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SATHYABAMA UNIVERSITY School of Mechanical Engineering SME1202 Fluid Mechanics and Machinery

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UNTT 1 FLUID PROPERTIES

Fluid Properties: Density, Specific weight ,Specific gravity, Viscosity ,Surface tension, Capillarity,

Vapour pressure and compressibility.

Fluid Statics: Hydrostatic Law - Pressure Variation in static fluid- Hydrostatic force on submerged

plane surfaces - Location of hydrostatic force.

Manometers- Simple U tube and differential manometers.

Buoyancy - Metacentric height; determination of stability of floating bodies and submerged bodies.

Fluids: Substances capable of flowing are known as fluids. Flow is the continuous deformation of

substances under the action of shear stresses.

Fluids have no definite shape of their own, but confirm to the shape of the containing vessel. Fluids

include liquids and gases.

Fluid Mechanics:

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

motion. Thus,it deals with the static, kinematics and dynamic aspects of fluids.

The study of fluids at rest is called fluid statics. The study of fluids in motion, where pressure

forces are not considered, is called fluid kinematics and if the pressure forces are also considered for

the fluids in motion, that branch of science is called fluid dynamics.

Fluid Properties:

1.DENSITY (or )MASS DENSITY:

Density or mass density of a fluid is defined as the ratio of the mass of the fluid to its volume.

Thus, Mass per unit volume of a fluid is called density.

fluidofVolume

fluidofMassdensityMass ,

S.I unit of density is kg/m3.

The value of density for water is 1000 kg/m3.

2.SPECIFIC WEIGHT (or) WEIGHT DENSITY (w ):

Specific weight or weight density of a fluid is the ratio between the weight of a fluid to its

volume.

The weight per unit volume of a fluid is called specific weight or weight density.

fluidofVolume

fluidofWeightdensityWeight

fluidofVolume

gXfluidofMassw

gw

S.I unit of specific weight is N/m3.

The value of specific weight or weight density of water is 9810N/m3 or 9.81 kN/m³.

3. SPECIFIC VOLUME (ʋ):


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Specific volume of a fluid is defined as the volume of a fluid occupied by unit mass.

Volume per unit mass of a fluid is called Specific volume.

1

fluidofMass

fluidaofVolumevolumeSpecific

Thus specific volume is the reciprocal of mass density. S.I unit: m3/kg.

4. SPECIFIC GRAVITY (s):

Specific gravity is defined as the ratio of the specific weight of a fluid to the specific weight

of a standard fluid.

waterof weightSpecific

liquidof weightSpecificgravitySpecific

Specific gravity is also equal to Relative density. Relative density = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟

5. VISCOSITY:

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

layer of fluid over adjacent layer of the fluid.

When two layers of a fluid, at distance ‘dy’ apart, move

one over the other at different velocities, say u and u+du as

shown in figure. The viscosity together with relative velocity

causes a shear stress acting between the fluid layers.

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

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

layer.

This shear stress is proportional to the rate of change of velocity

with respect to y.

dy

du

or dy

du

where, μ is the constant of proportionality and is known as the co-efficient of dynamic viscosity or

viscosity and dy

du represents the rate of shear strain or rate of shear deformation or velocity gradient.

dy

du

Thus the viscosity is also defined as the shear stress required to produce unit rate of shear

strain.

S.I unit: Ns/m². It is still expressed in poise (P) as well as centipoises (cP).

;

Kinematic Viscosity(ν):

It is defined as the ratio between the dynamic viscosity and density of the fluid

ν =.𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦

𝐷𝑒𝑛𝑠𝑖𝑡𝑦 =

𝜇

𝜌


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SI unit: m2/s; CGS unit ‘stoke’. 1 stoke = 1 cm2/ sec = 10-4 m2/s

Newton's Law of Viscosity:

It states that the shear stress (τ) on a fluid element layer is directly proportional to the rate of shear

strain. The constant of proportionality is called the co-efficient of viscosity.

dy

du

6.COMPRESSIBILITY:

Compressibility is the reciprocal of the bulk modulus of elasticity, K, which is defined as the

ratio of compressive stress to volumetric strain.

Compression of fluids give rise to pressure with the decrease in volume.

If dv is the decrease in volume and dp is the increase in pressure, Volumetric Strain = V

dV

(- ve sign indicate the volume decreases with increase of pressure)

Bulk modulus K = StrainVolumetric

pressureofIncrease

=

V

dV

dp

Compressibility = K

1

7.SURFACE TENSION:

Surface tension is defined as the tensile force acting on the surface of a liquid in contact with a

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

membrane under tension.

Surface Tension on Liquid Droplet:

Consider a small spherical droplet of a liquid of diameter ‘d’. On the entire surface of the droplet,

the tensile force due to surface tension will be acting.

Let σ = Surface tension of the liquid

p = Pressure intensity inside the droplet (in excess of the outside pressure intensity)

d = Dia. of droplet.

Let the droplet is cut into two halves. The forces acting on one half will be


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i) Tensile force (FT)due to surface tension acting around the circumference of the cut portion as

shown in fig. and this is equal to = σ x Circumference = σ x π d

ii) Pressure force (Fp) on the area 𝜋𝑑2

4 is = p x

𝜋𝑑2

4 as shown in the figure.

These two forces are equal under equilibrium conditions. i.e.,

p x 𝜋𝑑2

4 = σ x π d

Therefore, 𝑝 = 4𝜎

𝑑

Surface Tension on a Hollow Bubble:

A hollow bubble like a soap bubble in air has two surfaces in contact with air, one inside and other

outside. Thus two surfaces arc subjected to surface tension.

