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 nearby lake. The water in the lake is typically maintained at a constant temperature of \(15^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). 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 \(20^{\circ} \mathrm{C}\).

First, we need to calculate the heat transfer area (\(A\)) of the copper tube. To do this, we can use the following formula: \(A = \pi D L\) Where \(D\) is the diameter of the tube (0.10m) and \(L\) is the length of the tube, which we will find in step 4.

The heat transfer rate (\(q\)) from the air to the lake water can be calculated using the formula: \(q = hA(T_{air,in} - T_{water})\) Where \(h\) is the convection heat transfer coefficient (1000 W/m²K), \(T_{air,in}\) is the inlet air temperature (30°C), and \(T_{water}\) is the lake water temperature (15°C). We will simplify the equation by substituting the area we derived in step 1: \(q = h(\pi DL)(T_{air,in} - T_{water})\)

In order to find the amount of heat (\(Q\)) that has to be removed to achieve the desired outlet air temperature (\(T_{outlet}\) = 20°C), we can use the following formula: \(Q = \dot{m} C_p (T_{air,in} - T_{outlet})\) Where \(\dot{m}\) represents the mass flow rate of the air, and \(C_p\) is the specific heat capacity of air at constant pressure, which is approximately 1006 J/kgK. To find the mass flow rate, we can use the following formula: \(\dot{m} = \rho AV\) Where \(\rho\) is the air density (approximately 1.2 kg/m³ at 1 atm and 20°C), \(A\) is the cross-sectional area of the tube, and \(V\) is the average velocity of the air: \(\dot{m} = \rho (\frac{\pi D^2}{4})V\)

To find the necessary tube length (\(L\)) to achieve the desired outlet air temperature, we can equate the heat transfer rate to the amount of heat to be removed: \(q = Q\) \(h(\pi DL)(T_{air,in} - T_{water}) = \dot{m} C_p (T_{air,in} - T_{outlet})\) Substitute the value of \(\dot{m}\) from step 3: \(h(\pi DL)(T_{air,in} - T_{water}) = \rho (\frac{\pi D^2}{4})V C_p (T_{air,in} - T_{outlet})\) Now, we can solve for \(L\) by dividing both sides by \(h(\pi D)(T_{air,in} - T_{water})\) \(L = \frac{\rho (\frac{\pi D^2}{4})V C_p (T_{air,in} - T_{outlet})}{h(\pi D)(T_{air,in} - T_{water})}\) After plugging in the given values, we have: \(L = \frac{(1.2)(\frac{\pi (0.1)^2}{4})(2.5)(1006)(30 - 20)}{(1000)(\pi (0.1))(30 - 15)}\) Finally, we can calculate \(L\): \(L \approx 4.52 \mathrm{m}\) So, the necessary copper tube length to achieve the desired outlet air temperature is approximately 4.52 meters.