Fundamental Concepts in Fluid Mechanics

ME 1304 Fluid Mechanics

VENKITARAJ K P DEPT. MECH. ENGG

Syllabus Module 1 Preliminaries, Concept of continuum, Properties of fluids – density – pressure – viscosity - surface tension - capillarity - vapour pressure, Fluid statics, Basic equations of fluid statics, Variation of pressure in a fluid, - Manometry - Forces on surfaces and bodies in fluids, Floatation - stability of bodies in fluid - metacentric height and its measurement, Fluids in rigid body motion, Fluid kinematics -Eulerian and Lagrangian description - local and material rates - deformation of a fluid element -strain rate-velocity relations, Graphical description of flow – streamlines - path lines - streak lines - stream tube, Fluid dynamics - concept of the control volume -Reynolds transport equation and its use to formulate fluid mechanics problems, Integral and differential forms of the continuity - momentum and energy equations, Illustrative examples. Module II One dimensional flow through pipes, Non viscous equation for the flow through a stream tube and along a stream line – Euler‘s equation – Bernoulli‘s equation, - Energy equation, Applications of the one dimensional equations - velocity and flow measuring devices and quasi steady problems, Laminar and turbulent flow through pipes - Hagen- Poiseuille equation - Darcy-Weisbach equation - pipe friction -Moody‘s chart - minor losses in pipes. Module III Two dimensional incompressible inviscid flows – Vorticity - Vortex tube - Irrotational flow – Velocity potential, Stream function - relation between stream function and potential function in ideal flows -Equation of a streamline - governing equations, Fundamental flow patterns, Combination of basic patterns - Rankine half body - Rankine oval - Doublet and flow over a cylinder, Magnus effect and the calculation of lift on bodies. Module IV Plane viscous flow past bodies, The boundary layer - Prandtl‘s boundary layer equations, Blasius solution for the boundary layer over a flat plate, Karman‘s Momentum Integral equations - Solutions using simple profiles for the boundary layer on flat plate - calculation of skin friction drag. References: 1)Shames, I.H., Mechanics of fluids’, Mc Graw Hill Book Co., 2) White, F.M., Fluid Mechanics’, 6 th Ed., Tata Mc Graw Hill, New Delhi 3) Cengel, Y.A, Cimbala, John, M., Fluid Mechanics, Fundamentals and Applications’, 7th Ed. Tata Mc Graw Hill, New Delhi, 4) Gupta, V., Gupta, S.K.., Fluid Mechanics and its applications’, New Age International, New Delhi, 5) Som, S.K., and Biswas, G., Fluid Mechanics and fluid Machines’, 2nd Ed., Tata Mc Graw Hill, New Delhi. 6) Cohen and Kundu - Fluid Mechanics, Elsevier 7) V. Babu, Fundamentals of Incompressible flows, Ane Publishers. 8) Massey, Fluid Mechanics,ELBS Type of Questions for University Exam Q 1.Eight short answer questions of 5 marks with two questions from each of the four modules. (8x5 = 4 0 marks) Q 2. to Q.5 : Two questions A & B of 15 marks from each modules with option to answer either A or B. ( 4x15 = 60 marks)



Fundamental Concepts in Fluid Mechanics

Introduction

Mechanics is the oldest physical science that deals with both stationary and moving bodies under

the influence of forces. The branch of mechanics that deals with bodies at rest is called statics,

while the branch that deals with bodies in motion is called dynamics.

The subcategory fluid mechanics is defined as the science that deals with the behavior of fluids

at rest (fluid statics) or in motion (fluid dynamics), and the interaction of fluids with solids or

other fluids at the boundaries.

Fluid mechanics is also referred to as fluid dynamics by considering fluids at rest as a special case

of motion with zero velocity.

Fluid mechanics itself is also divided into several categories. The study of the motion of fluids

that are practically incompressible (such as liquids, especially water, and gases at low speeds) is

usually referred to as hydrodynamics.

A subcategory of hydrodynamics is hydraulics, which deals with liquid flows in pipes and open

channels.

Gas dynamics deals with the flow of fluids that undergo significant density changes, such as the

flow of gases through nozzles at high speeds.

The category aerodynamics deals with the flow of gases (especially air) over bodies such as

aircraft, rockets, and automobiles at high or low speeds.

Some other specialized categories such as meteorology, oceanography, and hydrology deal with

naturally occurring flows.

What is a fluid?

In everyday life, we recognize three states of matter: solid, liquid and gas. Although different in

many respects, liquids and gases have a common characteristic in which they differ from solids:

they are fluids, lacking the ability of solids to offer a permanent resistance to a deforming force.

A fluid is a substance which deforms continuously under the action of shearing forces, however

small they may be. Conversely, it follows that: If a fluid is at rest, there can be no shearing forces

acting and, therefore, all forces in the fluid must be perpendicular to the planes upon which they

act.

The differences between the behaviours of solids and fluids under an applied force are as follows:

i. For a solid, the strain is a function of the applied stress, providing that the elastic limit is not

exceeded. For a fluid, the rate of strain is proportional to the applied stress.

ii. The strain in a solid is independent of the time over which the force is applied and, if the elastic

limit is not exceeded, the deformation disappears when the force is removed. A fluid continues to



flow as long as the force is applied and will not recover its original form when the force is

removed.

iii. A further difference between solids and fluids is that a solid has a fixed shape whereas a fluid

owes its shape at any particular time to that of the vessel containing it.

