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Basic Fluid Mechanics by Prof. V.Sundar

BASIC FLUID MECHANICS

Prof. V. Sundar, Ocean Engineering Centre, IIT Madras

1.1 GENERAL

Matter exists in three states namely solid, liquid and gas. Liquid and gas are commonly

called fluids. Fluids can be defined as a substance, which undergoes continuous

deformation under the action of shear forces regardless of their magnitude. The maindistinction between a liquid and a gas lies in their rate of change in the density. The

density of gas changes more readily than that of liquid. However, they can be treated in

the same way without taking into account the change of density, provided that the speedof flow is low as compared with the speed of sound propagation in the fluid. The fluid is

called incompressible if the change of the density is negligible.

Ideal Fluids

Fluids with no viscosity, no surface Tension and are incompressible are called as ideal

fluids. That is, for ideal fluids no resistance is encountered as the fluid moves. Ideal fluidsdo not exist in nature. They are imaginary fluids. Fluids, which have, low viscosity such

as air, water may however be treated as ideal fluid without much error.

Practical or Real Fluids

Exist in nature. Has Viscosity, surface tension and are compressible.

1.2 TYPES OF FLOW:

1.

Steady and Unsteady flow

2. Uniform and Non-uniform

3. Rotational and Irrotational4. Laminar and Turbulent.

Steady flow: Fluid characteristics such as Velocity, u, pressure, p, density, ρ, temp, T,

etc., at any point do not change with time.

i.e. At(x,y,z), .0t

,0t

p,0

t

u=

∂ρ∂

=∂∂

=⎟ ⎠

⎞⎜⎝

⎛ ∂∂

Unsteady Flow: Fluid characteristics do change with time

i.e. At(x,y,z), 0t

T,0

t,0

t

p,0

t

u≠

∂∂

≠∂ρ∂

≠∂∂

≠⎟ ⎠

⎞⎜⎝

⎛ ∂∂

Most of the practical problems of engineering involve only steady flow conditions and is

simpler to solve than problems of unsteady flow.

1




Eg. Wave motion.

Uniform Flow:

When fluid properties does not change both in magnitude and direction, from point to

point in the fluid at any given instant of time, the flow is said to be uniform.

0

1tts

u=

=⎟ ⎠

⎞⎜⎝

⎛ ∂∂

Velocity with respect to distance

Eg. Flow of liquids under pressure through long pipelines of constant diameter is uniformflow.

Non-Uniform Flow:

If velocity of fluid changes from point to point at any instant, the flow is non-uniformEg. Flow of liquid under pressure through long pipelines of varying diameter.

Steady (or unsteady) and uniform (or nonuniform) flow can exist independently of eachother, so that any of four combinations is possible. Thus the flow of liquid at a constant

rate in a long straight pipe of constant diameter is steady uniform flow, the flow of liquid

at a constant rate through a conical pipe is steady nonuniform flow, while at a changingrate of flow these cases become unsteady uniform and unsteady nonuniform flow

respectively.

Rotational Flow:

A flow is said to be rotational if the fluid particles while moving in the direction of flow

rotate, about their mass centres.Eg. Liquid in a rotating tank.

Irrotational Flow:

The fluid particles while moving in the direction of flow do not rotate about their mass

centres. This type of flow exists only in the case of Ideal Fluid for which no tangential or

Shear Stresses occur. But the flow of Real fluids may also be assumed to be irrotational ifthe viscosity of the fluid has little significance. For a fluid flow to be irrotational the

following conditions are to be satisfied.

It can be proved that the rotation components about the axes parallel to x and y axes can

be obtained as

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

∂∂

−∂∂

=z

v

y

w

2

1xW

⎟ ⎠

⎞⎜⎝

⎛ ∂∂

−∂∂

=x

w

z

u

2

1yW

2




and if at every point in the flowing fluid the rotation components W x, Wy, and Wz areequal to zero then

y

u

x

v;0W

x

w

z

u;0W

z

v

y

w;0W

z

yx

∂∂

=∂∂

=

∂∂

=∂∂

=

∂

∂=

∂

∂=

(1.1)

where u, v and w are velocities in the x, y, and z directions respectively.

