m303,fluid mechanics fm - m iii ; p.p. binu & sunand c

7/29/2019 M303,Fluid MEchanics FM - M III ; P.P. Binu & Sunand C

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MODULE -3

Equations of Motion :

According to Newtons Second Law of Motion, the net force acting on a fluid element in the

directionx is equal to mass m of the fluid element multiplied by the acceleration ax, in thex-direction.

Mathematically,

Fx = m. ax

In the fluid flow, the following forces are present:

(i) Fg ,gravity force

(ii) Fp,the pressure force

(iii) Fv, force due to viscosity

(iv) Ft, force due to turbulence(v) Fc, force due to compressibility

The net force

Fx = (Fg )x + (Fp )x + (Fv )x + (Ft )x + (Fc )x

(i) For flow, where (Fc) is negligible, the resulting equations of motion are called ReynoldsEquations of Motion.i.eFx = (Fg )x + (Fp )x + (Fv )x + (Ft )x

(ii) For flow, where (Ft) is negligible, the resulting equations of motion are called Navier-Stokes

Equation.i.eFx = (Fg )x + (Fp )x + (Fv )x + (Fc )x

(iii) For flow, where (Fv) is negligible, the resulting equations of motion are called Eulers Equation

of Motion.i.eFx = (Fg )x + (Fp )x + (Fv )x + (Fc )x

Eulers Equation for Motion

Consider a stream-line flow taking place in the direction as shown in the figure below. Consider a

small cylindrical element LM of Cross-Section dA and length ds. The forces acting on the cylindrical

element are:

P the pressure of the fluid element at section L

P+ dp pressure at M , dA - area of crossection of the fluid element
http://www.docudesk.com/


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21. Net pressure force in the direction of flow,

p. dA - (p+dp)dA = - dp.dA

2. Componenet of the weight of the fluid element in the direction of flow

= - g. dA. ds cos

dz= - g. dA. ds =

ds

- g. dA. dz

mass of the fluid element = .dA.ds

dVAcceleration of the fluid element = = .

dt

According to Netons law, F = m. ,

-dp. dA - . dA. dz = .dA.ds . Deviding by . dA;

dv ds dvv

ds dt ds

a

dvg v

ds

=

dp- = v. dv

dpi.e. v. dv + = 0 , is Euler's equation

gdz

gdz

+

Let

Bernoullis Equation

Bernoullis Theorem states that in a steady, ideal flow of an incompressible fluid, the total energy at a

point of the fluid is constant. The total energy consists of pressure energy, kinetic energy, andpotential energy or datum energy.

2

2

p Vconstant

w 2

p VWhere, - pressure energy( head), Kinetic energy( head) , z - Datum head

w 2

Mathematically

zg

g

+ + =

Bernoullis Equation is obtained by integrating Eulers Equation Of Motion.

2

2

dpi.e. v. dv + = Constant

p Vi.e = Constant

2

p VConstant

2g

gdz

z g

zg

+

+ +

+ + =

Assumptions

The following assumptions are made in the derivation ofBernoullis Equation:

(i) The Fluid is Ideal ie. Viscosity is zero.

(ii) The flow is Steady

(iii) The flow is Incompressible(iv) The flow is Irrotational(v) The flow is Streamline.


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3(vi) The only forces acting on the fluid are gravity forces and pressure forces.

Orifice

An Orifice is an opening in the wall or base of a vessel through which the fluid flows.

Classification of Orifices

Orifices are classified as follows:

1. According to size

(i) Small Orifice (ii) Large Orifice

2. According to shape

(i) Circular Orifice (ii) Rectangular Orifice (iii) Square Orifice (iv) Triangular Orifice

3. Shape of the upstream edge

(i) Sharpe-edge Orifice (ii) Bell-Mouthed Orifice.

4. According to discharge conditions

(i) Free Discharge Orifice (ii) Drowned or Submerged Orifice(a) Fully Submerged Orifice (b) Partially Submerged Orifice

Flow Through an Orifice

Consider a small circular Orifice with sharp edge in the sidewall of a tank discharging free in

to the atmosphere.

Let the center of the Orifice be at a depth ofHbelow the free surface. As the fluid flows

through the Orifice it contracts and attains a parallel form at distance d/2 from the plane of the Orifice.

The point at which the streamlines first become parallel is called Vena Contracta. The cross-

sectional area at the Vena Contracta is less than that at the Orifice.

