Question

Find 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 is $10N/m^2$.

• A. $3.14\times {10}^{-4}{m}^3/s$
• B. $6.28\times {10}^{-4}{m}^3/s$
• C. $12.56\times {10}^{-4}{m}^3/s$
• D. $1.57\times {10}^{-4}{m}^3/s$

Answer 1

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

Here $(\frac{v_2}{v_1}{)}^2=(\frac{A_1}{A_2}{)}^2=(\frac{0.1}{0.04}{)}^2=6.25$ ..........(i)

Also, $P_1+1/2\rho v_1^2=P_2+1/2\rho v_2^2$

Hence, $v_2^2-v_1^2=\frac{2(P_1-P_2)}{\rho }=\frac{2\times 10}{1.25\times {10}^3}=16\times {10}^{-3}$ .....(2)

Now using eq. (i) and (ii)

$6.25v_1^2-v_1^2=16\times {10}^{-3}\Rightarrow v_1\equiv 0.0205m/s$

So rate of flow $R=A_1{v}_1=A_2{v}_2$

$=\pi {\left(0.1\right)}^2\times 0.02=6.28\times {10}^{-4}{m}^3/s$