pressure and head
Post on 07-Apr-2018
222 Views
Preview:
TRANSCRIPT
-
8/6/2019 Pressure and Head
1/37
1
FLUID MECHANICS (CHE 203)
CHAPTER 2
PRESSURE AND HEAD
-
8/6/2019 Pressure and Head
2/37
2
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
-
8/6/2019 Pressure and Head
3/37
3
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
-
8/6/2019 Pressure and Head
4/37
4
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
-
8/6/2019 Pressure and Head
5/37
5
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.
2.2 PASCALS LAW FOR PRESSURE AT A POINT
-
8/6/2019 Pressure and Head
6/37
6
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
2.2 PASCALS LAW FOR PRESSURE AT A POINT
-
8/6/2019 Pressure and Head
7/37
7
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
2.2 PASCALS LAW FOR PRESSURE AT A POINT
-
8/6/2019 Pressure and Head
8/37
8
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
2.2 PASCALS LAW FOR PRESSURE AT A POINT
-
8/6/2019 Pressure and Head
9/37
9
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.
2.2 PASCALS LAW FOR PRESSURE AT A POINT
-
8/6/2019 Pressure and Head
10/37
10
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,
-
8/6/2019 Pressure and Head
11/37
11
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
-
8/6/2019 Pressure and Head
12/37
12
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
-
8/6/2019 Pressure and Head
13/37
13
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.
2.4 EQUALITY OF PRESSURE AT THE SAME
LEVEL IN A STATIC FLUID
zz
Q
R
P
L
Two tanks of different cross-section connected by a pipe
-
8/6/2019 Pressure and Head
14/37
14
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.
2.4 EQUALITY OF PRESSURE AT THE SAME
LEVEL IN A STATIC FLUID
-
8/6/2019 Pressure and Head
15/37
15
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
-
8/6/2019 Pressure and Head
16/37
16
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
2.5 PRESSURE AND HEAD
-
8/6/2019 Pressure and Head
17/37
17
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.
2.5 PRESSURE AND HEAD
-
8/6/2019 Pressure and Head
18/37
18
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
-
8/6/2019 Pressure and Head
19/37
19
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
-
8/6/2019 Pressure and Head
20/37
20
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
-
8/6/2019 Pressure and Head
21/37
21
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
-
8/6/2019 Pressure and Head
22/37
22
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
-
8/6/2019 Pressure and Head
23/37
23
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
-
8/6/2019 Pressure and Head
24/37
24
2.6.2 THE U-TUBE MANOMETER
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
-
8/6/2019 Pressure and Head
25/37
25
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
2.6.2 THE U-TUBE MANOMETER
-
8/6/2019 Pressure and Head
26/37
26
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
-
8/6/2019 Pressure and Head
27/37
27
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?
EXAMPLE 2.4 - U-TUBE MANOMETER
-
8/6/2019 Pressure and Head
28/37
28
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
-
8/6/2019 Pressure and Head
29/37
29
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
-
8/6/2019 Pressure and Head
30/37
30
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?
EXAMPLE 2.5 - U-TUBE MANOMETER
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,
-
8/6/2019 Pressure and Head
31/37
31
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
-
8/6/2019 Pressure and Head
32/37
32
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
2.6.4 ADVANCES TO THE U-TUBE
MANOMETER
( ) 22
2
2
2
1
/dz=4/
)4/d(z=
sideleftofarea
movedvolume=z
-
8/6/2019 Pressure and Head
33/37
33
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
2.6.4 ADVANCES TO THE U-TUBE
MANOMETER
-
8/6/2019 Pressure and Head
34/37
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.
34
2.6.4 ADVANCES TO THE U-TUBE
MANOMETER
d
p1
z2
z1
1
2
p2
x
Datum line
D
-
8/6/2019 Pressure and Head
35/37
35
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.
2.6.4 ADVANCES TO THE U-TUBE
MANOMETER
( )
2 2
2
1 1
2
D d
z = x or z = x d / D4 4
z = x sin
-
8/6/2019 Pressure and Head
36/37
36
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
-
8/6/2019 Pressure and Head
37/37
37
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
top related