# Example Problem with Complete Solution

1E-1 : Pressure Measurement Using a Multi-Fluid Manometer 6 pts
A pressurized vessel contains water with some air above it, as shown below. A multi-fluid manometer system is used to determine the pressure at the air-water interface, point F. Determine the gage pressure at point F in kPa gage. Data:
h1 = 0.24 m, h2 = 0.35 m and h3 = 0.52 m
Assume the fluid
densities are water: 1000 kg/m3, oil: 790 kg/m3 and mercury(Hg): 13,600 kg/m3.

Read: Use the barometer equation to work your way through the different fluids from point 1 to point 2.
Remember that gage pressure is the difference between the absolute pressure and atmospheric pressure.
Given: h1 0.24 m rw 1000 kg/m3
h2 0.35 m roil 790 kg/m3
h3 0.52 m rHg 13600 kg/m3
P2 101.325 kPa
Find: P1,gauge ??? kPa gauge
Assumptions: 1- The fluids in the system are completely static.
2- The densities of the liquids are uniform and constant.
3- The acceleration of gravity is: g 9.8066 m/s2
gC 1 kg-m/N-s2
Equations / Data / Solve:
Gage pressure is defined by : Eqn 1
If we assume that P2 is atmospheric pressure, then Eqn 1 becomes : Eqn 2
The key equation is the Barometer Equation : Eqn 3
Now, apply Eqn 1 repeatedly to work our way from point 1 to point 2. Eqn 4
Some key observations are: Eqn 5
These are true because the points are connected by open tubing, the fluid is not flowing in this system and no change in the composition of the fluid occurs between A & B or C & D or D & E.
PA > P2, therefore : Eqn 6
PE > P1, therefore : Eqn 7
PB > PC, therefore : Eqn 8
Combine Eqns 2, 5 & 6 to get : Eqn 9
Use Eqns 3 & 5 to eliminate PC from Eqn 7 : Eqn 10
Now, solve for P1 - P2 : Eqn 11
Combining Eqns 10 & 2 yields : Eqn 12
Plugging values into Eqn 12 yields : P1,gage 64287 Pa gage
P1,gage 64.29 kPa gage
Answers: P1,gage 64.3 kPa gage
If you are curious : P1 165.61 kPa PA = PB 170.68 kPa
P2 101.325 kPa PC = PD = PE 167.97 kPa 