Pressure & it’s Measurement. Pressure & it’s Measurement Pressure & Measurement...

Pressure & it’s Measurement

Pressure & it’s Measurement

Pressure & Measurement -Pascal’s law -Piezo-meter & Manometer -Atmospheric - Absolute - Gauge - Negative

Buoyancy force -Meta Center - Equilibrium

Section I

Pressure: It is define as the ration of Force to areaunit: kgf/m2 , kgf/cm2 , N/m2, N/mm2 1 pa = 1 N/m2

Pascal’s Law:

px. dy. 1

py. dx. 1

pz. ds. 1

dydx

ds

B

A C

θ θ

px, py and pz are the pressure or intensity of pressure acting on the face AB, AC and BC.

Let ∟ABC = θ

Then the forces acting on the element are:

1. Pressure force normal to the surfaces.2. Wight of element in the vertical direction.

Force on the face AB = px . Area of the face AB = px. dy. 1 face AC = py. dx. 1 face BC = pz. ds. 1

Weight of element

Resolving the forces in x-direction, we have

(1)

px. dy. 1

py. dx. 1

pz. ds. 1

dy d

x

ds

B

A C

θ θ

Similarly, resolving the force in y-direction we get

But and also elements is very small and hence weight is negligible.

(2)

1 & 2

Pressure variation in a fluid at rest: (Hydrostatic Law)

Z

A

CD

B

Z

ΔZ

pΔA

P+ dp/dz. ΔZ) ΔA

Hydrostatic Law: the rate of increase of pressure in a vertically downward direction must be equal tothe specific weight of the fluid atThat point.

ΔA = Area of Element

ΔZ = Height of fluid element

p = Pressure on face AB

Z = Dis. Of fluid element from free Surface

Then the forces acting on the element are:

1. Pressure force on AB = p . ΔA

2. Pressure force on CD =

3. Weight of fluid element

4. Pressure force on surface BC and AD are equal and opposite

For equilibrium of fluid elements,

Ex 1 A hydraulic pressure has a ram of 30 cm diameterAnd a plunger of 4.5 cm diameter.Find the weight lifted by the hydraulic press when the force appliedAt the plunger is 500N.

Ans: Pressure = 314465.4 N/m2, W = 22222N

Ex 2 Calculate the pressure due to a column of 0.25 ma)Waterb)Oil of Sp.gr 0.8c)Mercury of Sp. Gr. 13.6Take density of water = 1000kg/m3.

Ans: a) 2452.5 N/m2, b) 1962 N/m2 c) 33354 N/m2

Ex 3 The pressure intensity any point in a fluid is given 3.924 N/cm2.Find the corresponding height of fluid when the fluid isa)Waterb)Oil of Sp.gr 0.9.

Ans: a) 4 m, b) 4.44m

1. Absolute Pressure: it is define as the pressure which is measured with reference to absolute vacuum pressure.

Various Types of Pressure:

2. Gauge pressure: it is defined as the pressure which is measured with help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum.

3. Vacuum pressure: it is defined as the pressure below the atmospheric pressure

Vacuum pressure = Atmospheric pressure- Absolute pressure

Measurement of Pressure:1) Manometers 2) Mechanical Gauge

1) Manometers: It is define as devices used for measurement the pressure at a point in fluid by balancing the column of fluid by the same or another column of fluid.

1) Simple Manometer 2) Differential ManometersSimple manometer consists of a glass tube having one of its ends connected to point where pressure is to be measured and other end remains open to atmosphere

Differential manometers are the Devices used for measuring the difference of pressure between two points in the pipe

1) Simple Manometer - Piezometer - U-tube Manometer - Single Column Manometer

1. Piezometer

2. U-tube Manometer

a) For Gauge pressure

Pressure above A-A left =

Pressure above A-A Right =

b) For Vacuum pressure

Pressure above A-A left =

Pressure above A-A Right = 0

Ex 1 The right limb of a simple U-tube manometer containing Mercury is open to the atmosphere while the left limb is connected To a pipe in which a fluid of Sp.gr. 0.9 is flowing.The centre of the pipe is 12 cm below the level of mercury in the Right limb.Find the pressure of fluid in the pipe if the difference of mercuryLevel in the two limbs is 20cm.

Ans: p = 2.587 N/cm2

Ex 2 A simple U-Tube manometer containing mercury is connectedTo a pipe in which a fluid of Sp.Gr. 0.8 and having vacuum pressureIs flowing. The other end of the manometer is open to atmosphere.Find the vacuum pressure in pipe, if the difference of mercuryLevel in the two limbs is 40 cm and the height of fluid in the leftFrom the centre of pipe is 15 cm below.