Consider a fluid between two parallel plates, which is subjected to a shear stress due to the

impulsive motion of the upper plate.

No slip condition: no relative motion between fluid and boundary, i.e., fluid in contact with lower

plate is stationary, whereas fluid in contact with upper plate moves at speed U.

Fluid deforms, i.e., undergoes rate of strain 𝜃 due to shear stress τ

For Newtonian fluid: τ ∝ 𝜃 = rate of strain

τ = 𝜇𝜃

μ = coefficient of viscosity

Such behavior is different from solids, which resist shear by static deformation (up to elastic limit

of material)



Elastic solid: τ ∝ γ = strain

τ = G γ

G = shear modulus

The differences between liquids and gases are as follows

i. Although liquids and gases both share the common characteristics of fluids, they have many

distinctive characteristics of their own. A liquid is difficult to compress and, for many purposes,

may be regarded as incompressible. A given mass of liquid occupies a fixed volume, irrespective

of the size or shape of its container, and a free surface is formed if the volume of the container is

greater than that of the liquid.

ii. A gas is comparatively easy to compress. Changes of volume with pressure are large, cannot

normally be neglected and are related to changes of temperature. A given mass of gas has no

fixed volume and will expand continuously unless restrained by a containing vessel. It will

completely fill any vessel in which it is placed and, therefore, does not form a free surface.

Continuum Hypothesis

The concept of a continuum is the basis of classic fluid mechanics. Matter is made up of atoms

that are widely spaced in the gas phase. Yet it is very convenient to disregard the atomic nature of

a substance and view it as a continuous, homogeneous matter with no holes, that is, a continuum.

As a consequence of the continuum, each fluid property is assumed to have a definite value at

every point in space. Thus fluid properties such as density, temperature, velocity, and so on, are

considered to be continuous functions of position and time.

However, under certain circumstances, the continuum hypothesis is inappropriate. Such

circumstances occur when the distance between the molecules or, more correctly, the mean free

path that they travel between collisions with other molecules, λ, is comparable with some

physical dimension (L) of the container of the flow. Naturally, this most often arises when the

density of the gas is very low (so that the mean free path is large) and when the gas interacts with

solid surfaces with a small-scale structure such as a porous solid or a capillary tube. In such

circumstances, the gas molecules may interact as frequently with the solid surface as they do with

other molecules and one has what is called a transition regime. As the density of the gas is



reduced further, the collisions of molecules with walls completely dominate the processes and

one reaches the free-molecule or Knudsen regime.

The ratio λ/L is termed the Knudsen Number, Kn. This is a non-dimensional number which is

utilized to judge whether fluids are continuous or not. Its definition is

Kn =λ

𝐿

It describes the degree of departure from continuum. Usually when Kn< 0.01, the concept of

continuum hold good. Beyond this critical range of Knudsen number, the flows are known as slip

flow (0.01 < Kn< 0.1), transition flow (0.1 < Kn< 10) and free-molecule flow (Kn> 10).

In the transition regime, the behaviour is naturally intermediate between the continuum and free-

molecule behaviour and is generally described in terms of slip at the walls so that the normal

boundary condition of continuum mechanics, that there is no relative motion between the wall

and the fluid, is abandoned.

In a problem such as rarefied gas flow (e.g. as encountered in flights into the upper reaches of the

atmosphere), we must abandon the concept of a continuum in favor of the microscopic and

statistical points of view. However, for the flow regimes considered in this course Kn is always

less than 0.01 and it is usual to say that the fluid is a continuum.

Properties of Fluids

Density: the density (also known as mass density or specific mass) of a fluid is defined as the

mass per unit volume at a standard temperature and pressure. It is usually denoted by Latin

character ρ (rho). Its unit is Kg/m3. Mass density of water at 4oC is 1000 kg/m3.

Weight Density: (also known as specific weight) is defined as the weight per unit volume at

the standard temperature and pressure; it is usually denoted as γ. Its unit is N/m3.

Where g gravitational acceleration = 9.81m/s2, γ of water = 9810 N/m3 at 4oC and 1atm.

T = 4°C, γ of air =12.01 N/m3 at standard atmosphere (T=15°C and p =101.33 kPa)

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Specific Volume: It is defined as a volume per unit mass of fluid, it is denoted by v. Its unit is

m3/Kg.

Specific Gravity: It is defined as the ratio of the specific weight of the fluid to the specific

weight of a standard fluid. For liquids the standard fluid is pure water at 4oC, and denoted by

S. For Gasses the standard fluid is air

Example 1: Calculate the Specific weight, specific mass, specific volume and specific gravity of a

liquid having a volume of 6m3 and weight of 44kN.

COMPRESSIBILITY: BEHAVIOR OF FLUIDS AGAINST PRESSURE

For most purposes a liquid may be considered as incompressible. The compressibility of a

liquid is expressed by its bulk modulus of elasticity. The mechanics of compression of a fluid

may be demonstrated by imagining the cylinder and piston of figure below to be perfectly

rigid (inelastic) and to contain a volume of fluid V. Application of a force, F, to piston will

increase the pressure, p, in the fluid and cause the volume decrease –dV. The bulk modulus

of elasticity, E, for the volume V of a liquid

Since dV/V is dimensionless, E is expressed in the units of pressure, p. For water at ordinary

temperatures and pressures, E = 2×104 kg/cm2.



