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CIVE1400: Fluid Mechanics Section 2: Statics 35
Pressure And Head
We have the vertical pressure relationship
dp dz
U g ,
integrating gives
p = - U gz + constant
measuring z from the free surface so that z = -h
x
y
z (^) h
p U gh constant
surface pressure is atmospheric, p atmospheric.
p
p gh p
atmospheric
atmospheric
constant so U
CIVE1400: Fluid Mechanics Section 2: Statics 36
It is convenient to take atmospheric pressure as the datum
Pressure quoted in this way is known as gauge pressure i.e. Gauge pressure is p (^) gauge = U g h
The lower limit of any pressure is the pressure in a perfect vacuum.
Pressure measured above a perfect vacuum (zero) is known as absolute pressure
Absolute pressure is p (^) absolute = U g h + p (^) atmospheric
Absolute pressure = Gauge pressure + Atmospheric
CIVE1400: Fluid Mechanics Section 2: Statics A gauge pressure can be given using height of any fluid. p U gh This vertical height is the head.
If pressure is quoted in head , the density of the fluid must also be given. Example: What is a pressure of 500 kNm-2^ in head of water of density, U = 1000 kgm -^3 Use p = U gh ,
h
p g
m
u
U u
3
.
. of water
In head of Mercury density U = 13.6u 103 kgm-^.
h m
u u u
3
. 3..^ of Mercury In head of a fluid with relative density J = 8.7. remember U = J u Uwater)
h m
u u u
3
..
. of fluid J= 8.
CIVE1400: Fluid Mechanics Section 2: Statics Pressure Measurement By Manometer
Manometers use the relationship between pressure and head to measure pressure
The Piezometer Tube Manometer
The simplest manometer is an open tube. This is attached to the top of a container with liquid at pressure. containing liquid at a pressure.
h1 h
A
B
The tube is open to the atmosphere, The pressure measured is relative to atmospheric so it measures gauge pressure.
CIVE1400: Fluid Mechanics Section 2: Statics 39
Pressure at A = pressure due to column of liquid h (^1)
p A = U g h 1
Pressure at B = pressure due to column of liquid h (^2)
p B = U g h 2
Problems with the Piezometer:
- Can only be used for liquids
- Pressure must above atmospheric
- Liquid height must be convenient i.e. not be too small or too large.
CIVE1400: Fluid Mechanics Section 2: Statics 40
An Example of a Piezometer. What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m? And if the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure?
CIVE1400: Fluid Mechanics Section 2: Statics Equality Of Pressure At The Same Level In A Static Fluid
Fluid density ρ
pl, A
Area A
weight, mg
Face L Face R
p (^) r, A
Horizontal cylindrical element cross sectional area = A mass density = U left end pressure = p (^) l right end pressure = p (^) r
For equilibrium the sum of the forces in the x direction is zero. p (^) l A = p (^) r A
p (^) l = p (^) r
Pressure in the horizontal direction is constant.
This true for any continuous fluid.
CIVE1400: Fluid Mechanics Section 2: Statics
P Q
L R
z z
We have shown p (^) l = p (^) r For a vertical pressure change we have
p l p p U gz
and
p r p q U gz
so p gz p gz p p
p q p q
U U
Pressure at the two equal levels are the same.
CIVE1400: Fluid Mechanics Section 2: Statics 45
Pressure difference measurement
Using a “U”-Tube Manometer.
The “U”-tube manometer can be connected at both ends to measure pressure difference between these two points
ha
A
B
h
hb
C D
E
Fluid density ρ
Manometric fluid density ρ man
CIVE1400: Fluid Mechanics Section 2: Statics 46
pressure at C = pressure at D
pC = pD
pC = pA + U g ha
pD = pB + U g (hb + h) + U man g h
pA + U g ha = pB + U g (hb + h) + U man g h
Giving the pressure difference
pA - pB = U g (hb - ha) + ( U man - U )g h
Again if the fluid is a gas U man >> U, then the terms involving U can be neglected,
pA - pB = U man g h
CIVE1400: Fluid Mechanics Section 2: Statics An example using the u-tube for pressure difference measuring In the figure below two pipes containing the same fluid of density U = 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?
ha = 1.5m
A
B
h = 0.5m hb = 0.75m
C D
E
Fluid density ρ
Manometric fluid density ρ man = 13.6 ρ
Fluid density ρ
CIVE1400: Fluid Mechanics Section 2: Statics
Advances to the “U” tube manometer
Problem: Two reading are required. Solution: Increase cross-sectional area of one side. Result: One level moves much more than the other.