In that case,

p x 𝜋𝑑2

4 = 2 x (σ x π d)

Therefore, 𝑝 = 8𝜎

𝑑

8. CAPILLARITY:

Capillarity is defined as a phenomenon of rise or fall of a liquid surface in a small tube relative to the

adjacent general level of liquid when the tube is held vertically in the liquid.

The rise of liquid surface is known as capillary rise while the fall of the liquid surface is known as

capillary depression. It is expressed in terms of cm or mm of liquid. Its value depends upon the

specific weight of the liquid, diameter of the tube and surface tension of the liquid.

Expression for Capillary Rise:

Consider a glass tube of small diameter ‘d' opened at both ends and is inserted in a liquid. The liquid

will rise in the lube above the level of the liquid.

Let, h = height of the liquid in the tube. Under

a state of equilibrium,

the weight of liquid of height h is balanced by

the force at the surface of

the liquid in the tube. But the force at the

surface of the liquid in the tube

is due to surface tension.

Let, σ = Surface tension of liquid

θ= Angle of contact between liquid and glass

lube.

The weight of liquid of height ‘h’ in the tube = (Area of tube x h) x ρ x g

where, ρ = density of liquid

Vertical component of the surface tensile force

= σ x Circumference x cos θ

= σ x πd x cos θ

Weight of liquid of height ‘h’ in the tube = Vertical component of the surface tensile force


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or h = 𝟒𝝈 𝒄𝒐𝒔ɵ

𝒘 𝒅

9.VAPOUR PRESSURE:

Vapour pressure is the pressure of the vapor over a liquid which is confined in a closed vessel at

equilibrium.

Vapour pressure increases with temperature.All liquids exhibit this phenomenon.

Types of fluid:

i. Ideal Fluid: A fluid, which is

incompressible and is having no viscosity, is

known as an ideal fluid. Ideal fluid is only an

imaginary fluid as all the fluids, which exist,

have some viscosity.

ii. Real Fluid: A fluid, which possesses

viscosity, is known as real fluid. All the fluids,

are real fluids in actual practice.

iii. Newtonian Fluid: A real fluid, in which the shear stress is directly proportional to the rate of

shear strain (or) velocity gradient, is known as a Newtonian fluid.

iv. Non-Newtonian Fluid: A real fluid, in which the shear stress is not proportional to the rate of

shear strain (or) velocity gradient, is known as a Non-Newtonian fluid.

v. Ideal Plastic Fluid: A fluid, in which shear stress is more than the yield value and shear stress is

proportional to the rate of shear strain (or) velocity gradient, is known as ideal plastic fluid

Fluid Pressure

Fluids exert pressure on surfaces with which they are in contact.

Fluid pressure is the force exerted by the fluid per unit area. Fluid pressure is transmitted with equal

intensity in all directions and acts normal to any plane.

In the same horizontal plane the pressure intensities in a liquid are equal.

Fluid pressure or Intensity of pressure or pressure, 𝑝 =𝐹

𝐴

S.I unit of fluid pressure are N/m² or Pa, where 1 N/m² = 1 Pa.

Many other pressure units are commonly used:

1 bar = 105 N/m²

1 atmosphere = 101325 N/m² = 101.325kN/m²

Some Terms commonly used in static pressure analysis include:

Pressure Head: The pressure intensity exerted at the base of a column of homogenous fluid of a

given height in metres.

Vacuum: A perfect vacuum is a completely empty space in which, therefore the pressure is zero.

Atmospheric Pressure: The pressure at the surface of the earth exerted by the head of air above the

surface.

At sea level the atmospheric pressure = 101.325 kN/m² = 101325 N/m² or pa


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= 1.01325 bar

= 760 mm of mercury

= 10.336 m of water

Atmospheric pressure is measured by a device called a barometer; thus, the atmospheric pressure is

often referred to as the barometric pressure.

Gauge Pressure: The pressure measured by a pressure gauge above or below atmospheric pressure.

Vacuum pressure: The gauge pressure less than atmospheric is called Vacuum pressure or negative

pressure.

Absolute Pressure: The pressure measured above absolute zero or vacuum.

Absolute Pressure = Atmospheric Pressure + Gauge Pressure

Absolute Pressure = Atmospheric Pressure – Vacuum pressure

Atmospheric, Gauge & Absolute pressure

Hydrostatic law

The hydrostatic law is a principle that identifies the amount of pressure exerted at a

specific point in a given area of fluid.

It states that, “The rate of increase of pressure in the vertically downward direction, at a point

in a static fluid, must be equal to the specific weight of the fluid.”

Pressure Variation in static fluid

Consider a small vertical cylinder of static fluid in equilibrium.

SATHYABAMA UNIVERSITY School of Mechanical Engineering SME1202 Fluid Mechanics and

Machinery

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Assume that the sectional area is dA and the pressure acting upward on the bottom surface is

p and the pressure acting downward on the upper surface (dz above bottom surface) is

(p + dp)dz.

Let the free surface of the fluid be the origin, i.e., Z = 0. Then the pressure variation at a

depth Z = - h below the free surface is

governed by

(p + dp) A + W = pA

dpA + ρgAdz = 0

dp = -ρgdz 𝑑𝑝

𝑑𝑧 = - ρg = - w

Therefore, the hydrostatic pressure increases linearly with depth at the rate of the specific weight,

w = ρg of the fluid.

If fluid is homogeneous, ρ is constant.

By simply integrating the above equation,

ʃdp = - ʃρg dz => p = - ρg Z + C

Where C is constant of integration

When z = 0 (on the free surface), p = C = po = the atmospheric pressure.

