Question

Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm and the areas of cross sections at A and B are 4 $c{m}^2$ and 2 $c{m}^2$ respectively, find the rate of flow of water across any section.

Answer 1

$v_A{a}_A=v_B\times a_B$

$\Rightarrow v_A\times 4v_B\times 2$

$\Rightarrow v_B=2v_A$ ...(i)

Again,

$\frac{1}{2}p{v}_A^2+\rho g{h}_A+p_A=\frac{1}{2}\rho v_B^2+\rho g{h}_B+p_B$

$\Rightarrow \frac{1}{2}\rho v_A^2+p_A=\frac{1}{2}\rho v_B^2+p_B$

$\Rightarrow p_A-p_B=\frac{1}{2}\rho (v_B^2-v_A^2)$

= $\frac{1}{2}\times 1\times (4v_B^2-v_A^2)$

$\Rightarrow 2\times 1\times 1000=\frac{1}{2}\times 1\times 3v_A^2$

$[p_A-p_B=2$ cm of water column = $2\times 1\times 1000\frac{\text{dyne}}{c{m}^2}]$

$\Rightarrow v_A^2=\sqrt{\frac{4000}{3}}=36.51$ cm/sec

So, Rate of flow = $V_a{a}_A=36.51\times 4$

= 146 $c{m}^3$ /sec