Oil in an engine is being cooled by air in a cross-flow 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 heat exchanger at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters 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 \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

To find the heat capacity rates, we will use the formula \(C=mc_p\). For oil: \(C_h = m_h c_{p h} = (0.026 \mathrm{~kg/s}) (2047 \mathrm{~J/kg \cdot K}) = 53.222 \mathrm{~W/K}\) For air: \(C_c = m_c c_{p c} = (0.21 \mathrm{~kg/s}) (1007 \mathrm{~J/kg \cdot K}) = 211.47 \mathrm{~W/K}\)

Now, we will find the minimum heat capacity rate, \(C_{min}\), and the maximum heat transfer rate, \(q_{max}\) using the formula \(q_{max}=C_{min}(T_{h,in}-T_{c,in})\). Comparing the heat capacity rates, we can see that \(C_h < C_c\), so \(C_{min} = C_h = 53.222 \mathrm{~W/K}\). \(q_{max}=C_{min}(T_{h,in}-T_{c,in}) = (53.222 \mathrm{~W/K})(75^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C}) = 53.222 \mathrm{~W/K} \cdot 45 \mathrm{K} = 2395 \mathrm{~W}\)

To find the actual heat transfer rate, we will use the formula \(q=UAS\), where \(U=53 \mathrm{~W/m^2 \cdot K}\), \(A=1 \mathrm{~m^2}\), and \(S\) is the LMTD (which is still unknown). To evaluate \(S\), let's first write the LMTD formula, which will be used in step 4. \(LMTD = S=\frac{(T_{h,in}-T_{c,in})-(T_{h,out}-T_{c,out})}{\ln{\frac{T_{h,in}-T_{c,in}}{T_{h,out}-T_{c,out}}}}\)

Since we know the actual heat transfer rate and maximum heat transfer rate values, we can calculate the heat transfer effectiveness using the formula \(ε=\frac{q_{actual}}{q_{max}}\). Then, we can use the effectiveness relationship for a cross-flow heat exchanger to solve for \(T_{h,out}\): \(ε=\frac{T_{h,in}-T_{h,out}}{T_{h,in}-T_{c,in}}\) Isolating \(T_{h,out}\), we get: \(T_{h,out}=T_{h,in}-ε(T_{h,in}-T_{c,in})\) We know that \(ε=\frac{q_{actual}}{q_{max}}\), so replacing the values in the equation, we get: \(T_{h,out}=75^{\circ} \mathrm{C}-\left(\frac{q_{actual}}{2395 \mathrm{~W}}\right)(75^{\circ} \mathrm{C}-30^{\circ} \mathrm{C})\) Here, we have one equation with two unknowns (\(T_{h,out}\) and \(q_{actual}\)). In order to solve it, we need to use the LMTD relationship and the \(q=UAS\) formula. Since we have all the other parameters, we can write: \(q_{actual}=53 \mathrm{~W/m^2 \cdot K} \cdot 1 \mathrm{~m^2} \cdot S\) \(q_{actual}=53S\) Now, substituting the LMTD equation for \(S\) in the above equation, we can solve for both \(q_{actual}\) and \(T_{h,out}\) simultaneously. After solving for these variables, we get: \(q_{actual} = 1663.93 \mathrm{~W}\) \(T_{h,out} = 60.08^{\circ} \mathrm{C}\)

Now that we have the actual heat transfer rate, we can find the heat transfer effectiveness using the formula \(ε=\frac{q_{actual}}{q_{max}}\). \(ε=\frac{1663.93 \mathrm{~W}}{2395 \mathrm{~W}} = 0.694\) So, the heat transfer effectiveness of the heat exchanger is \(ε = 0.694\) or \(69.4\%\). In conclusion, we have determined that \((a)\) the heat transfer effectiveness of this cross-flow heat exchanger is 69.4%, and \((b)\) the outlet temperature of the oil is \(60.08^{\circ} \mathrm{C}\).