Laminar Flow:

A flow is said to be laminar if the fluid particles move along straight parallel paths in

layers, such that the path of the individual fluid particles do not cross those of theneighbouring particles. In other words, the fluids appear to move by the sliding oflaminations of infinitesimal thickness relative to adjacent layers. This type of flow occurs

when the viscous forces dominate the inertia forces at low velocities. Laminar flow canoccur in flow through pipes, open channels, and through porous media.

Turbulent Flow:

A fluid motion is said to be turbulent when the fluid particles move in an entirely randomor disorderly manner, that results in a rapid and continuous mixing of the fluid leading to

momentum transfer as flow occurs. A distinguishing characteristic of turbulence is its

irregularity, there being no definite frequency, as in wave motion, and no observable

pattern, as in the case of large eddies. Eddies or Vortices of different sizes and shapes are present moving over large distances in such a fluid flow. Flow in natural streams,

artificial channels, Sewers, etc. are a few examples of turbulent flow.

1.3 CONTINUITY EQUATION CONSERVATION OF MASS :

3




4

In a real fluid, mass must be conserved; It cannot be created nor destroyed.

Consider an elementary rectangular parallel piped with sides of length z,y,x ∆∆Λ as

shown in Fig.1.1

Fig.1.1. Definition sketch of coordinate system.

Let the centre of the fluid medium be at a point P (x, y, z) where the velocity components

in the x, y, z directions are u, v, w respectively and ρ be the mass density of the fluid.

The mass of fluid passing per unit time through the face of area zy∆∆ normal to the x-

axis through point P is (ρu ∆ ).zy∆

Then mass of fluid flowing per unit time into the parallelepiped through the face ABCD

is

( ) ( ) ⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ ∆−∆∆ρ∂∂

+∆∆ρ2

xz.y.u

xz.y.u

In the above expression (-ve) sign has been used since face ABCD is on the left of

point P.

Similarly the mass of fluid flowing per unit time out of the fluid medium through the face

A' B' C' D' is

( ) ( ) ⎥⎦

⎤⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ ∆−∆∆ρ∂∂

+∆∆ρ2

xz.y.u

xz.y.u




5

Therefore, the net mass of fluid that has remained in the fluid medium per unit time

through the pair of faces ABCD and A’B’C’D’ is obtained as = ( ) .xz.y.u.x

∆∆∆ρ∂∂−

= ( ) z.y.x.ux

∆∆∆ρ∂∂−

The area has been taken out of the parenthesis since it is not a function of x.( z.y∆∆ )By applying the same procedure the net mass of fluid that remains in the cube per unit

time through the other two pairs of faces of the cube may also be obtained as

( ) ( z.y.x.vy

∆∆∆ρ∂∂−

= ) through pair of faces A A’D’D and B B’C’C

( ) ( z.y.x.wZ

∆∆∆ρ∂ ∂−= )

( )

through pair of faces D D’C’C and A A’ B’B.

By adding all these expressions the net total mass of fluid that has remained in the cube

per unit time is

( ) ( )z.y.x

z

w

y

v

x

u∆∆∆⎥

⎦

⎤⎢⎣

⎡∂ρ∂

+∂ρ∂

+∂ρ∂

− (1.2)

Since the fluid is neither created nor destroyed in the cube, any increase in the mass of

the fluid contained in this space per unit time, is equal to the net total mass of fluid,

that has remained in the cube per unit time, which is expressed by the aboveexpression.

( )z.y.x ∆∆∆ρThe mass of the fluid in the cube is and its rate of increase with time is

( ) ( )z.y.x.t

z.y.x.t

∆∆∆∂ρ∂

=∆∆∆ρ∂∂

(1.3)

Equating the two expressions Eq. (1.2) = Eq.(1.3)

( ) ( ) ( ) ( )z.y.x.t

z.y.xzw

yv

xu ∆∆∆

∂ρ∂=∆∆∆⎥

⎦⎤⎢

⎣⎡

∂ρ∂+

∂ρ∂+

∂ρ∂−

Dividing both sides of above expression by the volume of the parallelopiped ( )z.y.x ∆∆∆

And taking the limit so that the fluid medium shrinks to the point P(x,y,z) the continuity

equation is obtained as




6

( ) ( ) ( )0

z

w

y

v

x

u

t=

∂ρ∂

+∂ρ∂

+∂ρ∂

+∂ρ∂

(1.4)

This equation represents the continuity eqn. In its most general form and is applicable forsteady as well as unsteady flow, Uniform and Non-Uniform flow and compressible as

well as incompressible fluids.

