LECTURES

in

GASDYNAMICS

IN THE NAME OF ALLAH

FOR 3rd YEAR OF MECHANICAL POWER ENGINEERING

Prepared by

Assoc. Prof. Dr. Eng. Mostafa Nasre-meil, mostafanasr@yahoo.com

Exhaust gasdirection

Rocket direction

Gas dynamics is a branch of fluid mechanics

which describes the flow of compressible fluids.

Fluids which show appreciable variation in density,as a result of the flow are called compressible fluids (such as gases).

Variation in density is due mainly to variation in pressure and temperature.

Pre-requests Courses

• Engineering mathematics

• Thermodynamics

• Fluid mechanics

• Computer programming languages for these applications (for example FORTRAN).

SAMPLE of REFERENCES:

2 .Liepman H.W. and Roshko A.:” Element of gas dynamics ”, John Wiley &Sons, Inc., New York, 1957.

1 .Shapiro, H.:” The Dynamic and thermodynamics of Compressible Fluid Flow”, Vol 1 and 2 ,The Ronald Press Company, New York, 1955

3 .Zucrow,M.J., and Hoffman,J.D. :” Gas Dynamics”, John Willey, New York,1976; Vol.2 reprinted 1985, Krieger Publishing Co. Melbourne,Fl.

4 .Michel A. Saad, :” Compressible Fluid Flow”, Prentic-Hill, Inc., 1985.

5 .Hodge,B.K., and Ketith Koenig :” Compressible Fluid Dynamics with Personal Computer Applications”, Prentice-Hill, Inc.,A Simon&Schuster company Englewood

cliffs, New Jersey 07632, 1995 .

• Applications:- • Aerodynamics at very high speeds• Transport of gases along considerable

distances at very low speeds.• Transport of gases along considerable

distances at very low speeds.• Many applications in aerospace.• Compressible fluid is playing the key role in

numerous non-aerospace devices.• Gas turbines, Compressors, Gas

transportation pipelines, Internal combustion engine, Combustors, rockets, missiles, Wave interactions, Tunnels, aero planes, ….., etc.)

*Aeronautical and astronautically engineering Design and construction of airplanes bodies

Modeling of the solid and liquid rocket motor chambers

Design and construction of aerospace vehicles

Design of the boosting satellites

*Pneumatics systems Exhaust gasdirection

Rocket direction

CHAPTER 1 Fundamental Concepts of Compressible Fluid Dynamics

CHAPTER 2 Isentropic Flow (Simple Area change)

CHAPTER 3 Waves in Compressible Fluid Flow

- Flow with Heat Interaction (Ralieigh Flow)

CHAPTER 4 - SimpleFrictional Flow (Fanno Flow)

CHAPTER 5 The Method of Characteristics.

CONTENTS(five chapters)

LectureTutorialLab.Total

4217

Fundamental Concepts of Compressible Fluid Dynamics

المائع لديناميكا األساسيه المفاهيماالنضغاطي

CHAPTER 1

Two techniques are available for fluid dynamics description as

follows-:ووصف امعالجة طريقتين توجد

ك المائع تى :-اآلحركة

Eullerian approach Lagrangian approach

*Macroscopic or phenomenological standpoint.

*Microscopic or molecular standpoint.

..

*Treat the fluid as an infinitely divisible sub-stance a continuum.

All fluid properties are continuous functions and time within the framework of the continuum assumption.

*Provides details of the fluid molecules interaction {consider a fluid as a collection of particles (molecules, atoms, ions, & electrons) which are in random motion}.

*Classical fluid mechanics and classical thermodynamics.

*Statistical thermodynamics

)Kinetic theory of gas (

• The kinetic theory of gases and the subject of statistical thermodynamics provide information that can be used in the macroscopic formulation

• Statistical thermodynamics is especially useful since it permits the evaluation of thermodynamics properties that can be used in classical thermodynamics.

Knudsen Number (Kn) Kn=l /L

- l the mean free path (for air at standard conditions l=10 7 m)

- L characteristic line dimension.Kn=l /L Flow regime

Kn > 3.0 Free molecular

Kn = 0.1 : 3.0 Transition

Kn = 0.01 : 0.1 Slip

Kn < 0.01Continuum

Note: Kn > 1.0 A rarefied gas flow

• Governing equations :- (5 equations)• The 1st Law of thermodynamics, which

relates to energy balance (conservation of energy).

• The 2nd law of thermodynamics, which relates heat interaction and irreversibility to entropy

• The flow is also affected by both kinetic and dynamic effects, which are described by Newton's law of motion (conservation of momentum).

• The flow fulfills the requirements of conservation of mass.

• The equation of state

• The 1st law of thermodynamics (conservation of energy):-

• For control volume

• Q – W + m (h+V2+gz) = dU

• de= q – w.D.

• w.D.= P dv

• du= q – w.D.

