Chapter 7, Problem 7.1P

To measure the mass flow rate of a fluid in a laminar flow through a circular pipe, a hot-wire-type velocity meter is placed in the center of the pipe. Assuming that the measuring station is far from the entrance of the pipe, the velocity distribution is parabolic:

$u(r)/U_{\max }=[1-{(2r/D)}^2]$

where $U_{\text{max}}$ is the centerline velocity (r = 0), r is the radial distance from the pipe centerline, and D is the pipe diameter.

1. Derive an expression for the average fluid velocity at the cross section in terms of $U_{\text{max}}$ and D. (b) Obtain an expression for the mass flow rate. (c) If the fluid is mercury at $\text{30°C}$, D = 10 cm, and the measured value of $U_{\text{max}}$ is 0.2 cm/s, calculate the mass flow rate from the measurement.

To determine

Expression for average velocity of the fluid.

Answer to Problem 7.1P

Average velocity of the fluid is $\overline{\text{u}}=\frac{2}{3}{\text{U}}_{\text{max}}$

Explanation of Solution

Given information:

Velocity distribution is parabolic, $\frac{\text{u}\left(\text{r}\right)}{{\text{U}}_{\text{max}}}=\left[1-{\left(\frac{2\text{r}}{\text{D}}\right)}^2\right]$

Let ${\text{r}}_{\text{o}}$ be radius of the pipe.

$\text{D}=2{\text{r}}_{\text{o}}$

$\frac{\text{u}\left(\text{r}\right)}{{\text{U}}_{\text{max}}}=\left[1-{\left(\frac{\text{r}}{{\text{r}}_{\text{o}}}\right)}^2\right]$

Average velocity of the fluid is calculated as follows,

$\text{Average velocity }\left(\overline{\text{u}}\right)=\frac{1}{{\text{r}}_{\text{o}}}\underset{\text{r}=0}{\overset{\text{r}={\text{r}}_{\text{o}}}{{\displaystyle \int }}}\text{u}\left(\text{r}\right)\text{dr}$

$\left(\overline{\text{u}}\right)=\frac{1}{{\text{r}}_{\text{o}}}\underset{\text{r}=0}{\overset{\text{r}={\text{r}}_{\text{o}}}{{\displaystyle \int }}}{\text{U}}_{\text{max}}\left(1-{\left(\frac{\text{r}}{{\text{r}}_{\text{o}}}\right)}^2\right)\text{dr}$

$\overline{\text{u}}=\frac{{\text{U}}_{\text{max}}}{{\text{r}}_{\text{o}}}\left[{\text{r}}_{\text{o}}-\frac{{\text{r}}_{\text{o}}^3}{3{\text{r}}_{\text{o}}^2}\right]$

$\overline{\text{u}}={\text{U}}_{\text{max}}\left[1-\frac{1}{3}\right]=\frac{2}{3}{\text{U}}_{\text{max}}$

$\overline{\text{u}}=\frac{2}{3}{\text{U}}_{\text{max}}$

$\text{Average fluid velocity is }\overline{\text{u}}=\frac{2}{3}{\text{U}}_{\text{max}}$

To determine

Expression for mass flow rate.

Answer to Problem 7.1P

$\text{Mass flow rate of the fluid is }\dot{\text{m}}=\frac{2}{3}{\text{πρr}}_{\text{o}}^2{\text{U}}_{\text{max}}$

Explanation of Solution

Mass flow rate $\left(\dot{\text{m}}\right)$

$\dot{\text{m}}=\text{ρA}\overline{\text{u}}$

$={\text{ρπr}}_{\text{o}}^2\left(\frac{2}{3}{\text{U}}_{\text{max}}\right)$

$\dot{\text{m}}=\frac{2}{3}{\text{πρr}}_{\text{o}}^2{\text{ U}}_{\text{max}}$

$\text{ρ}=\text{denisty of fluid }$

$\text{mass flow rate of the fluid is }\dot{\text{m}}\frac{2}{3}{\text{πρr}}_{\text{o}}^2{\text{U}}_{\text{max}}$

To determine

Mass flow rate of mercury.

Answer to Problem 7.1P

Mass flow rate of mercury is 0.14185 kg/s.

Explanation of Solution

Given information:

Diameter of tube $\left(\text{D}\right)=10\text{cm}=0.1\text{m}=2{\text{r}}_{\text{o}}$

${\text{U}}_{\text{max}}=0.2\text{cm}/\text{s}=0.002\text{m}/\text{s}$

From Appendix-2, Table 12, density of mercury $\left(\text{ρ}\right)=13546\text{ kg}/{\text{m}}^3$

Mas flow rate of mercury $\left(\dot{\text{m}}\right)$

$\dot{\text{m}}=\frac{2}{3}\text{π}\left(13546\right){\left(\frac{0.1}{2}\right)}^2\left(0.002\right)$

$\dot{\text{m}}=0.14185\text{ kg}/\text{s}$

Mass flow rate of mercury is 0.14185 kg/s.