pressure and head

8/6/2019 Pressure and Head

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FLUID MECHANICS (CHE 203)

CHAPTER 2

PRESSURE AND HEAD


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OBJECTIVE

To acquire fundamental concepts pressure and head

LEARNING OUTCOMES

At the end of this chapter, student should be able to:

i. Define and derive Pascals Law

ii. Derive pressure variation with height in a fluid at rest

iii. Determine the pressure at various locations in a fluid at rest

iv. Differentiate between absolute and gage pressurev. Explain the concept of manometers

vi. Apply appropriate equations to determine pressures or pressure

difference using different types of manometers

vii. State advantages and disadvantages of manometer

viii. Convert pressure in terms of head and vice versa


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Pressure is defined as a compressive stress, or compressive force per unitarea.

In a stationary fluid (liquid or gas) the compressive force per unit area is

the same in all directions.

In a solid or moving fluid, the compressive force per unit area at some

point is not necessarily the same in all directions.

Units: newtons per square meter (Nm-2 or kgm-1s-2)

The same unit is also known as a Pascal Pa where 1 Pa = 1 Nm-2

Also frequently used is the alternative SI unit the bar, where 1 bar = 105

Nm-2.

Dimensions: ML-1T-2

A=por

boundaryofArea

exertedorce=essurePr

2.1 PRESSURE


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By considering a small element of fluid in the form of a triangular prism which

contains a point p, we can establish a relationship between the three pressures pxin the x direction, py in the y direction and ps in the in the direction normal to the

sloping face.

2.2 PASCALS LAW FOR PRESSURE AT A POINT


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The fluid is at rest, so we know that all force are acting at right angles to

the surfaces i.e

ps acts perpendicular to surface ABCD

px acts perpendicular to surface ABFE and

py acts perpendicular to surface CDEF

And, as the fluid is at rest, in equilibrium, the sum of the forces in any

direction is zero.



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Considering the x-direction:Force due to px

Fxx = px x area ABFE = pxHyHz;

Component of force in the x-direction due to ps,

Fxs = -(ps x area ABCD) x sin U

= -ps = -psHyHz

Since sin U = Hy/Hs

Component of force in the x-direction due to py,

Fxy = 0To be at rest (in equilibrium)

Fxx + Fxs + Fxy = 0

pxHyHz + -psHyHz = 0

px = ps

syzs



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Similarly, in y-direction:

Force due to pyFyy = py x area CDEF = pyHxHz;

Component of force due to psFys = -(ps x area ABCD) cos U

= -ps = - psHxHz

CosU = Hx/Hs

Component of force due to px,

Fyx = 0

Force due to gravity,Weight of element = -specific weight x volume

= -

s

xzs

zyx2

1g



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To be at rest (in equilibrium)

Fyy + Fys + Fyx + weight = 0

pyHxHz +(-psHxHz) + 0 + - = 0

Since Hx, Hy, Hz are all small quantities, HxHyHz is very small andconsider negligible, hence

py = psThus, ps = px = py

zyx2

1g



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Considering the prismatic element again, ps is the pressure on a plane

at any angle U, the x, y, and z directions could be at any orientation.

The element is so small that it can be considered a point so the derived

expression ps = px = py indicates that pressure at any point is the same

in all directions. (The proof may be extended to include the z axis). Pressure at any point is the same in all directions. This is known as

Pascals law and applies to fluid at rest.



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In the figure we can see an element of fluid

which is a vertical column consists of constant

cross sectional area, A, surrounded by the

same fluid of mass density .

The pressure at the bottom of the cylinder is p1at level z1, and at the top is p2 at level z2.

The fluid is at rest and in equilibrium so all the

forces in the vertical direction sum to zero.

The forces acting are

Force due to p1 on area A acting up = p1A,Force due to p2 on area A acting down = p2A

2.3 VARIATION OF PRESSURE VERICALLY IN A

FLUID UNDER GRAVITY

Vertical elemental cylinder of fluid

p2 A

p1 A

Area A

z2

z1

Fluid

Density,


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Force due to weight of the element = mg

= Mass density x g x Volume

= gA(z2-z1).

Taking upward forces as positive, in equilibrium we havep1A - p2A - gA(z2-z1) = 0

p1p2 = g(z2-z1)

p = g(z2-z1)

Thus in any fluid under gravitational attraction, pressure decreases with

increase of height z.

2.3 VARIATION OF PRESSURE VERICALLY IN A

FLUID UNDER GRAVITY


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Consider the horizontal cylindrical element of fluid in the figure below,with cross sectional area A, in a fluid of density , pressure p1 at the left

hand end and pressure p2 at the right hand end.

