Example Problem with Complete Solution

 1E-1 : Pressure Measurement Using a Multi-Fluid Manometer 6 pts The water in a tank is pressurized by air, and the pressure is measured by a multi-fluid manometer, as shown in the figure. Determine the gage pressure of air in the tank if h1 = 0.2 m, h2 = 0.3 m and h3 = 0.46 m. Take the densities of water, oil and mercury to be 1000 kg/m3, 850 kg/m3 and 13,600 kg/m3, respectively. 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 0.20 m ρw 1000 kg/m3 h2 0.30 m ρoil 850 kg/m3 h3 0.46 m ρHg 13600 kg/m3 P2 101.325 kPa Find : P1,gage ??? Solution : 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. Some key observations are: Eqn 4 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 11 yields : g 9.8066 m/s2 P1,gage 56888 Pa gage gC 1 kg-m/N-s2 P1,gage 56.89 kPa gage Answers : P1,gage 56.9 kPa gage If you are curious : P1 158.21 kPa P2 101.325 kPa PA = PB 162.68 kPa PC = PD = PE 160.17 kPa