• h=u+Pv=u+P/

• dh+dV2/2=0.0

• The 2nd law of thermodynamics :-

•

• S2 – S1 = dq /T dq =T dS

Equation of state for a perfect gas

(thermal equation of state)-:

Functional form Name

P=P(v,T) or P(,T)Thermal equation of state

Internal energy & Entropy U=u (v ,T) or u(,T)s=s(v,T) or s(,T)

Caloric equation of state

U= u(s,v) or u (s, ) Enthalpy h=h(s,P)

Canonical equation of state

• Thermal equation of state P = R T / v = RT• Where: R = R' / (molecular weight, MW)• Universal gas constant R' = 8.3135 Kj/Kg.mol.• R' is the same for all perfect gases. MW for

air=28.966 Kg/(Kg.mol)• The thermal equation of state is an accurate

representation for gases and mixtures of gases except for conditions at high pressure (or density) and/or low temperature. (15000 psia, 100R) Under these conditions the behavior of real gas deviates from that of a perfect gas. The extent of deviation is indicated by means of a compressibility factor, Z.

• compressibility factor, Z, which is defined by: Z = Pv/RT approximately.

• For a perfect gases:• U=u(T), H=u+Pv= u(T)+RT=h(T)• Specific heat C=dq/dT• Cv=(q/T)v=(u/T)v • Cp=(q/T)p =(h/T)p • R=Cp- Cv = Cp / Cv• Cp=R/( -1) Cv =R/( -1)• For air under P=1 Bar & T=300 K• Cp=1.0035 Kj/Kg.K , Cv=0.7165 Kj/Kg.K,

=1.4

• For isentropic flow, S1 = S2• P2/P1= (v1/v2)= (2/1)= (T2/T1)/(-1)

• The isentropic approximation is common in compressible-flow theory. The entropy change is computed from the 1st and 2nd laws of thermodynamics for pure substance.

• TdS=dh-dp/ dS=Cp dT/T-RdP/P

• S2-S1=Cp {Ln(T2/T1)}-R{Ln(P2/P1)}

• =Cv {Ln(T2/T1)}-R{Ln(1/2)}

• These equations are used to compute the entropy change across an irreversible process.

• Conservation of mass

• m. = A u ( 1-D )

• General form

/t + u/x + v/y + w/z = 0

• Conservation of momentum• Newton's 2nd law of motion:- F=m*a• F (pressure force, body force, …….….. etc

• General form of the Navier-Stokes equation:

DV/Dt = -P+((4/3)V)-* (*V)+ F

• In x-direction: Du/Dt=-P/x+/x[(2u/x+(-(2/3)V)]

+/y[ (u/y+v/x)] +/z[ (w/x+u/z )]+X

• In y-direction: Du/Dt=-P/y+/y[(2v/y+(-(2/3)V)]

+/z[(v/z+w/y)] +/x[(u/y+v/x )]+Y

• In z-direction: Du/Dt=-P/z+/z[(2w/z+((2/3)V)]

+/x[(w/x+u/z)] +/y[(v/z+w/y )]+Z

• Euler's equation:-

u/t+uu/x=-(1/)p/x (1-D)

At u/t=0 udu=-dp/

• In integral form

U2/2 + dp/= const. (Bernoulli's equation for compressible fluid flow)

/t(uA)dx+(2u22A2-1u1

2A1)=(P1A1-P2A2)+Pm(A2 -A1)

• For steady flow in a duct of constant cross-sectional area:

• 2u22-1u1

2 = P1-P2

• Conservation of energy

• General form:Dh / Dt = DP / Dt + q. +F+ chemical

reaction energy+…………..+etc.

• F sometimes called the viscous dissipation function.

F Mechanical work is entirely dissipated by heat transfer only if the boundary condition permits (not isolated system)

• In the case of the system is insulated, the work mFis not really dissipated but is retained by the system as internal energy.

• In an adiabatic steady-state flow process, the following equation expresses energy relationships for a perfect gas:

• Q – W + m (h+V2+gz) = dU • ho= h+ V2/2 = constant• dh + v dv = 0 • Cp To = CP T +V2/2 (The stagnation or the reservoir

enthalpy)

Reservoir condition

h‘o

T‘o

P‘o

ho

To

Po

hTP‘

Flow between two reservoirs

Gas flow direction

If there is no heat addition to the flow between the two reservoirs, then ho = ho

'

(throttling process)Cp To = CP T +V2/2,

To = T o'

From 2nd of thermodynamic : S'o-S0 =R { Ln (Po /P'o )}- Cp{Ln (T'o /To)}

For a perfect gas

S'o-S0=R{ Ln (P o /P 'o)}

S'o-S0 0

Po/P'o 1 Po P'o

dS = Q / T = [ dh - dP / ] / T

(S / P ) h =const. = - ( 1/ ( T ) ) 0

An increase in entropy (ho =constant), must result in a decrease of stagnation pressure.

In an adiabatic flow To = To'

Isentropic flow To = To' , So = So', and Po = Po'

The speed of sound ( c )

Sound Waves

Molecules in the air vibrate about some average position creating the compressions and rarefactions. We call the frequency of sound the pitch.

Sound Waves Measurement

Sound waves propagate in any medium that can respond elastically and thereby transmit vibration energy.

