LECTURE 3

FLUID STATICS

By definition, a fluid must deform continuously when a shearing stress of any magnitude

is applied.

3.1 THE BASIC EQUIATION OF FLUID STATICS

For a deferential fluid element, the body force, , is

Where is the local gravity vector, ρ the density, and is the volume of the element.

In Cartesian coordinates, , so

By use of the Taylor series representation, the pressure at the left face of the differential

element is

(Terms of higher order omitted because in the limit they vanish.) The pressure on the

right face of the deferential element is

1

x

y

z

dx

dy

dz

jdxdzyypp

2

jdxdzyypp

20

Pressure, pyyL yR

Stress and forces on the other faces of element are obtained in the same way. Combining

all such forces gives the surface force acting on the element. Thus

Collecting and canceling terms, we obtain

or, (3.1a)

The term in parentheses is called the gradient of the pressure or simply the gradient, and

is designated as grad p or . In rectangular coordinates

Using the gradient designation, Eq. 3.1a can be written as

(3.1b)

From Eq. 3.1b,

Total force act on a fluid element,

or on a unit volume basis

(3.2)

For a fluid particle, Newton’s second law gives . For a static fluid, =

0. Thus from Eq. 3.2, becomes

Substituting for

2

rear front

left right

lower upper

Let us review briefly our derivation of this equation. The physical significance of each

term is

+ = 0

+

= 0

This is a vector equation, which means that it really consists of three component

equations that must be satisfied individually. Expanding into components, we find

x direction

y direction (3.4)

z direction

under this condition, the component equations become

(3.5)

(3.6)

3.1.1 Pressure Variation in a Static Fluid

a. Incompressible Fluid

For incompressible fluid, ρ = ρo = constant. Then for constant gravity,

If the pressure at the reference level, zo, is designated as po, then pressure, p, at location z

is found by integration

or

With h measured positive downward, then

3

= 0

= 0

and (3.7)

Fig. Coordinate for determination of pressure variation in a static liquid.

Example 3.1

Water flows through pipes A and B. oil, with specific gravity 0.8, is in the upper portion

of the inverted U. Mercury (specific gravity 13.6) is in the bottom of the manometer

bends.

FIND:

Determine the pressure difference, pA – pB, in units of lbf/in2.

SOLUTION:

Basic equations:

For γ = constant

4

x

y

h

g

z

zo

z

po

p

Beginning at point A and applying the equation between successive point around the

manometer gives

Substituting

in.

in.

Example 3.2

A reservoir manometer is built with a tube diameter of 10 mm and a reservoir diameter of

30 mm. The manometer liquid is Meriam red Oil with SG = 0.827. Determine the

manometer deflection in millimeters per millimeter of applied pressure deferential.

FIND

5

C

pA

12 dγ OH

Liquid deflection, h, in millimeter per millimeter of water applied pressure

SOLUTION:

Basic equations:

and

For ρ = constant

or

To eliminate H, note that volume of manometer liquid must remain constant. Thus the

volume displaced from the reservoir must be the same as that which rises into the tube.

or

Substituting gives

This equation can be simplified by expressing the applied pressure differential as an

equivalent water column

and noting that . Then

or

6

D

d

z

p2

p1

h

2

1

Oil, SG = 0.827

H

Equilibrium

liquid level

Evaluating

This problem illustrates the effect of manometer design and choice of gage liquid on

sensitivity.

b. Compressible Fluid

Pressure variation in any static fluid is described by the basic pressure-height relation

For many liquids, density is only a weak function of temperature. Pressure and

density of liquids are related by the bulk compressibility modulus, or modulus of

elasticity,

(3.8)

If the bulk modulus is assumed constant, then density is only a function of pressure.

The density of gases generally depends on pressure and temperature. The ideal

gas equation of state

(3.9)

where: R = the gas constant

T = the absolute temperature

Example 3.3

The maximum power output capability of an internal combustion engine decreases with

altitude (sea level) because the air density and hence the mass flow rate of fuel and air

decrease. A truck leaves Denver ( jenenge kutho ing monconegoro ) (elevation 5,280 ft).

Determine the local temperature and barometric pressure are 80oF and 24.8 in. of

mercury, respectively. It travels through Vail Pass ( jenenge kutho ing monconegoro )

(elevation 10,600 ft). The temperature decreases at the rate of 3 oF/1000 ft of elevation

change. Determine the local barometric pressure at Vail Pass and the percent decrease in

maximum power available, compared to that at Denver.

GIVEN:

Truck travels from Denver to Vail Pass. Engine power output is directly proportion to air

density.

7

Denver: z = 5,280 ft Vail Pass: z = 10,600 ft

ρ = 24.8 in. Hg

T = 80o FFIND:

a) Atmosphere pressure at Vail Pass.

b) Percent engine at Vail Pass compared to Denver.

SOLUTION:

Basic equations:

Assumptions: 1) Static fluid

2) Air behaves as an ideal gas

By substituting into the basic pressure-height relation,

or

But temperature varies linearly with elevation, dT/dz = – m, so T = To – m(z– zo)

=

By integrating from po in Denver to p at Vail,

or

Evaluating gives

and

8

Note that To must be expressed as an absolute temperature because it came from the ideal

gas equation.

Thus

and

The percentage change in power is equal to the change in density, so that

By substituting from the ideal gas equation,

or

3.2 THE STANDARD ATMOSPHERE

Several International Congresses for Aeronautics have been held so that aviation experts

around the world might better be able to communicate.

Table 3.1 Sea Level Condition of the U.S. Standard Atmosphere

Property Symbol SI English

Temperature T 288 oK 59 oFPressure p 101.3 k Pa (abs) 14.696 psiaDensity ρ 1.225 kg/m3 0.002377 slug/ft3

Specific weight γ - 0.7651 lbf/ft3

Viscosity μ 1.781 x 10-5 kg/m sec 3.719 x 10-7 lbf/ft2

3.3 ABSOLUTE AND GAGE PRESSURES

9

Pabsolute

Pressure level

Atmospheric Pressure101.3 kPa (14.696 psia)at standard sea level conditions

Vacuum

Absolute pressures must be used in all calculations with the ideal gas or other equations

of state. Thus

BIBLIOGRAPHY:

1. Fox & Mc Donald, Introduction to fluid mechanics, 2nd edition, John Wiley &

Sons, Canada.

2. Irving H. Shames, Mechanics of Fluids, Fourth Edition, Mc Graw Hill, Singapore.

10

11