dr. md. quamrul islam, sir’s class notes (05 batch) qis

Dr. Md. Quamrul Islam, Sir’s Class notes (05 Batch)

Made by S. Ehtesham Al Hanif (0510035)

QIS Class – 01 Date: 29-03-09

Five sections:

1. Dimensional analysis a. Prototype b. Model c. Dimensions

2. Compressible fluid a. Mach number b. Flow through nozzles

3. Real fluid flow (actual fluid and viscous fluid flow) a. Frictional losses b. Minor losses

4. Boundary layer theory a. Skin friction

5. Open channel flow a. Different types of channels

Fluid equation:

Energy equation

Continuity equation

Momentum equation

Assumptions of Bernoulli’s equation:

# Irrotational - free form angular rotation

# Steady – no parameter changes with respect to time



QIS

Dimensional analysis:

Dimensional analysis are reduces the number ofsome variables to from non dimensional parameters. Instead of observing the effeparameters the effect of the non

Advantages of dimensional analysis:

i) ii)

iii) iv)

Example:

Models are used for ships, airplanes, pumps, turbines, river channels, rockets, missiles etc

Model:

May be bigger or smaller or same as prototype

Limits and dimensions:

Theoretical analysis




Dimensional analysis are reduces the number ofsome variables to from non dimensional parameters. Instead of observing the effeparameters the effect of the non


By applying dimensional analysis no. of experiment can be reduced Dimensional analysis helps us to make experiments to air or water

results to a fluid which is less convenient to work with such as gas, steam or oil Cost can be reduced by doing experiments with models of the full size apparatus Performance of the prototype can be determined from test on the model

xample:



# Flow in a carburetor may be studied in a large model

# model same as prototype: racing car, ten



Physical quantities




Dimensional analysis are reduces the number ofsome variables to from non dimensional parameters. Instead of observing the effeparameters the effect of the non-dimensional parameters are studied.


ing dimensional analysis no. of experiment can be reducedDimensional analysis helps us to make experiments to air or water results to a fluid which is less convenient to work with such as gas, steam or oilCost can be reduced by doing experiments with models of the full size apparatusPerformance of the prototype can be determined from test on the model







Physical quantities



Class – 02

Dimensional analysis are reduces the number ofsome variables to from non dimensional parameters. Instead of observing the effe

dimensional parameters are studied.







Fluid flow problems

Experimental analysis

Physical quantities


02

Dimensional analysis are reduces the number of variables in a fluid phenomena by combining some variables to from non dimensional parameters. Instead of observing the effe






# model same as prototype: racing car, tennis racket etc

Fluid flow problems

Experimental analysis

Primary/Fundamental (ID)

Secondary or Derived (more than ID)

variables in a fluid phenomena by combining some variables to from non dimensional parameters. Instead of observing the effe





nis racket etc

Combination of both


Secondary or Derived (more than ID)

variables in a fluid phenomena by combining some variables to from non dimensional parameters. Instead of observing the effect of individual

ing dimensional analysis no. of experiment can be reduced Dimensional analysis helps us to make experiments to air or water and then applying the results to a fluid which is less convenient to work with such as gas, steam or oilCost can be reduced by doing experiments with models of the full size apparatusPerformance of the prototype can be determined from test on the model


Combination of both


Secondary or Derived

Page

Date: 30-03

variables in a fluid phenomena by combining ct of individual

and then applying the results to a fluid which is less convenient to work with such as gas, steam or oil Cost can be reduced by doing experiments with models of the full size apparatus


Combination of both

03-09

variables in a fluid phenomena by combining

and then applying the



Mass (M), length (l) and time (τ) or force (F), length (L) and time (τ) are considered as fundamental quantities. Other physical quantities can be expressed by this expression / quantities.

Example:

Velocity (V) can be obtained by dividing length (L) with time (τ).

ρ V d µ Head loss per unit length of a pipe depends on viscosity (µ), density (ρ), average velocity (v) of flowing fluid and diameter (d) of pipe. This statement can also be stated by saying that for

incompressible flow to a pipe friction factor depends on Reynolds number

Physical quantities are expressed by FLT or MLT systems. These systems of dimensions are related by Newton’s second law of motion.

