91Ó°ÊÓ

We are given the mass flow rate of water \(\dot{m}=0.015kg/s\), inlet water temperature \(T_{m,i}=97^{\circ}C\), tube length \(L=1m\), and inside diameter \(D_i=3mm\). The tube surface temperature is maintained at \(T_s=27^{\circ}C\). Considering a steady-state heat transfer, the outlet temperature can be calculated using the following formula: \[q = \dot{m}C_p(T_{m,o}-T_{m,i})\] where \(q\) is heat transfer rate, \(C_p\) is specific heat of water, and \(T_{m,o}\) is outlet temperature of water. First, let's calculate the heat transfer rate, which is the conduction of heat through the metal tube: \[q = h \pi D_i L (T_s - T_m)\] where \(h\) is the heat transfer coefficient and \(T_m\) is the average temperature of water. First, we need to find the heat transfer coefficient and then use it to find the outlet temperature of water. \(Nu = \frac{hd_i}{k}\), where \(Nu\) is Nusselt number and \(k\) is thermal conductivity of water. Assuming that the flow is fully developed, the value of Nusselt number for a circular tube can be taken as: \(Nu = 3.66\) Now, plugging the values for \(Nu\), \(D_i\), and thermal conductivity of water \(k=0.6W/m\cdot K\) at \(T_m=62^{\circ}C\), we get the heat transfer coefficient, \(h\). Next step is to calculate the heat transfer rate, \(q\), and then use it to find the outlet temperature of water.

In this part, we need to calculate the outlet temperature of the water with the \(0.5mm\) thick insulation layer applied. First, let's calculate the overall heat transfer coefficient, \(U\): \[U = \frac{1}{\frac{1}{h}+\frac{t_{ins}}{k_{ins}}} \] where \(t_{ins}\) is the insulation thickness, and \(k_{ins}\) is the thermal conductivity of the insulation material. Now we can use this overall heat transfer coefficient to find the heat transfer rate, \(q\), with the presence of insulation: \[q = U \pi D_i L (T_s - T_m)\] Determine the new heat transfer rate, \(q\), and use it in the energy balance equation to find the outlet temperature of water with insulation: \[q=\dot{m}C_p(T_{m,o}-T_{m,i})\]

In this part, the insulation material is allowed to exchange heat with the ambient air at \(27^{\circ}C\). The free convection heat transfer coefficient is given as \(h_c=5W/m^2\cdot K\). We have to find the new outlet temperature of the water. First, we need to write an energy balance including conduction through insulation and convection to the ambient air for the heat transfer rate: \[q = h_c A (T_{ins} - T_a)\] where \(A\) is area and \(T_{ins}\) is the temperature at the outer surface of insulation, and \(T_a\) is the ambient temperature. Use the equation for thermal resistance due to conduction and convection, and the overall heat transfer rate to find the new heat transfer rate, \(q\): \[q = U \pi D_{ins} L (T_{ins} - T_m)\] Now, we have two equations and two unknowns, \(T_{ins}\) and \(q\). Solve this system of equations to find the new heat transfer rate, and use it in the energy balance equation to find the outlet temperature of water.