Chapter FLUID PROPERTIES 1 - Eduzphere

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Fluid mechanics is that branch of science which deals with the behavior of fluid at rest as well as in

motion.

The study of fluid at rest is called fluid statics. In fluid Statics fluid particles do not move with respect to

one another.

The studies of fluid in motion where external force are not considered are called fluid kinematics.

If external force is also considered for the fluid to be in the motion it is called fluid dynamics.

Matters

Solid Fluid

Liquid Gases

Particulars Solid Liquid Gas

1. Intermolecular bond Very strong

Weak

Very weak

2. Shape define Indefinite Indefinite

3. Volume define Define Indefinite

4. Nature Rigid/ incompressible Incompressible/ but can be

compressible as in water

hammer.

compressible

In case of solid, deformation due to normal & tangential force within elastic limit disappear if

external force is removed but fluid in rest condition sustain only normal stress & deform continuously

when subjected to shear stress.

Chapter

1 FLUID PROPERTIES

Syllabus: Properties, Newton‟s law of viscosity, Types of Fluid,

Compressibility & Bulk Modulus, Surface Tension and Capillarity,

Vapour pressure and cavitation. Weightage : 5%


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FM + OCF & MACHINES

Fluid can offer no permanent resistance to shear force & possess ability to flow & change its shape.

The tendency of continuous deformation of a fluid is called fluidity & act of continuous deformation

is called flow.

Liquid Gas

1. Liquid have a definite volume irrespective of size

of container hence it easily acquire shape of

container.

No fixed volume

2. Free surface is formed if volume of container >

volume of liquid

No free surface is formed

Free Surface

Gas

1. Liquid can be regarded as incompressible

(𝜌 constant)

It is compressible.

2. Pressure & temperature changes have practically no

effect on volume of liquid.

e.g. water, kerosene etc.

Gas expand infinitely

e.g. air, ammonia, CO2 etc.

Continuum

Fluid consist of discrete molecules, for the analysis of fluid flow problem fluid is treated as continuous

media which means all void or cavities (microscopic and macroscopi(C) which may be occur in fluid are

ignored. Hence a continuous & homogenous fluid medium is called continuum.

Properties of fluid

(1) Density or mass density:

It is represented by 𝜌. The ratio of mass of fluid to its volume.

Mathematically 𝜌 =𝑀

𝑉

Unit

Dimensional formula 𝑀1𝐿−3𝑇0 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

𝑀

𝑉 ×

𝑎

𝑎=

𝐹

𝑉 × 𝑎=

𝐹

𝑚3 . 𝑉

𝑇

=𝐹

𝑚3 . 𝑑

𝑇2

= 𝐹𝐿−4𝑇2 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

S.I. = kg/m3

C.G.S. = gm/cm3


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FM + OCF & MACHINES

Mass density of water = 1000 kg/m3 or 1g/cm

3.

1. Mass density of air = 1.24 kg/m3.

2. Density of liquid may be considered as constant value while that of gas changes with

variations of pressure & temperature.

3. Density of water is maximum at 4oC if temperature increases 0 to 4

oC, density of water

increases but if we further increases temperature beyond 4oC, then density starts decreasing.

(2) Weight density or specific weight

It is represented by w.

The ratio of weight of fluid to its volume.

Mathematically, w = 𝑊

𝑉

Where, 𝑊 = weight of fluid

V = volume

Unit kN/m3

Dimensional formula - 𝑘𝑁

𝑚3 = 𝑀1𝐿1𝑇−2

𝑀0𝐿3𝑇0 = 𝑀1𝐿−2𝑇−2 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

w = 𝑤

𝑉=

𝐹

𝐿3 = 𝐹𝐿−3𝑇0 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

1. The value of specific weight of water is = 9810 N/m3 = 9.81 kN/m

3

2. Relationship between specific weight & density

w m gw g

v v

w = g

3. As temperature increases ↑ 𝑇 , specific weight decreases ↓ 𝑤 = 𝑤 ↓

𝑉𝑐𝑜𝑛𝑠 .=

↓ 𝑚 × 𝑔

𝑉𝑐𝑜𝑛𝑠 .

Temp.

4. If pressure increases ↑ 𝑃 , specific weight increases ↑ 𝑤 .

Pressure

(3) Specific gravity or relative density (S or G) –

The ratio of density (Specific weight) of liquid to density (specific weight) of water.

Mathematically, S = 𝜌 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑

𝜌 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑜𝑟

𝑤 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑

𝑤 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟


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FM + OCF & MACHINES

Where 𝜌 = density

𝑤 = unit weight

1. Specific gravity of water = 1.

2. Specific gravity of mercury = 13.6.

3. Specific gravity of petrol = 0.9.

4. Liquid whose specific gravity less than 1, float over water whereas liquid whose specific gravity

greater than 1, sinks in water.

5. Specific gravity have no units & dimensionless.

(4) Specific volume

Reciprocal of density. (Volume per unit mass)

Mathematically 1

𝑑𝑒𝑛𝑠𝑖𝑡𝑦=

1

𝜌

Unit = 𝑚3

𝑘𝑔

Dimensional formula 𝑉

𝑚= 𝑀−1𝐿3𝑇0 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

𝑉 ×𝑎

𝑚 ×𝑎=

𝐿3𝐿 𝑇−2

𝐹= 𝐹−1𝐿4 𝑇−2 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

Example 1:– 1 liter of liquid whose weight is 7N. Calculate density, specific weight, specific gravity &

specific volume.

Solution:- 1 liter = 1

1000 𝑚3

W = 7N

1. Specific weight, w = 7

1 1000 = 7000

𝑁

𝑚3

2. We know that

w = 𝜌g

7000 = 𝜌 x 9.81

Density, 𝜌 = 7000

9.81 =

7000 × 100

981 = 713.5 𝑘𝑔/𝑚3

3. Specific gravity

= 𝑤𝑒𝑖𝑔 𝑕𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑

𝑤𝑒𝑖𝑔 𝑕𝑡 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 =

7000

9810 = 0.7135

4. Specific volume = 1

𝜌=

1

713.5 𝑘𝑔

𝑚 3

= 1

713.5

𝑚3

𝑘𝑔= 0.0014

𝑚3

𝑘𝑔

(5) Viscosity

Property of a fluid which offer resistance to the movement of one layer of fluid to the adjacent layer

of fluid.


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FM + OCF & MACHINES

Mathematically 𝜏 ∝ 𝑑𝑢

𝑑𝑦

𝜏 = 𝜇 𝑑𝑢

𝑑𝑦

y

u

u

dy

+ du

u

du

Where 𝜏 = shear stress

𝑑𝑢

𝑑𝑦= Rate of change velocity with respect to depth.

𝜇 = Coefficient of viscosity, absolute viscosity, dynamic viscosity, simple viscosity.

Unit :- S.I unit = τ

𝑑𝑢

𝑑𝑦

= Ns/m2 CGS unit = poise

Relationship between 𝑺. 𝑰. & CGS unit:-

1 poise = 1

10 𝑁𝑠/𝑚2

1 centipoise = 1

1000

𝑁𝑠

𝑚2

Dimension Formula –

𝜇 = 𝜏𝑑𝑢

𝑑𝑦

= 𝐹 𝐴

𝑑𝑢 𝑑𝑦 =

𝑚𝑎𝐴𝑑𝑢𝑑𝑦

= 𝑀1𝐿1𝑇−2 𝐿

𝐿𝑇−1 𝑋𝐿2 = 𝑀1𝐿−1𝑇−1 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

𝜇 =𝜏𝑑𝑢

𝑑𝑦

= 𝐹 .𝑑𝑦

𝐴 .𝑑𝑢=

𝐹 𝐿

𝑚2 .𝐿 𝑇−1 = 𝐹1𝐿−2𝑇1 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

1. 𝜇 for water = 1 centipoise = 10−3 𝑁𝑠

𝑚2

for air = 0.0181 centipoise = 0.0181 10−3 𝑁𝑠

𝑚2

2. 𝜇 water = 55 𝜇 𝑎𝑖𝑟

3. Viscosity of liquid is due to cohesion & adhesion for gases viscosity is due to molecular momentum

transfer.

4. If fluid flowing passed a fixed surface, due to adhesion between fixed surface & fluid at the surface

of contact, velocity of fluid particles become zero this condition is called no slip condition.

EFFECT OF VISCOSITY WITH TEMPERATURE


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FM + OCF & MACHINES

1. Mathematical aspect –

(i) For liquid 𝜇 ∝ 1

𝑇

It means with increase in temperature, viscosity of liquid decreases.

(ii) For gases → 𝜇 ∝ 𝑇

𝑀 M= Molecular weight

It means with increase in temperature, viscosity of gas increases.

2. Conceptual Aspect / Theoretical Aspect –

(i) Viscosity of liquid decreases with increase in temperature because in liquid adhesive forces

predominate over molecular momentum transfer. Due to closely packed molecules with

increase in temperature cohesive forces decreases which results in decreasing viscosity.

(ii) In case of gas viscosity of gas increasing with increase in temperature because molecular

momentum transfer as compare to cohesive forces hence with increase in temperature molecular

momentum transfer increases. Hence viscosity of gas increases.

(6) Specific Viscosity –

The ratio of viscosity of fluid to viscosity of water is called specific viscosity

𝜇 of fluid

𝜇 of water

(7) Kinematic Viscosity

Ratio of dynamic viscosity / density

Mathematically (𝜐) = dynamic viscosity

density =

𝜇

𝜌

Unit – S.I. = 𝑚2/s

C.G.S. = c𝑚2/s or stoke

Dimension Formula –

𝜐 = 𝑚2

𝑠= 𝑀0𝐿2𝑇−1 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

𝐹0𝐿2𝑇−1 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

Relationship between 𝑺. 𝑰. & CGS unit:-

1 stoke = 1c𝑚2/ sec = 10−4 𝑚2/𝑠

1 centistoke = 1 10−6 𝑚2/𝑠

1. 𝜐 water = 10−6 𝑚2/𝑠

2. 𝜐 air = 15 x 10−6 𝑚2

𝑠𝑒𝑐


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FM + OCF & MACHINES

3. 𝜐 air = 15 (𝜐)water

(8) NEWTON’S LAW FOR VISCOSITY

Shear stress (𝜏) on a fluid element layer is directly proportional to the rate of shear strain.

Mathematically 𝜏 ∝ 𝑑𝑢

𝑑𝑦 𝜏 = 𝜇 .

𝑑𝑢

𝑑𝑦

Following observations helps to appreciate the interaction between viscosity & velocity distribution:-

1. Maximum shear stress occurs where the velocity gradient is largest & shear stresses are zero

when velocity gradient is zero.

2. Velocity gradient at the solid boundary has finite value.

3. Velocity gradient become less steep with distance from the boundary.

0Velocity profile

u + du

u

(9) NON - NEWTON’S LAW FOR VISCOSITY –

𝜏 = 𝐶 𝑑𝑢

𝑑𝑦 𝑛

+ 𝐵

Where n = power index

C & B = constants called consistency index & additive index respectively showing flow behavior.

DIFFERENT TYPES OF FLUID

1. Ideal fluid - Fluid which is incompressible (𝜌= constant) & inviscid (non-viscous 𝜇 = 0) are called

ideal fluid. An ideal fluid has no surface tension (𝜍 = 0)

Ideal fluid is imaginary fluid which does not exist in nature. However most common fluid such as air

& water has low viscosity, so treated as ideal fluid.

Bulk modulus of ideal fluid is infinite. B. Curve (A) corresponds ideal fluid curve


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FM + OCF & MACHINES

2. Real fluid – Fluid which possesses viscosity, compressibility & surface tension are termed as real

fluid.

Real fluids are also known as practical fluid.

3. Newtonian’s fluids – Fluid which obey newton‟s law of viscosity.

A. Curve (B) & (C) are Newtonian fluid curve.

B. Fluid represented by line (B) is more viscous than line (C).

C. Example – Air, water, kerosene oil, petrol, thin lubricating oil, ethanol benzene etc.

4. Non – Newtonian fluid – fluid which do not obey Newton‟s law of viscosity.

Non – Newtonian fluids are of two types:-

(i) Pseudo plastic fluid –

(D) curve represents pseudo plastic fluids.

Example – Blood, milk, liquid cement, clay.

(ii) Dilatant fluid –

(e) curve represents the dilatant fluid.

Example – Concentrated Solution of sugar, aqueous suspension of rice starch.

For non – Newtonian law of viscosity i.e.

n

duB C

dy

Where n < 1, then it is pseudo plastic fluids

n > 1, then it is dilatant fluid.


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5. Ideal plastic fluid – Fluid in which shear stress more than yield value & later proportional to shear

strain called ideal plastic fluid. Ideal plastic fluid is also known as Bingham‟s fluids (f) Curve

corresponds ideal plastic fluid curve.

6. Thixotropic fluid – Curve (g)

7. Rheophatic fluid – Curve (h)

8. Ideal solid – Curve (i)

(10) Compressibility (β) & Bulk Modulus (K)

Compressibility (β) of a fluid is its ability to change its volume under pressure.

The relative change of volume per unit pressure is given by coefficient of compressibility β.

Mathematically β = −𝑑𝑉

𝑉 ×

1

𝑑𝑝

Where – ve sign represents decrease in volume when pressure is applied.

Bulk Modulus (K) – Reciprocal of compressibility is called bulk modulus.

Mathematically K = 1

β = −

𝑑𝑝

𝑑𝑉/𝑉

Where 𝑑𝑝 = change in pressure

dV= change in volume

Example 2:- Determine Bulk modulus of a liquid if pressure of liquid is increased from 70 N/cm2 to 130

N/cm2. Volume of liquid is decreases by 0.15%.

Solution:- K = −𝑑𝑝

𝑑𝑉/𝑉 =

𝑐𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒𝑐𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒

𝑜𝑟𝑖𝑔𝑖𝑛𝑒 𝑣𝑜𝑙𝑢𝑚𝑒

𝑑𝑉

𝑉= 0.15% =

0.15

100 =

130−70

0.15 × 100

= 60

15 × 100 × 100 = 4 10

4 N/cm

2

(11) Surface tension & Capillarity concept

Liquid have characteristic properties of cohesion & adhesion. Cohesion refers to force with which the

neighboring fluid molecule are attracted toward each other whereas, Adhesion refers to adhering or

clinging of fluid molecules to the solid surface with which they comes in contact.

1. Forces between like molecules (water to water) are cohesion.

2. Forces between unlike molecules (water to glass) are adhesion.


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FM + OCF & MACHINES

Wetting liquids:- Liquids in which adhesive forces predominate than cohesive forces.

For wetting liquids, angle of contact should be less than 90o.

Example –Water.

Gas

Liquid

Solid

Wetting Liquid(water)

Wetting liquid (Water)

<2

Liquid

Solid

Gas

Non-wetting Liquid (Mercury)

<2

Non – wetting liquid:- Liquids in which cohesive forces predominate to adhesive forces.

For non-wetting liquids, angle of contact > 𝜋

2.

Example –Mercury.

Surface Tension:- Surface tension is defined as the tensile force acting on the surface of a liquid in

contact with gas or on the surface between two immiscible liquid such that contact surface behaves like a

membrane under tension.

e

AB

Unit:- N/m

Dimensional formula:- 𝑁

𝑚=

𝐹

𝐿=

𝑚𝑎

𝐿=

𝑀1𝐿1𝑇−2

𝑀0𝐿1𝑇0

= 𝑀1𝐿0𝑇−2 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚 =𝑁

𝑚=

𝐹

𝐿= 𝐹1𝐿−1𝑇0 → 𝐹𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

1. Surface tension only depends upon cohesion.

2. With rise in temperature, surface tension decreases.

3. For water-air interface, surface tension 𝜍 = 0.073 𝑁/𝑚

4. For air-mercury interface, surface tension 𝜍 = 0.48 𝑁/𝑚

5. Surface tension depends upon

a) Nature of liquid

b) Nature of surrounding material

c) Kinetic Energy hence temperature of liquid molecule.

6. Growth of temperature results in reduction in intermolecular cohesive forces hence reduction in

surface tension force.


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FM + OCF & MACHINES

Pressure inside a water droplet:-

p = 4𝜍

𝑑

Where p = pressure

d = diameter of water droplet.

𝜍 =Surface tension

Pressure inside a soap bubble:-

Air

p = 8𝜍

𝑑

Pressure inside a soap bubble:-

𝑝 =2𝜍

𝑑

Example 3:- Find the surface tension (𝜍) in a soap bubble of a 40 mm diameter when inside pressure is

2.5 N/m2 above atmospheric pressure.

Solution – p = 8𝜍

𝑑

𝜍 = 𝑝𝑑

8 = 2.5

𝑁

𝑚2 × 0.040 𝑚

= 2.5 × 0.04

8

𝑁

𝑚 =

0.100

8

= 0.100

8 = .0125

𝑁

𝑚


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FM + OCF & MACHINES

Example 4:- Surface tension of water in contact with air is 0.0725 𝑁

𝑚 pressure inside the droplet of water

is 0.02 N/cm2 above atm. Calculate diameter of droplet of water.

Solution:- For a water droplet –

p = 4𝜍

𝑑

d = 4𝜍

𝑝 =

4 ×0.0725

0.02

𝑁

𝑚 .

𝑐𝑚 2

𝑁 [1m = 100cm]

= 1.45 10-3

m = 1.45 mm

Capillarity:- The phenomena of rise or fall of liquid surface in small tube relative to the adjacent general

level of a liquid when the tube is held vertically in the liquid.

hFree water level

Capillary rise

h

Capillary fall

The rise in liquid surface is known as capillary rise.

Capillary Rise = 4𝜍 cos 𝜃

𝜌𝑔𝑑 or

4𝜍 cos 𝜃

𝑤 .𝑑

In case of water, As adhesion between glass &

water molecule is greater than cohesion between

molecules. Hence water level is raised in the

narrow tube. Consequently water molecules spread

over glass surface & form concave meniscus with

small angle of contact (𝜃 = 0o).

h = 4𝜍

𝜌𝑔𝑑 or

4𝜍

𝑤 .𝑑

Fall in liquid surface is known as capillary

depression or capillary fall.

h = 4𝜍 cos 𝜃

𝜌𝑔𝑑 or

4𝜍 cos 𝜃

𝑤 .𝑑

In case of mercury, cohesion between mercury

molecules is greater than adhesion of mercury to

glass consequently mercury is depressed at the

point of contact & display convex meniscus with

angle of contact greater than 90o. (𝜃 = 128o)

For mercury h = 4𝜍 cos 𝜃

𝜌𝑔𝑑

Unit:- mm of Hg. or cm of Hg.

1. Capillary action depends upon both cohesion & adhesion.

2. Capillary depends upon

(A) Surface tension (B) Cosine of angle of contact

(C) Weight (𝜌𝑔 𝑜𝑟 𝑤) (D) diameter of tube.

3. As h ∝1

𝑑 so small diameter of tube are to be avoided for precise work. The recommended value of d

for water & mercury is 6 mm. When the diameter of tube is large (above 12 mm) capillary rise & fall

is negligible.


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FM + OCF & MACHINES

Example 5:- Calculate Capillary rise in a glass tube of 2.5 mm diameter when immersed vertically take

𝜍 = 0.0725 𝑁

𝑚 for water

Solution – h = 4𝜍

𝜌𝑑𝑔=

4 × 0.0725

1000× 9.81 × 2.5 𝜌𝑔 =

𝑁

𝑚3

= 4 × 0.0725

1000× 9.81 × 2.5

= 0.0118 m = 1.18 cm.

(12) Vapour pressure & cavitation:-

Vapour pressure – A change from liquid state to gaseous state is called vaporization. Vaporization

(depend upon prevailing pressure & temperature condition) occur because of continuous escaping of

molecules through the free surface.

Liquid will vaporized & molecules start escaping from the free surface of liquid these vapour

molecules get gathered the surface between free liquid surface & top of vessel. These vapours exert

pressure on liquid surface. This pressure is called vapour pressure.

Free surface

Heat up

Vapour pressure

Closed vessel

Vapour

If the pressure above the liquid surface is reduced, boiling temperature also reduced. If pressure is

reduced to such extent that it becomes equal to or less than vapour pressure, boiling of liquid will

start.

Cavitation: - In a flowing liquid system if the pressure at the any point becomes equal to or less than

vapour pressure, vaporization of liquid start. The bubbles of this vapour are carried out in the region

of high pressure where they collapse. This phenomenon is called cavitation.

Cavitation is the phenomena of formation of vapour bubbles of flowing liquid at pressure lower than

vapour pressure & sudden collapse of these vapour bubble in region of high pressure.

1. Vapour pressure increases with increase in temperature.

2. Vapour pressure of mercury is very low due to its low vapour pressure; it is suitable to be

used as barometric fluid.


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FM + OCF & MACHINES

1. A fluid is substance that :

(A) Is practically incompressible and inviscid

(non-viscous)

(B) Cannot be subjected to shear forced

(C) Cannot remain at rest under the action of

any shear force

(D) Always expands until it fills any container

2. When subjected to shear force, a fluid

(A) Deforms continuously only for large shear

forces

(B) Deforms continuously only for large shear

stresses

(C) Deforms continuously no matter how

small the shear stress may be

(D) Undergoes static deformation.

3. An ideal fluid

(A) as infinite viscosity

(B) Statisfies the relation pv = RT

(C) Obey‟s the Newton‟s law of viscosity

(D) Is both incompressible and non-viscous.

4. A fluid which obeys the relation 𝜇 =𝜏

(𝑑𝑢/𝑑𝑦 ) is

called the

(A) Real fluid (B) Perfect fluid

(C) Newtonian fluid (D) Pseudo plastic

5. The general relation between shear stress 𝜏 and

velocity gradient du/dy for a fluid can be

written as 𝜏 = 𝐴 𝑑𝑢

𝑑𝑦 𝑛

+B

Which of the following is then a false

statement?

(A) For ideal fluids : A= B =0

(B) For Newtonian fluids : n = 1 and B = 0

(C) For dilatant fluids: n < 1 and B = 0

(D) For ideal plastic or Bingham fluid : n = 1

and B = 𝜏0

6. If the relationship between the shear stress 𝜏

and the rate of shear strain du/dy is given by 𝜏

= K ( 𝑑𝑢/𝑑𝑦)𝑛

The fluid with the exponent n < 1 is known as

(A) Bingham fluid

(B) Dilatants fluid

(C) pseudoplastic fluid

(D) Newtonian plastic

7. If shear stress 𝜏 and shear rate du/dy

relationship of a material is plotted with 𝜏 on

the y-axis and du/dy on the x-axis, the

behavior of an ideal fluid is exhibited by

(A) A straight line passing through the origin

and inclined to x-axis

(B) The positive x- axis

(C) The positive y-axis

(D) A curved line passing through the origin .

Practice Problem Level -1


15

FM + OCF & MACHINES

8. These fluids exhibit a certain shear stress at

zero shear strain rate followed by a straight

line relationship between shear stress and shear

strain rate

(A) Newtonian fluids

(B) Ideal plastic

(C) Pseudoplastic fluids

(D) Dilatant fluids.

9. Paper pulp can be regarded as

(A) Newtonian fluid

(B) Dilatant fluid

(C) Pseudoplastic fluid

(D) Bingham plastic.

10. Typical example of a Non- Newtonian fluid of

pseudoplastic variety is

(A) Air (B) Blood

(C) Water (D) Printing ink.

11. Choose the correct statement about the

viscosity of a liquid

(A) Considerably influenced by molecular

momentum transfer

(B) Remains practically constant with

temperature rise of fall

(C) Decreases with increase in temperature

(D) Fairly large as compared to viscosity for

gases.

12. Newton‟s law of viscosity relates

(A) Stress and strain in a fluid

(B) Pressure, velocity and viscosity of a gas

(C) Shear stress and rate of angular

deformation in a fluid

(D) Yield stress, viscosity and rate of angular

deformation.

13. The coefficient of viscosity is a property of

(A) The fluid

(B) The boundary condition

(C) The body over which flow occurs

(D) The flow velocity

14. The dimensions of a dynamic viscosity are

(A) MLT (B) ML–1

T–2

(C) ML–1

T–1

(D) ML–2

T–2

15. Poise is the unity of :

(A) Density

(B) Velocity gradient

(C) Kinematic viscosity

(D) Dynamic viscosity

16. Correct units for kinematic viscosity are:

(A) m2/s (B) Ns/m

2

(C) m/kg s (D) kg/m2s

17. The viscosity of water with respect to air is

about

(A) 50 (B) 55

(C) 60 (D) 65 times

18. SI unit of viscosity is

(A) Equal to poise

(B) 9.81 times of poise

(C) 10 times of poise

(D) 98.1 times of poise.

19. The multiplying factor for converting one

stoke into m2/s is

(A) 102

(B) 104

(C) 10-2

(D) 10-4


16

FM + OCF & MACHINES

20. At a certain point in castor oil, the shear stress

is 0.022 kgf/m2 and the velocity gradient 0.22

per second. The dynamic viscosity of castor oil

in poises is numerically equal to

(A) 0.1 g (B) 1 g

(C) 10 g (D) 100g

Where g equals 9.81 m/s2.

21. For a liquid having specific gravity 0.93 and

dynamic viscosity 0.012 poise, the kinematic

viscosity in centistokes will be about

(A) 1.29 (B) 0.129

(C) 12.9 (D) 0.0129

22. The density of a fluid is sensitive to changes

temperature and pressure. The fluid will be

known as

(A) Newtonian fluid

(B) Perfect fluid

(C) Real fluid

(D) Compressible fluid.