Newton’s 2nd law:

Force = mass x acceleration

F = M x

M =

Dimensions of various quantities:

Geometrical characteristics FLT systems MLT systems Length (dia/height/breadth etc) L L Area 퐿 퐿 volume 퐿 퐿

Fluid mechanics:

Geometrical characteristics FLT systems MLT systems Mass (m) 퐹푇

퐿

M

Density (ρ) 퐹푇퐿

푀퐿



Specific weight (γ) 퐹퐿

푀퐿 푇

Kinematic viscosity (υ) 퐿푇

퐿푇

Dynamic viscosity (µ) 퐹푇퐿

푀퐿푇

Velocity (V) 퐿푇

퐿푇

Acceleration (a) 퐿푇

퐿푇

Pressure (P) 퐹퐿

푀퐿푇

Force (F) L 푀퐿푇

Shear Stress (τ) 퐹퐿

푀퐿푇

Flow rate (Q) 퐿푇

퐿푇

Torque and moment FL 푀퐿푇

Mass flow rate 퐹푇퐿

푀푇

There are two methods for solving the problems with dimensional analysis

1. Rayleigh’s method’s 2. Buckingham π – theorem





a) Rayleigh’s method:

Example: (problem solved by dimensional analysis)

Let, 퐹 is the drag force of a smooth sphere of diameter D, moving through a viscous incompressible fluid with velocity V. others variables involved are m, ρ, µ.

ρ

V 푭푫

D

µ

Solution:

퐹 : drag force

D: diameter of sphere

µ: viscosity of fluid

ρ: density of fluid

V: velocity of fluid

Drag force may be considered as the function of these variables:

i.e.: 퐹 = f (D, V, ρ, µ)

To find relations between the variables the dimensions of both sides of the equations will have to be considered.

퐹 = 퐶 퐷 푉 휌 µ

Where 퐶 is a dimensional constant putting correct dimensions according to MLT systems:

= 1. 퐿 ( ) ( ) ( )



=

To satisfy dimensional homogeneity the exponents of the each dimension must be identical of both sides of the equations.

Then,

For M: 1= c + d

For L: 1= a + b -3c –d

For T: -2 = -b –d

Solving for a, b, c in terms of d we get:

a = 2 – d

b = 2 – d

c = 1 – d

Thus, 퐹 = 퐶 퐷 푉 휌 µ

And grouping variables according to their exponent:

퐹 = 퐶 휌(퐷 푉 )( )

It may be noted that the quantity is the Reynolds number. Thus the previous equation can

be written as:

퐹 =푓/(푁 ) 휌퐷 푉

Or, = 푓/(푁 )

Coefficient

The results indicate that the drag on a sphere is equal to some coefficient times 휌퐷 푉 where the coefficient is the functions of the Reynolds numbers. This process of analysis is known as Rayleigh’s method.



Without dimensional analysis With dimensional analysis ρ,µ - const ρ-const,µ - vary ρ-vary,µ-const D D D

푁

b) Buckingham π- theorem :

This theorem states that if there are n dimensional variables in a dimensionally homogenous equation described by m fundamental dimensions, they may be grouped in (n-m) dimensionless groups.

Thus in the previous example:

n=5 and m=3 (M, L and T) and (n-m) = 2

Parameters --- (a)

Parameters --- (b)

These dimensionless groups as π- terms.

Advantages of π- theorem are that it tells one ahead of time, how many dimensionless groups are to be expected.

Applying the π- theorem to the previous example one would proceed as follows:

퐹 = f (D, V, ρ, µ) here n= 5

m= 3

So, n-m = 2

Thus we can write Q (휋 ,휋 ) = 0.

퐹

퐹

퐹

퐹휌푉 퐷



QIS Class – 04 date: 05-04-09

Buckingham π- theorem:

Q (휋 ,휋 ) = 0

The problems in to find π’s by arranging the fine parameters into 2D groups. Taking ρ, D, V as the primary (repeating) variables, the π-terms are:

(퐹 ,휌,퐷,푉, µ) five parameters

휋 = 휌 퐷 푉 µ

휋 = 휌 퐷 푉 µ

Since the π’s are dimensionless, they can be replaced by 푀 퐿 푇

Considering 휋 , 푀 퐿 푇 = ( ) (퐿) ( )

