Chapter 2

Pressure & Fluid Statics

STATIC FLUIDS

Static fluids

Objectives &

Learning outcomes

PressurePressure

Measurement

Pressure

Pascal’s Law

Variation of Pressure

in a static fluids

Pressure & Head

Barometer

Piezometer

Manometer

(u-tube)

OBJECTIVES- Introduce the concept of pressure

- Prove it has a unique value at any particular elevation

- Show it varies with depth according to the hydrostatic equation

can be expressed in terms of head of fluid

LEARNING OUTCOMES- Understand the concept of pressure

- Able to derive an equation for variation of pressure

- Solve the simple problems related to hydrostatic

characteristics such as pressure, pressure distributions,

pressure measurements, hydrostatic force on a plane

and curved surface immersed in a liquid

STATIC OF FLUID SYSTEMS

Static of fluid = Fluid in a rest condition

• No shearing occurs

• In equilibrium conditions

• Moment at any point is zero

• All forces exerted between the fluid and solid boundary must

act at right angle to the boundary

PRESSURE• Pressure is defined as a normal force exerted by a fluid per unit

area.

• Pressure only deal with fluid and gas. The counterpart of pressure

in solids is normal stress.

PressureappliedisforcethewhichoverArea

Force

A

Fp

• Units: Newton's per square meter (N/m2) which call a Pascal (Pa).

1 N/m2 = 1 Pa

• Other units commonly used in practice are bar and atm.

1 bar = 105 Pa

1 atm = 101.325 kPa

*English system: 1 atm = 14.696 psi (pound force per square inch)

PASCAL’S LAW (pressure at a point)

Px

Ps

Py

δy

δs

δz

δx

A

F

D

B

C

E

θ

• Consider the equilibrium of a small fluid element in the form of

triangular prism below.

• By assuming a fluid is at rest, px will act at right angles to the

plane ABEF, py at right angles to CDEF and ps at right angles to

ABCD.

δyδzxp

ABFEAreaxpxptodueForce

δyδzsp

δs

δyδsδzsp

θsinABCDAreasp

:sptodueforceofComponent

zpxp

0δyδzspδyδzxp

• Considering the force in x-direction:

• The element will be in equilibrium if:

(2.1)

δxδzyp

CDEFAreaypyptodueForce

δxδzsp

δs

δxδsδzsp

θcosABCDAreasp

:sptodueforceofComponent

zp

yp

0δxδyδz2

1ρgδxδz

spδxδz

xp

• Considering the force in y-direction:

• The element will be in equilibrium if:

δxδyδz2

1ρg -

VolumeeightSpecific welement of Weight

* Since δx, δy and δz are all very small quantities, δxδyδz is negligible in comparison with the other two terms.

(2.2)

• Thus from equation 2.1 and 2.2 :

• From the equation 2.3 we can conclude that:

“ Pressure at any point is the same in all directions. This known as

Pascal’s Law (Blaise Pascal-France Philosophy, 1632-1662), and

applies to a fluid at rest”

zp

yp

xp (2.3)

Atmospheric pressure, Gauge pressure and Absolute pressure.

• Absolute pressure - The actual pressure at a given position. It is measured

relative to absolute vacuum (i.e., absolute zero pressure).

• Gauge pressure - Difference between the absolute pressure and the local

atmospheric pressure. Most pressure-measuring devices are calibrated to read

zero in the atmosphere, and so they indicate the gauge pressure.

• Vacuum pressures and are measured by vacuum gages that indicate the

difference between the atmospheric pressure and the absolute pressure.

Absolute, gage, and vacuum pressures are all positive quantities and are

related to each other by:

atmP

absP

gaugeP

absP

atmP

vacP

(2.4)

(2.5)

Patm

Pvac

Pabs

Vacuum

Absolute Pabs=0

Patm

Pgauge

Pabs

Patm

Vacuum

Absolute

Illustrated of atmospheric pressure, gauge pressure and absolute pressure.

VARIATION OF PRESSURE IN A STATIC FLUIDS

z

δs

area, A

p

area, A

p+δp

θ

mg

z+δz

2

1

• The axis of the prism is inclined at right angle θ to the vertical, the

height of 1 above horizontal datum is z and that of 2 is z+δz.

• The force acting on the element are:

i) F1= pA → acting at right angles to the end face at 1 along the axis of

prism.

ii) F2= (p+δp).A → acting at 2 along the axis in the opposite directions.

iii) FB = mg → weight of the element due to gravity

• For equilibrium; ΣF = 0

ie: F1+F2+FB = 0

cos

0

sgA

mgA

pA-

pp-pA

cosgds

dp

gdz

dp

• In differential form:

• Taking the (x,y) plane as horizontal:

• If 1-2 parallel with z-axis, i.e θ=00, ds=dz

0dz

dp

2

1

12

z

z

dzgppordzgp

• If 1-2 perpendicular to z-axis, i.e θ=900, ds=dz

• The actual pressure variation with elevation is found by integrating the

above equation.

