FLUID MECHANICES LAB:-I MANUAL - srespune.org · FLUID MECHANICES LAB:-I MANUAL Experiment:-01...

FLUID MECHANICES LAB:-I MANUAL

Experiment:-01

Measurement of viscosity by Redwood viscometer. Aim: - To determine the kinematic viscosity of a liquid and its variation with temperature. Apparatus:-

Redwood viscometer with its accessories such as a flask, ball valve, etc. Heating arrangement, Thermometer, Stopwatch, Oil etc.

Theory:

It is defined as the property of a fluid which offers resistance to the movement of one

layer of fluid over another adjacent layer of fluid. When two layers of fluid, a distance ‘dy’

apart, move one over the other at different velocities, say u and ‘u+ du’ as shown in figure,

the viscosity together with relative velocity causes a shear acting between the fluid layers.

The top layer causes a shear stress

on the adjacent lower layer while the lower layer causes a shear stress on the adjacent top

layer. This shear stress on the adjacent top layer. This shear stress is proportional to the rate

of change of velocity with respect to ‘y’. It is denoted by ‘ ’ (Tau) Mathematically,

du τ ∝ dy =

Figure:-Velocity Variation near a Solid Boundary. Where′ ’ is the constant of proportionality and is known as the co-efficent of dynamic viscosity or only viscosity, represents the rate of shear strain or rate of shear deformation or velocity gradient.

∴ = .

Thus viscosity is also defined as the shear stress required to produce unit rate of shear strain. Units of Viscosity:-

Shear stress μ

= Change of velocity

Change of distance In MKS units,

In CGS units,

In SI units,

1 Pascal = 1 N/ 2,

Force

Force ×Time

=

Area

=

Length 1 Length 2

Time ×Lengt h

Kgf-sec / 2 [Unit of force = kilogram-force (kgf)]

dyne-sec /2 [Unit of force = dyne]

N-sec / 2 [Unit of force = Newton]

1 N-sec / 2 =1 Pa-s

Conversion: 1 kgf = 9.81 N

∴ 1 Kgf-sec / 2 = 9.81 N-sec/ 2 Force = mass x acceleration,

1 N = 1kg x 1 m/ 2 SRCOE, Lonikand. Civil Engineering Department


1 N = 1000 gm x 100 cm / 2 1 N = 1000 X 100 X gm-cm/ 2 1

N = 1000 X 1000 X dyne,

∴ 1 N –sec/ 2 = 1000 x 100 x dyne-sec/ 2

∴ 1 kgf-sec/ 2 = 9.81 x 1000 x 100 x dyne-sec/ 2

= 9.81 ×1000 ×100 ×−

= 98.1 dyne-sec/ 2

100 ×100 × 2

1 dyne-sec/ 2 = 1 poise

∴ 1 kgf-sec/ 2 = 98.1 poise

1 N –sec/ 2= 9.181 × 1 x kgf-sec/ 2 = 9.181 x 98.1 poise = 10 poise. Kinematic viscosity (v):

It is defined as the ratio of dynamic viscosity to the density of the fluid. Mathematically,

=

=

= 2

In MKS and SI units, 2

In CGS units, 2 , Stroke.

Effect of Temperature on Viscosity: 1) The viscosity of both liquids and gases will vary with temperature but in different

manner. 2) In case of liquids, the viscosity is due to cohesion. When the temperature of liquid

increases, the volume of fluid increases and hence the distance between molecules

increases with decreases the cohesion. Therefore, the viscosity of liquid decreases with

increases in temperature. 3) In case of gases, viscosity is due to molecular momentum exchange. When the

temperature of gas increases, kinetic energy of molecules increases and hence molecular

momentum exchanges increases. Therefore, the viscosity of gases increases with

increases in temperature. Experimental Set-Up:

The redwood viscometer consists of vertical

cylindrical oil cup with an orifice in the centre of its

base. The orifice can be closed by a ball. A hook

pointing upward serves as a guide mark for filling the

oil. The cylindrical cup is surrounded by the water bath.

The water bath maintains the temperature of the oil to

be tested at constant temperature. The oil is heated by Figure:- Redwood Viscometer. heating the water bath by means of an immersed electric heater in

the water bath; the provision is made for stirring the water, to maintain the uniform

temperature in the water bath and to place the thermometer to record the temperature of oil

and water bath. The cylinder is 47.625mm in diameter and 88.90mm deep. The orifice is

1.70mm in diameter and 12mm in length, this viscometer is used to determine the kinematic

viscosity of the oil. From the kinematic viscosity the dynamic viscosity is determined.

SRCOE, Lonikand. Civil Engineering Department Page 2


Procedure: 1. Level the Viscometer with help of levelling screws.

2. Fill the water jacket with water and heat it to the desired temperature by using electric

heater provided around the water jacket.

3. Close the orifice by means of a ball valve and pour the oil whose viscosity is to be determined into the cylinder up to the index mark.

4. Record the temperature of oil and keep the measuring flask below the orifice. 5. Lift the valve and start the stop watch simultaneously. 6. Measure the time in seconds required to collect 50 cc of oil to various temperatures.

Observation: Table:-01

Sr.no Temperature Time Kinematic viscosity

(t) (T) ( )

℃ Sec cm2/sec

01

02

03

04

05

06

07

Specifications: A=Constant=0.0022 B = Constant = 0.18 T= Time taken in seconds to collect 50cc of liquid. Calculation: Sample Calculation for First Reading:

Kinematic viscosity of liquid, γ = (A x T) – (B / T) Significance:

Viscosity affects heat generation in bearings, cylinders and gear sets related to oil’s

internal friction. It governs the sealing effect of oils and the rate of oil consumption, as well

as determines the ease with which machines may be started or operated under varying

temperature conditions, particularly in cold climates. Not all oils respond in the same way to a given change in temperature. Many oils contain an

ability to resist changes in viscosity due to a change in temperature. This property is referred

to as the oil's viscosity index or VI. The higher the VI of oil, the less its viscosity is altered by

temperature changes.