23. Choose the correct statement in the context of

bulk modulus and coefficient of

compressibility

(A) The bulk modulus of elasticity of a fluid

is the same as its coefficient of

compressibility

(B) Most of the liquids have a low value of

bulk modulus

(C) The bulk modulus is not influenced by

changes in pressure and temperature

(D) The relative change of volume per unit

pressure is called the coefficient of

compressibility.

24. The bulk modulus of elasticity of a fluid is

defined as

(A) d /

dp

(B)

dp

d /

(C) dp

dp (D)

dp / dp

25. Which one of the following is an example of

non – Newtonian fluid

(A) Blood (B) Air

(C) Water (D) Kerosene oil

26. A unit cubic metre of water is subjected to a

pressure of 10 bar. Then a change in the

volume of water amounts to

(A) 31m

20 (B) 31

m200

(C) 31m

2000 (D) None of these

For water, bulk modulus k= 2× 109Pa

27. Measure of the effect of compressibility in

fluid flow is the magnitude of a dimeasionless

parameter know as

(A) Mach number (B) Newton‟s number

(C) Weber number (D) Euler number.

28. The bulk modulus of water with respect to air

is about

(A) 500 (B) 1000

(C) 10,000 (D) 20,000 times.

29. Select the correct statement :

(A) Higher is the bulk modulus. lower is

compressibility

(B) For an adiabatic process the bulk modulus

equals the pressure


17

FM + OCF & MACHINES

(C) The bulk modulus, is independent of both

pressure and temperature

(D) The bulk modulus of a liquid is less than

that of a solid.

30. All liquid surfaces tend to stretch. This

phenomenon is called

(A) Cohension (B) Adhesion

(C) Surface tension (D) Cavitation.

1. C

2. C

3. D

4. C

5. C

6. C

7. B

8. B

9. C

10. B

11. C

12. C

13. A

14. C

15. D

16. A

17. B

18. C

19. D

20. B

21. A

22. D

23. D

24. B

25. A

26. C

27. A

28. D

29. A

30. C

[Sol] 5. (C) 𝜏 = 𝐵 + 𝐴 𝑑𝑢

𝑑𝑦 𝑛

For ideal fluid ;

𝜏 = 0 𝑆𝑜 𝑖𝑓 𝐴 = 𝐵 = 0

𝑛 = 1; 𝐵 = 0; 𝜏 =

𝐴 𝑑𝑢

𝑑𝑦 𝑛𝑒𝑤𝑡𝑒𝑛𝑖𝑎𝑛 𝐹𝑙𝑢𝑖𝑑

𝑛 < 1; 𝐵 = 0; 𝜏 =

𝐴 𝑑𝑢

𝑑𝑦 𝑛

𝑃𝑠𝑢𝑒𝑑𝑜𝑝𝑙𝑎𝑠𝑡𝑖𝑐

𝐵𝑖𝑛𝑔𝑕𝑎𝑚 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 ; 𝑛 = 1; 𝐵 = 𝜏0

𝜏 = 𝜏0 + 𝐴 𝑑𝑢

𝑑𝑦

[Sol] 7. (B)

Explanations

Answer key

𝜏 ↑


18

FM + OCF & MACHINES

[Sol] 14. (C)

Dynamic Viscosity (𝜇)

𝜏 = 𝜇𝑑𝑢

𝑑𝑦

𝑁

𝑚2 = 𝜇 𝑚

𝑠𝑒𝑐×𝑚

𝜇 =𝐾𝑔 .𝑚

𝑆𝑒𝑐2 ×𝑠𝑒𝑐

𝑚2

𝜇 =𝐾𝑔

𝑆𝑒𝑐 .𝑚

𝜇 = 𝑀𝐿−1𝑇−1

[Sol] 21. (A)

Specific Gravity = 0.93

So 𝜌 = 0.93 × 1000 = 030 𝐾𝑔/𝑚3

Dynamic Viscosity 𝜇 = 0.012 𝑃𝑜𝑖𝑠𝑒

1 𝑃𝑜𝑖𝑠𝑒 = 0.1𝑁𝑆

𝑚2 = 0.0012 𝑁𝑆/𝑀2

Kinematic Viscosity 𝜗 =𝜇

𝜌

𝜗 =0.0012

930= 1.29 × 10−6𝑚2/𝑆𝑒𝑐

1 𝑐𝑒𝑛𝑡𝑖𝑠𝑡𝑜𝑘𝑒 = 10−6 𝑚2/𝑆𝑒𝑐.

𝜗 = 1.29 𝑐𝑒𝑛𝑡𝑖𝑠𝑡𝑜𝑘𝑒]

[Sol] 26. (C)

Bulk modulus (k) = 𝑑𝑝𝑑𝑣

𝑣

= 𝑣𝑑𝑝

𝑑𝑣

𝑘 = 2 × 109 𝑃𝑎

𝑑𝑝 = 10 × 105𝑃𝑎

𝑉𝑜𝑙. = 1𝑚3

2 × 109 = 1 10×105

𝑑𝑣

[𝑑𝜗 =1

2000𝑚3]


Pressure :-

In case of fluid at rest, there is no relative motion between the layers of fluids.

So velocity gradient is zero 𝑑𝑢

𝑑𝑦= 0 . Hence, shear stress (τ = 0) is zero. Consequently, there is no

tangential or shear force. Hence for static fluid, the force exerted is normal to the surface of containing

vessel. This normal surface force is called pressure force.

P= 𝑁𝑜𝑟𝑚𝑎𝑙 𝑓𝑜𝑟𝑐𝑒

𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎=

𝐹

𝐴

Unit:- N/m2

Dimensional Formula

p = 𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

𝐴𝑟𝑒𝑎 =

𝑀1𝐿1𝑇−2

𝐿2 = 𝑀1𝐿−1𝑇−2 → 𝑀𝐿𝑇 𝑠𝑦𝑠𝑡𝑒𝑚

Pascal’s law:-

Its states that the pressure or the intensity of pressure at the point in a static fluid is equal in all direction.

Mathematically, 𝑝𝑥 = 𝑝𝑦 = 𝑝𝑧

py

pz

px

Hydrostatic law:-

It states that the rate of increase of pressure in a vertically downward direction must be equal to the

specific weight of fluid at that point.

A B

CD

W

Z

Free Space

𝜕𝑝

𝜕𝑧= 𝑤 ∴ 𝑤 = 𝑔

On integrating, we get

Chapter

2 PRESSURE & ITS

MEASUREMENTS Syllabus: Pascal‟s Law, Hydrostatic law, Atmospheric Gauge and

vacuum pressure, Manometer methods. Weightage :5%


20

FM + OCF & MACHINES

p = 𝜌𝑔𝑧

Also z = 𝑝

𝜌𝑔

Where z = pressure head

Example 1:-

Calculate pressure due to column of 0.3 m of water

Z = 0.3m

Solution – p = 𝜌𝑔𝑧 = 1000 0.3 9.81 = 2943 𝑁

𝑚2

Example 2:-

Calculate pressure due to column of 0.3 m of mercury whose specific gravity is 13.6.

Solution – s = 13.6

s = 𝜌𝑚𝑒𝑟𝑐𝑢𝑟𝑦

𝜌𝑤𝑎𝑡𝑒𝑟

s = 1000 13.6 = 𝜌𝑚𝑒𝑟𝑐𝑢𝑟𝑦

𝜌𝑚𝑒𝑟𝑐𝑢𝑟𝑦 = 13600

p = 13600 9.81 0.3 = 40024.8 𝑁

𝑚2

Example 3:- Pressure intensity at a point is 3.924𝑁

𝑚2. Find corresponding height of fluid of water.

Solution – p = 𝜌g𝑧

𝑧 =𝑝

𝜌𝑔 =

3.924 𝑥 104

1000 ×9.81 𝑁

𝑚 3

𝑧 =4 m of water column

p = 3.92 N/m2


21

FM + OCF & MACHINES

Example 4:- An open tank contain water upto a depth a 2 m & oil with specific 0.9 for a depth of 1 m.

Calculate pressure intensity

(A) At bottom of the tank

(B) At interface of liquid surface.

Solution- p = 𝜌𝑔𝑧

0.9 = 𝜌𝑜𝑖𝑙

𝜌𝑤𝑎𝑡𝑒𝑟

𝜌𝑜𝑖𝑙 = 0.9 1000 = 900

Water

Oil

Free surface pressure = 0

1m

2m

3m

(A) p = 𝜌𝑔𝑧𝑜𝑖𝑙 + 𝜌𝑔𝑧𝑤𝑎𝑡𝑒𝑟

p = 900 𝑥 9.81 𝑥 1 + 1000 𝑥 9.81 𝑥 2

p = 28449 𝑁

𝑚2

(B) p = 𝜌𝑔𝑧𝑜𝑖𝑙

p = 900 𝑥 9.81 𝑥 1

p = 8829 𝑁

𝑚2

ABSOLUTE, ATMOSPHERIC, GAUGE & VACUUM PRESSURE

Atmospheric Pressure – Pressure exerted by the envelop of air surrounding the earth surface.

Atmospheric pressure is determined by mercury column barometer.

Atmospheric pressure varies with altitude because air near the earth surface is compressed by air above.

At sea level, Value of atmospheric pressure = 1 atm.

1 atm = 1.01325 bar = 760 mm of Hg column = 10.33 m of water column.

Absolute pressure – Pressure intensity measured from the state of zero pressure is called absolute

pressure.

Gauge pressure – Pressure which is measured more than 1 atm. Pressure is called gauge pressure.

Vacuum pressure – Pressure reading below the atmospheric pressure is vacuum pressure.


22

FM + OCF & MACHINES

1. Vacuum pressure is also known as negative gauge pressure or Rarefaction pressure or negative

pressure.

2. pabs. = patm + pgauge

3. pabs = patm - pvacuum

+Vegauge

Pgauge

PAbsolute

Patm

–Vegauge

Absolute Zero Presure

METHODS TO DETERMINE PRESSURE

1. MANOMETER METHOD

Simple manometer

PiezometerManometer

u-tubeManometer

SingleColumn

Vertical singleColumn Manometer

Inclined singleColumn Manometer

MANOMETER METHOD

U-tubedifferentialManometer

Inverted u-tubedifferentialManometer

Differential Manometer

2. Mechanical Gauge Method –

Manometer – Devices which are used for measuring the pressure at a point in a fluid by balancing

the column of fluid by same or another column of fluid.

1) Simple manometer – It is a glass tube having one end connecting to point by other end open to

atmosphere.

i. Piezometer –

𝑝 = 𝜌𝑔𝑕

𝑝 = 𝑤𝑕

h

It is a simplest form of manometer which is used to calculate gauge pressure.


23

FM + OCF & MACHINES

2) Tube manometer

FOR GAUGE PRESSURE FOR VACCUM PRESSURE

1.

A A

h1

h2

B

Atm.

𝑝 = 𝜌2𝑔𝑕2 − 𝜌1𝑔𝑕1

1.

A A

h1h2

B

𝑝 = − 𝜌2𝑔𝑕2 − 𝜌1𝑔𝑕1

2. pB > patm

2 = density of liquid U tube (mercury)

1 = density of fluid in pipe.

h2 = height of liquid above datum line which is

exposed to atmosphere.

h1 = height of liquid above the datum line.

2. 𝑝𝐵 > 𝑝𝑎𝑡𝑚

Example 5:- The right limb of u-tube manometer containing mercury is open to the atmosphere which

flow of fluid with specific gravity = (0.85). The centre of the pipe is 15cm below the level of

mercury in right limb. If the difference of mercury level in two limb is 25m calculate

pressure in pipe.

Solution – Gauge pressure exist because 𝑝𝐵 > 𝑝𝑎𝑡𝑚

𝑝 = 𝜌2𝑔𝑕2 − 𝜌1𝑔𝑕1

𝑝 = 0.85 ×25

100× 1000 − 1000 × 136 ×

10

100× 9.81𝐴

A A

h1

h2

pB

ch =

10

cm1

25cm = b2

15cm

Specific gravity = 𝜌 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝜌 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 1000 0.85 = 𝜌 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑= 850 = 𝜌 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑

𝑝 = 13600 × 9.81 ×15

100− 850 × 9.81 ×

10

100

𝑝 = 19178.55 𝑁/𝑚2


24

FM + OCF & MACHINES

Single Column Manometer – Modified form of U-tube manometer.

Vertical Inclined

x hx

h1 h2

Reservoir(A)

𝑝 =𝑎𝑕2

𝐴 𝜌2𝑔 − 𝜌1𝑔 + 𝜌2𝑔𝑕2 − 𝜌1𝑔𝑕1

𝑎 = area of cross – section of right limb

A = Area of Reservoir

x

y

xy

h2

L

𝑝 = 𝜌2𝑔𝑕2 − 𝜌1𝑔𝑕1

Where 𝑕2 = 𝐿 sin 𝜃

Sensitivity of inclined manometer improves by

1

𝑆𝑖𝑛𝜃

2. DIFFERENTIAL MANOMETER – These are used to measure the differential of pressure

between two points in a pipe in two different pipes.

U-tube differential Inverted U-tube

g

y

h

x

x x

B

𝑝𝐴 − 𝑝𝐵 = 𝑕𝑔 𝜌𝑔 − 𝜌1 + 𝜌2𝑔 𝑔 − 𝜌1𝑔 𝑥

When two pipes are in different level.

When at same level –

𝑥 = 𝑦

𝑝𝐴 − 𝑝𝐵 = 𝑕𝑔 𝜌𝑔 − 𝜌1 + 𝜌2 − 𝜌1 𝑔𝑥

A

2

y

h

g

𝑝𝐴 − 𝑝𝐵 = 𝜌1 × 𝑔 × 𝑕1 − 𝜌2 × 𝑔 × 𝑕2 − 𝜌𝑠

× 𝑔 × 𝑕1


25

FM + OCF & MACHINES

1. The pressure intensity at a point in a fluid is

the same in all directions only when

(A) The fluid is frictionless

(B) The fluid is frictionless and

incompressible

(C) The fluid has zero viscosity and is at rest

(D) There is no motion of one fluid layer

relative to an adjacent layer.

2. One atmospheric pressure equals

(A) 1.0132 kgf/cm2

(B) 760 mm of mercury

(C) 101.35 KN/m2

(D) Any of the above.

3. An open tank contains I m deep water with 50

cm depth of oil of specific gravity 0.8 above it.

The intensity of pressure at the bottom of tank

will be (g=10m/sec2)

(A) 4 kN/m2

(B) 10 kN/m2

(C) 12 kN/m2

(D) 14 kN/m2

4. Gauge pressure is

(A) absolute pressure – atmospheric pressure

(B) absolute pressure + atmospheric pressure

(C) atmospheric pressure – absolute pressure

(D) none of these

5. Atmospheric pressure varies with

(A) altitude (B) temperature

(C) weather conditions (D) all of these

6. Atmospheric pressure is equal to water

column head of

(A) 9.81 m (B) 5.0 m

(C) 10.30 m (D) 7.5 m

7. The pressure less than atmospheric, pressure,

is known

(A) suction pressure

(B) vacuum pressure

(C) negative gauge pressure

(D) all the above

8. Pick out the incorrect statement(s).

An inverted differential manometers is most

suitable if

1. The pressure difference is small

2. the pressure difference is large

3. the fluid in the pipe is a gas

On the above statements

(A) 1 alone is correct

(B) 2 alone is correct

(C) 1 and 3 are correct

(D) 2 and 3 are correct

9. Gauge pressure is

(A) pressure measured above complete

vacuum

(B) absolute pressure + local atmosphere

pressure

(C) Local atm pressure

(D) Absolute pressure



26

FM + OCF & MACHINES

10. Piezometers are used to measure

(A) Pressure in water channels, pipes etc.

(B) Difference in pressure at two points

(C) Atmospheric pressure

(D) Very low pressure.

11. Manometers are used to measure


(B) Difference in pressure at two points



12. Differential manometers are used to measure


(B) Different in pressure at two points



13. One kilo Pascal is equivalent to : -

(A) 1N/mm2 (B) 1000N/m

2

(C) 1000 N/mm2 (D) 1000N/cm

2

14. U- tube manometer measures

(A) Local atmospheric pressure

(B) Difference in pressure between two point

(C) Difference in total energy between two

points

(D) Absolute pressure at a point

15. Bourdon gauge measures :

(A) Absolute pressure

(B) Gauge pressure

(C) Local atmospheric pressure

(D) None of these

16. The avg. value of atmospheric pressure at sea

level is:

(A) 76mm of Hg (B) 7.6mm of Hg

(C) 760 mm of Hg (D) 0.76 mm of Hg

17. Bourden‟s pressure gauge measures the

(A) Liquid pressure (B) Gas pressure

(C) Velocity pressure (D) Height pressure

18. A piezo meter cannot be used for pressure

measurement in pipes when

(A) The velocity is high

(B) The fluid in pipes is a gas

(C) The pressure is less

(D) None of the above

19. The continuity equation:

(A) Expresses the relation between energy

and work

(B) Required Newton‟s law to be satisfied at

every point

(C) Relates the mass flow rate


20. 1 pascal pressure is equal to

(A) 1 m water pressure

(B) 10.01 m water pressure

(C) 0.0001 m water pressure

(D) none of the above

21. Mercury is suitable for manometer because of

it

(A) Has high density

(B) It can easily be seen in tube

(C) Does not stick to tube walls

(D) It is generally not used in manometers.


27

FM + OCF & MACHINES

1. C

2. D

3. D

4. A

5. D

6. C

7. D

8. B

9. C

10. D

11. A

12. B

13. B

14. B

15. B

16. C

17. A

18. B

19. C

20. C

21. A

[Sol] 2. (D) 1 atm pressure = 101.35 KPa

760 mm Hg; P = 𝜌𝑔𝑕

P = (13600)(9.81)(.76)

[P=101.3 KPa]

10,3m of H2O : P = 𝜌𝑔𝑕

P = (1000× 9.81)(10.3)

[P=101.2 KPa]

Hence for 1 atm pressure

760 mm of Hg = 101.35 KPa

= 10.3m of water column

[Sol] 3. (D)

𝑃𝑏𝑜𝑡𝑡𝑜𝑚 = 𝜌1𝑔𝑕1 + 𝜌2𝑔𝑕2

𝑃𝑏𝑜𝑡𝑡𝑜𝑚 = 1000 × 9.81 × 1 +

800 × 9.81 × 0.5

𝑃𝑏𝑜𝑡𝑡𝑜𝑚 = 14 𝐾𝑃𝑎

[Sol] 20. (C)

Continuity 𝑒𝑞𝑛 − 𝜌𝐴𝑉 = 𝐶

𝜌1𝐴1𝑉1 = 𝜌2𝐴2𝑉2

Mass flow rate

[Sol] 21. (C)

1 Pascal ⇒ P = 𝜌𝑔𝑕

𝑕 =1

100×10⇒ 0.0001𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

Explanations

Answer key


HYDROSTATIC FORCES ON SUBMERGED SURFACES

Total pressure (F) = Forces exerted by static fluid either on plane or curved when the fluid in contact

with the surface. Total pressure force always act normal to the surface.

Centre of Pressure (h):- The point of application of total pressure force on surface.

Case 1:- Horizontal plane surface submerged in liquid.

Total pressure force =𝜌𝑔𝐴𝑕 or 𝑤𝐴𝑕

Where 𝑤 = specific Weight

A = area of cross-section of plane

𝑕 = distance of centre of gravity of area from the free surface of liquid.

Centre of pressure (h) = 𝑕

Free surface

h

Case 2:- Vertical plane surface submerged in liquid-

F

h*h

Total pressure force (F) = 𝜌𝑔𝐴𝑕 or 𝑤𝐴𝑕

Centre of pressure (h) = 𝑕 + 𝐼𝐺

𝐴𝑕

Where 𝐼𝐺 = M.O.I. about centre of gravity

Chapter

3 HYDROSTATIC FORCES

ON SURFACES

Syllabus: Hydrostatic Forces on submerged surfaces on horizontal

plane surface, inclined surfaces and Curved surfaces.

Weightage: 10%


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FM + OCF & MACHINES

From above equation h it is clear that centre of pressure lie below the C.G. of the vertical surface.

1. Deeper the surface is lowered into the liquid i.e. greater is the value of 𝑕 , centre of pressure comes

closer to the centroid of an area.

2. Depth of centre of pressure is independent of specific weight of liquid & which is same for all fluid.

3. If a vertical plane surface has width (B) & depth (D) coincide with free surface, then centre of

pressure (h*) is

Free surface

d

h = d2

b

h* = 𝑕 + 𝐼𝐺

𝐴𝑕

h* = 𝑕 + 𝑏𝑑2

12 ×𝑏𝑑 × 𝑕

h* = 𝑕 + 𝑑2

12 × 𝑕 𝑕 =

𝑑

2 𝐴𝑠 𝑖𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑕 =

𝑑

2

= 𝑑

2 +

𝑑 ×2

12 ×𝑑 =

𝑑

2 +

𝑑

6

h* = 3𝑑+𝑑

6 =

2𝑑

3 =

2

3 d

h* = 2

3 d

It means centre of pressure in case when free surface coincide with wide of a vertical plane which is at a

depth of 2

3 of depth plane section.

Case 3:- Inclined plane surface submerged in liquid:-

h* h–

FCG

Free surface

F = 𝜌𝑔𝐴𝑕

Centre of pressure h* = 𝑕 + 𝐼𝐺

𝐴𝑕 . 𝑆𝑖𝑛2𝜃

Case 4:- Curved surface submerged in liquid

Fx = Total pressure force on the projected area of the curved surface on vertical plane.


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FM + OCF & MACHINES

Fy = Weight of liquid supported by curved surface upto free surface of liquid.

FR = Resultant force

Fx

FRFy

Curved surface

FR = 𝐹𝑥2 + 𝐹𝑦

2

Angle made by resultant with horizontal = tan 𝜃 = 𝐹𝑦

𝐹𝑥

Example 1:- Rectangular Plane surface to 2 m wide & 3 m deep it lie in vertical plane in water

Determined total pressure (P) & Centre of pressure on the plane surface when upper edge coincide with

water surface.

x

h* h–

3m

2m

Solution – F = 𝜌𝑔𝐴𝑕

𝑕 = 32 = 1.5𝑚

F = 1000 9.81 3 2 1.5 𝑁

𝑚3 𝑚3

= 88920 N

h* = 2

3 3 = 2𝑚

Example 2:- Rectangular Plane surface to 2 m wide & 3 m deep it lie in vertical plane in water.

Determined total pressure (F) & Centre of pressure on the surface when upper edge is 2.5m below the free

surface.

h–

3m

2.5m

2m

Solution – F = 𝜌𝑔𝐴𝑕


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FM + OCF & MACHINES

F = 1000 9.81 3 2 (2.5 + 1.5)

= 235440 N

h* = 𝑕 +𝐼𝐺

𝐴𝑕 = 4 +

2 × 3

12 × 6 × 4 = 4 +

2 × 27

72 × 4

= 4.1875 m 𝐼𝐺 =𝑏𝑑 3

12

Example 3:- Determined total pressure & Centre of pressure of a circular plate where diameter = 1.5 m

which is placed vertically in water. In which a way that centre of plate is 3m below the full surface

0.75 1.5m

3m

Solution – F = 1000 9.81 𝜋

4 1.5 2 (1.5 +

1.5

2)

= 39005.4 N

h* = 3 + 𝜋

64 ×

4 × 1.54

𝜋 1.75 2 × 3 (M.O.I =

𝜋

64 𝐷4)

= 3.06249 m

Example 4:- Determined total pressure & Centre of pressure on an isosceles triangular plate of base 4 m

& altitude h = 4m when immersed vertically in an oil of specific gravity = 0.9 & base of plate coincide

with free surface of oil.

4mh–

4m

(MOI = )bd

3

36

Solution – 𝜌 = 0.9 × 1000 = 900 𝑘𝑔

𝑚3

F = 900 9.81 1

2 4 4

4

3= 94176 N

h* = 4

3+

4 × 43 × 2 × 3

36 × 4 × 4 × 4

= 1.33 + 0.66

h* = 1.993 m


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Example 5:- Rectangular Plane surface 3m wide & 4m deep submerged in water at angle 30o with free

surface of water. Determine total pressure & Centre of pressure when upper edge is 2m below the free

surface.

Solution –F = 1000 9.81 𝑕 = 𝐴𝐵 = 𝐴𝐶 + 𝐶𝐵 AC = 2m 𝐶𝐵

2𝑚= 𝑠𝑖𝑛30°

G

4m

3m

30°

A

C2m

h*30°

B

CB = 2 1

2

CB = 1m

𝑕 = 2 + 1 = 3m

F = 9.81 1000 4 3 3

= 353160 N

h* = 3 + 3 ×4

12 × 𝑠𝑖𝑛 2 30°

× 4 ×3 ×3

= 3 + 16 ×

12 ×3 1

4 = 3 +

1

9

= 23 + 1

9=

28

9 = 3.11 m.

Example 6:- Compute horizontal & vertical component of the total force acting on a curved surface AB

which is in the form of a quadrant of a circle of Radius 2, as shown in fig. Take the width of

the gate unity.

Solution – Width of gate = 1m

Radius = 2m

OA = OB Distance = 2m

D

1.5

2m

A O h*

B

Fx

C


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FM + OCF & MACHINES

Fx = 𝜌𝑔𝐴𝑕

= 1000 9.81 (2 1) (1.5 + 2

5)

= 49050 N

h* = 𝐼𝐺

𝐴𝑕 + 𝑕

𝐼𝐺 = 𝑏𝑑 3

12 =

1 × 23

12 =

2

3 m

4

h* = 2/3

2 ×1 2.5 + 2.5 = 2.633 from free surface

Fy = weight of DAOC + weight of AOB

= 𝜌𝑔 vol. of DAOC + 𝜌𝑔 vol. of AOB

= 1000 9.81 (AD AO 1 + 𝜋

4 𝐴𝑂 2 1)

= 60249.1 N


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FM + OCF & MACHINES

1. For a fluid a rest

(A) The shear stress depends upon the

coefficient of viscosity

(B) The shear stress is zero

(C) The shear stress is zero only on horizontal

planes

(D) The shear stress is maximum on a plane

inclined at 45- degree to the horizontal.