For liquids, the changes in pressure occurring in many fluid mechanics problems are not

sufficiently great to cause appreciable changes in specific mass. It is, therefore, usual to

ignore such changes and to treat liquids as incompressible. ρ = Constant

VISCOSITY: BEHAVIOR OF FLUIDS AGAINST SHEAR STRESS

Viscosity is a property of a real fluid (an ideal fluid has no viscosity) which determine its

resistance to shearing stresses. When real fluid motions are observed carefully, two basic

types of motion are seen. The first is a smooth motion in which fluid elements or particles

appear to slide over each other in layers; this motion is called laminar flow. The second

distinct motion that occurs is characterized by a random or chaotic motion of individual

particles; this motion is called turbulent flow.

Now consider the laminar motion of a real fluid along a solid boundary as in Figure below.

Observations show that, while the fluid has a finite velocity, u, at any finite distance from the

boundary, there is no velocity at the boundary. Thus, the velocity increases with increasing

distance from the boundary. These facts are summarized on the velocity profile, which

indicates relative motion between adjacent layers.

When two layer of fluid at the distance of y apart, move one over the other at different

velocities, say u and u+du, the viscosity together with relative velocity causes shear stress

acting between layers. With respect to the distance between these two layers dy, the shear

stress, τ, this becomes a relation between shear strain rate and velocity gradient

It is evident that a frictional or shearing force must exist between the fluid layers; it may be

expressed as a shearing or frictional stress per unit of contact area. This stress, designated by

τ, has been found for laminar (non-turbulent) motion to be proportional to the velocity

gradient, du/dy, with a constant of proportionality, μ, defined as coefficient of viscosity or

dynamic viscosity. Thus,



All real fluids possess viscosity and therefore exhibit certain frictional phenomena when

motion occurs. Viscosity results fundamentally from cohesion and molecular momentum

exchange between fluid layers and, as flow occurs, these effects appear as tangential or

shearing stresses between the moving layers. This equation is called as Newton’s law of

viscosity.

Causes of Viscosity

To understand the causes of viscosity of a fluid, consider the observed effects of temperature

on the viscosity of a gas and a liquid. It has been noted that for gases, viscosity increases with

increasing temperature and for liquids, viscosity decreases with increasing temperature. The

reason for this is that viscosity appears to depend on two phenomena, namely the transfer of

momentum between molecules and the intermolecular (cohesive) forces between molecules

of the fluid.

Consider a fluid consisting of two layers aa and bb as shown below, with the layer aa moving

more rapidly than bb. Some molecules in aa owing to their thermal agitation will migrate to

bb and take with them the momentum they have as a result of the overall velocity of aa.

These molecules on colliding with molecules in the bb layer transfer their momentum

resulting in an overall increase in the velocity of bb. In turn, molecules from bb, also owing

to thermal agitation cross over to layer aa and collide with molecules there. The net effect of

the crossings is that the relative motion between the two layers is reduced: layer aa is slowed

down because of the collision with the slower molecules; layer bb is accelerated because of

collision with faster molecules.

Now use this to explain why it is observed that viscosity of a gas increases with increasing

temperature. With a liquid, transfer of momentum between layers also occurs as molecules

move between the two layers. However, what is different from the gas is the strong

intermolecular forces in the liquid. Relative movement of layers in a liquid modifies these

intermolecular forces, thereby causing a net shear force that resists the relative movement.

The effect of increasing the temperature is to reduce the cohesive forces while

simultaneously increasing the rate of molecular interchange. The net effect of these two in

liquids is a decrease in viscosity.

The dimensions of the (dynamic) viscosity μ may be determined from dimensional

homogeneity as follows:



In SI units, N/m2.s or (Pa.s).

Dynamic viscosity may also be expressed in the metric CGS (centimeter-gram-second)

system as g/(cm s), dyne s/cm2 or poise (p) where

1 poise = 1 dyne s/cm2 = 1 g/(cm s) = 0.1 Pa s = 0.1 N s/m2

For practical use the Poise is normally too large and the unit is often divided by 100 - into the

smaller unit centiPoise (cP) - where

1 p = 100 cP

1 cP = 0.01 poise = 0.001 Pascal second = 0.001 N s/m2

Kinematic Viscosity

Kinematic viscosity is the ratio of absolute (or dynamic) viscosity to density. Kinematic

viscosity can be obtained by dividing the absolute viscosity of a fluid with the fluid mass

density.

ν = μ / ρ

where ν = kinematic viscosity (m2/s), μ = absolute or dynamic viscosity (N s/m2), ρ = density

(kg/m3)

In the SI-system the theoretical unit of kinematic viscosity is m2/s or Stoke (St) where

1 St (Stokes) = 10-4 m2/s

Since the Stoke is a large unit it is often divided by 100 into the smaller unit Centistoke (cSt)

- where

1 St = 100 cSt

1 cSt (centiStoke) = 10-6 m2/s



Newtonian vs. Non-Newtonian Fluids

Fluids that follow Newton‟s viscosity law are commonly known as Newtonian fluids. Other

fluids are classed as non-Newtonian fluids. For non-Newtonian fluids, the relation between

shear stress and shear strain rate is non-linear. The science of Rheology, which broadly is

the study of the deformation and flow of matter, is concerned with plastics, blood,

suspensions, paints, and foods, which flow but whose resistance is not characterized by

Newton‟s law of viscosity.