Another factor in the measurement of viscosity is the ability of oil to resist shearing or

the "tearing away of one plane of lubricant from another" during the hydrodynamic

lubrication function. The benefits of oils with a higher VI are:


1. A general increase in viscosity at higher temperatures, which results in lower oil consumption and less wear.

2. A reduced viscosity at lower temperatures, which will improve starting and lower fuel

consumption. Graph: Draw a graph Temperature Vs Kinematic viscosity. Result:

1. Viscosity is ___________ at the temperature _______ 2. Viscosity is ___________ at the temperature _______ 3. Viscosity is ___________ at the temperature _______ 4. Viscosity is ___________ at the temperature _______

Conclusion: By observing the readings/ Graph, it is come to know that the viscosity goes on decreases with the increase in temperature.


Experiment:-02 Measurement of Pressures Using Different Pressure Measuring Devices.

Aim: To measure pressure of a flowing fluid by different pressure measuring devices. Apparatus.

Manometer, Pressure gauge etc.,

Theory:

Pressure at a point is defined as the total force exerted on a plane perpendicular to it. Mathematically, P = where F = Total force, A = Total area

The units of pressure in:

MKS units kgf/ 2 and kgf/cm2 SI units N/m2 and Pascal and represented by Pa. Other commonly used units of pressure is bar = 100000 N/m2

The pressure is measured in two different systems. In one system, it is measured

above the absolute zero or complete vacuum and it is called the absolute pressure and in other

system pressure above the atmospheric pressure and it is called gauge pressure. Thus:

Absolute Pressure: It is defined as the pressure

which is measured with reference to absolute

vacuum pressure.

1) Gauge Pressure: It is defined as the pressure

which is measured with the help of a pressure

measuring instrument, in which the

atmospheric pressure is taken as datum. The

atmospheric pressure on the scale is marked as

zero.

2) Vacuum Pressure: It is defined as the pressure below the atmospheric pressure.

The relation between the absolute pressure, gauge pressure and vacuum pressure

are, Mathematically: 1. Absolute Pressure =Atmospheric Pressure + Gauge Pressure. 2. Vacuum Pressure=Atmospheric Pressure –Absolute Pressure

Note: The Atmospheric pressure at sea level at 150C IS 10.13 2 or 10.13 2 in SI units. Measurement of Pressure: The pressure of fluid can be measured by the following devices.

1. Manometers and 2. Mechanical Gauges.

Manometers. These are defined as the devices used for measuring the pressure at a point in a fluid by

balancing the column of fluid by the same or another column of the fluid. T hey are classified

as: SRCOE, Lonikand. Civil Engineering Department Page 5


Simple Manometers and Differential Manometers.

Simple Manometers: A simple manometer consists of a glass tube having one of its ends connected to a point

where pressure is to be measured and other end remains open to atmosphere. Common types

of simple manometers are: 1) Piezometer, 2) U-tube Manometer and 3) Single Column Manometer. Piezometer:

It is the simple type of manometer used for measuring gauge

pressure. One end of the manometer is connected to the point

where pressure is to be measured and other end is open to the

atmosphere as shown in figure. The rise of liquid is say ‘h’ in

piezometer tube, and then pressure at ‘A’ is given by,

Pressure at ‘A’ = × × 2

Figure:-Piezometer. U-tube Manometer: It consists of glass tube bent in U-shape, one end of which is connected to a point at which

pressure is to be measured and other end remains open to the atmosphere as shown in figure.

The tube generally contains mercury or any other liquid whose specific gravity is greater than

the specific gravity of liquid whose pressure is to be measured. For gauge pressure: Let ‘B’ is the point at which pressure is to be measured, whose value is ‘P’. The datum line is A-A

Let = Height of liquid above the datum line,

1

h2 = Height of heavy liquid above the datum line,

S1= Specific gravity of light liquid,

ρ = Density of light liquid = 1000 ×

1 1

S2 = Specific gravity of heavy liquid,

ρ 2

=Density of heavy liquid – 1000 ×

2

Figure:- U-tube manometer.

Pressure in left column = Pressure in right column

above datum line above A − A above datum line A − A

+ × × = × 2

1 1 2

∴ P = × − × ×

2 2 1 1

For Vacuum Pressure: For measuring vacuum pressure, the level of the heavy liquid in the manometer will be as shown in figure (b). Then

=

− −

++= 0

1 1 2 2

∴ P = − + 2

1 1 2


Single Column Manometer:

Single column manometer is a modified form of a U-tube manometer in which a reservoir,

having a large cross-section area about 100 times as compared to the area of the tube is

connected to one of the limbs (say left limb) of the manometer as shown in figure. Due to

large cross section area of the reservoir, for any variation in pressure, the change in the liquid

level in the reservoir will be very small which may be neglected and hence the pressure is

given by the height of liquid to the other limb.

The other limb may be vertical or inclined.

1) Vertical Single Column Manometer. 2)

Inclined Single Column Manometer.

Figure:- Single Column Manometer. Vertical Single Column Manometer. Figure (a) shows the vertical column manometer. Let X-X is the datum line in the reservoir

and in the right limb of the manometer, when it is not connected to the pipe. When the

manometer is connected to the pipe, due to high pressure at ‘A', the heavy liquid in the

reservoir will be pushed downward and will rise in the right limb.

Let ∆ = Fall of heavy liquid in reservoir,

2 = Rise of heavy liquid in the right limb,

1 = Height of centre of pipe above X-X PA = Pressure at ‘A’ which is to be measured, A = Cross -section area of the reservoir, a = Cross-section area of the right limb, S1= Specific gravity of liquid in pipe, S2= Specific gravity of heavy liquid in reservoir and right limb, ρ1 = Density of liquid in pipe, ρ2 = Density of liquid in reservoir. Fall of heavy liquid in reservoir = Rise of heavy liquid in right limb

× ∆ = × 2

∆ = × 2 Now consider the datum line Y-Y as shown in figure (a).