2. All fluids exert

(A) Pressure in the direction of flow only

(B) Pressure in the direction of force of

gravity

(C) Equal pressure in all directions

(D) Equal pressure in x, y and Z- plane.

3. For warships, metacentric height of a ship

should vary between

(A) 0 – 1 m (B) 1 – 2m

(C) 5 – 10m (D) more than 10m

4. The time period of oscillation of a floating

body

(A) Is a function of buoyant force

(B) Is independent of radius of gyration of the

body about its centre of gravity

(C) Decreases with increase in metacentric

height

(D) Is not affected by acceleration due to

gravity.

5. The position of Centre of pressure on a plane

surface immersed vertically in a static mass of

fluid is

(A) At the centroid of the submerged area

(B) Always above the centroid of the area

(C) Always below the centroid of the area


6. A vertical triangular area with vertex

downward and altitude „h‟ has its base lying on

the free surface of a liquid. The Centre of

pressure below the free surface is at a distance

of

(A) h

4 (B)

h

3

(C) h

2 (D)

2h

3

7. The total pressure on a plane surface inclined

at an angle θ with the horizontal is equal to

(A) pA (B) pA sin

(C) pA cos (D) pA tan

Where p is pressure intensity at centroid of

area and A is area of plane surface.

8. A vertical rectangular plane surface is

submerged in water such that its top and

bottom surfaces are 1.5 m and 6.0 m

respectively below the free surface. The



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FM + OCF & MACHINES

position of centre of pressure below the free

surface will be a distance of

(A) 3.75 m (B) 4.0 m

(C) 4.2 m (D) 4.5 m

9. The horizontal component of force on a curved

surface is equal to the

(A) Product of pressure intensity at its

centroid and area

(B) Force on a vertical projection of the

curved surface

(C) Weight of liquid vertically above the

curved surface

(D) Force on the horizontal projection of the

curved surface

10. On an inclined plane, Centre of pressure is

located

(A) at the centroid

(B) above the centroid

(C) below the centroid

(D) anywhere

11. The hydrostatic force acts through

(A) centre of pressure

(B) centre of top edge

(C) centre of bottom edge

(D) metacentre

12. If H is depth of rectangle & B is width then,

center of pressure below the free surface of

water is?

(A) H

2 (B)

H

3

(C) 2H

3 (D)

H

5

13. The centre of pressure of vertical plane

immersed in a liquid is at

(A) centre of higher edge

(B) centre of lower edge

(C) centroid of the area

(D) none of these

1. B

2. C

3. B

4. C

5. C

6. C

7. A

8. C

9. B

10. C

11. A

12. C

13. D

Answer key


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FM + OCF & MACHINES

[Sol] 4. (C)

Time of oscillation = 2𝜋𝐾 1

𝐺𝑀 .𝑔

𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑤𝑖𝑡𝑕 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑚𝑒𝑡𝑒𝑟 𝑐𝑒𝑛𝑡𝑟𝑖𝑐 𝑕𝑒𝑖𝑔𝑕𝑡

[Sol] 6. (C)

h/3

𝐼𝑎 =𝑏𝑕3

36 𝑕 =

𝑕

3

𝑎 =1

2𝑏𝑕

𝑕∗ = 𝑕 +𝐼𝑎

𝐴𝑕

𝑕∗ =𝑕

3+

𝑏𝑕3

36

1

2𝑏𝑕

𝑕

3

𝑕∗ =𝑕

3+

𝑕

6

𝑕∗ =𝑕

2

[Sol] 8. (C)

𝑕 = 1.5 +4.5

2

𝑕 = 1.5 + 2.25

𝑕 = 3.75𝑚

𝑕∗ = 𝑕 +𝐼𝑎

𝐴𝑕

𝐼𝑎 =𝑏𝑕3

12; 𝐴 =

1

2𝑏𝑕

𝑕∗ = 𝑕 +𝑏𝑕3

12× 𝑏𝑕 𝑕

𝑕∗ = 𝑕 +𝑕2

12𝑕

𝑕∗ = 3.75 + 4.5 2

12 3.75

𝑕∗ = 3.75 + 4.5 2

12×3.75

[𝑕∗ = 4.2𝑚]

[Sol] 12. (C)

𝑕 =𝐻

2

𝑕∗ = 𝑕 +𝐼𝑎

𝐴𝑕

𝑕∗ =𝐻

2+

𝐵𝐻3

12

𝐵𝐻×𝐻

2

𝑕∗ =𝐻

2+

𝐻

6

𝑕∗ =2𝐻

3

Explanations


BUOYANCY & FLOATATION

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

Archimede’s Principle:- According to this principle, resultant upward thrust on the body is equal to

weight of fluid displaced by the submerged body.

C.G

W

Fb

Free Surface

Centre of buoyancy (B):- It is the point through which the force due to buoyancy is supposed to act. The

centre of buoyancy will be the centre of gravity of the fluid displaced & line of action of buoyant force Fb

is vertical upward & passes through the centre of gravity of the displaced fluid i.e. the centroid of the

displaced fluid.

Three possibilities between weight of body (W) & buoyant force 𝐹𝑏 .

1. If 𝑊 > 𝐹𝑏 , body tends to more downward & sink.

2. If 𝑊 = 𝐹𝑏 , body float & is only partially submerged.

3. If 𝑊 < 𝐹𝑏 , body is lifted upward & rise to the surface.

Example 1:- A stone weight 400 kN in air & when immersed in water it weight is 225 kN.

Calculate Volume of the stone.

Solution – Buoyant force = 400 – 225 = 175 kN

Buoyant force = specific weight of water x volume of water displaced

175 = 9810 V 𝑁

𝑚3 𝑃

103

V = 0.01783 m3

Example 2:- A solid metallic piece whose relative specific gravity is 7.2 float above the surface of

mercury whose specific gravity is 13.6 contained in a tank. What fraction of volume of metallic solid lie

above the mercury surface.

Solution – Smetallic = 7.2

Chapter

4 BUOYANCY &

FLOATATION Syllabus: Archimedes Principle, Center of buoyancy, Metacentric

height, Stability criteria for submerged and floating bodies.

Weightage:10%


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FM + OCF & MACHINES

𝜌𝑚𝑒𝑡𝑎𝑙𝑙𝑖𝑐 = 7200

V = total volume of solid piece

V = volume of solid piece immersed in mercury

According to Archimedes principle,

weight of solid piece = weight of mercury displaced by immersed portion of solid piece.

7.2 9810 V = 13.6 9810 V

V

𝑉=

7.2

13.6

V

𝑉 = 0.529

Fraction of volume of solid piece below mercury= 1 – V

= 1 – 0.529

= 0.471

= 47.10%

Metacentre (M):- The point about which body start oscillating when the body is tilted by small angle.

CG

B

(Metacentre M)

B1

On the other hand the point at which the line of action of force of buoyancy will meet the normal axis of

the body when a body is given by some angular displacement.

GM metacentric height:- GM = distance Between Centre of gravity of body & metacentric of a

floating body is called metacentric height.

Determination of metacentric height:- Two methods –

1. Analytical method

GM = BM ± BG

GM = 𝐼

𝑉 ± BG

When G lie above B, GM = BM – BG.

When G lie below B, GM = BM + BG.

Where I = M.O.I. of floating object or body object its longitudinal axis.


40

FM + OCF & MACHINES

2. Experimental Method:-

G

B

G

B

G

B B

θ

xM

GM = 𝑤 × 𝑥

𝑊𝑥 𝑡𝑎𝑛𝜃

𝑤 = weight added W = weight of ship + 𝑤

𝑥 = distance between normal axis & point where 𝜔 is placed.

𝜃 = angle of heel

1. With increase in metacentric height, stability of floating body increases but comfort level of

passenger decreases.

STABILITY CONDITION:

Case 1:- For submerged body –

1. Stable equilibrium – if W = Fb & B is above G.

2. Neutral equilibrium – if W = Fb & B coincide with G.

3. Unstable equilibrium – if W = Fb & B below G.

Stable equilibrium

B

G

G & B

Neutral equilibrium Unstable equilibrium

B

G

Case 2:- For floating bodies –

1. Stable equilibrium. – if M above G. 3. Neutral equilibrium – if M coincide G.

2. Unstable equilibrium – if M below G.

Stable equilibrium

M

G

M & G

Neutral equilibrium Unstable equilibrium

M

G


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FM + OCF & MACHINES

1. Which is the necessary condition for a body to

float in a liquid?

(A) Density of the material of the body is less

than that of the liquid

(B) Weight of the liquid displaced by the

body is greater than. The weight of the

body

(C) The liquid is at rest

(D) The liquid has a high surface tension

force.

2. The centre of buoyancy is

(A) Centre of gravity of the body

(B) Centre of the displaced fluid volume

(C) Point of intersection of the body

(D) Point of intersection of the buoyant force

and the gravitational force.

3. The centre of buoyancy of a submerged body

(A) Coincides with the centre of gravity of the

body

(B) Coincides with the centroid of the

displaced volume of the fluid

(C) Is always below the centre of gravity of

the body

(D) Is always above the centroid of the

displaced volume of liquid.

4. For a floating body, the line of action of

buoyant force acts through

(A) Centroid of liquid displaced by the body

(B) Metacentre of the body

(C) Centre of gravity of the body

(D) Centre of gravity of submerged part of the

body.

5. The metacentre is

(A) Centroid of the displaced fluid volume

(B) Mid point between centre of gravity and

centre of buoyancy

(C) The point of intersection of the line of

action of buoyant force and the centre line

of the body

(D) The point of intersection of the line of

action of buoyant force and that of

gravitational force.

6. A floating body is in stable equilibrium when

(A) Its centre of gravity is below the centre of

buoyancy

(B) Its metacentric height is zero

(C) Its metacentric height is positive

(D) Its metacentric height is negative.

7. It G is the centre of gravity, B is centre of

buoyancy and M is metacentre of a floating

body, then for the body to be in unstable

equilibrium

(A) MG = 0 (B) BG = 0

(C) M is below G (D) M is above G



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FM + OCF & MACHINES

8. For stable equilibrium of a floating body

(A) 1

BG MGv

(B) 1

MG BGv

(C) 1/ V

MGBG

(D) 1

BG MGV

Where I is moment of inertial of water line

area about the longitudinal axis, V is volume

of displaced liquid, G is centre of gravity, B is

centre of buoyancy and M is metacentre

9. A pontoon has displacement of 2000 metric

tons whilst floating in sea water. When a load

of 25 metric tons is moved through a distance

of 8 m across the deck, the pontoon heels over

1/20. The metacentric height of pontoon is

(A) 1.2 m (B) 2 m

(C) 1.5 m (D) 2.5 m

10. If the weight of a body immersed in a fluid

exceeds the buioyant force, then the body will

(A) Rise until its weight equals the buoyant

force

(B) Tend to move downward and it may

finally sink

(C) Float


11. Metacentric height for small values of angle of

heel is the distance between the

(A) Centre of gravity and centre of buoyancy

(B) Centre of gravity and metacentre.

(C) Centre of buoyancy and metacentre

(D) Free surface and centre of buoyancy

12. A floating body is said to be in a state of stable

equilibrium

(A) When its metacentric height is zero

(B) When the metacentre is above the centre

of gravity

(C) When the metacentre is below the centre

of gravity

(D) Only when its centre of gravity is below

its centre of buoyancy

13. A rectangular block 2 m long, I m wide and I

m deep floats in water, the depth of immersion

being 0.5 m. If water weights 10 kN/m2, then

the weight of the block is

(A) 5 kN (B) 10 kN

(C) 15 kN (D) 20 kN

14. The point in the immersed body through

which the resultant pressure of the liquid may

be taken to act is known as

(A) Centre of gravity

(B) Centre of buoyancy

(C) Centre of pressure

(D) Metacentre

15. When a body is totally or partially immersed in

a fluid, it is buoyed up by a force equal to

(A) weight of the body

(B) weight of the fluid displaced by the body

(C) weight of the body and fluid displaced by

the body


43

FM + OCF & MACHINES

(D) difference of weights of the fluid

displaced and that of the body

(E) None of these

16. Water displaced by a floating wooden block of

density 0.75, 5 m long, 2 m wide and 3 m high,

is

(A) 17.5 m3

(B) 20.0 m3

(C) 22.5 m3 (D) 25 m

3

17. For exerting a pressure of 4.8 kg/cm2, the

depth of oil (specific gravity 0.8), should be

(A) 40 cm (B) 41 cm

(C) 56 cm (D) 60 cm

18. A floating body attains stable equilibrium if its

metacentre is

(A) at the centroid

(B) above the centroid

(C) below the centroid

(D) anywhere

19. The radius of gyration of the water line of a

floating ship is 6.25 m and its metacentric

height is 72.5. The period of oscillation of the

ship, is

(A) (B) 2

(C) 3 (D) /2

1. B

2. B

3. B

4. A

5. C

6. C

7. C

8. D

9. B

10. B

11. B

12. B

13. A

14. C

15. B

16. C

17. D

18. D

19. D

Answer key


44

FM + OCF & MACHINES

[Sol] 9.(B)

Weight of pantaloon (W) = 2000 metric tons

Load (w) = 25 metric tons

Distance moved (x) = 8m

𝑡𝑎𝑛𝜃 =1

20 𝑊 = 𝑤 + 2000

𝐺𝑀 =𝑤𝑥

𝑤 .𝑡𝑎𝑛𝜃 𝑊 = 2025 𝑀𝑒𝑡𝑟𝑖𝑐 𝑇𝑜𝑛𝑛𝑒𝑠

𝐺𝑀 = 25 8

2000+25 ×1

20

𝐺𝑀 =200

2025 ×1

20

𝐺𝑀 =4000

2025

𝐺𝑀 = 2𝑚

[Sol] 14. (A)

Weight of water displaced = weight of block

𝜌𝑔𝑣 𝑤𝑎𝑡𝑒𝑟 = 𝜌𝑔𝑣 𝑏𝑙𝑜𝑐𝑘

10000 0.5 × 1 × 2 = 𝑊𝑏𝑙𝑜𝑐𝑘 1 × 1 × 21

𝑊𝑏𝑙𝑜𝑐𝑘 = 5000𝑁

𝑊𝑏𝑙𝑜𝑐𝑘 = 5 𝐾𝑁

[Sol] 16. (C)

density of wooden block = 750 Kg/m3

Density of water = 1000 Kg/m3

Weight of water displaced = weight of block

𝜌𝑔𝑣 𝑤𝑎𝑡𝑒𝑟 = 𝜌𝑔𝑣 𝑏𝑙𝑜𝑐𝑘

1000 × 9.81 × 𝑉𝑤 = . 750 × 9.81 × 5 × 2 × 3

1000 × 𝑉𝑤 = −750 × 30

Explanations


45

FM + OCF & MACHINES

𝑉𝑤 =750×30

1000

𝑉𝑤 =225

10

𝑉𝑤 = 22.5 𝑚3

[Sol] 18. (D)

According to hydrostatic law (P) = 𝜌𝑔𝑕

𝜌 = 800 𝐾𝑔/𝑚3

[𝜌 =800𝐾𝑔

100 3𝑐𝑚3 ⇒ 800 × 10−6 𝐾𝑔/𝑐𝑚3]

𝑕 =4800

8→ 600𝑐𝑚

𝑃 = 𝜌𝑔𝑕

4.8 = 800 × 10−6 × 1000 × 𝑕

4.8 = .8 𝑕

𝑕 = 60𝑐𝑚

[Sol] 19. (D)

Radius of gyration (K) = 4m

Metacentric height (GM) = 72.5

Period of oscillation = 2𝜋.𝐾

𝐺𝑀 .𝑔

𝑇 = 2𝜋 ×6.25

72.5×10

𝑇 =13.5𝜋

27

𝑇 = 𝜋

2


FLUID MECHANICS

Fluid kinematics is a branch of science that describe the geometry of fluid motion in terms of

displacement, velocity & acceleration & any other quantity derived from displacement & time.

On the other hand the branch of science which deals with the motion of particle without considering force

causing motion.

MEHTODS TO DESCRIBE FLUID MOTION

1. Langrangian Method –

In this method single fluid particle is followed during its motion & its characteristics are describe.

t = 0 t = 2 t = 4

2. Eulerian Method –

Fluid characteristics are described at a particular point in fluid flow.

In fluid mechanics, we deal with Eulerian method.

TYPES OF FLUID FLOW

1. Steady & unsteady flow

t = time

A. Steady flow – When fluid parameters at any point in a flow changes with respect to time.

𝛿𝑣

𝛿𝑡= 0 ,

𝛿𝜌

𝛿𝑡= 0 ,

Chapter

5 FLUID KINEMATICS

Syllabus: Velocity and Acceleration, Potential function, stream-

function, Stream Line equation, Vorticity, Circulation, Flow net,

Stream line, Pathline, streak line. Weightage : 15%


47

FM + OCF & MACHINES

Example – Flow of water in a pipe line due to centrifugal pump being run at uniform rotational

speed.

B. Unsteady flow – When fluid parameters at any point 1m a flow changes w.r.t time.

Mathematically, 𝛿𝑣

𝛿𝑡≠ 0 ,

𝛿𝜌

𝛿𝑡≠ 0

Example – Liquid falling under gravity out of an opening in a bottom of vessel.

2. Uniform & Non – uniform Flow space coordinator

A. Uniform flow – If parameters remains constant w.r.t. space co-ordinate (x,y,z) at any given time.

Mathematically, 𝛿𝑣

𝛿𝑠= 0

= 2S = 2

B. Non – uniform flow – flow in which fluid parameters changes w.r.t. space co-ordinate at any given

time.

𝛿𝑣

𝛿𝑠≠ 0

Steadiness refers to no change w.r.t. time. So steadiness & uniformity of flow can co-exist independently

under following possibilities.

a) Steady uniform flow – Flow at constant rate with constant cross-sectional area. (Q = constant & A =

constant)

10 10

s

10

b) Unsteady Uniform flow – Flow at increasing or decreasing rate with constant cross – sectional area.

(Q = variable & A = Constant)

10

s

9

c) Steady non uniform flow – Flow at constant rate with converging or diverging pipe in cross –

sectional area. (Q = constant & A = Variable)


48

FM + OCF & MACHINES

AA = 5V = 2

BV = 2A = 5

B

A V = A V1 1 2 2

At individual section velocity always same e.g. A – A.

d) Unsteady non uniform flow – Flow increasing or decreasing with converging & diverging pipe (Q

= variable & A = variable)

A = varyV = varyQ = vary A = vary

V = varyQ = vary

t = 0, v = 2t = 2, v = 6

3. Laminar & Turbulent flow –

A. Laminar flow - Fluid particles moves along a well defined path or stream line path or known as

laminar flow or viscous flow or stream lime flow.

B. Turbulent flow – Fluid particle moves in zig – zag way due to which eddies formation takes place

which are responsible for high energy loss.

Laminar & turbulent flow is explained on the bases of Reynold‟s no. (Re)


49

FM + OCF & MACHINES

Transition flow

Case 1:- For pipe flow –

𝑅𝑒 < 2000 → Laminar Flow

𝑅𝑒 → 2000 − 4000 → Transition flow

𝑅𝑒 > 4000 → Turbulent flow

Case 2:- For parallel plate flow –

𝑅𝑒 > 1000 → Laminar flow



Case 3:- For open channel flow

𝑅𝑒 < 500 → Laminar flow



4. Compressible & incompressible flow

A. Compressible – If density changes due to pressure & temperature variation.

On the other hand, density do not remains constant for fluid.

Mathematically, 𝜌 ≠ constant

Example – gas.

B. Incompressible – When density remains constant 𝜌 = constant.

Example – water on liquids.

Mach number is generally taken as a measure of compressibility.

(i) If Mach no. < 0.3 → No compressibility effect.

(ii) If Mach no. < 1 → Subsonic flow.

(iii) If Mach no. = 1 → Sonic flow.

(iv) If Mach no. > 1 → Supper sonic flow.

(v) If Mach no. > 5 → Hypersonic flow.


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FM + OCF & MACHINES

5. Rotational & Ir-rotational Flow

A. Rotation flow:- Fluid particle which flow along laminar flow stream line & rotate their own mass

axis is called rotational flow.

A B B A

B

A

A

B

B. Ir-rotational flow:- Fluid particles which flow along stream line & do not rotate their own mass axis

is called ir-rotational flow.

A

B

A

B

A

B

A

B

6. Rate of flow or discharge (Q)

It is a product of area of cross – section & velocity at that point.

𝑄 = 𝐴𝑉 = 𝑚2 .𝑚

𝑠

Unit – m3/sec.

In MLT – = 𝑀𝑜𝐿3𝑇−1

In FLT – = 𝐹𝑜𝐿3𝑇−1

7. Continuity equation

It based upon law of conservation of mass.

According to continuity, (mass of flow)1 – 1 = (mass of flow)2 – 2

𝜌1𝑉1 . 𝐴1 = 𝜌2 . 𝑉2. 𝐴2

1 2

1 2

Where 𝜌1 & 𝜌2 are densities at section 1 – 1 & 2 – 2.


51

FM + OCF & MACHINES

𝐴1 & 𝐴2 are area of cross – section at 1 – 1 & 2 – 2.

𝑉1 & 𝑉2 are velocities at 1 – 1 & 2 – 2.

Continuity equation is applicable for both compressible & incompressible flow.

For compressible flow –

𝜌1 𝐴1 𝑉1 = 𝜌2 𝐴2 𝑉2

For incompressible flow –

𝜌 = Constant

So, 𝜌1 = 𝜌2

𝐴1 𝑉1 = 𝐴2 𝑉2

8. Continuity equation in three dimension

𝜕𝜌

𝜕𝑡+

𝜕

𝜕𝑥 ( 𝜌. 𝑢) +

𝜕

𝜕𝑦 (𝜌. 𝑉) +

𝜕

𝜕𝑧 𝜌𝜔 = 0

For steady flow,

𝜕𝜌

𝜕𝑡= 0

𝜕

𝜕𝑥 𝜌. 𝑢 +

𝜕

𝜕𝑦 ( 𝜌. 𝑉) +

𝜕

𝜕𝝆 𝜌𝜔 = 0

For incompressible flow,

𝜌 = Constant

𝜌 𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑌+

𝜕𝜔

𝜕𝑥 = 0

𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑌+

𝜕𝜔

𝜕𝑥 = 0

Continuity equation in 2D –

𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑌 = 0

Where 𝑢 is velocity component in x direction

𝑣 is velocity component y in y direction

𝑤 is velocity component z in z direction

9. Velocity & Acceleration

Let 𝑢, v & 𝑤 are the velocity component along x, y & z direction respectively.

A. Velocity vector:-

𝑉 = 𝑢. 𝑖 + 𝑣𝑗 + 𝑤. 𝑘

Resultant velocity:-

𝑉𝑅 = 𝑢2 + 𝑣2 + 𝑤2

B. Acceleration:- 𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 – are acceleration along x, y, z direction.


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FM + OCF & MACHINES

𝑎𝑥 = 𝑑𝑢

𝑑𝑡=

𝜕𝑢

𝜕𝑥 .

𝜕𝑥

𝜕𝑡+

𝜕𝑢

𝜕𝑦 .

𝜕𝑦

𝜕𝑡+

𝜕𝑢

𝜕𝑧 .

𝜕𝑧

𝜕𝑡+

𝜕𝑢

𝜕𝑡

𝑎𝑦 = 𝑑𝑣

𝑑𝑡=

𝜕𝑣

𝜕𝑥 .

𝜕𝑥

𝜕𝑡+

𝜕𝑣

𝜕𝑦 .

𝜕𝑦

𝜕𝑡+

𝜕𝑣

𝜕𝑧 .

𝜕𝑧

𝜕𝑡+

𝜕𝑣

𝜕𝑡

𝑎𝑧 = 𝑑𝑤

𝑑𝑡=

𝜕𝑤

𝜕𝑥 .

𝜕𝑥

𝜕𝑡+

𝜕𝑤

𝜕𝑦 .

𝜕𝑦

𝜕𝑡+

𝜕𝑤

𝜕𝑧 .

𝜕𝑧

𝜕𝑡+

𝜕𝑤

𝜕𝑡

𝜕𝑢

𝜕𝑡, 𝜕𝑉

𝜕𝑡,𝜕𝜔

𝜕𝑡 are the rate of change of velocity w.r.t. time & is called local acceleration.

Components other than 𝜕𝑢

𝜕𝑡, 𝜕𝑉

𝜕𝑡,𝜕𝜔

𝜕𝑡 is called convective acceleration.

Acceleration Vector:-

𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦 . 𝑗 + 𝑎𝑧 . 𝑘

Resultant Acceleration –

𝑎𝑅 = 𝑎𝑥2 + 𝑎𝑦

2 + 𝑎𝑧2

Example 1:- Water flowing through pipe of diameter 0.5m with velocity 1m/s. Calculate rate of

discharge of water.

Solution – Q = AV

Q = 𝜋

4 0.5 2 × 1𝑚/𝑠

Q = 0.196 m3/s.

Example 2:- Water is flowing through a pipe at section 1 – 1 diameter is 0.5m & velocity 1m/s

while at section 2 diameter is 1m calculate. Velocity at section 2 – 2.