Example 3:

Oil has dynamic viscosity (μ = 0.01 Pa.s) filled the space between two concentric cylinders, where

the inner one is movable and the outer is fixed. If the inner and outer cylinders has diameters 15mm

and 15.6mm respectively and the height of both cylinders is 250mm, determine the value of the

torque (T) that necessary to rotate the internal cylinder with 12 rpm?

Example 4:

The velocity distribution for flow over a plate is given by u= 2y - y2 where u is the velocity in m/s at

a distance y meters above the plate surface. Determine the velocity gradient and shear stresses at the

boundary and 15cm from it. Take dynamic viscosity of fluid as 0.9 N.s/m2



VAPOR PRESSURE AND CAVITATION

All liquids tend to evaporate or vaporize, which they do by projecting molecules into the

space above their surfaces. If this is a confined space, the partial pressure exerted by the

molecules increases until the rate at which molecules reenter the liquid is equal to the rate at

which they leave. For this equilibrium condition, we call the vapor pressure the saturation

pressure.

When the pressure of a liquid falls below the vapor pressure it evaporates, i.e., changes to a

gas. If the pressure drop is due to temperature effects alone, the process is called boiling. If

the pressure drop is due to fluid velocity, the process is called cavitation. Cavitation is

common in regions of high velocity, i.e., low pressure such as on turbine blades and marine

propellers.

Cavitation can cause serious problems, since the flow of liquid can sweep this cloud of

bubbles on into an area of higher pressure where the bubbles will collapse suddenly. If this

should occur in contact with a solid surface, very serious damage can result due to the very

large force with which the liquid hits the surface. Cavitation can affect the performance of

hydraulic machinery such as pumps, turbines and propellers, and the impact of collapsing

bubbles can cause local erosion of metal surfaces.

Vapor pressure of water at 100°C =101 kPa

The dimensionless parameter describing flow‐induced boiling is called cavitation number:

where, p is ambient pressure, pv is vapor pressure and V∞ is the fluid velocity.



Surface Tension

Surface tension is a property of liquids which is making what is like a thin tensioned

membrane at the interface between the liquid and another fluid (typically a gas). Surface

tension has dimensions of force per unit length and denoted as, σ (Sigma), and its unit is

N/m. Dimensional Formula: [MLT-2] / [L] = [MT-2]

At the interface of two immiscible fluids (e.g., a liquid and a gas), forces develop to cause the

surface to behave as if it were a stretched membrane. Molecules in the interior at-tract each

other equally, whereas molecules along the surface are subject to a net force due to the

absence of neigh-bor molecules. The intensity of the molecular attraction per unit length

along any line in the surface is call the surface tension.

Surface tension is the surface force that develops at the interface between two immiscible

liquids or between liquid and gas or at the interface between a liquid and a solid surface.

Because of surface tension, small water droplets, gas bubbles and drops of mercury tend to

maintain spherical shapes.

If a line is imagined drawn in a liquid surface, then the liquid on one side of the line pulls that

on the other side. The magnitude of surface tension is defined as that of the tensile force

acting across and perpendicular to a short, straight element of the line drawn in the surface

divided by the length of that line.

Effects of surface tension: Capillarity

The capillarity phenomenon is due to the rise or depression of the meniscus of the liquid due

to the action of surface tension forces.

Capillary action in small tubes which involve a liquid-gas-solid interface is caused by surface

tension. The fluid is either drawn up the tube or pushed down.



Contact angle θ: appears when a liquid interface intersects with a solid surface. θ < 90o Wetting, e.g., Water, θ ≈ 0°

θ > 90o, Non-wetting, e.g., Mercury, θ≈130°

The forces of attraction binding molecules to one another give rise to cohesion, the tendency

of the liquid to remain as one assemblage of particles rather than to behave as a gas and fill

the entire space within which it is confined.

On the other hand, forces between the molecules of a fluid and the molecules of a solid

boundary give rise to adhesion between the fluid and the boundary.

It is the interplay of these two forces that determine whether the liquid will “wet” the solid

surface of the container. If the adhesive forces are greater than the cohesive forces, then the

liquid will wet the surface; if vice versa, then the liquid will not.

It is rare that the attraction between molecules of the liquid exactly equals that between

molecules of the liquid and molecules of the solid and so the liquid surface near the boundary

becomes curved.

Weight of fluid column = Surface tension pulling force.



Expression above calculates the approximate capillary rise in a small tube. The meniscus lifts

a small amount of liquid near the tube walls, as r increases this amount may become

insignificant. Thus, the equation developed overestimates the amount of capillary rise or

depression, particularly for large r.

For r > ¼ in ( 7 mm), capillarity is negligible.

Pressure jump across curved interfaces

(d) Bubble



(π R2) Δp = 2 (2 π R) σ

Δp = 4 σ /R

Example 5

Find the pressure inside a water droplet having diameter of 0.5 mm at 20 0

C if the outside pressure is

1.03N/cm 2

and the surface tension of water at that temperature is 0.0736 N/m.

Example 6

Compare the capillary rise of water and mercury in a glass tube of 2 mm diameter at 200 C .Given

that the surface tension of water and mercury at 200 C are 0.0736 N/m and 0.051N/m respectively.

Contact angles of water and mercury are 00 and 130

0 respectively.



