Question

Calculate the speed (in $\text{mm/s}$) of flow of glycerin at the inlet of the conical section of a pipe. If its inlet and outlet radius are $0.2\text{m}$ and $0.05\text{m}$ respectively. The pressure drop across its length is $10{\text{N/m}}^2$ and density of glycerin is $1.25\times {10}^3{\text{kg/m}}^3$.

• A. $64.4$
• B. $32.6$
• C. $10$
• D. $2.6$

Answer 1

The correct option is B $32.6$


Let speed of flow at inlet and outlet is $v_1$ and $v_2$. From the equation of continuity,
$A_1{v}_1=A_2{v}_2$
$\Rightarrow \frac{v_1}{v_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}=\frac{r_2^2}{r_1^2}$
Inlet and outlet radius $r_1=0.2\text{m}$
$r_2=0.05\text{m}$
$\frac{v_1}{v_2}=\frac{(0.05{)}^2}{(0.2{)}^2}=(0.25{)}^2=0.0625$
$\Rightarrow v_2=16v_1....\left(1\right)$
Applying Bernoulli's equation,
$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
$\Rightarrow v_2^2-v_1^2=\frac{2}{\rho }\times (P_1-P_2)$
$\because h_1=h_2$, as tube is horizontal.
$\Rightarrow v_2^2-v_1^2=\frac{2\times 10}{1.25\times {10}^3}=\mathrm{0.016....}\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, we get
$16v_1^2-v_1^2=0.016$
$\Rightarrow v_1^2=\frac{0.016}{15}$
$\Rightarrow v_1=0.0326\text{m/s}=32.6\text{mm/s}$