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Chapter 8: Problem 19

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

Expert verified
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

01

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, \(A = \pi R^2 = \pi(0.025)^2 \mathrm{~m^2}\)
02

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: \(U = \frac{1}{A}\int_0^R 0.05\left[\left(1-(r/R)^{2}\right)\right] 2\pi r ~dr\)
03

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.
04

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: \(T_{m} = \frac{1}{A}\int_0^R \left[400+80(r/R)^{2}-30(r/R)^{3}\right] 2\pi r ~dr\)
05

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.

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Most popular questions from this chapter

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