Question

Water at an average temperature of \(180^{\circ} \mathrm{F}\) and an average velocity of \(4 \mathrm{ft} / \mathrm{s}\) flows through a thin-walled \(\frac{3}{4}\)-indiameter tube. The water is cooled by air that flows across the tube with a velocity of \(12 \mathrm{ft} / \mathrm{s}\) at an average temperature of \(80^{\circ} \mathrm{F}\). Determine the overall heat transfer coefficient.

Short answer

Use the provided steps in the solution to determine the heat transfer coefficients for the fluids, dimensions of the tube, and calculate the overall heat transfer coefficient.

Step-by-step solution

Compute the heat transfer coefficient of the water

To compute the heat transfer coefficient of the water, we can use the Sieder-Tate equation in the form: \$ h_w = k_w * Nu \$, where \$h_w\$ is the heat transfer coefficient of the water, \$k_w\$ is the thermal conductivity of the water, and \$Nu\$ is the dimensionless Nusselt number. The thermal conductivity of the water at the given average temperature will have to be found from a reference table or a correlation. Similarly, the Nusselt number depends on the dimensionless Reynolds and Prandtl numbers. Use available correlations to compute these dimensionless numbers and the corresponding heat transfer coefficient for water.

Compute the heat transfer coefficient of the air

We can also compute the heat transfer coefficient of the air using the same Sieder-Tate equation for air: \$ h_a = k_a * Nu \$, where \$h_a\$ is the heat transfer coefficient of the air, \$k_a\$ is the thermal conductivity of the air, and \$Nu\$ is the Nusselt number for air. Again, the thermal conductivity of the air at the given average temperature can be found from reference tables or correlations. Similarly, use appropriate correlations for determining the dimensionless numbers and corresponding heat transfer coefficient for air.

Calculate the overall heat transfer coefficient

The overall heat transfer coefficient (\$ U \$) can be found using the following equation for a concentric pipe arrangement: \$ \frac{1}{U} = \frac{1}{h_w} + \frac{D_o}{D_i*h_a} \$, where \$D_i\$ is the inner diameter of the tube, and \$D_o\$ is the outer diameter of the tube (assumed to be the same as the inner diameter in a thin-walled tube). Using the heat transfer coefficients obtained in Step 1 and Step 2, along with the given diameter for the tube, we can calculate the overall heat transfer coefficient. Calculate the overall heat transfer coefficient using the values obtained in the previous steps and the given tube diameter, and that will be the final answer.