Question

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 \(k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, v=0.801 \times\) \(\left.10^{-6} \mathrm{~m}^{2} / \mathrm{s} .\right)\)

Short answer

Answer: The tube surface temperature at the exit is approximately \(48.7^{\circ} \mathrm{C}\) (option (a)).

Step-by-step solution

Finding the cross-sectional area of the tube

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: \(A = \pi \cdot (5 \times 10^{-3} / 2)^{2} = 1.963 \times 10^{-5} \mathrm{m}^2\)

Finding the mass flow rate of water

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: \(\dot{m} = 999 \cdot 0.3 \cdot 1.963 \times 10^{-5} = 5.882 \times 10^{-3} \mathrm{kg/s}\)

Finding the overall heat transfer rate

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: \(Q = 2000 \cdot 0.2042 = 408.4 \mathrm{W}\)

Finding the change in temperature of water

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: \(\Delta T = \frac{408.4}{5.882 \times 10^{-3} \cdot 4190} = 21.1 \mathrm{K}\)

Calculating the tube surface temperature at the exit

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)).

Key concepts

Convective Heat Transfer

Convective heat transfer pertains to the transfer of heat between a surface and a fluid moving over it. The movement of fluid can be natural due to density differences, or forced by a pump or a fan. This process is crucial in numerous applications such as heating systems, cooling of electronic devices, and industrial processes.

In our exercise, water flows through a tube and gains heat from the tube walls due to a prescribed heat flux. Here, the motion of water is a forced convection scenario because the water is pushed through the tube at a known velocity. The rate at which heat is transferred from the tube's surface to the water depends on several factors, including the properties of the fluid, the flow characteristics, and the surface temperature.

To enhance student understanding, it's essential to visualize convective heat transfer through practical examples, like a radiator heating a room or a fan blowing over a wet surface causing it to cool. These examples can help make the connection between abstract concepts and real-world applications clearer.

Reynolds Number

The Reynolds number (Re) is a dimensionless quantity that helps predict flow patterns in different fluid flow situations. It is an essential factor in determining whether the flow will be laminar or turbulent. Laminar flow is characterized by smooth, constant fluid motion, typically occurring at lower Reynolds numbers, while turbulent flow involves chaotic changes in pressure and flow velocity, generally occurring at higher values.

In our exercise's context, the Reynolds number is calculated using the formula \( Re = \frac{VD}{u} \) where \( V \) is the velocity of water, \( D \) is the diameter of the tube, and \( u \) is the kinematic viscosity of the water. With \( Re \) found to be 1869.5 and below the critical value of 2300, we determined that the water flow within the tube is laminar. This information is vital in choosing the right correlation for the heat transfer coefficient. Students should grasp the significance of the Reynolds number for predicting the heat transfer characteristics in a system, which could be further illustrated using the example of ink dropping into water – dispersing evenly in a laminar flow versus chaotically in turbulent flow.

Dittus-Boelter Equation

The Dittus-Boelter equation is an empirical correlation used to estimate the convective heat transfer coefficient in a fluid flowing within tubes under certain flow conditions. It is specifically designed for turbulent flow but can also be applied under laminar flow conditions, as seen in our exercise. The equation is given by \( h = \frac{k}{D} \cdot 1.86 \cdot Re \cdot Pr^{\frac{1}{3}} \) where \( h \) is the average heat transfer coefficient, \( k \) is the thermal conductivity of the fluid, \( D \) is the diameter of the tube, \( Re \) is the Reynolds number, and \( Pr \) is the Prandtl number, which correlates the fluid's velocity and thermal diffusivity.

In our scenario, we used the Dittus-Boelter equation to find \( h \) because the flow is laminar and the Reynolds number is less than 2300. By knowing \( h \) and the other given parameters, we were able to determine the tube surface temperature at the exit. This equation allows students to relate the fluid’s physical properties and flow dynamics to its heat transfer capability, and a practical way to visualize it might involve comparing the cooling rate of various metal rods with different diameters and thermal conductivities when immersed in moving water.