Example 7

Find the excess pressure inside a cylindrical jet of water 4 mm diameter than the outside atmosphere?

The surface tension of water is 0.0736 N/m at that temperature.

Example 8

Air is forced through a tube of internal diameter of 1.5 mm immersed at a depth of 1.5 cm in a mineral

oil having specific gravity of 0.85. Calculate the unit surface energy of the oil if the maximum bubble

pressure is 150 N/m2.



Example 9

Determine the minimum size of a glass tube, which can be used to measure pressure in water

flowing system. The capillary rise in the tube must not exceed 10 mm and surface tension of

water- air - glass interface is 0.001 N/m.

Example 10

The inside diameters of the two arms of a U-tube are 1.0 mm and 1.5 mm respectively. Now if it is

partially filled with water having surface tension of 0.0736 N/m and zero contact angle what will be

the difference in the level of menisci between the two arms.(shown in the figure below)



Exercise 1

1. A thin film of liquid flows down an inclined channel. The velocity distribution in the flow is

given by

where, h = depth of flow, α = angle of inclination of the channel to the horizontal, u = velocity at

a depth h below the free surface, ρ = density of liquid, μ = dynamic viscosity of the fluid.

Calculate the shear stress: (a) at the bottom of the channel (b) at mid-depth and (c) at the free

surface. The coordinate y is measured from the free surface along its normal

[(a) α, (b) α , (c) 0]

2. Two discs of 250 mm diameter are placed 1.5 mm apart and the gap is filled with an oil. A

power of 500 W is required to rotate the upper disc at 500 rpm while keeping the lower one

stationary. Determine the viscosity of the oil.

[ 0. 71 kg/ms]

3. Eight kilometers below the surface of the ocean the pressure is 100 MPa. Determine the

specific weight of sea water at this depth if the specific weight at the surface is 10 kN/m3 and the

average bulk modulus of elasticity of water is 2.30 GPa. Neglect the variation of g.

[ 10. 44 kN/m3 ]

4. The space between two large flat and parallel walls 20 mm apart is filled with a liquid of

absolute viscosity 0.8 Pas. Within this space a thin flat plate 200 mm × 200mm is towed at a

velocity of 200 mm/s at a distance of 5 mm from one wall. The plate and its movement are

parallel to the walls. Assuming a linear velocity distribution between the plate and the walls,

determine the force exerted by the liquid on the plate.

[1. 71 N]

5. What is the approximate capillary rise of water in contact with air (surface tension 0.073

N/m) in a clean glass tube of 5mm in diameter?



Fluid Statics Fluid statics concerned with the balance of forces which stabilize fluids at rest.

Fluids at rest cannot resist a shear stress; in other words, when a shear stress is applied

to a fluid at rest, the fluid will not remain at rest, but will move (flow) because of the

shear stress.

The shear force is zero for any fluid element at rest and hence the only surface force on a

fluid element is the normal component.

Pressure may be defined as the normal force that‟s applied toward the unit area, and

denoted by P. Its units are N/m2 or what is called Pascal, Pa. Dimensions of pressure is

ML-1

T-2

When the fluid velocity is zero, called the hydrostatic condition , the pressure variation is

due only to the weight of the fluid.

Pascal’s Law for Pressure

To prove this, a small wedge-shaped free body element is taken at the point (x,y,z) in a

fluid at rest.

The fluid is at rest, so we know there are no shearing forces, and we know that all force

are acting at right angles to the surfaces .i.e.

ps acts perpendicular to surface ABCD,

px acts perpendicular to surface ABFE and

py acts perpendicular to surface FECD.

And, as the fluid is at rest, in equilibrium, the sum of the forces in any direction is zero.

Summing forces in the x-direction:

px δy δz - ps δs δz sin θ = 0

But from the figure δs sin θ = δy



px δy δz - ps δy δz = 0

px = ps

Similarly, summing forces in the y-direction.

py δx δz - ps δs δz cos θ- ρg 1

2𝛿𝑥 𝛿𝑦 𝛿𝑧 = 0

The element is small i.e. δx , δy and δz are small, and so 𝛿𝑥 𝛿𝑦 𝛿𝑧 is very small and

considered negligible.

py δx δz - ps δs δz cos θ=0

But from the figure δs cos θ = δx

py δx δz - ps δx δz=0

py = ps

Thus we have,

px = ps =py

Considering the prismatic element again, ps is the pressure on a plane at any angle q ,

the x, y and z directions could be any orientation. The element is so small that it can be

considered a point so the derived expression px = ps =py indicates that pressure at any

point is the same in all directions.

Pressure at any point is the same in all directions. This is known as Pascal’s Law and applies

to fluids at rest

General Equation For Variation Of Pressure In A Static Fluid



Consider the cylindrical element of fluid in the figure below, inclined at an angle θ to the

vertical, length δs , cross-sectional area A in a static fluid of mass density ρ . The pressure at the

end with height z is p and at the end of height z +δz is p +δp .

For equilibrium of the element the resultant of forces in any direction is zero.

Resolving the forces in the direction along the central axis gives

Or in the differential form

If 90� then s is in the x or y directions, (i.e. horizontal),so

Confirming that pressure change in any horizontal plane is zero.

If 0_ then s is in the z direction (vertical) so

For static fluid, pressure varies only with elevation (depth, head) change within fluid.

