IIAandB Hydrostatics 2014

Southern Methodist University

Bobby B. Lyle School of Engineering

CEE 2342/ME 2342 Fluid Mechanics

Roger O. Dickey, Ph.D., P.E.

II. HYDROSTATICS

A. Pressure Measurement and Units

B. Pressure Variation in a Static Fluid

Reading Assignment:

Chapter 2 Fluid Statics, Sections 2.1 through 2.7

A. Pressure Measurement and Units

Pressure is expressed as a difference from a

reference datum. Common reference datums are:

• Complete vacuum – absolute zero pressure

• Local atmospheric pressure

Definitions,

Absolute pressure the difference between the

pressure at the point and absolute zero pressure

Gauge Pressure the difference between the

pressure at the point and local atmospheric

pressure

Most pressure measuring devices, called pressure gauges, measure the fluid pressure relative to local atmospheric pressure, hence the term gauge pressure.

• Absolute pressures > 0

• Gauge pressure < 0 a partial vacuum exits

• Gauge pressure > 0 pressurization above atmospheric

pressure

Common pressure measuring devices include:

• Piezometer tube

• U-tube manometer

• Differential U-tube manometer

• Inclined tube manometer

• Bourdon tube pressure gauge

• Diaphragm-strain gauge pressure transducers

Figure 2.9 (p. 52)Piezometer tube.

Figure 2.10 (p. 53)Simple U-tube manometer.

Figure 2.11 (p. 54)Differential U-tube manometer.

Figure 2.12 (p. 56)Inclined-tube manometer; used for accurate measurement of low differential pressures.

Figure 2.13 (p. 57)(a) Liquid-filled Bourdon pressure gages for various pressure ranges. (b) Internal elements of Bourdon gages. The “C-shaped” Bourdon tube is shown on the left, and the “coiled spring” Bourdon tube for high pressures of 1000 psi and above is shown on the right. (Photographs courtesy of Weiss Instruments, Inc.)

Figure 2.14 (p. 58)Pressure transducer which combines a linear variable differential transformer (LVDT) with a Bourdon gage. (From Ref. 4, used by permission.)

Figure 2.15 (p. 59)Diaphragm-strain gaugepressure transducer – (a) photo, (b) schematic diagram

(b)

(a)

Local atmospheric pressure is measured with a

device called a barometer (mercury barometer

invented by Toricelli in 1644).

StandardAtmosphericPressure(absolute pressure)

14.7 psi2,116 lb/ft2

29.92 inches of mercury33.91 ft of water1 atm760 mm of mercury101.325 kPa10.34 m of water

Figure 2.8 (p. 51)Mercury barometer.

Aneroid barometers are similar in design to Bourdon tube pressure gauges, but with the interior of the sensing element evacuated to near absolute zero pressure. The sensing element then deflects in response to changing ambient atmospheric pressure. Aneroid barometer

Any gauge pressure measurement can be

converted to absolute pressure as follows,

Convention:

All pressures are gauge pressures unless

specifically indicated as absolute (e.g., psia).

atmospheregaugeabsolute PPP

B. Pressure Variation in a Static Fluid

Pascal’s Law (stated in 1653 by Blaise Pascal),

The pressure is the same in all directions at

any point in a fluid with no shear stresses

(i.e., = 0).

From Newton’s Law of Viscosity,

= 0

dydu

du/dy = 0, i.e., the fluid is at rest

= 0, i.e., the fluid is non-viscous

Since pressure at a point in a static fluid is

independent of direction, pressure is a scalar

quantity. Conversely, pressure forces exerted on

surfaces by fluids are vectors, always acting in a

direction normal to the surface.

Vector ≡ a physical quantity having both

magnitude and direction, and that

obeys the rules of vector algebra.

Corollary to Pascal’s Law relating to pressure

transmission,

In a closed fluid system, a pressure change

applied at one point in the system will be

transmitted throughout the system.

The phenomena of pressure transmission in a fluid system is widely applied in hydraulic systems:

• Brakes

• Aircraft control surfaces

• Jacks and lifts

• Presses•••

Auto Hydraulic Brakes

Aircraft Hydraulic Systems – Control Surfaces, Landing Gear

Manual Hydraulic Press Hydraulic Jack

Construction EquipmentCherry Pickers

In analysis of pressure transmission in hydraulic

systems, the variation in pressure with elevation

is usually neglected.

Refer to Handout – II.B. Hydraulic Lift

Example for a simple example problem.

