Water enters a \(5-\mathrm{mm}\)-diameter and \(13-\mathrm{m}\)-long tube at \(15^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\), and leaves at \(45^{\circ} \mathrm{C}\). The tube is subjected to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\) on its surface. The temperature of the tube surface at the exit is (a) \(48.7^{\circ} \mathrm{C}\) (b) \(49.4^{\circ} \mathrm{C}\) (c) \(51.1^{\circ} \mathrm{C}\) (d) \(53.7^{\circ} \mathrm{C}\) (e) \(55.2^{\circ} \mathrm{C}\) (For water, nse \(k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, v=0.801 \times\) \(\left.10^{-6} \mathrm{~m}^{2} / \mathrm{s} .\right)\)

First, we will find the cross-sectional area of the tube using its diameter (in meters), given by \(A = \pi \cdot (D / 2)^{2}\) where \(D = 5 \times 10^{-3} \mathrm{m}\) (diameter of the tube) Calculating the cross-sectional area: \(A = \pi \cdot (5 \times 10^{-3} / 2)^{2} = 1.963 \times 10^{-5} \mathrm{m}^2\)

Next, we need to find the mass flow rate of water through the tube, given by \(\dot{m} = \rho \cdot V \cdot A\) where \(V = 0.3 \mathrm{~m} / \mathrm{s}\) (velocity of water) For water at \(15^{\circ} \mathrm{C}\), we use \(\rho = 999 \mathrm{~kg/m^3}\) (density of water) Calculating the mass flow rate: \(\dot{m} = 999 \cdot 0.3 \cdot 1.963 \times 10^{-5} = 5.882 \times 10^{-3} \mathrm{kg/s}\)

Now we can calculate the overall heat transfer rate, \(Q\), using the heat flux, \(\phi\), given by \(Q = \phi \cdot A_s\) where \(\phi = 2000 \mathrm{~W} / \mathrm{m}^{2}\) (heat flux) \(A_s = \pi D L\) (surface area of the tube) Calculating the surface area of the tube: \(A_s = \pi (5 \times 10^{-3})(13) = 0.2042 \mathrm{m}^2\) Calculating the overall heat transfer rate: \(Q = 2000 \cdot 0.2042 = 408.4 \mathrm{W}\)

Once we have the heat transfer rate, we can find the change in temperature of water, given by: \(\Delta T = \frac{Q}{\dot{m} \cdot C_p}\) where \(C_p\) is the specific heat capacity of water (approximated as a constant), which is roughly \(4190 \mathrm{J/kg\cdot K}\) Calculating the change in temperature: \(\Delta T = \frac{408.4}{5.882 \times 10^{-3} \cdot 4190} = 21.1 \mathrm{K}\)

Finally, we can calculate the tube surface temperature at the exit, \(\theta_s\), using the water temperature at the exit and the average heat transfer coefficient, given by \(\theta_s = \theta_{out} + \frac{Q}{h \cdot A_s}\) To find the average heat transfer coefficient, \(h\), we need to use the properties given: \(k = 0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (thermal conductivity of water) \(\operatorname{Pr} = 5.42\) (Prandtl number of water) \(\nu = 0.801 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) (kinematic viscosity of water) The Reynolds number, \(Re\), can be calculated as: \(Re = \frac{VD}{\nu}\) Calculating the Reynolds number: \(Re = \frac{0.3 \cdot 5 \times 10^{-3}}{0.801 \times 10^{-6}} = 1869.5\) Since \(Re < 2300\), we have a laminar flow and can use the Dittus-Boelter equation for the average heat transfer coefficient, \(h\): \(h = \frac{k}{D} \cdot 1.86 \cdot Re \cdot \operatorname{Pr}^{\frac{1}{3}}\) Calculating the heat transfer coefficient: \(h = \frac{0.615}{5 \times 10^{-3}} \cdot 1.86 \cdot 1869.5 \cdot 5.42^{\frac{1}{3}} = 3024.25 \mathrm{W/m^{2}\cdot K}\) Now, we can find the tube surface temperature at the exit: \(\theta_s = 45 + \frac{408.4}{3024.25 \cdot 0.2042} = 45 + 3.7\) \(\theta_s = 48.7^{\circ} \mathrm{C}\) Thus, the tube surface temperature at the exit is approximately \(48.7^{\circ} \mathrm{C}\) (option (a)).