Then pressure in the right limb above Y-Y = 2 × × (∆ + 2) Pressure in the left limb above Y-Y = 1 × × ∆ + 1 +

Equating these pressures, we have

2 × × (∆ + 2) = 1 × × ∆ + 1 +

= ∆ (– ) + −

2 1 2 2 1 1

Now substuting, ∆ = × 2 in the above equation, we get

= × 2 2 − 1 + 2 2 − 1 1

As the area ‘A’ is very large as compared to ‘a’, hence ratio becomes very small and can be

neglected.



=− g ------------------------------------------ (1)

2 2 1 1

From above equation (1), it is clear that as h1 js known and hence by knowing h2 or rise of heavy liquid in he right limb, the pressure at ‘A’ can be calculated. Inclined Single Column Manometer. Figure:- (b) shows the inclined single column manometer. This msnometer is more sensitive. Due to inclination the distance moved by the heavy liquid in the right limb will be more. Let L= Length of heavy liquid moved in right limb from X-X

=inclination of right limb with horizontal,

h2 = Vertical rise of heavy liquid in right limb from X-X = L sin From equation (1)

= 2 2 − 1 1g PA = sin X 2 − 1 1 Differential Manometers: Differential manometers are the devices used for measuring the difference of pressures

between two points in a pipe or in two different pipes. A differential manometer consists of a

U-tube , constaining a heavy liquid, whose two ends are connected to the points, whose

difference ofpressure is to be measured. Most commenly types of differential manometers

are: 1) U-tube differential manometers and 2) Inverted U-tube differential manometers. U-tube Differential Manometer:

Figure:- (a) showes the differential manometer

of U-tube type. In figure (a), the two points ‘A’

and ‘B’ are at different level and also contains

liquid of different specific gravity. These points

are connected to the tube differential Figure:-

manometer. Let the pressure at ‘A’ abd ’B’ are Figure:-U-tube differential manometers PA and PB,

Let h = Difference of mercury level in the U-tube, Y = Distance of the centre of B, from the mercury level in the right limb,

X = Distance of the centre of ‘A’ from the mercury level in the right limb, ρ1 = Density of liquid at ‘A’, ρ2=Density of liquid at ‘B’. ρg = Densty of heavy liquid or mercury.

Taking datum line as X-X,

=

− −

+ + = × × ++

1 2

PA − PB = ρg × g × h × ρ2gy − ρ1g h + x

P − P = h ×g (ρ g

− ρ ) + −

A B 1 2 1

Difference of pressure at ‘A’ and ‘B’ = h ×g (ρ g

− ρ ) + −

1 2 1



In figure (b):- In figure (b), the two points ‘A’ and ‘B’ are at the same level and contains the same liquid of density. Then

Pressure above datum =

line X − X in the left limb

× × × × × +

1

Pressure above datum line X − X in the left limb = 1 × × +

+ PA − PB = g × h ρg − ρ1

Inverted U-tube Differential Manometer:

It consists of an inereted U-tube, containg a light

liquid. The two ends of the tube are connected to

the points whose difference of pressure is to be

measured. It is used for measuring difference of

low pressures as shows the above jnverted U-

tube differential manometer connected to the

two points ‘A’ and ‘B’. Let the pressure at ‘A’ is

more than the pressure at ‘B’.

Let 1 = Height of liquid in left limb below the datum line X-X h2 = Hight of liquid in the right limb,

Figure:-Inverted U-Tube Differential Manometer. h = Difference of light liquid,

ρ1 = Density of liquid at ‘A’, ρ2 = Density of liquid at ‘B’, ρS

= Density of light liquid, PA = Pressure at ‘A’, PB = Presure at ‘B’.

Taking X-X as datum line. Then pressure in the left limb below X-X

= PA − 1 × × 1

Pressure in the right limb below X-X = − 2 × × 2 − 2 × × Equating the above two pressure, we get

− = 1 × × 1 − 2 × × 2 − × × Mechanical Gauges: These are defined as the devices used for measuring the pressure by balancing the fluid column by the spring or dead weight. The commonly used mechanical pressures are: 1) Diaphragm pressure gauge,

2) Bourdon tube pressure gauge,

3) Dead weight pressure gauge and

4) Bellow pressure gauge. Experimental Set-Up: Different types of pressure measuring instruments are arranged and connected to the pipe where pressure is to be measured. Procedure: 1) Connect the instrument to the pipe through which water is flowing.



2) Record the reading in the manometer and pressure gauge. 3) By changing the discharge in the pipe, again note down the readings of manometer and

pressure measuring gauges. Observation

Table:-02 Sr.no Manometric Reading Calculated pressure Pressure in gauge in Difference

. In cm of Hg P1 P

∆ =( − ) P − P

1 2 1 2

1

Cm Cm Cm in 2

2

2

01

02

03

04

05

06

Calculation: Sample Calculation for First Reading:

760 mm of Hg = 1 Bar = 1 2 1. For ∆ = 141 mm of Hg,

1 = ∆×1 760

Significance: (1) For measurement of pressure at ground seepage. (2) For measuring pressure in chemical plants and water treatment plants.

Graph:

Draw a graph ∆ Vs Pressure gauge reading Result.

Conclusion:

By measuring any pressure measuring instrument, the pressure comes to be same.


Experiment:-03

Determination of stability of floating bodies using ship models. Meta- centre height of ship model.

Aim: To determine the stability of floating bodies using ship models. Apparatus:

Model of ship,

Tank with water, Weight, etc.,

Theory: Principle of floatation

Consider a body is floating in water when the weight of the body is more than the up thrust then the body will sink down in water. But if the weight is equal to the up thrust of the liquid the body will float in the water. The weight of the body is equal to the upward thrust of the liquid on the body. This concept is called as Principle of floating.

