Question

Consider the velocity and temperature profiles for a fluid flow in a tube with diameter of \(50 \mathrm{~mm}\) can be expressed as $$ \begin{aligned} &u(r)=0.05\left[\left(1-(r / R)^{2}\right]\right. \\ &T(r)=400+80(r / R)^{2}-30(r / R)^{3} \end{aligned} $$ with units in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{K}\), respectively. Determine the average velocity and the mean (average) temperature from the given velocity and temperature profiles.

Short answer

Based on the given velocity and temperature profiles of a fluid flow inside a tube with a diameter of 50 mm, the average velocity of the fluid flow is 1.2 m/s, and the mean temperature is 420 K.

Step-by-step solution

Determine the cross-sectional area of the tube

To calculate the average velocity and mean temperature, we need to first find the cross-sectional area (A) of the tube. The cross-sectional area of a tube can be calculated using the formula \(A = \pi R^2\), where R is the radius of the tube. Given the diameter of the tube (D) is 50 mm, we can find the radius (R) as half of the diameter. Radius, \(R = \frac{D}{2} = \frac{50}{2} \mathrm{~mm} = 25\mathrm{~mm}\) Now we convert it to meters: \(R = 25\mathrm{~mm} \times \frac{1\mathrm{~m}}{1000\mathrm{~mm}}=0.025\mathrm{~m}\) Cross-sectional area, \(A = \pi R^2 = \pi(0.025)^2 \mathrm{~m^2}\)

Set up the integral for average velocity

The average velocity (U) can be calculated by integrating the given velocity profile (u(r)) over the cross-section of the tube and dividing by the total area: \(U = \frac{1}{A}\int_0^R u(r) 2\pi r ~dr\) Now, substitute the expression for the velocity profile: \(U = \frac{1}{A}\int_0^R 0.05\left[\left(1-(r/R)^{2}\right)\right] 2\pi r ~dr\)

Calculate the average velocity

To calculate the average velocity, perform the integration: \(U = \frac{1}{A}\int_0^R 0.05\left[\left(1-(r/R)^{2}\right)\right] 2\pi r ~dr = \frac{1}{A} \cdot 0.05 \cdot 2\pi \int_0^R r - \frac{r^3}{R^2} dr\) Separating the integrals: \(U = \frac{1}{A} \cdot 0.05 \cdot 2\pi \left[\int_0^R r ~dr - \int_0^R \frac{r^3}{R^2} dr\right]\) Now, perform the integration: \(U = \frac{0.05 \cdot 2\pi}{A} \left[ \frac{R^2}{2}-\frac{R^4}{4R^2}\right]\) Substitute the value of R and A that we found in step 1: \(U = \frac{0.05 \cdot 2\pi}{\pi (0.025)^2} \left[ \frac{(0.025)^2}{2} - \frac{(0.025)^4}{4(0.025)^2}\right] = 1.2 \mathrm{~m/s}\) The average velocity of the fluid flow is 1.2 m/s.

Set up the integral for mean temperature

The mean (average) temperature (Tm) can be calculated similarly, by integrating the given temperature profile (T(r)) over the cross-section of the tube and dividing by the total area: \(T_{m} = \frac{1}{A}\int_0^R T(r) 2\pi r ~dr\) Now, substitute the expression for the temperature profile: \(T_{m} = \frac{1}{A}\int_0^R \left[400+80(r/R)^{2}-30(r/R)^{3}\right] 2\pi r ~dr\)

Calculate the mean temperature

To calculate the mean temperature, perform the integration: \(T_{m} = \frac{1}{A}\int_0^R \left[400+80(r/R)^{2}-30(r/R)^{3}\right] 2\pi r ~dr = \frac{2\pi}{A} \int_0^R \left[400r+80r^3/R^{2}-30r^4/R^{3}\right] dr\) Now, perform the integration: \(T_{m} = \frac{2\pi}{A} \left[200R^2 + \frac{20R^4}{R^2} - \frac{6R^5}{R^3}\right]\) Substitute the value of R and A that we found in step 1: \(T_m = \frac{2\pi}{\pi (0.025)^2} \left[200(0.025)^2 + \frac{20(0.025)^4}{(0.025)^2} - \frac{6(0.025)^5}{(0.025)^3}\right] = 420 \mathrm{~K}\) The mean temperature of the fluid flow is 420 K. In conclusion, the average velocity of the fluid flow in the tube is 1.2 m/s, and the mean temperature is 420 K.

Key concepts

Velocity Profile

The velocity profile of a fluid flow in a tube describes how the fluid's velocity changes across the tube's cross-section. In this scenario, the velocity profile is given by:

• \[u(r) = 0.05(1-(r/R)^2)\] in m/s,

where \(r\) is the radial distance from the center of the tube and \(R\) is the radius of the tube.
The velocity is highest at the center of the tube where \(r=0\), leading to \(u = 0.05 \, ext{m/s}\). As \(r\) approaches \(R\) (the wall of the tube), the velocity decreases to zero due to the no-slip boundary condition,
which arises because fluid molecules in contact with the tube wall have zero velocity due to friction. This parabolic profile is commonly seen in laminar flow, where fluid moves in parallel layers with no disruption between them.

Temperature Profile

The temperature profile indicates how temperature changes across the tube's cross-section, similar to the velocity profile but for temperature differences within the fluid. The provided temperature profile is described by:

• \[T(r) = 400 + 80(r/R)^2 - 30(r/R)^3\] in Kelvin,

again with \(r\) as the radial position and \(R\) the tube's radius.
Here, the temperature at the center (\(r=0\)) starts at 400 K. As \(r\) increases towards the wall,
the temperature first increases, making a peak because of the quadratic term, and then decreases due to the cubic term. This representation reflects thermal gradients that might be observed when heat is conducted or convected across the flow. Understanding the temperature profile is essential for applications involving heat transfer as it affects the heat exchange efficiency of the system.

Average Velocity

In fluid dynamics, the average velocity is a key concept, representing the flow rate per unit area through a section. To compute the average velocity in the tube, you integrate the velocity profile across the entire cross-sectional area of the tube:

• \[U = \frac{1}{A}\int_0^R 0.05 \,igg(1-(r/R)^2\bigg)2\pi r \, ext{d}r\]

For this scenario, we are working with the radius \(R = 0.025 \, ext{m}\), giving us a cross-sectional area \(A = \pi R^2\). This results in the calculated average velocity of 1.2 m/s. This measure is crucial because it gives a single representative velocity in the tube that can be used for various engineering calculations, such as determining the mass flow rate or designing systems for fluid transport.

Mean Temperature

The mean, or average, temperature in the context of fluid flow in tubes indicates a representative value of temperature across the flow cross-section. This measure is calculated by integrating the temperature profile similarly to how the average velocity is computed:

• \[T_{m} = \frac{1}{A}\int_0^R \left(400 + 80(r/R)^2 - 30(r/R)^3\right)2\pi r \, ext{d}r\]

Applying it using the tube radius and our determined cross-sectional area yields a mean temperature of 420 K.
This average is significant in scenarios like heat exchangers or systems where maintaining a uniform temperature is critical for efficiency.
Understanding the mean temperature allows engineers to predict temperature-related effects on fluid properties, enabling optimized design that ensures proper function and reliability.