fluid motion

Complimentary course notes for

Fluid Mechanics III, MFM31

Central University of Technology, Free State,

South Africa

Dr. Eng. Karanja Kibicho (PhD., R.Eng.,MIEK)Senior Lecturer

Department of Mechanical EngineeringJomo Kenyatta University of Agriculture & Technology.

CUT: MFM31 FLUID MECHANICS III NOTES BY DR. KARANJA KIBICHOJAN-MAY 2010

Contents

1 GENERAL INFORMATION 4

2 Fluid Motion 62.1 Types of Fluid Flows . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 Steady and unsteady flows . . . . . . . . . . . . . . . . . . . 62.1.2 Uniform and non-uniform flows . . . . . . . . . . . . . . . . 6

3 Types of Flows 73.1 Reynolds Demonstration of Different Kind of Flows . . . . . . . . . 7

4 Viscous Flows 104.1 Laminar Flow in Pipes . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 Laminar Flow Between Parallel Plates . . . . . . . . . . . . . . . . 144.3 The Simple Cylindrical Dashpot . . . . . . . . . . . . . . . . . . . . 18

5 Turbulent Flow in Pipes 215.1 Darcy’S Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Velocity Distribution in Turbulent Flows . . . . . . . . . . . . . . . 265.3 Minor Losses in Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . 285.4 Loss due to Sudden Expansion . . . . . . . . . . . . . . . . . . . . . 295.5 Loss due to Sudden Contraction . . . . . . . . . . . . . . . . . . . . 315.6 Power Transmission through Pipes . . . . . . . . . . . . . . . . . . 335.7 Pipe in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.8 Pipes in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.9 Pipe Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.10 Branching-Pipe System of 3 Reservoirs . . . . . . . . . . . . . . . . 40

6 Momentum Equation 446.1 Linear Momentum Equation . . . . . . . . . . . . . . . . . . . . . . 446.2 Angular Momentum Equation . . . . . . . . . . . . . . . . . . . . . 476.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7 Dimensional Analysis and Similitude 527.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 527.2 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 527.3 Geometric Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . 557.4 Kinematic Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . 557.5 Dynamic Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . 557.6 Buckingham Pi Method . . . . . . . . . . . . . . . . . . . . . . . . 557.7 Determination of Pi Groups . . . . . . . . . . . . . . . . . . . . . . 567.8 Common Dimensionless Groups in Fluid Mechanics . . . . . . . . . 57

7.8.1 Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . 587.8.2 Froude Number . . . . . . . . . . . . . . . . . . . . . . . . . 587.8.3 Mach Number & Cauchy Number . . . . . . . . . . . . . . . 597.8.4 Weber Number . . . . . . . . . . . . . . . . . . . . . . . . . 597.8.5 Pressure, stress and force coefficients . . . . . . . . . . . . . 59

2


7.8.6 Power coefficients . . . . . . . . . . . . . . . . . . . . . . . . 607.8.7 Strouhal number . . . . . . . . . . . . . . . . . . . . . . . . 60

7.9 Modeling and Similitude . . . . . . . . . . . . . . . . . . . . . . . . 607.9.1 Theory of models . . . . . . . . . . . . . . . . . . . . . . . . 60

3

1 GENERAL INFORMATION

Course Title: Fluid Mechanics III

Course Code: MFM31BI

Lecturer: Dr. Karanja Kibicho

Office: BHP 152

Email: [email protected]

Lectures: Tuesdays: 13:30-14:55pm and Wednesdays: 9:25-11:00am

Practicals: Thursdays: 13:30-15:40pm

Learning Units:

(i) Viscous flow

(ii) Turbulent flow in pipelines

(iii) Flow under variable head

(iv) Hydrodynamic Force of a Jet

(v) Free and Forced Vortexes

(vi) Dimensional analysis

(vii) Reciprocating pumps

ExaminationTest 1 date to be announced cover unit 1 and 2Test 2 Main Scheduled) cover unit3, 4 and 5Test 3 date to be announced cover unit 6 and 7

Marks Distribution:40% CA (Semester mark)60% Main exam

Prescribed Text Books:

(i) Douglas J F & Matthews R D, 1996, Solving Problems in Fluid Mechanics(Volume 1), 3rd Ed., Longman Group Limited,

(ii) Douglas J F & Matthews R D, 1996, Solving Problems in Fluid Mechanics(Volume 2), 3rd Ed., Longman Group Limited,


Reference BookDouglas J F, Gasiorek JM, & Swaffield JA, 2001, Fluid Mechanics, 4th Ed., Pren-tice Hall

Course Aims

(i) solve any problem related to flow under varying head,

(ii) determine which type of flow occurs in a pipeline. The student must alsobe able to use all the losses in a pipeline to determine the discharge or therequired pipe diameters,

(iii) solve any problem relating to positive displacement pumps,

(iv) solve any problem relating to free and forced vortexes. The students mustalso know the application of vortexes,

(v) know the difference between laminar and turbulent flow to solve practicaldirected problems,

(vi) consider and compare the relation between certain equations and units bymeans of the basic dimensions (M,L,T).

5


2 Fluid Motion

A fluid is made up of millions of particles. In general, when a fluid flows, there is net

motion of molecules from one point in space to another point as a function of time.

To be able to describe the motion, then it is apparent that we have to account for

the motion of individual molecules. Considering the number of particles involved,

then this approach is totally unrealistic.

Rather, we employ the continuum hypothesis and consider fluids to be made up

of fluid particles that interact with each other and the surroundings only.

The fluid particles contain numerous molecules. The motion can thus be described

in terms of the velocity and acceleration of these fluid particles.

With continuum hypothesis, a description of any fluid flow property eg. pressure,

temperature, velocity and density, can be given as a function of the fluid’s location.

2.1 Types of Fluid Flows

2.1.1 Steady and unsteady flows

The flow parameters that describe a fluid flow are generally velocity, pressure,

temperature and density (known as the dependent variables). Experience shows

that these parameters are not constant based on both time and space variations.

A flow is considered to be steady if the dependent fluid variables at any point in

the flow do not change with time, i.e. the rate of change of the dependent variables

at a position is zero.

∂

∂t(dependent variables) = 0 (1)

A flow is said to be unsteady if the dependent fluid variables alter with passage of

time at a position in the flow, thus,

∂

∂t(dependent variables) 6= 0 (2)

2.1.2 Uniform and non-uniform flows

A flow is said to be uniform at an instant of time if the dependent variable (in

magnitude and direction sense) is identical throughout the flow-field. Thus the

6


dependent variable components at different positions of the flow are the same.

Thus, the space rate of change of the dependent variable components is zero. i.e.

∂

∂s(dependent variables) = 0 (3)

If the velocity components at different locations are different, the flow is said to

be non-uniform.i.e

∂

∂s(dependent variables) 6= 0 (4)

Steadiness therefore refers to ‘no change with time’ while uniformity refers

to ‘no change in space’. Therefore a flow can be steady or unsteady quite

independent of its being uniform or non-uniform. Thus we have four possibilities;

(i) Steady-uniform flow - e.g. in a long straight pipe

(ii) Steady-non-uniform - e.g. in a tapering pipe

(iii) Uniform-non-steady - e.g. Accelerating a fluid in a long straight pipe (think

of flow during start up of a pump)

(iv) Non-uniform-non-steady - e.g. Accelerating a fluid in a tapered pipe (think

of flow during start up of a pump)

3 Types of Flows

3.1 Reynolds Demonstration of Different Kind of Flows

Sketch on Douglas text page 100

7


Reynolds set up an experiment to investigate the types of flow that exist. The

apparatus was as shown above. The velocity of the water inside the glass tube

could be controlled by a valve. A fine nozzle connected to a small reservoir of a

liquid dye discharged a coloured filament into the inlet of the glass tube. At low

velocities, Reynolds observed that the filament of dye would pass down the tube

without mixing with the water and often so steadily as almost to seem stationary.

As he opened the valve further, therefore increasing the velocity of the water, this

type of flow persisted until a velocity was reached at which the stream of the dye

began to waver. Further increase in the velocity of water made the fluctuations

in the stream of dye more intense, particulary towards the outlet end of the tube

until a state was reached in which the dye no longer travelled the whole of the tube

as a distinct unbroken thread but quite suddenly mixed more or less completely

with the water in the tube.

In the first kind of flow, the particles of the fluid are evidently moving entirely in

straight lines. This kind of flow is called ‘laminar’ flow.

The second type of flow is known as turbulent flow. Turbulent motion is essentially

irregular motion with the movements of individual particles having no definite fre-

quency.

Consider a fluid particle in motion. Assume that the fluid is disturbed. The iner-

tial forces possessed by the moving fluid particle will tend to carry it in the new

direction while the viscous forces due to the surrounding fluid will tend to make

it conform to the motion of the rest of the stream.

