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 $0.1m$ and $0.04m$ 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

Answer: (A)

From Bernoulli's theorem
$p_1+\frac{1}{2}\rho v_1^2=P_2\frac{1}{2}\rho v_2^2$
$\therefore p_1-p_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$
$\therefore 10=\frac{1}{2}\times 1.25\times 10^3\left(v_2^2-v_1^2\right)$
$\therefore v_2^2-v_1^2=\frac{10\times 2}{1.25\times 10^3}=16\times 10^{-3}...$(i)
Also, from equation of continuity
$=A_1{v}_1=A_2{v}_2$
Or $\pi r_1^2{v}_1=\pi r_2^2{v}_2$
$\therefore \frac{v_1}{v_2}={\left[\frac{r_2}{r_1}\right]}^2=\frac{0.04}{0.1}=0.4$
ie, $v_1=0.4v_2....$ (ii)
Substituting this value in Eq. (i)
$v_2^2-{\left(0.4v_2\right)}^2=16\times 10^{-3}$
$\Rightarrow v_2=1.38\times 10^{-1}=0.138m{s}^{-1}$
Rate of flow of glycerine $v=A_2{v}_2$
$=\pi r_2^2{v}_2$
$=6.93\times 10^{-4}{m}^3{s}^{-1}$