A liquid moves in a smoothh tube of uniform cross section. Find the pressure `P_(2)` at point `2` of the tube if the pressure at point `1` of the extended tube is `P_(1)` and the vertical distance between points `1` and `2` is `h`. brgt

Explanation

Since the cross section of the tube is uniform.
`A_(1)=A_(2)=A`
Then applying equation of continuity at points `1` and `2`, we have
`v_(1)=v_(2)`……i
Substuting `v_(1)=v_(2)(=v)` in Bernoulli's equation we have
`P_(1)+1/2rhov^(2)+rhogh_(1)=P_(2)+1/2rhov^(2)+rhogh_(2)`
This gives `P_(1)+rhogh_(1)+P_(2)+rhogh_(2)`
Since `h_(1)-h_(2)=h`, we have `P_(2)=P_(1)+rhogh`