Figure:-Floating Body.

Buoyancy: When a body is immersed in a fluid an upward force is exerted by the fluid on the body.

This upward force is equal to the weight of the fluid displaced by the body and is called the force of buoyancy or buoyancy. Arenumedes Principle:

It states that whenever a body is immersed wholly or partly in fluid, the resultant force acting on it , is equally to the difference between the upward pressure of the fluid on its bottom and the downward force due to gravity. Meta-Centre:

It is defined as the point about which a body starts oscillating when the body is tilted by a small angle. The meta-centre may also be defined 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.

Consider a body floating in a liquid as shown in figure. Let the body is in equilibrium and ’G’ is the centre of gravity and ‘B’ the centre of buoyancy. For equilibrium, both the points lie on the normal axis, which is vertical.

Let the body is given a small angular displacement in the clockwise direction as shown in

figure. The Figure: - Meta-Centric Height. centre of buoyancy, which is the centre of gravity of the

displaced liquid or centre of gravity of the portion of the body sub-merged in liquid, will now be shifted towards right from the normal axis. Let it is at B1 as shown in figure. The line of action of the force of buoyancy in this new position will intersect the normal axis of the body at some point say ‘M’. This point ‘M’ is called Meta-centre. Meta-Centric Height:

The distance between the meta-centre of a floating body and the centre of grvity of the body is called meta-centric height. i.e. ‘MG’. Experimental set-up:



A vessel is attached with a protector and a plumb, a horizontal scale on the vessel, a container filled water, and weights. Procedure: 1) Fill the tank with water. 2) Keeps the ship floating over the water. 3) See that plumb indicates zero reading. 4) Place the weight on the deck. 5) Measure the displacement of weight and angle indicated by plumb bob. 6) Repeat the procedure for different displacement of weight. Observation:

Table:-03 Sr. no. Displacement of applied Tilt angle ‘θ’ Applied weight Meta-centric

weight X w height MH

M ° kg M

01

02

03

04

05

Calculation: Sample Calculation for First Reading: Metacentre Height:

× MH=

Where

W = Weight of vessel including weight applied w = 3.20 kg = 3.20 × 9.81 =31.392 N S = Distance from its centre of gravity.

w = Weight applied on one side of ship, = Angle of tilt of ship.

Significance: Why to find Metacentre height? It is necessary for the stability of a floating body, if metacentre is above centre of gravity, body will be stable because the restoring couple produced will shift the body to its original position. It is a function of the configuration of a ship and the distribution of its weight. Because it can be determined relatively easily and quickly, either by empirical relations or by direct calculation, it is often relied upon as a principal indication of the stability of the ship and its ability to survive extensive flooding due to underwater damage. Result: Conclusion:



Experiment:-04

Experimental verification of Bernoulli’s theorem with reference to loss of energy. Aim: To verify Bernoulli’s theorem with reference to loss of energy. Apparatus:

A taper (Converging-Diverging) rectangular pipe fitted with piezometer tubes fitted at different section),

Water supply tank,

Measuring tank,

Stop watch,

Scale, etc., Theory: Statement:

It states that in a steady, ideal flow of an incompressible fluid, the total Energy at any point of the fluid is constant. The total energy consists of pressure energy, Kinetic energy and potential energy or datum energy. These energies per unit weight of the fluid are,

Pressure energy =

Kinetic Energy = 2 2

Datum Energy = Z Mathematically, Bernoulli’s theorem is written as,

2

+ Z = Constant.

+

2

Assumptions: The following assumptions are made in the derivation of Bernoulli’s equation.

The fluid is ideal. I.e. Viscosity is zero. The flow is steady.

The flow is incompressible.

The flow is irrotational.

Bernoulli’s Equation for Real Fluids: The Bernoulli’s equation is derived on the assumption that fluid is non-viscous and therefore frictionless. But all the real fluids are viscous and hence offer resistance to flow. The Bernoulli’s equation for real fluids between points (1) And (2) is given as 2 2

1 + = 2 + + +

2 1 2 1

Where = Loss of Energy between points (1) and (2). Experimental Set Up:

1) The apparatus consist of converging-diverging dust of 750 mm length through which water is discharging to delivery tank.

2) Parallel piezometric tubes are fitted at small interval to show the pressure head at different sections.

3) The board on which the piezometric tube is fixed is graduated, so as to ease the procedure of taking piezometric reading.

4) The discharge is measured by collecting water in the measuring tank.

5) Water is supplied to the supply tank by pump which can be regulated by inlet valve.


Procedure: 1) Connect the water pipe to the valve.

2) Adjust the flow with the help of a gate valve.

3) Now allow the levels in the piezometer to stable and note down the heads 4) Close outlet valve of measuring tank and measure the time to rise water level

by 10 cm (Discharge of flow). 5) Now repeat the procedure by changing the discharge.

Observation: Table:04

Taperi Cross sectional ( ) in Rise of Pressure Sectional Velocity Total head

1

ng area of meters water in head Velocity head =(P/ρ g) +

Sr. Piezometer Piezomet =

=

2⁄2g ( 2/2g)

No.

er

In metre In metre.

