Many fluid problems do not involve motion.They concern the pressure distribution in a staticfluid and its effect on solid surfaces and onfloating and submerged bodies.
When the fluid velocity is zero, denoted as thehydrostatic condition, the pressure variation is dueonly to the weight of the fluid.
Pressure always acts inward normal to any surface
Pressure
If the force exerted on each unit
area of a boundary is the same,
the pressure is said to be uniform.
π =πΉ
π΄
Units: Newton's per square meter, Nm-2, Kg m-1s-2
(The same unit is also known as a Pascal, Pa, i.e. 1bar = 105 Nm-2 )
Dimensions: M L-1T-2
PASCAL'S LAW FOR PRESSURE AT A POINT
Proof that pressure acts equally in all directions.Summing forces in the x-direction
Fxx +Fxs+Fxy= 0
Fxs = - ps ΓAreaABCD sin Σ¨= -ps s z( y/ s) = -ps y z
px y z+ (-ps y z)=0
px = ps
Summing forces in the y-direction
Fyy +Fys+Fyx+ weight = 0
Weight = - specific weight Γvolume of element
= - g Γ 0.5 x y z
Fys = - ps ΓAreaABCD cos Σ¨= -ps s z( x/ s) = -ps x z
py x z + (-psx z)=0
py = ps
px = py = ps
VARIATION OF PRESSURE VERTICALLY IN A FLUID UNDER GRAVITY
Taking upward as positive, in equilibrium we have
p1 A - p2 A- g A(z2-z1) = 0
p2 - p1 = - g (z2-z1)
Thus in a fluid under gravity, pressure decreases with
increase in height.
EQUALITY OF PRESSURE A T THE SAME LEVEL IN A STATIC FLUID
The fluid is at equilibrium so the sum of the
forces acting in the x direction is zero.
pl A = pr Apl = pr
Example
pl = pp + g z
and
pr = pq + g z
so
pp + g z = pq + g z
pp = pq
This shows that the pressures at the two equal levels, P and Q are
the same.
p1 = p2 and p4 = p5 since these points are at the same elevation in the
same fluid. However, p2 does not equal p3 even though they are at the
same elevation, because one cannot draw a line connecting these
points through the same fluid. In fact, p2 is less than p3 since mercury
is denser than water.
β’
The shape of a container does not matter in hydrostatics.
Hydraulic lifting of heavy objects
GENERAL EQUATION FOR VARIATION OF PRESSURE I N A STATIC FLUID
p A - ( p+ Ξ΄p)A - gA s cos Σ¨ = 0
Ξ΄p = - gs cos Σ¨
πΏπ
πΏπ = β π cos π
ππ
ππ = βππ cos π
If π = 90π
ππ
ππ
90π=
ππ
ππ₯=
ππ
ππ¦= 0
ππ
ππ
0π=
ππ
ππ§= βπ π
π2 β π1
π§2 β π§1= βππ
π2βπ1=βππ π§2 β π§1
PRESSURE AND HEADIn a static fluid of constant density we have the relationshipas shown above. This can be integrated to give
π =βπππ§ + ππππ π‘πππ‘
In a liquid with a free surface the pressure at any depth zmeasured from the free surface so that z = -h
π = ππβ + ππππ π‘πππ‘
π = ππβ + πππ‘πππ π βππππ
ππππ’ππ = ππβ
ππππ πππ’π‘π = ππβ + πππ‘πππ π βππππ
Absolute pressure = Gauge pressure + Atmospheric pressure
vertical height is known as head of fluid.
Since atmospheric pressure changes constantly, it may be difficult to pinpoint the
gauge pressure zero point. Therefore, we use standard atmospheric pressure
set at 101.3 kPa (a).
One commonly used scale is the absolute scale. It starts at the point of no
pressure at all, i.e., the absolute zero pressure. Readings taken on this scale are
called absolute pressure and have suffix (a)
A scale with zero at atmospheric pressure is known as the gauge scale. Readings
made on this scale are called gauge pressure. The name reflects the fact that most
gauges read zero at the atmospheric pressure. To distinguish readings on this
scale, we use suffix (g).
Absolute pressure = Gauge pressure + Atmospheric pressure
Example
If instrument air gauge pressure is 580-kPa (g), what is its absolute value?
p(a) = 580 kPa (g) + 101.3 kPa
= 681.3 kPa (a)
Example Calculate a pressure 500 KNm-2 of water in terms of theheight of a column of water of density = 1000 kg m-3,and in terms of mercury with density, =13.6Γ103 kg m-3.
Sol.
head in terms of water
head in terms of mercury
p = Οgh
β =π
ππ=
500 Γ 103
1000 Γ 9.81= 50.95π ππ π€ππ‘ππ
β =π
ππ=
500 Γ 103
13.6 Γ 103 Γ 9.81= 3.75π ππ πππππ’ππ¦
Pressure in layered fluid
water
mercury
Pressure Measurement and Manometers The Piezometer Tube Manometer
The simplest manometer is a tube, open at the top, which is
attached to a vessel or a pipe containing liquid at a pressure
(higher than atmospheric) to be measured. 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
ππ΄ = ππβ1
ππ΅ = ππβ2
h1 h2
Example of a PiezometerWhat 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 0.85 what would the maximum measurable gauge pressure?
