Question

To cool a storehouse in the summer without using a conventional air- conditioning system, the owner decided to hire an engineer to design an alternative system that would make use of the water in the nearby lake. The engineer decided to flow air through a thin, smooth, 10 -cm-diameter copper tube that is submerged in the lake. The water in the lake is typically at a constant temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). If air (1 atm) enters the copper tube at a mean temperature of \(30^{\circ} \mathrm{C}\) with an average velocity of \(2.5 \mathrm{~m} / \mathrm{s}\), determine the necessary copper tube length so that the outlet mean temperature of the air is \(20^{\circ} \mathrm{C}\).

Short answer

The necessary length of the copper tube to achieve the desired outlet temperature is approximately \(1.45 \mathrm{~m}\).

Step-by-step solution

Calculate the heat transfer rate

Air is entering the copper tube at a mean temperature of \(30^{\circ} \mathrm{C}\) and leaving the tube at \(20^{\circ} \mathrm{C}\). We will use the specific heat capacity of air to calculate the heat transfer rate. Let the mass flow rate of air be \(m\). Then the specific heat capacity, \(c_p\), for air at constant pressure is approximately \(1000 \mathrm{~J} / \mathrm{kg}·\mathrm{K}\). The heat transfer rate can be calculated as follows: \(\dot{Q} = m c_p (\Delta T)\) where \(\Delta T = T_{in} - T_{out} = 30 - 20 = 10 \mathrm{~K}\). Let us find the mass flow rate of air, \(m\), using the mass flow equation: \(m = \rho A_c v\), where \(\rho\) is the density of air, \(A_c\) is the cross-sectional area of the copper tube, and \(v\) is the average velocity of air. We are given that the diameter of the copper tube is \(10 \mathrm{~cm}\), the average velocity of air is \(2.5 \mathrm{m}/\mathrm{s}\), and at \(30^{\circ} \mathrm{C}\) temperature, the density of air is approximately \(1.165 \mathrm{kg} / \mathrm{m}^3\). The cross-sectional area of the copper tube is: \(A_c = \pi d^2 / 4 = \pi(0.1)^2 / 4 = 0.00785 \mathrm{m}^2\) Thus, the mass flow rate of air is: \(m = \rho A_c v = 1.165 × 0.00785 × 2.5 = 0.0228\mathrm{~kg/s}\) Now, we can calculate the heat transfer rate: \(\dot{Q} = m c_p (\Delta T) = 0.0228 × 1000 × 10 = 228 \mathrm{~W}\)

Calculate the surface area of the copper tube

The surface area, \(A_s\), of the tube is given by the formula: \(A_s = \pi d L\) where \(d\) is the diameter of the tube and \(L\) is the length of the tube. We are solving for the length \(L\).

Calculate the difference in temperature between the air and the lake

The difference in temperature between the air and the lake, \(\Delta T_s\), is given by: \(\Delta T_s = T_{air} - T_{lake} = 20 - 15 = 5 \mathrm{~K}\)

Solve for the necessary length of the copper tube

The convection heat transfer equation, rearranged for the length of the tube, is: \(L = \frac{\dot{Q}}{h A_s (\Delta T_s)}\) Using the values obtained from previous steps, we can calculate \(L\): \(L = \frac{228}{1000 \times \pi \times 0.1 \times 5} = 1.45 \mathrm{~m}\) Therefore, the necessary length of the copper tube to achieve the desired outlet temperature is approximately \(1.45 \mathrm{~m}\).