Waves

The speed of sound ( c ) ( gas dynamics )

V=0 P,T

Piston

Cylinder with piston at rest

Velocity propagation of a plan infinitesimal pressure pulse, along a pipe of constant cross-sectional area, is calculated as follows:-

V=0 P,T, a

Cylinder with piston in motion (in the absence of viscosity and wall effects)Note: This sound wave propagates ahead of the piston at speed c . (c is the speed of sound)

Flow properties behind a planer sound wave dP+P, dV+V

dV dV

Movable Piston (It's impulsively accelerated) at very small velocity dV

PT..C

P+dPT+dT+ d..c-dV

A

Sound wave in stationary reference frame

Controlsurface

The continuity equation:

C2=dP/d ….……….(3)

cA= (+d)(c-dV)Ac d - dV = 0 ……………..(1)

The momentum equation (with no shear forces):F=m*a =m. *V

PA-(P+dP)A= (+d) (c-dV)2A - (cA)c

dP = 2cdV - c2d

From Eq.(1) dV=c dρ/ρ

dP = c2 d …………………(2)

P - P - dP = (+d) (c2-2cdV+ dV2 ) - c2

From isentropic flow and perfect gas:

[S1 = S2 &

P2/P1= (v1/v2) = (2/1)= (T2/T1)

/(-1)]

C2 = RT ………….(4)

P2/ 2 = P1 /1 = constant = P /

dP = constant * -1 * d

dP = P / * -1 * d

dP /d = P / = RT

P / = constant

For air (=1.4,

M.W.=38.966Kg/Kg.mol, and R=287 J/kg.K

C2 = 20.046 T (T in K)

C2 = 49.021 T (T in R)

The speed of sound is the speed at which information about an infinitesimal disturbance or pressure pulse is transmitted through a medium. Waves involving finite disturbances also exist and can likewise move or propagate through a medium. Such finite waves are called shock wave.

For incompressible fluid (r=const.)C2=E/r where E is Bulk modulusC=1415 m/Sec for water at 15oC and E=2*109 N/m

Speed of sound depends on the following:

• Depends on the material of the vibrating medium

• Sound can vibrate water, wood (speaker enclosures, pianos), metal, plastic, etc.

• Sound speed in dry air is 330 meters/second at 0o C

• Faster in warm air, slower in cold

• Water 4 times faster, steel 15 times faster

speedgases < speedliquids < speedsolids

Medium Speed of sound(m s 1)

air (at 0C) 330air (at 20C) 340

water 1400concrete 5000

steel 6000

Wave is a traveling disturbance.Wave carries energy from place to place. There are two basic types of waves:

Transverse and longitudinal waves.

Transverse waves

A transverse wave is one in which the disturbance is perpendicular to the direction of travel of the wave.

Examples: Light wave, waves on a guitar string.

Longitudinal Waves

Longitudinal wave is one in which the disturbance is parallel to the line of travel of the wave.

Example: Sound wave in air is a longitudinal wave.

Wave ParametersWavelength () length or size of one oscillation

Amplitude (A) strength of disturbance (intensity)

Frequency (f) repetition / how often they occur per

second

The frequency (ƒ) is the number of complete oscillations made in one second. Unit : Hz The period (T) is the time taken for one complete oscillation. It is related to frequency by T = 1/ƒ Unit : s

Sound waves propagate in any medium that can respond elastically and thereby transmit vibration energy. sound waves in gases and liquids are longitudinal (alternating compression and rarefaction); in solids, both longitudinal and transversal; speed of sound is independent of frequency; speed of sound in air 340m/s at 20o C; increases with temperature; 1500m/s in water; three frequency ranges of sound waves:

** below 20 Hz: infrasonic ** 20 Hz to 20 kHz: audible, i.e. sound proper ** above 20 kHz: ultrasonic, “ultrasound”

pitch is given by frequency e.g. “standard a” corresponds to 440 Hz intervals between tones given by ratio of frequencies (e.g. doubling of frequency - one octave) male voice range 80 Hz to 240 Hz for speech, up to 700 Hz for song; female voice range 140 Hz to 500 Hz for speech, up to 1100 Hz for song.

When you move away from a fixed source of sound, the frequency of the sound you hear...

(a) is greater than what the source emits

(b) is less than what the source emits

(c) is the same as what the source emits

(b) is less than what the source emits

Question

Then why?

Mach Number (Ernst Mach 1870s)

The ratio of flow speed to sound speed is defined as Mach number. M=V/c

The classification of fluid flow

M<0.25Incompre-ssibleflow

0.25<M<0.8Subsonicflow

0.8<M<1.2Transonicflow

1.2<M<5.0Supersonic flow

M>5.0Hypersonicflow

Compressible fluid flow

M = 1Transonicflow

0.25 < M >1Subsonic flow

1<M <5Supersonicflow

2 C2o /( -1) = 2 C2 /( -1) +V2= V2

max …....(5)

Where:Co is the speed of sound at stagnation temperature

(M= 0.0)Vmax is the maximum attainable speed which is

attained when the static temperature is 0 K. (T=0.0 K)

In an adiabatic-steady flow process, the following equation expresses energy relationships for a perfect gas:

CpTo= CpT+V2/2Cp=R/( -1),

C2

o =RTo, and C2

=RTSubstituting these relations into energy equation gives:

2 C2 /(( -1) V2max) +V2 /V2

max = 1

C2 / C2o +V2 /V2

max = 1 ……………(6)

This equation describes an ellipse as shown in the next figure.