For Steady flow 0t=

∂ρ∂

i.e, the above eq. Becomes

( ) ( ) ( ).0

z

w

y

v

x

u=

∂ρ∂

+∂ρ∂

+∂ρ∂

Further for an incompressible fluid the mass density ‘ρ’ does not change with x,y,z and

t, hence, the above equation simplifies to

0z

w

y

v

x

u=

∂∂

+∂∂

+∂∂

(1.5)

1.4 FORCES ACTING ON FLUIDS IN MOTION

The different forces influencing the fluid motion are due to gravity, pressure, viscosity,turbulence, surface tension and compressibility and are listed below.

Fg (Gravity Force) Due to Wt. of Fluid

= (Mass * gravitational constant)

F p (Pressure Force) Due to pressure gradient

Fv (Viscous Force) Due to Viscosity

Ft (Turbulent Force) Due to Turbulence

Fs (Surface Tension Force) Due to Surface Tension

Fe (Compressibility Force) Due to elastic property of the fluid

If a certain mass of fluid in motion is influenced by all the above mentioned forces then

according to Newton’s Second law of motion the following equation of motion may bewritten

Ma = Fg + F p + Fv + Ft + Fs + Fe (1.6)

Further resolving these forces in x,y,z direction




7

Max = Fgx + F px + Fvx + Ftx + Fsx + Fex

May = Fgy + F py + Fvy + Fty + Fsy + Fey

Maz = Fgz + F pz + Fvz + Ftz + Fsz + Fez

Where M is the mass is of fluid and ax , ay , az are fluid acceleration in the x, y and z

directions respectively.In most fluid problems Fe and Fs may be neglected

Hence Ma = Fg + F p +Fv + Ft (1.7)

Eq.(1.7) is known as Reynold’s Equation of Motion

For laminar Flows Ft is negligible

Hence Ma = Fg + F p + Fv (1.8)

Eq.(1.8) known as Navier Stokes equation

In case of Ideal Fluids Fv = o

Hence Ma = Fg + F p (1.9)

Eq.(1.9) is known as the Euler’s Equation of Motion

1.5 EULER’S EQUATION OF MOTION

Only pressure forces and the fluid weight or in general, the body force, are assumed to be

acting on the mass of fluid in motion. Consider a point P(x,y,z) in a flowing mass of fluid

at which let u, v and w be the velocity components in directions x,y and z respectively. ρ

mass density; p be pressure intensity. Let X, Y, and Z be the components of the body

force per unit mass at the same point.

Mass of fluid in the fluid medium considered in Fig. 1.2 is ( )z.y.x ∆∆∆ρ . Therefore, total

component of the body force acting on the cube in x direction = ( )z.y.xX ∆∆∆ρ . Similarly

in y and z direction are ( )z.y.xY ∆∆∆ρ and ( )z.y.xZ ∆∆∆ρ ’p’ is pressure intensity at

point ‘P’. Since the lengths of the edges of the fluid medium are extremely small, it may be assumed that the ‘p’ on the face PQR’S is uniform and equal to p.




8

zyxx

p p ∆∆⎟

⎠

⎞⎜⎝

⎛ ∆∂∂

+

Fig.1.2. Definition sketch of coordinate system.

Therefore, Total pressure force acting on face

PQR’S in x direction = p. z.y ∆∆ .

Since the ‘p’ vary with x, y and z the pressure intensity on the face

RS’P’Q’ will be = ⎟ ⎠

⎞⎜⎝

⎛ ∆∂∂

+ x.x

p p

Therefore, the total pressure force acting on the face RS’P’Q’ in the

z.yx.x

p p ∆∆⎟

⎠

⎞

⎝ ∆∂∂

+⎜⎛

x direction =

Net pressure force F px acting on the fluid mass in the x direction, the magnitude of which

is obtained as

z.y.x.z

p pzF

z.y.x.