In laminar flow, these viscous forces due to the viscous shear stresses are sufficient

to eliminate the effects of any deviation, but in turbulent flow, they are not.

Consequently, the criterion which determines whether flow will be laminar or tur-

bulent is the ratio of the inertial forces to the viscous forces acting on the particle.

By use of dimensional analysis, this ratio may be shown to be:

⇒ Inertial forces

Viscous forces= constant× ρv`

µ(5)

The quantityρv`

µis known as the Reynolds number. It is denoted by either R or

8


Re. It is the one that determines whether the flow will be laminar or turbulent.

Hence

Re =ρv`

µ(6)

and since it is a ratio of two forces, it is dimensionless.

Introducing the kinematic viscosity

ν =µ

ρ

Then

⇒ Re =v`

ν(7)

It is clear that Re is dependent on v. From the Reynolds experiment, we saw

that as the valve was opened further and further, a certain velocity was reached

when the flow transited from laminar to turbulent.That velocity is known as the

critical velocity and the Reynolds number associated with it is known as the crit-

ical Reynolds number. The critical Reynolds number is largely dependent on the

shape of the surfaces on which the fluid flows.

For a circular pipe Recritical = 2000

For a flat plate Recritical = 3.2× 105

It is important to also note the Recritical is dependent on surface roughness, stabil-

ity of the flow, the flow surface, etc.

Example 1

If the critical velocity of water in a 50mm diameter pipe is 0.049m/s, find the

critical velocity of air in m/s in a pipe of 150mm diameter given that the density

of water is 1000kg/m3 and of air is 1.2kg/m3. The coefficient of viscosity of water

is 1.2× 10−3kg/ms and of air is 1.8× 10−5kg/ms.

9


4 Viscous Flows

4.1 Laminar Flow in Pipes

Although not all conduits used to transport fluid from one location to another are

round in cross-section, most of the common ones are. Consider a general case of a

pipe inclined to the horizontal plane so that the effects of gravity are not ignored

Douglas book page 342

Let us consider equilibrium of a small cylindrical laminar layer of thickness δr at

a radius r.

If the flow is assumed steady, then,∑Fs = 0

Assume also that the fluid is incompressible.

The element is acted upon by the following forces;

(i) pressure forces

(ii) viscous forces

(iii) gravitational forces

10


∴ in the direction of motion, the forces can be summed up as follows;

P2πrδr−(P +

∂P

∂x

)2πrδr+τ2πrδx−

(τ2πrδx+

∂

∂r(τ2πrδx)

)δr+W sin θ = 0

Note that τ = τ(r) by definition

Where P = flow static pressure, W = mg = Weight of element, and τ = shear

stress at radius r

But sin θ = −δzδx, and W = 2πrδrδxρg

∴ The above equation simplifies to:-

∂

∂x(P + ρgz) +

1

r

∂

∂r(τr) = 0 (8)

The term (P + ρgz) in equation 8 is called the ‘piezometric pressure’ usually de-

noted as P ∗. For a fully developed steady laminar flow P ∗ is constant at a given

cross-section and is therefore independent of r.

Consequently, the piezometric pressure gradient is also independent of r.

∂P ∗

∂x6= f(r)

Equation 8 may be written as;

∂P ∗

∂x+

1

r

∂

∂r(τr) = 0 (9)

We can now integrate equation 9 wrt r to get;

r2

2

∂P ∗

∂x+ τr + C1 = 0 (10)

Recall the definition of the viscous shear stress τ

τ = −µ∂u∂r

The (−)ve implies that r is measured from the center of the pipe and

not from the pipe wall as the definition of shear stress would require!

Equation 10 can be written as

r2

2

∂P ∗

∂x− µr∂u

∂r+ C1 = 0 (11)

But we know that at r = 0, u = umax (a boundary condition, what is the value

of u when r = R?)

11


Mathematically, points at a maximum, a minimum or a point of inflexion are

known as turning points where the gradients must be zero. Consequently;

∂u

∂r= 0 at r = 0

Applying the above boundary condition to equation 11, it is easy to see that the

constant of integration C1 = 0, hence equation 11 becomes;

r2

2

∂P ∗

∂x− µr∂u

∂r= 0

which can be rearranged as;

r

2

∂P ∗

∂x− µ∂u

∂r= 0 (12)

Further integration of equation 12 wrt r gives;

r2

4

∂P ∗

∂x− µu+ C2 = 0 (13)

You must now know the ‘no slip on the wall’ boundary condition. This requires

that at r = R, u = 0.

Applying this boundary condition to equation 13, we get

R2

4

∂P ∗

∂x+ C2 = 0

⇒ C2 = −R2

4

∂P ∗

∂x

and therefore equation 13 can now be written as;

r2

4

∂P ∗

∂x− µu− R2

4

∂P ∗

∂x= 0

From which the velocity distribution can be worked out as;

u = −(R2 − r2)

4µ

∂P ∗

∂x(14)

At a given cross-section, equation 14 defines a parabolic velocity distribution pro-

file.

In the direction of flow (when u is (+)ve), since the term (R2 − r2) is always (+)ve,

the (−)ve in equation 14 can only mean that the pressure gradient is (−)ve in the

direction of flow.

12


The maximum velocity occurs when r = 0 i.e. at the center of the pipe and

therefore;

umax = −R2

4µ

∂P ∗

∂x(15)

For a small annular element, the discharge is;

δQ = u2πrδr

∴ Q =

∫ R

0

u2πrδr

Writing the expression for u from equation 14 and simplifying

⇒ Q =

∫ R

0

− π

2µ

∂P ∗

∂x

(R2r − r3

)dr

=π

2µ

∂P ∗

∂x

[R2r2

2− r4

4

]R0

And hence;

Q =π

8µ

∂P ∗

∂xR4 (16)

Now, for a pipe flow with a pressure drop of ∆P over a pipe length L, if we assume

a constant pressure drop gradient in the direction of flow, then

−∂P∗

∂x=

∆P ∗

L

Application of this assumption to equation 16 gives,

Q =∆P ∗πR4

8µL=

∆P ∗πd4

128µL(17)

Equation 17 is known as the ‘Hagen-Poisseiule’ equation and can be used to

evaluate a flow rate by simply measuring a pressure drop in a pipe flow.

The mean velocity is given by

u =Q

A=

∆P ∗πd4

128µL÷ πd2

4=

∆P ∗R2

8µL(18)

From equation 15 we can now write for the maximum velocity

umax =∆P ∗R2

4µL= 2×

(∆P ∗R2

8µL

)(19)

Thus, by comparing equations 18 and 19, we can deduce that;

u =1

2umax (20)

13


Example 1

An oil having a viscosity of 0.0098kgs/m2 and specific gravity of 1.59 flows through

a horizontal pipe of 5cm diameter with a pressure drop of 0.06kg/cm2 per metre

length of pipe. Determine;

(i) the rate of flow in kilogram per minute

(ii) the shear stress at the pipe wall

(iii) the total drag for 100m length of pipe

(iv) power required for 100m length of pipe to maintain the flow.

4.2 Laminar Flow Between Parallel Plates

We wish to establish the velocity profile between two infinitely long plate. To make

the analysis simple, we assume that:-

(i) the fluid incompressible

(ii) the flow is uniform and laminar

(iii) the plates are parallel and infinitely long so that end effects may be ignored

Consider the flow shown below:-


14


A small element at a distance y from the lower plate is of length δx and breadth

δy and unit width. It is acted upon by static pressure forces and shearing forces.

Since the flow is steady, we expect that in the axial direction,∑Fs = 0

and hence we can sum the forces in the figure above as;

Pδy −(P +

∂P

∂xδx

)δy +W sin θ − τδx+

(τδx+

∂

∂y(τδx)δy

)= 0 (21)

But

W = ρδxδy × 1× g and sin θ = −δzδx

Knowing that

limδx→0

δz

δx=dz

dx

and applying it to equation 21 and simplifying we get;

−∂P∂x

δxδy − ρgδxδy dzdx

+∂

∂y(τδx)δy = 0

Since x 6= x(y), the above equation simplifies to

− d

dx(P + ρgz) +

dτ

dy= 0

⇒ −dP∗

dx+dτ

dy= 0

Recall that dP ∗/dx 6= f(y), hence the above equation may be integrated wrt y as;∫dτ =

∫dP ∗

dxdy

⇒ τ =dP ∗

dxy + C1 (22)

and since

τ = µdu

dy

then

µdu

dy=dP ∗

dxy + C1

15


Which may be re-arranged as;

du =1

µ

dP ∗

dxydy +

C1

µdy

Integration of this equation wrt y gives;

u =1

µ

dP ∗

dx

y2

2+C1

µy + C2 (23)

The boundary conditions are such that at;

y = 0, u = 0

y = c, u = V

Applying these boundary conditions to equation 23 we get;

C2 = 0

C1 = µV

2− C

2

dP ∗

dx

Which when substituted to equation 23 gives;

u =1

2µ

dP ∗

dx

(y2 − yc

)+yV

c(24)

Equation 24 represents the velocity profile across the gap between the two plates,

and is a general equation from which a number of restricted cases may be consid-

ered.

