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
Text Solution
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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`
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