Working Under Pressure An Examination of a Two

Working Under Pressure? An Examination of a Two Phase System

Agenda n Introduction/Theory - Arbab Khan n Problem Statement - Jeremy Jones n Solution Procedure - Alex Ibañez n Results/Applications - Steve Merchant

Hydrostatics n Hydrostatics is the study of motionless fluid. n Important parameters: n P 2 - P 1 = rgh - Density - Geometry - Pressure

Pressure Versus Depth P 0 P 1

Initial Setup § An open tank contains water and an immiscible oil. § The oil has a specific gravity of 0. 92. § The density of water is given as r. H 20 = 62. 4 lbm / ft 3.

Initial Setup § d 1 = 8 ft ; d 2 = 5 ft § Gradient of pressure: § Law of hydrostatics for an incompressible fluid: P = rg Pb = Pa + rgh D

Where Do We Start? d. P i + d. P j + d. P k = r(g i + g j + g k) x y z dx dy dz d. P = 0 For this situation: dx dz and: gx = gz = 0 Separate/Integrate: d. P dy = rg

Final Equation d. P = rg dy Pb – Pa = rg(yb – ya) Pb = Pa + rg(yb – ya)

Physical Description

Calculations roil = (s. g. oil)(rwater) = (. 92)(62. 4 lbm/ft 3) roil = 57. 4 lbm/ft 3 (roil)(g) = 57. 4 lbf/ft 3 Pinterface = 14. 7 lbf + 57. 4 lbf (8 ft – 0 ft) (1 ft 2) = 17. 9 lbf (144 in 2) in 2 ft 3 in 2 Pbottom = 17. 9 lbf + 62. 4 lbf (13 ft – 8 ft) (1 ft 2) = 20. 1 lbf (144 in 2) in 2 ft 3 in 2

Solutions Pinterface = 17. 9 lbf in 2 Pbottom = 20. 1 lbf in 2 * We cannot use the hydrostatic equation to solve directly for the pressure at the bottom of the tank because the density changes in the tank at the interface.

Other Situations Compare to Oil/Water Pbottom = 20. 1 lbf in 2

Useful Applications Situations Involving the Hydrostatic Equation

Open-tube Manometer Po Atmospheric pressure P (measured pressure) h P = Po + (rgh)

Mercury Barometer Po = 0 P = Po + (rgh) r = 13. 6 E 3 kg/m 3 P = 1 atm 76 cm g = 9. 80 m/s 2 h =. 760 m Mercury

What We Have Learned n n n Derivation of the Basic Hydrostatic Equation Use of Basic Hydrostatic Equation to find pressure at interface Other applications of Basic Hydrostatic Equation