Question

Calculate the mass flow rate of glycerin of density $1.25\times {10}^3{\text{kg/m}}^3$ through the conical section of a horizontal pipe, if the radii of its ends are $0.1\text{m}$ and $0.04\text{m}$ and pressure drop across its length is $10{\text{N/m}}^2$.

• A. $0.96\text{kg/s}$
• B. $0.785\text{kg/s}$
• C. $0.6\text{kg/s}$
• D. $0.34\text{kg/s}$

Answer 1

The correct option is B $0.785\text{kg/s}$

Applying equation of continuity at both ends of the pipe,
$\frac{v_2}{v_1}=\frac{A_1}{A_2}=\frac{r_1^2}{r_2^2}=\frac{(0.1{)}^2}{(0.04{)}^2}=\frac{25}{4}$
Applying Bernoulli's equation for the horizontal tube $(h=\text{constant})$,
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
or, $v_2^2-v_1^2=2\left(\frac{P_1-P_2}{\rho }\right)$
$\Rightarrow v_2^2-v_1^2=\frac{2\times 10}{1.25\times {10}^3}=16\times {10}^{-3}$
Since, $v_2=\frac{25}{4}{v}_1=6.25v_1$
$[{\left(6.25\right)}^2-1^2]v_1^2=16\times {10}^{-3}$
$\Rightarrow v_1\approx 0.0205\text{m/s}$
The rate of volume flow
$Q=A_1{v}_1=\pi (0.1{)}^2\times (0.02)=6.28\times {10}^{-4}{\text{m}}^3/\text{s}$
The rate of mass flow is $\frac{dm}{dt}=\rho Av=\rho Q$
$\Rightarrow \frac{dm}{dt}=(1.25\times {10}^3)\times (6.28\times {10}^{-4})$
$\therefore \frac{dm}{dt}=0.785\text{kg/s}$