Pressure Variation for a Uniform-Density Fluid The basic equation for pressure variation with elevation can be integrated depending on whether

ρ = constant or ρ = ρ (z), i.e., whether the fluid is incompressible or compressible.



Pressure Measurements.

Barometer The simplest application of the hydrostatic formula is the barometer ,which measures

atmospheric pressure. A tube is filled with mercury and inverted while submerged in a

reservoir. This causes a near vacuum in the closed upper end because mercury has an

extremely small vapor pressure at room temperatures (0.16Pa at 20°C). Since

atmospheric pressure forces a mercury column to rise a distance hi into the tube, the

upper mercury surface is at zero pressure.

The unit of pressure is the pascal (Pa), but it is also expressed in bars or metres of water

column. In addition, in some cases atmospheric pressure is used. At sea level standard,

1 atm = 760 mm Hg (at 273K) =101325 Pa

There are two methods used to express the pressure: one is based on the perfect vacuum

and the other on the atmospheric pressure. The former is called the absolute pressure and

the latter is called the gauge pressure.

Gauge pressure = Absolute pressure – Atmospheric pressure

In gauge pressure, a pressure under one atmospheric pressure is expressed as a negative

pressure.

Most gauges are constructed to indicate the gauge pressure.



In a static fluid of constant density we have,

In a liquid with a free surface the pressure at any depth z measured from the free surface so that z

= -h (see the figure below)

This gives the pressure

At the surface of fluids we are normally concerned with, the pressure is the atmospheric

pressure, patm . So



1. Piezometer tube

The simplest manometer is a tube, open at the top, which is attached to a vessel or

a pipe containing liquid at a pressure (higher than atmospheric) to be measured.

This simple device is known as a piezometer tube.

As the tube is open to the atmosphere the pressure measured is relative to

atmospheric so is gauge pressure: PA =ρgh1 = γ h1

This method can only be used for liquids (i.e. not for gases) and only when the

liquid height is convenient to measure.

2. U-tube manometer

Using a U-Tube enables the pressure of both liquids and gases to be measured with the

same instrument.



The U- tube is connected as in the figure below and filled with a fluid called the

manometric or gauge fluid.

The fluid whose pressure is being measured should have a mass density less than that of

the manometric fluid and the two fluids should not be able to mix readily - that is, they

must be immiscible.

Pressure in a continuous static fluid is the same at any horizontal level so,

pressure at B = pressure at C

pB =pC

For the left hand arm

pressure at B = pressure at A + pressure due to height h of fluid being measured

pB =pA + ρ g h1

For the right hand arm

pressure at C = pressure at D + pressure due to height h of manometric fluid

pC =pD + ρman g h2

pC =patm + ρman g h2

As we are measuring gauge pressure we can subtract patm giving

pC = ρman g h2

pB =pC

pA + ρ g h1 = ρman g h2

pA = ρman g h2 – ρ g h1



If the fluid being measured is a gas, the density will probably be very low in comparison to the

density of the manometric fluid i.e. man >> . In this case the term gh1 can be neglected, and

the gauge pressure give by

pA = ρman g h2

3. Differential manometer

If the U-tube manometer is connected to a pressurized vessel at two points the

pressure difference between these two points can be measured.

If the manometer is arranged as in the figure above, then

Pressure at C = pressure at D

pC =pA + ρ g ha

pD =pB + ρ g (hb – h) + ρmano g h

pC =pD

pA + ρ g ha = pB + ρ g (hb – h) + ρmano g h

pA - pB = ρ g (hb – ha) + (ρmano – ρ) g h



Again, if the fluid whose pressure difference is being measured is a gas and man , then the

terms involving can be neglected, so

pA - pB = ρmano g h

4. Inclined-tube manometer

As shown above, the differential reading is proportional to the pressure

difference. If the pressure difference is very small, the reading may be too small

to be measured with good accuracy. To increase the sensitivity of the differential

reading, one leg of the manometer can be inclined at an angle θ, and the

differential reading is measured along the inclined tube.

pA - pB = γ3 h3 + γ2 h2– γ1 h1

pA - pB = γ3 h3 + γ2 l sin θ – γ1 h1

5. Inverted U-tube manometer

Inverted U-Tube manometer consists of an inverted U - Tube containing a light

liquid. This is used to measure the differences of low pressures between two

points where better accuracy is required. It generally consists of an air cock at top

of manometric fluid type.

Inverted U-tube manometer is used for measuring pressure differences in liquids.

The space above the liquid in the manometer is filled with air which can be

admitted or expelled through the tap on the top, in order to adjust the level of the

liquid in the manometer.

For inverted U - tube manometer the manometric fluid is usually air.



Equating the pressure at the level XX'(pressure at the same level in a continuous

body of static fluid is equal)

Pressure at X = pressure at X‟

pX =p1 - ρ g (h + a)

pX‟ =p2 - ρ g a – ρmano g h

p1 - ρ g (h + a) = p2 - ρ g a – ρmano g h

p1 – p2 = (ρ – ρmano) g h

If the manometric fluid is choosen in such a way that m << then,

p1 – p2 = ρ g h



6. Micromanometer

Micro Manometer is is the modified form of a simple manometer whose one limb is made of

larger cross sectional area. It measures very small pressure differences with high precision

Micro-manometer is also known as single column manometers. The construction of a micro-

manometer is as follows:

a. One limb of manometer is a tank of large cross sectional area as compared to the cross

sectional area of the other limb.

b. This tank acts as a reservoir to hold the manometer fluid.

c. The cross sectional area of the tank is 100 times greater than that of the other limb.

d. When there is change in the pressure in the pipe, there is negligible change in the level of

fluid in tank. This change can be neglected and pressure can be measured as height of liquid in

the other column.