For M: 0 = 푎 + 푑

L: 0 = −3푎 + 푏 + 푐 − 푑

T: 0= - 푐 - 푑

Solving for 푎 , 푏 푎푛푑 푐 푖푛 푡푒푟푚푠 표푓 푑

푎 = −푑 , 푏 = −푑 , 푐 = −푑

Thus 휋 = 휌 퐷 푉 µ

=

Or, (휋 ) =

Any π-term may be replaced by any power of that term including negative as well as fractional powers. For example, 휋 may be replaced by (휋 ) or 휋 may be replaced by (휋 ) , 휋 may be

replaced by etc

So, 휋 = = 푁



Similarly, 휋 =

Finally, Q (휋 ,휋 ) = 0 may be expressed as

휋 = 푄′ (휋 )

휋 = 푄" (휋 )

Or, = 푄" (푁 )

Or, 퐹 = 푄" (푁 )휌푉 퐷

It should be emphasized that dimensional analysis does not provide a complete solution to fluid problems. It provides partial solutions only.

The most suitable repeating variables are those which contain the flow characteristics (such as velocity, mass density etc).

A geometric property (such as length), fluid property (such as mass, density) and a flow characteristic (such as velocity) are generally most suitable as reciprocating variables.

Dimensionless groups in fluid mechanics:

The following variables are important in the fluid phenomenon:

1. Length, L 2. Acceleration due to gravity, g 3. Mass density, ρ 4. Velocity, V 5. Pressure, P 6. Viscosity, µ 7. Surface tension, σ 8. Velocity of sound, c

The following dimensionless groups can be formed from these variables:

1. Reynolds number, Nre 2. Froude number, Fr 3. Euler number, E 4. Mach number, M 5. Weber number, W

(i) Reynolds number (Nre): The ratio of inertia force and viscous force. Inertia force = mass x acceleration



= ρ퐿 푥 = ρ퐿 푉 = ρ퐿 푉

Viscous force = area x shear stress

= A.τ = 퐿 µ = µ푉퐿

Nre = =

Here L is knows characteristics length (hydraulic diameter)

This number is important where viscous force is predominant.

Hydraulic diameter, D =

=

=d

Example:

Incompressible fluid flow through pipes, flow through venturimeter, orifice meter, nozzle etc.

(ii) Froude number, Fr:

The ratio of inertia force and gravity force.

Gravity force = mass x acceleration due to gravity

=ρ퐿 g

And inertia force = ρ퐿 푉

So, Fr = =

This number is important when gravitational force is predominant.

Example:

Open channel flow, flow over notches and weirs (weirs motion created by waters etc)

(iii) Euler number, E:

The ratio of pressure force and inertia force

Pressure force = pressure x area = P퐿




E = = =

This number is important when pressure predominant.

Example: flow through pipe, flow through submerged body



QIS Class – 05 date: 06-04-09

(iv) Mach number, M: Square root of the ratio of inertia force and elastic force Elastic force = E퐿 = ρ퐿 퐶 where C is the velocity of sound (The pressure waves moves at a velocity equal to that of velocity of sound through the fluid

So, C = or E = ρ퐶 )

M= ( ) = =

Mach number is important when the value of M exceeds 0.4

Examples: supersonic aircraft, rocket, missiles etc.

(v) Weber number, W:

The ratio of inertia force and surface tension force Surface tension force = σL


W = =

Examples: capillary tube flow, human blood flow etc.

Similitude:

D’

D

1 C 1’ C’

2 2’

Model prototype

The similarity relation between a prototype and its known as similitude.



3 types of similarities exist for complete similitude between a model and its prototype.

These are:

1. Geometric similarity 2. Kinematic similarity 3. Dynamic similarity

Two flows will be similar if they are geometrically, kinematically and dynamically similar.

Geometric similarity:

A model and its prototype are said to be in geometric similarity if the ratios of their corresponding linear dimensions are equal (such as: length, breadth, width etc)

For geometric similarity, the corresponding areas are related by the square of the length scale ratio and corresponding volume by the cube of the length scale ratio.

Length scale ratio = = = =

(퐿푒푛푔푡ℎ 푠푐푎푙푒 푟푎푡푖표) =

= ( ) = ( ) =( )

(퐿푒푛푔푡ℎ 푠푐푎푙푒 푟푎푡푖표) =

= ( ) = ( ) =( )

Where 푙 , 푏 ,푑 are linear dimensions of a model and 푙 , 푏 ,푑 are dimensions of prototype.