Solution – A1 V1 = A2 V2

𝜋

4 × 0.5 2 × 1𝑚 =

𝜋

4 × 1 2 × 𝑉2

V2 = 0.25 m/s

Example 3:- Velocity vector is given by 𝑉 = 4𝑥3𝑖 − 10𝑥2𝑦𝑗 + 2𝑡𝑘 find velocity of a fluid particle

at (2, 1, 3) at t = 1.

Solution – 𝑉 = 4𝑥3𝑖 − 10𝑥2𝑦𝑗 + 2𝑡𝑘

𝑢 = 4𝑥3 𝑣 = −10𝑥2𝑦 𝑤 = 2𝑡

𝑢 = 32 𝑣 = −40 𝑤 = 2

𝑉𝑅 = 322+ −40 2 + 22

𝑉𝑅 = 51.26 unit.

Acceleration –


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FM + OCF & MACHINES

𝐴𝑅 = 𝑎𝑥 2 + 𝑎𝑦

2 + 𝑎𝑧 2

𝑉 = 4𝑥3𝑖 − 10𝑥2𝑦𝑗 + 2𝑡𝑘

𝑢 = 4𝑥3 𝑣 = −10𝑥2𝑦 𝑤 = 2𝑡

𝜕𝑢

𝜕𝑥= 12𝑥2

𝜕𝑣

𝜕𝑥= −20𝑥𝑦

𝜕𝑤

𝜕𝑥= 0

𝜕𝑢

𝜕𝑦= 0

𝜕𝑣

𝜕𝑦= −10𝑥2

𝜕𝑤

𝜕𝑦= 0

𝜕𝑢

𝜕𝑧= 0

𝜕𝑣

𝜕𝑧= 0

𝜕𝑤

𝜕𝑧= 0

𝜕𝑢

𝜕𝑡= 0

𝜕𝑣

𝜕𝑡= 0

𝜕𝑤

𝜕𝑡= 2

𝑎𝑥 = 𝜕𝑢

𝜕𝑥 .

𝜕𝑥

𝜕𝑡+

𝜕𝑢

𝜕𝑦 .

𝜕𝑦

𝜕𝑡+

𝜕𝑢

𝜕𝑧 .

𝜕𝑧

𝜕𝑡+

𝜕𝑢

𝜕𝑡

= 𝜕𝑢

𝜕𝑥 . 𝑢 +

𝜕𝑢

𝜕𝑦 . 𝑣 +

𝜕𝑢

𝜕𝑧 . 𝑤 +

𝜕𝑢

𝜕𝑡

= 12𝑥2 . 4𝑥3 put 𝑥 = 2

= 12 × 4 . 4 × 8

= 1536 units.

𝑎𝑦 = −20𝑥𝑦 . 4𝑥3 − 10𝑥2 . −10𝑥2𝑦

= −20 × 2 × 1 × 4 × 23 + 10 × 22 . −10 × 22 × 1

= −1280 + 1600

= 320 units.

𝑎 = 𝑎𝑥 𝑖 + 𝑎𝑦. 𝑗 + 𝑎𝑧. 𝑘

= 1536 𝑖 + 320. 𝑗 + 2. 𝑘

𝑎𝑅 = 1536 2+ 320 2 + 2 2

𝑎𝑅 = 1568.98 unit

Example 4:- Velocity component are given as –

𝑢 = 𝑥2 + 𝑦2 + 𝑧2 𝑣 = 𝑥𝑦2 − 𝑦𝑧2 + 𝑥𝑦

Determine 3rd

component by satisfying continuity equation.

Solution – By continuity equation, 𝜕𝜇

𝜕𝑥+

𝜕𝑣

𝜕𝑦+

𝜕𝜔

𝜕𝑧= 0 …..(1)

𝑢 = 𝑥2 + 𝑦2 + 𝑧2 𝑣 = 𝑥𝑦2 + 𝑦𝑧2 + 𝑥𝑦 put in (1)

𝜕𝜇

𝜕𝑥= 2𝑥 2𝑥 + 2𝑥𝑦 − 𝑧 2 + 𝑥 +

𝜕𝜔

𝜕𝑧

𝜕𝑦

𝜕𝑡= 2𝑥𝑦 − 𝑧 2 + 𝑥

𝜕𝜔

𝜕𝑧= − 3𝑥 − 2𝑥𝑦 + 𝑧2

𝜕𝜔

𝜕z= 𝜔 = 3𝑥𝑧 − 2𝑥𝑦𝑧 +

𝑧3

3


54

FM + OCF & MACHINES

Example 5:- Fluid vector is given by 𝑉 = 𝑥2𝑦 𝑖 + 𝑦2 . z𝑗 − 2𝑥𝑦𝑧 + 𝑦𝑧2 𝑘 . Prove that it is a

possible case of study of incompressible fluid flow.

Solution – 𝑉 = 𝑥2𝑦 𝑖 + 𝑦2 . 𝑧𝑗 − 2𝑥𝑦𝑧 + 𝑦𝑧2 𝑘

Continuity equation,

𝑢 = 𝑥2𝑦 𝜕𝜇

𝜕𝑥= 2𝑥𝑦

𝑉 = 𝑦2𝑧 𝜕𝑉

𝜕𝑦= 2𝑦𝑧

𝜔 = − 2𝑥𝑦𝑧 + 𝑦z2 𝜕𝜔

𝜕z= −2𝑥𝑦 − 2𝑦𝑧

Put in (1)

𝜕𝜇

𝜕𝑥+

𝜕𝑦

𝜕𝑡+

𝜕𝜔

𝜕𝑧= 0

2𝑥𝑦 + 2𝑦𝑧 − 2𝑥𝑦 − 2𝑦𝑧

10. Velocity potential function ():-

𝑢 =−𝜕

𝜕𝑥

𝑣 =−𝜕

𝜕𝑦

𝑤 =−𝜕

𝜕𝑧

According to continuity equation in three dimension –

𝜕𝜇

𝜕𝑥+

𝜕𝑦

𝜕𝑡+

𝜕𝜔

𝜕𝑧= 0

𝜕

𝜕𝑥 . −

𝜕

𝜕𝑥 . +

𝜕

𝜕𝑦 . −

𝜕

𝜕𝑦 . +

𝜕

𝜕𝑧 . −

𝜕

𝜕𝑧 . = 0

− 𝜕2

𝜕𝑥2 . +𝜕2

𝜕𝑦2 . +𝜕2

𝜕z2 . = 0

It is known as Laplace equation.

1. If exist, the flow should be ir-rotational.

2. If hold laplace equation then the flow is steady, incompressible and ir-rotational.

11. Stream function ():-

𝑢 =−𝜕

𝜕𝑦

𝑣 =𝜕

𝜕𝑥

According to continuity equation in two dimension –

𝜕𝜇

𝜕𝑥+

𝜕𝑦

𝜕𝑡= 0

𝜕

𝜕𝑥 −𝜕

𝜕𝑦 +

𝜕

𝜕𝑦 −𝜕

𝜕𝑥 = 0


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FM + OCF & MACHINES

−𝜕2

𝜕𝑥 . 𝜕𝑦+

𝜕2

𝜕𝑦 . 𝜕𝑥= 0

𝜕2

𝜕𝑥 . 𝜕𝑦=

𝜕2

𝜕𝑦 . 𝜕𝑥

1. If exist, flow may be rotational or may be irrotational.

2. If hold Laplace equation then the flow is irrotational.

Example 6:- which of the following potential function satisfied continuity equation.

(A) 𝑥2𝑦 (B) 𝑥2 − 𝑦2 (C) 𝑥2 + 𝑦2 (D) cos 𝑥

Solution – = 𝑥2 . 𝑦

(A) − 𝜕2

𝜕𝑥2 . +𝜕2

𝜕𝑦2 . +𝜕2

𝜕𝑧2 . = 0

− 𝜕2

𝜕𝑥2 . 𝑥2 . 𝑦 +𝜕2

𝜕𝑦2 . 𝑥2 . 𝑦 +𝜕2

𝜕𝑧2 . 𝑥2 . 𝑦 = 0

−𝜕

𝜕𝑥

𝜕

𝜕𝑥 . 𝑥2 . 𝑦 +

𝜕

𝜕𝑦

𝜕

𝜕𝑦 . 𝑥2 . 𝑦 +

𝜕

𝜕𝑧

𝜕2

𝜕𝑧2 . 𝑥2 . 𝑦 = 0

− 𝜕

𝜕𝑥 2𝑥𝑦 +

𝜕

𝜕𝑦𝑥2 + 0

= −2𝑦

(C) − 𝜕

𝜕𝑥

𝜕

𝜕𝑥 𝑥2+𝑦2 +

𝜕

𝜕𝑦

𝜕

𝜕𝑦 𝑥2+𝑦2 + − −

= 𝜕

𝜕𝑥 2𝑥 +

𝜕

𝜕𝑦 2𝑦

= 2 + 2 = 4

(B) − 𝜕

𝜕𝑥

𝜕

𝜕𝑥 𝑥2−𝑦2 +

𝜕

𝜕𝑦

𝜕

𝜕𝑦 𝑥2−𝑦2

− 𝜕

𝜕𝑥 2𝑥 +

𝜕

𝜕𝑦 −2𝑥

− 2 − 2 = 0

Example 7:- The magnitude of component of velocity at point (1, 1) for stream function 𝜑 = 𝑥2 −

𝑦2.

(A) 2 (B) 4 (C) 2 2 (D) 4 2

Solution – 𝜕2𝜑

𝜕𝑦 .𝜕𝑥=

𝜕2𝜑

𝜕𝑥 .𝜕𝑦

𝜕2 𝑥2−𝑦2

𝜕𝑦 .𝜕𝑥=

𝜕2 𝑥2−𝑦2

𝜕𝑥 .𝜕𝑦

𝜕

𝜕𝑦

𝜕

𝜕𝑥 𝑥2 − 𝑦2 =

𝜕

𝜕𝑥

𝜕

𝜕𝑦 𝑥2 − 𝑦2

𝑣 =−𝜕𝜑

𝜕𝑦=

−𝜕

𝜕𝑦 𝑥2−𝑦2 = − −2 = 2𝑦


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FM + OCF & MACHINES

𝑉 =𝜕𝜑

𝜕𝑦=

𝜕

𝜕𝑦 𝑥2−𝑦2 = 2

𝑉 = 𝑢𝑖 + 𝑣𝑗

𝑉𝑅 = 𝑢2+𝑣2

= 22+22

= 4 + 4

= 8

𝑉𝑅 = 2 2

12. Rotational components 𝝎 :- With the help of rotation components we may identify flow is

rotational or ir-rotational. If rotational components are zero, flow is ir-rotational.

𝜔𝑥 = 𝜔𝑦 = 𝜔𝑧 = 0 (Irrotational)

If rotation components are non-zero, then flow is rotational.

𝜔𝑥 = 𝜔𝑦 = 𝜔𝑧 ≠ 0 (Rotational flow)

𝜔𝑥 =1

2

𝜕𝜑

𝜕𝑦−

𝜕𝑉

𝜕z

𝜔𝑦 =1

2

𝜕𝜑

𝜕z−

𝜕𝑉

𝜕𝑦

𝜔𝑧− =1

2

𝜕𝑉

𝜕𝑥−

𝜕𝑢

𝜕𝑦

13. 𝛏 Vorticity:- Twice the rotational component 𝝎 .

ξx = 2𝜔𝑥

ξy = 2𝜔𝑦

ξz = 2𝜔z

ξ𝑥 = ∂w

𝜕𝑦−

∂v

𝜕z

ξ𝑦 = ∂u

𝜕z−

∂w

𝜕𝑥

ξ𝑧 = ∂v

𝜕𝑥−

∂u

𝜕𝑦

If Vorticity = 0 then flow ir-rotational.

∴ ξ𝑥 = ξ𝑦 = ξ𝑧 = 0 (Irrotational)

ξ𝑥 = ξ𝑦 = ξz ≠ 0 (Rotational)

14. Circulation (⎾ ):- It is line integral of tangential velocity around a close contour in flow field.

Vorticity; Circulation per unit area is vorticity.


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FM + OCF & MACHINES

𝜕⎾

𝜕𝑥 𝜕𝑦=

𝜕𝑣

𝜕𝑥−

𝜕𝑢

𝜕𝑦

Circulation/unit area = vorticity of flow.

For irrotational motion, vorticity = 0

So, circulation around any closed path is irrotational.

Example :- Determine rotational component if 𝑢 = 𝐴𝑧𝑥 𝑣 = 𝐴𝑥𝑦 𝜔 = 𝐴𝑦𝑧 & compute

flow.

Solution:- 𝜔𝑥 =1

2

𝜕𝑢

𝜕𝑦−

𝜕𝑣

𝜕𝑧

=1

2 𝐴𝑥 − 𝐴𝑥 = 0

𝜔𝑦 =1

2

𝜕𝑢

𝜕𝑧−

𝜕𝑤

𝜕𝑥 =

1

2 𝐴𝑦 − 𝐴𝑦 = 0

𝜔𝑧 =1

2 𝐴𝑧 − 𝐴𝑧 = 0

∴ 𝜔𝑥 , 𝜔𝑦 , 𝜔𝑧 = 0 then flow is irrotational.

15. Equipotential Lines:- Line along which velocity potential function () is constant.

Mathematically, = Constant

Slope of equipotential lines

𝑑𝑦

𝑑𝑥=

−𝑢

𝑣 = m1

16. Stream lime:- Line along which function is constant.

𝜑 = Constant

17. Slope of stream line:-

𝑑𝑦

𝑑𝑥=

𝑣

𝑢= m2

𝑚1 × 𝑚2 = −1

The product of slope of equipotential lines & stream line is –

It means equipotential lines are orthogonal to stream lines. Orthogonility between stream lines &

equipotential lines serves to draw a flow net.

18. Flow Net:- A grid obtained by drawing a series of equipotential lines & stream line all called flow

net.


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FM + OCF & MACHINES

3

2

1

1

2

3

Stream Line

Equipotential Line

Limitation of flow net –

1. Fluid has negligible viscosity & incompressible.

2. Flow is irrotational (𝜔 = 0).

3. Fluid flow is confined with 2D boundary.

4. There is no flow across fixed boundaries. (stream lines).

BASIC DEFINITIONS

1. Stream line – A stream line is an imaginary line drawn through the flow field in such a manner that

velocity vector of fluid at each & every point on the stream line is longest to the stream line at that

instant.

For 2D:-

𝑑𝑥

𝑢=

𝑑𝑦

𝑣

𝑑𝑦

𝑑𝑥=

𝑣

𝑢

Slope of stream line is the ratio of velocity component.

Important Characteristics :-

i. Stream line do not cross, if cross fluid particle will have two velocities at the point of inter section

i.e. physically not possible.

ii. There cannot be any movement of fluid mass across stream line i.e. flow is within stream line not

across.


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FM + OCF & MACHINES

iii. Stream line spacing varies inversely as the velocity i.e. converging of stream line in any particular

direction shows accelerated flow in that direction. As cross – section of CD > AB so according to

continuity equation flow of velocity at AB > CD.

15 dropStream

10 drop

Group of neigoubing stream lines forming a cylindrical passage with elementary area of cross –

section called stream filament. When no. comprises consist stream tube.

2. Path line:- It represent the trace or trajectory of fluid particle over a period of time.

Path line show direction of the velocity of the same fluid particle at successive instant of time. Path

line can be intersect itself at different time.

Path Line

3. Streak line:- It is instantaneous picture of the position of all fluid particle that pass through a fixed

point in the flow field.

Line formed by smoke particle into an atmosphere from a fixed nozzle (Hukk(A).

1. For steady flow, there is not geometrical distinction between stream line, path line & streak

line, they are coincident if its originate at the same point.

2. For unsteady flow, path line, stream line & streak lines are different.


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FM + OCF & MACHINES

1. A streamline is a line

(A) Connecting mid points of a flow Cross-

section

(B) Connecting points of equal velocity in a

flow field

(C) Tangent to which at any point gives the

direction of velocity vector at that point

(D) Drawn normal to the velocity vector at

any-point

2. A pathline represents

(A) Mean direction of a number of particles at

the same instant of time

(B) Trace made by a single particle over a

period of time

(C) Instantaneous picture of positions of all

particles in the flow which passed a given

point

(D) The line of flow, normal to which at any

point gives the velocity vector.

3. The path traced by a single particle of smoke

issuing from a cigarette is a

(A) streamline (B) flow line

(C) path line (D) streak line.

4. There is no geometrical distinction between

the streamline, path line and streak line in case

of

(A) Steady flow (B) Uniform flow

(C) Laminar flow (D) irrigational flow.

5. Mark the wrong statement :

(A) Streamlines cannot start or end anywhere

except at the interface or infinity

(B) Streamline spacing varies directly as the

flow velocity

(C) Streamlines can meet at a stagnation point

where the velocity is zero

(D) The flow is only along the streamline and

not across it.

6. For a two-dimensional flow field, the equation

of a streamline is given as

(A) u dy

dx v

(B)

du dy0

dx dy

(C) dy dx

u v

(D)

dx dy

u v

7. One dimensional flow means

(A) Uniform flow

(B) Straight line flow

(C) Non Straight line flow

(D) Flow which neglects changes in

transverse direction.

8. Steady flow occurs when

(A) Conditions change steadily with time

(B) Conditions donot change with time at any

point



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FM + OCF & MACHINES

(C) Conditions are the same at adjacent points

at any instant

(D) Only the velocity vector at any point does

not change with time, i.e., 𝜕𝑉/ 𝜕𝑡 is

constant.

9. Uniform flow occurs when

(A) There is constant discharge through a

pipeline

(B) Conditions at any point in the flow field

practically remain constant as the time

elapses

(C) The velocity vector at any point remains

constant

(D) At any given instant, the velocity vector

at every point in the flow field is identical

in magnitude and direction.

10. In Uniform flow, the velocities of fluid

particles are

(A) Equal at all sections

(B) Always dependent on time

(C) Mutually perpendicular to each other

(D) The fluid particles move in well-defined

paths.

11. The flow of liquid through a tapering pipe at

constant rate is

(A) Steady uniform

(B) Steady non- uniform

(C) Unsteady uniform

(D) Unsteady non- uniform.

12. Which of the following represents steady

uniform flow?

(A) Flow through a diverging duct at constant

rate

(B) Flow through a diverging duct at any

increasing rate

(C) Flow through a long pipe at constant rate

(D) Flow through a long pipe at decreasing

rate.

13. Which of the following represents steady non-

uniform flow?

(A) Constant discharge through a long straight

pipe

(B) Steadily decreasing flow through a

reducing section

(C) Steadily increasing flow through a long

straight pipe of same section

(D) Flow through an expanding tube at

constant rate.

14. Which of the following represents unsteady-

uniform flow?

(A) Steadily decreasing flow through a

reducing section

(B) Flow through an expanding tube at any

increasing rate

(C) Flow through a long pipe at decreasing

rate

(D) Flow through a long pipe at constant rate.

15. Which of the following represents unsteady

non-uniform flow?

(A) Discharge through a river at a bridge site

during flood

(B) Flow through an expanding tube at any

increasing rate

(C) Flow through a reducing section at

constant rate


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FM + OCF & MACHINES

(D) Flow through a long pipe at decreasing

rate

16. Identify the statements pertaining to laminar

flow

(A) Fluid particles exhibit a irregular pattern

of flow

(B) Fluid flows through a narrow passage

(C) Momentum transfer is on macroscopic

level

(D) The injection of smoke or dye shows

considerable lateral dispersion.

17. For a streamline flow, velocity at a certain

point is

(A) constant

(B) A function of time only

(C) Constant but depends on time

(D) Constant and independent of time.

18. Indicate the turbulent flow conditions amongst

the following:

(A) Rise of water in plants through roots

(B) Flow of water through pipes

(C) Flow of oil in measuring instruments

(D) Movement of blood in the arteries of a

human body

19. If ϕ is the potential function in two-

dimensional flow field, then the velocity

components u and v are defined as

(A) u= 𝜕ϕ

𝜕𝑥 and v=-

𝜕ϕ

𝜕𝑦 (B) u=

𝜕ϕ

𝜕𝑦 and v=

𝜕ϕ

𝜕𝑥

(C) u= −𝜕ϕ

𝜕𝑥 and v=

𝜕ϕ

𝜕𝑦 (D) u=

𝜕ϕ

𝜕𝑦 and =

−𝜕ϕ

𝜕𝑥

20. A velocity potential function exists only for

(A) Steady flow (B) Uniform flow

(C) Irrotational flow (D) Compressible flow

21. Which is not true in the context of velocity

potential function?

(A) Is defined as the integral of the tangential

velocity component along a closed

contour

(B) exists for irrotational motion of fluids

whether compressible or incompressible

(C) Satisfies the Laplace equation

(D) Lines of constant velocity potential

function are normal to the streamlines.

22. The existence of velocity potential in a flow

field is indicative of the fact that

(A) vorticity must be non-zero

(B) Circulation around any closed contour

must have a finite value

(C) Flow satisfies the conditions of

irrotational motion

(D) Flow is continuous irrespective of it being

rotational or irrotational.

23. Select the correct statement about equipotential

line

(A) Has a constant dynamic pressure

(B) Connects the mid points of a flow cross-

section

(C) Has no velocity component tangential to it

(D) Lies orthogonal to streamlines for the

flow pattern

24. For irrotational flow

(A) x y

(B) y x

(C) y x

(D) x y


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FM + OCF & MACHINES

25. The relation ∂2ϕ

𝜕𝑥2 + ∂2ϕ

𝜕𝑦2 =0 for an irrotational

flow is referred to as

(A) Euler‟s equation

(B) Laplace equation

(C) Reynolds equation

(D) Cauchy- Reimann‟s equation.

26. Which of the following statements are true for

two-dimensional flow field?

(A) If ϕ exists, Ψ will be also exist

(B) If Ψ exists, ϕ will be also exist

(C) If ϕ exists, the flow will be rotational

(D) If Ψ exists, flow will be either rotational

of irrotational.

27. A control volume refers to

(A) An isolated system

(B) A closed system

(C) A specified mass

(D) A fixed region in space

28. The property of steam function is :

(A) is constant everwhere on any streamline

(B) the flow around any path in the fluid is

zero for continuous flow

(C) the rate of change of with distance in an

arbitrary direction, is proportional normal

to that direction

(D) all the above

29. The continuity equation

(A) expresses the relationship between work

and energy

(B) relates the momentum per unit volume

between two points on a stream line

(C) relates mass rate of flow along a stream

line

(D) requires that Newton‟s second law of

motion be satisfied at every point in fluid.

30. If u, v, w are the components of the velocity v

of a moving particle, the equation 𝑢

𝑑𝑥=

𝑣

𝑑𝑦=

𝑤

𝑑𝑧 represents

(A) one dimensional flow

(B) two dimensional flow

(C) three dimensional flow

(D) Impossible

31. A steady uniform flow is through

(A) a long pipe at decreasing rate

(B) a long pipe at constant rate

(C) an expanding tube at constant rate

(D) an expanding tube at increasing rate

32. Uniform flow is said to occur when

(A) size and shape of the cross section in a

particular length remain constant

(B) size and shape of the cross section change

along a length

(C) frictional loss in the particular length of

the channel will the more than the drop in

its elevation

(D) frictional loss in the particular length of

the channel, will be less than the drop in

elevation.


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FM + OCF & MACHINES

33. A fluid particle may possess the displacement

of

(A) translation (B) rotation

(C) distortion (D) all the above

34. A non-uniform steady flow is through

(A) a long tube at a decreasing rate

(B) an expanding tube at constant rate

(C) an expanding tube at increasing rate

(D) a long pipe at increasing rate

35. In steady flow, which one of the following

changes with time

(A) velocity (B) pressure

(C) density (D) none of these

36. The flow in which each liquid particle has a

definite path and the paths of adjacent particles

do not cross each other, is called

(A) stream line flow (B) uniform flow

(C) steady flow (D) turbulent flow

37. The flow in a channel is said to be non-

uniform, if

(A) free water surface of an open channel is

not parallel to the bed of channel

(B) head needed to overcome frictional

resistance is less than the drop in

elevation of channel bed

(C) head needed to overcome friction

resistance is more than the drop in

elevation of channel bed

(D) all the above

38. „Flow net‟ can be drawn only if the flow is

(A) turbulent (B) rotational

(C) distortion (D) none of these

39. For the two dimensional flow, the stream

function is given by = 2xy. The velocity at a

point (3, 4) is

(A) 6 m/sec (B) 8 m/sec

(C) 10 m/sec (D) 12 m/sec

40. Equation of continuity of fluids is applicable

only if

(A) flow is steady

(B) flow is compressive

(C) flow is one dimensional

(D) all the above

41. The imaginary line drawn such that the

tangents at its all points indicate the direction

of the velocity of the fluid particles at each

point, is called

(A) path line (B) stream line

(C) potential line (D) streak line

42. The equation 𝑝

𝑤+

𝑉2

2𝑔+ 𝑍 = Constant is

based on the following assumptions regarding

the flow of fluid:

(A) Steady, frictionless, incompressible and

along a streamline

(B) Steady, frictionless, uniform and along a

streamline

(C) Steady, incompressible, uniform and

along a streamline

(D) Steady, frictionless, incompressible and

uniform


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FM + OCF & MACHINES

43. The continuity equation

𝜌1 V1A1 = 𝜌2 V2 A2

Is based on the following. assumption

regarding flow of fluid

(A) steady flow

(B) uniform flow

(C) incompressible flow

(D) frictionless flow

Where p1 and p2 are mass densities.

44. Irrotational flow means

(A) the fluid does not rotate while moving

(B) the fluid moves in straight lines

(C) the net rotation of fluid-particles about

their mass centers is zero

(D) none of the above.