(i) Horizontal plate with no movement of the upper plate

Vdz

dx= 0, V = 0 and P ∗ = P

Thus equation 24 becomes

u =1

2µ

dP

dx

(y2 − yc

)(25)

(ii) Horizontal plates with upper plate in motion

Vdz

dx= 0, V = V and P ∗ = P

which makes equation 24 to become

u =1

2µ

dP

dx

(y2 − yc

)+yV

c(26)

16


(iii) Horizontal plates with upper plate in motion and no pressure gradient

Vdz

dx= 0, V = V and

dP ∗

dx= 0

and hence from equation 24 we get;

u =yV

c(27)

Equation 27 tells us that there will be a fluid flow even if there is no pressure

gradient. This is called coutte flow.

Let us now use the general equation 24 to obtain the flow rate between the two

plates.

For the element under analysis,

δQ = uδy per unit width.

∴ Q =

∫dQ =

∫ y=c

y=0

udy

⇒ Q =

∫ c

0

[1

2µ

dP ∗

dx(y2 − yc) +

yV

c

]dy

=

[1

2µ

dP ∗

dx

(y3

3− cy2

2

)+y2V

2c

]c0

∴ Q =cV

2− 1

12µc3dP ∗

dx(28)

Example 2

Fluid of density 1260kg/m3 and viscosity 0.9Ns/m2 passes between two infinitely

long parallel plates 2cm separation. If the flow rate is 0.5litres/sec per unit width,

calculate the pressure drop per unit length if;

(i) both plates are stationary

(ii) plates are inclined at an angle 25◦ to the horizontal

(iii) the upper plate in case (ii) moves at a velocity o.75m/s relative to the lower

plate in the direction of flow.

17


4.3 The Simple Cylindrical Dashpot


A dashpot is a device for damping vibrations of machines or rapidly reciprocating

motion.

It uses a cylinder and a fluid of relatively high viscosity.

If a force is applied on the piston downwards, oil is displaced from underneath it

and passes through the clearance.

For high viscosity, the flow in the clearance may be considered as laminar and

occurs simply as a result of the pressure developed underneath the piston.

This action may be considered as laminar flow between parallel plates with relative

velocity with the piston as the moving plate.

Recall equation 28;

∴ Q =cV

2− 1

12µc3dP ∗

dxper unit width of the plates

18


The plate may be considered to be bent around the circle of width πD

∴ Q =

(cV

2− 1

12µc3dP ∗

dx

)× πD

The fluid velocity is in opposite direction to that of the plate (piston). If up is the

velocity of the piston, then up = −V

⇒ Q =

[c(−up)

2− 1

12µc3dP ∗

dx

]× πD

=

[−cup

2− 1

12µc3dP ∗

dx

]× πD

Recall;

−∂P∗

∂x=

∆P ∗

L

Q =

[1

12µc3

∆P ∗

L− cup

2

]× πD

The term cup/2 is usually small and may be neglected.

Considering the displacement of fluid caused by motion of the piston;

Q =πD2

4× up

⇒ πD2

4× up = Q =

[1

12µc3

∆P ∗

L− cup

2

]× πD

Therefore;

up

(D

2+ c

)=

1

6

c3

µ

∆P ∗

L(29)

Example 3

The radial clearance between an hydraulic plunger and the cylinder wall is 0.1mm,

the length of the plunger is 0.3m and the diameter 100mm. Find the velocity and

the rate of leakage past the plunger at an instant when the difference of the pressure

between the two ends of the plunger is 9m of water. Take µ = 1.31× 10−3kg/ms.

19


TUTORIAL SHEET 3 - VISCOUS FLOW BETWEEN SOLIDBOUNDARIES

(i) A hydraulic system operates at a gauge pressure of 20MPa and 55◦C. The hy-

draulic fluid is SAE10W oil of specific gravity 0.92 and viscosity 0.018kg/ms.

A control valve consists of a piston 25mm in diameter fitted to a cylinder

with a mean radial clearance of 0.005mm. Determine the leakage flow rate

if the gauge pressure on the low pressure side of the piston is 1Mpa and the

length of the piston is 15mm.

(ii) A viscous oil flows steadily between parallel plates. The flow is laminar and

fully developed. The velocity profile is given by;

u =−h2

8µ

∂P

∂x

[1−

(2y

h

)2]

where the total gap width between the plates, h=3mm, and y is measured

from the centerline of the gap. The oil viscosity is 0.5Ns/m2 and the pressure

gradient is -1200N/m2/m. Find the magnitude and direction of the shear

stress on the upper plate and the volume flow rate through the channel, per

meter width.

(iii) (a) The figure below shows a fluid coupling for power transmission between

input and output shafts. It comprises two flat circular discs of 300mm

diameter D, separated by a 1.2mm space, h, filled with oil of viscosity

µ, 0.5Ns/m2. Assuming that the velocity variation is linear, show that

the torque, T available at the output shaft is

T =πD4µ(ω1 − ω2)

32h

ω1, ω2 are input and output shaft speeds respectively in rad/s.

(b) The input shaft is rotating at 900rpm and is supplied with power of

500W. Calculate speed, torque and power available at the output shaft.

(c) If the input speed remains constant at 900rpm, determine the maximum

output power available from the coupling and the corresponding output

speed.

20


5 Turbulent Flow in Pipes

5.1 Darcy’S Equation

Turbulent flow in a pipe was classified as that flow whose Re > 2000. The flows

are characterized by mixing and random motion of fluid particles which involve ex-

change of momentum between these particles. At macroscopic levels, the average

velocity due to this exchange of momentum is constant apart from a region near

the wall where velocity grows from zero to this constant average velocity, known

as the boundary layer.


Recall the definition of shear stress: Fluid layers must move at different velocities

for the shearing force and hence shear stress to exist. Basically, a velocity gradient

must exist for a shear stress to exist.

Going by our description of a turbulent pipe flow, then it is only in the region next

to the wall (boundary layer) that the velocity gradient exists. Thus we can only

talk of a shearing retarding force that is confined to the region near the wall. We

shall call this, the ‘wall shear stress’ and denote it by τ0 (some people prefer to

denote it by τw).

21


Recall the definition of Re

Re =u`

υ

Most flows in engineering involve fluids of low kinematic viscosity and large charac-

teristic lengths. Consequently, turbulent flows are the most common in engineering

applications. Unfortunately, due to the complex nature in turbulent flows (inter-

action of fluid particles through exchange of energy and momentum of the mean

and fluctuating flow-fields is quite complex), analytical solutions to turbulent flow

problems are unavailable. To make progress, we result in using empirical relation-

ships, that borrow heavily from the experimental data, to carry out any meaningful

analysis.

The following assumptions are made;

(i) the fluid is incompressible

(ii) the flow is fully developed, uniform and steady

(iii) the velocity gradient existing within the boundary layer introduces a shearing

stress τ0 at the wall

Consider a general case of a pipe inclined at an angle θ,


22


where;

S=Wetted perimeter (contact with fluid)

A=X-sectional area

τ0=Wall shear stress.

For steady flow

∑Fs = 0

∴ P1A− P2A− τ0s`+W sin θ = 0 (30)

But W = ρA`g and sin θ =−∆z

`

Noting that ∆P = P2 − P1

Equation 30 can be written as;

−∆PA− τ0S`− ρA�g∆z

�= 0

Dividing through by A and defining the hydraulic mean depth m, as;

m =A

s

We get

−∆(P + ρgz)− τ0m` = 0

and thus;

−∆P ∗

`− τ0m

= 0 (31)

Recalling that∆P ∗

`= −∂P

∗

∂xequation 31 then becomes

∂P ∗

∂x=τ0m

(32)

Let us now define a new parameter called ‘flow friction factor’, f , as;

f =τ0

ρv2/2(33)

where v is the average velocity.

By using equations 32 and 33, we can write

fρv2

2= m

∂P ∗

∂x

23


⇒ ∂P ∗

∂x= f

ρv2

2m(34)

If we denote the frictional head loss by hf for a length `, then if the pressure drop

is purely out of frictional losses (read shear stress losses), then;

∂P ∗

∂x=ρghf`

(35)

Combining equations 34 and 35, we get

fρv2

2m=ρghf`

∴ hf =f`v2

2gm(36)

For a circular pipe;

m =A

s=πd2

4× 1

πd=d

4

Substituting this to equation 36

hf =f`v2

2gd/4=

4f`

d· v

2

2g(37)

Equation 37 is called the ‘Darcy’s equation’.

