Question

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil $\left(c_{p k}=2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( flowing with a flow rate of \)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 \(53 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). Determine \((a)\) the heat transfer effectiveness and \((b)\) the outlet temperature of the oil.

Short answer

a) To calculate the heat transfer effectiveness, first determine the maximum possible heat transfer rate using $$Q_{max} = \min{\left(\dot{m}_c c_{p_c}, \dot{m}_k c_{p_k}\right)} (T_{h,i} - T_{c,i})$$ Then, divide the actual heat transfer rate (Q) by the maximum possible heat transfer rate (Q_max): $$\epsilon = \frac{Q}{Q_{max}}$$ b) To find the outlet temperature of the oil, use the heat transfer rate (Q) and the formula $$T_{h,o} = T_{h,i} - \frac{Q}{\dot{m}_k c_{p_k}}$$

Step-by-step solution

Calculate the heat transfer rate

To calculate the heat transfer rate, we will use the formula $$Q = UA \Delta T_{lm} $$ where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the total surface area, and \(\Delta T_{lm}\) is the Log Mean Temperature Difference. The Log Mean Temperature Difference is calculated by $$ \Delta T_{lm} = \frac{T_{h,i} - T_{h,o}}{\ln{\frac{T_{h,i} - T_{c,i}}{T_{h,o} - T_{c,i}}}} $$ We know the inlet temperatures (\(T_{h,i}\) and \(T_{c,i}\)) and all other given values, but we need to find the outlet temperature of the oil \((T_{h,o})\) first to calculate the LMTD.

Calculate the heat transfer rate using the energy balance

To find the outlet temperature of the oil, we can use the energy balance, which states that the heat gained by the cold fluid is equal to the heat lost by the hot fluid, expressed as $$Q = \dot{m}_c c_{p_c} (T_{c,o} - T_{c,i}) = \dot{m}_k c_{p_k} (T_{h,i} - T_{h,o}) $$ where \(\dot{m}_c\) and \(\dot{m}_k\) are the mass flow rates of the cold and hot fluids, \(c_{p_c}\) and \(c_{p_k}\) are the specific heat capacities of the cold and hot fluids, and \(T_{c,o}\) is the outlet temperature of the cold fluid. We can express the outlet temperature of the cold fluid in terms of the heat transfer rate: $$T_{c,o} = T_{c,i} + \frac{Q}{\dot{m}_c c_{p_c}}$$ Substituting this expression into the energy balance equation, we get $$Q = \dot{m}_k c_{p_k} (T_{h,i} - T_{h,o}) + \frac{Q^2}{\dot{m}_c c_{p_c} \dot{m}_k c_{p_k}}$$ This is a quadratic equation in Q, which can be solved to find the heat transfer rate.

Find the outlet temperature of the oil

After solving the quadratic equation in step 2, we can find the outlet temperature of the oil using the formula $$T_{h,o} = T_{h,i} - \frac{Q}{\dot{m}_k c_{p_k}}$$

Calculate the heat transfer effectiveness

The heat transfer effectiveness is the ratio of the actual heat transfer rate to the maximum possible heat transfer rate, expressed as $$\epsilon = \frac{Q}{Q_{max}}$$ The maximum possible heat transfer rate can be found using the formula $$Q_{max} = \min{\left(\dot{m}_c c_{p_c}, \dot{m}_k c_{p_k}\right)} (T_{h,i} - T_{c,i})$$ Finally, we can find the heat transfer effectiveness by dividing the actual heat transfer rate by the maximum possible heat transfer rate. Now, we have the heat transfer effectiveness and the outlet temperature of the oil, which are the answers to \((a)\) and \((b)\), respectively.