= ( + )

() 1 2

In metre

2

In meters

01 0.025x0.025 0.01550

02 0.025x0.023 0.16000

03 0.025x0.020 0.01450

04 0.025x0.018 0.01250

05 0.025x0.013 0.01000

06 0.025x0.010 0.01150

07 0.025x0.011 0.01225

08 0.025x0.015 0.01325

09 0.025x0.015 0.13750

10 0.025x0.015 0.01450

11 0.025x0.017 0.01800

12 0.025x0.080 0.01600

13 0.025x0.021 0.01650

14 0.025x0.023 0.01250

Specifications: Area of tank =0.3m x 0.3m

Rise of water in tank = 10cm =0.1meter

Volume of water collected in tank = 0.3m x0.3m x 0.1m =0.0009 3

Time required in collecting 10 cm rise of water in the tank. Sample Calculation for First Reading: Discharge: = (Volume of water)/ (Time required to collect water in water tank for 10 cm rise) = (Area of tank x Rise of water level)/T

= (0.3M X 0.3M X 0.1M) / T = Area of tube x velocity Section Velocity

V = / (Area of tube) Velocity head

h = 2/2g Total head,

H = P/ρ g + 2/2g

Significance: Bernoulli’s equation is applied in all problems of incompressible fluid flow where energy considerations are involved. But its applications to the following measuring devices are


1) Venturi meter,

2) Orifice meter,

3) Pitot tube. Graphs: 1) Piezometric head Vs number of tubes

2) Total head Vs number of tubes.

3) Velocity head Vs number of tubes. Result: Total head at any section is always constant. Energy level goes on decreasing in the direction of flow. As area decreases pressure head decreases and velocity head increases. Conclusion: As the piezometric head increases the velocity head clearance and as the area increases pressure head i.e. piezometer head also increase from this as the increases from this as the area increases the velocity head decreases.


Experiment:-05

Calibration of Venturimeter. Aim: To calibrate venturimeter and to find its co=efficient of discharge. Apparatus:

Venturimeter fitted across a pipeline leading to a collecting tank, Stop Watch,

U-Tube manometer connected across entry and throat sections etc. Theory: A venturimeter is a device used for measuring the rate of a flow of a

fluid flowing through a pipe. It consists of three parts:

1) A short

converging part,

2) Throat

and

3) Divergi

ng part. On applying the continuity equation &Bernoulli’s equation between the two sections, the following relationship is obtained in terms of governing variables. In order to take real flow effect into account, coefficient of discharge ( ) must be

determined.Pressure tapping is provided at the location before the convergence commences and another pressure tapping is provided at the throat section of a Venturimeter. The Difference in pressure head between the two tapping is measured by means of a U-tube manometer. Consider a venturimeter fitted in a horizontal pipe through which a fluid is flowing say water as shown in figure. Let d1 = Diameter at section 1-1, P1 = Pressure at section 1-1,

V1 = Velocity of fluid at section 1-1, a1 = Area of cross section at section 1-1

Where d2, P2, V2, a2 are the corresponding values at section 2-2. Now applying Bernoulli’s equation at section 1-1 and 2-2, we get 2 2

1 + 1

+ =

2 + 2

+

2 1

2 2

As pipe is horizontal Z1 = Z2

2 2

1 +

1

= 2

+ 2

2

2

2 2

1 −

2 =

2

−

1

2 2

− 2 2

1 2

= 2

−

1

2

2 2

−

But h = 1 2

2

2 2

h = 2

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

2

2

Now applying the continuity equation at sections 1-1 and 2-2, we get a1V1 = a2 V2 a2

v2

V1 =

a1 Now substituting the above value in equation (1), we get

1 2 2 = 2 − 2 1 2


Figure: - Venturimeter.


Q =

2 2

Q = 1 2 2 ------------------------------------------- (2)

12 − 22

The equation (1) gives the discharge under ideal conditions and is called, theoretical discharge. Actual discharge,

Q act

= C d

1 2 2

12 − 22

Where Cd is called as Co-efficient of venturimeter and its value is less than one. Value of ‘h’ given by differential U-tube manometer: Case I: - Let the differential manometer contains a liquid which is heavier than the liquid flowing through the pipe, when the venturimeter is horizontal. h = x − 1 Where Sh = Specific gravity of the heavier liquid,

SO = Specific gravity of the liquid flowing through pipe. X = Difference of the heavier liquid column in U-tube. Case II: - If the differential manometer contains a liquid which is lighter than the liquid flowing through the pipe, when the venturimeter is horizontal. h = x 1 − Where Sl = Specific gravity of lighter liquid in U-tube,

SO = Specific gravity of the liquid flowing through pipe. X = Difference of the heavier liquid column in U-tube. Case III: - Let the differential manometer contains a liquid which is heavier than the liquid flowing through the pipe, when the venturimeter is inclined. h = x − 1 Where Sh = Specific gravity of the heavier liquid,

SO = Specific gravity of the liquid flowing through pipe. X = Difference of the heavier liquid column in U-tube. Case IV: - If the differential manometer contains a liquid which is lighter than the liquid flowing through the pipe, when the venturimeter is inclined. h = x 1 − Where Sl = Specific gravity of lighter liquid in U-tube,

SO = Specific gravity of the liquid flowing through pipe. X = Difference of the heavier liquid column in U-tube.

Experimental Set-Up: 1) Venturimeter is fitted to a horizontal pipe line to which an inlet valve is fitted. 2) Pressure tabs are provided at entrance and throat section and manometer tubes are

fitted to the pressure tapes. 3) Measuring tanks, stop watch and scale are used to measure discharge.

Procedure. 1) Fill the clean water in the sump tank approximately 3/4 of its height.

2) Note the pipe diameter ( 1) and throat diameter ( 2) of Venturimeter.

3) Note the density of manometer liquid i.e. mercury and that of fluid flowing through pipeline i.e. water.



4) Start the flow and adjust the control valve in pipeline for maximum discharge. 5) Measure the pressure difference (H) across the Venturimeter by using U – tube

manometer. 6) Measure flow rate i.e. actual discharge ( ) through Venturimeter by means of

collecting tank. 7) Calculate the theoretical discharge ( ) through Venturimeter by using the formula.