In terms of pressure head, Darcy’s equation may be written as;

ρghf = ∆P =4f`

d· v

2

2g�× ρg�

⇒ ∆P =4f`

d· ρv

2

2(38)

Some authors prefer to write (4f) as f ′ so that

hf =f ′`

d· v

2

2g(39)

Though equation 37 is more frequently used. Experimentally, it has been found

out that;

f = φ(v, d, ρ, µ, k) (40)

24


Figure 1: Moody’s Chart

By using dimensional analysis, it can be shown that

f = φ

(Re,

k

d

)(41)

Experimentally, Blasius in 1913, showed that for turbulent flows in smooth pipes

f = 0.079Re−1/4 for Re > 105 (42)

Other researchers have shown that

f = 0.316Re−1/4 for 4× 103 < Re < 105 (43)

However, a reference fvsRe plots are most convenient. The carpet plot takes

into account the variation of f wrt Re as well as the relative roughness of the

pipe. Nikuradse compiled these plots after extensive experimentation into what is

frequently referred to as Stanton or Moody diagram shown in the figure 1.

For laminar flow, the Hagen-Poiseuille equations.

Q =∆Pπd4

128µ`and since Q = v

πd2

4

∴ vπd2

4=

∆Pπd4

128µ`but ∆P = ρghf

⇒ v =∆Pd2

32µ`=ρghfd

2

32µ`

25


and thus,

hf =32µ`v

ρgd2=

4f`

d· v

2

2gfrom Darcy’s equation

∴ f =16µ

ρvd=

16

Re(44)

Example 1

Water at a density of 998kg/m3 and kinematic viscosity 1×10−6m2/s flows through

a smooth tubing at a mean velocity of 2m/s. If the tube diameter is 30mm,

calculate the pressure gradient per unit length necessary. Assume that friction

factor for turbulent flow is given by 0.079Re−1/4. If instead of water, oil of density

800kg/m3 and viscosity 0.027kg/ms was used, what would be the pressure gradient

per unit length?

5.2 Velocity Distribution in Turbulent Flows

Let us use the dimensional analysis to build up a case. For a fully developed

turbulent flow, we may assume the local velocity u at a distance y is related to

only the following parameters;

u = u(ρ, µ, τ0, R, y, k) (45)

By using dimensional analysis, we may show that

u√τ0/ρ

= u

(ρR

µ

√τ0ρ,y

R,k

R

)(46)

The term√τ0/ρ = u∗ is referred to as the shear stress velocity.

∴u

u∗= u

(ρu∗R

µ,y

R,k

R

)(47)

The term k/R is significant close to the wall, while the term ρu∗R/µ is only im-

portant in the laminar sub-layer.

Equation 47 can therefore be written as;

∴u

u∗= u

( yR

)(48)

26


Since equation 48 is a result of dimensional analysis and a ratio of two velocities,

if we consider the centreline velocity to be the maximum, umax, we can generally

write equation 48 as

∴u

umax

= φ( yR

)(49)

Prandtl proposed that the velocity distribution can be represented by

∴u

umax

=( yR

)n(50)

Experimentally, n = 1/7 for Re > 105, leading to equation 50 being referred to as

the one-seventh power law.

For a symmetrical flow however, we expect that at the centre when y = R, then

∂u/∂y = 0 , which obviously cannot be satisfied by equation 50.

Near the wall, we know also that the viscous effects are dominant and hence;

u

u∗= φ(Re∗) where Re∗ =

ρu∗y

µ

Further, at y = 0, the velocity must be zero.

To satisfy these two requirements then the velocity profile must be defined such

that

u

u∗= A lnRe∗ + A1 (51)

The velocity profile given by this equation is known as the universal velocity dis-

tribution. Both A and A1 are determined experimentally.

In the laminar sub-layer viscosity is predominant and since

τ = µ∂u

∂y

Hence by integration

u = τ

(y

µ

)+ E

At y = 0, u = 0 and thus E = 0 which results in

u =τy

µ

27


∴u√τ/ρ

=u

u∗=yρτ 1/2

µρ1/2=ρu∗y

µ= Re∗ up to Re∗ ∼ 8 (52)

Nikuradse experimentally showed that for Re∗ > 30, A = 2.5 and A1 = 5.5 and

thus;

u

u∗= 2.5 lnRe∗ + 5.5 (53)

Empirical relationships that include f are found in literature. An example is the

Colebrook-White equation.

1√f

= −4 log

{k

3.71d+

1.26

Re√f

}(54)

Example 2

(i) Assuming the velocity distribution in a circular pipe is given by the one-

seventh power law, calculate the ratio between mean velocity and maximum

velocity.

(ii) the radius at which the actual velocity is equal to mean velocity.

5.3 Minor Losses in Pipe Flow

Consider flow in the following pipeline;

Draw a pipe line with all sorts of pipe fittings

28


The flow can be analyzed by taking Bernoulli’s equation between any 2 points.

Thus between points 1 and 2;

Energy at 1/wt+energy added by pump/wt=Energy at 2/wt+losses of energy/wt+energy

removed by turbine/wt.

The losses in the pipeline are due to

(i) Major losses due friction (out of fluid viscosity and pipe roughness)

(ii) Minor losses due to

(a) Entrance (tank-pipe entry; velocity being created)

(b) Exit (pipe-tank-atmosphere; velocity being destroyed)

(c) Sudden expansion (change of velocity)

(d) Sudden contraction (change of velocity)

(e) Bends, elbows, fittings and valves (change in magnitude and direction

of velocity)

(f) Gradual expansion and contraction (change of velocity) e.t.c.

Quantitative expression for minor losses are hard to develop except for some cases

and since these losses are small compared to the frictional losses, it has been found

convenient to express these losses as a fraction of velocity head i.e.

hl = kv2

2g(55)

Where k is the loss coefficient and can only be determined experimentally.

5.4 Loss due to Sudden Expansion

Due to sudden enlargement, the flow is decelerated abruptly, separates from the

walls develops eddies and loss of energy (pressure) results. The flow then reat-

taches downstream.

A lot of experimental evidence exists about this behavior (Read backward facing

steps).

29



Applying Bernoulli’s equation between points 1 and 2, we get

P1

ρg+v2

1

2g=P2

ρg+v2

2

2g+ he (56)

From the conservation of momentum across the expansion, we expect that;

Force on expansion=Rate of change of momentum.

Thus;

P1A1 + P0(A2 − A1)− P2A2 = ρQ(v2 − v1)

Experimentally, it has been found that P0 ' P1.

P1A1 + P1(A2 − A1)− P2A2 = ρQ(v2 − v1) = ρv2A2(v2 − v1)

⇒ P1 − P2 = ρv2(v2 − v1) (57)

From equation 56 we can show that

P1 − P2 =ρv2

2 − ρv21

2+ ρghe (58)

Thus comparing equations 57 and 58, it is easy to show that;

he =1

2g(v2 − v1)

2 (59)

and since

v2A2 = v1A1

then we again can easily show that

he =v2

1

2g

(A1

A2

− 1

)2

(60)

30


For a given expansion the term(A1

A2

− 1

)2

= constant = k

which then makes equation 60 become

he = kv2

1

2g(61)

If A2 � A1 for example when discharging to the atmosphere or a large tank;(A1

A2

− 1

)2

' (0− 1)2

⇒ k ' 1

and hence

he 'v2

1

2g(62)

5.5 Loss due to Sudden Contraction

Due to sudden contraction, the streamlines converge to a minimum cross-section

called the ‘vena contracta’ and then expand to fill the downstream pipe.


31


Loss due to sudden contraction=Loss up vena contracta+Loss due sudden expan-

sion beyond vena contracta

hc = Negligibly small +(vc − v2)

2

2g

⇒ hc '(vc − v2)

2

2g

But Acvc = A2v2 and if Ac = Cc · A2 where Cc is referred to as coefficient of

contraction, then

hc =

(1

Cc− 1

)2v2

2

2g(63)

Again for a given contraction;

k =

(1

Cc− 1

)2

hc = kv2

2

2g(64)

Experimentally, it has been found that

Cc = 0.62 + 0.38

(A2

A1

)3

(65)

Typical values of loss coefficient are available in the literature.

Example 3

A 150mm diameter pipe reduces in diameter abruptly to 100mm diameter. If the

pipe carries water at 30liters/sc, calculate the pressure loss across the contraction

and express this as a percentage of the loss to be expected if the flow was reversed.

Take the coefficient of contraction as 0.6.