Hence, p = - ρgZ + po

Pressure given by this equation is called ABSOLUTE PRESSURE, i.e., measured above perfect

vacuum.

However, it is more convenient to measure the pressure as gauge pressure by setting atmospheric

pressure as datum pressure. By setting po = 0,

p = -ρgz+0 = -ρgz=ρgh

p = wh

The equation derived above shows that when the density is constant, the pressure in a liquid at rest

increases linearly with depth from the free surface.

The above expression can be rearranged as, h = 𝒑

𝒘

Here, h is known as pressure head or simply head of fluid.

In fluid mechanics, fluid pressure is usually expressed in height of fluids or head of fluids.

Hydrostatic force

Hydrostatic pressure is the force exerted by a static fluid on a plane surface, when the static

fluid comes in contact with the surface. This force will act normal to the surface. It is also known as

Total Pressure.

The point of application of the hydrostatic or total pressure on the surface is known as Centre of

pressure.

The vertical distance between the free surface of fluid and the centre of pressure is called depth of

centre of pressure or location of hydrostatic force.


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Total Pressure on a Horizontal Immersed Surface

Consider a plane horizontal surface immersed in a liquid as shown in figure.

Let, w = Specific weight of the liquid, kN/m³

A = Area of the immersed surface in m²

x = Depth of the horizontal surface from the liquid level in m

We know that,

Total pressure on the surface, P = Weight of the liquid above the immersed surface

P = Specific weight of liquid x Volume of liquid

= Specific weight of liquid x Area of surface x Depth of liquid

P = wA x kN

Total Pressure and depth of centre of pressure on a Vertically Immersed Surface

Consider an irregular plane vertical surface immersed in a liquid as shown in figure .

Let,

w = Specific weight of liquid

A = Total area of the immersed surface

x = Depth of the center of gravity of the immersed surface from the liquid surface

Now. consider a strip of width ‘b’, thickness ‘dx’ and at a depth x from the free surface of the liquid


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Moment of pressure on the strip about the free surface of liquid = x b dx X x = x² b dx

Total moment on the entire plane immersed surface = ∫ x² b dx

M = ∫ 𝑥² 𝑏 𝑑𝑥

But, ∫ 𝑥² 𝑏 𝑑𝑥 = second moment of area about free liquid surface = Io

therefore, M = Io

Io = IG + A x², according to parallel axis theorem.

Therefore, M = (IG + A x²) ------------------------------ (1)

Also M = P x h = Ax x h ---------------------------------- (2)

Since equations 1 & 2 are equal,

Ax x h = (IG + A x²)

Depth of centre of pressure, h = (IG + A x²) / Ax

Therefore, h = x + 𝐼𝐺

A𝑥

Total Pressure and depth of Centre of Pressure on an Inclined Immersed Surface

Consider a plane inclined surface, immersed in a liquid as shown in figure. Let,

w = Specific weight of the liquid

A = Total area of the immersed surface

x = Depth of the centroid of the immersed plane surface from the free surface of liquid.

θ = Angle at which the immersed surface is inclined with the liquid surface

h = depth of centre of pressure from the liquid surface

b = width of the considered thin strip

dx = thickness of the strip

O = the reference point obtained by projecting the plane surface with the free surface of liquid

l = distance of the strip from O

Area of the strip = b dx

Pressure of liquid on the strip = w x sinθ

For Vertically immersed plane surfaces,

Total pressure, P = A x and Depth of centre of pressure, h = x + 𝐼𝐺

𝐴𝑥


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Total pressure on the strip = w x sinθ X b dx

Total pressure on the entire area of plane surface, P = ∫𝑤 𝑥 sinθ X b dx

P = w sinθ ∫ 𝑥. 𝑏𝑑𝑥, and we know from parallel axis theorem, ∫ 𝑥. 𝑏𝑑𝑥 = 𝐴 x

sin 𝜃

Therefore, P = wAx Moment of pressure on the strip about O = wl sinθ b dx x x = wx² sinθ b dx

Total moment of pressure on the entire area = ∫𝑤 𝑥² sin 𝜃 𝑏 𝑑𝑥 = 𝑤 sin 𝜃 ∫ 𝑥2 𝑏 𝑑𝑥

But, ∫ 𝑥2 b dx = second moment of area about O-Oʹ = Io

oo o M.I = 𝑤 sin 𝜃 Io --------------------------------------- (1)

but, Io = IG + A 𝑥 ²

sin²𝜃

Also,M.I = P.y; y = h

𝑠𝑖𝑛𝜃

ooo M.I = P.

h

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

Equating 1 & 2, P.h = 𝑤 sin² 𝜃 Io

P.h = 𝑤 sin² 𝜃 [IG + A 𝑥 ²

sin²𝜃]

ooo h = 𝑤 sin² 𝜃 [IG + A x²]/ P =

𝑤sin²𝜃 [IG + A 𝑥 ²

sin²𝜃]

𝑤𝐴x =

𝐼𝐺

𝐴

sin²𝜃

x + x

or h = x + 𝐼𝐺 sin²𝜃

𝐴x

M.I and Geometric Properties of some plane surfaces

For inclined immersed surfaces,

Total pressure, P = A x

Depth of centre of pressure , h = 𝐼𝐺𝑠𝑖𝑛

2 ɵ

Ax + x


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Pascal's law

Pascal’s law states that “Intensity of pressure at a point in a fluid at rest is same in all directions”. The basic property of a static fluid is pressure.