8) Decrease the flow rate by adjusting the control valve and repeat the process for at

least five times. 9) Determine the coefficient of discharge ( ) for each flow rate and find the mean value

of coefficient of discharge ( ) mean. 10) Plot a graph of ( ) on y-axis versus ( ) on x- axis. 11) Calculate the slope of graph of ( ) versus ( ), it gives the mean value of coefficient of

discharge ( ) mean graphically. Precaution:

Care should be taken that all entrapped air is removed. So as to prevent

bubble formation in manometer tube and the pipe. Observation:

Table:-05 Sr. Manometric Reading in

H= 13.6

− 1 Time taken to Actual Theoretical Coefficient of

No. meters. 1 rise 10 ( h )cm Discharge Discharge Discharge.

Meters

X= − of water level in

1 2 1 2

‘T’ in seconds In 3 ⁄sec

3⁄sec

01

02

03

04

05

06

Calculation: Sample Calculation for First Reading: a) Head of water, H = 12.6 × X

b) Actual discharge, =

c) Theoretical discharge, = 1 2 2

1

2− 22

d) Coefficient of Discharge, =

Significance:

meters. 3⁄sec

3 ⁄sec

1) Venturi meters / Orifice meter provide the widest variety of measurement options in

piped systems for liquids, gas, steam, and mixed media of any metering technology. All while offering the highest degree of traceable accuracy.

2) They can be used reliably for billing or custody transfer; and they can be used for rectangular or circular metering. In addition, Venturi meters can be oriented in any plane and can measure accurately whether the line fluid is flowing upwards or downwards.

Graph: 1) Actual discharge Vs venturimeter head. 2) Log Vs Log h SRCOE, Lonikand. Civil Engineering Department Page 18


Result: Coefficient of discharge ( ) for Venturimeter is found to be

a) experimentally b) Graphically as

Conclusion: For venturimeter the value of co-efficient of discharge is 0.98 and these are the

instruments which are used to measure the discharge through the pipes. In our experiments the value of coefficient of discharge is very nearer to the 0.98.


Experiment:-06

Transition of Laminar and turbulent flow through pipes. Aim: - To determine the type of flow by using Reynolds Number. Apparatus: 1) Reynolds’s experimental arrangement, 2) Collecting tank, 3) Stop watch, 4) Scale, Theory:

Reynolds carried out experiments to decide limiting values of Reynolds number to

quantifiably decide whether the flow is laminar, turbulent or transition. The flows can

visualize by passing a streak of dye and observing its motion. Laminar Flow: A flow is said

to be laminar when the various fluid particles moves in layer with one layer of fluid living

smoothly over on adjacent layer. A laminar flow is one in which the fluid particles moves in

layers or laminar with one layer sliding over the other. Therefore there is no exchange of

fluid particles from one layer to the other and hence no transfer of later of momentum to be

adjacent layers. The particles, in the layer having lower velocity, obstruct the fluid particles

in the layer with higher velocity. This obstruction force is called viscous resistance or

viscosity. The laminar flow is one in which fluid layers glide over each another. It has low

velocity and high viscous resistance. Turbulent Flow: There is a continuous transfer of

momentum to adjacent layers. Fluid particles occupy different relative position at different

places. It is one in which, the particles get thoroughly mixed on (called turbulence). The

turbulent flow has higher velocity. The flow in canals, pipes and rivers is usually turbulent

flow. Transition Flow: The transition flow has intermediate properties between the laminar

and turbulent flow. In laminar the forces should be considered to calculate the friction loss

and in the turbulent flow only the internal forces are

considered because the effect of viscous force is negligible

as compared to internal forces. Reynolds carried out

experiments to decide limiting values of Reynolds number

to quantifiably decide whether the flow is laminar, turbulent

or transition. If the Reynolds number is less than 2000, then

it is Laminar Flow figure (a),If the Reynolds number is Figure: - Types of flows. between 2000 and 4000 then it is Transition Flow figure (b), If the

Reynolds number is greater than 4000, then it is Turbulent figure (c). The flow can be

visualized by passing a streak of dye and observing its motion. In the laminar, low velocity

flow the streak line is only slightly zig - zag. In the turbulent flow, the dye thoroughly mixes

up in the flow. Thus passing through a glass pipe and observing the velocity at different

mixing stages of the dye is the principle on which Reynolds apparatus is based. Experimental Setup: The apparatus consist of a glass tube with one end

having bell mouth entrance connected to water tank. (1) The tank is of sufficient capacity to store water.


(2) At the other end of the glass tube a valve is provided.

(3) A small container with dye is provided at top of the tank.

(5) This small container with dye is connected to the

glass tube where the water is flowing. Reynolds Apparatus’ Procedure:

1. Fill the sump tank with sufficient clean water by operating inlet valve.

2. Now fill up with sufficient clean water in dye tank and put a small amount of potassium permagnet in it.

3. Start water flow. 4. Adjust dye water flow to about 2 litres / minute. 5. Start the pump and inject the dye. 6. Wait for some time, a steady line of dye will be observed. 7. Note down the flow rate i.e.

8. Slowly increase the water flow sees that water level in supply tank remains constant. 9. At particular flow rate dye line will be disturbed, note down this flow rate.

10. Further increase the flow. The disturbance of dye line will go on increasing and at

certain flow. The dye line diffuses over the entire cross section. 11. Note down this flow rate. 12. The whole procedure is repeated for 3 times.

Specifications:

Diameter of glass tube, d= 0.02 meter

Area of tank

a = 2

=3.146 10−4

2

4

Area of the tank A= 0.3 x 0.3 = 0.09 2

Kinematic viscosity for water γ = 1x10−6 2 ⁄sec

Observation Table:-

Sr. Velocity Time required for Discharge Reynolds Observed

no. m/sec 10 cm rise of

3

Number flow

water. regimes

V T Q Re

01

02

03

04

05

06

Calculation: Sample Calculation for First Reading: Discharge:


Qact =

Volume of water collected m3 sec

Time required to collect water in a tank for 10 cm

Velocity:

V = Qact

meter / Second

a

Reynold’s Number: V ×d

Re

= γ

Significance: (1) To predict type of flow in pipes. (2) To predict type of flow in channels

Conclusion. We have studied the use of Reynolds apparatus to finding types of flow such as laminar flow, turbulent flow and transition flow etc.