Example 4

The difference in water surface elevations in two reservoirs A and B is 10m and

the gauge pressure of air space in A is 50kN/m2. They are connected by a single

pipe 200m long and 20cm in diameter. If the friction factor for the pipe is 0.02,

calculate the discharge.

Example 5

Water flows from a reservoir through a series of pipes joined as shown in the

figure below. If the exit is 20m below the free level in the reservoir, calculate the

discharge. Find the percentage error in discharge if the minor losses are neglected.

32


5.6 Power Transmission through Pipes

We employ pipes while transmitting hydraulic power in conjunction with hydraulic

machines.

Consider the pipe shown below used for transmitting power

Draw a pipe line to suit the notations

H2 = H1 − hf = H1 −4f`

d

u2

2g

But u = 4Q/πd2, where H1 is the head available given as

P1 +u2

1

2g+ z1 = H1

∴ H2 = H1 −4f`

d

16Q2

π2d4 · 2g= H1 − 0.33

Q2f`

d5(66)

If P = Power available at exit = ρgH2Q then,

P = ρg

(H1Q− 0.33

f`

d5Q3

)(67)

33


For a given value of H1, we can thus plot a curve of P against Q as shown the

figure below.

and we can also plot for efficiency against flow rate as;

Clearly from the two graphs, the maximum efficiency does not occur at the max-

imum Q. It is thus important to establish the conditions under which a pipeline

will transmit maximum power for a given pipe size. And also to determine for a

given condition what the smallest pipe size to transmit maximum power will be.

To maximize power at the exit, then

∂P

∂Q= 0 = ρg

(H1 − 0.99

f`

d5Q2

)= ρg(H1 − 3hf )

⇒ hf =H1

3(68)

Therefore, to get maximum power at exit, we must ensure that head lost in friction

is one-third the head available at entry.

34


By definition

Efficiency of transmission, η =Power output

Power input× 100

Thus

η =H2

H1

× 100 =H1 −H1/3

H1

× 100 = 66.67% (69)

If a nozzle is fitted at the exit, discharging to the atmosphere at a velocity un,

then

H2 =u2n

2g

and ∴ H1 =u2n

2g+ 0.33

Q2f`

d5=u2n

2g+ 0.20

f`

du2

We know that ud2 = und2n ⇒ u2 = (dn/d)4u2

n, thus

H1 =u2n

2g+ 0.20

f`

d

(dnd

)4

u2n

∴ u2n =

H1{0.05 + 0.20f`

d

(dn

d

)4} (70)

As stated earlier, the power, P transmitted =ρgH2Q, which we can write as a

function of the nozzle geometry for a given head H1 as;

P = ρgu2n

2g· πd

2n

4un =

ρπd2n

8

H1{0.05 + 0.20f`

d

(dn

d

)4}3/2

(71)

Thus for a given available head H1 and pipe, the power can be maximized with

respect to nozzle diameter dn by setting ∂P/∂dn = 0 in equation 71.

With this in mind, it is easy to show that power will be maximum when

dn =

(d5

8f`

)1/4

(72)

Example 6

A power transmission pipe 10cm diameter and 500m long is fitted with a nozzle at

the exit. The inlet is from a lake with water level 60m above the discharge nozzle.

Taking f = 0.0075, calculate the maximum power which can be transmitted, and

the diameter of the nozzle required.

35


5.7 Pipe in Series

If different pipes of different diameters are connected end-end, then the pipes are

said to be in series. We are usually interested in designing for a known flow rate;

(i) Pipe diameters

(ii) Pipe lengths and materials

(iii) Determining the pipe pressures


Total loss of energy per unit weight = Loss due friction in each pipe + minor losses

in each pipe

Taking energy per unit weight between the two ends of the combination we get

H1 −H3 = KQ2

With Q1 = Q2 = . . . = Qn

Example 7

A pipeline 30m long connects two tanks which have a difference of water level of

12m. The first 10m of pipeline from the upper tank is 40mm diameter, and the

next 20m is 60mm diameter. At the change of section a valve is fitted. Calculate

36


the rate of when the valve is fully open assuming that the resistance is negligible

and that f for both pipes is 0.0054. In order to restrict the flow, the valve is then

partially closed. If k for the valve is now 5.6, find the percentage reduction in flow.

5.8 Pipes in Parallel

Consider two tanks connected by several pipes that run parallel to each other.

Flow of fluid may be through either pipe. The head causing flow however will

be the same in each pipe and this is what will overcome frictional and secondary

losses. Bernoulli’s equation can therefore be applied to each pipe independently

and continuity requires that

QT = Q1 +Q2 = . . . = Qn sum of the volumetric flow rate in each pipe


Example 8

Two reservoirs are connected by 0.45m diameter pipeline 1km long. In order to

augment the discharge, another pipeline is introduced parallel to the first

(i) for the entire length

(ii) in the second half of the length

37


Neglecting minor losses, find the percentage increase in discharge for each proposal,

if f = 0.03 for both pipes and the difference between the reservoir water levels is

20m and remains the same.

5.9 Pipe Networks

The most common pipe networks are water distribution systems for municipalities.

These systems have one or more sources (discharges of water into the system) and

numerous loads. One for each household or commercial establishment.

Consider junction J on the network shown below.


Flow can be through numerous paths. Whichever is the path, it requires that the

value of pressure at J be unique. It is impossible to have two values of pressure

for the same point!

Thus for example;

hJ = hA − hf1 = hB − hf2 = hC − hf3 = hD − hf4 (73)

Also

hf1 =4f1`1Q

21

12d51

(74)

38


and similarly for pipes 2, 3, 4 and 5. So we may write generally,

hfi =4fiìQ

2i

12d5i

(75)

which will give n equations for n junctions.

Continuity requires that the net mass outflow from a junction be zero. Thus

Q1 +Q2 +Q3 −Q4 = 0 (76)

which gives an extra 1 equation. Hence we have a total of n + 1 equations. For

this example, we have 5 equations.

The engineer is faced with either of the following problems;

(i) To determine di and head hJ for prescribed values Qi, fi, and ì e.g. a new

network (requires n+1 equations)

(ii) To determine Qi and head hJ for prescribed values di, fi, and ì e.g. expan-

sion of an existing network (requires n+1 equations)

In either of the above cases, we need n+1 equations. These equations are available

from the head and continuity equations above and the problem is closed.

In general we can achieve these equations if we are able to satisfy the following 2

conditions

(i) Continuity: The flow into a junction of the network must be equal to the

flow out of the junction and must be satisfied for all junctions

(ii) The head loss between any two junctions in a closed loop must be the same

regardless of the path in the series of pipes taken to get from one junction to

the other. This means that the algebraic sum of head losses around a given

loop must be equal to zero -(ve) or +(ve) head loss being given by the sense

of the flow wrt loop, eg clockwise or counterclockwise


39


∑hlclockwise

= hlAB+ hlBC∑

hlanticlockwise= hlAD

+ hlDC

∴ hlAB+ hlBC

= hlAD+ hlDC

(77)

These equations are written for all junctions and loops in the network. We can

obviously see how cumbersome it is to simultaneously solve this huge number of

equations. Iteration is in this case used where an initial guess is made and the two

conditions checked for an improved guess

5.10 Branching-Pipe System of 3 Reservoirs

Consider flow in the following three reservoirs.


Assume that the pipe lengths, diameters and friction factors are pre-determined.

Normally minor losses are neglected in this complex flow. We can not say by visual

inspection whether junction J is fed by tank B or it feeds tank B.It is better and

40


usual to assume that the highest point feeds point J and the lowest point is fed by

J. We can determine the direction of flow in pipe if we know the pressure head at

J and at B. We will normally be required to evaluate one of the following 3 types

of problems;

(i) If elevations A, B and C are known, evaluate Q1, Q2 and Q3

(ii) If elevations A and B and Q1 are known, evaluate elevation C, Q2 and Q3

(iii) If elevations of A and C and Q2 are known determine elevation of B, Q1 and

Q3.

Whichever condition, we wish to solve for, we need to evaluate hJ before flow

direction in pipe 2 can be established.

Thus we are posed with the problem of evaluating 4 parameters (three parameters

plus hJ). We therefore need 4 equations to be able to solve this flow problem.

They are set up as follows: (recall; neglect minor losses)

hf1 = hA − hJ =4f1`1Q

21

12d51

hf2 =| hB − hJ |=4f2`2Q

22

12d52

hf3 = hJ − hC =4f3`3Q

23

12d53

Q1 ±Q2 −Q3 = 0

(78)

Example 9

A system of 3 interconnected reservoirs A, B, and C by pipes 1,2 and 3 to form a

junction J has the following data:

Pipe 1; 0.8m dia; 2000m long, f=0.005

Pipe 2; 0.5m dia; 500m long, f=0.00625

Pipe 3; 0.6m dia; 1000m long, f=0.0075

The elevations of water surfaces in reservoirs A, B and C are 30m, 20m and 5m

respectively. Determine the discharge through each pipe.

