Question

The rate of flow of glycerene (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 $10N/m^2$, is

• A. $5.28\times {10}^{-4}{m}^3/s$
• B. $6.28\times {10}^{-4}{m}^3/s$
• C. $6.28\times {10}^4{m}^3/s$
• D. $5.28\times {10}^4{m}^3/s$

Answer 1

The correct option is B $6.28\times {10}^{-4}{m}^3/s$

Density of glycerine$=1.25\times {10}^{3}kg/m^3$

Pressure drop across its length$=10N/m^2$

Applying bernoullis eqn across 1 and 2,

$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2$

$p_1-p_2=\frac{1}{2}\rho (v_2^2-v_1^2)$

$10N/m^2=\frac{1}{2}\times 1.25\times {10}^3\left(v_2^2(1-{\left(\frac{0.04}{0.1}\right)}^4)\right)$

$\Rightarrow \frac{20\times {10}^{-3}}{1.25\times 0.9\times 744}=v_2^2\Rightarrow v_2^2=0.016$

$\Rightarrow v_2=0.12814m/s$

So,

Rate of flow$=\pi r^2{v}_2=A_2{v}_2=\pi \times {\left(0.04\right)}^2\times 0.12814m^3/s$

$=6.44\times {10}^{-4}{m}^3/s$

$\approx 6.28\times {10}^{-4}{m}^3/s$