Question

If the rate of flow of glycerine of density $1.25\times {10}^{3}kg/m^3$ through the conical section of a pipe is $x\times {10}^{-5}{m}^3/s$ and the radii of its ends are $0.1m$ and $0.04m$ and the pressure drop across its length is $10N/m^2$. Find $x:$

Answer 1

Applying Bernoulli's principle on a horizontal pipe, we have

$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$

And $Q=A_1{v}_1=A_2{v}_2\Rightarrow v_1=\frac{Q}{A_1}$ & $v_2=\frac{Q}{A_2}$

$\Rightarrow P_1+\frac{1}{2}\rho {\left(\frac{Q}{A_1}\right)}^2=P_2+\frac{1}{2}\rho {\left(\frac{Q}{A_2}\right)}^2$

$\Rightarrow Q=A_1{A}_2\sqrt{\frac{2(P_1-P_2)}{\rho (A_1^2-A_2^2)}}$

$\Rightarrow Q=\pi (0.1{)}^2\pi \times (0.04{)}^2\times \sqrt{\frac{2\times 10}{1.25\times {10}^3\times \left[\right(\pi (0.1{)}^2{)}^2-(\pi \left(0.04{)}^2{)}^2\right]}}=64\times {10}^{-5}{m}^3/s$