As will be demonstrated through this worked example, obviously when the network

involves more number of loops, the procedure becomes more laborious and calls

for elaborate iterative procedures. Numerical iteration is more commonly used.

41


TUTORIAL SHEET 4 - FLOW IN CIRCULAR PIPES,PIPE SYSTEMS AND PIPE NETWORKS

(i) Consider fully developed laminar flow due to gravity in a vertical tube. As-

sume that the pressure is atmospheric at both the tube inlet and outlet.

Show that the relationship between tube diameter and Reynolds number

may be expressed as;

D =

(32Reν2

g

)1/3

(ii) Consider fully developed laminar flow in the annulus between two concentric

pipes. The inner pipe is stationary and the outer pipe moves in x-direction

with a speed v0. Assume the axial pressure gradient to be zero. Obtain

a general expression for the shear stress, the velocity distribution and the

volumetric flow rate.

(iii) The velocity profile for a turbulent flow through smooth pipes is often rep-

resented by the empirical equation;

u

U=(

1− r

R

)1/n

Show that the ratio of the average velocity U to the centerline velocity U is

given by;

U

U=

2n

(n+ 1)(2n+ 1)

(iv) If the pump efficiency is 70%, what power must be supplied to the pump in

order to pump fuel oil of relative density 0.94 at a rate of 1m3/s to the upper

reservoir. Assume that the conduit is a steel pipe and ν = 5× 10−5m2/s.

42


(v) Two reservoirs are connected by three pipes in series;

` d 4fFirst Pipe 300m 30cm 0.02

Second Pipe 250m 25cm 0.025Third Pipe 200m 20cm 0.03

Calculate the discharge to the lower tank if the elevation difference of levels

is 20m.

(vi) Two reservoirs are connected by three pipes in parallel;` d 4f

First Pipe 100m 20cm 0.02Second Pipe 250m 20cm 0.02Third Pipe 100m 10cm 0.02

The total discharge is o.2m3/s. Calculate the difference in elevation of free

surfaces in the reservoirs for each pipe.

(vii) Three reservoirs with water elevations 20m, 30m and 40m respectively above

a level datum are connected by 200m, 300m and 400m long pipes forming a

common junction. Take f=0.02 for each pipe. Calculate; (a)the discharge

in each pipe (b)the piezometric head at the junction

43


6 Momentum Equation

6.1 Linear Momentum Equation

Sometimes, it is necessary to determine/know forces on bodies due to flowing flu-

ids or know the forces required to give a flowing fluid its flowing velocities. The

momentum equation which helps us to determine these forces is derived from the

Newton’s second law of motion.

Newton’s law of motion;“The net force acting on a body in any fixed direction is

equal to the rate of increase of momentum of the body in that direction.”

The analysis is made under the following assumptions;

(i) that the fluid is ideal

(ii) that the fluid is incompressible

(iii) that the flow is steady

(iv) the velocities of all particles for a given cross-section are the same.

Consider a stream tube shown below;

Assume that after time δt the fluid occupying stream tube ABCD now occupies

stream tube A′B′C ′D′

Taking the velocity and density at planes AB and CD to be v1, ρ1 and v2, ρ2,

respectively.

Then for a small δt, distance travelled by particles at AB, i.e.

AA′ = v1δt (79)

44


Particles at CD will travel,

CC ′ = v2δt (80)

Since continuity still holds (i.e. mass of fluid remains unchanged)

Mass in AA′BB′ = Mass in CC ′DD′

∴ ρ2A2v2δt = ρ1A1v1δt

As the fluid changes the velocity from v1 at AB to v2 at CD, the total momentum

must also change.

(Note: This is equivalent to replacing the fluid in AA′BB′ of velocity v1 with fluid

in CC ′DD′ of velocity v2)

Thus, momentum of fluid in AA′BB′;

M1 = ρ1A1v1δt× v1 = ρ1A1v21δt

And momentum in fluid CC ′DD′

M2 = ρ2A2v2δt× v2 = ρ2A2v22δt

Hence the change of momentum of the fluid between AB and CD

∆M = M2 −M1 = ρ2A2v22δt− ρ1A1v

21δt

⇒ The rate of change of momentum between AB and CD is;

∆M

δt= ρ2A2v

22 − ρ1A1v

21 (81)

Recall continuity requirement for flow within a stream tube

m = ρ1A1v1 = ρ2A2v2

Equation 81 can be written as;

∆M

δt= ρ2A2v2.v2 − ρ1A1v1.v1 = m(v2 − v1) (82)

And therefore

∆M

δt= Mass flowing per unit time× velocity change

45


But from Newton’s 2nd law of motion, the rate of change of momentum is equiva-

lent to the force acting on the fluid element ABCD in the direction of motion, F .

Thus;

F = m(v2 − v1) (83)

Equation 83 is known a momentum equation for one dimensional flow.

By applying Newton’s third law, we expect that the fluid will exact an equal but

opposite reaction to the surrounding. Thus, in x direction;

Fx = −Rxm(v2x − v1x) (84)

For a two-dimensional flow, consider the following flow,

Let Fx and Fy be the components of the resultant force F in x and y directions,

respectively.

Fx = Rate of change of momentum of fluid in x-direction

∴ Fx = m(v2x − v1x) (85)

Similarly;

Fy = m(v2y − v1y) (86)

These components will give a resultant force, F ,

F =√F 2x + F 2

y (87)

F is assumed positive in the positive direction of v.

46


In differential form, the momentum equation can be presented as;

F =D(mu)

Dt=

∫s

uρu.dA+∂

∂t

∫V

uρdV (88)

For a steady flow;

∂{}∂t

= 0

F=External surface and body forces∫s

uρu.dA = Rate of efflux of momentum across a control surface

∂

∂t

∫V

uρdV = Rate of change of momentum within the control volume

6.2 Angular Momentum Equation

An alternative manifestation of the Newton’s 2nd law when expressed as ‘net ex-

ternal moment on a system equals the rate of change of angular momentum’ is the

angular momentum equation.

For steady flows:-

T = r2uθ2ρ2A2 − r1uθ1ρ1A1 = m(r2uθ2 − r1uθ1) (89)

where

T = Torque about an axis of rotation

uθ1, uθ = absolute circumferential velocities of the fluid entering and leaving at

radial locations r1 and r2, respectively.

6.3 Examples

Example 1

A uniform pipe 75m long containing water is fitted with a plunger. The water

is initially at rest. If the plunger is forced into the pipe in such a way that the

water is accelerated uniformly to a velocity of 1.7m/s, in 1.4s, what will be the

pressure on the face of the plunger assuming that the water and pipe are not elastic.

Example 2

A jet of water strikes a flat plate inclined to it. Work out the distribution of

discharge over the plate and the force of impact on it, if

(i) the plate is stationary

47


(ii) the plate moves in the same direction as the incoming flow.

Example 3

A jet of water 50mm diameter with a velocity 18m/s strikes a flat plate inclined

at an angle 25◦ to the axis of the jet. Determine the normal force exerted on the

plate

(i) when the plate is stationary

(ii) when the plate is moving at 4.5m/s in the direction of the jet and

(iii) determine the work done and efficiency for case (ii)

(iv) determine the distribution of flow rates for case (i)

Example 4

Consider a water jet that is deflected by a stationary vane such as is shown in the

figure below. If the jet speed and diameter are 30.5m/s and 25cm respectively and

the jet is deflected 60◦, what force is exerted on the vane by the jet?

48


Example 5

A jet of water delivers 85dm3/s at 36m/s onto a series of vanes moving in the

same direction as the jet at 18m/s. If stationary the water which enters tangen-

tially would be diverted through and angle of 135◦. Friction reduces the relative

velocity at exit from the vanes to 0.80 of that at entrance. Determine the mag-

nitude of the resultant force on the vanes and the efficiency of the arrangement.

Assume no shock at entry.

Example 6

A reducing bend is incorporated in a pipeline so that the direction of flow is turned

through 60◦ in the horizontal plane and the pipe diameter is reduced from 0.25m

to 0.15m. The velocity and pressure at the entry to the bend are 1.5m/s and

300kN/m2 gauge respectively.At the exit the pressure is 287.2kN/m2 gauge. Find

the magnitude and direction of the reaction force on the bend in the horizontal

plane due to the flowing water.

Example 7

Water enters a rotating lawn sprinkler through its base at the steady rate of

1000m/s as is shown in the figure below. The exit area of each of the nozzles

is 30mm2 and the flow leaving each nozzle is 200mm.