Considering points at (1) and (2) and applying Bernoullis Theorem,

p1 + v12 + z1 =p2 + v2

2 + z2

w 2g w 2g

But p1 =p2 =pa =Atmospheric Pressure

z1 =z2 +H


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4and v1 = 0 (Liquid at Section (1) is still)

Thus

v22 = H

2g

v2 = 2gH

Flow Through mouth pieceMouth pieces classified according to

1. Position : Internal and external2. Shape : cylindrical, convergent and convergent divergent3. Nature of discharge : mouth piece running full and running free.

Hydraulic Co-efficients

1. Co-efficient of contraction, Cc2. Co-efficient of velocity, Cv

3. Co-efficient of Discharge, Cd

4. Co-efficient of Resistance, Cr

1. Co-efficient of contraction, Cc

It is the area of the jet at Vena Contracta to the area of the Orifice. The value of Cc varies from 0.61 to

0.69 depending up on the size, shape of the orifice and head under which the flow takes place.

Let ac= Area of the Jet at Vena Contracta

a = Area of the Orifice

Then

Cc = ac

A

2. Co-efficient of velocity, Cv

It is the ratio of Actual Velocity (V) of the jet at Vena Contracta to the Theoretical Velocity (Vth). The

value of Cv varies from 0.95 to 0.99 depending up on the size, shape of the orifice and head underwhich the flow takes place.

Cv = Actual Velocity of the jet at Vena Contracta (V)

Theoretical Velocity(Vth)

3. Co-efficient of Discharge, Cd

It is the ratio of Actual Discharge (Q) through an Orifice to the Theoretical Discharge(Vth). The valueof Cd varies from 0.61 to 0.65 depending up on the size, shape of the orifice and head under which the

flow takes place.

Cd = Actual Discharge(D)

Theoretical Discharge(Dth)


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5

= Actual Area Actual Velocity

Theoretical Area Theoretical Velocity

Cd= Cc Cv

4. Co-efficient of Resistance, Cr

The loss of head (or loss of Kinetic Energy) in the orifice to the head of water (actual Kinetic Energy)available at the exit of the Orifice is known as Co-efficient of Resistance.

Cr = Loss Of Head in the Orifice

Head Of Water

Problems :

1.An orifice 50mm in diameter is discharging water under a head of 10m. If Cd = 0.6 and Cv= 0.97,find:

1. Actual Discharge 2.Actual Velocity of Jet at Vena ContractaSol

n

:Cd = 0.6

Cv = 0.97

D1 = 50cm=.05m

Area,A1 = (.05)2 = . 001963 m

2

4

Cv = Actual Discharge(D) = 0.6

Theoretical Discharge(Dth)

Theoretical Discharge = Area of Orifice Theoretical Velocity

= a (2gH).5 = 0.02749 m3/s

Actual Discharge = 0.6 0.02749 = 0.01649 m3/sCv = Actual Velocity

Theoretical Velocity

Actual Velocity = Cv Theoretical Velocity = 0.97 (2gH).5 = 13.58m/s

Notch

It is an opening provided in the side of a tank or a vessel such that the liquid surface in the tank is

always below the top edge of the opening.

Types of Notches

1. Rectangular Notch2. Triangular Notch

3. Trapezoidal Notch4. Stepped Notch

Weir

A Weir is a regular obstruction in an open stream over which the flow takes place.

Discharge Over a Rectangular Notch or Weir


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6

d

Let

H = Head of water above the apex of the notch

= length of the Notch

c = Co-efficient of Discharge

Area of strip = L dh

Theoretical Velocity of the water flowing through the strip = 2gh

L

d

d

d

discharge through the strip dQ = c x Area of the strip x Theoretical velocity

= c L. dh 2gh

Q = c

H

0

3/2

d

L. 2gh dh

2

= c L. 2g H3

Problems :

1.A rectangular weir 2.0m wide has a constant head of 500m. Find the discharge over the notch if

Cd=0.62.

Soln:

Q = 2 Cd (2g)1/2 LH3/2

3

= 1.294 m3

/s

Discharge Over a Triangular Notch or Weir


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7

d

Let

H = Head of water above the apex of the notch

= angle of the Notch

c = Co-efficient of Discharge

Consider a horizontal strip of water of thickness dh at a depth h from free surface

From f

d

ig. LN = (H-h) tan /2Area of the strip = 2 (H-h) tan /2 . dh

Theoretical Velocity of the water flowing through the strip = 2gh

discharge through the strip dQ = c x Area of the strip x Theore

d

H

d

0

tical velocity

= c 2 (H-h) tan /2 . dh 2gh

Q = c 2 (H-h) tan /2 . 2gh dh

5/2

d

8= c 2g tan /2 H

15

Problems :

1.Find the discharge over a triangular notch of angle 60 when the head over the triangular notch is

0.2m. Assume Cd=0.6.