Kinematic similarity:

A model and its prototype are said to be kinematically similar if the flow patterns in the model and the prototype for any fluid motion as geometric similarity and if the ratio’s of the velocity as well as accelerations at all corresponding points in the flow is the same.

Velocity ratio = =

Acceleration ratio = =

V1, V2 ---- velocities of fluid in prototype in points 1 and 2

v1, v2 ---- velocities of fluid in model at corresponding points 1 and 2

A1, A2 ----acceleration of fluid in prototype in points 1 and 2

a1, a2 ---- acceleration of fluid in model at corresponding points 1 and 2



Dynamic similarity:

A model and its prototype are said to be dynamically similar if the ratio of the forces acting at the corresponding points are equal.

Geometric and kinematic similarities exist for dynamic similarity systems.

i.e.: =

Where, F1, F2 ---- forces acting in prototype at points 1 and 2

f1, f2 ---- forces of fluid in model at corresponding points 1 and 2

Problem 1:

Show that power supplied by pump depends on specific weight, flow rate and total head of water.

Solution:

Let, P: - power supplied by pump

Γ: - specific weight of water

Q: - water flow rate

H: - total head

Now, let,

P = f (γ, Q, H)

Or, P = K훾 푄 퐻 ---------------------- (i)

Where k is a constant.

Putting the dimensions according to FLT systems in equation (i)

= ( ) ( ) 퐿 ------------------ (ii)

Satisfy dimensional homogeneity the exponents of each dimension must be identical on both sides of equation (ii).

Thus,



For F: 1 = a

L: 1 = -3a + 3b + c

T: -1 = -b

Solving equation,

a = 1, b = 1, c=1.

So, from equation (i) we have

P = KQγH (Proved)




Problem 2:

Velocity of sound depends on air pressure, density and viscosity. Find an expression of velocity of sound.

Solution:

Let, C: velocity of sound

P: air pressure

ρ: density of air

µ: viscosity of air

Now, velocity of sound

C = f (P, ρ, µ)

Or, C = K’푃 휌 µ ------- (i)

According to MLT system putting the dimensions in equation (i)

= ( ) ---------------- (ii)

To satisfy dimensional homogeneity the exponents of each dimension must be identical on both sides of equation (ii)

For M: 0 = a + b +d

L: 1 = -a -3b –d

T: -1 = -2a –d

Solving the equation,

d = 0, b = - , a =

Putting the values in equation (i)

C = K’

Or, C = , where K is a constant.



Problem 3:

Force (P) required to drive a propeller is known to depend on

D: Diameter of propeller]

V: velocity

ρ: fluid density

N: RPM

µ: viscosity

Prove that, P =ρ푉 퐷 푓 ,

Solution:

Let, P = f (D, V, ρ, N, µ)

Or, P = f1 퐷 푉 휌 푁 µ

= f1 퐿 ( ) ( ) ( ) ( )

To satisfy dimensional homogeneity the exponents of each dimension must be identical on both sides of equation. Thus, now:

For M: 1 = c + e

L: -1 = a + b – 3c – e

T: -2 = -b –d – e

Solving the equation with respect to e and d:

So, c = 1 – e, b = 2 – e – d, a = 2 – e + d

Putting the values, now:

P = f1 퐷 푉 휌 푁 µ

Or, P = f1 휌푉 퐷 ( )

Or, P = 휌푉 퐷 푓 , (proved)



Problem 4:

Kinematic viscosity of 1st liquid, 휗 = 0.02 x 10

Velocity of 1st liquid, 푉 = 5 m/s

Diameter of pipe for 1st liquid, 퐷 = 20 mm

Kinematic viscosity of 2nd liquid, 휗 = 0.029 x 10

Diameter of pipe for 2nd liquid, 퐷 = 20 mm

To find:

Velocity of 2nd liquid, 푉 if dynamic similarity is to be observed.

Solution:

When fluid flows through pipes, viscous and inertia forces are important and Reynolds number is the criteria for similarity.