(i) Determine the resisting torque required to hold the sprinkler head stationary

(ii) Determine the resisting torque associated with the sprinkler rotating with a

constant speed of 500rpm

(iii) Determine the speed of the sprinkler if no resisting torque is applied

49


TUTORIAL SHEET 2 - MOMENTUM EQUATION ANDITS APPLICATIONS

(i) Water flows through a 180◦ vertical reducing bend as shown in the figure

below. The discharge is 0.25m3/s and the pressure at the centre of the inlet

of the bend is 150kPa. If the bend volume is 0.1m3 and it is assumed that

Bernoulli’s equation is valid, what reaction is required to hold the bend in

place? Assume that the bend weighs 500N.

(ii) Determine the power produced by a sprinkler-like turbine that rotates in a

horizontal plane at 500rpm. The radius of the turbine is 0.5m. Water enters

the turbine from a vertical pipe that is co-axial with the axis of rotation

and it exits through two nozzles, each of which has across-sectional area of

10cm2. The exit velocity of water is 50m/s with respect to the nozzle. The

water density is 1000kg/m3 and the pressure at the exit is atmospheric.

(iii) Two large tanks containing water have small smoothly contoured orifices of

equal area. A jet of liquid issues from the left tank. Assume the flow is

uniform and unaffected by friction. The jet impinges on a flat plate covering

the opening of the right tank. Obtain an expression for the height h required

to balance the hydrostatic force on the plate from water in the right tank.

50


(iv) A jet of water is directed against a vane which could be a blade of a turbine

or in any piece of hydraulic machinery. The water leaves the stationary,

50mm in diameter nozzle at a speed of 20m/s and enters a vane tangent

to the surface A. The inside surface B makes an angle θ = 150◦ with the

x-direction. Compute the force that must be applied to maintain the vane

at constant speed u = 5m/s.

(v) A square plate, mass 12.7kg of uniform thickness and 300mm edge is hung

so that it can freely swing about its upper horizontal edge. A horizontal jet

20mm in diameter strikes the upper plate with a velocity of 15m/s and the

centerline of the jet is 150mm below the upper edge of the plate so that when

the plate is vertical, the jet strikes the plate normally at the center. Find

what force must be applied to the lower edge to keep the plate vertical and

what inclination to the vertical the plate will be if allowed to swing freely

under the action of the jet.

51


7 Dimensional Analysis and Similitude

7.1 Introduction

Due to the complexities involved in analyzing fluid mechanics problems, the ana-

lytical tools presently available are not capable of yielding exact solutions to many

of the problems. It is possible to get exact solutions for all hydrostatic problems

and for many laminar-flow problems.

However, the most general equations solved on the largest computers yield only

fair approximations for turbulent flow problems. Hence the need for experimental

evaluation and verification.

Again, since most flows of engineering applications are turbulent, experimental

analysis are complex. An obvious goal of any experiment is to make the results as

widely applicable as possible. We use the concept of similitude so that measure-

ments made on one system (models) can be used to describe the behavior of other

systems called prototype.

7.2 Dimensional Analysis

A qualitative description of any physical quantity can be given in terms of basic

dimensions such as mass, M , length, L, and time, T . Alternatively we could use

force, F , L and T as basic dimensions.

We shall use the symbol.= to indicate dimensional equality. For example if we use

the MLT system of measurement then;

∆P.= ML−2T−2

ρ.= ML−3

µ.= ML−1T−1

u.= LT−1

(90)

Consider experimentation for an incompressible steady flow through a long, smooth-

walled, horizontal circular pipe. Of interest to an engineer designing this pipeline

would be the pressure drop that develops per unit length as a result of friction.

When planning for this experiment, one needs to decide on the factors (variables)

that would affect the pressure drop per unit length. We can thus write;

∆P` = φ(D, ρ, µ, u) (91)

The nature of the function φ is unknown and the purpose of experiment is to

determine the nature of this function. We would thus be expected to perform the

52


following experiments:

These experiments are only valid for a specific pipe and fluid. Furthermore, some

of the experiments are difficult to carry out, eg. how can one vary the density of

the fluid while holding the viscosity constant?

The other difficulty faced in this approach is;-after we have obtained the various

curves, how do we combine them so that the functional relationship would be valid

for any similar pipe system.

Dimensional analysis helps us eliminate these difficulties, if we can be able to group

these variables into dimensionless groups such that;

D∆P`ρu2

= φ

(ρuD

µ

)(92)

53


Thus instead of working with five variables, we end up with only two variables.

We can get this curve by choosing a pipe of a convenient diameter, a fluid easy to

work with and only vary the velocity, while measuring D∆P`.

These are easily measurable quantities. Furthermore, it is very easy to vary the

velocity experimentally. These results can however be used to interpret data for

any other combination of pipe and fluid. Obviously saving cost of experimentation

and time. This type of forming dimensionless groups is called dimensional analysis.

There are several techniques of forming the dimensionless groups such as;

(i) Buckingham Pi Method

(ii) Rayleigh Method

(iii) Bridgman Method

(iv) Indices Method

(v) Matrix-Tensor Method

The logic in any method selected in forming the dimensionless groups is the same.

We shall however only discuss the Buckingham Pi Method. But before we do

that, we recognize that the relative importance of the dimensionless groups such

as Reynolds, Froude or Mach numbers can only be determined through model

testing. This is carried out in wind tunnels as it might be experimentally difficult

to carry out such tests on the prototypes such as aircrafts, trains, cars, buildings,

ships etc.

To apply dimensional analysis in model testing there are two types of similarity

that must be observed

54


(i) Geometric similarity

(ii) Dynamic similarity

7.3 Geometric Similarity

This requires that between a model and a prototype, the ratio of ALL correspond-

ing linear dimensions be the same. For any model testing, geometric similarity

must be observed.

7.4 Kinematic Similarity

This requires that between a model and prototype, the velocities of all correspond-

ing points are in the same direction and their scale factors are constant throughout.

7.5 Dynamic Similarity

This requires that between a model and a prototype, the ratio of all the forces for

corresponding points remain the same which ensures that flow pattern between

model and prototype remain the same. Whenever the condition of dynamic simi-

larity is met, geometric similarity is also met.

7.6 Buckingham Pi Method

Consider a situation where a dependent parameter is a function k−1 independent

parameters.

We may express the relationship as;

q1 = φ(q2, q3, . . . . . . , qk) (93)

The Buckingham Pi theorem states:

“Given a relation among k parameters of the form:

g(q1, q2, q3, . . . . . . , qk) = 0,

then if the variables are dimensionally homogeneous, the k parameters can be

grouped into k − r independent dimensionless ratios called Π parameters such

that;

Π1 = φ(Π2,Π3, . . . . . . ,Πk−r)

55


where r is the minimum number of reference (basic) dimensions required to de-

scribe the variables”

In our case, we shall use the MLT system throughout and hence r = 3 at all times.

7.7 Determination of Pi Groups

The following steps are taken in determining the Pi groups;

(i) List all the variables that are involved in the problem.

It is important that the experimenter has adequate knowledge of the phys-

ical laws that govern the phenomenon. Incorrect decision at this point will

make the analysis wrong. As a guide, choose variables that are necessary to

describe geometry of the system, variables that define any fluid properties,

and variables that indicate any external effects.

(ii) Express each of the variables in terms of basic dimensions: For typical fluid

mechanics problems, the basic dimensions we shall use are M , L and T

(iii) Determine the required number of Pi groups. Use the Buckigham theorem

such that

Number of Pi groups =k − rwhere k is determined in step (i) and r is determined in step (ii)

(iv) Select a number of repeating variables, where the number required is equal

to the number of reference dimensions.

From the original list, select variables that you think have the greatest in-

fluence on the dependent variables such that

(a) the list does not include the dependent variable

(b) all the reference dimensions must be included within the group

(c) the dimensions of each repeating variable are different and cannot be re-

produced by for example (product of the remaining repeating variables)

so that they themselves cannot form a dimensionless group

(v) form a Pi group by multiplying one of the non-repeating variables by the

product of the repeating variables, each raised to an exponent that will make

the combination dimensionless.

(vi) repeat step (v) for all the remaining non-repeating variables

(vii) check all the resulting Pi groups to make sure they are dimensionless

56


(viii) express the final form as a relationship among the Pi groups and think about

their physical meaning such that

Π1 = φ(Π2,Π3, . . . . . . ,Πk−r)

Π1 should contain the dependent variable in the numerator.

The functional relationship can however only be determined experimentally.

Example 1

A thin rectangular plate having a width w and a height h is located so that it

is normal to a moving stream of fluid. Assume the drag force FD, that the fluid

exerts on the plate is a function of w and h, the fluid viscosity and density, µ and

ρ, respectively, and the velocity u of the fluid approaching the plate. Determine a

suitable Pi groups to study this problem experimentally.

