Physics - Fluid Question with Solution | TestHub

Using the equation of continuity, $A_1{v}_1=A_2{v}_2=Q$, where $Q$ is the rate of flow.

$v_1=\frac{Q}{A_1}$ and $v_2=\frac{Q}{A_2}$

Using Bernoulli's theorem for a horizontal pipe:

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

$P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P=\frac{1}{2}\rho Q^2\left(\frac{1}{A_2^2}-\frac{1}{A_1^2}\right)$

Given:

$\Delta P=3{\text{ N/m}}^2$

$\rho =1.25\times 10^3{\text{ kg/m}}^3$

$A_1=10{\text{ cm}}^2=10^{-3}{\text{ m}}^2$

$A_2=5{\text{ cm}}^2=5\times 10^{-4}{\text{ m}}^2$

Substituting the values:

$3=\frac{1}{2}\times 1.25\times 10^3\times Q^2\left(\frac{1}{(5\times 10^{-4}{)}^2}-\frac{1}{(10^{-3}{)}^2}\right)$

$3=\frac{1}{2}\times 1.25\times 10^3\times Q^2\left(\frac{1}{25\times 10^{-8}}-\frac{1}{10^{-6}}\right)$

$3=\frac{1}{2}\times 1.25\times 10^3\times Q^2(4\times 10^6-1\times 10^6)$

$3=\frac{1}{2}\times 1.25\times 10^3\times Q^2\times 3\times 10^6$

$1=\frac{1}{2}\times 1.25\times 10^9\times Q^2$

$Q^2=\frac{2}{1.25\times 10^9}=\frac{1.6}{10^9}=16\times 10^{-10}$

$Q=4\times 10^{-5}{\text{ m}}^3/\text{s}$

Answer: $4\times 10^{-5}{\mathrm{m}}^3/\mathrm{s}$