2.4 EQUALITY OF PRESSURE AT THE SAME

LEVEL IN A STATIC FLUID

p2 Ap1 A

Area A

Fluid density,

Weight, mg

Face L Face R

Horizontal elemental cylinder of fluid


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The fluid is at equilibrium so the sum of the forces acting in the x-directionis zero.

p1A = p2A

p1 = p2***Pressure in the horizontal direction is constant

This result is the same for any continuous fluid. It is still true for twoconnected tanks which appear not to have any direct connection, forexample consider the tank in the figure below.



zz

Q

R

P

L

Two tanks of different cross-section connected by a pipe


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We have show above that pL= pRand from the equation for a vertical

pressure change we have

pL = pP + gz and pR= pQ + gz

so pP + gz = pQ + gz

pP = pQ

This shown that the pressures at the two equal levels, P and Q are the

same.




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In a static fluid of constant density we have the relationship,

This can be integrated to give, p = - gz + constant

In a liquid with a free surface the pressure at any depth z measured fromthe free surface so that z = - h (see the figure below)

g-=dz

dp

2.5 PRESSURE AND HEAD

Fluid head measurement in a tank

y

z

x

h


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This gives the pressure

p = gh + constant

At the surface of fluids we are normally concerned with, the pressure is the

atmospheric pressure, patmospheric.

So, p = gh +patmospheric

As we live constantly under the pressure of the atmosphere, and everything

else exists under this pressure, it is convenient (and often done) to take

atmospheric pressure as the datum. So we quote pressure as above or below

atmospheric.

Pressure quoted in this way is known as gauge pressure

Gauge pressure is

pgauge = gh



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The lower limit of any pressure is zero - that is the pressure in a perfectvacuum. Pressure measured above this datum is known as absolutepressure

Absolute pressure is

pabsolute = gh + patmosphereAbsolute pressure = Gauge pressure + Atmospheric pressure

As g is (approx.) constant, the gauge pressure can be given by statingthe vertical height of any fluid of density which is equal to thispressure.

p = gh

The vertical height is known as head of fluid.

Note: If pressure is quoted in head, the density of the fluid must alsobe given.



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A cylinder contains a fluid at a gauge pressure of 350 kNm-2

. Express thispressure in terms of a head of,

(a) water (H20 = 1000 kgm-3),

(b) mercury (relative density 13.6).

(c) what would be the absolute pressure in the cylinder if the

atmospheric pressure is 101.3 kNm-2

?

EXAMPLE 2.1 - PRESSURE AND HEAD


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The relationship between pressure and head is used to measure pressure

with a manometer (also know as a liquid gauge).

2.6 PRESSURE MEASUREMENT BY

MANOMETER


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The simplest manometer is a tube, open at the top, which is attached to the

top of a vessel containing liquid at a pressure (higher than atmospheric) to

be measured. An example can be seen in the figure below. This simple

device is known as a Piezometertube. As the tube is open to the

atmosphere the pressure measured is relative to atmospheric so is gauge

pressure.

2.6.1 THE PIEZOMETER TUBE MANOMETER

h1

A

B

h2

A simple piezometer tube manometer


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Pressure at A = pressure due to column of liquid above A

pA = gh1

Pressure at B = pressure due to column of liquid above B

pB

= gh2

This method can only be used for liquids (i.e. not for gases) and only when

the liquid height is convenient to measure. It must not be too small or too

large and pressure changes must be detectable.

2.6.1 THE PIEZOMETER TUBE MANOMETER


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What is the maximum gauge pressure of water that can be measured by meansof a piezometer tube 2 m high?

And if the liquid has a relative density of 8.5, what would the maximum

measurable gauge pressure?

EXAMPLE 2.2 - PIEZOMETER


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Using a U-Tube enables the pressure of both liquids and gases to be

measured with the same instrument. The U is connected as in the figure below and filled with a fluid called

the 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 - that is, they must be immiscible.

2.6.2 THE U-TUBE MANOMETER

Fluid P, mass density,

A U-Tube manometer

Manometric fluid Q, mass density, man

h2

h1

A

B C

D


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Pressure in a continuous static fluid is the same at any horizontal level so,

Pressure at B = Pressure at CpB = pC

For the left hand arm

pB = Pressure pA at A + Pressure due to depth h1of fluid P.

= pA + gh1For the right hand arm

pC = Pressure pD at D + Pressure due to depth h2 of

manometric fluid Q.

But pD = Atmospheric pressure = Zero gauge pressure

and so pC = 0 + mangh2since pB= pC,

pA + gh1 = mangh2pA = mangh2 - gh1


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If the fluid is being measured is a gas, the density will probably be

very low in comparison to the density of the manometric fluid i.e. man>> . In this case the term can be neglected, and the gauge pressure

is given by

pA = mangh2



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A U-tube manometer similar to that shown in figure is used to measure thatgauge pressure of a fluid P of density = 800 kgm-3. If the density of the

liquid Q is 13.6 x 103 kgm-3, what will be the gauge pressure at A if,

(a) h1 = 0.5 m and D is 0.9 m above BC?

(b) h1 = 0.1 m and D is 0.2 m below BC?