Datum line z 1
p 1 p 2
z 2
diameter D diameter d
If the manometer is measuring the pressure difference of a gas of (p 1 - p 2 ) as shown,
we know
p 1 - p 2 = U man g h
CIVE1400: Fluid Mechanics Section 2: Statics 49
volume of liquid moved from
the left side to the right
= z 2 u ( S d^2 / 4)
The fall in level of the left side is
z
z d D
z
d D
1
2
2 2
2
2
©¨^
Volume moved Area of left side
S
S
Putting this in the equation,
p p g z z
d D
gz
d D
1 2 2 2
2
2
2 1
U
U
If D >> d then ( d/D)^2 is very small so
p 1 p 2 U gz 2
CIVE1400: Fluid Mechanics Section 2: Statics 50
Problem: Small pressure difference, movement cannot be read.
Solution 1: Reduce density of manometric fluid.
Result: Greater height change - easier to read.
Solution 2: Tilt one arm of the manometer.
Result: Same height change - but larger movement along the manometer arm - easier to read.
Inclined manometer
CIVE1400: Fluid Mechanics Section 2: Statics
Datum line z 1
p 1 p^2
z 2
diameter D
diameter d
Scale Reader
x
θ
The pressure difference is still given by the height change of the manometric fluid.
p 1 p 2 U gz 2
but,
z x p p gx
2 1 2
sin sin
T
U T
The sensitivity to pressure change can be increased further by a greater inclination.
CIVE1400: Fluid Mechanics Section 2: Statics Example of an inclined manometer. An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of +/- 3%. The inclined arm is 8mm in diameter and the larger arm has a diameter of 24mm. The manometric fluid has density Uman = 740 kg/m^3 and the scale may be read to +/- 0.5mm. What is the angle required to ensure the desired accuracy may be achieved?
CIVE1400: Fluid Mechanics Section 2: Statics 57
Resultant Force and Centre of Pressure on a general plane surface in a liquid.
P
Q
D
z z
O Fluid^ θ density ρ
C
RForce Resultant
d
G G
elementalarea δ A
area A
x
area δ A
s
Sc
O
x
Take pressure as zero at the surface.
Measuring down from the surface, the pressure on
an element G A , depth z ,
p = U gz
So force on element
F = U gz G A
Resultant force on plane
R U g ¦ z A G
(assuming U and g as constant).
CIVE1400: Fluid Mechanics Section 2: Statics 58
¦^ z A^ G is known as
the 1 st^ Moment of Area of the
plane PQ about the free surface.
And it is known that
¦^ z A^ G^ Az
A is the area of the plane z is the distance to the centre of gravity (centroid)
In terms of distance from point O
¦^ z A^ G^ Ax sin^ T
= 1 st^ moment of area u sin T
about a line through O
(as z x sin T )
The resultant force on a plane R gAz gAx
U
U sinT
CIVE1400: Fluid Mechanics Section 2: Statics This resultant force acts at right angles through the centre of pressure, C, at a depth D.
How do we find this position?
Take moments of the forces.
As the plane is in equilibrium: The moment of R will be equal to the sum of the moments of the forces on all the elements G A about the same point.
It is convenient to take moment about O
The force on each elemental area:
Force on G U G
U T G
A gz A g s sin A
the moment of this force is:
Moment of Force on G about O U T G U T G
A g s A s g As
sin u sin 2
U , g and T are the same for each element, giving the total moment as
CIVE1400: Fluid Mechanics Section 2: Statics
Sum of moments U g sin T ¦ s^2 G A
Moment of R about O = R uS (^) c = U g Ax sinT Sc
Equating
U gAx sin T S c U g sin T ¦ s^2 G A
The position of the centre of pressure along the plane measure from the point O is:
S
s A c Ax
2 G
How do we work out the summation term?
This term is known as the 2 nd^ Moment of Area , I (^) o , of the plane (about the axis through O)
CIVE1400: Fluid Mechanics Section 2: Statics 61
2nd moment of area about O I (^) o ¦ s^2 G A
It can be easily calculated for many common shapes.
The position of the centre of pressure along the plane measure from the point O is:
S (^) c
2 Moment of area about a line through O 1 Moment of area about a line through O
nd st
and
Depth to the centre of pressure is
D S (^) c sin T
CIVE1400: Fluid Mechanics Section 2: Statics 62
How do you calculate the 2nd^ moment of area?
2 nd^ moment of area is a geometric property.
It can be found from tables - BUT only for moments about an axis through its centroid = IGG.
Usually we want the 2 nd^ moment of area
about a different axis.
Through O in the above examples.
We can use the
parallel axis theorem
to give us what we want.
