Question

Water flows in a horizontal tube as shown in figure. The pressure of water changes by $600N/m^2$ between $A$ and $B$ where the area of cross-sectional are $30c{m}^2$ and $15c{m}^2$ respectively. Find the rate of flow of water through the tube.

Answer 1

Let velocity at $A$ & $B$ are $V_A$ and $V_B$ respectively.

Using equation of continuity,

$\frac{V_B}{V_A}=\frac{30}{15}=2$

Using Bernoulli equation

$P_A+\frac{1}{2}p{V}_A^2=P_B+\frac{1}{2}p{V}_B^2$

$P_A-P_B=\frac{1}{2}\rho \left(2V_A^2\right)-\frac{1}{2}\rho V_A^2$

$=\frac{3}{2}\rho V_A^2$

$600=\frac{3}{2}\times 1000V_A^2$

$V_A=0.63=m/s=63cm/s$

Rate of flow $=A\times V_{A}c{m}^3/s$

$\therefore$ Rate of flow $=300\times 63c{m}^3/s$

$=18900c{m}^3{s}^{-1}$