EXAMPLE 2.3 - U-TUBE MANOMETER


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Using a U-tube manometer to measure gauge pressure of fluid density = 700kgm-3 and the manometric fluid is mercury, with a relative density of 13.6.

What is the gauge pressure if:

(a) h1 = 0.4 m and h2 = 0.9 m?

(b) h1 stayed the same but h2 = -0.1 m?



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

2.6.3 MEASUREMENT OF PRESSURE

DIFFERENCE USING U-TUBE MANOMETER

Fluid density,

h b

h a

h

A

B

C D

E

Manometric fluid density, man


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If the manometer is arrange as in the figure above, then

Pressure at C = Pressure at D

pC = pDpC = pA + ghapD = pB + g(hb- h) + mangh

pA + gha = pB + g(hb- h) + mangh

Giving the pressure difference

pA - pB = g(hb- h) + mangh gha

Again, if the fluid whose pressure difference is being measured is a gas andman >> , then the terms involving can be neglected, so

pA - pB = mangh

2.6.3 MEASUREMENT OF PRESSURE

DIFFERENCE USING U-TUBE MANOMETER


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Two pipes containingthe same fluid of density = 990 kg m-3 areconnected using a U-tube manometer. What

is the pressure betweenthe two pipes if themanometer containsfluid of relative density13.6?


Fluid density,

C

Manometric fluid density, man

h = 0.5 m

h a= 1.5m

h b = 0.75 m

E

A

B

D

Fluid density,


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The "U"-tube manometer has the disadvantage that the change in height of

the liquid in both sides must be read. This can be avoided by making thediameter of one side very large compared to the other. In this case the side

with the large area moves very little when the small area side move

considerably more.

2.6.4 ADVANCES TO THE U-TUBE

MANOMETER


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Assume the manometer is arranged as above to measure the pressuredifference of a gas of (negligible density) and that pressure difference is

p1-p2. If the datum line indicates the level of the manometric fluid when

the pressure difference is zero and the height differences when pressure is

applied is as shown, the volume of liquid transferred from the left side to

the right = z2 x (d2 / 4)

And the fall in level of the left side is


MANOMETER

( ) 22

2

2

2

1

/dz=4/

)4/d(z=

sideleftofarea

movedvolume=z


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We know from the theory of the "U" tube manometer that the height

different in the two columns gives the pressure difference so,

if D is very much larger than d then (d/D)2 is very small so p1 p2 = gz2

So only one reading need to be taken to measure the pressure difference.

p - p g(z z )

g z d / z

gz d /

1 2 1 2

2

2 2

2

21

! V

! V ! V


MANOMETER


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If the pressure to be measured is very small then tilting the arm provides

a convenient way of obtaining a larger (more easily read) movement ofthe manometer. The above arrangement with a tilted arm is shown in the

figure below.

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MANOMETER

d

p1

z2

z1

1

2

p2

x

Datum line

D


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The pressure difference is still given by the height change of the

manometric fluid but by placing the scale along the line of the tilted arm

and taking this reading large movements will be observed. The pressure

difference is then given by,

p1 - p2 = g (z1 + z2)

= g [[x (d/D)2 ]+ x sin]

= gx[(d/D)2 + sin)]

if D is very much larger than d then (d/D)

2

is very small so p1 p2 = g xsin

The sensitivity to pressure change can be increased further by a greater

inclination of the manometer arm, alternatively the density of the

manometric fluid may be changed.


MANOMETER

( )

2 2

2

1 1

2

D d

z = x or z = x d / D4 4

z = x sin


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An inclined tube manometer consists of a vertical cylinder 35 mm diameter.

At the bottom of this is connected a tube 5 mm in diameter inclined upwardat an angle of 15 to the horizontal, the top of this tube is connected to an air

duct. The vertical cylinder is open to the air and the manometric fluid has

relative density 0.785. Determine the pressure in the air duct if the

manometric fluid moved 50 mm along the inclined tube.

EXAMPLE 2.6 - ADVANCES U-TUBE

MANOMETER

D

d

p1

z2

z1

=15

1

2X = 50 mm

Datum line

p2


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Care must be taken when attaching the manometer to vessel, no burrs must be

present around this joint.B

urrs would alter the flow causing local pressurevariations to affect the measurement.

Some disadvantages of manometers:

Slow response - only really useful for very slowly varying pressures - no

use at all for fluctuating pressures;

For the "U" tube manometer two measurements must be takensimultaneously to get the h value. This may be avoided by using a tube

with a much larger cross-sectional area on one side of the manometer

than the other;

It is often difficult to measure small variations in pressure - a different

manometric fluid may be required - alternatively a sloping manometer may

be employed; It cannot be used for very large pressures unless several

manometers are connected in series;

Some advantages of manometers:

They are very simple.

No calibration is required - the pressure can be calculated from first

principles.

2.7 CHOICE OF MANOMETER

pressure and head

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