V

C

Vmax

Cmax=Co

C*

V*

M=1

M>1

M<1

A plot of C versus V Prandtl velocity ellipse.

2 C2 /[( -1) (2 C2o /( -1))] +V2 /V2

max = 1

The sonic velocity at Mach (M=1) is calculated as follow:-From equation (6)C*2/C2

o + C*2/ V2max =1

C*2/C2o = 1/(1+ (Co/ Vmax)

2) ………..(7)

Where C=C* at M=1 Then C= V

If the fluid is generated sound waves incompressible, the move at an infinite speed and can be considered as a series of concentric spheres of pressure disturbances.In a compressible fluid, the speed of sound is finite. As shown in the next figure:

Plane flights with M<1 at time t

A

C*t

Disturbed gas

Undisturbed gas

Plane flights with M<1 at time 2t

B

C*t

A

C*2t

L= V*t

Disturbed gas

Undisturbed gas

ABCD

(C)t

(C)3t

(C)2t

Undisturbed gas at rest

Initial wave

V

Disturbed gas

L2=V2tL1=V3t

L3=Vt

(a) V < CPlane flights with M<1 at time 3t

( a ) suppose a particle moves at constant subsonic velocity V, and it reaches each of locations A,B,C, and D at times 0,t,2t, and 3t, it will emit spherical waves.

ABCDUndisturbed gas at rest

Initial wave

V

Disturbed gas

L2=V2t

L1=V3t

L3=Vt

(b) V = C

Mach wave

( b ) If the velocity of source is sonic ,the source moves at the same speed as the wave that it propagates. The sound waves can be represented as a series of spheres that touch each other tangentially at the disturbing point.

Supersonic FlightSupersonic Flight

ABC

Undisturbed gas (Zone of Silence).

Initial wave

V

Disturbed gas

L2=V2t

L1=V3t

L3=Vt

(c) V > C (V is a constant speed)

Mach wave line

Mach angle

D

Apex of the cone.

C(3t)C(2t)Ct

Zone of dependence

The Mach angle

L2

2ct

sin = Ct / Vt = C/V =1/M

( C) distance Ct while the particle moves a During an interval time t the wave propagates a distance Vt.From the geometry of the figure (c ), it is evident that:

sin = Ct / Vt = C/V =1/M …………(8)It applies to supersonic flow only, and a<90o

As M increases, a decreases, and Mach cone becomes narrower.

Shock Wave

CompressionMach Wave

StreamlineD b

Streamline

D b

Expansion Mach Wave

Mach waves past a concave wall and a convex wall in the supersonic flow condition.

Sonic boom

Flight Path

Incident wave

Ground level

Reflected wave

Wave shape

Pressure wave produced by supersonic airplane in steady flight

Flight path

After shock front

Ground track

Forward shock front

Boom carpet

Shock cones produced by supersonic airplane

EFFECTIVE PARAMETERS OF GAS FLOW WHICH ARE

CONSIDERED IN THE PRESENT COURSE..

VARIABLE CROSS-SECTIONAL AREA DUCT EFFECT

WALL FRICTION EFFECT.

HEAT ADD OR REJECT EFFECT

CHAPTER 2Isentropic Flow (Simple

Area change)

Adiabatic flow through a variable-area duct approaches isentropic flow if the walls of the duct are

smooth (frictionless) and if the fluid has zero viscosity.

Assumption-1-D-Compressible fluid-The flow is adiabatic and frictionless (isentropic)To= constant,Po= constant

ThroatInlet

Outlet

Controlvolume

According to steady-state continuity relationships:-

m. = Au=constantLn + Ln A+ Ln u = Ln const.d/ +dA/A +du/u =0. (2.1)According to Euler's equation:

udu=-dp/ (2.2)

From equations (2.2), and (2.1) -udu/dp =1 /

(-udu/dp) d+dA/A +du/u =0

-u2 (d/dp) (du/u)+dA/A +du/u =0

du/u(-(u2/c2) +1 )+dA/A =0

(du/u) ( 1 - M2 ) = - dA/A

du/u =[1/ (M2-1)] dA/A (2.3)

From equation (2.2) du/u=-dp/( u2)

dp/( u2)=[1/ (1-M2)] dA/A (2.4)

udu=-dp/ (2.2)

d/ =[ M2/ (1-M2)] dA/A (2.5)

1/ =-(du/u) (u2/ c2) / d

d/ = - M2 (du/u)

From equation (2.2) 1/ =- (du/u)* u2/ dp * (d/d)

Flow direction

dA=-vedA=+ve

Equation (in the case of dA/A= -ve )

M<1

M>1

[du/dA]*[ A/u] =[1/ (M2-1)]du= +ve

du= -ve

[dp/ dA]*[A /( u2)]=[1/ (1-M2)]

dp= -ve

dp= +ve

d/dA]*[A/] =[ M2/ (1-M2)]

d= -ve

D= +ve

Equation (in the case of dA/A=+ve )