CIVE1400: Fluid Mechanics Section 2: Statics
The parallel axis theorem can be written I (^) o I (^) GG Ax^2
We then get the following equation for the position of the centre of pressure
S
I
Ax
x
D
I
Ax
x
c
GG
GG
sin T ¸
(In the examination the parallel axis theorem and the I (^) GG will be given)
CIVE1400: Fluid Mechanics Section 2: Statics
The 2 nd^ moment of area about a line
through the centroid of some common
shapes.
Shape Area A 2 nd^ moment of area, I (^) GG , about an axis through the centroid Rectangle
G G
b
h
bd (^) bd^3
12
Triangle G h h/3G b
bd 2
bd^3 36 Circle
G G
R S R^2 S R^4
Semicircle G R (4R)/(3 π )
S R^2
. R^4
CIVE1400: Fluid Mechanics Section 2: Statics 69
Check this against the moment method:
The resultant force is given by: R gAz gAx
g H
H
gH
u
U U T
U T
U
sin
1 sin 2 1 2
2
and the depth to the centre of pressure by:
D
I
Ax
§ o ©
sin T ¸
and by the parallel axis theorem (with width of 1)
I I Ax
H H
H H
o GG u u § ©¨^
2
1 3 2 3 12
Depth to the centre of pressure
D
H
H
H
3 2
CIVE1400: Fluid Mechanics Section 2: Statics 70
The same technique can be used with combinations of liquids are held in tanks (e.g. oil floating on water). For example:
R
0.8m
1.2m
oil ρ o
water ρ
ρ g0.8 ρ g1.
D
Find the position and magnitude of the resultant force on this vertical wall of a tank which has oil floating on water as shown.
CIVE1400: Fluid Mechanics Section 2: Statics
Submerged Curved Surface
If the surface is curved the resultant force must be found by combining the elemental forces using some vectorial method.
Calculate the
horizontal and vertical components.
Combine these to obtain the resultant
force and direction.
(Although this can be done for all three dimensions we will only look at one vertical plane)
CIVE1400: Fluid Mechanics Section 2: Statics
In the diagram below liquid is resting on
top of a curved base.
FAC R (^) H
Rv R
O
G
A
C B
E D
The fluid is at rest – in equilibrium.
So any element of fluid
such as ABC is also in equilibrium.
CIVE1400: Fluid Mechanics Section 2: Statics 73
Consider the Horizontal forces
The sum of the horizontal forces is zero.
FAC R (^) H
A
C B
No horizontal force on CB as there are no shear forces in a static fluid
Horizontal forces act only on the faces AC and AB as shown.
FAC , must be equal and opposite to R (^) H.
AC is the projection of the curved surface AB onto a vertical plane.
CIVE1400: Fluid Mechanics Section 2: Statics 74
The resultant horizontal force of a fluid above a curved surface is: R (^) H = Resultant force on the projection of the curved surface onto a vertical plane.
We know
- The force on a vertical plane must act horizontally (as it acts normal to the plane).
2. That R H must act through the same point.
So: R (^) H acts horizontally through the centre of pressure of the projection of the curved surface onto an vertical plane.
We have seen earlier how to calculate resultant forces and point of action.
Hence we can calculate the resultant horizontal force on a curved surface.
CIVE1400: Fluid Mechanics Section 2: Statics
Consider the Vertical forces
The sum of the vertical forces is zero.
Rv
G
A
C B
E D
There are no shear force on the vertical edges, so the vertical component can only be due to the weight of the fluid.
So we can say The resultant vertical force of a fluid above a curved surface is:
R (^) V = Weight of fluid directly above the curved surface.
It will act vertically down through the centre of gravity of the mass of fluid.
CIVE1400: Fluid Mechanics Section 2: Statics
Resultant force
The overall resultant force is found by combining the vertical and horizontal components vectorialy,
Resultant force
R R (^) H^2 RV^2
And acts through O at an angle of T.
The angle the resultant force makes to the horizontal is
T
tan ^1 R R
V H
The position of O is the point of interaction of the horizontal line of action of RH and the vertical line of action of RV.
The resultant force and direction of application are calculated in the same way as for fluids above the surface:
Resultant force
R R (^) H^2 RV^2
And acts through O at an angle of T. The angle the resultant force makes to the horizontal is
T
tan ^1 ¸
R
R
V H
An example of a curved sluice gate which experiences force from fluid below. A 1.5m long cylinder lies as shown in the figure, holding back oil of relative density 0.8. If the cylinder has a mass of 2250 kg find a) the reaction at A b) the reaction at B C
D A
B
E