45. If the velocity, pressure, density etc. , do not

change at a point with respect to time, the flow

is called

(A) uniform (B) incompressible

(C) non-uniform (D) steady

46. If the velocity, pressure, density etc., change

at a point with respect to time, the flow is

called

(A) uniform (B) compressible

(C) unsteady (D) incompressible

47. If the velocity in fluid flow does not change

with respect to length of direction of flow, it is

called

(A) steady flow (B) uniform flow

(C) incompressible flow (D) rotational flow

48. If the density of a fluid is constant from point

to point in a flow region, it is called

(A) steady flow (B) incompressible flow

(C) uniform flow (D) rotational flow

49. If the fluid particles move in straight lines and

all the lines are parallel to the surface, the flow

is called

(A) steady (B) uniform

(C) compressible (D) laminar

50. If the fluid particles move in a zig-zag way,

the flow is called

(A) unsteady (B) non-uniform

(C) turbulent (D) incompressible

51. Study of fluid motion with the forces causing

the flow is known as

(A) kinematics of fluid flow

(B) dynamics of fluid flow

(C) statics of fluid flow


52. Stream lines and Equipotential lines

(A) Can be drawn graphically for viscous

flow around any boundary

(B) Form meshes of perfect squares

(C) Are orthogonal wherever they meet

(D) Can be determined mathematically for all

boundary conditions

53. The magnitude of the buoyant force can be

determined by:

(A) Newton‟s law of viscosity

(B) Archimede‟s principle


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FM + OCF & MACHINES

(C) Principles of moments


54. Which of the following equations will be

satisfied by irrotational flow of an

incompressible fluid?

1. 𝜕𝑢

𝜕𝑥+

𝜕𝑢

𝜕𝑦+

𝜕𝑢

𝜕𝑧= 0

2. 𝜕𝑢

𝜕𝑦+

𝜕𝑣

𝜕𝑥+

𝜕𝑢

𝜕𝑧+

𝜕𝑤

𝜕𝑥+

𝜕𝑤

𝜕𝑦+

𝜕𝑣

𝜕𝑧= 0

3. 𝜕2𝑢

𝜕𝑥 2 +𝜕2𝑢

𝜕𝑦 2 +𝜕2𝑢

𝜕𝑧 2 = 0

4. 𝜕𝑣

𝜕𝑥=

𝜕𝑢

𝜕𝑦,

𝜕𝑢

𝜕𝑧=

𝜕𝑤

𝜕𝑥,

𝜕𝑣

𝜕𝑧=

𝜕𝑣

𝜕𝑧

Select the correct answer from the codes given

below:

(A) 3 and 4 (B) 1 and 2

(C) 1 and 3 (D) 1 and 4

1. C

2. B

3. C

4. A

5. B

6. D

7. D

8. B

9. D

10. A

11. B

12. C

13. D

14. C

15. B

16. B

17. C

18. B

19. A

20. C

21. A

22. C

23. D

24. A

25. B

26. D

27. D

28. D

29. C

30. D

[Sol] 5.(B) Streamline is a line tangent to which at any point gives the direction of velocity vector at

that point. Streamline never intersects each other, otherwise it will have different fluid

velocity.

[Sol] 6. For streamline flow : equation is

𝑑𝑥

𝑢=

𝑑𝑦

𝑉

Explanations

Answer key


67

FM + OCF & MACHINES

[Sol] 21. Line of constant potential velocity & streamline are normal to each other.

[Sol] 22. Velocity potential exists for irrotational flow only & Laplace equation is satisfied.

[Sol] 29. (C) Mass flow rate 𝑚 = 𝜌𝐴𝑉

𝜌1𝐴1𝑉1 = 𝜌2𝐴2𝑉2 (𝐶𝑜𝑛𝑡𝑖𝑛𝑢𝑡𝑖𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)

If fluid is incompressible ; (𝜌1 = 𝜌2)

𝐴1𝑉1 = 𝐴2𝑉2


In fluid dynamics study of fluid motion involve force of attraction & reaction i.e. force which cause

acceleration to flow.

Dynamic behavior of fluid flow is analyzed on the bases of newton‟s 2nd

law of motion.

According to Newton‟s 2nd

law of motion Fx = m ax

In fluid flow various types of forces are act such as gravity force (Fg), pressure force (FP), Viscosity force

(FV), turbulent force (FT) & compressibility force (F(C).

𝐹𝑡𝑜𝑡𝑎𝑙 = 𝐹𝑔 + 𝐹𝑃 + 𝐹𝑉 + 𝐹𝑡 + 𝐹𝑐

𝐹𝑐 = 0 𝑅𝑒𝑦𝑛𝑜𝑙𝑑′𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛

𝐹𝑡 = 0, 𝐹𝑐 = 0 (𝑁𝑎𝑣𝑖𝑒𝑟 𝑠𝑡𝑜𝑘𝑒𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)

𝐹𝑡 = 0; 𝐹𝑐 = 0; 𝐹𝑣 = 0 (𝐸𝑢𝑙𝑒𝑟′𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)

EULER’S EQUATION OF MOTION:-

𝑑𝑝

𝜌+ 𝑔𝑑𝑧 + 𝑣𝑑𝑣 = 0

Where 𝑑𝑝 is the change in the pressure, 𝑑𝑧 change in datum, 𝑑𝑣 is change in velocity for elementary

element.

dy

dz

dx

BERNOULLI’S EQUATION FROM EULER EQUATION OF MOTION:-

𝑑𝑝


On integrating both sides

𝑑𝑝

𝜌+ 𝑔𝑑 𝑧 + 𝑣𝑑 𝑣 = 0

𝑝

𝜌 + 𝑑 𝑧 +

𝑣2

2= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Chapter

6 FLUID DYNAMICS

Syllabus: Euler‟s equation of motion, Bernoulli‟s equation,

Venturimeter, Orifice meter, Pitot tube, Orifice and Mouthpiece,

Notch and weir, Correction factor. Weightage 10%


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FM + OCF & MACHINES

Diving both side 𝑝𝑦𝑔

𝑝

𝜌𝑔+

𝑔.𝑧

𝑔+

𝑣2

2𝑔=

𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑔

𝑝

𝜌𝑔+ 𝑧 +

𝑣2

2𝑔= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Where 𝑝

𝜌𝑔=

𝑝

𝜔 (pressure energy/ unit wt. or pressure hea(D)

𝑣2

2𝑔= K.E per unit wt. or kinematic head or velocity head

𝑧 = potential energy per unit wt. or potential head or datum head or elevation head or

position or geodetic head.

Bernoulli‟s equation is based upon law of conservation of energy.

Assumptions of Bernoulli’s Equation

1. Fluid is ideal. 𝑣𝑖𝑠𝑜𝑐𝑖𝑡𝑦 𝑢 = 0

2. Flow is steady.

3. Flow is incompressibile.

4. Flow is irrotational (𝜔𝑥 = 𝜔𝑦 = 𝜔𝑧 = 0).

Total Head:- Sum of velocity head, pre. head & elevation head.

𝐻𝑇𝑜𝑡𝑎𝑙 =𝑝

𝜌𝑔+

𝑣2

2𝑔+ 𝑧 ∴ H = Total head

Piezometric head:- Sum of pre. head & elevation head.

𝐻𝑃𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑕𝑒𝑎𝑑 =𝑝

𝜌𝑔+ 𝑧

1. Total head is represented by total energy line.

2. Piezometric head is represented by hydraulic gradient line.

LINEAR MOMENTUM EQUATION

Or impulse momentum relationship.

In fluid mechanics, there occur a change in velocity of a steadily moving fluid, this change may be

magnitude or direction or in both. Magnitude of force required to effected this change can be calculated

by using momentum principle.

According to momentum principle time rate of change of momentum ∝ force takes place in the direction

in which force act.

Mathematically, 𝐹 = 𝑑

𝑑𝑡(𝑝 ) =

𝑑

𝑑𝑡(𝑚𝑣)

= 𝑚𝑑𝑣

𝑑𝑡+ 𝑣

𝑑𝑚

𝑑𝑡 𝑚 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡


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FM + OCF & MACHINES

𝐹 = 𝑚𝑑𝑣

𝑑𝑡

𝐹 = 𝑚𝑑𝑣

𝑑𝑡 𝐹 . 𝑑𝑡 = 𝑚. 𝑑𝑣

Where 𝐹 . 𝑑𝑡 represent impulse of applied force.

𝑚. 𝑑𝑣 is change in momentum.

Equation (i) is called impulse momentum equation.

1. Impulse momentum equation is based upon law of conservative of momentum

2. Theoretically change in momentum is given as

𝜌 𝑄 𝑣2 − 𝑣1 . 𝑑𝑡 …..(2)

Hence, use (2) in (i)

𝐹. 𝑑𝑡 = 𝜌 𝑄 𝑣2 − 𝑣1 . 𝑑𝑡

𝐹 = 𝜌𝑄 𝑣2 − 𝑣1

This equation is called momentum flux equation where quantity 𝜌 𝑄 is mass flow per second called mass

flux.

Force external by flowing fluid in pipe bend:-

p

v1

Fy

v2

Fx

v sin2 p = A sin2

p A2 2

v cos2

p = A cos2

𝐹𝑥 = 𝜌𝑄 𝑣1 − 𝑣2 cos 𝜃 + 𝑝1𝐴1 − 𝑝2𝐴2 cos𝜃

𝐹𝑦 = 𝜌𝑄 −𝑣2 sin𝜃 − 𝑝2𝐴2 sin𝜃

𝐹𝑅 = 𝐹𝑥2 + 𝐹𝑦

2

Angle mady resultant force with horizontal direction tan 𝜃 =𝐹𝑦

𝐹𝑥

VENTURIMETER

U-tube differentialmanometer

(2)

1 23

(1)

d1 Inlet Converging

(1)(2)

= 21°

Throat d2

5° to 7°

DivergingOutlet


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FM + OCF & MACHINES

A device which is used for measuring the rate of flow (discharge) of a fluid flowing through the

pipes.

Converging angle = 21o ± 2

o

Cylindrical section with minimum area of cross-section is of throat.

Throat diameter is usually between 1

2𝑡𝑜

1

4 times of inlet diameter.

Length of throat = it diameter of throat.

𝑙2 = 𝑑2

Diverging angle = 5o to 7

o

Small diverging angle is required in order to avoid flow separation from the wall & eddies

formation which may result in excessive loss in energy.

𝑄𝑡𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 =𝑎1𝑎2

𝑎1 2−𝑎2

2 . 2𝑔𝑕

𝑄𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶𝑑 .𝑎1𝑎2

𝑎1 2−𝑎2

2 . 2𝑔𝑕

𝐶𝑑 = Coefficient of discharge

Coefficient of discharge for Venturimeter = 0.97 to 0.98

Value of h by differential u-tube manometer.

Case 1:- Manometer contain liquid heavier than liquid in pipe.

𝑕 = 𝑆𝑕

𝑆𝐿− 1 . 𝑥

𝑆𝑕 = specific gravity of heavier liquid

𝑆𝐿 = specific gravity of lighter liquid

𝑥 = difference of heavier lighter than liquid in u-tube.

Case 2:- Manometer contain liquid lighter than liquid in flowing pipe.

𝑕 = 1 −𝑆𝑙

𝑆𝑕 𝑥 .

ORIFICEMETER OR ORIFICE PLATE

(1) (2) (v)Venacontractor

(1) (2) (v)

A device used for measuring rate of flow (discharge) of a fluid through a pipe.


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FM + OCF & MACHINES

Coefficient of Contraction

𝐶𝑐 =𝑎𝑣

𝑎2

𝑎𝑣 = 𝑎2 . 𝐶𝑐

According to continuity equation,

𝑎1𝑣1 = 𝑎2𝑣2 = 𝑎𝑣𝑣𝑣

𝑎1𝑣1 = 𝑎𝑣𝑣𝑣

𝑣1 =𝑎𝑣𝑣𝑣

𝑎1

𝑣1 = 𝑎2 . 𝑐𝑐 .𝑉𝑣

𝑎

𝑣1 =𝑎2

𝑎1. 𝑐𝑐 . 𝑉𝑣

𝑄𝑡𝑕𝑒 𝑑𝑖𝑠𝑐𝑕𝑎𝑟𝑔𝑒𝑡𝑕𝑒 =𝑎2 .𝑐𝑐 .

1− 𝑎2𝑎1

2𝑐𝑐

2 . 2𝑔𝑕

But actual discharge is less than 𝑄𝑡𝑕𝑒 .

Then 𝑄𝑎𝑐𝑡 =𝑐𝑑 .𝑎2 .𝑐𝑐

1− 𝑎2𝑎1

2𝑐𝑐

2

Or

𝑄𝑎𝑐𝑡 = 𝑐𝑑 𝑎1 . 𝑎2

𝑎1 2−𝑎2

2

For orificemeter coefficient of discharge is 0.6. Orifice diameter, is kept 0.5 times the diameter of

inlet.

𝑐𝑑 = 0.64 − 0.76

1. On the basis of lost, Orifice plate is cheaper than venturimeter.

2. On the bases of discharge Venturimeter is quite better as compare to orifice meter.

3. The point at which stream lines become parallel is called vena contracta & it is located at a distance

of half the diameter of orifice i.e. (𝑑2).

Pitot tube – A device used to measure the velocity of a flow at any point in pipe or open channel.

v

Flow

Static pre.

Static head

Total pre. = Static pre +

Dynamic pre.

Static Pressure –

It is defined as the force per unit area acting on the wall by a fluid at rest or flowing parallel to wall in

pipe flow. Static pressure of a moving fluid is measured with an instrument which is rest relative to fluid.

Static pressure is measured by inserting a tube into path flow.


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FM + OCF & MACHINES

Total Pressure or stagnation pressure –

The pressure that would be obtain if fluid stream were brought to rest isentropically. The difference

between total pressure & static pressure gives the pressure due to fluid refers dynamic pressure.

Pitot tube is based on principle of conversion of kinetic head into pressure head.

Pitot tube is used to measure stagnation. Pressure or total pressure because head so obtained is the

sum of static head & dynamic head.

𝑉𝑡𝑕𝑒 = 2𝑔𝑕 But 𝑉𝑎𝑐𝑡 = 𝐶𝑣 . 2𝑔𝑕

𝐶𝑣 = coefficient of velocity = 0.98

The point at which velocity reduces to zero is called stagnation point.

Pitot Static Tube – Pitot static tube is combination of static tube & stagnation tube which is called pitot

static tube. It is also used to measure velocity of liquid flowing through a pipe. It is used to measure

dynamic pressure.

Stagnation tube

Static tube

Flow

Venturimeter, orifice meter & pitot tube all the application of Bernoulli‟s equation.

Orifice – It is an instrument used to measure discharge through the vessel which contain liquid. Orifice is

a geometric opening in the side or bottom of a thin walled tank or vessel.

Classification of Orifice –

1. On the basis of size –

(i) Small Orifice – Orifice is said to be small if the depth of upstream liquid on the top of orifice is

> the depth or diameter of orifice itself.

i.e. If d > do then it is small orifice

(ii) Large Orifice – If the depth of upstream liquid on the top of orifice < diameter of orifice it is

called large orifice.

2. On the bases of shape –

(i) Circular (generally use(D)

(ii) Rectangular

(iii) Square


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FM + OCF & MACHINES

3. One the bases of upstream edge of orifice –

(i) Sharp edge orifice (ii) Square edge orifice (iii) Bell mouth

Flow

Sharp Edge

Flow

Sharp Edge

For circular orifice sharp edge is adopted.

4. On the bases of discharge condition –

(i) Free discharge

(ii) Submerged discharge

Submerged discharge

Fully Submerged partially Submerged

Hydraulic Coefficient –

1. Coefficient of velocity (𝐶𝑣) –

𝐶𝑣 =𝑎𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑡 𝑣𝑒𝑛𝑎 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑎

𝑡𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦=

𝑉

2𝑔𝑕

Coefficient of Construction (𝑪𝒄) –

𝐶𝑐 =𝐴𝑟𝑒𝑎 𝑜𝑓 𝑗𝑒𝑡 𝑎𝑡 𝑣𝑒𝑛𝑎𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑜𝑟

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑟𝑖𝑓𝑖𝑐𝑒

It varies from 0.61 – 0.69

General value – 0.65

Coefficient of discharge (D) –

𝐶𝑑 =𝐴𝑐𝑡𝑢𝑎𝑙 𝑑𝑖𝑠𝑐𝑕𝑎𝑟𝑔𝑒

𝑡𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑐𝑕𝑎𝑟𝑔𝑒=

𝐴𝑐𝑡𝑢𝑎𝑙 𝐴𝑟𝑒𝑎 × 𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑜𝑐𝑖𝑡𝑦

𝑇𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑎𝑟𝑒𝑎 × 𝑡𝑕𝑒𝑜𝑟𝑒𝑐𝑡𝑖𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

=𝐴𝑐𝑡𝑢𝑎𝑙 𝐴𝑟𝑒𝑎

𝑇𝑕𝑒𝑜𝑟𝑒𝑐𝑡𝑖𝑐𝑎𝑙 𝐴𝑟𝑒𝑎 .

𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑇𝑕𝑒𝑜𝑟𝑒𝑐𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝐶𝑑 = 𝐶𝑐 . 𝐶𝑣

Value of 𝐶𝑑 varies from 0.61 – 065

& general value → 0.62

Discharge through small orifice –

𝑄 = 𝐶𝑑 . 𝐴 . 2𝑔𝑕


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FM + OCF & MACHINES

h

Discharge for large orifice –

𝑄 =2

3𝐶𝑑 . 𝑏 . 2𝑔 (𝐻2

3 2 − 𝐻1

3 2 )

h2

h1

Mouthpiece –

Pipe

Mouthpiece

A mouthpiece is an attachment in the form of small tube or pipe fixed to orifice.

Length of pipe = 2 – 3 times orifice diameter.

Measure discharge.

Classification –

1. External mouthpiece 2. Internal mouthpiece

Running free Running free

Internal Mouthpiece

External Running Free Running Full

L=3d

L=3d

L=3d


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FM + OCF & MACHINES

𝐶𝑣 = 0.855 𝐶𝑣 = 1 𝐶𝑣 = 0.707

𝐶𝑐=1 𝐶𝑐 = 0.5 𝐶𝑐 = 1

𝐶𝑑 = 0.855 𝐶𝑑 = 0.5 𝐶𝑑 = 0.707

𝑄 = 0.855 𝐴 2𝑔𝑕 𝑄 = 0.5 𝐴 2𝑔𝑕 𝑄 = 0.707 𝐴 2𝑔𝑕

Example 1:- Internal mouthpiece is also known as Borda‟s mouthpiece & Re-entrant mouthpiece. In

case of internal Borda Re-entrant mouthpiece.

(i) If jet of liquid comes out from mouthpiece without touching the side of pipe is called running free.

(ii) If jet of liquid come out from mouthpiece with touching its side is called running full.

1. A mouthpiece is cylindrical mouthpiece it is of uniform cross-section i.e. when its diameter from

inlet to outlet is uniform.

2. A mouthpiece is called convergent & divergent mouthpiece when diameter of pipe decreases or

increases.

3. 𝐶𝑑 of mouthpiece is more than 𝐶𝑑 of orifice. So for a same diameter of mouthpiece & orifice the

discharge through mouthpiece is greater than orifice.

NOTCH – It is a device used for measuring the rate of flow through a small channel with sharp edge.

CLASSIFICATION OF NOTCH


(i) Rectangular (ii) Triangular

(iii) Trapezoidal (iv) Stepped

2. On the bases of edge –

(i) Sharp edge (ii) Beveled edge

Discharge for Notches –

a) Rectangular Notch :-

𝑄 =2

3 𝐶𝑑 . 𝐿 . 2𝑔 𝑕

32

Discharge for notch varies with depth 𝑕3

2 .

b) Triangular Notch :-

𝑄 =8

15𝐶𝑑 𝑡𝑎𝑛

𝜃

2 2𝑔 𝑕

32

For 𝜃 = 90°, we get maximum discharge.


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FM + OCF & MACHINES

c) Trapezoidal Notch :-

𝑄 =2

3 𝑐𝑑1 . 𝐿 . 2𝑔 𝑕

32 +

8

15 𝑐𝑑2 𝑡𝑎𝑛

𝜃

2 2𝑔 𝑕

32

Cipolletti Notch – It is a special class of trapezoidal notch having side slope fixed one horizontal to 4

vertical.

14°

76°

Angle with vertical = 14o

𝑐𝑑 = 0.632

Q = 1.86 L . H3/2

WEIR – Weir is a concrete or masonry structure placed in an open channel over which flow occur.

CLASSIFICATION


(i) Rectangular (ii) Triangular

(iii) Trapezoidal (iv) Stepped

2. On the bases of edge –

(i) Sharp edge (ii) Beveled edge

DISCHARGE

a) Rectangular Weir :-𝑄 =2

3 𝐶𝑑 . 𝐿 . 2𝑔 𝑕

32

Discharge for notch varies with depth 𝑕3

2 .

b) Triangular Weir :-

𝑄 =8

15𝐶𝑑 𝑡𝑎𝑛

𝜃

2 2𝑔 𝑕

32

For 𝜃 = 90°, we get maximum discharge.

c) Trapezoidal Weir :-

𝑄 =2

3 𝑐𝑑1 . 𝐿 . 2𝑔 𝑕

32 +

8

15 𝑐𝑑2 𝑡𝑎𝑛

𝜃

2 2𝑔 𝑕

32

1. Napper – vein:- The sheet of water flowing through notch or over a weir.

2. Crust or sill:- Bottom edge of notch or top of weir over which water flows.

3. Triangular notch or weir is preferred over rectangular notch or weir because it is free from weir.

Sill or crust

Napper or vein


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FM + OCF & MACHINES

Sr. No. Devices Measurement

1. Viscometer Dynamic viscosity

2. Piezometer Pressure

3. Differential manometer Pressure difference

4. Pitot tube Velocity, stagnation pressure

5. Pitot static tube Dynamic pressure

6. Hot weir anemometer Velocity of gas or air velocity

7. Current turbine Velocity of moving water

8. Orifice Discharge

9. Mouthpiece Discharge

10. Weir Discharge

11. Notch Discharge

12. Venturimeter Discharge

13. Flow nozzle Discharge

14. Rota meter & elvometer Discharge

15. Venture flume Discharge

16. current meter Velocity f open channel

17. Bend meter Discharge

CORRECTION FACTORS

1. Kinematic energy correct factors(∝) – Velocity distribution across a section depends upon nature

of flow & smoothness & roughness of pipe. Kinematic energy correction factor is represented by ∝.

Total kinematic energy at a particular section is obtained by integrating the kinematic energy of an

element area within the local velocity. Kinematic energy correction factor, ∝= kinematic

energy/second based upon local velocity.

Kinematic energy/ unit based upon average velocity ∝= 𝑢 . 𝑑𝐴

𝑢𝑎𝑣 . 𝐴

𝑅

0

𝑢 = local velocity

𝑑𝐴 = small elementary area where velocity is 𝑢

𝑢𝑎𝑣 = area of cross section


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FM + OCF & MACHINES

∝= 1

𝐴

𝑢 . 𝑑𝐴

𝑢𝑎𝑣𝑔 .

𝑅

0

Minimum value of ∝= 1

1. For laminar flow ∝= 2

2. For Turbulent flow ∝= 1.02 − 1.15

General value = 1 for turbulent flow.

2. Momentum energy correction factors (𝜷) –

𝛽 =𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚/ 𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑎𝑠𝑒𝑑 𝑢𝑝𝑜𝑛 𝑙𝑜𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚/ 𝑠𝑒𝑐𝑜𝑛𝑑 𝑏𝑎𝑠𝑒𝑑 𝑢𝑝𝑜𝑛 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝛽 = 𝑢2 . 𝑑𝐴

𝑢𝑎𝑣𝑔 2 . 𝐴

𝑅

0

(i) Minimum value for 𝛽 = 1

(ii) For laminar flow – 1.33

(iii) For turbulent flow – 1.01 – 1.07

General value for turbulent flow – 1.0


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FM + OCF & MACHINES

1. The Euler‟s equation of motion

(A) Is a statement of energy balance

(B) Is a moment of momentum equation

(C) Relates various forces with change in

momentum

(D) Is a preliminary step to derive the

Bernoulli‟s equation.

2. The Euler‟s equation of motion

1

2 𝑑 𝑣2 +

𝑑𝑝

𝜌+ 𝑔 𝑑𝑦 = 0

Is based on the assumption of

(A) Frictionless fluid only

(B) Frictionless, incompressible and steady

flow

(C) Frictionless and steady flow

(D) Motion along a streamline, frictionless

and steady flow.

3. The Euler‟s equation of motion can be

integrated only when

(A) The fluid is compressible

(B) The continuity equation is satisfied

(C) The flow is steady and irrigational

(D) The flow is non-viscous and

Incompressible.

4. The Bernoulli‟s equation refers to conservation

of

(A) mass (B) momentum

(C) force (D) energy

5. The expression 𝑑𝑝

𝑝 +

𝑣2

2+ gy = constant, is

Derived with the assumption that the flow is

(A) Steady, frictionless, ρ a function of p

along a streamline

(B) Uniform, frictionless, ρ a function of p

along a streamline

(C) Steady, frictionless, incompressible, along

a streamline

(D) Steady, uniform, incompressible, along a

streamline.

6. Each term of Bernoulli‟s equation stated in the

form 𝑝

𝑤 +

𝑣2

2𝑔+ 𝑦 = constant has units of

(A) N (B) mN/kg

(C) mN/N (D) mN/s.