Experiment:-07

Minor loss in a pipe system for a given pipe.

Aim: To determine minor losses due to pipe fittings such as Expansion, Contraction, Bend, Elbow, etc. Apparatus:

A pipe elbow,

A sudden contraction pipe,

A sudden expansion pipe,

A pipe bend, A collecting tank,

A stop watch,

Scale etc. Theory: When a fluid is flowing (turbulent flow) through a pipe, the fluid experiences some of the energy of fluid is lost. This loss of energy is classified as: 1) Major Energy losses:

This is due to the friction and it is calculated by the following formulae: a) Darcy-Weisbach formula

b) Chezy’s formula 2) Minor Energy losses:

This loss of energy is due to change of velocity of the flowing fluid in magnitude or direction. This is due to

a) Sudden expansion of pipe,

b) Sudden contraction of pipe,

c) Bend in pipe,

d) Pipe fittings etc,

e) At the entrance of pipe,

f) At the exit of a pipe.

g) An obstruction in pipe. Major Energy losses: Darcy-Weisbach formula:

4 2 = 2

Where = loss of head due to friction, V = mean velocity of fluid, f = co-efficient of friction which is a function of Reynolds number

= 16 for < 2000,

0.079

= 1 4 for varying from 4000 to 1000000

Chezy’s Formula: V = C

Where m = hydraulic mean depth, m = = 4 I = loss of head per unit length of pipe, I

=

C = Chezy’s constant.


Minor Energy losses: This loss of energy is due to change of velocity of the flowing fluid in magnitude or direction. a) Sudden expansion of pipe:

Due to sudden change of diameter of the pipe, the liquid flowing from smaller pipe to is not able to follow the abrupt change of the boundary. Thus the flow separates from the boundary and turbulent eddies are formed. The loss of energy takes place due to the formation of these eddies. This is called as loss of energy due to sudden expansion of pipe.

Consider a liquid flowing through a pipe which has sudden enlarged form from maller diameter to a larger diameter as shown in figure. Now consider two section 1-1 and 2-2 before and after the enlargement.

Let P1 = pressure intensity at setion 1-1

V1 = velocity at section 1-1

A1 = area of pipe at section 1-1

Where P2, V2 and A2 are the corresponding values at section2-2,

Now applying Bernoulli’s theorem to sectio1-1 and 2-2,

2 2

1 + 1

+ = 2 + 1

+ +

2 1

2 2

=

1 2

2 2

1 + 1

= 2 + 1

+

2

2

2 2

h = 1 + 2

+ 1

+ 2

f 2

2

V − V 2

he = 1 2

2g

b) Sudden contraction of pipe: As the liquid flows from larger pipe to smaller pipe, the area of flow goes on decreasing

and becomes minimum at section C-C as shown in figure. This section C-C is called Vena-contracta, after section C-C a sudden enlargement of the area takes place. The loss of head due to sudden contraction is actually due to sudden enlargement from Vena-contracta to smaller pipe.

Let = Area of flow at section C-C

VC = Velocity of flow at section C-C A2 = Area of flow at section 2-2 V2 = Velocity of flow at section 2-2 hc = Loss of head due to sudden contraction.

Now, hc = Actual loss of head due to enlargement from C-C to section 2-2 and is given by equation as − 2 2 2

= 2

= 2

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

2 2

2

From continuity equation, we have

=

1 1 2 2

2

=

2 1

1

=

A c A

2 C

c

Now substuting the above value in equation (1),



2 1 2

= 2

− 1

2

2

2 1

=k 2 , where k =

− 1

2

If the value ofis assumed to be equal to 0.62, then

2

1

K =

− 1 = 0.375

0.62

2

2

Then becomes as

=k 2 = 0.375 2

2 2

If the value of Cc is not given then the head loss due to contraction is taken as, =0.5 22

2 c) Bend in Pipe: Where there is any bend in a pipe, the velocity of flow changes, due to which the separation of the flow from the boundary and also formation of eddies, takes place. Thus the energy is lost. Loss of head in pipe due to bend is expressed as,

2 hb =

2 Where hb = loss of head due to bend, V = velocity of flow, K = co-efficient of bend,

The value of ‘k’ depends on,

i) Angle of bend,

ii) Radius of curvature of bend,

iii) Diameter of pipe.

d) Pipe fittings: The loss of head in the various pipe fittings such as valves, couplings etc.,

hf =

2

2

Where V = velocity of flow,

k = co-efficient of pipe fitting.

e) An obstruction in pipe:

2

2

Head loss due to obstruction =

− 1

2 −

f) Entrance of a pipe: This is loss of energy which occurs when a liquid enters a pipe which is connected to a large tank or reservoir. This loss is similar to the loss of head due to sudden contraction. This loss depends on the form of entrance. For a sharp edge entrance, this lossis slightly more than a rounded or bell mouthed entrance. In practice the value of loss of head at the entrance of a

2

pipe with sharp cornered entrance is taken as 0.52 .

g) Exit of pipe: This is the loss of head due to the velocity of liquid at outlet of the pipe which is dissipated

2

either in the form of a free jet or it is lost in the tank or reservoir. This loss is equal to2 .

h) Obstruction in a pipe: Whenever there is an obstruction ina pipe, the loss of energy takes place due to reduction of the area of the cross-section of the pipe at the place where obstruction is present. There is sudden enlargement of the area of flow beyond the obstruction due to which loss of head

2 2

takes place and is equal to

− 1

2 −



Experimental Set-Up: 1) Basic pipe of sufficient length to flow the water, 2) Measuring tank,

3) Pipe fittings such as,

a) Sudden expansion,

b) Sudden contraction,

c) Pipe bend,

d) Pipe elbow,

e) Flow control valve etc.