Pressure is the surface force exerted by a fluid against the walls of its container.

Pressure also exists at every point within a volume of fluid.

For a static fluid, as shown by the following analysis, pressure turns to be independent direction.


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Consider a triangular prism of small fluid element ABCDEF in equilibrium. Let Px is the

intensity of pressure in the X direction acting at right angle on the face ABFE, Py is the intensity of

pressure in the Y direction acting at right angle on the face CDEF, and Ps is the intensity of pressure

normal to inclined plane at an angle θ as shown in figure at right angle to ABCD..

For a fluid at rest there will be no shear stress, there will be no accelerating forces, and

therefore the sum of the forces in any direction must be zero.

Thus the forces acting on the fluid element are the pressures on the surrounding and the gravity force.

Force due to px = px x Area ABFE = px dydz

Horizontal component of force due to pN = - (pN x Area ABCD) sin(θ) = - pNdNdz dy/ds = -PNdydz

As Py has no component in the x direction, the element will be in equilibrium, if

px dydz + (-pNdydz) = 0

i.e. px = pN

Similarly in the y direction, force due to py = pydxdz

Component of force due to pN = - (pN x Area ABCD) cos(θ) = - pNdsdz dx/ds = - pNdxdz

Force due to weight of element is negligible and the equation reduces to,

py = pN

Therefore, px = py = pN

Thus, Pressure at a point in a fluid at rest is same in all directions.

Piezometer :The simplest form of manometer is the piezometer. The height of the fluid in the tube

gives the difference between pressure in the pipe and atmosphere.

The piezometer is only useful when the pressure to be measured is greater than

atmospheric (otherwise) air would be sucked back into system.

ℎ𝐴 = h .s1 m of water

pA = w.hA kN/m2

where, w = specific weight of fluid in kN/m³

Simple U tube manometer:

A very common form of manometer is the U tube manometer. In this version one of the tubes is open

to the atmosphere.


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Manometer equation

Let the lower level of manometer liquid be the datum

Pressure of liquid above datum in the left limb = Pressure of

liquid above datum in the right limb

ℎ𝐴 + h1.s1 = h2. s2 m of water

ℎ𝐴 = h2. s2 - h1.s1 m of water.

PA = w.hA kN/m2

where , w = specific weight of water.

Differential U-tube manometer

Let,

A and B are the two pipes carrying liquids of specific gravity s1 and s3 & s2 = specific gravity of

manometer liquid.

h1 = height of pipe A liquid in left limb & h2 = height of manometer liquid in right limb above datum;

h3 = height of pipe B liquid right limb as shown in figure.

Let, 𝑝𝐴 and 𝑝𝐵 be the pressure of liquids in the corresponding pipes A and B.

If 𝑝𝐴 > 𝑝𝐵 , then the manometer equation is,

𝑝𝐴 + 𝑤1ℎ1= 𝑝𝐵 + 𝑤2ℎ2+ 𝑤3ℎ3

𝑝𝐴 − 𝑝𝐵 = 𝑤2ℎ2 + 𝑤3ℎ3 − 𝑤1ℎ1 N/ m2

Dividing both sides by specific weight of water, w 𝑝𝐴−𝑝𝐵

𝑤=

𝑤2ℎ2

𝑤+

𝑤3ℎ3

𝑤−

𝑤1ℎ1

𝑤 m of water

ℎ𝐴 − ℎ𝐵 = 𝑠2ℎ2 + 𝑠3ℎ3 − 𝑠1ℎ1 m of water [ ᶱ ᶱᶱ 𝑝

𝑤 = pressure head ‘h’, in m of water &

𝑤𝑙

𝑤𝒘 = sl (sp.gravity of liquid)]


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= [𝑠2ℎ2 + 𝑠3ℎ3 − 𝑠1ℎ1 ] x 9.81 kN/m²

Buoyant force: The upward force exerted by a liquid on a body when the body is immersed in the

liquid is known as buoyancy or buoyant force.

The point through which force of buoyancy is supposed to act is called centre of buoyancy.

The buoyant force acting on a body is equal to the weight of the liquid displaced by the body.

For a fluid with constant density, the buoyant force is independent of the distance of the body from the

free surface. It is also independent of the density of the solid body.

Archimedes' principle: The buoyant force acting on a body immersed in a fluid is equal to the weight

of the fluid displaced by the body, and it acts upward through the centroid of the displaced volume.

For floating bodies, the weight of the entire body must be equal to the buoyant force, which is the

weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body.

ρ - density of body; ρf – density of fluid

A solid body dropped into a fluid will sink, float,

or remain at rest at any point in the fluid,

depending on its average density relative to the

density of the fluid.

Stability of Immersed and Floating Bodies

A floating body possesses vertical stability, while an immersed neutrally buoyant body is neutrally

stable since it does not return to its original position after a disturbance.

An immersed neutrally buoyant body is (a) stable if the center of gravity G is directly below the center of

buoyancy B of the body, (b) neutrally stable if G and B are coincident, and (c) unstable if G is directly above B.

Stability of floating bodies: A floating body is stable if the body is bottom-heavy and thus the center

of gravity G is below the centroid B of the body, or if the metacentre M is above point G. However,

the body is unstable if point M is below point G.


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Metacentre: The point about which a body starts oscillating when the body is tilted is known meta-

centre.

Metacentric height GM: The distance between the center of gravity G and the metacenter M is

known as Meta centric height. It is the point of intersection of line of action of buoyant force with the

line passing through centre of gravity, when the body is slightly tilted.

The length of the metacentric height GM above G is a measure of the stability: If the metacentric

height increases, then the floating body will be more..