y

p pyF

z.y.x.x

p pxFor

z.yx.x

p pz.y p pxF

∆∆∆∂∂

−=

∆∆∆

∂

∂−=

∆∆∆∂∂

−=

∆∆⎟ ⎠

⎞⎜⎝

⎛ ∂∂∂

+−∆∆=




9

Further, Pressure Force per unit volume are

z

p pzF

y

p pyF

x

p pxF

∂∂

−=

∂∂−=

∂∂

−=

Now adding the pressure force and Body Force and equating to mass* Acceleration in ‘x’

direction in accordance with the Newton’s Second law of motion we get

( ) ( ) xaz.y.xz.y.x.x

pz.y.xX ∆∆∆ρ=∆∆∆∂

∂−∆∆∆ρ (1.10)

i.e. X -x

p1

∂∂

ρ = ax

Similarly (1.11)

Y -y

p1

∂∂

ρ = ay

Z -z

p1

∂∂

ρ = az

The above equations are known as Euler’s equation of motion. Here ax =

dt

du, ay =

dt

dv,

and az =dt

dwmay be expressed in terms of the velocity components u,v and w as

z

uw

y

uv

x

uu

t

u

∂∂

+∂∂

+∂∂

+∂∂

ax =

z

vw

y

vv

x

vu

t

v

∂∂

+∂∂

+∂∂

+∂∂

ay =

az =z

ww

y

wv

x

wu

t

w

∂

∂+

∂

∂+

∂

∂+

∂

∂

dt

u∂,

dt

v∂,

dt

w∂are know as local or temporal accel. and




10

x

u

∂∂

,y

u

∂∂

,z

u

∂∂

,x

v

∂∂

,y

v

∂∂

,z

v

∂∂

,x

w

∂∂

,y

w

∂∂

,z

w

∂∂

are Convective accel.

In these derivations no assumptions has been made that ‘ρ ’ is a constant. Hence, these

equations are applicable to compressible or incompressible, nonviscous fluid in steady or

unsteady state of flow.

1.6 PATHLINES AND STREAMLINES:

A Pathlines is the trace made by a single particle over a period of time. The pathlineshows the direction of the velocity of the fluid particle at successive instants of time.

Streamlines show the mean direction of a number of particles at the same instant of time.

If a camera were to take a short time exposure of a flow in which there were a large

number of particles, each particle would trace a short path, which would indicate its

velocity during that brief interval. A series of curves drawn tangent to the means of thevelocity vectors are Streamlines. Pathlines and Streamlines are identical in the Steady

flow of a fluid in which there are no fluctuating velocity components, in other words, for

truly steady flow. The equation of a Streamline is represented as

dz

w

dy

v

dx

u== (1.12)

1.7 VELOCITY POTENTIAL :

Velocity Potential, is defined as a scalar function of space and time such that its

derivative with respect to any direction yields velocity in that direction. Hence, for anydirection S, in which the velocity is V

φ

s

sVs=

∂φ∂

yv,

x ∂φ∂

=∂φ∂

u =

when substituted in continuity eq. (1.5) we get

∇ 02 =φ

1.8 STREAM FUNCTION:

Stream function, ψ is defined as a scalar function of space and time, such that its partial

derivative with respect to any direction gives the velocity component at right angles (inthe counter clockwise direction) to this direction.

xv,

yu

∂ψ∂

=∂ψ∂

−=




11

For irrotational flow 02 =ψ∇

1.9 BERNOULLI EQUATION:

Let potentialforce be''Ω

i.e. X=z

Z,y

Y,x ∂

Ω∂−=∂Ω∂−=

∂Ω∂

u=z

w,y

v,x ∂

φ∂=

∂φ∂

=∂φ∂

Substituting these in Euler’s equation (1.11) and using the irrotational flow conditions we

get the following set of equations.

z

p1

zz

ww

z

vv

z

uu

tz

2

y

p1

yy

ww

y

vv

y

uu

ty

2

x

p1

xx

w

wx

v

vx

u

utx

2

∂∂

ρ−

∂Ω∂−

=∂∂

+∂∂

+∂∂

+∂∂φ∂

∂∂

ρ−

∂Ω∂−

=∂∂

+∂∂

+∂∂

+∂∂φ∂

∂

∂

ρ−∂

Ω∂−

=∂

∂

+∂

∂

+∂

∂

+∂∂

φ∂

(1.13)

If is constant integration with respect to x, y, and z of the above set of equations yieldsρ

