Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell side at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). If the correction factor is \(F=\) \(0.96\), determine the outlet temperatures of the oil and air.

We need to find the heat capacities of the oil and air. Heat capacity can be calculated using the following formula: $$C = mc_p$$ For oil: $$C_h = 0.026 \mathrm{~kg/s} \times 2047 \mathrm{~J/kg\cdot K}$$ For air: $$C_c = 0.21 \mathrm{~kg/s} \times 1007 \mathrm{~J/kg\cdot K}$$

To proceed with the LMTD method, we need to find the limiting heat capacity rates between the two fluids. The lower heat capacity rate fluid will be the one that undergoes the most significant temperature change. $$C_{min} = \min(C_h, C_c)$$ $$C_{max} = \max(C_h, C_c)$$

We will now calculate the heat transfer using the LMTD method with the known specific heat capacities, heat exchanger area, overall heat transfer coefficient, and inlet temperatures. First, we need to find out the LMTD. Using the expression: $$\Delta T_{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)},$$ $$\Delta T_1 = T_{h,i} - T_{c,i} = 75^\circ\mathrm{C} - 30^\circ\mathrm{C}$$ $$\Delta T_2 = T_{h,o} - T_{c,o}$$ We need to analyze to find the outlet temperature changes. Since we have a correction factor, we can modify the LMTD method by multiplying the LMTD by the correction factor: $$q = U A F \Delta T_{LMTD}$$, Now we replace the LMTD with the new expression: $$q = UAF\cdot\frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$$ Since the heat transfer is the same for both fluids, we can write two expressions for it using the heat capacities of the oil and air: $$q = C_{min}(T_{c,i}-T_{c,o})$$ $$q = C_{h}(T_{h,i}-T_{h,o})$$ Now we have two equations with two unknowns, \(T_{h,o}\) and \(T_{c,o}\).

With our two equations for the heat transfer, we can now solve for the outlet temperatures of the oil and air. We will use the given specific heat capacity, heat transfer coefficient, correction factor, and cross-sectional area to solve the unknowns using simultaneous equations. The final expressions should represent the outlet temperatures of the oil and air as follows: $$T_{h,o} = f(T_{h,i},C_h,C_{min}, U, A,F, T_{c,i})$$ $$T_{c,o} = g(T_{h,o},C_h,C_{min}, U, A,F, T_{c,i})$$ By solving these equations, we will have the outlet temperatures of the oil and air, which is the objective of this exercise.