7. Identify the Bernoulli‟s equation where each

term represents energy per unit mass

(A) 𝑃

𝑊+

𝑣2

2𝑔+ 𝑦 = constant

(B) 𝑃

𝜌+

𝑣2

2+ 𝑔𝑦 = constant

(C) 𝑃 +𝜌 𝑣2

2+ 𝑤𝑦 = constant

(D) None of these.

8. Bernoulli‟s equation is applicable between any

two points in

(A) Rotational flow of an incompressible fluid

(B) Irrotational flow of compressible or

incompressible fluid



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FM + OCF & MACHINES

(C) Steady rotational flow of an

incompressible fluid

(D) Steady, irrotational flow of an

incompressible fluid.

9. Which is not the assumption made in writing

the linear momentum equation in x-direction as

Σ𝐹𝑥 = 𝜌𝑄(𝑉𝑥2− 𝑉𝑥1

)

(A) Steady flow

(B) Incompressible flow

(C) Uniform flow

(D) Velocity uniform and normal to the inlet,

and outlet areas.

10. A change in angular momentum of fluid

flowing in a curved path results in a

(A) Change in pressure

(B) Change in total energy

(C) Dynamic force passing through its centre

of curvature

(D) Torque.

11. In the most general form of Bernoulli‟s

equation 𝑝

𝑤+

𝑉2

2𝑔+ 𝑍 = constant, each term

represents

(A) Energy per unit mass

(B) Energy per unit weight

(C) Energy per unit volume


12. The motion of air mass in a tornado is a

(A) Free vortex motion

(B) Forced vortex motion

(C) Free vortex at centre and forced vortex

outside

(D) Forced vortex at centre and free vortex

outside

13. In a forced vortex motion, the velocity of flow is

(A) Directly proportional to its radial distance

from axis of rotation

(B) Inversely proportional to its radial

distance from the axis of rotation

(C) Inversely proportional to the square of its

radial distance from the axis of rotation

(D) Directly proportional to the square of its

radial distance from the axis of rotation

14. Bernoulli‟s equation assumes that

(A) fluid is non-viscous

(B) fluid is homogeneous

(C) flow is steady

(D) all the above

15. While applying the Bernoulli‟s equation

𝑝

𝜔+ 𝑧 +

𝑣2

2𝑔 any section = total head, the work

any section done on the flow system, if any

(A) is added on the right side of the equation

(B) is added on the left side of the equation

(C) is ignored

(D) none of these

16. Euler‟s equation for motion of liquids, is given

by

(A) 𝑑𝜌

𝜌− 𝑔𝑑𝑧 + 𝑣𝑑𝑣 = 0

(B) 𝑑𝜌

𝜌+ 𝑑𝑔𝑧 − 𝑣𝑑𝑣 = 0


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FM + OCF & MACHINES

(C) 𝑑𝜌


(D) 𝜌𝑑𝜌 − 𝑔𝑑𝑧 + 𝑣𝑑𝑣 = 0

17. The most familiar form of Bernoulli‟s

equation, is

(A) 𝑃1

𝜔+ 𝑍1 +

𝑣1 2

2𝑔=

𝑃2

𝜔+ 𝑍2 +

𝑣2 2

2𝑔

(B) 𝑑𝜌

𝜌+ 𝑔 . 𝑑𝑧 + 𝑣𝑑𝑣 = 0

(C) 𝑃1

𝜔+ 𝑍1 +

𝑣2

2𝑔 𝑎𝑛𝑦 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑕𝑒𝑎𝑑

(D) none of these

18. Euler‟s equation for the motion of liquids

assumes that

(A) fluid is viscous

(B) fluid is homogeneous and incompressible

(C) velocity of flow is non – uniform over the

section

(D) flow is unsteady along the stream line

19. The main assumption of Bernoulli‟s equation

is:

(A) the velocity of energy of liquid particle,

across and cross – section of a pipe is

uniform

(B) no external force except the gravity acts

on the liquid

(C) there is no loss of energy of the liquid

while flowing

(D) all the above

20. If the forces are due to inertia and gravity, and

frictional resistance plays only a minor role,

the design of the channels is made by

comparing

(A) Reynold number (B) Froude number

(C) Weber number (D) Mach number

21. The differential equation 𝑑𝑝


for a fluid motion is suggested by

(A) Bernoulli (B) Cauchy-Riemann

(C) Laplace (D) Leonard Euler

22. Consider the following situations:

1. The energy correction factor is unity

2. the flow is laminar

3. the flow is incompressible

4. the flow is rotational

5. the flow is ideal and ir-rotational

Bernoulli‟s Equation is true only if conditions

(A) 1, 2 and 5 are satisfied

(B) 1, 3 and 5 are satisfied

(C) 2, 3 and 5 are satisfied

(D) 1, 3 and 4 are satisfied

23. Consider the following conditions:

1. fluid is ideal

2. flow is steady

3. fluid is laminar

4. fluid is Newtonian and flow is turbulent

5. flow is along a streamline

For 𝑝

𝛾+ 𝑧 +

𝑈2

2𝑔= constant, the conditions to be

satisfied are

(A) 1, 2, 3 and 5 (B) 2, 3 and 4

(C) 1, 3 and 4 (D) 2, 3 and 5


83

FM + OCF & MACHINES

24. The co-efficient of discharge (Cd)

(A) for an orifice is more than that for a

mouthpiece

(B) for internal mouthpiece is more than that

external mouthpiece

(C) for a mouthpiece is more than that for an

orifice.


25. Bernoulli‟s equation assumes that

(A) Fluid is non-viscous

(B) Fluid is homogeneous

(C) Flow is steady

(D) Flow is along the stream line

26. The most common device for measuring

discharge through channels is

(A) Venture flume

(B) Current meter

(C) Pitot tube

(D) All the above

27. Navier-Stroke equation

(A) no force is neglected

(B) only force of compressibility is neglected

(C) both force of compressibility and force of

turbulence are neglected

(D) forces of compressibility, turbulence and

velocity are neglected

28. The term z in total energy expression

𝑃

𝜌𝑔+

𝑣2

2𝑔+ 𝑧

(A) potential energy

(B) pressure energy

(C) pressure energy per unit weight


29. 29. Bernoulli equation finds its application in

(A) pitot tube

(B) Venturimeter

(C) orifice meter

(D) all the above

30. 30. Which one of the following is not used for

measuring flow in a pipe

(A) nozzle meter

(B) orificemeter

(C) bend meter

(D) venturiflume


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FM + OCF & MACHINES

1. D

2. D

3. D

4. D

5. A

6. C

7. B

8. D

9. C

10. D

11. B

12. D

13. A

14. D

15. B

16. C

17. C

18. B

19. D

20. B

21. D

22. B

23. A

24. C

25. C

26. A

27. A

28. C

29. D

30. D

[Sol] 1. With the help of Euler‟s equation,

we derive Bernouli‟s equation

[Sol] 2. Assumptions in Bernoulli‟s

equation.

Flow is steady

Incompressible

Frictionless

Streamline flow

Laminar Flow

[Sol] 4. equation states that total energy

remains constant.

𝑃

𝜌𝑔+

𝑉2

2𝑔+ 𝑔𝑧 = 𝐶𝑜𝑛𝑠𝑡

𝑃

𝜌𝑔⇒ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑒𝑛𝑒𝑟𝑔𝑦 𝑕𝑒𝑎𝑑

𝑣2

2𝑔⇒ 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑕𝑒𝑎𝑑

𝑔2 ⇒ 𝐷𝑎𝑡𝑢𝑚 𝑕𝑒𝑎𝑑

[Sol] 10. Tarque = Change in angular

momentum.

Tarque = 𝑚𝜗𝑟2 − 𝑚𝜗𝑟

𝑇𝑎𝑟𝑞𝑢𝑒 ⇒ 𝑚𝜗 𝑟2 − 𝑟1

[Sol] 12. For forced vortex flow 𝑣 ∝ 𝑟

For free vortex flow 𝑣 ∝1

𝑟

Explanations

Answer key


FUNDAMENTAL UNITS

QUANTITY UNIT DIMENSION

1. Mass Kg M

2. Length M L

3. Time sec T

4. Luminous of intensity candela C

5. Current Ampere A

6. Temperature kelvin K

7. Amount of substance mole mol.

Derived Units:- Which are derived from fundamental quantities.

QUANTITY FORMULA UNIT MLT FLT

1. Area L x B m2 𝑀0𝐿2𝑇0 𝐹0𝐿2𝑇0

2. Volume L x B x H m3 𝑀0𝐿3𝑇0 𝐹0𝐿3𝑇0

3. M.O.I. Ar2 m

4 𝑀0𝐿4𝑇0 𝐹0𝐿4𝑇0

4. Section Mod (I/Y) m4/m 𝑀0𝐿3𝑇0 𝐹0𝐿3𝑇0

5. Radius of gyration r M 𝑀0𝐿1𝑇0 𝐹0𝐿1𝑇0

6. Velocity d/T m/s 𝑀0𝐿1𝑇−1 𝐹0𝐿1𝑇−1

7. Acceleration v/T m/s2 𝑀0𝐿1𝑇−2 𝐹0𝐿1𝑇−2

8.

9. Angular Velocity

10. Angular

Acceleration (𝜔)

𝑑𝑎

𝑑𝑦

𝑚

𝑠2 .

1

𝑚

𝑀0𝐿0𝑇−2 𝐹0𝐿0𝑇−2

11. Density 𝑚

𝑣 𝑘𝑔/𝑚3 𝑀1𝐿−3𝑇0 𝐹0𝐿−4𝑇−2

12. Specific weight 𝜌𝑔 𝑁/𝑚3 𝑀1𝐿−2𝑇−2 𝐹1𝐿−3𝑇0

Chapter

7 DIMENSION ANALYSIS Syllabus: Buckingham-pie theorem, Reynolds number, Froude‟s

number, Weber‟s number, Euler‟s number, Mach number,

Applications of dimensionless number.

Weightage: 5%


86

FM + OCF & MACHINES

13. Specific volume 𝑀−1𝐿3𝑇0 𝐹−1𝐿4𝑇−2

14. Dynamic viscosity 𝜏

𝑑𝑢 𝑑𝑦 𝑁𝑠/𝑚2 𝑀1𝐿−1𝑇−1 𝐹1𝐿−2𝑇1

15. Kinematic

viscosity

𝑚2/𝑠 𝑀1𝐿2𝑇−1 𝐹0𝐿2𝑇−1

METHODS OF DIMENSIONAL ANALYSIS

1. Rayleigh’s method – It is suitable to derive expression upto four variables.

2. Buckingham’s 𝝅 method – Suitable to derive a relationship between variables greater than four.

In Buckingham’s 𝝅 method of selecting repeating variables should be chosen in any way such that it

contain one variable from geometrical property one from kinematic factor & one from dynamic

factor.

Geometrical Variables – Length, breadth, height, depth, diameter.

Flow property variable - velocity, acceleration, angular velocity.

Fluid property variable - viscosity, kinematic viscosity.

Modals Analysis –

1. Modal – Small scale replica of actual structure or machine.

2. Prototype – Actual structure is called prototype.

Similarities – Similarities between modal & prototypes with respect to any parameter, similarity is

also known as similitude.

Types of forces acting in moving fluid.

When the fluid is moving forces acting on a fluid element are may be any one or combination of

the various forces:-

1. Inertia Force (Fi) 2. Viscous Force (Fv)

3. Gravity Force (Fg) 4. Pressure Force (Fp)

5. Surface Tension Force (Fs) 6. Elastic Force (Fe)

Inertia Force:- Inertia is the resistance of any physical object to any change in its state of motion,

including changes to its speed and direction. It is the tendency of object to keep moving in a straight line

at constant velocity.

Dimensionless Numbers-

These are the number which is obtained by dividing the inertia force by any one of other forces. As this is

ratio of one force to the other force, it will be a dimensionless number.

These numbers are also called Non-dimensional parameters.


87

FM + OCF & MACHINES

The important dimensionless numbers are:

1. Reynold‟s Number 2. Froude‟s Number

3. Euler‟s Number 4. Weber‟s Number

5. Madi‟s Number.

1. Reynold’s Number (Re)

Re = Inertia Force

Viscous Force {RIV}

As, we know that

Inertia Force= Mass ×Acceleration to flowing fluid

= 𝜌 × Volume × Velocity

Time

= 𝜌 ×Volume

Time× Velocity

{Volume per second = Area × Velocity = Av}

Inertia Force = A × V × 𝜌 × V

=𝜌 𝐴v2 (Velocity)

Viscous Force = Shear Stress × Area

= Z × A

= 𝜇𝑑𝑢

𝑑𝑦 × A = 𝜇 .

𝑉

𝐿 × A

𝑑𝑢

𝑑𝑦 =

𝑉

𝐿

Renold‟s Number

Re = 𝐹𝑖

𝐹𝑣=

𝜌𝐴𝑣2

𝜇 .𝑉

𝐿 ×A

= 𝜌𝑉𝐿

𝜇

= 𝑉 × 𝐿

𝜇 𝜌 =

𝑉 × 𝐿

𝑣

𝜇

𝜌 = 𝑢 = kinematic Viscous

Re = 𝑉𝐿

𝜈

Example of flow situation –

(i) Incompressible flow through small size pipe.

(ii) Low velocity motion around automobiles & aeroplane.

(iii) Flow through low speed turbo machines.

(iv) Motion submarines completely under water.

(v) Open channel flow so long as wave & hydraulic jump do not occur.

2. Froude’s Number (FR)

The Froude‟s Number is defined as the square root of the ratio of an Inertia Force to the gravity

force.

Fe = Inertia

Gravity


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FM + OCF & MACHINES

Fe = Fi

Fg {FIG}

As we know Fi = 𝜌𝐴𝑣2

Fg = Force due to gravity

= Mass × Acceleration due to gravity

= 𝜌 × Volume ×g = 𝜌𝐿3 × g

= 𝜌 × 𝐿2 × 𝐿 × 𝑔 = 𝜌 𝐴 𝐿 g

Fe = ρAv 2

ρALg =

V2

Lg

FR = V2

Lg

Example –

(i) Flow through open channel where wave & hydraulic jump are consider.

(ii) Flow of liquid jet from orifice.

(iii) Flow over spillway of dam.

(iv) Flow over notch & weir.

(iv) Motion of ship in rough & turbulent sea.

3. Euler’s Number (Eu)

Eu = Inertia Force

Pressure Force {EIP}

As, we know that

Fi = ρAv2

Fp = Intensity of Pressure × Area = p × A

Eu = ρAv 2

p × A =

ρV2

p

Example –

(i) Flow through pipe (any size).

(ii) Flow over submerged body.

(iii) Water hammer created in penstock.

4. Weber’s Number (We)

We = Inertia Force

Surface Tension Force {WIS}

As, we know that

Fi = ρAv2


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FM + OCF & MACHINES

Fs = 𝜍 × 𝐿

= ρL2V2

σ × L =

ρL2V2

σ × L

= ρLV 2

σ =

V2

σ / ρ L =

V

σ / ρ L

W𝑒 = V

σ / ρ L

Example –

(i) Capillary tube flow.

(ii) Capillary movement of water in soil.

(iii) Flow of blood in veins.

5. Mach’s Number (M)

M = Inertia Force

Elastic Force {MIE}

Where, Fi = ρAV2

Elastic Force = Elastic Stress × Area

= K A

M = 𝜌AV 2

K A =

𝜌V2

K =

V

𝐾 / 𝜌

M = V

𝐾 / 𝜌

Velocity of sound in the fluid C = K

𝜌

M =V

C

Examples –

(i) Aerodynamic testing.

(ii) Water hammer problem.

(iii) Flow of gases whose velocity exceeding velocity of sound.

Fluid Condition Pipe Flow Open Channel Flow

Laminar Re ≤ 2000 Re ≤ 500

Turbulent Re > 4000 Re > 1000

Transitional 2000 < Re < 4000 500< Re < 1000


90

FM + OCF & MACHINES

(i)

F e w man

G P S E

Sr. No. Dimensionles

s No.

Symbol

&

Formula

Significant Field of use

1 Reynold‟s No. 𝜌𝑣𝑙

𝜇 𝑜𝑟

𝑣𝑙

𝜐

𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑓𝑜𝑟𝑐𝑒

Laminar viscous flow in

confined passage where

viscosity effect

predominate.

2 Froude No. 𝑣

𝐿𝑔


𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒

Free surface flow where

gravity force predominate.

3 Mach No.

𝜌𝑣2

𝑘


𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒

High speed flow where

compressible effect are

important.

4 Weber No.

𝜌𝐿𝑣

𝜍


𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠 𝑓𝑜𝑟𝑐𝑒

capillary where surface

tension predominate.

5 Euler No.

𝜌𝑣2

𝑝


𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒

Conduits flow where

pressure variation

predominates.


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FM + OCF & MACHINES

1. Which of the following is not a dimensionless

number?

(A) the coefficient of lift 𝐶𝑙

(B) the pipe friction factor 𝑓

(C) the Manning‟s coefficient 𝑛

(D) the coefficient of discharge 𝐶𝑑

2. If there are 𝑛 variables in a particular flow

situation, and these variables contain 𝑚

primary dimensions, then the number of

dimensionless groups relating the variables

will be

(A) n + m (B) n – m

(C) n

m (D)

m

n

3. The repeated variables in dimensional analysis

should

(A) form the non-dimensional parameters

among themselves

(B) not include the dependent variables

(C) have two variables with the same

dimensions

(D) must contain jointly all the fundamental

dimensions involved in the phenomenon.

4. Dimensional analysis is useful in

(A) checking the correctness of a physical

equation

(B) determining the number of variables

involved in a particular phenomenon

(C) determining the dimensionless groups

from the given variables

(D) the exact formulation of a physical

phenomenon.

5. Kinematic similarity between model and

prototype is

(A) the similarity of streamline pattern

(B) the similarity of discharge

(C) the similarity of force influencing the

flow

(D) the use of same model scale throughout

6. Dynamic similarity between model and

prototype implies that

(A) the forces acting at corresponding

locations are same

(B) the flow pattern is similar

(C) there is point to point correspondence

between the two systems

(D) both the systems undergo similar rates of

change of motion.

7. Principles of similitude form the basis of

(A) performing acceptance tests

(B) comparing two identical equipments

(C) comparing similarity between design and

actual equipment

(D) designing and testing models so that the

results can be worked out for the

prototype.



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FM + OCF & MACHINES

8. Select the situation in which Reynolds model

law is applicable

(A) flow over the spillway of a dam

(B) flow of blood in veins and arteries

(C) water hammer created in penstocks

(D) flow through low speed turbomachines

9. The square root of inertia force to gravity

force is known as

(A) pressure coefficient

(B) Froude‟s number

(C) Weber number

(D) Mach number

10. The Froude‟s model law would not be

applicable for the analysis of

(A) pressure rise due to sudden closure of

valves

(B) flow over the spillway of a dam

(C) flow of liquid jets from orifices

(D) motion of ship in rough and turbulent seas

11. Euler‟s dimensionless number relates

(A) inertia and gravity force

(B) viscous and inertia force

(C) pressure and inertia force

(D) buoyant and viscous force

12. Mach number is significant in

(A) flow of highly viscous fluids

(B) motion of rocket

(C) water waves breaking against a wall

(D) motion of submarine completely in water

13. Weber number will be an important

consideration in the study of

(A) flow of water through a pipeline

(B) surface wave generated liquids

(C) steel balls dropping through oil

(D) formation of spherical drops – rain drops

14. Indicate the criterion to be applied when a

river model is to be tested in a laboratory

(A) Reynolds number

(B) Froude‟s number

(C) Weber number

(D) Euler number

15. Reynold number is the ratio of initial force

and

(A) viscosity (B) elasticity

(C) gravitational force (D) surface tension

16. Mach number is the ratio of inertia force to

(A) viscosity (B) surface tension

(C) gravitational force (D) elasticity

17. For the flow of liquid from an open ended

tube (or nozzle) leading to the formation of

spray of liquid drops, the number generally

applied, is

(A) Froude number (B) Weber number

(C) Reynold number (D) Mach number

18. Weber number is the ratio of inertia force to

(A) surface force (B) gravitational force

(C) elasticity (D) viscosity


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FM + OCF & MACHINES

1. C

2. B

3. D

4. C

5. A

6. A

7. D

8. D

9. B

10. A

11. C

12. B

13. B

14. B

15. A

16. D

17. B

18. A

[Sol] 9. Froude‟s Number = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑜𝑟𝑐𝑒

[Sol] 11. Euler‟s Number = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒

[Sol] 15. Reynold‟s Number = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝐹𝑜𝑟𝑐𝑒

[Sol] 16. Mach Number = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒

[Sol] 18. Weber Number = 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 𝑓𝑜𝑟𝑐𝑒

𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒

Explanations

Answer key


FLOW THROUGH PIPE

Major losses

1. Frictional losses

Flow losses in pipe

Minor losses

1. Sudden Enlargement2. Sudden Contraction3. Losses due to bend

4. Pipe fitting5. Load of head at entrance6. Load of head at unit7. obstacle in pipe

Renold’s Number (Re)

Re = Inertia Force

Viscous Force =

𝜌𝑉𝐿

𝜇 =

𝑉𝐿

𝜇

𝜌 =

𝑉𝐿

𝜗

Re = 𝑉𝐿

𝜗

It is the ratio of Inertia Force to Viscous Force, for pipe flow L = D

Re = 𝑉𝐷

ν

Re < 2000 (Laminar)

Re > 4000 (Turbulent)

2000 < Re < 4000 (Transition)

𝑃 = 𝜌 𝑔 𝑕

For pipe flow the “Characteristic Length” (L) is taken as diameter of the pipe.

Chapter

8 FLOW THROUGH

PIPE Syllabus: Reynolds number, Major losses, Darcy Weisbech equation,

Chezy‟s constant, Minor losses, HGL & EGL, Pipe arrangements,

Equivalent pipe. Weightage: 10%


95

FM + OCF & MACHINES

𝑕 = 𝑃

𝜌𝑔

𝑕 = 𝑃

𝑤

𝑕1 = 𝑃1

𝑤 , 𝑕2 =

𝑃2

𝑤

𝑄 = 𝐴1𝑉1 = 𝐴2𝑉2

But 𝐴1 = 𝐴2

𝑉1 = 𝑉2

𝑃1

𝑤+

𝑉12

2 𝑔 =

𝑃2

𝑤+

𝑉22

2 𝑔 + 𝑕𝐿 {𝑉1 = 𝑉2}

𝑃1

𝑤−

𝑃2

𝑤 = 𝑕𝐿

In the direction of flow pressure ↓es in order to overcome losses i.e. pressure gradient in the div‟n of flow

is –ve.

Darcy – Weisbach Equation

This equation is used for finding out Head Loss due to friction (hL) in a steady Laminar and

Turbulent flow.

F‟ = Friction Coefficient

hL = 𝐹 𝐿 𝑉2

2 𝑔 𝐷 {F = 4 F‟}

F = Friction Factor

Flow through pipes:–

When fluid is flowing through a pipe it has to overcome various losses and these losses are

termed as “Major Loss” and “Minor Loss”.

Friction Factor always depends only on Reynold‟s Number (Re)

F = 16

Re

Pb – Find the minimum value of friction factory that can occur in Laminar flow through

circular pipe.

Sol – For Laminar flow

F = 16

Re

F = 16

2000

F = 0.032


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FM + OCF & MACHINES

Major Loss – Head Loss due to friction is known as Major Loss. It can be calculated by

1. Darcy-Weisbach Equation

According to this

hL = 𝐹𝐿𝑉2

2 𝑔 𝐷

As, we know that Q = AV

A = π

4D2 . V

V = 4Q

πD2

hL = 𝐹𝐿

2 𝑔 𝐷

4𝑄

π D2 2

hL = 16 𝐹𝐿Q2

2 𝑔 𝜋2𝐷5

hL = 𝐹𝐿Q2

2 𝑔 𝜋2

16 𝐷5

hL = 𝐹𝐿Q2

2 ×9.81 × 3.14 2

16 𝐷5

F = Darcy′s Friction Factor

hL = 𝐹𝐿Q2

12 𝐷5

2. Chezy’s Formula

Used for find out velocity of flow in open channel.

V = C m. i

C = Chezy‟s Constant

m = Hydraulic mean depth

m = Area

Perometer =

π

4D2 =

D

4

m =D

4

i = Hydraulic Slope

i = hL

L

→→

Head LossCharacteristic Length

V= C m. i

V = C D

4 ×

hL

𝐿

V2 = 𝐶2 𝐷 hL

𝐶2 𝐿

hL =4 LV 2

𝐶2 𝑑

Also, we know that

hL =𝐹𝐿𝑉2

2 𝑔 𝐷


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FM + OCF & MACHINES

Equating both side, we get

𝐹𝐿𝑉2

2 𝑔 𝐷 =

4 LV 2

𝐶2 𝑑

𝐶2 = 8𝑔

𝐹

𝐶 = 8𝑔

𝐹

Dimension of chezy’s constant

𝐿1/2𝑇−1

Minor Losses–

(A) Looses due to sudden expansion:- By applying Bernoulli‟s equation, continuity equation and

momentum equation, we get

hL exp = 𝑉1−𝑉2

2

2𝑔 =

𝑉1 2

2𝑔 1 −

𝑉2

𝑉1

2

Also, 𝐴1𝑉1 = 𝐴2𝑉2

𝑉2

𝑉1 =

𝐴1

𝐴2

hL exp = 𝑉1

2

2𝑔 1 −

𝑉2

𝑉1

2

(B) Losses at the exit of pipe:- It is similar to sudden expansion with 𝐴2 = ∞

Exit Loss = 𝑉2

2 𝑔

V = Velocity in pipe.