(2.6)

(2.7)

(2.8)

(2.9)

PRESSURE AND HEAD

- In a fluid of constant density;

gdz

dp

constant ghp

- Integrate immediately to give;

constant gzp

- The pressure ‘p’ at any depth ‘z’ measured downwards from the

free surface so that;

hz i.e,

- Since the pressure at the free surface will normally be

atmospheric pressure, patm ;

atmpghp

(2.10)

(2.11)

p

h

patm

Liquid with

density ρ

PRESSURE AT LAYERED FLUID

The pressure at the bottom of the tank in Fig. 3–13 can be

determined by starting at the free surface where the pressure is

Patm, moving downward until we reach point 1 at the bottom, and

setting the result equal to P1. It gives

Patm + ρ1gh1 +ρ 2gh2+ ρ 3gh3 = P1

PRESSURE MEASUREMENT

• Barometer- Use to measure atmospheric pressure, thus the atmospheric pressure

referred to as the barometric pressure.

- Pressure at point B is equal to the atmospheric pressure.

- Pressure at point C can be taken to be zero (only mercury vapor above the point C and the pressure is very low relative to patm).

ghpatm

h

B

C

hA

mercury

W=ρghA

patm

Evangelista Torricelli (1608-1647)

prove that the atmospheric pressure

can be measured by inverting a

mercury-filled tube into a mercury

container that is open to the

atmosphere.

(2.12)

Example 2

A mountain lake has an average temperature of 10 °C

and a maximum depth of 40 m. For a barometric

pressure of 598 mm Hg, determine the absolute

pressure (in pascals) at the deepest part of the lake.

Example 1

Determine the atmospheric pressure at a location

where the barometric reading is 740 mm Hg and the

gravitational acceleration is g = 9.81 m/s2. Assume

the temperature of mercury to be 10°C, at which its

density is 13,570 kg/m3.

• Piezometer

- The simplest manometer.

- Consist of tube that open at top which is attached to the vessel containing liquid at a pressure (higher than atmospheric pressure) to be measured.

- As the tube is open at atmosphere, the pressure measured is relative to atmospheric so is gauge pressure.

- Pressure at point A

rhgpA

(2.13)

r

h

A

• U-tube Manometer

- U-tube manometer enables to measure pressure for both liquids and gases with the same instrument.

- The U-tube is connected as in figure and filled with a fluid called manometric fluid.

- The fluid whose pressure is being measured should have a mass density less than that of the manometric fluid and the two fluids should not be able to mix readily.

- For the left hand arm:

1ghpp

AB

- For the right hand arm:

2ghpp

mDC

- Pressure are same at same level, i.e:

D

A

fluid density, ρ

B C

h1

h2

manometric fluid

density, ρm

*As we measuring gauge pressure we can subtract patm (pD)

CBpp

- Pressure at point A:

12ghghp

mA (2.14)

Example 3

A manometer is used to measure the pressure in a tank. The

fluid used has a specific gravity 0.85, and the manometer

column height is 55 cm, as shown in Fig. 3–12. If the local

atmospheric pressure is 96 kPa, determine the absolute

pressure within the tank.

Example 4

A closed tank contains compressed air and oil (SGoil = 0.90) as is

shown in Fig. E2.4. A U-tube manometer using mercury (SGoil = 13.6)

is connected to the tank as shown. For column heights h1 = 36 in.,

and h2 = 6 in, and h3 = 9 in, determine the pressure reading (in psi)

of the gage.

• U-tube Manometer (measurement of pressure difference)

A

CB

D

E

h2

h3

h1

ρ1

ρ2

ρ3

- If the U-tube is connected to a pressurized vessel at two points, the pressure difference between these two points can be measured.

- For the manometer that arranged as in figure:

- For the left hand arm:

CBpp

11ghpp

AB

332211ghghpghp

EA

- For the right hand arm:

- Since PB=PC;

3322ghghpp

EC

- Pressure difference between these two

points (point A and point E) in the vessel is:

113322ghghghpp

EA (2.15)

Example 5

The water in a tank is pressurized by air, and the pressure is

measured by a multifluid manometer as shown in Fig. 2.10. The

tank is located on a mountain at an altitude of 1400 m where

the atmospheric pressure is 85.6 kPa. Determine the air

pressure in the tank if h1 = 0.1 m, h2 = 0.2 m, and h3 = 0.35 m.

Take the densities of water, oil, and mercury to be 1000 kg/m3,

850 kg/m3, and 13,600 kg/m3, respectively

Example 6

• Choice of Manometer

Care must be taken when attaching the manometer to vessel. No burrs must be present around this joint. Burrs will alter the flow causing local pressure variations to affect the measurement.

Some disadvantages of Manometers:

Slow response-only really useful for very slowly varying pressure.

For U-tube manometer, two measurements must be taken simultaneously to get h value.

For very accurate work the temperature and relationship between temperature and density must be known.

Some advantages of Manometers:

Very simple

No calibration is required.

PRESSURE MEASURING DEVICES

Bourdon pressure gages

Pressure transducers

TUTORIAL