U tube manometer


by Physicist3
Tags: manometer, tube
Physicist3
Physicist3 is offline
#1
Nov27-12, 11:10 AM
P: 72
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?
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pukb
pukb is offline
#2
Nov28-12, 11:10 AM
P: 90
Exactly, p1-p2 = gh(density(manometer fluid) - density(water))
Physicist3
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#3
Nov29-12, 09:36 PM
P: 72
Quote Quote by pukb View Post
Exactly, p1-p2 = gh(density(manometer fluid) - density(water))
In the case im 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 ρmanometergh?

pukb
pukb is offline
#4
Nov29-12, 11:03 PM
P: 90

U tube manometer


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.


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