It is of two types:

i. Vertical column micro-manometer

ii. Inclined column micro-manometer

Vertical Column Micro-manometer

Initially, when there is no fluid flowing in the pipe the level of manometer liquid is at XX‟. But

due to liquid pressure in the pipe the level of manometer liquid moves down from XX‟ to YY‟ in

the tank and rises in the right limb to the distance h2 from XX‟.

Let,

ρ1 = density of liquid for which pressure has to be determined

ρ2 = density of manometer liquid (assume mercury)

S1 = Specific gravity of liquid for which pressure has to be determined

S2 = Specific gravity of manometer liquid

δh = Fall in the level of liquid in the tank

A = Area of cross-section of the tank

a = Area of cross-section of the right limb



h = Pressure head of fluid in the pipe (as head of water)

The rise in the manometer fluid in the right limb will be equal to the fall of level in the tank.

Therefore by mass conservation

δhA=a.h2

Pressure in the left limb at YY‟ = p + (h1 + δh) ρ g

Pressure in the right limb at YY‟ = (h2 + δh) ρmano g

Pressure at the same level in a continuous body of static fluid is equal)

p + (h1 + δh) ρ g = (h2 + δh) ρmano g

Pressure in the pipe,

p = (h2 + δh) ρmano g - (h1 + δh) ρ g

p = (h2 + 𝑎

𝐴 ℎ2) 𝜌 g − ℎ1 +

𝑎

𝐴 ℎ2 𝜌𝑚𝑎𝑛𝑜 g

p = ℎ2 𝜌 − ℎ1 𝜌𝑚𝑎𝑛𝑜 g + 𝑎

𝐴 (𝜌 − 𝜌𝑚𝑎𝑛𝑜 )ℎ2 g

If the cross sectional area of the tank A is very large compared to cross sectional area of the right

limb then i.e. A >> a, then, Ratio of a/A will be zero and the above equation can be re-written as

p = ℎ2 𝜌 − ℎ1 𝜌𝑚𝑎𝑛𝑜 g

Inclined column Micro-manometer



Inclined manometers are more sensitive than vertical column manometers. Inclined tube SCM is

used to measure small intensity pressure. It consists of a large reservoir to which an inclined U –

tube is connected as shown in fig. For small changes in pressure the reading „l‟ in the inclined

tube is more than that of SCM. Knowing the inclination of the tube the pressure intensity at the

gauge point can be determined.

Initially, when there is no fluid flowing in the pipe the level of manometer liquid is at XX‟. But

due to liquid pressure in the pipe the level of manometer liquid moves down from XX‟ to YY‟ in

the tank and rises in the right limb to the distance h2 from XX‟.

p = ℎ2 𝜌 − ℎ1 𝜌𝑚𝑎𝑛𝑜 g + 𝑎

𝐴 (𝜌 − 𝜌𝑚𝑎𝑛𝑜 )ℎ2 g

ℎ2 = 𝑙 𝑠𝑖𝑛𝛼

If the cross sectional area of the tank A is very large compared to cross sectional area of the right

limb then i.e. A >> a, then, Ratio of a/A will be zero and the above equation can be re-written as

p = ℎ2 𝜌 − ℎ1 𝜌𝑚𝑎𝑛𝑜 g

Desirable properties of manometric fluids Commonly used manometric liquids are mercury, water or alcohol. Some of the important and

desirable properties of the manometric liquids are:

High chemical stability

Low viscosity

Low capillary constant

Low coefficient of thermal expansion

Low volatility

Low vapour pressure



7. Mechanical Gauges

Pressure gauges are the devices used to measure pressure at a point. They are used

to measure high intensity pressures where accuracy requirement is less.

Pressure gauges are separate for positive pressure measurement and negative

pressure measurement. Negative pressure gauges are called Vacuum gauges.

Mechanical gauge consists of an elastic element which deflects under the action

of applied pressure and this deflection will move a pointer on a graduated dial

leading to the measurement of pressure. Most popular pressure gauge used is

Burdon pressure gauge.

The arrangement consists of a pressure responsive element made up of phosphor

bronze or special steel having elliptical cross section. The element is curved into a

circular arc, one end of the tube is closed and free to move and the other end is

connected to gauge point. The changes in pressure cause change in section

leading to the movement. The movement is transferred to a needle using sector

pinion mechanism. The needle moves over a graduated dial.



Example 1.

Determine the pressure at A for the U- tube manometer shown in fig. Also

calculate the absolute pressure at A in kPa.



Example 2

Find the pressure at the center of the pipe A. Assume a temperature T = 200C



Example 3



Example 4

An inverted U-tube manometer is shown in the figure. Determine the pressure

difference between the gauge points A and B.



Example 5

Water and oil flow in horizontal pipelines. A double U -tube manometer is

connected between the pipelines as shown in the sketch below. Calculate the

pressure difference between the water pipe and the oil pipe. Note that the densities

for water, carbon tetrachloride, air, and oil are 1000kg/m3, 1590kg/m

3, 1.23kg/m

3

, and 912 kg/m3, respectively. The gravity acceleration is 9.81 m/s

2



Example 6

Petrol of specify gravity 0.8 flows up through a vertical pipe. A and B are the two

points in the pipe, B being 0.3 m higher than A. Connection are led from A and B

to aU–tube containing Mercury. If the pressure difference between A and B is 18

kPa, find the reading of manometer.