M<1M>1

[du/dA]*[ A/u] =[1/ (M2-1)]

du= -ve

Du= +ve

[dp/ dA]*[A /( u2)]=[1/ (1-M2)]

dp= +ve

Dp= -ve

d/dA]*[A/] =[ M2/ (1-M2)]

d= +ve

D= -ve

The energy equation:ho= h+V2/2

To=T+V2/2 Cp

To/T=1+(-1)M2/2

(2.6) When the flow is isentropically, its relationships as follows:

(P2/P1)= (v1/v2)= (2/1)

= (T2/T1)/(-1) Co

2/C2 = To/T

Dividing by T To/T=1+V2/2 CpT To/T=1+V2(-1)/2 RT =1+V2(-1)/2c2

o/=[1+(-1)M2/2] 1/(-1) (2.8)

co/c=[1+(-1)M2/2] )1/2 (2.9)

Po/P=[1+(-1)M2/2] /(-1) (2.7)

Equations describing critical properties referred to stagnation

properties as obtain from equations)2.6),(2,7),(2,8 ,(and(2.9) ,by substituting M=1 and

when =1.4:T*/T0 = c*2/c0

2=2/(+1)=0.8333

P*/P0 =[2/(+1)] /(-1) =0.5283

*/0 =[2/(+1)] 1/(-1) =0.6339

c*/c0 =[2/(+1)] 1/2 =0.912

Mass flow rate in terms of Mach number :

RT

P

G = m. /A = V = V

c

VRT

G =P

To

Po

Po

PT

ToR

m. /A = M

To

Po

R

m. /A =

[1+ (-1) M2 / 2 ]1/2 [M / [1+(-1)M2/2] /(-1) ]

To

Po

R

m. /A = [M/[1+(-1)M2/2] (+1)/(2(-1))]

m. /A = To

Po

R

[M/[1+(-1)M2/2] (+1)/(2(-1))]

(2.10)

To

Po

R

m. /A* =

[2/(g+1)] (+1)/(2(-

1))]

)2.11(

A/A*= (m. /A*) / (m. /A)

A/A*= [ (2/(g+1))(1+(-1)M2/2) (+1)/(2(-1))] / M

A/A* >1 (2.12)

APo

Tom.

APo

Tom.R

APo

Tom.

Equations (2.6) to (2.12) values of To/T, Po/P, o/, co/c, A/A*,

and

can be plotted and tabulated for several g and M.From equation (2.10)G'=m'/A

= [M/[1+(-1)M2/2] (+1)/(2(-1))]

=constant = C1

M0.10.20.60.81.01.5

C10.6941.3643.4023.8934.0423.436

APo

Tom.In the next figure the parameter

is ploted as a function of M, for g=1.4, and R=287.04 J/kg.

APo

Tom.

M

( )max APo

Tom.

M=1

To

Po

R

To

Po

Gmax occure at the section of minimum flow area

when M=1

For g=1.4, and R=287.04 J/kg.K

m.(kg/sec), A(m2), Po(Pa), and To(K)

G*=(m. /A)max= ]2)/+1 [()+1)/(2(-1)(

G*=(m. /A)max = 0.04042

*T

T

)2

11(2

1

2M

*P

P)1(

2

])

2

11(2

1[

M

* )1(

1

2

])

2

11(2

1[

M

*c

c 2

1

2

])

2

11(2

1[

M

From equations (2.6) to (2.9):

=

=

=

=

To/T=1+(-1)M2/2 (2.6)

o/=[1+(-1)M2/2] 1/(-1) (2.8)co/c=[1+(-1)M2/2] )1/2 (2.9)

Po/P=[1+(-1)M2/2] /(-1) (2.7)

TTTT

o

o*

)2

11(

))1(2

11(

2

2

M

V= )(2 TToCp

= )(1

2 TToR

12

RTo

o

Po

12

Vmax =

=

The energy equation:ho= h+V2/2 To=T+V2/2 Cp

Vmax =

1

2

Note that Vmax is always finite, at Vmax however, the Mach number is infinite, because the sonic

velocity at that temperature is zero.

Co

For =1.4 Vmax= 2.24 Co (2.13)

Note that Vmax is always finite, at Vmax however, the Mach number is infinite, because the sonic velocity at that temperature is zero. The highest velocity attainable with 293 K air is therefore:

oT14.1

04.287*4.1*2

oTVmax = = 44.82 = 767.2 m/s

Critical Mach number M*=V/c*

M*/M = (V/c*) / (V/c) = c/c*

C*/C0 = [2 / (+1) ] 1/2

Co/C = [1 + (-1) M2 / 2 ] 1/2

2

2

11

2

1

*

M

MM

2*1

11

1

2*

M

M

M

).……..…2.14 (

).……..…2.15 (

M*/M = { [(+1)/2] 1/2 } / [1+(-1)M2/2] 1/2

At M=1 for g=1.4

1

1

A plot of M* versus M is shown in figure.

M=* =2.4495

V1V2

P1 A1P2 A2

F

Impulse function Force acting a control volume

The momentum equation is:-F+ P1A1- P2A2= m. V2-m

. V1

F is the wall force exerted on the fluid by inner walls of the duct in the direction of flow.

Note that the force F acts in a direction

opposite to thrust.

The thrust produced by the stream between section 1 and 2 is :-F= (P2A2+m. V2) - (P1A1+m. V1)=I2-I1

I is the impulse function.