The meta-centric height (GM) is.given by, GM = 𝐼

V - BG

Where, I = Moment of Inertia of the floating body (in plan) at water surface about the axis Y- Y

V = Volume of ihe body sub merged in water

BG = Distance between centre of gravity and centre of buoyancy.

Conditions of equilibrium of a floating and submerged body are :

Equilibrium Floating Body Sub-merged Body

(i) Stable Equilibrium

(a) Unstable Equilibrium

(Hi) Neutral Equilibrium

M is above G

M is below G

Af and G coincide

B is above G

B is below G

B and G coincide


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The value of meta-cenlric height GM, experimentally is given as GM = 𝑤1 𝑥

𝑊 tan𝜃

Where

w1 = Movable weight

x = Distance through which w1 is moved

W = Weight of the ship or floating body including w1

θ = Angle through the ship or floating body is tilted due to the movement of w1

Problems

1. 5000 litres³ of an oil weighs 45 kN. Find its Specific weight, mass density and relative density.

Given: To find:

Volume, V = 5000 lit = 5000/1000 = 5 m³ i)Density,ρ

Weight, W= 45 kN = 45000 N ii)relative density, s

Specific Weight, w = W/V = 45000 / 5 = 9000 N/m³ = 9 kN/m³

Specific Weight, w = ρg

Mass density, ρ = w/g = 9000/ 9.81 = 917.43 kg/m³

Relative density = Density of oil/density of water = 917.43/ 1000 = 0.917

2. The density of an oil is 850 kg/m³. Find its relative density and Kinematic viscosity if the

dynamic viscosity is 5 x 10-3 kg/ms

Density of oil, ρoil = 850 kg/m³

Density of water, ρwater = 1000 kg/m³

Relative density of oil = 850/1000 = 0.85

Dynamic viscosity = µ = 5 x 10-3 kg/ms = 5 x 10-³ N s/m²

Kinematic viscosity = ν = µ / ρ = 5 x 10-³/ 850 = 5.882 x 10-6 m²/s

3. The space between two large inclined parallel planes is 6mm and is filled with a fluid. The planes

are inclined at 30° to the horizontal. A small thin square plate of 100 mm side slides freely down

parallel and midway between the inclined planes with a constant velocity of 3 m/s due to its

weight of 2N. Determine the viscosity of the fluid.

Given: Gap between plane and plate =6/2 = 3 mm = 0.003 m

Inclination of plane, θ = 30°

Velocity, u = 3 m/s

Weight of plate, W = 2 N

To find: viscosity, µ

Area of plate, A = 0.1 x 0.1 = 0.01 m²

Force due to the weight of the sliding plate along the direction of motion = 2 sin 30 = 1 N

Viscous force, F = shear stress x A = µ x (du/dy) x (2 x A) [. .. viscous force both sides of plate]


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Substituting the values,

1 = µ x [3/0.003] x [2 x 0.01]

Solving for viscosity, µ = 0.05 Ns/m² or 0.5 Poise

4. Determine the resistance offered to the downward sliding of a shaft of 400 mm dia and 0.1 m

length by the oil film between the shaft and a bearing of internal diameter 402 mm. The

kinematic viscosity is 2.4 x 10-4 m²/s and density is 900 kg/m³. The shaft is to move centrally

and axially at a constant velocity of 0.1 m/s.

Force, F opposing the movement of the shaft = shear stress x area

F = µ(du/dy) (π x D x L)

µ = ν x ρ = 2.4 x 10-4 x 900 = 0.216 Ns/m²

du = 0.1 m/s, L = 0.1 m, D= 0.4 m

dy = (402 - 400)/(2 x 1000) m = 0.001m,

Substituting,

F = 0.216 x {0.1 /0.001} ( π x 0.4 x 0.1) = 2714 N

5. A square plate of size 1m x 1m and weighing 350 N slides down an inclined plane with a

uniform velocity of 1.5 m/s.The inclined plane is laid on a slope of 5 vertical to 12 horizontal

and has an oil film of 1 mm thickness. Calculate the dynamic viscosity of oil.

6. Two large surfaces are 2.5 cm apart. This space is filled with an oil of absolute viscosity 0.82

NS/m². Find what force is required to drag a plate of area 0.5m² between the two surfaces at a

speed of 0.6m/s. (i) When the plate is equidistant from the surfaces, (ii) when the plate is at

1cm from one of the surfaces.


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Case (i) When the plate is equidistant from the surfaces,

Let F1 and F2be the force required to overcome viscous resistance of oil above and below the plate

respectively.

In this case F1 = F2, Since the liquid is same on both side and the plate is equidistant from the

surfaces.

u = du = 0.6 m/s

F1 = F2 = A x µ x du/dy = 0.5 x 0.82 x 0.6/0.0125 = 19.68 N

Total force required, F = F1 + F2 = 19.68+19.68 = 39.36 N

Case (ii) when the plate is at 1cm from one of the surfaces.

Here, F1 ≠ F2

dy1 = 1.5 cm = 0.015 m; dy2= 1cm = 0.01m

F1= A x µ x du/dy1= 0.5 x 0.82 x 0.6/0.015 = 16.4 N

F2 = A x µ x du/dy2= 0.5 x 0.82 x 0.6/0.01= 24.6 N

Total force required, F = F1 + F2 = 16.4+24.6 = 41 N

7. Determine the power dissipated to rotate a shaft of 300 mm diameter at 400 rpm supported at

two journal bearings of 300 mm length with uniform oil thickness of 1 mm. Take viscosity for

oil as 0.03 Ns/m².

