Chapter 10, Problem 10.1P

To determine

Mass flow rate

Answer to Problem 10.1P

Mass flow rate $m=4.5*?10{?}^{(-5)}kg/s=0.16kg/hr$

Explanation of Solution

Given:

Air flow through a pipe

Velocity profile: $U(r)=25[1-{(r/r_1)}^2]cm/s$

p = 1 bar abs = 100,000 N/m²abs

T = 5^oC = 278 K

d1 = 2*r1 = 5 cm

Concept Used:

This for steady, incompressible, axis-symmetric flow becomes:

Mass flow rate through a pipe:

  $\dot{m}=\underset{0}{\overset{2\pi }{{\displaystyle \int }}}\underset{0}{\overset{r_1}{{\displaystyle \int }}}\rho U\left(r\right)rdrd\theta$

Calculation:

As per the given problem

For a perfect gas, the density obtained by the relation:

  $\rho =\frac{p}{RT}=\frac{100000}{287*278}=1.16 kg/m^3$

We know that, Mass flow rate through pipe:

  $\dot{m}=\underset{0}{\overset{2\pi }{{\displaystyle \int }}}\underset{0}{\overset{r_1}{{\displaystyle \int }}}\rho U\left(r\right)rdrd\theta$

  $\dot{m}=\underset{0}{\overset{2\pi }{{\displaystyle \int }}}\underset{0}{\overset{r_1}{{\displaystyle \int }}}\rho 25\left[1-{\left(\frac{r}{r_1}\right)}^2\right]rdrd\theta$

  $\dot{m}=39.3\rho \frac{c{m}^3}{s}$

Putting the value of density to calculate mass flow rate:

  $\dot{m}=39.3*1.16\frac{kg}{m^3}* \frac{c{m}^3}{s}$

  $\dot{m}=4.5*{10}^{-5} \frac{kg}{s}=0.16 kg/hr$