Question

Consider the velocity and temperature profiles for airflow in a tube with a diameter of \(8 \mathrm{~cm}\) that can be expressed as $$ \begin{aligned} &u(r)=0.2\left[\left(1-(r / R)^{2}\right)\right] \\ &T(r)=250+200(r / R)^{3} \end{aligned} $$ with units in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{K}\), respectively. If the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\), determine the mass flow rate and surface heat flux using the given velocity and temperature profiles. Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short answer

Question: Determine the mass flow rate and surface heat flux using the given velocity and temperature profiles for airflow in a tube with a diameter of 8 cm and convection heat transfer coefficient of 100 W/m²-K. Answer: Using the step-by-step solution above: 1. Calculate the cross-sectional area (A) of the tube: \(A = \pi \dfrac{(0.08 \text{ m})^2}{4}\) 2. Evaluate the mass flow rate (\(\dot{m}\)) by integrating the velocity profile and using the air properties: \(\dot{m} = \rho A \int_{0}^{R} 0.2\left[\left(1-(r / R)^{2}\right)\right] dr\) 3. Determine the surface temperature (\(T_s\)) at r = R: \(T_s = 250 + 200\left(1\right)^{3}\) 4. Calculate the surface heat flux (q) using the convection heat transfer coefficient (h) and temperature difference at the surface: \(q = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (T_s - 293 \text{ K})\) After computing the values, we have the mass flow rate and surface heat flux.

Step-by-step solution

Find cross-sectional area of the tube

To calculate the mass flow rate, we need to find the cross-sectional area (A) of the tube. The tube has a diameter of 8 cm, which needs to be converted into meters first: Diameter in meters: \(D = 8 \mathrm{~cm} \cdot \dfrac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.08 \mathrm{~m}\) Now we can find the cross-sectional area (A) of the tube using the formula: \(A = \pi \dfrac{D^2}{4}\) Substitute the value of D: \(A = \pi \dfrac{(0.08 \text{ m})^2}{4}\)

Calculate the mass flow rate

The mass flow rate (\(\dot{m}\)) can be calculated using the formula: \(\dot{m} = \rho A \int u(r) dr\) Where \(\rho\) is the density of air, \(u(r)\) is the given velocity profile, and A is the cross-sectional area. First, we need to evaluate the air properties at the given temperature and pressure. At \(20^{\circ} \text{C}\) and 1 atm, using air properties tables or the ideal gas law, we have: \(\rho = 1.205 \mathrm{~kg} / \mathrm{m}^3\) We know that \(u(r) = 0.2\left[\left(1-(r / R)^{2}\right)\right]\) and \(A = \pi \dfrac{(0.08 \text{ m})^2}{4}\). Now, integrate the velocity profile with respect to r from 0 to R: \(\dot{m} = \rho A \int_{0}^{R} 0.2\left[\left(1-(r / R)^{2}\right)\right] dr\) Upon solving the integral and substituting the values of A and \(\rho\), we obtain the mass flow rate \(\dot{m}\).

Calculate the temperature difference at the surface

We can now determine the temperature difference at the surface of the tube (\(r = R\)). The surface temperature is given by the temperature profile: \(T_s = 250 + 200(r / R)^{3}\) At \(r = R\), the surface temperature becomes: \(T_s = 250 + 200\left(1\right)^{3}\)

Calculate the surface heat flux

The surface heat flux (q) can be calculated using the convection heat transfer coefficient (h) given as 100 W/m²-K and the temperature difference at the surface: \(q = h (T_s - T_\infty)\) Substitute the values of h, \(T_s\), and average air temperature (\(T_\infty = 20^{\circ} \text{C} = 293 \text{ K}\)): \(q = 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot (T_s - 293 \text{ K})\) By calculating the mass flow rate and surface heat flux using the given velocity and temperature profiles, air properties at the given temperature and pressure, and the convection heat transfer coefficient, you have successfully solved the problem!