Question

Consider the flow of engine oil $\left(c_{p}=2048 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( through a thin-walled copper tube at a rate of \)0.3 \mathrm{~kg} / \mathrm{s}$. The engine oil that enters the copper tube at an inlet temperature of \(80^{\circ} \mathrm{C}\) is to be cooled by cold water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at a temperature of \(20^{\circ} \mathrm{C}\). We wish to have the exit temperature of the engine oil not exceed \(40^{\circ} \mathrm{C}\). The individual convective heat transfer coefficients on oil and water sides are \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and $350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, respectively. If a thermocouple probe installed on the downstream side of the cooling water measures a temperature of \(32^{\circ} \mathrm{C}\), for a double-pipe parallelflow heat exchanger determine \((a)\) the mass flow rate of the cooling water, \((b)\) the log mean temperature difference, and (c) the area of the heat exchanger.

Short answer

Answer: (a) The mass flow rate of the cooling water is approximately 0.419 kg/s. (b) The log mean temperature difference is approximately 31.69°C. (c) The area of the heat exchanger is approximately 2.86 m².

Step-by-step solution

Energy balance equation for engine oil and cooling water

An energy balance equation states that the heat gained by the cooling water must equal the heat lost by the engine oil: \(m_o c_{p_o}(T_{o,inlet} - T_{o,exit}) = m_w c_{p_w}(T_{w,exit} - T_{w,inlet})\) We know the values for engine oil mass flow rate (\(m_o = 0.3 kg/s\)), specific heat capacities (\(c_{p_o} = 2048 J/kg·K\) and \(c_{p_w} = 4180 J/kg·K\)), and inlet temperatures (\(T_{o,inlet} = 80°C\), and \(T_{w,inlet}=20°C\)). We also know the required exit temperature for engine oil (\(T_{o,exit}=40°C\)). We need to find the mass flow rate of cooling water (\(m_w\)) and the exit temperature of cooling water (\(T_{w,exit}\)). We can rewrite the energy balance equation in terms of mass flow rate of cooling water: \(m_w = \frac{m_o c_{p_o}(T_{o,inlet} - T_{o,exit})}{c_{p_w}(T_{w,exit} - T_{w,inlet})}\)

Temperature measurement of cooling water outlet

We are given that a thermocouple probe measures a temperature of \(32°C\) on the downstream side of the cooling water. Therefore, the temperature of cooling water at the outlet (\(T_{w,exit}\)) can be assumed to be \(32°C\).

Calculate the mass flow rate of cooling water

Substitute the given values in the formula for mass flow rate of cooling water and calculate \(m_w\): \(m_w = \frac{0.3 kg/s * 2048 J/kg\cdot K * (80°C - 40°C)}{4180 J/kg\cdot K * (32°C - 20°C)}\) \(m_w \approx 0.419~kg/s\) Thus, the mass flow rate of the cooling water is approximately \(0.419~kg/s\).

Calculate the log mean temperature difference (LMTD)

The log mean temperature difference (LMTD) is used in heat exchanger calculations and is calculated as follows: \(LMTD = \frac{(ΔT_1 - ΔT_2)}{\ln(\frac{ΔT_1}{ΔT_2})}\) where \(ΔT_1\) is the temperature difference between the inlet temperatures of hot and cold fluids, and \(ΔT_2\) is the temperature difference between the outlet temperatures of hot and cold fluids. \(ΔT_1 = T_{o,inlet} - T_{w,inlet} = 80°C - 20°C = 60°C\) \(ΔT_2 = T_{o,exit} - T_{w,exit} = 40°C - 32°C = 8°C\) Now, we can calculate the LMTD: \(LMTD = \frac{60°C - 8°C}{\ln(\frac{60°C}{8°C})} \approx 31.69°C\) The log mean temperature difference is approximately \(31.69°C\).

Determine the area of the heat exchanger

To calculate the area of the heat exchanger, we will use the following formula: \(Q = U A LMTD\) where \(Q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, \(A\) is the area of the heat exchanger, and the log mean temperature difference (LMTD) was calculated in Step 4. First, let's calculate the heat transfer rate, \(Q\): \(Q = m_o c_{p_o}(T_{o,inlet} - T_{o,exit})\) \(Q = 0.3 kg/s \cdot 2048 J/kg\cdot K \cdot (80°C - 40°C) = 24576~W\) Now, we need to calculate the overall heat transfer coefficient (\(U\)), using the convective heat transfer coefficients on oil and water sides (\(750~W/m^2\cdot K\) and \(350~W/m^2\cdot K\)): \(\frac{1}{U} = \frac{1}{h_o}+\frac{1}{h_w}\) \(\frac{1}{U} = \frac{1}{750}+\frac{1}{350}\) Solving for \(U\) gives us: \(U \approx 262.96~W/m^2\cdot K\) Now, we can calculate the area of the heat exchanger using the formula \(Q = U A LMTD\): \(A = \frac{Q}{U \cdot LMTD}\) \(A = \frac{24576~W}{262.96~W/m^2\cdot K \cdot 31.69°C} \approx 2.86~m^2\) The area of the heat exchanger is approximately \(2.86~m^2\). In conclusion, we have determined (a) the mass flow rate of the cooling water to be approximately \(0.419~kg/s\), (b) the log mean temperature difference to be approximately \(31.69°C\), and (c) the area of the heat exchanger to be approximately \(2.86~m^2\).