(C) Losses due to sudden contraction


98

FM + OCF & MACHINES

𝐶𝑐 = Area of Vena Contracta

Area of Orifice

𝐶𝑐 =𝐴𝐶

𝐴2

𝐴𝐶𝑉𝐶 = 𝐴2𝑉2

𝐴𝐶

𝐴2 =

𝑉2

𝑉𝐶 = 𝐶𝑐

hL Contraction = 𝑉𝐶−𝑉2

2

2 𝑔

= 𝑉2

2

2𝑔 𝑉𝐶

𝑉2− 1

2

hL Contraction = 𝑉2

2

2𝑔

1

𝐶𝑐− 1

2

(D) Velocity at the entrance of pipe - It is similar to sudden contraction

Therefore

Entrance Loss = 0.5 𝑉2

2 𝑔

V = Velocity in pipe

(e) Losses due to Bend in pipe –

hL = 𝐾 𝑉2

2 𝑔

K = Constant (Depend on angle of bend and radius of curvature of ben(D)

K ≅ 1.2 for 90°bend K ≅ 0.4 for 45°bend.

Syphon

It is a long head pipe which is used to transfer liquid from a reservoir at a when the two reservoir

separated by hill or high level ground.

Hill

C

Reservoir at higher level

Reservoir at Lower level

Syphon Arrangement

B

A

Highest point of syphon is called summit. As point C is above the free surface of water. The pressure at C

will be less than atmospheric pressure.

If „𝐶𝑐‟is not given them Losses due to sudden contraction are taken as

hL Contraction = 0.5 𝑉2

2

2𝑔

𝑉2 = Velocity in smaller pipe


99

FM + OCF & MACHINES

Theoretically the pressure at C may be reduce to 10.3 m of water.

But in actual this pressure is only – 7.6 m of water or 10.3 – 7.6 = 2.7m of absolute.

If this pressure at C become than 2.7m of water absolute, dissolved air & other gases would come out

from water & collect at the summit.

The flow of water will be obstructed.

Pipe Arrangement –

1. Series arrangement – In series arrangements two or more popes of different diameter are connected

with one another to form a single pipe line.

Q Q , v1 1

hf1

1

Q , V1 2

hf2

2

Q , V3 3

hf3

3

Q

i. Discharge through all pipes are same

𝑄 = 𝑄1 = 𝑄2 = 𝑄3

ii. Total loss of head through the entire system is given as

𝑕𝑓 = 𝑕𝑓1+ 𝑕𝑓2

+ 𝑕𝑓3

2. Parallel arrangement –

Q2

Q1

Q

Q3

Q

(i) 𝑄 = 𝑄1 + 𝑄2 + 𝑄3

(ii) 𝑕𝑓 = 𝑕𝑓1= 𝑕𝑓2

= 𝑕𝑓3

Concept of equivalent pipe

Pipe flow by replacing the series combination by a single pipe with uniform diameter would have same

head loss & discharge. This pipe is called equivalent pipe.

𝑙1

𝑑1 5 +

𝑙2

𝑑25 +

𝑙3

𝑑35 =

𝑙

𝑑5


100

FM + OCF & MACHINES

1. To replace a pipe of diameter D by n parallel

pipes of diameter d, the formula is

(A) 𝑑 =𝐷

𝑛 (B) 𝑑 =

𝐷

𝑛1/2

(C) 𝑑 =𝐷

𝑛3/2 (D) 𝑑 =𝐷

𝑛2/5

2. Hydraulic gradient equal to difference in

water surfaces

(A) 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑤𝑎𝑡𝑒𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠

𝑡𝑜𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 𝑕𝑜 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑕𝑎𝑛𝑛𝑒𝑙

(B) 𝑕𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛

𝑡𝑜𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 𝑕 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑕𝑎𝑛𝑛𝑒𝑙

(C) 𝑤𝑒𝑡𝑡𝑒𝑑 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟


(D) 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡𝑕𝑒 𝑐𝑟𝑜𝑠𝑠 −𝑠𝑒𝑐𝑡𝑖𝑜𝑛


3. The distance y from pipe boundary, at which

the point velocity is equal to average velocity

for turbulent flow, is

(A) 0.223 R (B) 0.423 R

(C) 0.577 R (D) 0.707 R

Where R is radius of pipe.

4. The losses are more in

(A) Laminar flow (B) Transition flow

(C) Turbulent flow (D) Critical flow

5. Chezy‟s formula is given as

(A) 𝑉 = 𝑖 𝑚𝐶 (B) 𝑉 = 𝐶 𝑚𝑖

(C) 𝑉 = 𝑚 𝐶𝑖 (D) none of the above.

6. The value of the kinetic energy correction

factor (α) for the viscous flow through a

circular pipe is

(A) 1.33 (B) 1.50

(C) 2.0 (D) 1.25

7. The value of the momentum correction factor

(β) for the viscous flow through a circular pipe

is

(A) 1.33 (B) 1.50

(C) 2.0 (D) 1.25

8. For solving network problem of pipes,

necessary condition is

(A) Continuity equation

(B) Energy equation

(C) Darcy-Weisbach equation

(D) All the above

9. Chezy‟s formula is used to determine

(A) Head loss due to friction in pipe

(B) Velocity of flow in pipe

(C) Velocity of flow in open channels

(D) None of these.

10. In Chezy‟s formula V=C 𝑚𝑖

(A) V is the mean velocity of flow

(B) m is the hydraulic mean depth

(C) i is the loss of head per unit length of pipe

(D) All the above.



101

FM + OCF & MACHINES

11. The major loss of energy in long pipes is due

to

(A) Sudden enlargement

(B) Sudden contraction

(C) Gradual contraction or enlargement

(D) Friction

12. In parallel pipes:

(A) The discharge is the same in all pipes

(B) The head loss is the same in all pipes

(C) The velocity is equal in all pipes


13. Identify the incorrect statement

(A) In laminar flow, the eddy viscosity is zero

(B) In turbulent flow, the molecular viscosity

is insignificant compare with eddy

viscosity

(C) In any given flow, the eddy viscosity is

constant across the fluid stream

(D) The eddy viscosity is dependent on the

state of turbulent flow

14. The flow of fluid through a pipe is laminar

when

(A) The fluid is ideal

(B) The fluid is viscous

(C) The Reynolds number is less than 2000

(D) There is considerable lateral dispersion of

smoke of dye injected into the flow

stream.

15. For flow through a pipeline, the lower critical

Reynolds number is

(A) The least Reynolds number for laminar

flow ever obtained

(B) The Reynolds number at which turbulent

flow changes to laminar flow

(C) Is always more than 2500

(D) Is different for different fluids.

16. In the Navier-Stokes equations, the forces

considered are :

(A) Pressure, viscous and turbulence

(B) Gravity, pressure and viscous

(C) Gravity, pressure and turbulence

(D) Pressure, gravity turbulence and viscous.

17. Navier-Stokes equations are associated with

(A) Buoyancy (B) Turbulence

(C) Viscosity (D) Compressibility

18. Navier-Stokes equations are associated with

(A) Turbulent flows (B) Vortex flows

(C) Viscous flows (D) Rotational flows.

19. The relation between pressure and shear stress

for one-dimensional laminar flow in x-

directions is :

(A) 𝑑𝑝

𝑑𝑥= 𝜇

𝑑𝜏

𝑑𝑦 (B)

𝑑𝑝

𝑑𝑥=

𝑑𝜏

𝑑𝑦

(C) 𝑑𝑝

𝑑𝑥=

1

𝜇

𝑑𝜏

𝑑𝑦 (D) None of these

20. For laminar flow between two fixed parallel

plates, the flow velocity:

(A) Is constant over the cross-section

(B) Varies parabolically across the section

(C) Varies as three-halves power of distance

from the mid-point

(D) Is maximum at the centres, zero at the

plates and the variation in-between is

linear.


102

FM + OCF & MACHINES

21. The shear stress between two fixed parallel

plates with a laminar flow between them:

(A) Varies directly as distance from the mid-

plane

(B) Varies inversely as distance from the mid-

plane

(C) Varies parabolically across the gap

(D) Remains constant across the gap.

22. When a fluid flows through a pipeline under

viscous flow conditions, the ratio of the

velocity at the axis of the pipe to the mean

velocity of flow is

(A) 0.5 (B) 0.707

(C) 1.67 (D) 2.0

23. According to Darcy-Weisbach, the loss of head

due to friction in the pipe is given by

(A) hf = 4𝑓𝐿𝑉 2

2𝑔𝑑 (B)

4𝑓𝐿𝑑 2

2𝑔𝑉

(C) hf = 4𝑓𝐿𝑉 2

𝑔𝑑 (D)

4𝑓𝐿𝑑 2

𝑔𝑉

where f = friction L = length

V = Mean velocity d = diameter of pipe

24. Models are known undistorted model if

(A) the prototype and model are having

different scale ratios

(B) the prototype and model are having same

scale ratio

(C) the model and prototype are kinematically

similar


25. Model analysis of pipes flow are based on


(C) Mach number (D) Euler number

26. Model analysis of free surface flows are based

on


(C) Mach number (D) Euler number.

27. Model analysis of aero planes and projectile

moving at super-sonic speed are based on


(C) Mach number (D) Euler number.

28. Match List 1 with List 2 and select the correct

answer using the codes given below the lists

List- I List- II

(i) Reynold

number

1. Gravity force

(ii) Froude

number

2. Surface energy force

(iii) Weber

number

3. Viscous force

(iv) Mach

number

4. Elastic force

5. Shear force

Codes:

(i) (ii) (iii) (iv)

(A) 1 2 3 4

(B) 1 2 4 5

(C) 3 1 2 4

(D) 1 2 3 5


103

FM + OCF & MACHINES

1. D

2. B

3. A

4. C

5. B

6. C

7. A

8. D

9. B

10. D

11. D

12. B

13. C

14. C

15. A

16. B

17. C

18. C

19. B

20. B

21. A

22. D

23. A

24. B

25. A

26. B

27. C

28. C

[Sol] 5. Chezy‟s Formula ;-

𝑉 = 𝑐 𝑚𝑖

C = Chezy‟s constant

V= Mean velocity of flow

M=Hydraulic mean depth

L = Loss of head per unit length of pipe

[Sol] 6. Kinetic energy correction factor ∝ = 2

Momentum correction factor 𝛽 = 1.33

[Sol] 12. for parallel pipes : -

𝑕𝑓 = 𝑕1 = 𝑕2 = 𝑕2 = 𝑕𝑢

For series pipes;-

𝑕𝑓 = 𝑕𝑓1 + 𝑕𝑓2 + 𝑕𝑓3 + 𝑕𝑓4 + ⋯……………

Explanations

Answer key


104

FM + OCF & MACHINES

[Sol] 20.

[Sol] 21.

[Sol] 24. Undistorted model, means model & prototype are having same scale ratio.


BOUNDARY LAYER FLOW

When a real fluid flow passed a solid body fluid particle adhere to boundary & no slip condition is occur.

This mean that velocity of fluid closed to the boundary will be same as that of boundary.

If boundary is stationary, the velocity of fluid at boundary will be zero.

Further away from the boundary velocity will be higher as a result of this variation of velocity, velocity

gradient exist.

The velocity of fluid increases from 0 on stationary boundary to free stream velocity () of fluid in the

direction normal to the boundary.

The variation of velocity from zero to free stream velocity in the direction normal to boundary takes place

in a narrow region in the vicinity of solid boundary. This narrow region is called boundary layer & theory

dealing this boundary layer flow is called boundary layer theory.

TransitionTurbulent

u

Laminar

laminar sub layerleading edge

According to boundary layer theory, the flow of fluid in the neighborhood of the solid boundary may be

divided into two region –

1. A very thin layer of fluid called boundary layer in immediate neighborhood of solid boundary where

variation of velocity zero to the stream velocity in the direction normal to the boundary takes place.

As velocity changes from zero to free stream velocity () with respect to 𝑦 (normal to the boundary

surface).

So, 𝑑𝑢

𝑑𝑦 exist hence shear stress also exist.

𝜏 =𝜇𝑑𝑢

𝑑𝑦

2. The remaining fluid outside the boundary layer velocity is constant i.t. free stream velocity ().

Chapter

9 BOUNDARY LAYER

THEORY Syllabus: Velocity distribution, Displacement thickness, Momentum

thickness, Energy thickness, Separation of boundary layer, Methods to

prevent Separation. Weightage : 10%


106

FM + OCF & MACHINES

So, 𝑑𝑢

𝑑𝑦= 0

𝜏 = 0

For turbulent boundary layer does not extend to the solid surface because underline it an layer called

laminar sub-layer in which flow is along stream line.

Flow pattern in boundary layer is judged by Reynold‟s number i.e. 𝑅𝑒 =𝑥𝑈0

Where 𝑥 = distancealong the plate from leading edge. For such case i.e. transition from laminar to

turbulent flow pattern occur at a value of 3 × 105 − 5 × 105.

VELOCITY DISTRIBUTION IN LAMINAR & TURBULENT BOUNDARY LAYER ON A FLAT

PLATE –

1. In laminar boundary layer velocity gradient become step, as one proceed along flow because change

in velocity from zero to

y

laminar

Turbulent

uU

2. In turbulent boundary layer there occur an interchange of momentum & energy among the individual

layer. Hence velocity profile is fuller & much stepper at the plate surface.

BOUNDARY LAYER PARAMETER

1. Boundary layer thick (𝜹) –The distance from the boundary of solid body measured in normal

direction to the point where velocity of fluid is approximately equal to 0.99 U free stream velocity.

2. Displacement thick (𝜹∗) –Distance measure perpendicular to the boundary of solid body by which

boundary should be displaced to compensate for the reduction in flow rate for the account of

boundary layer formation.

𝛿∗ = 1 −𝑢

𝑈 𝑑𝑦

𝛿

0

3. Momentum thick (𝜽) –Thick of flow moving at a free stream velocity & having momentum flux =

deficiency of momentum flux in the region of boundary layer.

𝜃 = 𝑢

𝑈 1 −

𝑢

𝑈 𝑑𝑦

𝛿

0


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FM + OCF & MACHINES

4. Kinematic Energy thick (𝜹∗∗) –Thick of flow moving at free stream velocity & having energy =

deficiency of the energy in the boundary layer region.

𝛿∗∗ = 𝑢

𝑈 1 −

𝑢

𝑈

2

𝑑𝑦𝛿

0

𝛿∗ > 𝛿∗∗ > 𝜃

SEPARATION OF BOUNDARY LAYER

Loss of kinematic energy is recovered from the immediate fluid layer in contact with layer adjacent to

solid surface through momentum exchange process. Thus velocity of layer goes on decreasing. Along the

length of solid body at a contain point a stage may occur when boundary layer may not be able to keep

striking to the solid body. If cannot provide kinematic energy to overcome the resistance offered by solid

body thus boundary layer separated from surface. This is called boundary layer separation. The point on

body at which separation of boundary is on the verge of from surface is called point of separation.

EFFECT OF PRESSURE GRADIENT ON BOUNDARY LAYER

1. Velocity as well as pressure remains constant, 𝑑𝑝

𝑑𝑥= 0 this condition prevails when fluid flow passed

a thin flat plate. Velocity profile changes from parabolic to logarithmic.

2. In a region ABC area of flow decreases, velocity increases, due to increase in velocity pressure

decreases in the direction of flow so, 𝜕𝑝

𝜕𝑥= −𝑣𝑒.

A

BC

D

E

Solid body

Hence, entire boundary moves forward.

3. At the point C pressure is minimum area of flow start increasing beyond C velocity decreases,

pressure increases. Hence pressure 𝜕𝑝

𝜕𝑥= +𝑣𝑒.This combined effect

𝜕𝑝

𝜕𝑥= +𝑣𝑒& surface resistance

reduces the momentum. Hence, momentum becomes unable to overcome the effect of surface

resistance. Hence boundary layer start separating.

1. 𝑑𝑢

𝑑𝑦= −𝑣𝑒, flow has separated.

2. 𝑑𝑢

𝑑𝑦= 0, flow is verge of separation.

3. 𝑑𝑢

𝑑𝑦= +𝑣𝑒, No. separation of flow.


108

FM + OCF & MACHINES

Method to prevent separation –

1. Stream line design of body i.e. by changing the radius of curvature of boundary.

2. Placing some disturbance near the boundary in the approach section.

3. Sunction, whereby the fluid whose momentum has been completely depleted by adverse pressure

gradient.

Sunction

4. Blowing whereby an additional energy is provided to slow moving fluid by injecting high velocity

fluid.

5. Providing slot near the leading edge.

Body

Without slot

BodyBody

With Blot

6. By provided rotating cylinder near the leading edge which induce Magnus effect.


109

FM + OCF & MACHINES

1. Who introduced the concept of the theory of

boundary layer?

(A) Osborne Reynolds (B) Leonard Euler

(C) Sir lssac Newton (D) Ludwig Prandtl.

2. A boundary layer

(A) Is set up near any boundary, wall or

centre line

(B) Exists only near a solid boundary

(C) May or may not grow along the flow

direction

(D) May not form for some fluids.

3. The nominal thickness of boundary layer

represents the distance from the surface to a

point where

(A) Flow ceases to be laminar

(B) The shear stress becomes maximum

(C) Velocity is 99 per cent of its asymptotic

limit

(D) The flow behaves as it were rotational.

4. The displacement thickness 𝛿 . of a

boundary layer is defined as

(A) 𝛿 . = 𝑢

𝑈𝑜

𝛿

0 𝑑𝑦

(B) 𝛿 . = 1 −𝑢

𝑈0 𝑑𝑦

𝛿

𝑜

(C) 𝛿 . = 1 −𝑢

𝑈0 𝑑𝑦

𝛿

𝑜

(D) 𝛿 . = 𝑢

𝑈0 1 −

𝑢

𝑈0 𝑑𝑦

𝛿

𝑜

Where 𝑈𝑜 is the free stream velocity.

5. The displacement thickness for a boundary

layer

(A) Is the distance from the boundary affected

by velocity gradients

(B) Represents the mass deficit in flow

(C) Must be less than the momentum

thickness and greater than the boundary

layer thickness

(D) May be conceived as the transverse

distance by which boundary should be

displaced to compensate for reduction in

momentum on account of boundary layer

formation

6. The momentum thickness θ of a boundary

layer is defined as

(A) θ = 1 −𝑢

𝑈0

𝛿

𝑜dy

(B) θ = 1 −𝑢

𝑈0

2𝛿

𝑜 𝑑𝑦

(C) θ = 𝑢

𝑈0 1 −

𝑢

𝑈0

𝛿

𝑜dy

(D) θ = 𝑢

𝑈0 1 − 𝑢 U0 2

𝛿

𝑜dy

7. The shape factor H of the boundary layer is

defined as

(A) H =𝛿 ∙

θ (B) H =

𝛿 ∙

δ

(C) H =𝛿 ∙

δ ∙∙ (D) H =θ

δ ∙∙

Where 𝛿 ∙ is displacement thickness, θ is

momentum thickness 𝛿 ∙∙ is energy thickness,



110

FM + OCF & MACHINES

and 𝛿 is the nominal thickness of boundary

layer.

8. The velocity distribution in laminar boundary

layer is prescribed by the relation 𝑢

𝑈𝑜=

𝑦

𝛿. The

displacement thickness for this laminar

boundary layer would be

(A) 𝛿 (B) 𝛿 2

(C) 𝛿 3 (D) 𝛿 4

9. For a linear distribution of velocity in the

boundary layer on a flat plate, the ratio of

displacement thickness to nominal thickness is

(A) 1/2 (B) 1/3

(C) 1/4 (D) 2/3

10. The velocity distribution in laminar boundary

layer is presumed to conform to the identity

𝑢

𝑈𝑜=

𝑦

𝛿 . The momentum thickness for this

laminar boundary layer would be

(A) 𝛿/2 (B) 𝛿/3

(C) 𝛿/6 (D) 2𝛿/5

11. Which of the following velocity distributions

do not satisfy the boundary conditions for

boundary layer flow on a flat plate

(A) 𝑢

𝑈𝑜=

𝑦

𝛿

(B) 𝑢

𝑈𝑜=sin

𝜋

2

𝑦

𝛿

(C) 𝑢

𝑈𝑜= 𝑒𝑦/𝛿

(D) 𝑢

𝑈0=

3

2 (𝑦/𝛿) −

1

2(𝛾/𝛿)2

12. The velocity distribution in a laminar boundary

layer over a flat plate has been prescribed by

the identity 𝑢

𝑈𝑜= sin

𝜋𝐴𝑦

𝛿 .

The factor A will have an approximate value

of

(A) 1

2 (B) −

1

2

(C) 1.0 (D) 2.0

13. Momentum integral equation for zero pressure

gradient (𝜕𝑝/𝜕𝑥 = 𝑜) flow is given by

(A) 𝜏0

𝜌= 𝑈0

2 dθ

dx (B)

𝜏0

𝜌=

1

𝑈02

dθ

dx

(C) 𝜏0

𝜌= 𝑈0

dθ

dx (D)

𝜏0

𝜌= 𝑈0

𝑑𝜃

𝑑𝑥

2

Where 𝜏0 is the wall shear stress. 𝑈0 is the

undisturbed flow velocity and θ is the

momentum thickness.

14. The results for laminar boundary layer for flow

past a flat plate are based on a velocity

distribution which has

(A) Linear variation

(B) Parabolic variation

(C) Logarithmic variation

(D) One-seventh power law variation.

15. In a laminar boundary layer, the nominal

thickness varies with the longitudinal distance

x as

(A) x (B) x1/2

(C) x1/3

(D) x1/4

16. A smooth two- dimensional flat plate is

exposed to wind velocity of 100 m/s

(kinematic viscosity 0.15 stokes). If the


111

FM + OCF & MACHINES

transition Reynolds number is 2× 105, the

maximum distance from the leading edge upto

which laminar boundary layer exists is

(A) 3 mm (B) 3 cm

(C) 30 cm (D) 3 m

17. A laminar sub layer

(A) Develops on a hydraulically smooth

surfaces

(B) Is the boundary layer that occurs when the

transition from laminar to turbulent

boundary layer is imminent]

(C) A marrow region in the immediate

vicinity of the wall where flow is

essentially of laminar character while the

rest of flow is turbulent

(D) A narrow region near the wall in which

there is predominance of wall roughness

18. Which one of the following is a wrong

statement?

(A) The growth of boundary layer thickness

with the longitudinal distance on a flat

with the longitudinal distance on a flat

plate is faster in a turbulent boundary

layer when compared to that in laminar

boundary layer

(B) The boundary layer growth increases with

increase in kinematic viscosity only if the

boundary layer is turbulent

(C) Flow within the boundary layer is

rotational and shear stresses are present

(D) The laminar sublayer exists in all

turbulent boundary layers./

19. Sometimes and under certain conditions, the

boundary layer will leave the surface and curl

up into a vortex or whirl pool. This

phenomenon is known as

(A) Cavitation (B) Separation

(C) Drag (D) Wake

20. Boundary layer separation is caused by :

(A) Adverse pressure gradient

(B) Laminar flow changing to turbulent flow

(C) Reduction In pressure to vapour pressure

(D) Decrease in boundary layer thickness to a

negligible value

21. Separation of boundary layer must occur

when:

(A) 𝑑𝑝

𝑑𝑥< 𝑜

(B) 𝑑𝑝

𝑑𝑥= 𝑜

(C) 𝑑𝑝

𝑑𝑥> 𝑜

(D) 𝑑𝑝

𝑑𝑥> 𝑜 and the velocity profile has a point

of inflection.


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FM + OCF & MACHINES

1. D

2. A

3. C

4. B

5. B

6. C

7. A

8. B

9. A

10. C

11. C

12. A

13. A

14. B

15. B

16. B

17. C

18. B

19. B

20. A

21. D

[Sol] 4. Displacement Pickiness

⇒ 𝛿 = 1 −𝑢

𝑣 𝑑𝑦

𝛿

0

Momentum Thickness = 𝑈

𝑣0 1 −

𝑢

𝑣0 𝑑𝑦

𝛿

0

Energy Thickness ⇒ 𝑢

𝑣 1 −

𝑢2

𝑣2 𝑑𝑦 𝛿

0

[Sol] 7. Shape factor (H) = 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠

𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑡𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠

𝐻 =𝛿

𝜃

[Sol] 8.