Soln:

Q = 8 Cd tan /2 (2g)1/2H5/2

15= 0.01462 m3/s

Reynolds Number (Re):

It is defined as the ratio of Inertia Force of a fluid and the viscous force of a fluid.

Inertia Force = Fi = Mass Acceleration of the flowing fluid.

= Volume Velocity

Time

= Volume Velocity [Volume per Sec =A V]Time

= A VV

= AV2

Viscous Force Fv = Shear Stress Area = A = VA

L

Re = Inertia Force = AV2

Viscous Force VAL

= VL


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8In case of pipe flow, the linear DimensionL is taken as diameter, d. Hence Reynolds Number for

pipe flow =

= Vd [ = Kinematic Viscosity = ]

Re = Vd

WhenRe < 2000 .then the flow is laminar

WhenRe > 4000 .then the flow is turbulent

WhenRe between 2000 and 4000 .then the flow is unpredictable

Problems :

1.Water is flowing through a pipe having diameters 30cm and 20cm at the bottom and upper end

respectively. The intensity of pressure at the bottom end is 24.5104N/m2 and the upper end is

9.8104 N/m2. Determine the difference in datum head if the rate of flow through the pipe is

40litre/sec.Soln:

1000 litre = 1 m3

At section 1

D1 = 30cm=.30m

Area,A1 = (.3)2 = . 07065 m

2

4

V1 = ?p1 = 24.510

4N/m2

At section 2

D2 = 20cm=.20m

Area,A2 = (.2)2 = .0314 m24

V2 = ?


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9

p2 = 9.8104 N/m2

A1 V1 =A2 V2 = 40litre/sec = 40 10-3 m3/sec

V1 = 40 10-3 = 0.5661 m/sec

.07065

V2 = 40 10-3 = 1.274 m/sec

.0314

Applying Bernoullis Equation,

p1 + v12 + z1 =p2 + v2

2 + z2

w 2g w 2g

24.5104 + (0.5661)2 +z1 = 9.8104+ (1.274)2 +z2

9800 2 9.8 9800 2 9.8

z2-z1 = 13.70 m.

2.Water is flowing through a tapering pipe having diameters 300mm and 150mm at sections 1 and 2

respectively. The discharge through the pipe is 40 litres/sec. The section 1 is 10 m above datum andsection 2 is 6m above datum. Find the intensity of pressure at section 2 if that at section 1 is

400kN/m2.Soln:

1000 litre = 1 m3

At section 1

D1 = 300mm=.3m

Area,A1 = (.3)2 = . 0707 m

2

4

z1 = 10m

V1 = ?p1 = 40010

3 N/m2

At section 2

D2 = 150cm=.15m

Area,A2 =(.15)

2

= .01767 m2

4

z2 = 6m


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10

V2 = ?p2 = ?

A1 V1 =A2 V2 = 40litre/sec = 40 10-3 m3/sec

V1 = 40 10-3 = 0.566 m/sec

.0707

V2 = 40 10-3 = 2.264 m/sec

.01767


p1 + v12 + z1 =p2 + v2

2 + z2

w 2g w 2g

400103 + (0.566)2 + 10 = p2 + (1.274)2 + 6

9800 2 9.8 w 2 9.8

p2 = 436.8 kN/m2

Applications of Bernaulis Principle

1. Venturi meter2. Orifice meter3. Pitot tube4. Rotameter

Venturimeter

Venturimeter is a devise used for measuring the rate of flow of a fluid through a pipe. It consists ofthree parts.

1. A short Converging part

2. Throat

3. Diverging part

Expression for Rate of Flow Through Venturimeter

Consider a Venturimeter fitted in a horizontal pipe through which a fluid is flowing.