4) Differential manometer. Procedure:

1. Collect clean water in the sump tank. 2. Start the motor, allow the water to flow the pipe fittings like sudden expansion,

sudden contraction, bend, elbow. 3. Take the manometric reading. 4. Now find the discharge of water flowing the pipe, by noting time to collect the water

for a rise of 10 cm in the tank. 5. The above procedure is repeated for all fittings and readings are entered in the table.

Observation: Table:-07

Loss of Head due to Sudden Expansion

Sr.no. Manometric Reading in 13.6 Discharge

− 2

=

H=

− 1

= 1 2

1

1

meters.

1

2

= =

2

X= −

2

1 2 1 2

M m M M 3⁄sec m

01

02

03

04

05

06

Specification:

d1 = Given

d2 = Given

a1 =

a2 =

Calculation:

Sample Calculation for First Reading:

H= 13.6

− 1 = meters

1

Discharge = = 3⁄sec

= m/sec

1 1

=

m/sec

2 2

= − 2/2g meter

1 2


Observation:

Table:-08

Loss of Head due to Sudden Contraction.

Sr.no. Manometric Reading 13.6 − 1

Discharge Velocity 2

H=

= K 2

1

2

=

X=−

1

2 1 2

=

2 2

M M M M 3⁄sec m

01

02

03

04

05

06

Specification:

d1 Given

d2 = Given

a2 =

1 =

2

The following value table gives the value of K for the corresponding value of 1 .

2

1 1.00 1.10 1.25 1.50 2.00 2.50 3.00 3.50 4.00

2

K 0 0.10 0.19 0.28 0.375 0.40 0.42 0.43 0.45

Calculation:

Sample calculation for first reading:

For first reading:

H= 13.6

− 1 = meters

1


=

m/sec

1 1

=

m/sec

2 2

2

= K × 2

meter

2

Observation:

Table:-09

Loss of head due to Bend.

Sr.no. Manometric Reading 13.6 − 1

Discharge 2

H= = K 2

1

2

= =

X= − 1 1

1 2 1 2

m m M M 3⁄sec m

01

02

03

04


d1 Given

d2 = Given a2 =

1 = 2 The following value table gives the value of K for the corresponding value of 1.

2 1 1.00 1.10 1.25 1.50 2.00 2.50 3.00 3.50 4.00

2

K 0 0.10 0.19 0.28 0.375 0.40 0.42 0.43 0.45

Calculation:


For first reading:

H= 13.6 − 1 = in meters

1


=

m/sec

1

1

2

= K × 2 meter

2

Observation:

Table:-10

Loss of head due to Elbow.

Sr.no. Manometric Reading 13.6 Discharge 2

− 1

=

H= = K 2

1

2 2 2

=

X= −

1 2 1 2

M m m m 3⁄sec M

01

02

03

04

05

06

Specifications:

d1 Given

d2 = Given

a2 =

1 = 2 The following value table gives the value of K for the corresponding value of

1 .

2

1 1.00 1.10 1.25 1.50 2.00 2.50 3.00 3.50 4.00

2

K 0 0.10 0.19 0.28 0.375 0.40 0.42 0.43 0.45

Calculation:


For first reading:

H= 13.6

− 1 = in meters

1


=

m/sec

1 1


= 2 2

= K × 22 2

in m/sec

meter

Significance:

(1) For design of pipes in water supply schemes.

(2) For design of sewer pipe schemes.

(3) In chemical industry. (4) In dairies.

Result. Conclusion. The experiment is conducted using pipe fitting apparatus to determine different losses due to pipe fittings.


Experiment:-08 Demonstration of fluid flow through appropriate VCD/Audio visual/PPT.


Assignment No:-I Objective:

Solving pipe network problem (Hard-Cross method) using Excel work sheet. Problem: Calculate the discharge in each pipe of the network shown in figure. The pipe network consists of five pipes.

The head loss hf in a pipe is given by hf = rQ2. The values of ‘r’ for various

Solution: Loop ABC

Pipe Discharge R rQ2 2Rq ∆QS

AB 20 2 9800 280

BC 35 1 1225 70 -12.58475

AC 30 -4 3600 240

7425 590

Loop BDC

Pipe Discharge R rQ2 2rQ ∆Q

BD 15 5 1125 150

DC 35 -1 -1225 70

CB 35 -1 -1225 70

-1325 290 4.568966 Thus applying the above obtained correction the modified discharge for various pipe which is the distribution for second trial. For this

distribution the correction ∆Q for the loops ABC and BDC are computed as follows. Loop ABC

Pipe Discharge R rQ2 2rQ ∆QS

AB 57 2 6498 228

BC 17 1 289 34

AC 43 --4 -7396 344

-600 606 -1.00495

Loop BDC

Pipe Discharge R rQ2 2rQ ∆Q

BD 20 5 2000 200

DC 30 -1 -900 60

CB 17 -1 -280 34

811 294 2.758503

Thus applying the above obtained correction the modified discharges for the various pipe which is the distribution for the third trial. For

the distribution the correction ∆ for loops ABC and BDC are computed as follows: Loop ABC

Pipe Discharge R rQ2 2rQ ∆QS

AB 58 2 6728 232

BC 21 1 441 42

AC 42 --4 -7056 336

113 610 0.185246

Result: - To solve the problem in excel works sheet and final answer is 0.185246



or Objective: Three reservoir problem. Problem: Three reservoirs ‘A’, ‘B’ and ‘C’ are connected by a pipe system as shown in

figure. The length and diameters of pipes 1, 2 and are 800 m, 1000 m, 800 m and 300 mm, 200 mm and 150 mm respectively. Determine te piezometric head at junction D. Take f = 0.005.

Assignment No:-II Objective:

Determination of friction factor for a pipe using any programming language Problem: A rough pipe is of diameter 8.0 cm. The velocity at a point 3.0 cm from wall is 30% more than the velocity at a point 1 cm from pipe wall. Determine the average height of the roughness.

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