Question

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

• A. $6.43\times {10}^{-4}{m}^3/s$
• B. $5.43\times {10}^{-4}{m}^3/s$
• C. $5.44\times {10}^{-3}{m}^3/s$
• D. $6.43\times {10}^{-3}{m}^3/s$

Answer 1

The correct option is A $6.43\times {10}^{-4}{m}^3/s$

From continuity equation,$A_1{V}_1=A_2{V}_2$

$\Rightarrow \frac{A_2}{A_1}=\frac{V_1}{V_2}=\frac{\pi r_1^2}{\pi r_2^2}={\left(\frac{0.04}{0.1}\right)}^2=\frac{4}{25}$ ....$\left(1\right)$

From Bernoulli's equation,

$p_1+\frac{1}{2}\rho V_1^2=p_2+\frac{1}{2}\rho V_2^2$

$V_2^2-V_1^2=\frac{2(p_1-p_2)}{\rho }=\frac{2\times 10}{1.25\times {10}^3}=1.6\times {10}^{-2}{m}^2{s}^{-2}$ .......$\left(2\right)$

Solving eqns$\left(1\right)$ and $\left(2\right)$ we get

$\frac{V_1}{V_2}=\frac{4}{25}\Rightarrow V_1=\frac{4}{25}{V}_2$

And $V_2^2-V_1^2=1.6\times {10}^{-2}{m}^2{s}^{-2}$

$\Rightarrow V_2^2-{\left(\frac{4}{25}{V}_2\right)}^2=1.6\times {10}^{-2}{m}^2{s}^{-2}$

$\Rightarrow (1-\frac{16}{625})V_2^2=1.6\times {10}^{-2}{m}^2{s}^{-2}$

$\Rightarrow \left(\frac{625-16}{625}\right)V_2^2=1.6\times {10}^{-2}{m}^2{s}^{-2}$

$\Rightarrow V_2=\sqrt{\frac{1.6\times {10}^{-2}\times 625}{609}}=0.128m/s$

Rate of volume flow through the tube is $Q=A_2{V}_2=\pi r_2^2{V}_2$

$=\frac{22}{7}\times {\left(0.04\right)}^2\times 0.128=6.43\times {10}^{-4}{m}^3{s}^{-1}$