Question

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

• A. $6.93\times {10}^{-4}{m}^3{s}^{-1}$
• B. $7.8\times {10}^{-4}{m}^3{s}^{-1}$
• C. $10.4\times {10}^{-5}{m}^3{s}^{-1}$
• D. $14.5\times {10}^{-5}{m}^3{s}^{-1}$

Answer 1

The correct option is A $6.93\times {10}^{-4}{m}^3{s}^{-1}$

From Bernoulli's theorem
$p_1+\frac{1}{2}p{v}_1^2=p_2+\frac{1}{2}p{v}_2^2$
$\therefore p_1-p_2=\frac{1}{2}p(v_2^2-v_1^2)$
$\therefore 10=\frac{1}{2}\times 1.25\times {10}^3(v_2^2-v_1^2)$
$\therefore v_2^2-v_1^2=\frac{10\times 2}{1.25\times {10}^3}=16\times {10}^{-3}$ ..........(i)
Also from equation of continuity
$=A_1{v}_1=A_2{v}_2$
$\pi r_1^2{v}_1=\pi r_2^2{v}_2$
$\therefore$ $\frac{v_1}{v_2}={\left[\frac{r_2}{r_1}\right]}^2=\frac{0.04}{0.1}=0.4$
$v_1=0.4v_2$ .............(ii)
Substituting this value in Eq. (i)
$v_2^2-(0.4v_2{)}^2=16\times {10}^{-3}$
$v_2=1.38\times {10}^{-1}=0.138m{s}^{-1}$
Rate of flow of glycrine v$=A_2{v}_2$
$=\pi r_2^2{v}_2$
$=6.93\times {10}^{-4}{m}^3{s}^{-1}$.