𝜇

𝑈0=

𝑦

𝛿

Displacement thickness = 1 −𝑢

𝑉0 𝑑𝑦

𝛿

0

= 1 −𝑦

6 𝑑𝑦

𝛿

0

= 𝛿 −1

𝛿 𝑦2

2

0

𝛿

= 𝛿 −𝛿

2

𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑇𝑕𝑖𝑐𝑘𝑛𝑒𝑠𝑠 ⇒𝛿

2

[Sol] 9. Displacement Thickness

⇒ 1 −𝑢

𝑣 𝑑𝑦

𝛿

0

⇒ 1 −𝑦

𝛿 𝑑𝑦

𝛿

0

= 𝛿 =𝛿

2

Displacement Thickness ⇒ 𝛿

2

Ratio of Displacement Thickness &

nominal thickness :-

=𝛿/2

𝛿⇒

1

2

[Sol] 10. Momentum thickness : 𝑢

𝑣 1 −

𝑢

𝑣 𝑑𝑦

𝛿

0

⇒ 𝑦

𝛿 1 −

𝑦

𝛿 𝑑𝑦

𝛿

0

= 𝑦

𝛿 𝑑𝑦 −

𝑦2

𝛿2

𝛿

0

𝛿

0

=𝛿

2−

𝛿

3 =

𝛿

6

[Sol] 13. For momentum integral equation;

Von-Karman relation ;-

Explanations

Answer key


113

FM + OCF & MACHINES

𝜏0

𝜌𝑈2 =𝑑𝜃

𝑑𝑥

[Sol] 15. 𝛿𝐿𝑎𝑚𝑖𝑛𝑎𝑟 =5𝑥

𝑅𝑒𝑥

𝛿𝐿𝑎𝑚𝑖𝑛𝑎𝑟 ∝ 𝑥1/2

[Sol] 16. Wind velocity 𝜇 = 100 𝑚/𝑠

Kinematic viscosity 𝑣 = 0.15 𝑆𝑡𝑜𝑘𝑒𝑠

⇒ 0.15 × 10−4 𝑚2/𝑠𝑒𝑐

Reynolds Number = 2 × 105

𝑅𝑒 ⇒𝜌𝑉𝜃

𝜇 𝜗 = 𝜇/𝜌

𝑅𝑒 ⇒𝑉𝜃

𝑣

2 × 105 = 100 𝑥

0.15×10−4

𝑥 =0.15×10−4×2×105

100

𝑥 =20×0.15

100

𝑥 = 0.3𝑚

𝑥 = 30𝑐𝑚

[Sol] 23. Separation of boundary layer for adverse

pressure gradient : 𝑑𝜌

𝑑𝑥> 0


DRAG & LIFT

When a solid body moves through a fluid a resistance is encountered by body, conversely when fluid flow

passed a fluid encountered resistance. The resistance face remains same when the body moves through

fluid around the body. So long as relative motion.

Example –An aeroplane flies through atmosphere or submarines move through water. Propulsion unit has

to exert a force sufficient enough to resistance offered by fluid such a force is called drag force.

TYPES OF DRAG FORCE

DRAG FORCE

DRAG FORCE

Skin - friction or

Shear drag

Pressure or Formor

Shape drag

Wave Induced

Skin – friction or Shear drag

If the boundary layer completely laminar in nature only viscous effect contribute to shear stress. For

turbulent boundary layer stress are primarily from velocity gradient. This stress leads to a drag force

called shear drag or skin – friction drag.

Pressure shape or form drag

Pressure difference front & rare side produce a drag called pressure drag.

Wave drag

Motion of a boat on the surface of water set up waves corresponding drag is called wave drag.

Induced drag

It is a drag is lift force in an aerfoil to produce a drag force is called induce drag.

Body Diagram Shear drag Pressure

drag

Total drag

1. Thin plate placed

parallel to the direction

of flow

U

Shear drag

exist

negligible Shear drag +

negligible =

shear drag

Chapter

10 DRAG & LIFT

Syllabus: Drag force, Lift force, Magnus Effect. Weightage: 5%


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FM + OCF & MACHINES

2. Thin plate placed

perpendicular to the

direction of flow

Negligible Exist

Pressure drag

3. Cylinder with axis

perpendicular to the

flow direction

Negligible Exist

Pressure drag

4. sphere Negligible Exist Pressure drag

5. well streamline bodies Exist Negligible Shear drag

Force exerted on a flowing fluid on stationary object –

Two forces:-

1. Drag Force – Force in the direction of motion.

𝐹𝐷 = 𝐶𝐷 𝜌 𝑈2𝐴

2

𝐶𝐷 = Coefficient of drag

2. Lift Force – Force in the direction perpendicular to direction of motion.

𝐹𝐿 = 𝐶𝐿𝜌 𝑈2𝐴

2

𝐶𝐿 =Coefficient of lift.

Streamline body–Body whose surface coincide with the streamline when placed in flow.

Bluff body–Body whose surface does not coincide with streamline when placed in flow.

Terminal velocity–Maximum constant velocity of a falling body with which body will travel.

Magnus effect–When a cylinder is rotating in a uniform flow a lift force is produced in a cylinder which

is a Magnus effect drag is possibly exist done but lift cannot exist without drag.

Consider a body placed in the free stream of a real fluid. The fluid exerts a force „F‟ on the body. In

general, the force „F‟ may act in a direction inclined at an angle „θ‟ with free stream.


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FM + OCF & MACHINES

The component of the forces acting in the direction of free stream is called Drag force denoted by

D.

The component of the force acting in a div‟n at right angle of the direction of the free stream is

called Lift force denoted by L.

The coefficient of drag, CD is defined as

CD =D

1

2ρU2A

Drag force „D‟ is given by

D =CD .1

2ρU2A

The coefficient of Lift „CL‟ is defined as

CL =L

1

2ρU2A

Lift force „L‟ is given by

L = CL .1

2ρU2A

MAGNUS EFFECT:

The phenomena of Lift production by a rotating object placed in a free stream is known as the Magnus

effect.


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FM + OCF & MACHINES

1. In a laminar boundary layer over a flat plate,

the skin friction coefficient 𝑐𝑓𝑥 is given by

(A) 𝐶𝑓𝑥 =0.664

𝑅𝑒𝑥 (B) 𝐶𝑓𝑥 =

1.328

𝑅𝑒𝑥

(C) 𝐶𝑓𝑥 =0.874

𝑅𝑒𝑥 (D) 𝐶𝑓𝑥 =

1.912

𝑅𝑒𝑥 0.25

2. The drag force 𝐹𝐷on a plate is given by

(A) 𝐹𝐷 = 𝐶𝑝 ×1

2 ρ𝑈0

2

(B) 𝐹𝐷 = 𝐶𝑝 ×1

2 ρ𝑈0

2

(C) 𝐹𝐷 = 𝐶𝑝 ×1

2 ρ𝑈0

2𝐴

(D) 𝐹𝐷 = 𝐶𝑝 × ρ𝑈0𝐴

Where 𝑐𝑝 is the drag coefficient, U0 is the free

stream velocity and A is the area of the plate.

3. The separation of boundary layer can be

reduced by

(A) Use of smooth boundaries

(B) Suction of accelerating fluid within the

boundary layer

(C) Blowing high velocity fluid into the

retarded fluid and in the direction of flows

(D) Using large divergence angles.

4. A body is called a streamline body when:

(A) It is symmetrical about the axis along the

free stream

(B) It produces no drag for flow around it

(C) the flow is laminar around it

(D) The surface of the body coincides with

the streamlines

5. Drag is defined as the force exerted by a

flowing fluid on a solid body

(A) in the direction of flow

(B) perpendicular to the direction of flow

(C) in the direction which is at an angle of

45°to the direction of flow


6. Lift force is defined as the force exerted by a

flowing fluid on a solid body

(A) in the direction of flow

(B) perpendicular to the direction of flow

(C) t an angle of 45° to the direction of flow


7. Lift for (FL) is expressed mathematically as

(A) 𝐹𝐿 =1

2𝜌𝑈2×𝐶𝐿

(B) 𝐹𝐿 =1

2𝜌𝑈2×𝐶𝐿 × 𝐴

(C) 𝐹𝐿 = 2𝜌𝑈2×𝐶𝐿 × 𝐴

(D) 𝐹𝐿 = 𝜌𝑈2×𝐶𝐿 × 𝐴

8. Total drag on a body is the sum of

(A) pressure drag and velocity drag

(B) pressure drag and friction drag

(C) friction drag and velocity drag




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FM + OCF & MACHINES

9. Terminal velocity of a falling body is equal to

(A) the maximum velocity with which body

will fall.

(B) the maximum constant velocity with

which body will fall.

(C) half of the maximum velocity


10. When a falling body has attained terminal

velocity, the weight of the body is equal to

(A) drag force minus buoyant force

(B) buoyant force minus drag force

(C) drag force plus the buoyant force


1. A

2. C

3. C

4. D

5. A

6. B

7. B

8. B

9. B

10. B

Answer key


TURBINES

INTRODUCTION

These are those machines which convert hydraulic energy into mechanical energy & mechanical energy

into hydraulic energy is called hydraulic energy.

TURBINES –

Turbines are those machines which convert hydraulic energy into mechanical energy. This mechanical

energy used to running electric generator. Hence mechanical energy is converted into electric energy.

Electric power so obtained from hydraulic energy is known hydro electric power.

Sugar Tank

Penstock

Turbine

Nozzle

Tail Race

Dam

Head race

Hg Reservoir

H

hf

Gross head–Difference between head race & tail race.

Net head or effective head (H)–

Head available at the inlet of turbines with the help of artificial reservoir is made by constructing a dam

across a river pipe of large diameter usually made of steel or reinforced concrete are employed for

carrying water under high pressure from reservoir to turbine is called penstock.

Water level in reservoir is called head race. Water level after discharging from turbine is called tail race.

Surge tank–

It is a storage reservoir fitted at some location on the penstock to receive excess flow moving towards

turbine. When there is a sudden reduction in flow. It serve the purpose of supply reserve water to turbine.

Chapter

11 HYDRAULIC

MACHINES Syllabus: Turbines, Pelton turbine, Francis turbine, Kaplan turbine, Draft

tube, Efficiencies, Specific speed. Pumps: Centrifugal pump,

Reciprocating Pump Weightage: 20%


120

FM + OCF & MACHINES

Order–Reservoir → penstock → surge tank → turbine.

CLASSIFICATION OF TURBINES

1. On the bases of method of energy conversion –

a) Impulse Turbine –

i. All energy is converted into kinematic energy means of nozzle at the end of penstock.

ii. Install above the tail race.

Example – Pelton wheel.

b) Reaction Turbine –

i. Only a part of available energy is converted into kinematic energy.

ii. Installed completely submerged in water.

Example – Francis turbine, Propeller turbine, Kaplan turbine.

2. According to direction of flow –

According to direction of flow

Tangential Flow Radial Flow Arial Flowor

Parallel flow

Inward Outward

Mixed Flow

Tangential flow turbine–Water flow along tangentially to runner. Example – Pelton wheel turbine.

Radial flow turbine–If water flow radially to runner.

Radial Flow Turbine

Inward flow Outward Flow

If flow from outward to inwardExample - Francis Turbine

If flow from inward to outwardExample - Fourneyorn

Axial flow turbine -

Example–Keplan & propeller water flow in parallel direction of rotation.

Mixed flow turbine–Flow initially moves through radial reaction but at outlet direction parallel to the

axis of rotation.

Example – Modern Francis turbine.

HYDRAULIC MACHINE


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FM + OCF & MACHINES

These are those hydraulic machine which convert mechanical energy into hydraulic energy in the form of

pressure energy.

PUMP

Centrifugal Pump or

Rota dynamic Pump

Reciprocating Pumpor

Displacement Pump

Centrifugal Pump–

If mechanical energy is converted into pressure energy by means of centrifugal force acting on fluid is

called centrifugal pump.

C.f act as reverse of an inward radial flow reaction turbine means working on radial outward direction.

(i) Impeller – Rotating part

(ii) Casing – Air tight passing surrounded the impeller & design in such a way that kinematic energy of

water at outlet is converted in pressure energy.

(iii) Sunction head – :

Suctionhead Hs

Free surface

HdDelivery head

Total heador

static headH = H + Hs d

Sunction head – Vertical height of centre line of C.f pump & free surface of water in lower tank.

Delivery head – Vertical distance between centre line of pump & free surface in upper tank.

Static head or total head – Hs + Hd

CLASSIFICATION OF PUMP BASED ON HEAD–

1. Low lift pump – head upto 15 m

2. High lift pump – head above 15 m

Specific speed of pump– 𝑁𝑠 =𝑁 𝑃

𝐻5 4


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FM + OCF & MACHINES

Priming of pump–Process of filling the impeller & suction pipe before staring the pump in order to

remove air & gas from the pump

Example – Centrifugal pump.

Multistaging of Centrifugal Pump

Impeller in series arrangement

Single impeller then series arrangement is adopted.

Impeller on the parallel arrangement–Parallel arrangement is adopted when high discharge is adopted.

Q 1Q nQ

It creates the lift & pressure by displaying liquid with moving member called plunger which is fitted

inside a cylinder arrangement.


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FM + OCF & MACHINES

Discontinuous Source

Fluid Machinery

Reciprocating Pump

Main Parts Of A Reciprocating Pump

Reciprocating pump are used to lift water against high head at lam discharge

Slip = Qth – Q actual

Negative slip is equal to the difference of theoretical discharge and actual discharge is move than the

theoretical discharge, the slip of the bump will become –ve. It occurs when

Delivery pipe is short

suction pipe is long


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FM + OCF & MACHINES

pump running at high speed.

Point to be Remember

When the pump is in series →use for high head

when the pump is in Parallel → used for high discharge

For Pelton Wheel

Jet Ratio (m) = Pi tch diameter of the Pelton Wheel

Diameter of Jet

M= D

d

No of Vanes = 15+0.5 m ≈18+25

Known as Tygan formula.

No. of jet = Total Discharge

Discharge of one Jet=

Q

q


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FM + OCF & MACHINES

1. In a reaction turbine

(A) flow can be regulated without loss

(B) there is only partial conversion available

head to velocity head before entry to

runner

(C) the outlet must be above the tail race

(D) water may be allowed to enter a part or

whole of wheel circumference

2. Indicate the wrong statement with respect to

reaction water turbine

(A) the water leaves the turbine at

atmospheric pressure

(B) the guide vanes direct the flow at proper

angle

(C) the spiral casing serves to uniformly

distribute water into guide blades

(D) the draft tube allows setting of turbine

above tail race with minimum reduction

in available energy

3. The installation of a draft tube in a reaction

turbine helps to

(A) increase the flow rate

(B) prevent air from entering

(C) transport water to downstream without

eddies

(D) convert the kinetic energy to pressure

head.

4. Compared to cylindrical draft tube, a tapered

draft tube

(A) prevents hammer blow and surges

(B) responds better to load fluctuations

(C) converts more of kinetic head into

pressure head

(D) prevents cavitation even under reduced

discharges

5. In a Francis turbine, maximum efficiency is

obtained when

(A) relative velocity is radial at the outlet

(B) absolute velocity is radial at the outlet

velocity of flow is constant

(C) velocity of flow is constant

(D) guide vane angle is 90 degree

(E) blade tip is radial at the outlet

6. In practice, the flow ratio of a Francis turbine

is found to lie in the range

(A) 0.15 to 0.3 (B) 0.42 to 0.46

(C) 0.55 to 0.65 (D) 0.7 to 0.85

7. The modern Francis turbine is essentially

(A) a tangential flow turbine

(B) a mixed flow turbine

(C) an axial flow turbine

(D) a radial flow turbine

8. A Kaplan turbine is a

(A) low head axial flow turbine



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FM + OCF & MACHINES

(B) high head axial flow turbine

(C) high head mixed flow turbine

(D) inward flow impulse turbine

9. A Kaplan turbine is suitable for

(A) low head and low discharge

(B) low head and high discharge

(C) high head and low discharge

(D) high head and high discharge

10. An adjustable blade propeller turbine is called

(A) Pelton turbine

(B) Banki turbine

(C) Kaplan turbine

(D) Francis-Pelton turbine

11. The runner vanes of a reaction turbine are

made adjustable, as a Kaplan turbine, to

(A) reduce the wear and tear of the runner

(B) allow running at different speeds of

rotation

(C) operate the machine at optimum

efficiency at part load conditions

(D) permit the machine to operate under

varying conditions of pressure and

discharge

12. Which amongst the following is false with

respect to Kaplan turbine

(A) fit has blades of small camber to prevent

separation

(B) it employs large guide vane angles than

those in Francis turbine

(C) it can adjust both guide vane and blade

angles according to rate of discharge

(D) it is designed for flow velocity of the

mixed flow type.

13. The value of speed ratio for a Kaplan turbine

is about

(A) 0.5 (B) 0.9

(C) 1.5 (D) 2.0

14. Usually the ratio of hub-diameter to the

outside diameter of the runner of a Kaplan is

between

(A) 0.15 to 0.3 (B) 0.4 to 0.6

(C) 0.7 to 0.85 (D) 0.1 to 0.2

15. Run away speed of a hydraulic turbine is the

speed

(A) corresponding to maximum overload

permissible

(B) at full load

(C) at which there would be no damage to the

turbine runner

(D) at which the turbine runner can be

allowed to run freely without load and

with wicket gates wide open.

16. Critical speed of a turbine is

(A) same as run away speed

(B) speed that will cause mechanical failure

of the shaft

(C) speed at which natural frequency of

vibrations equals the number of

revolutions in the same time

(D) speed equal to synchronous speed of the

generator


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FM + OCF & MACHINES

17. Specific speed of a turbo machine

(A) is the speed of a machine having unit

dimensions

(B) related the shape rather than size of the

machine

(C) remains the same under different

conditions of operation

(D) depends only upon the head under which

the machine operates

18. The specific speed of a water turbine is

expressed as

(A) 𝑁𝑠 =𝑁 𝑃

𝐻5/4 (B) 𝑁𝑠 =𝑁 𝑃/𝜌

𝐻5/4

(C) 𝑁𝑠 =𝑁 𝑃

𝑔𝐻 5/4 (D) 𝑁𝑠 =𝑁 𝑃/𝜌

𝑔𝐻 5/4

19. Specific speed of an impulse turbine (Pelton

wheel) ranges from

(A) 10 – 20 (B) 25 – 40

(C) 60 – 250 (D) 300 – 800

20. Francis turbines are available in the following

range of specific speeds

(A) 1 – 40 (B) 50 – 250

(C) 150 – 500 (D) 250 – 850

21. High specific speed (250 – 850) and low

heads (below 30 m) indicate that the turbine is

(A) Pelton wheel (B) Francis

(C) Kaplan (D) Propellor

22. Suggest the turbine that works under a head of

30 m, runs at 400 rpm and develops 15 MW

power

(A) Pelton (B) Kaplan

(C) Francis (D) Propeller

23. Francis, Kaplan and propeller turbines fall

under the category of

(A) impulse turbine

(B) reaction turbine

(C) impulse – reaction combined

(D) axial flow

24. The degree of reaction of a Kaplan turbine is

(A) equal to zero

(B) greater than zero but less than 1/2

(C) greater than ½ but less than 1

(D) equal to 1

25. Mark the false statement:

(A) for a Pelton wheel, degree of reaction R = 0

(B) for the Francis turbine 0 < R < 1

(C) for the Kaplan turbine 0.5 < R < 1

(D) the specific speed decreases if the degree

of reaction in a turbine is increase

26. All of the following statements are correct,

except

(A) a slow runner turbine has a degree of

reaction greater than 50 percent

(B) for a 100 percent reaction turbine, the

energy transfer becomes zero

(C) in decreasing order of specific speed, the

water turbines can be put as propeller

turbine, Francis turbine and Pelton wheel

(D) a bulb or tubular turbine belongs to the

category of Kaplan turbine.

27. The energy conversion process in the outward

radial flow turbine is:

(A) purely by impulse only

(B) purely by reaction only


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FM + OCF & MACHINES

(C) may be by impulse or reaction

(D) partly by impulse and partly by reaction

28. Cavitations in a hydraulic turbine may occur

in all likelihood

(A) in the spiral casing

(B) in the guide passages

(C) at the inlet to runner

(D) at the inlet to draft tube

29. Amongst the following, which turbine is least

efficient under part load conditions?


(C) Kaplan (D) Propeller

30. Indicate the turbine that is most efficient at

part load operation.


(C) Kaplan (D) Propeller

31. Which of the following turbine has not only a

high design efficiency but that efficiency

practically remains constant too over a wide

range of regulation from the design condition?

(A) Pelton wheel (B) Francis turbine

(C) Kaplan turbine (D) Tubular turbine

32. Operating characteristic curves of a turbine

means the curves drawn at constant

(A) head (B) discharge

(C) speed (D) efficiency

33. Which one of the followings is a wrong

statement?

(A) only the tangential component of absolute

velocity is considered in the estimation of

theoretical head of a turbo machine

(B) a high head turbine has a high value of

specific speed

(C) for the same power, a turbo machine

running at high specific speed will be

small in size

(D) Pelton wheel is the tangential flow turbine

whereas the Propeller and Kaplan turbines

are axial flow units.

34. A dimensionless form of specific speed,

known as shape number, for a hydraulic

turbine may be written as

(A) 𝑁 𝑃

𝐻5/4 (B) 𝑁 𝑃/𝜌

𝐻5/4

(C) 𝑁 𝑃 𝜌𝑔

𝐻5/4 (D) 𝑁 𝑃

𝜌 𝑔𝐻 5/4

35. The function of surge tank is to

(A) avoid reversal of flow

(B) relieve the pipeline of excessive pressure

transients

(C) act as a reservoir for emergency

conditions

(D) prevent occurrence of hydraulic jump

36. A surge tank is provided to protect the

(A) turbine runner (B) spiral casing

(C) draft tube (D) penstock

37. In general, the vanes of a centrifugal pump are

(A) curved forward (B) curved backward

(C) radial (D) twisted


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FM + OCF & MACHINES

38. An impeller with backward curved vanes

(A) is easier to fabricate

(B) cannot run a speeds other than design

speed

(C) has a falling head discharge characteristic

(D) has greater absolute velocity at outlet than

that with forward curved vanes

39. A fast centrifugal pump impeller will have

(A) forward facing blades

(B) radial blades

(C) backward facing blades

(D) propeller type blades

40. In a centrifugal pump, the inlet angle will be

designed to have

(A) relative velocity vector in the radial

direction

(B) absolute velocity vector in the radial

direction

(C) velocity of flow to be zero

(D) peripheral velocity to be zero

41. The flow in the volute casing outside the

rotating impeller of a centrifugal pump is

(A) radial flow (B) axial flow

(C) free vortex flow (D) forced vortex flow

42. Centrifugal pumps dealing with muds have an

impeller of the type

(A) open (B) double suction

(C) one-side shrouded (D) two-sides shrouded

43. The specific speed of a pump is defined as the

speed of unit of such a size that it

(A) requires unit power to develop unit head

(B) delivers unit discharge at unit power

(C) delivers unit discharge at u..it head

(D) produces unit power with unit head

available.

44. The specific speed of centrifugal pump is

given by

(A) 𝑁 𝑄

𝐻3/4 (B) 𝑁 𝑃

𝐻3/4

(C) 𝑁 𝑄

𝐻5/4 (D) 𝑁 𝐻

𝑄3/4

45. The specific speed 𝑁𝑠 = 𝑁 𝑄 𝐻3/4 for a

double suction pump is to be evaluated. The

discharge would be taken as

(A) half the actual discharge

(B) actual discharge

(C) double the actual discharge

(D) square of the actual discharge

46. Higher specific speeds (160 to 500) of

centrifugal pump indicate that the pump is of

the type

(A) axial flow (B) radial flow

(C) mixed flow (D) any one of these types

47. For a given centrifugal pump.

(A) head varies inversely as square of speed

(B) discharge varies directly as speed

(C) discharge varies directly as square of

speed

(D) power varies directly as fifth power of

speed


130

FM + OCF & MACHINES

48. If the diameter of a centrifugal pump impeller

is doubled but the discharge is to remain same,

then the head needs to be reduced by

(A) 2 times (B) 4 times

(C) 8 times (D) 16 times

49. For a centrifugal pump, the cavitation

parameter 𝜍 is defined in terms of 𝑁𝑃𝑆𝐻 (net

positive suction hea(D) and net head 𝐻 as, 𝜍 =

(A) 𝑁𝑃𝑆𝐻

𝐻 (B)

𝑁𝑃𝑆𝐻

𝐻

(C) 𝑁𝑃𝑆𝐻

𝐻3/2 (D) 𝑁𝑃𝑆𝐻

𝐻2

50. Cavitation in centrifuganl pumps can be

reduced by

(A) Reducing the discharge

(B) Reducing the suction head

(C) Throuttling the discharge

(D) Increasing the flow velocity

1 B

2 A

3 D

4 C

5 B

6 A

7 B

8 A

9 B

10 C

11 C

12 D

13 C

14 B

15 D

16 C

17 B

18 A

19 A &

B

20 B

21 C

22 B

23 B

24 C

25 D

26 A

27 C

28 D

29 D

30 C

31 C

32 A

33 B

34 B

35 B

36 D

37 B

38 C

39 C

40 B

41 C

42 A

43 C

44 A

45 A

46 A

47 B

48 D

49 B

50 B

51

Answer key


131

FM + OCF & MACHINES

[Sol] 3. Draft tube is a tapered draft tube

having increasing diameter used

to convert KE to pressure head.

[Sol] 7. The modern francis turbine is a

mixed flow turbine.

[Sol] 8. Kaplan turbine is low head axial

flow turbine, which is suitable

for low head 4 high discharge

[Sol] 9. Kaplan turbine have adjustable

blade.

[Sol] 18. Specific speed of turbine

𝑁𝑠 =𝑁 𝑃

𝐻5/4

[Sol] 22. Specific speed of turbine

𝑁𝑠 =𝑁 𝑃

𝐻5/4

𝑁𝑠 =400 15000

30 5/4

𝑁𝑠 = 700

So; turbine is Kaplan turbine

[Sol] 24. RD=1 for Kaplan turbine, because it is

reaction turbine.

[Sol] 35. A surge tank is a reservoir that would

discharge the water when turbine is

started & filled the reservoir for

emergency.

[Sol] 37. For backward curved vanes, these is

maximum efficiency of Pump, so curved

backward vanes are used.

[𝑺𝒐𝒍] 𝟒𝟏. In impeller of centrifugal pump, the

mechanism is free vortex flow.

[Sol] 44. 𝑁𝑠 𝑝𝑢𝑚𝑝 =𝑁 𝑄

𝐻3/4

[Sol] 49. Cavitation No. ⇒𝑁𝑃𝑆𝐻

𝐻

Explanations

Chapter FLUID PROPERTIES 1 - Eduzphere

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Transcript of Chapter FLUID PROPERTIES 1 - Eduzphere