( )

( )

( )t,y,x2F p

t

22v2u

2

1

t,x,z2F p

t

22v2u2

1

t,z,y1F p

t

22v2u2

1

=

ρ

+Ω+

∂

φ∂+⎟

⎠

⎞⎜

⎝

⎛ ω++

=ρ

+Ω+∂φ∂

+⎟ ⎠ ⎞⎜

⎝ ⎛ ω++

=ρ

+Ω+∂φ∂

+⎟ ⎠ ⎞⎜

⎝ ⎛ ω++

(1.14)

If u, v, ω are resolvents of V

( )tF p

t

2v2

1=

ρ+Ω+

∂φ∂

+ (1.15)




12

For steady flow ‘t’ disappears

We also have –g = - if h∂

Ω∂h is positive upwards

Hence =gh.Ω

Total head =ρ

++ p

gh2

2v= constant

(Kinetic head) + (Potential head) + (Pressure head)

SELECTED REFERENCES

1. Daugherty, R.L., Franzini, J.B. and Finnemore, E.J. “Fluid Mechanics with

Engineering Applications”, McGraw Hill, Inc. 1985.

2.

Garde, R.J. “Fluid Mechanics through Problems”, Wiley Eastern Limited, 1989.

3. Modi, P.N. and Seth, S.M. “Hydraulics and Fluid Mechanics”, published by StandardBook House, 1985.

Problem1.

The rate at which water flows through a horizontal 25cm pipe is increased linearly from30to150 liters/sec in 3.5secs. What pressure gradient must exist to produce this

acceleration? What difference in pressure intensity will prevail between sections 8m

apart? Take ρ = 102m slug/m3 (1000 kg/m

3 in SI unit).

Solution

Euler’s Equation along the pipe axis may be written as

x

uu

t

u

x

p1X

∂∂

+∂∂

=∂∂

ρ−

since the pipe has a constant diameter 0x

u=

∂∂

and since it is horizontal, the body force per unit volume, X along the flow direction is

also zero. The above equation of motion, therefore, reduces to

x.

1

t

u

∂ρ∂

ρ−=

∂∂

The changes in velocity as the flow changes from 30 to 150 liters/sec in




( ) ( )

s\m445.2

225.0*4

*1000

30

225.0*4

*1000

150u

=

π−

π=∆

2s/m698.0

5.3

445.2

t

u==

∂∂

∴

and the pressure gradient

MKSin698.0*102t

u

x−=

∂∂

ρ−=∂ρ∂

= -71.25 Kg/m2 /m in MKS= -1000 * 0.698 in SI= -698 N/m

2/m

Difference in pressure between sections 8m apart

= 8*x∂ρ∂

= -71.25*8

= -570.0 Kg/m2 (or)

= -698* 8N/m2

= -5.58 KN/m2

Problem2.

The wind velocity in a cyclone may be assumed to vary according to free Vortex law. Ifthe velocity is 16 Km/ hr, 50 km from the centre of the cyclone, what pressure gradient

should obtain at this point? What reduction in barometric pressure should occur over a

radial distance of 10 km from this point towards the centre of the storm? Take massdensity of air as 1.208 Kg (mass ) per m

3.

Solution

In a cyclone the wind velocity is given to vary according to the free Vortex law. Thevelocity distribution in a free Vortex is given by

V.r = C (1)At a radial location defined by

13




14

r 1 = 50km, V1 = 16kmph

Substituting this in Equation (1)

C= 16 * 50 = 800 km2 / hr

Velocity at a radial distance (50-10) = 40 km

From the centre of cyclone, V2

V2 = C/r 2 = 800 / 40 = 20 kmph

From Bernoulis Equation

=ρ+2

V p

2constant

By differentiating with respect to r

dr

dvV

dr

dp

)or (0dr

dvV

dr

dp

ρ−=

=ρ+

Differentiating Equation (1)

r

V

dr

dv)or (0V

dr

dvr −==+

r

2V

dr

dvV

dr

dpρ=ρ−=

( )

The pressure gradient at a radial distance of 50km where the velocity is 16 kmph

1000*50

2

3600

1000*16

*208.1km50r dr /dp ⎥⎦

⎤

⎢⎣

⎡

==

= 4.77 * 10-4

kg\m2 \m

= 0.477 kg\m2 per km

In SI units = 0.477 * 9.81 N/m2 per m




15

= 4.679 N/m2 per km.