Given: Dia of Shaft, D= 300 mm= 0.3 m

Speed of rotation, N = 400 rpm

Bearing length, L=300 mm = 0.3 m

No.of bearings = 2

Thickness of oil,dy = 1 mm=0.001 m

Viscosity of oil, µ = 0.03 Ns/m²

Solution:

By Newton’s law of viscosity,

Shear stress, dy

du

For a rotating shaft, u = πDN

60 =

𝜋 x 0.3 x 400

60 = 6.283 m/s

°°° shear stress, τ = 0.03 x

6.283

0.001 = 188.49 N/m²

Area of contact between shaft and 2 bearings, A = 2 x πDL = 2 x π x 0.3 x 0.3 = 0.5655 m²

Shear force, F = τ x A = 188.49 x 0.5655 = 106.59 N

Torque, T = F x r = F x D/2 = 106.59 x 0.3/2 = 15.989 Nm

Power dissipated, P = 2πNT

60 =

2π x 400 x 15.989

60 = 669.75 Nm/s or Watts


Page | 19

8. Convert a pressure of 500 kN/m² in terms of i) height of a column of water of density 1000

kg/m³ and ii) height of mercury with specific gravity 13.6 .

We know that,

Pressure p = wh and w = ρ g

h = p/w

Specific weight of water = 1000 x 9.81 = 9810 N/m³ = 9.81 kN/m³

Pressure head of water = hw =500/9.81 = 50.97 m of water

Pressure head, hw = sm x hm

Therefore, hm = hw/sm = 50.97/13.6 = 3.75 m of mercury

9. A Capillary tube having an inside diameter 5mm is dipped in water at 20° C. Determine the

rise of water in the tube. Take σ =0.0736N/m at 20° C.

Dia of glass tube, d = 5 mm = 0.005 m

θ for water = 0°

Capillary rise, h = 𝟒𝝈 𝒄𝒐𝒔ɵ

𝒘 𝒅 = (4 x 0.0736 x cos 0) / (9810 x 0.005) = 0.006 m = 6 mm

10. Calculate capillary rise in a glass tube when immersed in Hg at 20° c. Assume

σ for Hg at 20°c as 0.51N/m. The diameter of the tube is 5mm. θ = 130°.

Dia of glass tube, d = 5 mm = 0.005 m

θ for Hg = 130° ; σ = 0.51N/m

Capillary rise, h = 𝟒𝝈 𝒄𝒐𝒔ɵ

𝒘 𝒅 = (4 x 0.51 x cos 130) / [(9810 x 13.6) x 0.005)] = -1.97x10-3 m

Negative sign indicates fall of mercury in the tube.

10. Convert the following absolute pressure to gauge pressure:

(a) 120kPa (b) 3kPa (c) 15m of H20 (d) 800mm of Hg.

Solution:

(a) Pabs = Patm + Pgauge

Pgauge = Pabs - Patm = 120- 101.3= 18.7 k Pa

(b) pgauge = 3-101.3 = -98.3 kPa

Pgauge = 98.3 kPa (vacuum)

(C) habs = hatm + hgauge

15 = 10.3+hgauge

hgauge = 4.7m of water

(d) habs = hatm + hgauge

800 =760 + hgauge

hgauge = 40 mm of mercury

11. The right limb of a simple U-tube manometer containing mercury is open to the atmosphere

while the left limb is connected to a pipe in which a fluid of specific Gravity, 0.9 is flowing.

The centre of the pipe is 12 cm below the level of mercury in the right limb. Find the pressure

of fluid in the pipe if the difference of mercury level in the two limbs is 22 cm.


Page | 20

Specific gravity of pipe liquid, S1 = 0.9;

Specific gravity of manometer liquid, S2 = 13.6;

Height of pipe liquid above datum, h1 = 22-12 = 10 cm = 0.1 m

Height of manometer liquid above datum, h2 = 22 cm = 0.22 m

Let, Pressure head of fluid in the pipe = hA m of water

Pressure of liquid above datum in the left limb = Pressure of liquid above datum in the right

limb

ℎ𝐴 + h1.s1 = h2. s2 m of water

ℎ𝐴 = h2. s2 - h1.s1 = (0.22 x 13.6) – (0.1 x 0.9) = 2 .902 m of water.

Pressure of fluid in the pipe, PA = w.hA = 9.81 x 2.902 = 28.469 kN/m²

where , w = specific weight of water = 9.81 kN/m²

12. A U tube differential mercury manometer is connected to two pipes A and B, both carrying

water. Pipe A lies 1.5 m above Pipe B. The level of mercury raised up in the left limb

connected to pipe A is leveling the centre of pipe B. If the difference in levels of mercury is 7.5

cm and the pressure in pipe A is 170 kN/m², find the pressure in Pipe B.

Given Data: Refer figure

To find: Pressure in lower pipe-B, pB = ?

