Engine oil \(\left(c_{p}=2100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated from \(20^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) in a \(2-\mathrm{cm}\)-diameter thinwalled copper tube by condensing steam outside at a temperature of \(130^{\circ} \mathrm{C}\left(h_{f g}=2174 \mathrm{~kJ} / \mathrm{kg}\right)\). For an overall heat transfer coefficient of \(650 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the length of the tube required to achieve it.

Given, the flow rate of engine oil is \(0.3\ kg/s\).

To find the heat transfer required to raise the temperature of engine oil from \(20^\circ C\) to \(60^\circ C\), we can use the formula: \(Q=mc_p\Delta T\) Where \(Q\) is the heat transfer, \(m\) is the mass flow rate of engine oil, \(c_p\) is the specific heat capacity and \(\Delta T\) is the change in temperature. Given \(m=0.3\ kg/s\), \(c_p=2100\ J/kg\cdot K\), and \(\Delta T = 60-20 = 40^\circ C\). Therefore, \(Q = (0.3)(2100)(40)= 25200\ W\).

We are given the overall heat transfer coefficient U as \(650\ W/m^2\cdot K\). Also, we know \(Q=U\cdot A\cdot\Delta T_{lm}\) where the area A is given by \(A = \pi d L\) and \(\Delta T_{lm}\) is the log mean temperature difference. First, we need to calculate the log mean temperature difference. The outlet temperature of the engine oil is given as \(60^\circ C\). Since the steam is at a constant temperature, the inlet temperature difference can be calculated as \(\Delta T_1 = 130 - 60 = 70^\circ C\). Similarly, the outlet temperature difference is \(\Delta T_2 = 130 - 20 = 110^\circ C\). Now, we can find the log mean temperature difference: \(\Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2/\Delta T_1)}=\frac{110-70}{\ln(110/70)} = 87.25^\circ C\)

We have the rate of heat transfer \(Q=25200\ W\) and \(\Delta T_{lm}=87.25^\circ C\). We can now solve for the area, \(A\) \(A = \frac{Q}{U\cdot \Delta T_{lm}}=\frac{25200}{650\cdot 87.25}=0.443\ m^2\) Now, we can find the length of the copper tube. We know the diameter of the tube is \(2\ \mathrm{cm} = 0.02\ \mathrm{m}\) and the area \(A=0.443\ \mathrm{m}^2\). \(L = \frac{A}{\pi d}=\frac{0.443}{\pi\cdot 0.02}=7.045\ m\) So, the length of the copper tube required to achieve the desired rate of heat transfer is approximately \(7.045\ \mathrm{m}\).