Ans: p = - 5.454 N/cm2

Single Column Manometer: a) Vertical Single Column Manometer

Rise of heavy liquid level in Right

Pressure in the right limb above Y-Y

Pressure in the Left limb above Y-Y

Ex 1 A Single column manometer is connected to a pipe containingA liquid of Sp.gr. 0.9 as shown in figure. Find the pressure in thePipe if the area of reservoir is 100 times the area of the tube for theManometer reading shown in figure. The specific gravity of Mercury is 13.6.

Ans: p = 5.12 N/cm2

A/a =100

b) Inclined Single column Manometer:

L = Length of heavy liquid moved in right limb from X-XΘ = Inclination of limbh2 = L sin θ

Ex 1 An inclined single column manometer is connected to a pipe Containing a liquid of sp.gr. 0.9 as shown in Fig. The area of theReservoir is 100 times the area of the tube. Calculate pressure in pipe.The manometric liquid is mercury of sp. Gr. 13.6.

Ans: P = 35.43 kN/m2

II) Differential Manometers: 1) U tube diff. Manometer

h = Difference of Mercury levely = Dis. Of centre B from the Mercury Right limbx = Dis. Of centre A, From Mercury, Left limbς1 = Density of liquid Aς2 = Density of liquid Bςg = Density of Mercury

a) Two pipes at different level:

Pressure X-X left limb=

Pressure X-X Right limb=

= Equating the press.

b) A and B at Same level:

Pressure X-X left limb=

Pressure X-X Right limb=

=

Ex 1 A pipe contains an oil of Sp. Gr. 0.9. A differential manometer Connected same level at the two points A and B shows a difference in mercury Level as 15 cm. Find the difference of pressure at thetwo points

Ans: p = 18688 N/m2

Ex 2 A differential manometer is connected at the two points A andB of two pipes as shown in figure. The pipe A contains a liquid ofSp. gr. 1.5 while pipe B contains a liquid of Sp.gr. 0.9.The pressure at A and B are 1 kgf/cm2 and 1.80 kgf/cm2 respectively.Find the difference in mercury level in the differential manometer.

Ans: h = 18.1 cm

II) Differential Manometers: 2) Inverted U tube diff. Manometer

h1 = Height of liquid in left limb below the datum line X-Xh2 = Height of liquid in right limbh = Difference of light liquidς1 = Density of liquid Aς2 = Density of liquid Bςg = Density of MercurypA = Pressure at ApB = Pressure at BPressure X-X left limb=

Pressure X-X Right limb=

Ex 1 Water is flowing through two different pipes to which an Inverted differential manometer having an oil Sp.gr. 0.8 is connected.The pressure head in the pipe A is 2 m of water.Find the pressure in the pipe B for the manometer readings as shownIn figure.

Ans: Pb = 1.8599 N/cm2

Ex 2 In figure an inverted differential manometer is connected toTwo pipes A and B which convey water. The fluid in manometerIs oil of Sp.gr. 0.8. for the manometer readings shown in figure,Find the pressure difference between A and B.

Ans: PA- PB = 4512.6-2943=1569.6 N/cm2

Section II

Total Pressure: Total pressure is defined as the force exerted by static fluid on a surface either plane or curved when the fluid comes in contact with surface

- This force always acts normal to the surface

Centre of Pressure: CP is define as the point of application of the total pressure on the surface.

4 cases Submerged surfaces :

- Vertical plane surface- Horizontal plane surface- Inclined plane surface- Curved surface

1. Vertical Plane Surface Submerged in Liquid:

A = Total area of the Surface

= Dis. Of C.G. of the area from free surface of liquid

G = Center of gravity of plane surfaceP = Center of Pressureh* = Dis. Of Centre of pressure from free surface

a) Total pressure: Consider a strip Thickness dh and width b at a depth of hPressure intensity on strip p = ς g h

Area of the strip dA = b . dh

Total pressure force on strip dF = p . Area = ς g h. b. dhTotal pressure force on the whole surface,

Total pressure

b) Centre of pressure (h*): by the “Principal of Moment” Moment of the force about free surface = F . h*

Moment of the force dF, acting on strip =dF . h (dF = ς g h. b. dh) = ς g h. b. dh. hSum of moments of all such forces about free surface of liquid

Sum of moments about free surface

Parallel axis

Ex 1 A rectangular plane surface is 2 m wide and 3 m deep. It lies inVertical plane in water. Determine the total pressure and positionOf centre of pressure on the plane surface when its upper edge isHorizontal and a) Coincides with water surfaceb) 2.5 m below the free water surface.

