2. The pipe with an outer diameter of d=35mm is cooled in an oil cross flow (oil is trans.oil). The oil flow velocity and the average oil temperature are respectively: w =0.5m/s and t_(1)=25^circ C Determine what temperature of the pips surface must be maintained so that the heat flux will be equal to q=30W/m^2 and what will be the value of the heat transfer coefficient?

To determine the temperature of the pipe's surface that must be maintained so that the heat flux will be equal to $q=30W/m^{2}$, we can use the formula for heat transfer by conduction:<br /><br />$q = h \cdot (t_{1} - t_{2})$<br /><br />where:<br />- $q$ is the heat flux (in $W/m^{2}$)<br />- $h$ is the heat transfer coefficient (in $W/m^{2}K$)<br />- $t_{1}$ is the average oil temperature (in $^{\circ}C$)<br />- $t_{2}$ is the temperature of the pipe's surface (in $^{\circ}C$)<br /><br />We need to find $t_{2}$, so we rearrange the formula:<br /><br />$t_{2} = t_{1} - \frac{q}{h}$<br /><br />To find the value of the heat transfer coefficient, we can use the convection heat transfer equation:<br /><br />$h = Nu \cdot h_{f}$<br /><br />where:<br />- $Nu$ is the Nusselt number (dimensionless)<br />- $h_{f}$ is the convective heat transfer coefficient (in $W/m^{2}K$)<br /><br />For turbulent flow, the Nusselt number can be estimated using the Dittus-Boelter equation:<br /><br />$Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}$<br /><br />where:<br />- $Re$ is the Reynolds number<br />- $Pr$ is the Prandtl number<br /><br />The Reynolds number can be calculated as:<br /><br />$Re = \frac{w \cdot d}{\nu}$<br /><br />where:<br />- $w$ is the oil flow velocity (in $m/s$)<br />- $d$ is the outer diameter of the pipe (in $m$)<br />- $\nu$ is the kinematic viscosity of the oil (in $m^{2}/s$)<br /><br />The Prandtl number can be estimated as:<br /><br />$Pr = \frac{\nu}{\alpha}$<br /><br />where:<br />- $\alpha$ is the thermal diffusivity of the oil (in $m^{2}/s$)<br /><br />Assuming typical values for oil, let's take $\nu = 10^{-5} m^{2}/s$ and $\alpha = 10^{-7} m^{2}/s$.<br /><br />First, calculate the Reynolds number:<br /><br />$Re = \frac{0.5 \cdot 0.035}{10^{-5}} = 2500$<br /><br />Next, calculate the Prandtl number:<br /><br />$Pr = \frac{10^{-5}}{10^{-7}} = 100$<br /><br />Now, calculate the Nusselt number:<br /><br />$Nu = 0.023 \cdot 2500^{0.8} \cdot 100^{0.3} \approx 1.38 \times 10^{3}$<br /><br />Finally, calculate the heat transfer coefficient:<br /><br />$h = 1.38 \times 10^{3} \cdot 10^{-5} = 13.8 W/m^{2}K$<br /><br />Now, we can find the temperature of the pipe's surface:<br /><br />$t_{2} = 25 - \frac{30}{13.8} \approx 25 - 2.17 \approx 22.83^{\circ}C$<br /><br />So, the temperature of the pipe's surface must be maintained at approximately $22.83^{\circ}C$ to achieve a heat flux of $30W/m^{2}$, and the value of the heat transfer coefficient is approximately $13.8W/m^{2}K$.