Question

A shell-and-tube heat exchanger is to be designed to cool down the petroleum- based organic vapor available at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) and at a saturation temperature of \(75^{\circ} \mathrm{C}\). The cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) used for its condensation is supplied at a rate of \(25 \mathrm{~kg} / \mathrm{s}\) and a temperature of \(15^{\circ} \mathrm{C}\). The cold water flows through copper tubes with an outside diameter of \(20 \mathrm{~mm}\), a thickness of $2 \mathrm{~mm}\(, and a length of \)5 \mathrm{~m}$. The overall heat transfer coefficient is assumed to be $550 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, and the latent heat of vaporization of the organic vapor may be taken to be \(580 \mathrm{~kJ} / \mathrm{kg}\). Assuming negligible thermal resistance due to pipe wall thickness, determine the number of tubes required.

Short answer

To design a shell-and-tube heat exchanger for cooling down a petroleum-based organic vapor using cold water, we first calculated the heat transfer rate, which was found to be $$2.9 \times 10^6\mathrm{~W}$$. We then calculated the Log-Mean Temperature Difference (LMTD), which was found to be $$44.6^{\circ}\mathrm{C}$$. Lastly, we used the overall heat transfer coefficient, tube dimensions, and LMTD to determine the number of tubes required. We found that 28 tubes were required for this heat exchanger design.

Step-by-step solution

Calculate the heat transfer rate

To calculate the heat transfer rate between the petroleum-based organic vapor and the cold water, we can use the latent heat of vaporization for the organic vapor and the flow rate of the organic vapor: $$Q = m_{vapor} \times L_{vapor}$$ where $$Q$$ is the heat transfer rate, $$m_{vapor}$$ is the flow rate of the organic vapor, and $$L_{vapor}$$ is the latent heat of vaporization of the organic vapor. Using the given values, $$Q = 5 \mathrm{~kg/s} \times 580 \times 10^3\mathrm{~J/kg} = 2.9 \times 10^6\mathrm{~W}$$

Calculate the Log-Mean Temperature Difference (LMTD)

To determine the number of tubes required, we need to calculate the Log-Mean Temperature Difference (LMTD) in the heat exchanger. The LMTD is calculated using the following formula: $$\Delta T_{LM} = \frac{\Delta T_1 - \Delta T_2}{\ln \left( \frac{\Delta T_1}{\Delta T_2} \right)}$$ where $$\Delta T_1$$ is the temperature difference between the inlet and outlet of the hot fluid, and $$\Delta T_2$$ is the temperature difference between the inlet and outlet of the cold fluid: $$\Delta T_1 = T_{vapor} - T_{water, in} = 75^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 60^{\circ} \mathrm{C}$$ $$\Delta T_2 = T_{vapor} - (T_{water, in} + \Delta T)$$ In this case, $$\Delta T$$ is the temperature increase of the cold water due to the heat transfer. We can calculate this by using the heat transfer rate and the cold water properties: $$Q = m_{water} \times c_p \times \Delta T$$ $$\Delta T = \frac{Q}{m_{water} \times c_p} = \frac{2.9 \times 10^6\mathrm{~W}}{25\mathrm{~kg/s} \times 4187 \mathrm{~J/kg \cdot K}} = 27.8^{\circ} \mathrm{C}$$ Now, we can calculate $$\Delta T_2$$: $$\Delta T_2 = 75^{\circ} \mathrm{C} - (15^{\circ} \mathrm{C} + 27.8^{\circ} \mathrm{C}) = 32.2^{\circ} \mathrm{C}$$ Finally, we can calculate the LMTD: $$\Delta T_{LM} = \frac{60^{\circ} \mathrm{C} - 32.2^{\circ} \mathrm{C}}{\ln \left( \frac{60^{\circ} \mathrm{C}}{32.2^{\circ} \mathrm{C}} \right)} = 44.6^{\circ} \mathrm{C}$$

Determine the number of tubes required

Now that we have the heat transfer rate ($$Q$$) and the LMTD ($$\Delta T_{LM}$$), we can use the overall heat transfer coefficient ($$U$$), the tube length ($$L$$), and the outside diameter of the tubes ($$D_{out}$$) to determine the number of tubes required: $$Q = n \times U \times A \times \Delta T_{LM}$$ where $$n$$ is the number of tubes, and $$A$$ is the surface area of a single tube: $$A = \pi D_{out} \times L$$ We can rearrange this equation to solve for the number of tubes: $$n = \frac{Q}{U \times A \times \Delta T_{LM}}$$ Substituting the given values, we get: $$n = \frac{2.9 \times 10^6\mathrm{~W}}{550\mathrm{~W/m^2 \cdot K} \times (\pi \times 0.02\mathrm{~m} \times 5\mathrm{~m}) \times 44.6^{\circ} \mathrm{C}} = 27.5$$ Since we cannot have a fraction of a tube, we round up to get the nearest whole number. So the number of tubes required is 28.