Ans: F = 88290 N, h* = 2.0 m B) F = 235440 N, h* = 4.1875 m

Ex 1 Determine the total pressure on a circular plate of diameter 1.5 m which is placed vertically in water in such a way that the Centre of the plate is 3 m below the free surface of water.Find the position of centre of pressure also.

2. Horizontal Plane Surface Submerge in liquid:

h* = Depth of centre of pressure from free surface = h

3. Inclined plane surface submerged in liquid:

Total pressure

Centre of pressure (h*)

Ex 1 A rectangular plane surface 2 m wide and 3 m deep lies in water in such a way that its plane makes an angle of 30o with the surface of water. Determine the total pressure and position of centre of pressure when the upper edge is 1.5 m below the free water surface.

Ans: F = 132435 N, h* = 2.33 m

(h*)

Ex b) A rectangular plane surface 3 m wide and 4 m deep lies in water in such a way that its plane makes an angle of 30o with the free surface of water. Determine the total pressure force and position of centre of pressure, when the upper edge is 2 m below the free water surface.

Ans: F = 353.16 N, h* = 3.11 m

(h*)

Ex A Circular plate 3 m dia. Is immersed in water in such a way that its greatest and least depth below the free surface are 4 m and 1.5 m respectively. Determine the total pressure on one face of the plate and position of the centre of pressure.

(h*)

Ans: F = 190621 N, h* = 2.89 m

4. Curved surface submerged in liquid:

Total pressure

Centre of pressure (h*)

Section III

Buoyancy and FloatationBuoyancy Force: When a body is immersed in a fluid, an upward

force is exerted by the fluid on the body. This upward forces is equal to the weight of the fluid displaced by the body and is called the buoyancy force

Centre of Buoyancy: It is defined as the point, through which the force of buoyancy is supposed to act.

Meta-Centre : It is define as the point about which a body starts oscillating when the body is tilted by small angle.

OrThe meta-centre may also be define as the point at which the line of

action of the force of buoyancy will meet the normal axis of the body when the body is given a small angular displacement

Meta-Centric height : The distance MG, i.re. the distance between the meta-centre of a floating body and the centre of gravity of the body is called meta-centric height.

Ex 1 Find the volume of the water displaced and position of centre of buoyance for a wooden block of width 2.5 m and of the depth 1.5 m, when it floats horizontally in water. The density of wooden block is 650 kg/m3 and its length 6 m.

W = ς g V

Volume of water Disp = Weight/Weight den. Of water

Ans: W = 143471 N, V = 14.625 m3, h = 0.975 m, CB = 0.48 m from B.

b h L = V

Analytical method for meta-centre height :

Wedge shape prism BOB’ (Emerges in)

Wedge shape prism AOA’ (Emerges out)

Couple Due to Wedges:The height of strip

Area of strip = Height x thickness

Volume of strip = Area x L

Weight of strip

Moment of the couple = Weight of each strip x Distance between these two waights

Moment of the couple for the whole wedge

(1)

Moment of the couple due to shifting of centre of buoyancy from B to B1

(2)

From Eq. (1) & (2)

W= Weight of the body = ς g x volume of the fluid displaced by the body = ς g x v

Ex 1 A rectangular pontoon is 5 m long, 3 m wide and 1.20 m high. The depth of immersion of the pontoon is 0.80 m in sea water. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the meta-centric height. The density for sea water = 1025 kg/m3.

Ans: GM= 0.7375 m.

Section.

Plan.

Condition of equilibrium of a floating and Sub merged bodies:

1) Stability of Sub-merge body:

a) Stable Equilibrium: When W = FB and point B is above G, the body is said to be stable equilibrium

b) Unstable Equilibrium: if W = FB , but the centre of buoyancy (B) is below centre of gravity (G), the body is in unstable equilibrium.

c) Neutral Equilibrium: if W = FB and G are at the same point, the body is said to be Neutral Equilibrium

1) Stability of Floating body:

a) Stable Equilibrium: if the point M is above G, the floating body will be in stable equilibrium

b) Unstable Equilibrium: if the point M is below G, the body will be in unstable equilibrium.

c) Neutral Equilibrium: if the point M is at Centre of Gravity (CG) of the body, the floating body will be in neutral equilibrium.

Experimental method of determination of meta-centre height :

Prepared by, Dr Dhruvesh Patel

Prepared by, Dr Dhruvesh Patel