Question

Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600 $N/m^2$ between x and y where the areas of cross - section are 3 $c{m}^2$ and 1.5$c{m}^2$ respectively. Find the rate of flow of water through the tube.

• A. $189c{m}^3/s$
  
• B. $159c{m}^3/s$
• C. $189c{m}^2/s$
• D. $159c{m}^2/s$

Answer 1

The correct option is A $189c{m}^3/s$

Let the velocity at x = $v_x$ and that at y = $v_y$.
By the equation of continuity, $\frac{v_y}{v_x}=\frac{3c{m}^2}{1.5c{m}^2}=2$
By Bernoulli's equation,
$P_x+\frac{1}{2}p{v}_x^2=P_y+\frac{1}{2}p{v}_y^2$
or,$P_x-O_y=\frac{1}{2}p(2v_y{)}^2-\frac{1}{2}p{v}_y^2=\frac{3}{2}p{v}_y^2$
or, $600\frac{N}{m^2}=\frac{3}{2}\left(1000\frac{kg}{m^3}\right)v_x^2$
or, $v_x=\sqrt{0.4m^2/s^2}=0.63m/s$
The rate of flow $=\left(3c{m}^2\right)\left(0.63m/s\right)=189c{m}^3/s$.