chapter 2 pressure and head

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FLUID MECHANICS (CHE 213)

CHAPTER 2PRESSURE AND HEAD

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OBJECTIVETo acquire fundamental concepts pressure and head

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

i. Define and derive Pascal’s Lawii. Derive pressure variation with height in a fluid at restiii. Determine the pressure at various locations in a fluid at restiv. Differentiate between absolute and gage pressurev. Explain the concept of manometersvi. Apply appropriate equations to determine pressures or pressure

difference using different types of manometersvii. State advantages and disadvantages of manometerviii. Convert pressure in terms of head and vice versa

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• Pressure is defined as a compressive stress, or compressive force per unit area.

• 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

AF

=porboundaryofAreaexertedForce

=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 px in the x direction, py in the y direction and ps in the in the direction normal to the sloping face.

2.2 PASCAL’S 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 = pxyz;Component of force in the x-direction due to ps,Fxs = -(ps x area ABCD) x sin = -ps = -psyz

Since sin = y/s• Component of force in the x-direction due to py,

Fxy = 0To be at rest (in equilibrium)Fxx + Fxs + Fxy = 0pxyz + -psyz = 0px = ps

sδyδ

zδsδ


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• Similarly, in y-direction:Force due to py

Fyy = py x area CDEF = pyxz;Component of force due to ps

Fys = -(ps x area ABCD) cos

= -ps = - psxzCos = x/sComponent of force due to px,Fyx = 0 Force due to gravity,Weight of element = -specific weight x volume

= -

sδxδ

zδsδ

zδyδxδ21

×gρ


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To be at rest (in equilibrium)Fyy + Fys + Fyx + weight = 0

pyxz +(-psxz) + 0 + - = 0

Since x, y, z are all small quantities, xyz is very small and consider negligible, hence

py = ps

Thus, ps = px = py

zδyδxδ21

×gρ


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• Considering the prismatic element again, ps is the pressure on a plane at any angle , 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 Pascal’s 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 ρ.

• The pressure at the bottom of the cylinder is p1 at 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 areForce 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, ρ

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• Force due to weight of the element = mg= Mass density x g x Volume= ρgA(z2-z1).

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

p1–p2 = ρg(z2-z1)

Δp = ρg(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 ρ, 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, ρ

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-direction is 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 two

connected tanks which appear not to have any direct connection, for example 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= pR and from the equation for a vertical pressure change we havepL = pP + ρgz and pR = pQ + ρgz

so pP + ρgz = pQ + ρgz

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 = - ρgz + constant

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

gρ-=dzdp

2.5 PRESSURE AND HEAD

Fluid head measurement in a tank

y

z

x

h

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• This gives the pressure p = ρgh + constant

• At the surface of fluids we are normally concerned with, the pressure is the atmospheric pressure, patmospheric.

So, p = ρgh + 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 = ρgh


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• The lower limit of any pressure is zero - that is the pressure in a perfect vacuum. Pressure measured above this datum is known as absolute pressure

Absolute pressure is pabsolute = ρgh + patmosphere

Absolute pressure = Gauge pressure + Atmospheric pressure

As g is (approx.) constant, the gauge pressure can be given by stating the vertical height of any fluid of density ρ which is equal to this pressure.p = ρgh

• The vertical height is known as head of fluid.Note: If pressure is quoted in head, the density of the fluid must also be given.


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A cylinder contains a fluid at a gauge pressure of 350 kNm-2. Express this pressure in terms of a head of,

(a) water (ρH20 = 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 Piezometer tube. 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 = ρgh1

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

pB = ρgh2

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 means of 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, ρ

A “U”-Tube manometer

Manometric fluid Q, mass density, ρman

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 C pB = pC

For the left hand arm pB = Pressure pA at A + Pressure due to depth h1

of fluid P. = pA + ρgh1

For 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 + ρmangh2

since pB= pC, pA + ρgh1 = ρmangh2

pA = ρmangh2 - ρgh1

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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. ρman >> ρ. In this case the term ρ can be neglected, and the gauge pressure is given by

pA = ρmangh2


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A U-tube manometer similar to that shown in figure is used to measure that gauge pressure of a fluid P of density ρ = 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 ρ = 700 kgm-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 two points, the pressure difference between these two points can be measured.

2.6.3 MEASUREMENT OF PRESSURE DIFFERENCE USING “U”-TUBE MANOMETER

Fluid density, ρ

h b

h a

hA

B

C D

E

Manometric fluid density, ρman

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

Pressure at C = Pressure at D pC = pD

pC = pA + ρgha

pD = pB + ρg(hb- h) + ρmangh pA + ρgha = pB + ρg(hb- h) + ρmangh

• Giving the pressure differencepA - pB = ρg(hb- h) + ρmangh – ρgha

• Again, if the fluid whose pressure difference is being measured is a gas and ρman >> ρ, then the terms involving ρ can be neglected, so pA - pB = ρmangh

2.6.3 MEASUREMENT OF PRESSURE DIFFERENCE USING “U”-TUBE MANOMETER

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Two pipes containing the same fluid of density ρ = 990 kg m-3 are connected using a U-tube manometer. What is the pressure between the two pipes if the manometer contains fluid of relative density 13.6?


Fluid density, ρ

C

Manometric fluid density, ρman

h = 0.5 m

h a= 1.5 m h b = 0.75 m

E

AB

D

Fluid density, ρ

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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 the diameter 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 pressure difference 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 (лd2 / 4)

• And the fall in level of the left side is


( ) 22

2

22

1

D/dz=4/Dπ

)4/dπ(z=

sideleftofareamovedvolume

=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 = ρgz2

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

p - p g(z z )

g z d / D z

gz d / D

1 2 1 2

22 2

22 1


• 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 of the manometer. The above arrangement with a tilted arm is shown in the figure below.

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d

p1

z2

z1

θ

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 = ρg (z1 + z2)

= ρg [[x (d/D)2 ]+ x sinθ]

= ρgx[(d/D)2 + sinθ)]

• if D is very much larger than d then (d/D)2 is very small so p1 – p2 = ρg x sinθ

• 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.


( )2 2

21 1

2

πD πdz = x or z = x d / D

4 4

z = x sin θ

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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 upward at 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

θ=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. Burrs would alter the flow causing local pressure variations 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 taken

simultaneously 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

chapter 2 pressure and head

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