Question

Water enters a 5-mm-diameter and 13 -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, use $k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, \nu=0.801 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$.)

Short answer

Answer: (c) 51.1 °C

Step-by-step solution

Determine the Reynolds number

To determine the heat transfer coefficient, we first need to find the Reynolds number (Re). The Reynolds number is calculated using the equation: \(Re = \frac{VD}{\nu}\) Where: \(V = 0.3\ m/s\) (velocity), \(D = 0.005\ m\) (diameter, converted from 5 mm), \(\nu = 0.801 \times 10^{-6}\ m^2/s\) (kinematic viscosity)

Calculate the Reynolds number

Now, plug the given values into the formula: \(Re = \frac{0.3 \times 0.005}{0.801 \times 10^{-6}}\) \(Re \approx 1869\)

Determine the Nusselt number

For laminar flow, as the Reynolds number is less than \(2300\), we can use the Sieder-Tate correlation to calculate the Nusselt number (Nu). The Sieder-Tate correlation is given as follows: \(Nu = 1.86 \times Re^{1/3} \times Pr^{1/3} \times \frac{L}{D}\) Where: \(L = 13\ m\) (length of the tube) and \(Pr = 5.42\) (Prandtl number)

Calculate the Nusselt number

Now, plug the given values into the formula: \(Nu \approx 1.86 \times 1869^{1/3} \times 5.42^{1/3} \times \frac{13}{0.005}\) \(Nu \approx 3127\)

Determine the heat transfer coefficient (h)

We can now use the Nusselt number to calculate the heat transfer coefficient as follows: \(h = \frac{Nu \times k}{D}\) Where: \(k = 0.615\ W/m\cdot K\) (thermal conductivity)

Calculate the heat transfer coefficient

Plug the given values into the formula: \(h \approx \frac{3127 \times 0.615}{0.005}\) \(h \approx 385\xa0W/m^2 \cdot K\)

Calculate the surface temperature at the exit

Now, using the heat flux (\(q''\)) given, let's find the temperature difference (\(\Delta T\)) between the water and tube surface at the exit: \(\Delta T = \frac{q''}{h}\) Where: \(q'' = 2000\ W/m^2\) (heat flux) Plug the given values into the formula: \(\Delta T \approx \frac{2000}{385}\) \(\Delta T \approx 5.2 \ ^\circ C\) Finally, add the temperature difference to the water temperature at the exit (\(45\ ^\circ C\)) to find the tube surface temperature: \(T_{surface} = T_{water} + \Delta T\) \(T_{surface} = 45 + 5.2\) \(T_{surface} \approx 50.2\ ^\circ C\) Based on the given options, the closest answer is (c) \(51.1\ ^\circ C\).