Question

Water flows in a horizontal tube. The pressure of water changes by 600 N m^-2between A and B where the areas of cross section are 30 cm² and 15 cm² respectively. Find the rate of flow of water through the tube. Given $\sqrt{0.4}$= 0.63 approx.

• A. 1890 m³/s
• B. 2260 cm³/s
• C. 1890 cm³/s
• D. 1664 cm³/s

Answer 1

The correct option is C

1890 cm³/s

Let the velocity at A=$v_A$ and that at B=$v_B$
By the equation of continuity,$\frac{v_B}{v_A}=\frac{A_A}{A_B}=\frac{30c{m}^2}{15c{m}^2}=2$
By Bernoulli equation,
$P_A+\frac{1}{2}\stackrel{,}{\rho }{v}_2^2=P_B+\frac{1}{2}\stackrel{,}{\rho }{v}_B^2$
or,$P_A-P_B=\frac{1}{2}\stackrel{,}{\rho }(v_B^2-v_A^2)=\frac{3}{2}\stackrel{,}{\rho }{v}_A^2,$
or,$600N{m}^{-2}=\frac{3}{2}\left(1000kgm{s}^{-3}\right)v_A^2$
or,$v_A=\sqrt{0.4m^2{s}^{-2}}=0.63m{s}^{-1}$
The rate of flow = Av = $\left(30c{m}^2\right)$ $\left(0.063m{s}^{-1}\right)$=1890 $c{m}^3{s}^{-1}$