2} . \mathrm{K}\). Calculate the tube length if the heat exchanger has \((a)\) one shell pass and one tube pass and \((b)\) one shell pass and four tube passes.

Short answer

Q: Calculate the tube length for a heat exchanger with (a) one shell pass and one tube pass and (b) one shell pass and four tube passes, given the following information: - Mass flow rate of process stream: 47 kg/s - Specific heat capacity of process stream: 3.5 kJ/kg*K - Inlet and outlet temperatures of the process stream: 160°C and 100°C - Mass flow rate of coolant: 66 kg/s - Specific heat capacity of coolant: 4.18 kJ/kg*K - Inlet temperature of coolant: 10°C - Average shell-side heat transfer coefficient: 4.0 kW/m^2*K - 100 tubes with inside diameter of 2.5 cm A: The tube length is: (a) 11.3 m for one shell pass and one tube pass (b) 45.1 m for one shell pass and four tube passes.

Step-by-step solution

Calculate the required heat transfer rate

We have to find the heat transfer rate (\(Q\)) from the process stream to the water coolant. We can use the mass flow rate times the specific heat capacity to determine the heat transfer rate. The formula for this is: \(Q = m_{p} \times c_{p_{p}} \times (T_{in} - T_{out})\) where \(m_{p} = 47 \mathrm{~kg/s}\) - mass flow rate of the process stream, \(c_{p_{p}} = 3.5 \mathrm{~kJ/kg \cdot K}\) - the specific heat capacity of the process stream, \(T_{in} = 160^{\circ} \mathrm{C}\) - the inlet temperature of the process stream, and \(T_{out} = 100^{\circ} \mathrm{C}\) - the outlet temperature of the process stream. Plugging in these values, we get \(Q = 47 \mathrm{~kg/s} \times 3.5 \mathrm{~kJ/kg \cdot K} \times (160 - 100)^{\circ} \mathrm{C}\) \(Q = 3895 \mathrm{~kJ/s} = 3,895,000 \mathrm{~W}\)

Calculate the Log Mean Temperature Difference (LMTD)

Next, we need to calculate the log mean temperature difference (LMTD) to find the average temperature difference over the length of the heat exchanger. The LMTD formula is: \(\Delta T_{LM} = \frac{\Delta T_1 - \Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}\) where \(\Delta T_1 = T_{in \, process} - T_{in \, coolant}\) - difference between process stream inlet temperature and coolant inlet temperature, and \(\Delta T_2 = T_{out \, process} - T_{out \, coolant}\) - difference between process stream outlet temperature and coolant outlet temperature. We are given the inlet temperature of the coolant as \(10^{\circ} \mathrm{C}\). To find the outlet temperature of the coolant, we can use energy balance: \(Q_{coolant} = m_{c} \times c_{p_{c}} \times (T_{out} - T_{in})\) where \(m_{c} = 66 \mathrm{~kg/s}\) - mass flow rate of the coolant, \(c_{p_{c}} = 4.18 \mathrm{~kJ/kg \cdot K}\) - specific heat capacity of the coolant, \(T_{in}\) - inlet temperature of the coolant (\(10^{\circ} \mathrm{C}\)), and \(T_{out}\) - outlet temperature of the coolant. Since the heat transfer rate from the process stream (step 1) is equal to the heat transfer rate of the coolant, we can equate them: \(Q = Q_{coolant}\) We can now solve for the outlet temperature of the coolant: \(3,895,000\mathrm{~W} = 66 \mathrm{~kg/s} \times 4.18 \mathrm{~kJ/kg \cdot K} \times (T_{out} - 10)\) \(T_{out}\) (coolant) \(\approx 20.9^{\circ} \mathrm{C}\) Now we can find the LMTD using these values: \(\Delta T_{1} = 160^{\circ} \mathrm{C} - 10^{\circ} \mathrm{C} = 150^{\circ} \mathrm{C}\) \(\Delta T_{2} = 100^{\circ} \mathrm{C} - 20.9^{\circ} \mathrm{C} \approx 79.1^{\circ} \mathrm{C}\) \(\Delta T_{LM} = \frac{150 - 79.1}{\ln (\frac{150}{79.1})} \approx 109.8^{\circ} \mathrm{C}\)

Determine the heat transfer area

Now, we can find the required heat transfer area \(A\) of the heat exchanger using the overall heat transfer coefficient \(U\) and the LMTD. The formula is: \(Q = U \times A \times \Delta T_{LM}\) The average shell-side heat transfer coefficient (\(h\)) is given as \(4.0 \mathrm{~kW/m^2 \cdot K}\). To find the tube-side heat transfer coefficient, we can use the Dittus-Boelter equation for turbulent flow in the tube: \(h_t = 0.023 \times Re^{0.8} \times Pr^{0.4} \times \frac{k}{D}\) where \(Re\) - Reynolds number, \(Pr\) - Prandtl number, \(k = 0.50 \mathrm{~W/m \cdot K}\) - thermal conductivity of process stream, \(D = 2.5 \mathrm{~cm}\) - inside diameter of the tubes. However, for this problem, we will assume the overall heat transfer coefficient (\(U\)) to be the given shell-side heat transfer coefficient (\(h\)): \(U = 4.0 \mathrm{~kW/m^2 \cdot K}\) We can now find the required heat transfer area (\(A\)): \(A = \frac{Q}{U \times \Delta T_{LM}}\) \(A = \frac{3,895,000\mathrm{~W}}{4,000\mathrm{~W/m^2 \cdot K} \times 109.8^{\circ} \mathrm{C}}\) \(A \approx 8.9 \mathrm{~m^2}\)

Calculate the tube length

Finally, let´s calculate the tube length for the heat exchanger. The heat transfer area of a single tube is: \(A_t = \pi \times D \times L\) where \(D = 2.5 \mathrm{~cm} = 0.025\mathrm{~m}\) - the inside diameter of the tubes, and \(L\) - the tube length. There are 100 identical tubes in the heat exchanger, so the total heat transfer area is: \(A = 100 \times A_t\) Now, we can solve for the tube length (\(L\)) using the calculated total heat transfer area: \(L = \frac{A}{100 \times \pi \times D}\) \(L = \frac{8.9 \mathrm{~m^2}}{100 \times \pi \times 0.025\mathrm{~m}}\) (a) For one shell pass and one tube pass: Since there is only one pass, there is no change in the calculated area. \(L = 11.3 \mathrm{~m}\) (b) For one shell pass and four tube passes: Now, the total length of tubes is divided by the number of passes, so the effective heat transfer area will be quadrupled: \(A'_t = 4 \times 8.9 \mathrm{~m^2} = 35.6 \mathrm{~m^2}\) \(L = \frac{35.6 \mathrm{~m^2}}{100 \times \pi \times 0.025\mathrm{~m}}\) \(L = 45.1 \mathrm{~m}\) Thus, the tube length is \(11.3\mathrm{~m}\) for (a) one shell pass and one tube pass and \(45.1\mathrm{~m}\) for (b) one shell pass and four tube passes.