Solution:

Let Z-Z be the datum (lower level of mercury in right limb connected to pipe-B) Pressure in Pipe- A, pA =170 kN/m² Specific weight of water in pipe A & B,

ѡ𝟏= 9.81 kN/m³

Difference in levels of pipe = h1 = 1.5 m

Difference in levels of mercury,

h2 = h3 =7.5 cm = 0.075 m


Page | 21

Specific gravity of water in pipe –A & pipe-B =

s1=1

Specific weight of mercury(manometer liquid),

ѡ𝟐 = s2 × ѡ𝒘𝒂𝒕𝒆𝒓 = 13.6 × 9.81 = 133.416

kN/m³ Specific gravity of manometer liquid, s2 = 13.6

Manometer equation:

Pressure of liquids above datum in Left limb = Pressure of liquids above datum in right

limb

𝒑𝑨 + ѡ𝟏 𝒉𝟏 + ѡ𝟐 𝒉𝟐 = 𝒑𝑩 + ѡ𝟏 𝒉𝟑

𝒑𝑩 = 𝒑𝑨 + ѡ𝟏 𝒉𝟏 + ѡ𝟐 𝒉𝟐 −ѡ𝟏 𝒉𝟑

= 𝒑𝑨 + ѡ𝟏 𝒉𝟏 + ѡ𝟐 𝒉𝟐 − ѡ𝟏 𝒉𝟐 [ since 𝒉𝟑 = 𝒉𝟐]

= 𝒑𝑨 + ѡ𝟏 𝒉𝟏 + [ѡ𝟐 − ѡ𝟏] × 𝒉𝟐 = 170 + [9.81 × 1.5] + [133.416 - 9.81] × 0.075

= 170 + 14.715 + 9.2705 = 193.9855 kN/m²

13. A rectangular plane surface is 2 m x 3 m is immersed vertically in water. Determine the total

pressure and location of centre of pressure on the plane surface when its upper edge is

horizontal and (a) coincides with water surface, (b) 2.5 m below the free water surface.

Case -1 Case -2

Area, A=2x3= 6 m² Area = 6 m²

x = 3/2 =1.5 m x = 2.5+3/2 =4 m

IG = bd³/12 = 2x3³/12 = 4.5 m4 IG = bd³/12 = 2x3³/12 = 4.5 m4

Total Pressure

P= wA x = 9.81 x 6 x 1.5 = 88.29 kN P = 9.81x 6x4 = 235.44 kN

Depth of centre of centre of pressure

h = x + IG/A x = 1.5+4.5/(6 x1.5) h = 4+4.5/(6x4)

= 2 m = 4.1875 m

14. An inclined circular plate, 3 m in diameter is submerged in an oil of specific gravity 0.8 in

such a way that its is greatest and least depth are 1 m and 2 m respectively. Find the total

pressure and the point where it acts.

𝒑𝑩 = 𝟏𝟗𝟑. 𝟗𝟖𝟓𝟓 𝒌𝑵/𝒎²


Page | 22

Given: Circular plate; Dia. D = 3 m

Specific gravity of oil = S = 0.8

Depth of point A = 1 m

Depth of point C = 2 m

To find:

Total pressure of oil, P

Depth of centre of pressure, h

Refer figure

From right angle triangle ABC,

Sin θ = BC/AC = (2-1)/ 3 = 1/3 m and

AG = 3/2 = 1.5 m

Depth of centre of gravity, x = DA + AG x Sinθ

= 1 + 1.5 x 1/3

= 1.5 m

Total pressure, P = wAx

= 7.848 x 7.0686 x 1.5

= 83.21 kN

Depth of centre of pressure, h = x + 𝐼𝐺 sin²𝜃

𝐴x

= 1.5 +3.9761 X (1 3⁄ )2

7.0686 X 1.5 = 1.5417 m

Specific weight of oil, w = Soil x ww

= 0 .8 x 9.81

= 7.848 kN/m³

Area of circular plate, A = 𝜋𝐷2

4 =

𝜋 𝑋 32

4

= 7.0686 m²

IG = 𝜋𝐷4

64 =

𝜋 𝑋 34

64 = 3.9761 m4

15. A rectangular block of wood is 5 m long, 3 m wide and 1.2 m thick. 0.8 m height is

submerged in sea water. If the centre of gravity is 0.6 m above the bottom of the wood block,

determine the metacentric height. Take density for sea water = 1025 kg/m³.

Given: Size of wood block = 5m x 3m x 1.2 m

Hight under water = 0.8 m

Density of sea water = 1025 kg/m³

Base to C.G distance, AG = 0.6 m

Solution:

Centre of buoyancy will lie at a distance of 0.8/2 = 0.4 m from base along the vertical axis

passing through C.G; ie., AB = 0.4 m

Moment of inertia about Y-Y, I = 5 x 3³/12 = 11.25 m4

Volume of wood block under water,V = 5x3x0.8 =12 m³

Distance between B and G, BG = AG-AB = 0.6 – 0.4 = 0.2 m

Metacentric height, GM = 𝐼

V – BG

= 11.25

12− 0.2 = 0.7375 m


Page | 23

16. An object weighing 2200 N in water has dimensions 1.75 rn x 1.25 m x 2.25 m, Find its

weight

in air and its specific gravity.

Given :

Size of the object = 1.75 m x 1.25 m x 2.25 m

Weight when in water = 2200 N

Solution:

Volume of the object = 1.75 m x 1.25 m x 2.25 m = 4.921875 m³

Volume of the water displaced - Volume of the object = 4.921875 m³

.-. Weight of water displaced = 9810 x 4.921875 = 48283.59375 N

For the equilibrium of the object

Weight of object in air - Weight of water displaced = Weight in water

Weight of object in air – 48283.59375 = 2200

Weight of object in air = 48283.59375 + 2200 = 50483.59375N

Weight in air = 50483.59375 N

Specific weight ot the object = W/V = 50483.59375 /4.921875 = 10256.9841 N/m³

Specific gravity = specic weight of object

specific weight of water =

10256.9841

9810 = 1.046

Questions for practice:

PART - A

1. Define fluid and fluid mechanics.

2. Define real and ideal fluids.

3. Define mass density and specific weight.

4. Distinguish between fluid statics and kinematics.

5. Define viscosity.

6. Define specific volume.

7. Define specific gravity.

8. Distinct b/w capillarity and surface tension.

9. Calculate the specific weight, density and specific gravity of 1 liter liquid which weighs

7N.