For deriving a general equation describing

pressure variation in a static fluid, consider the

free-body diagram for an arbitrary cylindrical

fluid element having constant cross-sectional

area, ΔA, and length, Δl, inclined at an arbitrary

angle from the horizontal within a static fluid

mass:

ll 1

l 2 l-axis

y

z

x

p ΔA

(p+Δp)ΔA

W

Free Surface

Apply Newton’s 2nd Law of Motion along the longitudinal axis of the cylindrical element, i.e., in the l-direction:

Thus,

maF 0, for static fluids

0 F

There are no shear stresses in a static fluid. Hence, the only forces acting on the fluid element are: (i) gravity, i.e., weight of the element, W, acting in the –z-direction, (ii) fluid pressure force, pΔA, acting normal to the lower end of the element in the +l-direction, and (iii) fluid pressure force, (p+Δp)ΔA, acting normal to the upper end of the element in the –l-direction. Summing the forces in the l-direction:

0 WAppAp

Wl is the component force of the weight acting

in the l-direction:

sin sin WWWW

W

+l-direction

W l

The total weight of the cylindrical fluid element is given by:

AW

Specific Weight of the Fluid

Cylinder Volume = Base Area × Height

Thus, Wl is:

Substituting into the equation resulting from application of Newton’s 2nd Law:

sin AW

0sin AAppAp

Dividing through by ΔA:

Subtract out p, divide through by Δl and rearrange:

0sin ppp

sinp

Now consider that:

By definition:

ΔlΔz

zsin

22 yx

l-axis

Substituting Δz/Δl for sin in the equation resulting from application of Newton’s 2nd Law:

Multiplying through by Δl and rearranging:

zp

zp

Taking the limit as Δz→0 yields,

where,

p = pressure [F/L2]

z = distance in vertical direction above a horizontal datum [L]

= specific weight of the fluid [F/L3]

Basic Equation of Fluid Statics, Equation (2.4), p. 44; applies to both compressible and incompressible fluids

dzdp

Given that the derivation was based on an arbitrary fluid element, several general conclusions may be drawn from this result:

• When moving along a path in a static fluid, pressure changes occur only when there is a change in elevation (i.e., movement in the vertical, or z-direction)

• There is no change in pressure along any horizontal plane through a static fluid

• The greatest pressure gradient is achieved

when following a vertical path (i.e., when

moving in the z-direction)

• dp/dz < 0 since > 0, revealing that moving

upward (z increasing) through a static fluid

results in a pressure decrease, and moving

downward (z decreasing) results in a

pressure increase

• Increasing pressure with greater depth results from the increasing weight of the overlying fluid column

• No limitations were placed on the compressibility of the fluid, therefore, the basic differential equation holds for both compressible and incompressible fluids;

= f(z), for compressible fluids

= constant, for incompressible fluids

Reconsider the Basic Equation of Fluid Statics,

Solve this first-order, linear, ordinary differential equation by separation of variables, integrating between two points at differing

elevations z1 and z2 having pressures p1 and p2,

respectively:

dzdp

2

1

2

1

z

z

p

p

dzdp

For incompressible fluids, γ is constant and may

be factored from the integral on the right-hand

side of the equation; integration then yields:

Where p1 and p2 are pressures at points located at

vertical elevations z1 and z2 , respectively, above

a horizontal reference plane (the xy-plane) as

illustrated in Figure 2.3, p. 44 in the textbook:

1212 zzpp Basic Equation of Hydrostatics; incompressible fluids only

Figure 2.3, p. 44 – Modified

Free Surface (pressure = p0)

12 zzz

Point 2

Point 1

Divide both sides of the Basic Equation of

Hydrostatics by and rearrange, yielding another

common form of the equation,

Each term has dimensions of [L]. For the p/γ terms:

22

11 z

γpz

γp

Alternate form of the Basic Equation of Hydrostatics; incompressible fluids only

LLFLF

γp

3

2

In this latter form of the equation:

1. p/ is referred to as the “pressure head,”

i.e., the equivalent depth of liquid having

specific weight that produces the

pressure, p.

2. z is referred to as the “elevation head,”

i.e., the gravitational potential energy per

unit weight of fluid.

* Important Points –

1. Both forms of the Basic Equation of

Hydrostatics apply between any two points

in a continuous, incompressible, static fluid

mass.

2. Fluids adhering to the Basic Equation of Hydrostatics are said to have a hydrostatic pressure distribution.

When applying the Basic Equation of

Hydrostatics where there is a free surface

exposed to atmospheric pressure, it is usually

convenient to choose the free surface as the

horizontal reference plane, defining an alternate

coordinate system based on depth, h, below the

surface. Notice that (z2 z1) = (h1 h2):

Figure 2.3, p. 44 – Modified

Free Surface (pressure = p0)

2112 hhzzz h

h2 h1 Point 2

Point 1

Write the Basic Equation of Hydrostatics starting

from an initial point located at an arbitrary depth h

below the surface, having pressure p and elevation

z, moving upward to a second point on the free

surface (i.e., at zero depth) having pressure p0 and

elevation z0,

as defined on the following sketch:

zzpp 00

z

z0

pressure = p

h = (z0 z)

hz

x

y

Free Surface (pressure = p0)

Transform to the new coordinate system by

substituting h for (z0 z) ,

Solve for the pressure, p, at depth h,

hpp 0

0php Equation (2.8), p.45

When a barometer reading is used to set p0 equal

to local atmospheric pressure, the previous

equation yields the absolute pressure at depth h

below the surface. Setting p0 = 0 yields the

gauge pressure at depth h:Gauge pressure at depth h below a free liquid surfacehp

*Important Point –

Pressure in a fluid having a hydrostatic pressure distribution depends only on the depth of the fluid relative to some horizontal reference plane, it is not influenced by the size or shape of the fluid’s container. For example, the pressure is constant at all points along horizontal Plane A-B through the irregularly shaped but continuous, homogenous fluid mass illustrated in the following figure:

Fluid equilibrium in a container of arbitrary shape.