Solution
gauge pressure p =πgh
π=Ο water Γrelative density
The maximum measurable pressure is
when the tube is completely full (h=1.5m).
Any higher and the tube will overflow. p = (0.85 x 9810) x 1.5
p = 12 508 N/m2 (or Pa)
p = 12.5 kN/m2 (or kPa)
U-tube manometer
ππ΅ = ππ΄ + ππβ1 ππΆ = ππ΄π‘πππ π βππππ + ππππ πβ2
ππ΅ = ππΆ
ππ΄ = πππ
πβ2 β ππβ1
If the fluid being measured is a gas, the density
will probably be very low in comparison to the
density of the manometric fluid.
ππ΄ = ππππ πβ2
ExampleUsing a u-tube manometer to measure gauge pressure of fluid
density = 700 kg/m3, and the manometric fluid is mercury, with a
relative density of 13.6. What is the gauge pressure if:
a. h1 = 0.4m and h2 = 0.9m?
b. h1 stayed the same but h2 = -0.1m?
pB = pC
pB = pA + g h1
pB = pAtmospheric+ man g h2
We are measuring gauge pressure so patmospheric = 0, pA
=man g h2 - g h1
a) pA = 13.6 x 103 x 9.81 x 0.9 - 700 x 9.81 x 0.4
= 117 327 N/m2 = 117.3 kN/m2 = (1.17 bar)
b) pA = 13.6 x 103 x 9.81 x (-0.1) - 700 x 9.81 x 0.4
= -16 088.4 N/m2 = -16 kN/m2 = (-0.16 bar)
The negative sign indicates that the pressure is below
atmospheric.
Solution
Differential manometer A differential manometer can beused to measure the difference inpressure between two containers ortwo points in the same system.Again, on equating the pressures atpoints labeled (2) and (3), we mayget an expression for the pressuredifference between A and B:
ππ΄ β ππ΅ = ππ βπ β βπ + ππππ β π πβ A
B
ππ¨ + ππππ = ππ©+ππ ππ β π + ππππππ
In the figure below two pipescontaining the same fluid ofdensity = 990 kg/m3 areconnected using a u-tubemanometer. What is the pressurebetween the two pipes if themanometer contains fluid ofrelative density 13.6?
solution
Example
pC = pD
ππ = ππ¨ + ππ ππ¨
ππ« = ππ© + ππ (ππ© β π) + ππππππ
ππ¨ - ππ© =πg (ππ© - ππ¨) + hg( π man -π)
= 990 x9.81x(0.75-1.5) + 0.5x9.81 x(13.6-0.99) x 103
= -7284 + 61852 = 54 568 N/m2 (or Pa or 0.55 bar).
Inclined-tube manometer
3
2L2
Advances to the "U" tube manometer
. 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
= π§2 Γ ππ2/4
π§1 =ππππ’ππ πππ£ππ
π΄πππ ππ ππππ‘ π πππ=
π§2 Γ ππ2/4
ππ·2/4 = π§2
π
π·
2
π1 β π2 = ππ π§2 + π§2 π
π·
2
= πππ§2 1 + π
π·
2
Clearly if D is very much larger than d then (d/D)2 is very small so
π1 β π2 = πππ§2
Tilted manometer
π1 β π2 = πππ§2 = πππ₯ sin π
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.
x
Example of an Inclined Manometer An inclined tube manometer consists of a vertical cylinder
35mm diameter. At the bottom of this is connected a tube5mm in diameter inclined upward at an angle of 15o to thehorizontal, the top of this tube is connected to an air duct.The vertical cylinder is open to the air and the manometricfluid has relative density 0.785. Determine the pressure inthe air duct if the manometric fluid moved 50mm alongthe inclined tube. What is the error if the movement of thefluid in the vertical cylinder is ignored?
x
ππ β ππ = πππ = ππ ππ + ππ
for a manometer where πman >>π
where ππ = ππ¬π’π§ π½ , A1 z1 = A2 x and ππ = π π /π« π
where x is the reading on the manometer scale.
p1 is atmospheric i.e. p1 = 0
ππ = βπππ π¬π’π§π½ + π
π« π
And x = -50mm = -0.05m,
βππ = π. πππ Γ πππ Γ π. ππ Γ (βπ. ππ) π¬π’π§ ππ + π.πππ
π.πππ π
ππ = πππ. ππ/π¦π
If the movement in the large cylinder is ignored the term (d/D)2 will
disappear: ππ β ππ = πππ π¬π’π§π½
ππ = π. πππ Γ πππ Γ π. ππ Γ π. ππ Γ π¬π’π§ ππ = ππ. ππ π/π¦π
So the error induced by this assumption is:
πππππ = πππ. π β ππ. ππ
πππ. ππππ = π. π %
FORCES ON SUBMERGED SURFACES IN STATIC FLUIDS
Pressure Distribution on Flat Surfaces
Pressure Distribution on Curved Surfaces
General Submerged Plane
Hydrostatic force on an arbitrary plane surface of area Ainclined at an angle below the free surface.