I = PA(1 + M2) (2.16)

I is the impulse function, is defined as :-

I = (PA + m. V) =PA (1+ V2 / P)

Specific impulse Is = F/ m. (2.17)

e

Patm Ae Pe Ae

F

Applying momentum equation at exit cross-section area:- F+ ( Patm - Pe ) Ae= m. Ve

)(2 TTCV op

))(1(1

21

o

eoe P

PRTV

eatmeo

eo APPP

PRTmF )())(1(

1

21

.

Applying momentum equation at exit cross-section area:-

(2.18)

F+ ( Patm - Pe ) Ae= m. Ve

F= m. Ve +(Pe –Patm) Ae From energy equation:- To=T+Ve

2/2Cp

)1(1

2

oo T

TT

R

][

))(1(1

2

21

1

1

.

o

e

o

eo

oe

P

P

PP

RT

RTmA

( F/ A e) P e=0 to determine Pe which provides maximum thrust. P e-P atm =0

Pe=Patm at maximum thrust (2.18)

( F/ P e) A e =0 to determine A e which provides maximum thrust.

)2.19(

eatmeo

eo APPP

PRTmF )())(1(

1

21

.

))(1(1

21

.

o

eooptimum P

PRTmF

))(1(1

2)(

1

.

o

eooptimums P

PRT

m

FI

0.0o

e

P

P

خروج ضغط يوجد مساحه أى عند أنه توضح السابقه المعادلةدفع ( Peمعين ) قوة أكبر اعطاء على يساعد الذى If Pe=Patmهو

(maximum thrust case)

(2.20)

(2.21)

يكون أن يجب دفع أقصى على للحصول()Pe/Po=0.0

أن ) يعنى الحصول( Pe وهذا يمكن ال وهذا صفر يساوى . جدا جدا كبيرة المساحة تكون حالةأن فى اال عليه

][

))(1(1

2

21

1

1

.

o

e

o

eo

oe

P

P

PP

RT

RTmA )2.19(

The maximum impulse is therefore obtained at zero exit pressure and is calculated from:-

1

2)( max

o

imums

RTI ( 2.22)

. نظريه ولكنها تتحقق أن يمكن ال المعادله هذهدالة بواسطة تصنيفه نستطيع الدفع نظام تأثير

( impulse functionالدفع )

I=PA + m. V =PA (1+ M)

At M=1 the critical condition is occured

I*=P*A* (1+ )

)1(

)1(

* **

2

AP

MPA

I

I

*P

P )1(

2

])

2

11(2

1[

M

*I

I)

2

11)(1(2

1

2MM

M

*I

I

=

A/A*=[(2/(+1))(1+(-1)M2/2) (+1)/(2(-1))]/M

= (2.23)

is tabulated as a function of M and .Where

Real nozzles and diffusers:According to frictional effects, the losses occur through nozzle and diffuser. From 1st and 2nd laws of thermodynamics:

To d So= d ho-v d PoIf no heat is transferred and if no work done (adiabatic), then:-

T0 d S=-v o d Po (d So=d S)Vo=R To/Po

d S/R= -d Po/Poby integration:- ln Po2- ln Po1=-(S2-S1)/Rln (Po2/Po1) =-(S2-S1)/R (2.24)األنتروبى للثروموديناميك الثانى للقانون طبقايقابلها ان يجب وبالتالى الجراء هذا فى تزيد دائما

. الكلى الضغط فى نقص

Nozzle efficiency:

h

ho

h2s

h1

V12/2

V2s2/2

h2

1

2

2s

o

V22/2

Po1 Po2

h-s diagram for nozzle

S

nozzle

nozzle

vC nozzlevC

isentropic

reald m

mC

)(

)(.

.

According to 1st law of thermodynamic:-

(2.25)

Velocity coefficient

(2.26)

The coefficient of discharge (2.27)

isentropics

real

V

V

)2/(

)2/(2

2

22

99.0:9.02

2

so

o

hh

hh

isentropics

real

V

V

)(

)(

2

2

Diffusers efficiency:Diffuser efficiency is defined as the isentropic enthalpy if the flow is decelerated to pressure equal to stagnation pressure at the diffuser exit divided by the decrease in kinetic energy if the flow entering is decelerated isentropically to the isentropic stagnation state

D

Po1 Po2

P2

P1

1

2s

3

o1o2

2V1

2 /2

h

Sh-S diagram for diffuser

D

2/21

13

V

hh

D

D

D

11

102

PP

PP

o

Pressure coefficient=

2/21

13

V

hh 1

13

hh

hh

o

pCV

TT

2/21

13

1

21

1

3

2

1

TCV

TT

p

1

21

)1

(

1

2

21

1)(

RTV

PPo

21

)1

(

1

2)

2

1(2

1

21

1)()2

11(

M

PP

Mo

o

Isentropic Flow through a NozzleOperation of A Convergent nozzle

under varying pressure ratios.