Reduction in barometric pressure over a radial distance of 10km from

r 1 = 50 kms to r 2 = 40km

r 1 = 50km, V1 = 16kmph

r 2 = 40km, V2 = 20kmph

using Bernoulis equation,

2

22

V

2 p2

21

V

1 p ρ+=ρ+

(or) ( ) ( ) ⎥⎦

⎤⎢⎣

⎡ −ρ=− 21

V22

V2/12 p1 p

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ 23

23

3600

10x16

3600

10x20

2

208. ⎢1

= 6.71N/m2

Problem 3.

A 25cm diameter pipe carries oil of specific gravity 0.8 at the rate of 120 liters per second

and the pressure at a point A is 0.2 kg/cm2 (gage). If the point a is 3.5 m above the

dantum line, calculate the total energy at point A in meters of oil.

Solution

Total energy in terms of meters of oil is given by

zg2

V p

2++γ

( )oilof m5.2

1000*8.0

410*2.0 p=

γ

By continuity equation




16

Q= AV

sec/3m12.0610

310*120Q ==

( )

( ) ( )

( )

5.3z

oilof m31.081.9*2

245.2

g2

2Vand

sec/m45.2

049.0/12.0A/QV

2m049.0

225.04

A

=

==

=

==∴

=

π=

Therefore Total energy = (2.5 + 0.31+ 3.5)

= 6.31 m of oil

Problem 4.

A 30cm pipe carries water at a velocity of 24.4 m\sec. At point A measurement of

pressure and elevation were 3.68 kg/cm2 and 30.5 m respectively. The pressure and

elevation at point B were 2.94 kg/cm2 and 33.5m respectively. For steady flow, find the

loss of head between A and B.

Solution

Total energy in terms of meters of water is given by




zg2

2V p++

γ

A point A

( p / γ ) = (3.68 * 104) /1000

= 36.80 m of water

(V2 / 2g) = (24.4)

2 / (2 * 9.81)

= 30.35 m of water

z=30.50 m

∴Total energy at

A= (36.80 + 30.35 + 30.50) = 97.65 m

At point B

(p / γ) = (2.94 * 104) / 1000

= 29.40m of water

(V2 / 2g) = (24.4)

2/ (2* 9.81)

= 30.35 m of water

z= 33.50 m

∴Total energy at

B = (29.40 +30.35 + 33.50 ) = 93.25 m

∴Loss of head = (97.65 –93.25)

= 4.40 m.

Problem 5.

A pipe 300m long has a slope of 1 in 100 and tapers from 1.2m diameter at the high end

to 0.60m diameter at the low end. Quantity of water flowing is 5400 litres per minute. If

17




18

the pressure at higher end is 0.70 kg\cm2 find the pressure at the lower end. Neglect

losses.

Solution

Discharge Q = (5400 * 10

3

) / (60 * 10

6

)

= 0.09 m3 /sec

Area of flow section at higher end

( ) sec/m13.12.14

2 =π

∴Velocity at higher end

= (0.09 / 1.13) = 0.0796 m / secArea of flow at lower end = (π/4)(0.60)

4 = 0.28m/s

Velocity at lower end

= (0.09 / 0.28 ) = 0.3184 m / sec

Applying Bernoulli’s equation between the higher and the lower ends of the pipe

=++γ 1z

g2

21v1 p

2zg2

22v2 p

++γ

Assuming datum to be passing through the lower end,

Z2 = 0

Z1 = (300/100) = 3.0 m

AndThus by substitution

081.9*2

2)3184.0(2 p0.3

81.9*2

2)0796.0(

1000

410*70.0++

γ=++

γ2 p

or 7.0 + 0.0003 +3.0 = +0.0048

∴γ2 p

= 9.996 m of water

or p2 = 9996kg/m2

Problem 6.




19

A conical tube is fixed vertically with its smaller end upwards. The velocity of flow down

the tube is 4.5 m/sec at the upper end and 1.5 m/sec at the lower end. The tube is 1.5mlong and the pressure head at the upper end is 3.1 m of the liquid. The loss in the tube

expressed as a head is 0.3g2

2)VV( 21− where V1 and V2 are the velocities at the upper

and lower ends respectively. What is the pressure head at the lower end?

