91Ó°ÊÓ

- Water flow rate: \(2.5 \ kg/s\) - Inlet water temperature: \(15^\circ C\) - Outlet water temperature: \(85^\circ C\) - Inlet oil temperature: \(160^\circ C\) - Outlet oil temperature: \(100^\circ C\) - Convection coefficient for oil: \(h_o = 400 \ W/m^2 K\) - Number of tubes: 10 - Diameter of each tube: \(D = 25 \ mm\) - Number of passes for each tube: 8

First, we need to find the heat transfer rate which can be found using the formula: \[Q = \dot{m} \times C_p \times ΔT\] Where: - \(Q\) is the heat transfer rate. - \(\dot{m}\) is the mass flow rate. - \(C_p\) is the specific heat of the fluid. - \(ΔT\) is the temperature difference. For water, we'll use \(C_p = 4.18 \times 10^3 \ J/kg K\). Find the heat transfer rate for the water: \[Q_{water} = \dot{m}_{water} \times C_p \times (T_{out} - T_{in})\] \[Q_{water} = 2.5 \mathrm{~kg/s} \times 4.18 \times 10^3 \mathrm{~J/kg \cdot K} \times (85 - 15)^\circ \mathrm{C}\] \[Q_{water} = 733750 \textrm{~W}\]

Since we are assuming steady-state operation, the heat gained by the water is the same as that lost by the oil. So, \(Q_{water} = Q_{oil}\). Now, let's find the oil mass flow rate (\(\dot{m}_{oil}\)). We'll assume the specific heat of oil is \(C_{p_{oil}} = 2.1 \times 10^3 \ J/kg \cdot K\). \[Q_{oil} = \dot{m}_{oil} \times C_{p_{oil}} \times (T_{in} - T_{out})\] \[\dot{m}_{oil} = \frac{Q_{oil}}{(C_{p_{oil}} \times (T_{in} - T_{out}))}\] \[\dot{m}_{oil} = \frac{733750 \textrm{~W}}{(2.1 \times 10^3 \mathrm{~J/kg \cdot K} \times (160 - 100)^\circ \mathrm{C})}\] \[\dot{m}_{oil} = 0.57031 \textrm{~kg/s}\]

To find the required tube length, we can use the formula for the overall heat transfer rate (\(Q\)): \[Q = UAΔT_{m}\] Where: - \(U\) is the overall heat transfer coefficient. - \(A\) is the surface area of the tubes. - \(ΔT_{m}\) is the Log Mean Temperature Difference (LMTD). We know the convection coefficient for oil, but we need the overall heat transfer coefficient (\(U\)). For simplified calculations, we can assume that the heat transfer coefficient for the water inside the tube is quite large, so the heat transfer is mainly limited by the oil outside the tubes. With this simplification, we can assume \(U \approx h_{o}\). Now, we need to calculate the LMTD using the given inlet and outlet temperatures: Let \(T_{H,i} = T_{water_{in}}\) , \(T_{H,o} = T_{water_{out}}\) , \(T_{C,i} = T_{oil_{in}}\) , \(T_{C,o} = T_{oil_{out}}\) \[\Delta T_{m} = \frac{(T_{C,i} - T_{H,o}) - (T_{C,o} - T_{H,i})}{\ln{\frac{T_{C,i} - T_{H,o}}{T_{C,o} - T_{H,i}}}}\] \[\Delta T_{m} = \frac{(160 - 85)^\circ \mathrm{C} - (100 - 15)^\circ \mathrm{C}}{\ln{\frac{(160 - 85)}{(100 - 15)}}}\] \[\Delta T_{m} \approx 61.79^\circ \mathrm{C}\] Next, we calculate the surface area of the tubes: \[A = 10 \times 8 \times \pi \times D \times L\] Where: - \(L\) is the length of the tube. Finally, we will solve for \(L\): \begin{align*} 733750 \textrm{~W} &= 400 \ \frac{\textrm{W}}{\textrm{m}^{2} \textrm{K}} \times 10 \times 8 \times \pi \times \frac{1}{1000} \textrm{~m} \times L \times 61.79 \textrm{~K} \\ L &= \frac{733750}{400 \times 10 \times 8 \times \pi \times 61.79 \times \frac{1}{1000}} \\ L &= 2.384 \textrm{~m} \end{align*} The oil flow rate is approximately \(0.57031 \textrm{~kg/s}\), and the required tube length is approximately \(2.384 \textrm{~m}\).