Question

A double-pipe heat exchanger is used to cool a hot fluid before it flows into a system of pipes. The inner surface of the pipes is primarily coated with polypropylene lining. The maximum use temperature for polypropylene lining is $107^{\circ} \mathrm{C}$ (ASME Code for Process Piping, ASME B31.32014, Table A323.4.3). The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be \(1400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The heat exchanger has a heat transfer surface area of \(2.5 \mathrm{~m}^{2}\), and the estimated fouling factor caused by the accumulation of deposit on the surface is \(0.0002\) \(\mathrm{m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The hot fluid \(\left(c_{p}=3800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the heat exchanger at \(200^{\circ} \mathrm{C}\) with a flow rate of $0.4 \mathrm{~kg} / \mathrm{s}\(. In the cold side, cooling fluid \)\left(c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ enters the heat exchanger at \(10^{\circ} \mathrm{C}\) with a mass flow rate of $0.5 \mathrm{~kg} / \mathrm{s}$.

Short answer

Answer: To find the outlet temperatures of the hot and cold fluids, follow the steps below: 1. Calculate the overall heat transfer coefficient (U) using the given heat transfer coefficients and fouling factor. 2. Calculate the heat exchange capacity (Q) using the given heat transfer surface area (A) and the overall heat transfer coefficient (U). 3. Use energy balance equations for both hot and cold fluids to calculate the outlet temperatures (T_h, out and T_c, out). After performing these calculations, you will obtain the outlet temperatures of the hot and cold fluids in the heat exchanger. Note that the specific values would require solving the equations for this specific problem.

Step-by-step solution

Calculate the overall heat transfer coefficient (U)

First, calculate the overall heat transfer coefficient (U) for the heat exchanger using the given heat transfer coefficients and fouling factor. The equation for the overall heat transfer coefficient is: \(U = \frac{1}{\frac{1}{h_{i}} + \frac{1}{h_{o}} + R_{f}}\) where \(h_{i} = 1400 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}}\) and \(h_{o} = 1100 \frac{\mathrm{W}}{\mathrm{m}^{2} \cdot \mathrm{K}}\) are the heat transfer coefficients inside and outside of the inner tube, respectively, and \(R_{f} = 0.0002 \frac{\mathrm{m}^{2} \cdot \mathrm{K}}{\mathrm{W}}\) is the fouling factor. \(U = \frac{1}{\frac{1}{1400} + \frac{1}{1100} + 0.0002}\) Compute the value of U.

Calculate the heat exchange capacity (Q)

Next, calculate the heat exchange capacity (Q) using the given heat transfer surface area (A) and the overall heat transfer coefficient (U) obtained in step 1. \(Q = U \cdot A \cdot \Delta T_{lm}\) where \(A = 2.5 \mathrm{~m}^{2}\) and \(\Delta T_{lm}\) is the logarithmic mean temperature difference. The logarithmic mean temperature difference can be found using the following formula: \(\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}}}\) where \(\Delta T_{1}\) and \(\Delta T_{2}\) are the temperature differences between the hot and cold fluids at the beginning and end of the heat exchanger, respectively. \(\Delta T_{1} = T_{h, in} - T_{c, in} = 200 - 10\) \(\Delta T_{2} = T_{h, out} - T_{c, out}\) Substitute the values of \(\Delta T_{1}\) and \(\Delta T_{2}\) in the formula for \(\Delta T_{lm}\) and compute the value of the heat exchange capacity (Q).

Calculate the outlet temperatures of the hot and cold fluids

Finally, calculate the outlet temperatures of the hot and cold fluids using an energy balance equation for both fluids. For the hot fluid: \(Q = \dot{m}_{h} \cdot c_{p, h} \cdot \left(T_{h, in} - T_{h, out}\right)\) For the cold fluid: \(Q = \dot{m}_{c} \cdot c_{p, c} \cdot \left(T_{c, out} - T_{c, in}\right)\) where \(\dot{m}_{h} = 0.4 \frac{\mathrm{kg}}{\mathrm{s}}\), \(c_{p, h} = 3800 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\), \(\dot{m}_{c} = 0.5 \frac{\mathrm{kg}}{\mathrm{s}}\), and \(c_{p, c} = 4200 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\) are the mass flow rates and specific heat capacities of the hot and cold fluids, respectively. Solve these two equations simultaneously to find the outlet temperatures of the hot and cold fluids.