Let


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11

d1 = Diameter at the inlet or Section (1)

p1 = Pressure at Section (1)

v1 = Velocity of fluid at section (1)

and d2,p2, v2 are corresponding values at Section (2)


p1 + v12 + z1 =p2 + v2

2 + z2

w 2g w 2g

As the pipe is horizontal,

z1 = z2

p1 + v12 =p2 + v2

2

w 2g w 2g

p1 p2 = v22

v12

----------------------(1)

w 2g 2g

p1 p2 is the difference of pressure heads at Sections 1 and 2 and is equal to h

w

p1 p2 = h

w

Substituting the value ofh in the above equation,

h = v22 v1

2 ----------------------(2)

2g 2g

a1 v1 = a2 v2

v1 = a2 v2

a1

Substituting the value ofv1 in the above equation,

h = v22 v1

2

2g 2g

= v22 (a2 v2)

2

2g 2g (a1)2

= v22 [ 1 - (a2)

2 ]

2g (a1)2

= v22 [ (a1)

2 - (a2)2 ]

2g (a1)2

v22 = 2gh (a1)

2

(a1)2-(a2)

2

v2 = 2gh (a1)

2


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12(a1)

2-(a2)2

v2 = 2gh a1

(a1)2-(a2)

2

Discharge Q = a2 v2

= 2gh a1 a2 ------------------(3)

(a1)2-(a2)

2

Equation (3) gives the discharge under ideal conditions and is called Theoretical Discharge. Actual

Discharge will be less that Theoretical Discharge.

Qact = Cd 2gh a1 a2

(a1)2-(a2)

2

Value ofh given by differential U-Tube Manometer

Case 1 :

If the differential manometer contains a liquid which is heavier than the liquid flowing through the

pipe.

Let

Sh = Specific Gravity of the heavier liquid in U-TubeSo = Specific Gravity of the liquid flowing through the pipe

x = Difference of the heavier liquid column in U-Tube

then

h =x [Sh 1 ]

SoProof(For Reference Only)


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13LetB andA be the throat and inlet of the Venturimeter which are at same levels from a horizontal

reference. Let P2 be the pressure at throat and P1 be the pressure at Inlet. Consider the datum line at X-

X.

Let

x = Difference of mercury level in the U-Tube

y = Distance of center ofB, from the mercury level in the right limb

Sl = Specific Gravity of Light liquid atA andB

Sh = Specific Gravity of Heavy Liquid

wl = Specific Gravity of liquid flowing through the venturimeter.

ww = Specific Gravity of water.

Pressure Head above X-X in the left limb

= x Sl ww +y Sl ww+ P2

Pressure Head above X-X in the right limb

= x Sh ww+y Sl ww+ P1

Equating the two pressure head ,x Sh ww+y Sl ww + P1 = x Sl ww +y Sl + P2

P1 - P2 = x Sh ww -x Sl ww

= x ww (Sh - Sl)

ie. Difference of pressure head atA andB

P1 - P2 = xww (Sh - Sl) [Sl = wl ]

wl wl ww

P1 - P2 = xww ( Sh - Sl) [wl = Sl ww]

wl wl wl

P1 - P2 = xww ( Sh - Sl )

wl Sl ww Sl ww

P1 - P2 = h = x(Sh - 1) [Meters of Flowing Liquid]

wl Sl

Note

Expression hA - hB = h[Sg - S1] is in Meters Of Water. But above expression is used for finding the

differential head whose value is in Meters of Flowing Liquid.

To find the difference in pressure using both the cases,

1. hA - hB = h[Sg - S1] in Meters Of Water,

so Pressure difference = h[Sg - S1] ww

2. P1 - P2 = h = x(Sh - 1) [Meters of Flowing Liquid]

wl Sl

so Pressure difference =x(Sh - 1) wliquidSl

Case 2 :


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14

Let the differential manometer contains a liquid which is heavier than the liquid flowing through the

pipe. Let

Slh = Specific Gravity of the light liquid in U-Tube

So = Specific Gravity of the liquid flowing through the pipe

x = Difference of the heavier liquid column in U-Tube

then

h =x[1- Sl ]

SoTypes of Venturimeters

Venturimeters are classified as

1. Horizontal Venturimeter

2. Vertical Venturimeter

3. Inclined Venturimeter

Vertical and Inclined Venturimeter

The same formula for discharge as used for Horizontal Venturimeter holds good in these cases as well.But

h =p1 p2 +z2-z1

w

Problems :

1.A horizontal venturimeter with inlet diameter 200mm, and throat diameter 100mm is used to

measure the flow of water. The pressure at inlet is 0.18N/mm2and the vacuum pressure head at the

throat is 280mm of mercury. Find the rate of flow. The value ofCd may be taken as 0.98.Sol

n:

Cd = 0.98

D1 = 200mm=.20m

Area,A1 = (.2)2 = . 0314 m

2

4

D2 = 100mm=.10m

Area,A2 = (.1)2 = . 00785 m

24

Pressure at Inlet =

p1 = 180103 N/m2

p1 = 180103 = 18.3m

w 9800

wwaterhwater = wmercuryhmercury

Dividing by wwater,

Specific Gravity of Waterhwater = Specific Gravity of Mercuryhmercuryhwater = Specific Gravity of Mercury hmercury

Specific Gravity of Water


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15

p2 = -280 mm of Mercury

w

= -28 m of Mercury = -0.28 13.6 = -3.8 m of water

h =p1 p2 = 18.3 (-3.8) = 22.1 m

w

Qact = Cd 2gh a1 a2

(a1)2-(a2)

2

Q = 0.165m3/s

2.A horizontal venturimeter with inlet diameter 200, and throat diameter 100mm is used to measure

the flow of water. The reading of the differential manometer connected to the inlet is 180mm of

Mercury. Find the rate of flow. The value ofCd may be taken as 0.98.Soln:

Cd = 0.98

D1 = 200mm=.20m

Area,A1 = (.2)2 = . 0314 m

2

4

D2 = 100mm=.10m

Area,A2 = (.1)2 = . 00785 m

2

4

h =p1 p2 = h =x [Sh 1 ] = 0.18 [ 13.6 1] = 2.268m

w So 1

Qact = Cd 2gh a1 a2

(a1)2-(a2)

2

Q = 0.0528m3/s

3.A 300150mm venturimeter is provided in a vertical pipeline carrying oil of specific gravity 0.9,flow being upward. The difference in elevation of the throat section and entrance section of the

venturimeter is 300mm. The differential U-Tube mercury manometer shows a gauge deflection of

250mm. Calculate 1) The Discharge of oil and 2)The pressure difference between entrance section

and throat section.

Soln:

Cd = 0.98

D1 = 300mm=.3m

Area,A1 = (.3)2 = . 07 m

2

4

D2 = 150mm=.15m

Area,A2 = (.15)2 = . 01767 m

2

4

Discharge

h =p1 p2 = h =x [Sh 1 ] = 0.25 [ 13.6 1] = 3.53m of oil


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16

w So .9

Qact = Cd 2gh a1 a2

(a1)2-(a2)

2

Q = 0.1489m3/s

Pressure Difference

p1 p2 +z2-z1 = 3.53

w

z2-z1 = 300

p1 p2 - 0.3 = 3.53

w

p1 p2 = 33.8 kN/ m2

Orifice Meter

Orifice Meter is a device used for measuring discharge of fluid through a pipe. It consists of a flat

circular plate having a circular hole which is concentric with the pipe. The orifice diameter may varyfrom 0.4 to 0.8 times the diameter of the pipe.

Let

d1 = Diameter at the inlet or Section (1)

p1 = Pressure at Section (1)

v1 = Velocity of fluid at section (1)

and d2,p2, v2 are corresponding values at Section (2)


p1 + v12 + z1 =p2 + v2

2 + z2

w 2g w 2g

(p1 + z1 ) (p2+ z2) = v22 v1

2 ----------------------(1)

w w 2g 2g


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17(p1 + z1 ) (p2+ z2) = h [Differential Head]

w w

Substituting the value ofh in the above equation,

h = v22 v1

2 ----------------------(2)

2g 2g

Cc = a2

a0Cc = Coefficient of contraction

a2 = Cc a0

Also a1 v1 = a2 v2v1 = a2 v2

a1 v1 = Cc a0 v2

a1

Substituting the value ofv1 in the above equation,

h = v22 (Cc a0 v2)

2

2g 2g (a1)2

h = v22 [ 1 - (a0)

2 Cc2 ]

2g (a1)2

v2 = 2gh

1 - a02

Cc2

a12

The Discharge Q = a2 v2= Cc a0 v2 ----------------------(4)

Cd = Cc 1 - a02

a12

1 - a02 Cc

2

a12

Cc = Cd 1 - a02 Cc

2

a12

1 - a02

a12

Substituting the value of Cc and v2 in equation (4),

Q = a0 Cd 1 - a02 Cc

2 2gh

a12

1 - a02 1 - a0

2 Cc2

a12 a1

2


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18

Q = a0 Cd 2gh = a0 a1Cd 2gh

1 - a02 Cc

2 a12- a0

2

a12

Problems :

1.Water is flowing through a pipeline of 50cm at 30C. An orifice is placed in the pipe line to measure

the flow rate. Orifice diameter is 20cm. If the manometer reads 30cm of Mercury, calculate the

water flow rate and velocity of the fluid through the pipe.

