Question

A double-pipe counterflow heat exchanger is used to cool a hot fluid before it flows into a pipe system. The pipe system is mainly constructed with ASTM F2389 polypropylene pipes. According to the ASME Code for Process Piping, the recommended maximum temperature for polypropylene pipes is $99^{\circ} \mathrm{C}$ (ASME B31.3-2014, Table B-1). The heat exchanger's inner tube has negligible wall thickness. The convection heat transfer coefficients inside and outside of the heat exchanger inner tube are $1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and 1000 \)\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. The fouling factor estimated for the heat exchanger is \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The hot fluid \(\left(c_{p}=3800\right.\) \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K})\) enters the heat exchanger at \(150^{\circ} \mathrm{C}\) with a flow rate of $0.5 \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 $0.75 \mathrm{~kg} / \mathrm{s}$. Determine the heat transfer surface area that the heat exchanger needs to cool the hot fluid to \(99^{\circ} \mathrm{C}\) at the outlet so that it flows into the pipe system at a temperature not exceeding the recommended maximum temperature for polypropylene pipes.

Short answer

Answer: The required heat transfer surface area is approximately 1.424 square meters.

Step-by-step solution

Calculate the heat transfer rate (Q)

To calculate the heat transfer rate, we will first find the change in temperature for the hot fluid as it passes through the heat exchanger. Given that the hot fluid needs to cool down to \(99^{\circ} \mathrm{C}\), the change in temperature will be the initial temperature \(150^{\circ} \mathrm{C}\) minus the required temperature \(99^{\circ} \mathrm{C}\). \(\Delta T = 150 - 99 = 51^{\circ} \mathrm{C}\) Next, we will use the mass flow rate, specific heat capacity, and temperature change to find the heat transfer rate (Q): \(Q = m_{h} \cdot c_{p,h} \cdot \Delta T\) where \(m_{h}\) is the mass flow rate of the hot fluid, \(c_{p,h}\) is the specific heat capacity of the hot fluid, and ∆T is the change in the temperature of the hot fluid. \(Q = 0.5 \mathrm{~kg/s} \times 3800 \mathrm{~J/kg \cdot K} \times 51 \mathrm{~K}\) \(Q = 96900 \mathrm{~W}\)

Calculate the overall heat transfer coefficient (U)

The overall heat transfer coefficient (U) accounts for the convection heat transfer coefficients on both sides of the heat exchanger tube and the fouling factor. In this case, we have inner (\(h_{i} = 1500 \mathrm{~W/m^{2} \cdot K}\)) and outer (\(h_{o} = 1000 \mathrm{~W/m^{2} \cdot K}\)) convection heat transfer coefficients and a fouling factor denoted by Rf (\(0.0001 \mathrm{~m^{2} \cdot K} / \mathrm{W}\)). To find the overall heat transfer coefficient, we will use the following formula: \(\frac{1}{U} = R_{f} + \frac{1}{h_{i}} + \frac{1}{h_{o}}\) Now, we will solve for U: \(U = \left(\frac{1}{0.0001 + \frac{1}{1500} + \frac{1}{1000}}\right)^{-1}\) \(U = 595.71 \mathrm{~W/m^{2} \cdot K}\)

Compute the required heat transfer surface area (A)

Finally, we will calculate the required heat transfer surface area (A) using the heat transfer rate (Q) and the overall heat transfer coefficient (U). The relationship between Q, U, and A is given as follows: \(Q= U \cdot A \cdot \Delta T_{lm}\) Here, we will consider the log mean temperature difference (LMTD) between the hot and cold fluids, \(\Delta T_{lm}\), since it is a counterflow heat exchanger. In a counterflow heat exchanger, the LMTD can be approximated as the average temperature difference between the two fluids. In this case, \(\Delta T_{1} = 150 - 10 = 140^{\circ} \mathrm{C}\) \(\Delta T_{2} = 99 - 10 = 89^{\circ} \mathrm{C}\) \(\Delta T_{lm} \approx \frac{\Delta T_{1} + \Delta T_{2}}{2} = \frac{140 + 89}{2} = 114.5^{\circ} \mathrm{C}\) Now, we have everything we need to solve for the required surface area (A): \(A = \frac{Q}{U \cdot \Delta T_{lm}}\) \(A =\frac{96900 \mathrm{~W}}{595.71 \mathrm{~W/m^{2} \cdot K} \times 114.5 \mathrm{~K}}\) \(A = 1.424 \mathrm{~m^{2}}\) The required heat transfer surface area for the heat exchanger is approximately 1.424 square meters to cool the hot fluid to \(99^{\circ} \mathrm{C}\) at the outlet.