Exercise 2

1. A U-tube manometer is connected to a closed tank containing air and water as

shown in Figure 2. At the closed end of the manometer the absolute air

pressure is 140kPa. Determine the reading on the pressure gage for a

differential reading of 1.5-m on the manometer. Express your answer in gage

pressure value. Assume standard atmospheric pressure and neglect the weight

of the air columns in the manometer.

2. A closed cylindrical tank filled with water has a hemispherical dome and is

connected to an inverted piping system as shown in Figure 3. The liquid in the

top part of the piping system has a specific gravity of 0.8, and the remaining

parts of the system are filled with water. If the pressure gage reading at A is 60

kPa, determine: (a) the pressure in pipe B, and (b) the pressure head, in

millimeters of mercury, at the top of the dome (point C).



3. A U-tube manometer contains oil, mercury, and water as shown in Figure 4.

For the column heights indicated what is the pressure differential between

pipes A and B?

4. A U-tube manometer is connected to a closed tank as shown in Figure 5. The

air pressure in the tank is 120 Pa and the liquid in the tank is oil (γ = 12000

N/m3). The pressure at point A is 20 kPa. Determine: (a) the depth of oil, z,

and (b) the differential reading, h, on the manometer.



5. The inverted U-tube manometer of Figure 6 contains oil (SG = 0.9) and water

as shown. The pressure differential between pipes A and B, pA − pB, is −5 kPa.

Determine the differential reading, h.

6. A piston having a cross-sectional area of 0.07 m

2 is located in a cylinder

containing water as shown in Figure 7. An open U-tube manometer is

connected to the cylinder as shown. For h1 = 60 mm and h = 100 mm, what is

the value of the applied force, P, acting on the piston? The weight of the

piston is negligible.

7. For the inclined-tube manometer of Figure 8, the pressure in pipe A is 8 kPa.

The fluid in both pipes A and B is water, and the gage fluid in the manometer

has a specific gravity of 2.6. What is the pressure in pipe B corresponding to

the differential reading shown?



Answer :

1. 12 kN

2. 66.8 kPa

3. (a) 103.2 kPa

(b) 229 mm

4. -20.7 kPa

5. z = 1.66 m h = 1.37 m

6. 0.46 mm

7. 892.7 N

8. 5508.3 Pa



Hydrostatic Forces on Submerged Surfaces and Bodies

Center of gravity: The center of mass also called the “center of gravity” (CG),

is a point in space or inside the body where the weight of a body acts as if it were

concentrated there.

For a plane figure (a two-dimensional shape) the “center of gravity” coincides

with the “geometric center”, also called centroid or barycenter.

A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid

storage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed

over its surface.

Hydrostatic Force on a submerged surface or body is the force due to the

pressure exerted on the surface by the fluid at rest.

The calculation of the magnitude, direction, and location of the total forces on

surfaces submerged in a liquid is essential in the design of dams, bulkheads,

gates, ships, and the like.

For a Plane surface submerged in a fluid, the hydrostatic force on the surface is

perpendicular to the surface since there can be no shear in static fluid

The fluid pressure is linearly dependent only to the vertical depth.

On a plane surface, the hydrostatic forces form a system of parallel forces, and

we often need to determine the magnitude of the force and its point of

application, which is called the center of pressure.

1. On a Horizontal surface (e.g. the bottom of a tank)

Pressure at the bottom, p = γh (Uniform on the entire plane)

Resultant force on the bottom area of container FR = pA = γhA



2. On a Vertical surface (e.g. the side wall of a tank)

3. On an Inclined surface

Consider a plane shown



At surface: p = patm

Angle θ between free surface & the inclined plane

y axis: Along the surface & x axis: Out of the plane

Along the vertical depth h, the pressure linearly changes & Hydrostatic force

changes

Differential Force acting on the differential area dA of plane,

dF = (Pressure) ⋅ (Area) = (γh) ⋅ (dA) (Perpendicular to plane)



Then, Magnitude of total resultant force FR

Since all the differential forces that were summed to obtain FR are perpendicular

to the surface, the resultant FR must also be perpendicular to the surface.

The point through which the resultant force acts is called the center of pressure.



.



3. Curved Surface (Nonplanar surface)

- FR: Change in both magnitude and direction

Special method using equilibrium of fluid

Hydrostatic force on curved surface can be analyzed best by resolving

the total pressure force on the surface into its horizontal & vertical

components.

Step 1. Isolate the section of fluid

(Section ABC with a unit perpendicular length)

Step 2. Draw a free-body diagram

F1(F2 ): Pressure force on Horizontal (Vertical) plane

W : Weight of fluid (γ × Volume)



FH (FV ): Horizontal (Vertical) comp. of reaction force by wall

Example 1



Example 2

The gate OA closes an opening 2.1 m deep and 1.6 m wide. What

torque must be applied at O to keep the gate closed?



Example3

Determine the total hydrostatic pressure and center of pressure on the 5 m long, 2

m high quadrant gate.

Fundamental Concepts in Fluid Mechanics

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