Solution

Applying Bernoulli’s equation between the upper and the lower ends,

=++γ 1z

g2

21v1 p

2zg2

22v2 p

++γ

+g2

2)2V1V(3.0 −

( ) ( )

( )

sec\m5.1V

sec\m5.4V

;m5.1zz.m1.3 p

g2

VV3.0

g2

V

g2

V

zz p p

2

1

211

22122

212112

=

=

=−=γ

−−−+−+

γ=

γ

or

by substitution we get

( ) ( ) ( )

( )

m38.5

14.011.003.15.11.32 por

81.9*2

25.15.43.0

81.9*2

25.1

81.9*2

25.45.11.32 p

=

−−++=γ

−−−++=

γ

∴Pressure head at lower end

.liquid of m38.52 p=

γ




20

Problem.7

In a three dimensional flow, velocity components in any two directions are as given

below. Find the velocity component in the the third direction such that the continuity

equation is satisfied.

(i) u = x3 = y

2+ 2z

2; v = -x

2y - yz -xy

(ii) u =222 )yx(

xyz2

+

− ; w =

)2y2x(

y

+

Solution:

(i) xz2xy

v,2x3

x

u−−−=

∂∂

=∂∂

Substituting the above in continuity equation (1.5) we get

3x2 - x

2 -z - z +

zw∂∂ = 0

z

w

∂∂

= x + z- 2x2

Integrating both sides, we get

W = (xz + )z2x22

2z− + c

Where C is constant of integration that could be a function of x and y

( ) ( )

422

22222

yx

x2*yx2*xyz2yz2yx

x

u

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ +−−−⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ +

=∂∂

(iii)

32y2x

z3y2yz2x6

⎟ ⎠ ⎞⎜

⎝ ⎛ +

− =

z

w

∂∂

= 0

Substituting the above in continuity equation (1.5) we get




21

32y2x

yz2x6z3y2

y

v

00y

v

32y2x

z3y2yz2x6

⎟ ⎠ ⎞⎜

⎝ ⎛ +

−=∂∂

=+∂∂

+⎟ ⎠ ⎞⎜

⎝ ⎛ +

−

Integrating both sides

1C2

2y2x

2y2xz

v +

⎟ ⎠ ⎞⎜

⎝ ⎛ +

⎟ ⎠ ⎞⎜

⎝ ⎛ −

=

Where C1 is integrating constant, a function of x and z.

Problem.8

The velocity components in a two-dimensional flow field for an incompressible fluid are

given as

3

3xy22xyv;y2xx2

3

3yu −−=−+=

i) Show that these functions represent an irrotational flow.

ii)

Obtain an expression for stream function. ψ

Solution:

2xy2y

v;xy22

x

u−=

∂∂

−=∂∂

i)

For a two dimensional flow of incompressible fluid, the continuity eq. (1.5) is expressed

as

0y

v

x

u

=∂∂

+∂∂

By substituting for the two terms

2-2xy +2xy – 2 = 0




22

Thus, continuity equation is satisfied and hence the functions represent a possible case of

fluid flow.Further,

2x2y

y

u;2x2y

x

v−=

∂

∂−=

∂

∂

On substituting the above in eq (1.1) conditions for irrotationality we get

02x2y2x2yy

u

x

v=+−−=

∂∂

−∂∂

Hence the flow is irrotational.

ii)3

3xy22xyv

x−−==

∂ψ∂

(i)

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −+−=−=

∂ψ∂

y2xx23

3yu

y (ii)

Integrating eq. (I) we get

( )yf 12

4xxy2

2

2y2x+−−=ψ (iii)

Differentiating eq. (iii) with respect to y we get

( )yf x2yxy

'2 +−=∂ψ∂

(iv)

Equating the values of ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ ∂ψ∂y

from eqs.(ii) and (iv) we get

- )y('f x2y2xy2xx23

3y+−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −−

or f’ (y) = C

12

4y+

−

Where C is a numerical constant of integration




12 b

2y

2a

2xisstreamlinetheof equationtheHence.

2

1cor 10

2a

2ac2 =+==+

1.basic fluid mechanics.pdf

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