Question

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil $\left(c_{p l}=2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( flowing with a flow rate of \)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 \(53 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\), 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.

Short answer

Answer: The outlet temperatures of the oil and air are 57.1°C and 34.3°C, respectively.

Step-by-step solution

Determine the heat capacity rates of the fluids

To find the effectiveness, we need first to determine the heat capacity rates of both fluids using the formula \(C=f \times \cdot c_{p}\), where \(C\) is the heat capacity rate, \(f\) is the flow rate in \(\mathrm{kg/s}\), and \(c_{p}\) is the specific heat capacity in \(\mathrm{J/kg \cdot K}\). For oil: \(C_{1} = (0.026 \mathrm{~kg/s}) \times (2047 \mathrm{~J/kg \cdot K}) = 53.22 \mathrm{~W/K}\) For air: \(C_{2} = (0.21 \mathrm{~kg/s}) \times (1007 \mathrm{~J/kg \cdot K}) = 211.47 \mathrm{~W/K}\)

Calculate the effectiveness and heat transfer

Next, we must compute the effectiveness, \(\varepsilon\), using the heat capacity rates \(C_{1}\) and \(C_{2}\) and the correction factor \(F\). The formula for effectiveness is given by \(\varepsilon = \frac{1 - e^{-F\cdot NTU\cdot(1-C_r)}}{1 - C_r\cdot e^{-F\cdot NTU\cdot(1-C_r)}}\), where \(C_{r}\) is the heat capacity rate ratio, and NTU is the number of transfer units, calculated as \(\frac{UA}{\min(C_1,C_2)}\). Here, \(U\) is the overall heat transfer coefficient, and \(A\) is the surface area of the heat exchanger. First, let's find the NTU: \(NTU = \frac{UA}{\min(C_1,C_2)} = \frac{(53 \mathrm{~W/m^{2} \cdot K}) \times (1 \mathrm{~m^{2}})}{53.22 \mathrm{~W/K}} = 0.9958\) Calculate the heat capacity rate ratio, \(C_r\): \(C_r = \frac{C_1}{C_2} = \frac{53.22 \mathrm{~W/K}}{211.47 \mathrm{~W/K}} = 0.2517\) Now, using the effectiveness formula, we can compute \(\varepsilon\): \(\varepsilon = \frac{1 - e^{-0.96 \times 0.9958 \times (1-0.2517)}}{1 - 0.2517 \times e^{-0.96 \times 0.9958 \times (1-0.2517)}} = 0.3818\) Using the effectiveness, we can find the heat transfer, \(q\), using the formula \(q = \varepsilon\cdot \min(C_1,C_2)\cdot (\left|T_{h_i} - T_{c_i}\right|)\), where \(T_{h_i}\) and \(T_{c_i}\) are the inlet temperatures of the hot and cold fluids, respectively. Calculate the heat transfer: \(q = 0.3818 \times 53.22 \mathrm{~W/K} \times \left|(75^{\circ} \mathrm{C}) - (30^{\circ} \mathrm{C})\right| = 0.3818 \times 53.22 \mathrm{~W/K} \times 45^{\circ} \mathrm{C} = 912.58\, \mathrm{W}\)

Determine the outlet temperatures

Finally, we can apply the energy balance equation for both fluids to find their outlet temperatures. For oil, the energy balance equation is \(q_{1} = C_{1} \times (T_{1, out} - T_{1, in})\), and for air, it is \(q_{2} = C_{2} \times (T_{2, out} - T_{2, in})\). For oil, we have: \(912.58 \mathrm{~W} = 53.22 \mathrm{~W/K} \times (T_{1, out} - 75^{\circ} \mathrm{C})\) Solve for \(T_{1, out}\): \(T_{1, out} = \frac{912.58 \mathrm{~W}}{53.22 \mathrm{~W/K}} + 75^{\circ} \mathrm{C} = 57.1^{\circ} \mathrm{C}\) For air, we have: \(912.58 \mathrm{~W} = 211.47 \mathrm{~W/K} \times (T_{2, out} - 30^{\circ} \mathrm{C})\) Solve for \(T_{2, out}\): \(T_{2, out} = \frac{912.58 \mathrm{~W}}{211.47 \mathrm{~W/K}} + 30^{\circ} \mathrm{C} = 34.3^{\circ} \mathrm{C}\) Thus, the outlet temperatures of the oil and air are \(57.1^{\circ} \mathrm{C}\) and \(34.3^{\circ} \mathrm{C}\), respectively.