Question

The section of a tube has height of 6m and it has a fluid with a density of 600 kg/m3, a

velocity of 77.1 m/s and a pressure of 900 Pa. Find the velocity at another section where

the height is 12 m, and the pressure is 400 Pa

Answer 1

In this problem, we will consider the general energy equation

$\frac{P_1}{\gamma}+ Z_1+\frac{v_1^2}{2g}=\frac{P_2}{\gamma}+ Z_2+\frac{v_2^2}{2g}$

$\frac{900}{600}+ 6+\frac{77.1^2}{2 \times 9.81}=\frac{400}{600}+ 12+\frac{v_2^2}{2 \times 9.81}$

$\frac{400}{600}+12+\frac{v^2}{2\cdot \:9.81}=\frac{900}{600}+6+\frac{77.1^2}{2\cdot \:9.81}$

$\frac{400}{600}+12+\frac{v^2}{2\cdot \:9.81}-\left(\frac{400}{600}+12\right)=\frac{900}{600}+6+\frac{77.1^2}{2\cdot \:9.81}-\left(\frac{400}{600}+12\right)$

$\frac{v^2}{2\cdot \:9.81}=\frac{3}{2}+\frac{5944.41}{19.62}-\left(\frac{2}{3}+12\right)+6$

$\frac{19.62v^2}{2\cdot \:9.81}=\frac{35058.24\cdot \:19.62}{117.72}$

$v^2=\frac{687842.6688}{117.72}$

$v=\sqrt{\frac{687842.6688}{117.72}},\:v=-\sqrt{\frac{687842.6688}{117.72}}$

$v=76.43978 , v=-76.43978$

Hence , $v_2=76.4 m/s$