Example applications of the Basic Equation of

Hydrostatics are presented in Handout – II.B.

Hydrostatics Examples.

Homework No. 4 Pressure, application of the Basic Equation of Hydrostatics, and manometry.

Manometry –

Manometers are pressure measuring devices that employ liquid columns in vertical or inclined tubes. Common types include:

1. Mercury barometers

2. Piezometers

3. U-tube manometers

4. Differential U-tube manometers

1. Mercury barometer,Figure 2.8 (p. 51)

γm

z

Apply the Basic Equation of Hydrostatics from Point B to Point A for the mercury barometer:

But here,

pA = pvapor the vapor pressure of

mercury in the head space at the ambient temperature

pB = patm local atmospheric pressure

h = (zA zB)

BAmBA zzpp

Substituting,

Then solving for patm yields:

Mercury has a very low vapor pressure at normal ambient temperatures, hence, it is often assumed negligible yielding,

hpp matmvapor

vapormatm php

hp matm

h1

2. Piezometer Tubes

Aγ1

z

Apply Equation (2.8), p. 45 from Point A to the free surface at the open end (Point 0) of either piezometer tube shown on the previous slide:

Set p0 = 0 to yield the gauge pressure at Point A,

Notice that Point A inside the pipe and Point (1) in the piezometer bend are at the same elevation,

hence, pA = p1 .

011 phpA

11hpA

Piezometer Wells

3. U-tube manometer,Figure 2.10 (p. 53)

*Manometer Problem Solving Hints –

When following a path through a continuous column of static manometer fluid having constant γ, use the following sign convention and assumptions for rapid problem solution:

1. γh > 0 when moving downward (pressure increases)

2. γh < 0 when moving upward (pressure decreases)

3. γh = 0 when moving horizontally (pressure unchanged because h unchanged)

4. γ ≈ 0 for gases without introducing significant error

Analyze the simple U-tube manometer in Figure 2.10, p. 53 using the rapid problem solving hints, starting with the pressure at Point A, pA, and then

proceeding along a path through the manometer tube to the free surface where the pressure is zero (i.e., atmospheric pressure):

• First, there is no change in pressure when moving horizontally along the path between Point A and Point (1), hence,

p1 = pA

• When moving downward along a path from Point (1) to Point (2), the pressure increases by an amount +γ1h1 such that,

p2 = pA + γ1h1

• There is no change in pressure when proceeding from Point (2) to Point (3) through the tube because these points lie in the same horizontal plane passing through the gauge fluid, thus p3 = p2, that is,

p3 = pA + γ1h1

• When moving upward from Point (3) to the free surface (Point 0), the pressure

decreases by an amount γ2h2 such that,

p0 = pA + γ1h1 γ2h2

• Set p0 = 0, and solve for the unknown

gauge pressure pA,

pA = γ2h2 γ1h1 Equation (2.14), p. 53

• If the fluid in the pipe at Point A is a gas,

then γ1 ≈ 0 when compared to γ2 for the

liquid gauge fluid, then,

pA ≈ γ2h2

4. Differential U-tube manometer, Figure 2.11 (p. 54)

Analyze the differential U-tube manometer in Figure 2.11, p. 54, using the rapid problem solving hints, starting with the pressure at Point A, pA, and then proceeding along a path through

the manometer tube to Point B where the pressure is pB:

• First, there is no change in pressure when moving horizontally along the path between Point A and Point (1)

• When moving downward along a path from Point (1) to Point (2), the pressure increases by an amount +γ1h1

• There is no change in pressure when proceeding from Point (2) to Point (3) through the tube because these points lie in the same horizontal plane

• When moving upward from Point (3) to Point (4), the pressure decreases by an

amount γ2h2

• When moving upward from Point (4) to Point (5), the pressure decreases by an

amount γ3h3

• There is no change in pressure when moving horizontally along the path between Point (5) and Point B

The pressure at Point B, pB, is determined by

summing the pressure changes while proceeding

along the path through the manometer tube

between Points A and B:

pA + γ1h1 γ2h2 γ3h3 = pB

The differential pressure, p = pA pB, is then,

p = γ2h2 + γ3h3 γ1h1

Refer to Handout – II.B. Manometry

Examples for additional example problems.

Homework No. 5 Pressure, pressure measurement, force due to pressure, and manometry.