Question

Ethylene glycol is heated from \(25^{\circ} \mathrm{C}\) to $40^{\circ} \mathrm{C}\( at a rate of \)2.5 \mathrm{~kg} / \mathrm{s}$ in a horizontal copper tube \((k=386 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with an inner diameter of \(2.0 \mathrm{~cm}\) and an outer diameter of \(2.5 \mathrm{~cm}\). A saturated vapor \(\left(T_{g}=110^{\circ} \mathrm{C}\right)\) condenses on the outside-tube surface with the heat transfer coefficient (in $\mathrm{kW} / \mathrm{m}^{2} \cdot \mathrm{K}\( ) given by \)9.2 /\left(T_{g}-T_{w}\right)^{0.25}\(, where \)T_{n}$ is the average outside-tube wall temperature. What tube length must be used? Take the properties of ethylene glycol to be $\rho=1109 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2428 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, \)k=0.253 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \mu=0.01545 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}$, and \(\operatorname{Pr}=148.5\).

Short answer

Answer: The main challenge in calculating the length of the tube in this scenario is that we have one equation with two unknowns (l and T_w), making it impossible to find an explicit value for the tube length without additional assumptions or simplification methods.

Step-by-step solution

Calculate the heat required to raise the temperature of the ethylene glycol

We will use the formula \(Q=m\cdot c_{p}\cdot \Delta T\), where \(Q\) is the heat required, \(m\) is the mass of the ethylene glycol, \(c_{p}\) is the specific heat capacity, and \(\Delta T\) is the difference in temperature. First, calculate the mass flow rate for \(2.5~kg/s\) and the temperature difference between \(25^{\circ}C\) and \(40^{\circ}C\). \(m=2.5~kg/s\) \(\Delta T=(40-25)^{\circ}C=15^{\circ}C\) Now, substitute these values into the formula to find the heat required: \(Q = (2.5~kg/s) \times (2428~J/kg\cdot K) \times (15~K) = 90825~W\)

Calculate the heat transfer between the ethylene glycol and the surrounding vapor

We are given the heat transfer coefficient \(h_{g}=\frac{9.2}{(T_{g}-T_{w})^{0.25}}\), where \(T_{g}\) is \(110^{\circ}C\) and \(T_{w}\) is the average outside-tube wall temperature. Let's find the heat transfer between the ethylene glycol and the surrounding vapor using the formula \(Q=h_{g}\cdot A\cdot \Delta T_{w_g}\), where \(A\) is the heat transfer area and \(\Delta T_{w_g}\) is the difference in temperature between inside and outside the tube. As we know the total heat required \(Q\), we can rearrange the formula to find the heat transfer area: \(A=\frac{Q}{h_{g}\cdot \Delta T_{w_g}}\) We need to express \(\Delta T_{w_g}\) in terms of \(T_{w}\), so we have: \(\Delta T_{w_g}=(T_{g}-T_{w})\) Now, substitute the values and rearrange the equation: \(A = \frac{90825~W}{\frac{9.2}{(110-T_{w})^{0.25}}\cdot (110-T_{w})}\)

Express heat transfer area in terms of the length of the tube

We can express the heat transfer area in terms of the length of the tube and the outer diameter of the tube: \(A=ld\pi\) where \(l\) is the length of the tube and \(d\) is the outer diameter of the tube. Substitute this expression into the equation obtained in step 2: \(l = \frac{90825~W}{\frac{9.2}{(110-T_{w})^{0.25}}\cdot (110-T_{w})\cdot (2.5\times10^{-2}m)\pi}\)

Calculate the length of the tube

As we only have one equation with two unknowns (\(l\) and \(T_w\)), it is not possible to find an explicit value for the tube lengths without some additional assumptions or simplification methods. It is possible to determine the tube length when given an appropriate assumption on \(T_w\).