Pressure Drop Calculation Using U-Tube Manometer

[Physicist3]

I have been told to calculate the pressure drop between two points in a pipe carrying water using a U-tube manometer. I understand that the pressure drop (P₁ - P₂) is given by pgh and that h is the difference in manometer fluid levels and g is gravity, but is p the density of the manometer fluid or the water flowing through the pipe?

 

[pukb]

Exactly, p1-p2 = gh(density(manometer fluid) - density(water))

 

[Physicist3]

Exactly, p1-p2 = gh(density(manometer fluid) - density(water))

In the case I am referring to, the manometer fluid has a lower density than the water. For a pipe, is the pressure drop measured using a utube manometer simply ρ_manometergh?

 

[pukb]

it is not a good idea to use a fluid of lower density in manometer than the density of fluid in the pipe. a higher density fluid is used to make measuring instruments smaller. for example, mercury will require a space 13.6 times smaller than water for same pressure rise or drop.