Now for dynamic similarity,

Reynolds number for 1st liquid = Reynolds number of 2nd liquid

= 표푟 =

Or, =

OR, 푉 = 푋

=

. 푋 .

= 1.81 m/s (answer)

Problem 5:

A 1:15 model of a boat is to be tested in testing containing salt water. If the prototype moves at a speed of 6 m/s, at what velocity should the model be toned for dynamic similarity. The prototype boat is subjected to wave resistance only.

Scale ratio, =



Speed of prototype, 푉 = 6

To find:

Velocity of model, 푉

Solution:

For wave resistance Fr number is to dominant. Equating Fr for model and prototype for similarity, we have,

푉퐿푔

=푉퐿푔

Or, = here 푔 = 푔

Or, =

Or, 푉 = 푉 푥

Or, 푉 = 푉

Or, 푉 = 6

Or, 푉 = 1.55 (answer)




Problem 6:

Given data:

Diameter of prototype pipe, 퐷 = 1.25 m

Viscosity of oil, µ = 3 Ns/푚

Specific gravity of oil = 0.85

Diameter of prototype pipe, 퐷 = 150 mm

Viscosity of oil, µ = 3 Ns/푚

Flow rate through prototype pipe, 푄 = 2900 l/s

To find:

- Velocity of water for model, 푉 - Flow rate of water for model, 푄

Solution:

Density of water, 휌 = 1000 kg/푚

Density of oil, 휌 = 0.85 x 1000 kg/푚

For flow through pipes viscous force is predominant; So Reynolds number is important.

Equating Reynolds number for model and prototype,

=

푉푉

= 휌 퐷 휇휌 퐷 휇

=850 푥 1.25 푥 1.11000 푥 .150 푥 3

= 2.597 = 2.6

Velocity of the pipe, 푉 = = 2.36

Therefore, 푉 = 2.6 푥 푉 = 2.6 x 2.36 = 6.136

Flew rate of water for model, 푄 = 퐴 푉 = 푑 푉 = 108.5 (answer)



Problem 7:

Given data:

Scale ratio: =

Drag force acting on the metal, 퐹 = 50 푁

Wind velocity in wind tunnel, 푉 = 45

Velocity of prototype, 푉 = 6

Density of air for model, 휌 = 1.2

Density of sea water for prototype, 휌 = 1030

To find:

- Drag force on the prototype, 퐹

A 1:60 model of a ship experiences a drag force of 50 N when testing in a wind tunnel at a velocity of 45 m/s. Calculate the drag force on the prototype if it moves at a velocity of 6 m/s

in sea water. Density of air and sea water is 1.2 and1030 respectively.

Solution:

To find the drag force on prototype equating Euler number for model and prototype,

=

Or, 퐹 = 퐹 = 2.75 푥 10 푁 (answer)

Problem 8:

The discharge over a spillway is 200 . Calculate the corresponding discharge over 1:30

scale ratio model.

Scale ratio: =

Discharge over prototype, 푄 = 200

To find:



- Discharge over the model, 푄

Solution:

For flow over spillway Froude number is predominant and considering Froude number for model and prortotype.

=

Or, =

Or, =

Or, =

Or, = 0.183

Now, =

=

= 0.041 (answer)




Problem 9:

Given data, 휌 = 1030 휗 = 0.012 푠푡표푘푒푠

휌 = 1.24 휗 = 0.016 푠푡표푘푒푠

Scale ratio: =

푉 = 10

To find:

- 푉

-

The ratio of length of submarine and its model is 30:1. The speed of submarine is 10 m/s. the model is to be tested in a wind tunnel. Find the speed of air in wind tunnel. Determine also the ratio of drag forces between the prototype and model.

Solution:

Considering Reynolds number between model and prototype.

=

Or, 푉 = 푥 푥푉

Or, 푉 = 400 (answer)

Considering Euler number,

=

Or, = = 467.24 (answer)



Problem 10:

Given data:

Scale ratio: =

휌 = 1024

Force on prototype, 퐹 = 5886 푁

Velocity of prototype, 푉 = 20

To find:

- Speed of model, 푉 - Resistance on model, 퐹

A 1:15 model of a flying board is load through water. The prototype is moving in sea water of density of a velocity of 20 m/s. find the corresponding speed of the model. Determine also the resistance due to waves on model of the resistance due to wave on the prototype is 5886 N.