10. State Newton’s law of viscosity.

11. Name the types of fluids.

12. Define compressibility.

13. Define kinematic viscosity.

14. Find the kinematic viscosity of oil having density 981 kg/m3. The shear stress at a point in

oil is 0.2452N/m2 and velocity gradient at that point is 0.2/sec.

15. Determine the specific gravity of a fluid having 0.05 poise and kinematic viscosity 0.035

stokes.

16. Find out the minimum size of glass tube that can be used to measure water level if the

capillary rise is restricted to 2 mm. Consider surface tension of water in contact with air

as 0.073575 N/m.

17. Write down the expression for capillary fall.


Page | 24

18. Explain vapour pressure .

19. Two horizontal plates are placed 1.25 cm apart. The space between them is being filled

with oil of viscosity 14 poises. Calculate the shear stress in oil if upper plate is moved

with a velocity of 2.5 m/s.

20. State Pascal’s law.

21. What is mean by absolute and gauge pressure and vacuum pressure?

22. Define Manometer and list out its types.

23. Define centre of pressure and total pressure.

24. Define buoyancy and centre of buoyancy.

25. Define Meta centre.

26. Define Hydro static Pressure.

PART – B

1. Calculate the capillary effect in a glass tube of 4.5 mm diameter, when immersed in (a) water

(b) mercury. The temperature of the liquid is 20o C and the values of the surface tension of

water and mercury at 20o C in contact with air are 0.073575 N/m and 0.51 N/m respectively.

The angle of contact for water is zero that for mercury 130o. Take specific weight of water as

9800 N/m3.

2. If the velocity profile of a liquid over a plate is a parabolic with the vertex 202 cm from the plate, where the velocity is 120 cm/sec. calculate the velocity gradients and shear stress at a distance of 0, 10 and 20 cm from the plate, if the viscosity of the fluid is 8.5 poise.

3. The dynamic viscosity of oil, used for lubrication between a shaft and sleeve is 6 poise. The shaft is of diameter 0.4 m and rotates at 190 rpm. Calculate the power lost in the bearing for a sleeve length of 90mm. the thickness of the oil film is 1.5 mm.

4. If the velocity distribution over a plate is given by u=2/3 y – y2 in which u is the velocity in

m/s at a distance y meter above the plate, determine the shear stress at y = 0 and y = 0.15 m.

5. Derive Pascal’s law.

6. Derive expression for capillary rise and fall. 7. Two large plane surfaces are 2.4 cm apart. The space between the gap is filled with glycerin.

What force is required to drag a thin plate of size 0.5 m between two large plane surfaces at a

speed of 0.6 m/sec. if the thin plate is (i) in the middle gap (ii) thin plate is 0.8 cm from one

of the plane surfaces? Take dynamic viscosity of fluid is 8.1 poise.

8. Calculate the capillary rise in a glass tube of 2.5 mm diameter when immersed vertically in (a) water

(b) mercury. Take surface tension = 0.0725 N/m for water and = 0.52 N/m for mercury in contact

with air. The specific gravity for mercury is given as 13.6 and angle of contact of mercury with glass =

130o.

9. A U - Tube manometer is used to measure the pressure of water in a pipe line, which is in

excess of atmospheric pressure. The right limb of the manometer contains water and mercury

is in the left limb. Determine the pressure of water in the main line, if the difference in level

of mercury in the limbs of U tube is 10 cm and the free surface of mercury is in level with over

the centre of the pipe. If the pressure of water in pipe line is reduced to 9810 N/m2,


Page | 25

Calculate the new difference in the level of mercury. Sketch the arrangement in both cases.

10. Calculate the total hydrostatic force and location of centre of pressure for a circular

plate of 2.5 m diameter when immersed vertically in an oil of specific gravity 0.8 with its top edge

1.5 m below the oil.

11. A rectangular plate 2.5m x 3.5 m is submerged in water and makes an angle of 60°

with the horizontal, the 2.5m sides being horizontal. Calculate the total force

on the plate and the location of the point of application of the force, when the top edge of

the plate is 1.6m below the water surface.

12. A rectangular plate 1.5 m x 3 m is immersed in an oil of specific gravity 0.82 such that its

upper and lower edge is at depths 1.5 m and 3 m respectively. Determine the total pressure

acting on the plate and its location.

13. In an open container water is filled to a height of 2.5m and above that an oil of Specific

gravity 0.85 is filled for a depth of 1.4 m. Find the intensity of pressure at the interface of

two liquids and at the bottom of the tank.

14. The pressure Intensity at a point is 40kPa. Find corresponding pressure head in (a) water (b)

Mercury (c) oil of specific gravity 0.9.

15. a)Calculate intensity of pressure due to a column of 0.3m of (a) water (b) Mercury (c) Oil of

specific gravity 0.8. Also express the same in absolute units.

b)Convert the following absolute pressure in to gauge pressure: (a) 110kPa (b) 7.3 kPa (c) 17

m of water (d) 860 mm of Mercury.

16. a)Convert a pressure head of 10 m of water column to kerosene of specific gravity 0.8 and carbon-

tetra-chloride of specific gravity of 1.62.

b)Determine (a) the gauge pressure and (b) The absolute pressure of water at a depth of 9 m from

the surface.

SATHYABAMA UNIVERSITY School of Mechanical Engineering SME1202 Fluid ... · School of Mechanical...

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