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.

𝑝 =𝐹

𝐴

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 ×AreaABCD sin Ө= -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 ×volume of element

= - g × 0.5 x y z

Fys = - ps ×AreaABCD cos Ө= -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.

•

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+ δp)A - gA s cos Ө = 0

δp = - gs cos Ө

𝛿𝑝

𝛿𝑠= − 𝑔 cos 𝜃

𝑑𝑝

𝑑𝑠= −𝜌𝑔 cos 𝜃

If 𝜃 = 90𝑜

𝑑𝑝

𝑑𝑠

90𝑜=

𝑑𝑝

𝑑𝑥=

𝑑𝑝

𝑑𝑦= 0

𝑑𝑝

𝑑𝑠

0𝑜=

𝑑𝑝

𝑑𝑧= −𝜌 𝑔

𝑝2 − 𝑝1

𝑧2 − 𝑧1= −𝜌𝑔

𝑝2−𝑝1=−𝜌𝑔 𝑧2 − 𝑧1

PRESSURE AND HEADIn a static fluid of constant density we have the relationshipas shown above. This can be integrated to give

𝑝 =−𝜌𝑔𝑧 + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

In a liquid with a free surface the pressure at any depth zmeasured from the free surface so that z = -h

𝑝 = 𝜌𝑔ℎ + 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑝 = 𝜌𝑔ℎ + 𝑝𝑎𝑡𝑚𝑜𝑠𝑝 ℎ𝑒𝑟𝑖𝑐

𝑝𝑔𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ

𝑝𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 = 𝜌𝑔ℎ + 𝑝𝑎𝑡𝑚𝑜𝑠𝑝 ℎ𝑒𝑟𝑖𝑐

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, =13.6×103 kg m-3.

Sol.

head in terms of water

head in terms of mercury

p = ρgh

ℎ =𝑝

𝜌𝑔=

500 × 103

1000 × 9.81= 50.95𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

ℎ =𝑝

𝜌𝑔=

500 × 103

13.6 × 103 × 9.81= 3.75𝑚 𝑜𝑓 𝑚𝑒𝑟𝑐𝑢𝑟𝑦

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

𝑝𝐴 = 𝜌𝑔ℎ1

𝑝𝐵 = 𝜌𝑔ℎ2

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 =𝜌gh

𝜌=ρ water ×relative 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

𝑝𝐵 = 𝑝𝐴 + 𝜌𝑔ℎ1 𝑝𝐶 = 𝑝𝐴𝑡𝑚𝑜𝑠𝑝 ℎ𝑒𝑟𝑖𝑐 + 𝜌𝑚𝑎𝑛 𝑔ℎ2

𝑝𝐵 = 𝑝𝐶

𝑝𝐴 = 𝑚𝑎𝑛

𝑔ℎ2 − 𝜌𝑔ℎ1

If the fluid being measured is a gas, the density

will probably be very low in comparison to the

density of the manometric fluid.

𝑝𝐴 = 𝜌𝑚𝑎𝑛 𝑔ℎ2

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:

𝑝𝐴 − 𝑝𝐵 = 𝜌𝑔 ℎ𝑏 − ℎ𝑎 + 𝜌𝑚𝑎𝑛 − 𝜌 𝑔ℎ A

B

𝒑𝑨 + 𝝆𝒈𝒉𝒂 = 𝒑𝑩+𝝆𝒈 𝒉𝒃 − 𝒉 + 𝝆𝒎𝒂𝒏𝒈𝒉

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

𝒑𝒄 = 𝒑𝑨 + 𝝆𝒈 𝒉𝑨

𝒑𝑫 = 𝒑𝑩 + 𝝆𝒈 (𝒉𝑩 − 𝒉) + 𝝆𝒎𝒂𝒏𝒈𝒉

𝒑𝑨 - 𝒑𝑩 =𝝆g (𝒉𝑩 - 𝒉𝑨) + hg( 𝝆 man -𝝆)

= 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

= 𝑧2 × 𝜋𝑑2/4

𝑧1 =𝑉𝑜𝑙𝑢𝑚𝑒 𝑚𝑜𝑣𝑒𝑑

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒=

𝑧2 × 𝜋𝑑2/4

𝜋𝐷2/4 = 𝑧2

𝑑

𝐷

2

𝑝1 − 𝑝2 = 𝜌𝑔 𝑧2 + 𝑧2 𝑑

𝐷

2

= 𝜌𝑔𝑧2 1 + 𝑑

𝐷

2

Clearly if D is very much larger than d then (d/D)2 is very small so

𝑝1 − 𝑝2 = 𝜌𝑔𝑧2

Tilted manometer

𝑝1 − 𝑝2 = 𝜌𝑔𝑧2 = 𝜌𝑔𝑥 sin 𝜃

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

𝒑𝟏 − 𝒑𝟐 = 𝝆𝒈𝒉 = 𝝆𝒈 𝒛𝟏 + 𝒛𝟐

for a manometer where 𝝆man >>𝝆

where 𝒛𝟐 = 𝒙𝐬𝐢𝐧 𝜽 , A1 z1 = A2 x and 𝒛𝟏 = 𝒙 𝒅/𝑫 𝟐

where x is the reading on the manometer scale.