7.8 Common Dimensionless Groups in Fluid Mechanics

Common variables in fluid mechanics are

(i) Acceleration due to gravity, g

(ii) Bulk modulus, Ev

(iii) Characteristic length, `

(iv) Density, ρ

(v) Frequency of oscillating flow, ω

(vi) Pressure, p or ∆p

(vii) Speed of sound, c or a

(viii) Surface tension, σ

(ix) Velocity, u

(x) Viscosity, µ

These variables are not encountered at the same time in all problems. By applying

any dimensional analysis technique, these variables can be grouped into various

dimensionless groups. It is often possible to provide the physical interpretation

to the dimensionless groups which can be helpful in assessing their influence in a

particular application.

57


To be able to provide the physical meaning of these groups, we think of the possible

forces that can occur in fluid mechanics problems (whether dynamic or static)

(i) Inertia forces ∝ ρu2`2

(ii) Viscosity forces ∝ µu`

(iii) Pressure forces ∝ ∆P`2

(iv) Gravitational forces ∝ gρu2`2

(v) Surface tension forces ∝ σ`

(vi) Compressibility forces ∝ Ev`2

Inertia forces are very important in most fluid mechanics problems. The ratio of

inertia forces to other forces lead to the several fundamental dimensionless groups

found in fluid mechanics.

7.8.1 Reynolds Number

The ratio between inertia forces of an element to the viscous forces. The criteria

of establishing whether a flow is laminar or turbulent.

Re =Inertial Forces

Viscous Forces=ρu`

µ(94)

In bounded flows, free surface effects are absent. If the flow is slow, then effects of

compressibility may also be ignored. In such cases Reynolds number equivalence

would be required. Sometimes the geometric similarity might make adherence of

Re very costly and required velocities to be unachievable. In such cases the Re

equivalence requirement is relaxed.

7.8.2 Froude Number

A measure of the ratio between the inertia forces of an element to its weight.

Important in flows involving a free surface like open channel flows since gravity

affects these flows. Used as a criteria of establishing whether a flow is sub-critical

or super-critical.

Fr =Inertial forces

Gravitational forces=

u√g`

(95)

58


7.8.3 Mach Number & Cauchy Number

A measure of the ratio between inertia and compressibility forces. Important where

compressibility is a significant factor. Used as a criteria to establish whether a flow

is subsonic or supersonic in cases of compressible flows.

Ma =Inertial forces

Compressibility forces=u

c(96)

and since c =√Ev/ρ

Ca =ρu2

Ev= u2 ρ

Ev= Ma2 (97)

Some people prefer to see Mach number as the ratio between the flow velocity to

the velocity of sound in the flowing fluid. Mach numbers becomes significant when

it exceeds 0.25 and compressibility effects are thus significant.

7.8.4 Weber Number

The ratio between inertia forces and surface tension forces. Important in problems

that involve an interface between two fluids. In cases where We >> 1, then surface

tension effects may be ignored.

We =Inertial forces

Surface tension forces=ρu2`

σ(98)

7.8.5 Pressure, stress and force coefficients

Measure of ratio of inertia forces to pressure forces where pressure forces are im-

portant.

Cp =Pressure forces

Inertial forces=

∆P

1/2ρu2(99)

The ratio 1/2 is introduced for convenience so that the denominator physically

means that we are non-dimensionalizing pressure head with the velocity head.

Sometimes when the pressure forces to inertial forces are to be used to determine

whether a certain physical phenomena occurs or not (e.g. to determine whether

cavitation takes place or not on the suction side of the pump), then the Euler’s

number is used:

Eu =Pressure forces

Inertial forces=

∆P

ρu2(100)

When we want to talk of force coefficients, then we introduce area into the pressure

coefficients. Such coefficients are lift and drag coefficients;

CD, CL =∆P`2

1/2ρu2`2(101)

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7.8.6 Power coefficients

Useful in definition of pump, fan and turbine characteristics in terms of other flow

variables included in other common dimensionless groups.

Power coefficients =P

ρu2`3(102)

7.8.7 Strouhal number

This number is important in unsteady oscillating flow problems in which oscillating

frequency is ω. It represents a measure of the ratio of inertial forces due to the

unsteadiness (local acceleration) to the inertial forces due to changes in velocity

from point to point in the flow field (convective acceleration). Examples are the

vortex sheet flows formed by a flow after a solid body.

St =Inertial forces (local)

Inertial forces (convective)=ω`

ν(103)

7.9 Modeling and Similitude

A model is a representative of a physical system that may be used to predict the

behavior of the system in some desired respect. The physical system for which the

predictions are to be made is called ‘prototype’.

Usually, the model is smaller than the prototype so that it can be easily handled

in a lab and less expensive to construct.

Occasionally, we have models bigger than the prototype, e.g., if we want to study

the motion of red blood cells.

A model therefore is a physical system that resembles the prototype but are gen-

erally of a different size, may involve different fluids, and often operate under

different conditions (pressure and velocities, etc)

7.9.1 Theory of models

By use of dimensional analysis, we showed that:

Π1 = φ(Π2,Π3, . . . . . . ,Πn) (104)

In formulating this relationship, only a knowledge of the general nature of the

physical phenomenon and the variables involved is required.

60


Specific values for variables (sizes of components, fluid properties and so on) are

not needed to perform the dimensional analysis. Thus equation 104 describes the

behavior of a particular prototype. A similar relationship can be written for a

model of this prototype. Thus

Π1m = φ(Π2m,Π3m, . . . . . . ,Πmn) (105)

where φ remains the same for both equations.

Therefore if;

Π2m = Π2

Π3m = Π3

......

Πnm = Πn

(106)

Then,

Π1m = Π1 (107)

These are the model design conditions also called similarity requirements or mod-

eling laws.

For this to happen, certain physical conditions (discussed earlier) must be met;

(i) geometric similarity

(ii) kinematic similarity

(iii) dynamic similarity

Example 2

A long structural component of a bridge has the X-section shown in the figure

below. It is known that when a steady wind blows past this type of bluff body,

vortices may develop on the downwind side that are shed in a regular fashion at

some definite frequency. Since these vortices can create harmful periodic forces

acting on the structure, it is important to determine the shedding frequency. For

the specific structure of interest, D=0.1m, H=0.3m and a representative wind

velocity is 50km/hr. Standard air can be assumed. The shedding frequency is

to be determined through the use of a small-scale model that is to be tested in

a water tunnel. For the model Dm=20mm and the water temperature is 20◦C.

Determine the model dimension, Hm and the velocity at which the test should be

performed. If the shedding frequency for the model is found to be 49.9Hz, what is

61


the corresponding frequency for the prototype?

62


TUTORIAL SHEET 5 - SIMILARITY LAWS AND DI-MENSIONAL ANALYSIS

(i) Show by dimensional analysis for a completely submerged body, the drag

force (resistance) is given by;

FD = ρ`2v2φ(Re)

(ii) An airplane wing of 1m chord moves through still air at 20◦C at 180km/hr.

A 1:15 scale model of this wing is placed in a wind tunnel with air blowing

at 75m/s at the same temperature as that in the flight. What should be the

pressure in the travel?

(iii) A sphere when dropped in water moving at a velocity of 1.6m/s experi-

ences a resistance of 22N. Another sphere twice the diameter is placed in

a wind tunnel. What should be the velocity of wind in the tunnel for dy-

namic similarity? What is the corresponding drag? ρair = 1.28kg/m3 and

νair = 13× νwater.

(iv) At its optimum point of operation, a given pump with an impeller diameter

of 50cm delivers 3.2m3/s of water against a head of 25m when rotating at

1450rpm.

(a) if its efficiency is 82%, what is the brake power of the diving shaft?

(b) If a homologous pump with an impeller diameter of 80cm is rotating at

1200rpm, what would be the discharge, head and shaft power? Assume

both pumps operate at the same efficiency.

(v) The following wind tunnel test data from a 1:16 scale model of a bus are

available;Air speed (m/s) 18.0 21.8 26.0 30.1 35.0 38.5 40.9 44.1 46.7Drag force (N) 3.1 4.41 6.09 7.99 10.7 12.9 14.7 16.9 18.9

Using the properties of standard air, calculate and plot the dimensionless

aerodynamic drag coefficient;

CD =FD

1/2ρu2Aversus Reynolds number Re =

ρu`

µ

where ` is the model width. Find the minimum test speed above which CD

remains constant. Estimate the aerodynamic force and power requirement

for the prototype vehicle at 100km/hr. The width and the frontal area of

the prototype vehicle are 8ft and 8ft2 respectively.

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fluid motion

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