Solution:

Considering Froude number between prototype and mode

=

Or, 푉 = 푉 푥 = 5.16 (answer)

Considering Euler number between model and prototype

=

Or, 퐹 = 퐹 = 1.7 푁 (answer)

Problem 11:

Scale ratio: =

Wave resistance on model, 퐹 = 0.36 푁



Velocity of model, 푉 = 1.25

To find:

- Prototype wave resistance, 퐹

- Power requirement for the prototype and model

A 1:60 model of a boat as a wave resistance on 0.36 N when operating 1.25 m/s. find the corresponding prototype wave resistance. Find also the power requirement for the prototype and model.

Solution:

Considering Froude number prototype and model,

=

Or, = = 60 --------------------(i)

Or, 푉 = 9.68

Considering Euler number, we have

=

Or, 퐹 = 퐹 = 77.76 퐾푁

Power required for the model = 퐹 푉 = 0.45 푤 (answer)

Power required for the prototype = 퐹 푉 =752.72 x 10 푤 = 752.72 퐾푊 (answer)

Compressible Flow

Compressibility or elasticity of a fluid, E:

dv



dp – increase in pressure in a unit volume of fluid

dv – decrease in volume

휌 =1푉

Or, 휌푉 = 1

Differentiating,

휌푑푣 + 푣푑휌 = 0

Or, = −

It is expressed by bulk modulus of elasticity.

Now, E =

=

=




Compressibility:

Example: consider the application of 100 psi to 1 푓푡 of water

퐸 = 300000 푝푠푖 (2068 푀푃푎)

d¥ = ¥ = = 푓푡

So, applications of 100 psi to water under ordinary conditions cause it, volume to decrease by 1 part in 3000.

Standard atmosphere:

Temperature, T = 15*C =28 K

Density, 휌 = 1.22

Specific gravity of air, 훾 = 휌푔 = 11.9682

R = 287 J/kgK

Pressure = 760 mm of Hg = 10.3 m of water = 101.31 KN/푚

K = = 1.4

훾 .

휌 = 1000

For, incompressible flow, density, 휌 = constant, flow in which the density of the fluid does not remain constant during flow.

i.e.: if ∆

> 0.05 the effect of compressibility must be consider.

The change of density is accompanied by the changes in pressure and temperature. So thermodynamic relation have to taken into account.

Classification of compressible:

Mach number, M =

If M ≤ 0.3 , the flow of fluid is considered as incompressible.



No Types of flow Mach no Examples a) Subsonic

incompressible flow M ≤ 0.3 Fan, blower’s

b) Subsonic compressible flow

0.3<M<1 Aircraft, turbo machine

c) Transonic flow 0.9<M<1 Compressor blades d) Sonic flow M = 1 e) Supersonic flow M > 1 upto 3 Mig 21 flight f) Hypersonic flow M > 3 Rockets, missiles

Fundamental equations for compressible fluid flow:

i) Continuity equation ii) Energy equation iii) Momentum equation iv) Thermodynamic relation (P,V,T relations) v) Equation of state (P = 휌 RT) vi) 퐶 − 퐶 = 푅

vii) = 퐾

Velocity of wave propagation in gaseous medium:

Y

Piston Rigid pipe Pressure wave

X x y

P Control volume

P+dp

x y

C c+dc

휌 휌 + 푑휌

푃푃

≈ 푓(푝푟푒푠푠푢푟푒 푟푖푠푒 푖푠 푣푒푟푦 푠푚푎푙푙)



Nomenclature:

Let us, consider a rigid long pipe of uniform cross sectional area fitted with a piston.

Let, the pipe is fit with a compressible fluid which is at rest initiative. The piston is moved towards right and a distribution is created in the fluid. Thus disturbance is in the form of pressure wave which travels in the fluid with the velocity of sound.

A: Cross sectional area of pipe

C: Velocity of fluid at section x-x

P: Pressure of fluid at section x-x

휌: Density of fluid at section x-x

c + dc: Velocity of fluid at section y-y

P + dp: Pressure of fluid at section y-y

휌 + d휌: Density of fluid at section y-y

Example:

1) Measuring of the depth of the ocean 2) Sound detection of submarine

dr. md. quamrul islam, sir’s class notes (05 batch) qis

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