Soln:

Cd = 0.6

D1 = 50cm=.5m

Area,A1 = (.5)2 = . 1963 m

2

4D0 = 20mm=.20mArea,A2 = (.2)

2 = . 0314 m

2

4

Qact = Cd 2gh a1 a0

(a1)2-(a0)

2

Q = 0.1655m3/s

Velocity ofwater through the pipe = Q = 0.1655 = 0.843 m/s

a1 0.1938

Difference Between Venturimeter and Orifice Meter

Venturimeter Orifice Meter

Used for Measuring the Flow Rate of all

Incompressible Fluids.

Used for Measuring the Flow Rate of Liquids.

Cd Value is More Cd Value is Less

More Accurate Less Accurate

Expensive Less Expensive

Pitot Tube

Pitot tube is a device used to measure the velocity of flow at any point in a pipe or channel. It

works on the principle that if the velocity of flow at any point becomes zero, the pressure there is

increased due to the conversion of the kinetic energy into pressure energy.


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19

Pitot Tube consists of a glass tube whose lower end is bent at right angle. The lower end is

directed in the upstream direction. The liquid rises in the tube due to conversion of kinetic energy intopressure energy. The velocity is determined by measuring the rise of liquid in the tube.

Consider two points (1) and (2) at the same level in such a way that point (2) is just at inlet of the Pitot

Tube and point (1) is far away from the tube.

Let

p1 = Intensity of Pressure at point (1)

v1 = Velocity of Flow at point (1)

p2 = Intensity of Pressure at point (2)

v2 = Velocity of Flow at point (2)

H = Depth of tube in the liquidh = Rise of liquid in the tube above the free surface.

Applying BernoullisEquation,

p1 + v12 + z1 =p2 + v2

2 + z2 ----------------------(A)

w 2g w 2g

z1 = z2 = 0 as (1) and (2) are on the same line

p1=H = Pressure Head at (1)

wp2= (h + H) = Pressure Head at (2)

w

Substituting these values in Equation (A)

H+ v12 = (h +H)

2g

h = v12

2g

v1 = 2gh


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20This is the theoretical velocity and the actual velocity is given by

( v1 )act = Cv 2gh

Cv is the Co-efficient of Pitot Tube.

Simple Pitot Tube (a) with static tube only (b) with Static and Stagnation Tube

Problems :

1.A sub-marine fitted with a pitot tube moves horizontally in sea. Its axis is 12m below the surface of

water. The pitot tube fixed infront of the sub-marine and along its axis is connected to the two limbs

of a U-Tube containing mercury, the reading of which is found to be 200mm. Find the speed of the

sub-marine. Take specific-gravity of sea water = 1.025 times fresh water.

Soln:

Reading of the manometer =y = 200 mm = 0.2m of mercury.Specific Gravity of mercury = Sh = 13.6

Specific Gravity of sea water = So = 1.025

h =x [Sh 1 ] = 0.2 [13.6 - 1 ] = 2.45

So 1.025

Velocity of sub-marine =

v1 = 2gh = 6.93 m/s

2.Petroleum oil of specific gravity 0.93 and viscosity flows through a horizontal 5 cm pipe. A pitot

tube is inserted at the center of the pipe and its leads are filled with the same oil and attached to a U-Tube containing water. The reading on the manometer is 10cm. Calculate the volumetric flow of oil.

The co-efficient of pitot tube is 0.98.

Soln:

Reading of the manometer =y = 10 cm = 0.1m of mercury.

Specific Gravity of mercury = Sh = 13.6

Specific Gravity of oil = So = .09

h =x [Sh 1 ] = 0.1 [13.6 - 1 ] = 1.411

So 0.9

Velocity of sub-marine =


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21

v1 = 2gh = 5.156 m/s

Q =A V= (.05)2 5.156 = 0.01m3/s

4

Rotameter :

Rotameter consists of a transparent tube kept in vertical position. Inside the tube there is a

float which moves up and down in the tube depending upon the intensity of the flow. A calibrated

scale on the side of the tube indicates the rates of flow. Slots on the head of the float keep the float invertical central position. Measurement of discharge using Rotameter is not accurate and so it is used

for small variations of discharge.

m303,fluid mechanics fm - m iii ; p.p. binu & sunand c

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