π = π1πΏπ΄1 + π2πΏπ΄2 + β―β¦ + πππΏπ΄π = π πΏπ΄
This resultant force will act through the centre of pressure, hence we
can say, if the surface is a plane, the force can be represented by one
single resultant force, acting at right-angles to the plane through the
centre of pressure.
Resultant Force and Centre of Pressure on a Submerged Plane Surface in a liquid
πΉ = ππ΄ =πππ§π΄
π = ππ π§πΏπ΄
π§πΏπ΄ = π΄π§ =1st moment of area about the line of the free surface
π§πΏπ΄ = π΄ π₯ sin π =1st moment of area about a line through O Γ sin Σ¨
πΉ = πππ¨ΰ΄₯π = πππ¨ΰ΄₯π π¬π’π§π½
π = πππ΄π = πππ΄π₯ sin π
This resultant force acts at right angles to the plane through the centre of
pressure, C, at a depth D. The moment of R about any point will be equal to the
sum of the moments of the forces on all the elements of the plane about the
same point. Take moments about point O in the figure
ππππππ‘ ππ πππππ πππΏπ΄ ππππ’π‘ π = πππ sin π πΏπ΄ Γ π = ππ sin π πΏπ΄ π 2
ππ’π ππ ππππππ‘ ππ ππππππ ππ πππ πππππππ‘π ππ πΏπ΄ ππππ’π‘ π == ππ sin π π 2 πΏπ΄
Moment of π ππππ’π‘ O = πππ΄π₯ sin π ππ Equating gives,
πππ΄π₯ sin π ππ = ππ sin π π 2 πΏπ΄
ππ = π 2 πΏπ΄
π΄π₯
2ππ ππππππ‘ ππ ππππ ππππ’π‘ O = πΌ0 = π 2 πΏπ΄
ππ =2ππ ππππππ‘ ππ ππππ ππππ’π‘ a line through O
1st Moment of area about a line through O
How do you calculate the 2nd moment of area?
πΌ0 = πΌπΊπΊ + π΄π₯ 2
where IGG is the 2nd moment of area about an axis though the centroid G of the plane.
ππ =πΌπΊπΊ
π΄π₯ + π₯
π· = sin π πΌπΊπΊ
π΄π₯ + π₯ =
πΌπΊπΊ
π΄π₯ sin π + π₯ sin π = π§ +
πΌπΊπΊ
π΄π₯ sin π
ππ =2ππ ππππππ‘ ππ ππππ ππππ’π‘ a line through O
1st Moment of area about a line through O
The parallel axis theorem can be written
π· = ππ sin π
The Second Moment of Area of Some Common Shapes.
Hydrostatic Force on Submerged Curved Surfaces
Horizontal component of force on surface
FH = F = resultant force of liquid acting on vertically projected area (BC) andacting through the centre of pressure of F.
Vertical component of force on surface
FV = W = weight of liquid vertically above the surface (ADEC) and through thecentre of gravity of the liquid mass.
Resultant force
Liquid below surface
FH = F = resultant force of liquid acting on vertically projected area (AB) and acting through the centre of pressure of F.
FV = W = weight of imaginary liquid (i.e., same liquid as on the other side of the surface) vertically above the surface (ADCB) and through the centre of gravity of the liquid mass.
ExampleA tank holding water has a triangular gate,
hinged at the top, in one wall. Find the moment
at the hinge required to keep this triangular
gate closed.
An Example of Force on a Curved Wall
Find the magnitude anddirection of the resultantforce of water on the gateshown in the figure.
Example The dam in figure is a
quarter-circle 50 m wide intothe paper. Determine thehorizontal and verticalcomponents of hydrostaticforce against the dam andthe point CP where theresultant strikes the dam.
Solution: The horizontal force acts as if the dam were vertical and 20 m high:
FH = Avert = (9790 N/m3 )(10 m)(20Γ 50 m2 ) = 97.9 MN Ans.π
This force acts 2/3 of the way down or 13.33 m from the surface, as in the figure
.
The vertical force is the weight of the fluid
above the dam:
FV = (Vol)dam = (9790 N/m3 )Ο
4 (20 m)2 (50 m) =
153.8 MN
This vertical component acts through the centroid of
the water above the dam, or 4R/3Ο = 4(20 m)/3Ο =
8.49 m to the right of point A, as shown in the figure.
The resultant hydrostatic force is
R = [(97.9 MN)2 +(153.8 MN)2]1/2 = 182.3 MN
acting down at an angle of 57.5o from the horizontal
The line of action of R strikes the circular-arc dam AB at the center of
pressure CP, which is 10.744 m to the right and 3.1316 m up from point A,
as shown in the figure.
π = tanβ1 153.8
97.9 = 57.5o
x
y 4.1959
X / 8.49 =4.1959/15.80408
X= 2.254 m
y / 13.33 = 4.1959/ 15.80408
y = 3.5384 m