• From equation (2.11) maximum mass flow rate per unit area (flow density) occurs at M=1 . Pe=P* (CHOKED FLOW)

To

Po

R

m. /A* =

[2/(g+1)] (+1)/(2(-

1))]

)2.11(

Po

To

Pb

Pe

Pb/Po

Distance along nozzle

1Pe/Po

12

3P*/Po

4d

Pe/Po

P*/Po

Pb/Po

P*/Po

1

23

d4

1

1

1

Pe/Po2

3

4

d

m.

m.max

Me=1

Illustration of expansion from a choked convergent nozzle ( Pb < P* )

Chapter 3

Waves in compressible fluid flow

Compression waves

Rarefaction (expansion) waves

Normal shock wavesNormal stationary

shock waves

Normal moving

shock waves Oblique and conical

shock waves

Prantle Meyer flow and the shock expansion procedure waves

Simple pressure waves

Normal stationary shock waves

• Governing equations:-

• Assumption:

• A shock wave is thin, steady, stationary, and one dimensional.

Normal stationary shock wave

FLOW DIECTION

Behind shock (1)Upstream shock

M1, V1, P1, T1, r1, h1, S1, Po1

Front of shock (2)Downstream shock

M2, V2, P2, T2, r2, h2, S2, Po2

)1( )2(

From continuity equation:-

P1(1+ M12)= P2(1+ M2

2) (3.2)

1 V1A1= 2 V2A2

1 V1= 2 V2 (3.1)

Conservation of momentum:-P1A1- P2A2 = 2 V2

2 A2 – 1 V1

2 A1

P1+ 1 V12= P2+ 2 V2

2

V2 = (P/RT) V2 = PM

From energy equation:-ho1 = ho2 To1 = To2

h1+V12/2= h2+V2

2/2

T1+V12/2Cp= T2+V2

2/2Cp

T1[1+V12/(2CpT1)] = T2[1+V2

2/(2CpT2)]

22

2

1

2M

TC

V

P

(3.3)2

1

22

2

1

21

1

21

1

M

M

T

T

2

2

1

1

2

1

P

RT

RT

P

1

2

2

1

2

1

T

T

P

P

(3.4)

222

211

1

1 RTMRT

PRTM

RT

P

2

2

21

1

1 MT

PM

T

P

)1()1( 222

22

11

1

MT

M

MT

M

From equation (3.1):-ρ1 V1= ρ2 V2

From equation (3.2)

2

1222

2

22

12

121

1 )2

11(

)1()

2

11(

)1(M

M

MM

M

M

)1(2

)1(22

1

212

2

M

MM

This is quadratic equation in M2 (or a quadratic

in M22).

, ± M12

M12= M2

2 is a simple solution, means that a

shock wave is not present. The negative roots are imaginary.

From 2nd law of thermodynamics:-

)1(2

)1(22

1

212

2

M

MM (3.5)

01

1

1

2

2

2

01

02

P

P

P

P

P

P

P

P o (3,6)

)()(1

2

1

212 P

PLnR

T

TLnCSS p

(3.7) )(01

0212

P

PLn

R

SS

1

1

1

2 21

1

2

M

P

P

)2

1

1

2(

)11

2)(

2

11(

21

21

21

1

2

M

MM

T

T

1)1(

)1(2

1

21

2

1

1

2

M

M

V

V

All equations from (3.2)to (3.7) can be formed as a function of M1 only (see gas dynamics table) as follows:-

(3.8)

(3.9)

(3.10)

)1(2

2)1(2

1

212

2

M

MM

12

1 o

o

T

T

1

1

21

1

21

21

2

1 )

11

12

1()

)1(21

1

)1(21

(

MM

M

P

P

o

o

)3.13(

) 3.12 (

)3.11(

)()(1

2

1

212 P

PLnR

T

TLnCSS p

)))1(

)1(2((

1

))1

1

1

2((

1

1

21

21

21

1

12

M

MLn

MLn

R

SS

1

2

1

12

o

o

P

PLn

R

SS

(3.14)

)3.16(

The next figure present the plot of ∆S/R versus M1

forg=1.4

Shocks are

possible

Shocks are not possible

1

2

0

-10 1 2 43

M1

∆S/R

Entropy changes across a normal stationary shock wave. (=1.4)

To

Po

R

M>1 ∆S/R= +veM<1 ∆S/R= -ve shocks are not possible

S2>S1 Po2<Po1

Stagnation pressure is always lost across a station-nary normal shock wave.

Shock wave is a discontinuous process involving large gradient in properties.

The pressure rise across a shock wave, (P2 / P1)

is called the strength of the shock wave.

G*=(m./A)max=m./A*=

[2/(+1)](+1)/(2(-1))]

To

Po

R

To

Po

R

1

2*2

*1

o

o

P

P

A

A

m.1 =

[2/(+1)] (+1)/(2(-1))] A*2

(3.17)

]2)/+1) [(+1)/(2(-1) [(A*1

m.2=

m.1=m.2

1

2

(3.17( ,)3.16معادله )عن المساحة تقل أال يجب الصدمه بعد أنه *Aتوضح

اذا 2 ألنهايؤدى مما التصرف معدل اقالل فى ستسبب ذلك عن قلت . والمفهوم الدخول عند االساسيه المشكله فى تناقض الى

. االبواق خالل السريان تطبيقات مناقشة عند مهم السابق

1

2

1

12

o

o

P

PLn

R

SS

1

2*2

*1

o

o

P

P

A

A

M1>1 M2<1

Stationary normal shock wave

P

X

A*1

A*2

By means of equations (3.2) to (3.7)and eqn.(3.17),

values of M2, P2/P1, 2/1, T2/T1, PO2/PO1, S2-S1/R, and A*2/A*1 can be tabulated

for several of and for any upstream Mach number

(M1).

