Question

Consider laminar flow of a fluid through a square channel maintained at a constant temperature. Now the mean velocity of the fluid is doubled. Determine the change in the pressure drop and the change in the rate of heat transfer between the fluid and the walls of the channel. Assume the flow regime remains unchanged. Assume fully developed flow, and disregard any changes in $\Delta T_{\mathrm{lm}^{-}}$

Short answer

Answer: When the mean velocity of the fluid doubles in a laminar flow through a square channel at a constant temperature, the change in pressure drop is equal to the original pressure drop, and the change in the rate of heat transfer can be calculated using the given formulas, considering that Nusselt number depends on the mean velocity.

Step-by-step solution

List the given information

We know the following information from the problem: 1. Laminar flow of a fluid through a square channel. 2. Constant temperature. 3. The mean velocity of the fluid is doubled. 4. Fully developed flow.

Use the Hagen–Poiseuille equation for pressure drop

The Hagen–Poiseuille equation for pressure drop in a square channel is given by: \(\Delta P = \frac{8\mu LQ}{h^3} \), where \( \Delta P \) is the pressure drop, \( \mu \) is the dynamic viscosity of the fluid, \( L \) is the length of the channel, \( Q \) is the volumetric flow rate, and \( h \) is the hydraulic diameter (which is equal to the side length for the square channel). Since we are given that the mean velocity of the fluid is doubled, we can write the new volumetric flow rate as: \( Q_{\mathrm{new}} = 2Q \) Now, we will find the new pressure drop (\(\Delta P_{\mathrm{new}}\)) using the Hagen–Poiseuille equation: \(\Delta P_{\mathrm{new}} = \frac{8\mu LQ_{\mathrm{new}}}{h^3} = \frac{8\mu L(2Q)}{h^3} \)

Calculate the change in pressure drop

To determine the change in pressure drop, we will subtract the original pressure drop (\(\Delta P\)) from the new pressure drop (\(\Delta P_{\mathrm{new}}\)): \(\Delta P_{\mathrm{change}} = \Delta P_{\mathrm{new}} - \Delta P = \frac{8\mu L}{h^3} (Q_{\mathrm{new}} - Q) = \frac{8\mu L}{h^3} (2Q - Q) = \frac{8\mu LQ}{h^3}\) We can conclude that the change in pressure drop is equal to the original pressure drop.

Determine the rate of heat transfer using Nusselt number

The Nusselt number is defined as the ratio of convective heat transfer to conductive heat transfer in a fluid and can be related to the heat transfer coefficient (h), hydraulic diameter (D), and thermal conductivity (k) as follows: \( Nu = \frac{hD}{k} \) For a square channel with constant wall temperature, the Nusselt number can be approximated as: \( Nu = 8.23 \) We can rearrange the Nusselt number formula and find the heat transfer coefficient (h): \( h = \frac{Nu \times k}{D} = \frac{8.23 \times k}{h} \) Next, we find the rate of heat transfer using the heat transfer coefficient: \( q = hA\Delta T_{\mathrm{lm}^{-}} \) Since we are assuming there is no change in \(\Delta T_{\mathrm{lm}^{-}}\), and the surface area (A) of the channel remains constant, the change in the rate of heat transfer can be calculated as: \(\Delta q = h_{\mathrm{new}}A\Delta T_{\mathrm{lm}^{-}} - hA\Delta T_{\mathrm{lm}^{-}} \)

Find the change in the rate of heat transfer

To determine the change in the rate of heat transfer, we will need to find the new heat transfer coefficient (\( h_{\mathrm{new}} \)). Since the Nusselt number depends on the mean velocity: \( Nu_{\mathrm{new}} \propto \frac{1}{\sqrt{Q_{\mathrm{new}}}} \) Solving for the new heat transfer coefficient: \( h_{\mathrm{new}} = \frac{Nu_{\mathrm{new}} \times k}{D} = \frac{\frac{Nu \times \sqrt{Q}}{\sqrt{Q_{\mathrm{new}}}} \times k}{D} = \frac{8.23 \times k \times \sqrt{Q}}{\sqrt{2Q} \times D} \) Now, calculate the change in the rate of heat transfer: \(\Delta q = \left( \frac{8.23 \times k \times \sqrt{Q}}{\sqrt{2Q} \times D} \right) A\Delta T_{\mathrm{lm}^{-}} - \frac{8.23 \times k}{D} A\Delta T_{\mathrm{lm}^{-}}\) Thus, we have determined the change in the pressure drop and the change in the rate of heat transfer when the mean velocity of the fluid is doubled in a square channel.