Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be heated by solarheated hot air \(\left(c_{p}=1010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a double-pipe counterflow heat exchanger. Air enters the heat exchanger at \(90^{\circ} \mathrm{C}\) at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\), while water enters at \(22^{\circ} \mathrm{C}\) at a rate of \(0.1 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient based on the inner side of the tube is given to be \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The length of the tube is \(12 \mathrm{~m}\) and the internal diameter of the tube is \(1.2 \mathrm{~cm}\). Determine the outlet temperatures of the water and the air.

In this step, we will calculate the heat transfer area of the heat exchanger. We know the length of the tube and the internal diameter which can be used to determine the inner surface area. The heat transfer area \(A\) can be calculated using the following formula: \(A = \pi D L\) where \(D\) is the internal diameter of the tube and \(L\) is the length of the tube. From the given values, we have \(D = 1.2 \times 10^{-2}\, \mathrm{m}\) (converting from cm to meters) and \(L = 12\, \mathrm{m}\). Plugging in these values, we find the heat transfer area: \(A = \pi (1.2 \times 10^{-2}) (12) = 0.4524\, \mathrm{m}^2\).

Now that we know the heat transfer area, we can calculate the heat transfer rate (\(q\)) between the water and the air, using the given overall heat transfer coefficient \(U\). The formula for the heat transfer rate is: \(q = U\, A\, \Delta T_{lm}\) where \(\Delta T_{lm}\) is the log mean temperature difference between the two fluids. The log mean temperature difference can be calculated using the formula: \(\Delta T_{lm} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln(\frac{\Delta T_{1}}{\Delta T_{2}})}\) We know the inlet temperatures for both fluids, so we can find the initial (\(\Delta T_{1}\)) and final (\(\Delta T_{2}\)) temperature differences: From the inlet temperatures, we have: \(\Delta T_{1} = T_{\mathrm{air, in}} - T_{\mathrm{water, in}} = 90 - 22 = 68\, ^{\circ}\mathrm C\) Now we need to find \(\Delta T_{2}\). We know the mass flow rates and specific heat capacities of the fluids, so we can write the energy balance for the water: \(q = m_{\mathrm{water}}\, c_{p, \mathrm{water}}\, (T_{\mathrm{water, out}} - T_{\mathrm{water, in}})\) Similarly, we can write the energy balance for the air: \(q = m_{\mathrm{air}}\, c_{p, \mathrm{air}}\, (T_{\mathrm{air,in}} - T_{\mathrm{air, out}})\) We can use these two equations to find the relationship between \(T_{\mathrm{water, out}}\) and \(T_{\mathrm{air, out}}\): \(m_{\mathrm{water}}\, c_{p, \mathrm{water}}\, (T_{\mathrm{water, out}} - T_{\mathrm{water, in}}) = m_{\mathrm{air}}\, c_{p, \mathrm{air}}\, (T_{\mathrm{air, in}} - T_{\mathrm{air, out}})\) Divide both sides by \(m_{\mathrm{water}}\, c_{p, \mathrm{water}}\): \((T_{\mathrm{water, out}} - T_{\mathrm{water, in}}) = \frac{m_{\mathrm{air}}\, c_{p, \mathrm{air}}}{m_{\mathrm{water}}\, c_{p, \mathrm{water}}} (T_{\mathrm{air, in}} - T_{\mathrm{air, out}})\) Plugging in the given values, we get: \((T_{\mathrm{water, out}} - 22) = \frac{0.3 \times 1010}{0.1 \times 4180} (90 - T_{\mathrm{air, out}})\) Simplify the equation: \((T_{\mathrm{water, out}} - 22) = 7.25 (90 - T_{\mathrm{air, out}})\) Now we can express \(\Delta T_{2}\) in terms of unknown outlet temperatures \(T_{\mathrm{water, out}}\) and \(T_{\mathrm{air, out}}\): \(\Delta T_{2} = T_{\mathrm{air, out}} - T_{\mathrm{water, out}}\)

Now we have two equations: 1. \((T_{\mathrm{water, out}} - 22) = 7.25 (90 - T_{\mathrm{air, out}})\) 2. \(\Delta T_{2} = T_{\mathrm{air, out}} - T_{\mathrm{water, out}}\) We can eliminate \(\Delta T_{2}\) from the second equation, and then use them to find \(T_{\mathrm{air, out}}\) and \(T_{\mathrm{water, out}}\). Substitute the second equation into the expression for log mean temperature difference and solve for \(q\): \(q = 80\, (0.4524)\, \frac{68 - (T_{\mathrm{air, out}} - T_{\mathrm{water, out}})}{\ln(\frac{68}{T_{\mathrm{air, out}} - T_{\mathrm{water, out}}})}\) Now use the energy balance for the water to find \(q\): \(q = 0.1\, (4180)\, (T_{\mathrm{water, out}} - 22)\) From these two equations, we numerically solve for \(T_{\mathrm{air, out}}\) and \(T_{\mathrm{water, out}}\): \(T_{\mathrm{air, out}} \approx 69.37\, ^{\circ}\mathrm C\) and \(T_{\mathrm{water, out}} \approx 60.49\, ^{\circ}\mathrm C\) Thus, the outlet temperatures of the water and the air are approximately \(60.49\, ^{\circ}\mathrm C\) and \(69.37\, ^{\circ}\mathrm C\), respectively.