1. Working Under Pressure?
An Examination of a Two Phase System

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

3. Hydrostatics Hydrostatics is the study of motionless fluid.
Important parameters: - Density Geometry Pressure
P2 - P1 = rgh

4. Pressure Versus Depth P0 P1

5. 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
rH20 = 62.4 lbm / ft3.

6. Initial Setup D d1 = 8 ft ; d2 = 5 ft Gradient of pressure: P = rg
Law of hydrostatics for an incompressible fluid:
Pb = Pa + rgh D

7. Where Do We Start? = r(gx i + gy j + gz k) dP dx i + dy j dz k
For this situation: dP dx dz =
and:
gx = gz = 0
Separate/Integrate: dP dy = rg

8. Final Equation
dP = rg dy
Pb – Pa = rg(yb – ya)
Pb = Pa + rg(yb – ya)

9. Physical Description

10. Calculations roil = (s.g. oil)(rwater) = (.92)(62.4 lbm/ft3)
roil = 57.4 lbm/ft3
(roil)(g) = 57.4 lbf/ft3
Pinterface = 14.7 lbf lbf (8 ft – 0 ft) (1 ft2)
in2
ft3
(144 in2)
= 17.9 lbf
Pbottom = 17.9 lbf lbf (13 ft – 8 ft) (1 ft2)
(144 in2)
in2
ft3
= 20.1 lbf

11. Solutions Pinterface = 17.9 lbf in2 Pbottom = 20.1 lbf *
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.

12. Other Situations
Compare to Oil/Water Pbottom = 20.1 lbf
in2

13. Situations Involving the Hydrostatic Equation
Useful Applications
Situations Involving the Hydrostatic Equation

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

15. Mercury Barometer Po = 0 P = Po + (rgh) r = 13.6 E3 kg/m3 76 cm
g = 9.80 m/s2
P =
1 atm
h = .760 m
Mercury

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