Question

Air flows in an isothermal tube under fully developed conditions. The inlet temperature is \(60^{\circ} \mathrm{F}\) and the tube surface temperature is \(120^{\circ} \mathrm{F}\). The tube is \(10 \mathrm{ft}\) long, and the inner diameter is 2 in. The air mass flow rate is \(18.2 \mathrm{lbm} / \mathrm{h}\). Calculate the exit temperature of the air and the total rate of heat transfer from the tube wall to the air. Evaluate the air properties at a temperature \(80^{\circ} \mathrm{F}\). Is this a good assumption?

Short answer

Based on the given problem and the step-by-step solution, provide a short answer to the question: The exit temperature of the air flowing through the tube is 77.5°F, and the total rate of heat transfer from the tube wall to the air is 652.8 W.

Step-by-step solution

Calculate the air properties at 80°F

To find the air properties, we can refer to the air properties table or use the ideal gas law. In this case, we are told to evaluate the air properties at \(80^{\circ} \mathrm{F}\). Convert this temperature to Kelvin: $$ T = \frac{80 - 32}{1.8} + 273.15 = 300.15K $$ Using the air properties table or an online calculator, find the specific heat (Cp), thermal conductivity (k), and dynamic viscosity (μ) at 300.15K for air: $$ \begin{aligned} C_p &= 1005 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \\ k &= 0.0262 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \\ \mu &= 1.983 \times 10^{-5} \frac{\mathrm{kg}}{\mathrm{m} \cdot \mathrm{s}} \end{aligned} $$

Calculate the tube's cross-sectional area and the air velocity

Given the tube inner diameter (D) is 2 in, first, convert it to meters: $$ D = \frac{2}{39.37} = 0.0508 \,\mathrm{m} $$ Calculate the cross-sectional area (A) of the tube using the formula: $$ A = \pi \frac{D^2}{4} = \pi \frac{(0.0508)^2}{4} = 0.00203 \,\mathrm{m}^2 $$ The air mass flow rate (m_dot) is given as \(18.2 \frac{\mathrm{lbm}}{\mathrm{h}}\). Convert it to \(\frac{\mathrm{kg}}{\mathrm{s}}\): $$ \dot{m} = \frac{18.2}{2.2046} \frac{1}{3600} = 0.00229 \frac{\mathrm{kg}}{\mathrm{s}} $$ Calculate the air velocity (V) using the formula: $$ V = \frac{\dot{m}}{A \cdot \rho} = \frac{0.00229}{0.00203} = 1.128 \mathrm{\frac{m}{s}} $$

Calculate the Reynolds number

Use the air velocity (V), tube inner diameter (D), and dynamic viscosity (μ) to find the Reynolds number (Re): $$ Re = \frac{VD\rho}{\mu} = \frac{1.128 \times 0.0508 \times \rho}{1.983 \times 10^{-5}} $$ We can assume air to be an ideal gas at low pressures. Therefore, we can use the Ideal Gas Law to find the density (ρ) of air at \(80^{\circ}F\): $$ ρ = \frac{P}{R T} = \frac{1.013 \times 10^5}{287 \times 300.15} = 1.183 \,\mathrm{kg/m^3} $$ Now, calculate the Reynolds number: $$ Re = \frac{1.128 \times 0.0508 \times 1.183}{1.983 \times 10^{-5}} = 3621 $$

Calculate the Nusselt number

For an isothermal tube under fully developed conditions, we can calculate the Nusselt number (Nu) using the Dittus-Boelter equation and considering the flow to be turbulent (as \(Re > 2300\)): $$ Nu = 0.023 Re^{0.8} Pr^{0.4} $$ The Prandtl number (Pr) can be calculated using the properties of air found in step 1: $$ Pr = \frac{\mu C_p}{k} = \frac{1.983 \times 10^{-5} \times 1005}{0.0262} = 0.756 $$ Using the Reynolds number, and Prandtl number, calculate the Nusselt number: $$ Nu = 0.023 \times 3621^{0.8} \times 0.756^{0.4} = 49.1 $$

Calculate the convection heat transfer coefficient and total heat transfer rate

Use the Nusselt number and the thermal conductivity of air (k) to find the convection heat transfer coefficient (h) using the formula: $$ h = \frac{k}{D} Nu = \frac{0.0262}{0.0508} \times 49.1 = 25.37 \,\mathrm{\frac{W}{m^2 \cdot K}} $$ The total heat transfer rate (Q) can be calculated as: $$ Q = h A_s \Delta T $$ where \(A_s\) is the surface area of the tube, and \(\Delta T\) is the temperature difference between the tube wall and the inlet air temperature. The tube length (L) is given as \(10 ft\), convert it to meters. $$ L = \frac{10}{3.281} = 3.048 \, \mathrm{m} $$ Calculate the surface area of the tube: $$ A_s = \pi D L = \pi (0.0508)(3.048) = 0.7724 \, \mathrm{m^2} $$ Calculate the temperature difference: $$ \Delta T = 120^{\circ} \mathrm{F} - 60^{\circ} \mathrm{F} = 60^{\circ} \mathrm{F} = 33.33^{\circ} \mathrm{C} $$ Finally, calculate the total heat transfer rate: $$ Q = 25.37 \times 0.7724 \times 33.33 = 652.8 \, \mathrm{W} $$

Calculate the exit air temperature

Use the heat transfer rate (Q), mass flow rate (m_dot), and specific heat (Cp) to find the exit air temperature (T_out) using the formula: $$ Q = \dot{m} C_p (T_{out} - T_{in}) $$ Substitute the values and solve for T_out. $$ T_{out} = T_{in} + \frac{Q}{\dot{m} C_p} = 60^{\circ} \mathrm{F} + \frac{652.8}{0.00229 \times 1005} = 77.5^{\circ} \mathrm{F} $$

Checking Assumptions

Now, we will check if evaluating the air properties at 80°F was a good assumption. The temperature of the air at the outlet is 77.5°F, which is close to 80°F. Thus, it is a reasonable assumption in this case. #Conclusion# The exit temperature of the air is \(77.5^{\circ} \mathrm{F}\). The total rate of heat transfer from the tube wall to the air is \(652.8 \, \mathrm{W}\). Evaluating the air properties at 80°F is a good assumption for this situation.