Question

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 -cm-diameter thin-walled copper tube by condensing steam outside at a temperature of \(130^{\circ} \mathrm{C}\) $\left(h_{f /}=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. Answers: $25.2 \mathrm{~kW}, 7.0 \mathrm{~m}$

Short answer

Answer: The rate of heat transfer is 25.2 kW, and the required tube length is 7.0 meters.

Step-by-step solution

Determine the heat required to raise the oil temperature.

Using the given information, we first need to calculate the heat required to raise the temperature of the oil from 20°C to 60°C using the formula: \(Q = mc_p\Delta T\) where \(Q\) = heat required in J \(m = 0.3 kg/s\) = mass flow rate of oil \(c_p = 2100 J/kg\cdot K\) = specific heat of oil \(\Delta T = (60-20) K = 40 K\) = change in temperature

Calculate the rate of heat transfer.

Now that we have the heat required, we can determine the rate of heat transfer (\(Q'\)) by dividing \(Q\) by the mass flow rate: \(Q' = \frac{Q}{\Delta t}\) So, using the values from step 1: \(Q' = \frac{0.3 kg/s \times 2100 J/kg\cdot K \times 40 K}{1s}\) Calculate \(Q'\) to obtain: \(Q' = 25.2 \times 10^3 W = 25.2 kW\)

Determine the length of the tube.

Next, we'll use the overall heat transfer equation to find the length of the tube required to achieve the heat transfer rate calculated in step 2: \(q' = U A \Delta T_m\) where \(q'\) = heat transfer rate per unit area = \(Q'/A\) \(U = 650 W/m^2\cdot K\) = overall heat transfer coefficient \(A = πDL\) = surface area of the tube \(\Delta T_m = 130°C - 60°C = 70K\) = temperature difference between steam and oil First, rearrange the equation as follows to find L: \(L = \frac{Q'}{U \times π \times D \times \Delta T_m}\) Inserting the given values: \(L = \frac{25.2\times 10^3 W}{650 W/m^2\cdot K \times π \times 0.02 m \times 70 K}\) Calculate \(L\) to obtain: \(L = 7.0 m\)

Report the results.

We've found the rate of heat transfer and the length of the tube needed to achieve it. So the final answer is: Rate of heat transfer: 25.2 kW Length of the tube: 7.0 m