U-Tube Manometer: What is the Difference in Pressure Between Points B and A?

[Doubell]

Homework Statement

An inverted U-tube monometer, as shown in Figure attached , has air at the top of the tube.
If the pipes contain oil (s.g. = 0.9), h1= 0.6 m, h2= 1.8 m and h= 0.45 m, determine
the difference in pressure between point B and point A.

Homework Equations

The Attempt at a Solution

for my analysis i dedicded to use the equation defining the pressures from point a to be as :
Pa - (900kg/m³*9.81m/s²*0.6m)- (101325Pa) +((0.6m+1.8m)*9.81m/²*900kg/m³)) = Pb =
which implied (Pa-Pb) = (900kg/m³*9.81m/s²*0.6m) + 101325Pa - ((0.6m+1.8m)*9.81m/²*900kg/m³)) = 133109.4Pa = 133Kpa. [/B]

the thing about the solution is i think atmospheric pressure of 101325Pa can be used as 0 gauge. which would define the equation as Pa - (900kg/m³*9.81m/s²*0.6m) +((0.6m+1.8m)*9.81m/²*900kg/m³)) = Pb.

also Could an analysis just be made from point A to point D then this pressure just equated to the pressure from ( Pb to point D ) excluding atmospheric pressure term (101325) and the term ((0.45m)*9.81m/²*900kg/m³)) since the fluid at these points are at the same horizontal height so equal pressure.

that is Pa - (900kg/m³*9.81m/s²*0.6m)= Pb -(900kg/m³*9.81m/s²*1.8m).
verification of the right analysis would be appreciated as I am a bit confused which is the most suited approach.

 

[Chestermiller]

Homework Statement

An inverted U-tube monometer, as shown in Figure attached , has air at the top of the tube.
If the pipes contain oil (s.g. = 0.9), h1= 0.6 m, h2= 1.8 m and h= 0.45 m, determine
the difference in pressure between point B and point A.

Homework Equations

The Attempt at a Solution

for my analysis i dedicded to use the equation defining the pressures from point a to be as :
Pa - (900kg/m³*9.81m/s²*0.6m)- (101325Pa) +((0.6m+1.8m)*9.81m/²*900kg/m³)) = Pb =
which implied (Pa-Pb) = (900kg/m³*9.81m/s²*0.6m) + 101325Pa - ((0.6m+1.8m)*9.81m/²*900kg/m³)) = 133109.4Pa = 133Kpa.[/B]

Where in the problem statement does it say that the air pressure is atmospheric? It could just as easily be some other pressure. Also, your equation implies that the pressure difference between points D and E is atmospheric. This certainly wouldn't be the case. So this answer is incorrect.

the thing about the solution is i think atmospheric pressure of 101325Pa can be used as 0 gauge. which would define the equation as Pa - (900kg/m³*9.81m/s²*0.6m) +((0.6m+1.8m)*9.81m/²*900kg/m³)) = Pb.

This equation is correct, but for the wrong reason. The implication of the problem statement is that the air is much less dense than the fluid in the manometer, so it can be neglected in the hydrostatic equilibrium balance.

also Could an analysis just be made from point A to point D then this pressure just equated to the pressure from ( Pb to point D ) excluding atmospheric pressure term (101325) and the term ((0.45m)*9.81m/²*900kg/m³)) since the fluid at these points are at the same horizontal height so equal pressure.

that is Pa - (900kg/m³*9.81m/s²*0.6m)= Pb -(900kg/m³*9.81m/s²*1.8m).
verification of the right analysis would be appreciated as I am a bit confused which is the most suited approach.

This approach is incorrect because of the assumption that fluid at the same horizontal heights implies equal pressures. This is only true if the bodies of fluid are directly connected to one another.

The easiest way to do this problem is to integrate vertically on each side of the manometer, starting both from A and B, and determine the air pressure, as reckoned from each side. Then set the two values of the air pressure equal to one another.

Chet

 

[Doubell]

after re looking at the image in the problem statement i saw the indication ( air pressure , 'p'). so i guess instead of atmospheric pressure , the pressure P should be the variable in the equation. Pa - (900kg/m3*9.81m/s2*0.6m)- (P )+((0.6m+1.8m)*9.81m/2*900kg/m3)) = Pb . i would now have a variable pressure P which i think would make it difficult to obtain a numerical solution. what's ur take on this other approach.

 

[Chestermiller]

after re looking at the image in the problem statement i saw the indication ( air pressure , 'p'). so i guess instead of atmospheric pressure , the pressure P should be the variable in the equation. Pa - (900kg/m3*9.81m/s2*0.6m)- (P )+((0.6m+1.8m)*9.81m/2*900kg/m3)) = Pb . i would now have a variable pressure P which i think would make it difficult to obtain a numerical solution. what's ur take on this other approach.

As I said, because the air density is low, the pressure of the air is constant in the u tube (independent of location). Your equation implies that there is a difference in air pressure of P between points D and E.

Chet