Question

A smooth pipe of uniform diameter 25 cm, a pressure of 50 KPa was observed atsection 1 which has an elevation of 10 m. At another section 2, at an elevation of 12m, the pressure was 20 KPa and the velocity was 1.25 m/s. Determine the directionof flow and the head loss between the two sections. The fluid in the pipe is water.​

Answer 1

Given:-

Pressure at section 1, $P_{1}$=50kPa

Elevation at section 1, $z_{1}$=10m

Pressure at section 2, $P_{2}$ =20kPa

Elevation at section 2, $z_{2}$ =12m

velocity, v=1.25m/s

Fluid is water => density, $\rho$=1000kg/$m^{3}$

To Find:-

Direction of flow and head loss.

Explanation:-

$\text{From Bernoulli's equation:-}\\\\\text{Total head at section 1},E_{1}=\frac{P_{1}}{\rho g}+\frac{v^{2}}{2g}+z_{1}\\\\E_{1}=\frac{50\times10^{3}}{10^{3}\times9.81}+\frac{1.25^{2}}{2\times9.81}+10\\\\E_{1}=15.176m\\\\\text{Now,}\\\\\text{Total head at section 2, }E_{2}=\frac{P_{2}}{\rho g}+\frac{v^{2}}{2g}+z_{2}\\\\E_{2}=\frac{20\times10^{3}}{10^{3}\times9.81}+\frac{1.25^{2}}{2\times9.81}+12\\\\E_{2}=14.118m$

We know that fluid goes from higher energy level to lower energy level and $E_{1}>E_{2}$. Therefore, fluid will go from section 1 to section 2.

Now,

$\text{Head loss,} h_{L}=E_{1}-E_{2}\\\\h_{L}=15.176-14.118=1.058m$