p1 is atmospheric i.e. p1 = 0

𝒑𝟐 = −𝝆𝒈𝒙 𝐬𝐢𝐧𝜽 + 𝒅

𝑫 𝟐

And x = -50mm = -0.05m,

−𝒑𝟐 = 𝟎. 𝟕𝟖𝟓 × 𝟏𝟎𝟑 × 𝟗. 𝟖𝟏 × (−𝟎. 𝟎𝟓) 𝐬𝐢𝐧 𝟏𝟓 + 𝟎.𝟎𝟎𝟓

𝟎.𝟎𝟑𝟓 𝟐

𝒑𝟐 = 𝟏𝟎𝟕. 𝟐𝐍/𝐦𝟐

If the movement in the large cylinder is ignored the term (d/D)2 will

disappear: 𝒑𝟏 − 𝒑𝟐 = 𝝆𝒈𝒙 𝐬𝐢𝐧𝜽

𝒑𝟐 = 𝟎. 𝟕𝟖𝟓 × 𝟏𝟎𝟑 × 𝟗. 𝟖𝟏 × 𝟎. 𝟎𝟓 × 𝐬𝐢𝐧 𝟏𝟓 = 𝟗𝟗. 𝟔𝟔 𝐍/𝐦𝟐

So the error induced by this assumption is:

𝒆𝒓𝒓𝒐𝒓 = 𝟏𝟎𝟕. 𝟐 − 𝟗𝟗. 𝟔𝟔

𝟏𝟎𝟐. 𝟐𝟏𝟎𝟎 = 𝟕. 𝟑 %

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.

𝑅 = 𝑝1𝛿𝐴1 + 𝑝2𝛿𝐴2 + ⋯… + 𝑝𝑛𝛿𝐴𝑛 = 𝑝 𝛿𝐴

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

𝐹 = 𝑝𝐴 =𝜌𝑔𝑧𝐴

𝑅 = 𝜌𝑔 𝑧𝛿𝐴

𝑧𝛿𝐴 = 𝐴𝑧 =1st moment of area about the line of the free surface

𝑧𝛿𝐴 = 𝐴 𝑥 sin 𝜃 =1st moment of area about a line through O × sin Ө

𝑹 = 𝝆𝒈𝑨ഥ𝒁 = 𝝆𝒈𝑨ഥ𝒙 𝐬𝐢𝐧𝜽

𝑅 = 𝜌𝑔𝐴𝑍 = 𝜌𝑔𝐴𝑥 sin 𝜃

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

𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒 𝑜𝑛𝛿𝐴 𝑎𝑏𝑜𝑢𝑡 𝑂 = 𝜌𝑔𝑠 sin 𝜃 𝛿𝐴 × 𝑠 = 𝜌𝑔 sin 𝜃 𝛿𝐴 𝑠2

𝑆𝑢𝑚 𝑜𝑓 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒𝑠 𝑜𝑛 𝑎𝑙𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑜𝑓 𝛿𝐴 𝑎𝑏𝑜𝑢𝑡 𝑂 == 𝜌𝑔 sin 𝜃 𝑠2 𝛿𝐴

Moment of 𝑅 𝑎𝑏𝑜𝑢𝑡 O = 𝜌𝑔𝐴𝑥 sin 𝜃 𝑆𝑐 Equating gives,

𝜌𝑔𝐴𝑥 sin 𝜃 𝑆𝑐 = 𝜌𝑔 sin 𝜃 𝑠2 𝛿𝐴

𝑆𝑐 = 𝑠2 𝛿𝐴

𝐴𝑥

2𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 O = 𝐼0 = 𝑠2 𝛿𝐴

𝑆𝑐 =2𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 a line through O

1st Moment of area about a line through O

How do you calculate the 2nd moment of area?

𝐼0 = 𝐼𝐺𝐺 + 𝐴𝑥 2

where IGG is the 2nd moment of area about an axis though the centroid G of the plane.

𝑆𝑐 =𝐼𝐺𝐺

𝐴𝑥 + 𝑥

𝐷 = sin 𝜃 𝐼𝐺𝐺

𝐴𝑥 + 𝑥 =

𝐼𝐺𝐺

𝐴𝑥 sin 𝜃 + 𝑥 sin 𝜃 = 𝑧 +

𝐼𝐺𝐺

𝐴𝑥 sin 𝜃

𝑆𝑐 =2𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎 𝑎𝑏𝑜𝑢𝑡 a line through O

1st Moment of area about a line through O

The parallel axis theorem can be written

𝐷 = 𝑆𝑐 sin 𝜃

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)(20× 50 m2 ) = 97.9 MN Ans.𝑍

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 )π

4 (20 m)2 (50 m) =

153.8 MN

This vertical component acts through the centroid of

the water above the dam, or 4R/3π = 4(20 m)/3π =

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.

𝜃 = tan−1 153.8

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