The Rankin-Hugoniot relations

From the pressure expression given by equation (3.8) and substituting that into the density ratio

given by equation (3.10), the following equations can be obtained:-

1

1

1

2 21

1

2

M

P

P (3.8)

1)1(

)1(2

1

21

2

1

1

2

M

M

V

V

(3.10)

1

2

1

2

1

2

1

1

11

1

P

P

1

2

1

2

1

2

1

1

11

1

p

PP

P

)3.18 (

)3.19 (

(3.21) )11

()1(

)1(21)(

22

22

1

2

SS M

MC

C

(3.22) )

11(

12

1

1

21

2

SM

(3.22) )1(1

21 2

1

2

SMP

P

(3.20) )1

(1

2

1

12

SS M

MC

UU

From Rankin Hoiguount relation

Equations (3.18), and (3.19) are called the Rankin-Hugoniot relations.For isentropic flow process:

(3.23) )(1

2

1

2

P

P

50

5

30

3

5 7 10

7

70

10

100

P2/P1

2/1

Normal shock wave

Isentropic

The normal shock wave (Rankine-hugniot) and isentropic curves for g =1.4.

3

Operation of a Convergent-divergent

nozzle under varying pressure

ratios (De-Laval nozzle)

1.0

P/Po

P*/Po

DC

BA

EF

GH

I0 X

Sonic throat

throat

Pe

PbPt

Po , T0

1.0

1.00 P*/Po

Design pressure ratio

m,/m.max

FI H G E D CB

A

(a) nozzle geometry with possible flow configuration

(b) Pressure distribution caused by various back pressure

(c) Mass flux versus back pressure

Pb/

Po

Curve A, and BThe back pressure is not low enough to induce sonic flow in

the throat, and flow in the nozzle is subsonic throughout. The pressure distribution is computed from subsonic area-change relation. The exit pressure Pe=Pb and the jet is

subsonic.Curve C Ae/At=Ae/A* for Me

-The throat becomes sonic and mass flux reach a maximum.

- The remainder of the nozzle flow is subsonic Pe=PbCurve H

-Here Pb is such that Pb/Po corresponds to the critical area ratio Ae/A* for a supersonic Me

-The diverging flow is entirely supersonic, including the jet flow, and Pe=Pb.

- this is called "THE DESIGN PRESSURE RATIO OF THE NOZZLE" . It is the back pressure suitable for:-

*operating a supersonic wind tunnel *an efficient rocket exhaust

Curve D, E, and F -Here, it is impossible according to purely

isentropic-flow calculations. - the throat remains choked at the sonic value.

-We can match Pe=Pb by placing a normal stationary shock wave at just the right place in the

diverging section to cause a subsonic diffuser flow - The mass flux remains at maximum.

-At back pressure (F) the required normal stationary shock wave stands in the duct exit.

Curve G -No signal normal stationary shock wave can do the job, and so the flow compresses outside the

exit in a complex series of oblique shocks until it matches Pb

Curve I at exit Pback<Pdesign -the nozzle is choked -the exit flow expands in a complex series of supersonic wave motions until it match the low back pressure.Note: Downstream of shock, the nozzle flow has an adverse pressure gradient, usually leading to wall boundary-layer separation. Blockage by the greatly thickened separated layer interacts strongly with the core flow.

Moving normal shock wave relation :-

Stationary

Moving

Vb>Va

Vb VaVs

(a)(b)

(1)(2)

V1>V2

Vs-Va=V1Vs-Vb=V2

P2

T2

.

.

.

P1

T1

.

.

.

Pb

Tb

.

.

.

Pa

Ta

.

.

.

Moving Normal shock waveStationary Normal shock wave

Pa=P1Pa =P1

Pb =P2Pb =P2

Ta =T1Ta =T1

Tb =T2Tb =T2

Ma =Va /caM1 =(Vs –Va )/c1 =V1 /c1

Mb =Vb /cbM1 =(Vs –Vb )/c2 =V2 /c2

Toa =Ta (1+(-1)Ma2 /2)To1 =T1 (1+(-1)M1

2 /2)

Tob =Tb (1+(-1)Mb2 /2)To2 =T2 (1+(-1)M2

2 /2)Poa =Ta (1+(-1)Ma

2 /2) /(-1)Po1 =T1 (1+(-1)M12 /2) /(-1)

Pob =Tb (1+(-1)Mb2 /2) /(-1)Pob =T2 (1+(-1)M2

2 /2) /(-1)

Toa Not equal TobT01 =To2

Super sonic wind tunnels

Heat ExchangerCompressor

Test Section

Throat

M<1 M>1

M=1

M<1

Flow Direction

Continuous Wind Tunnel (no diffuser)

Heat ExchangerCompressor

Test Section

Throat

M<1 M>1

M=1

M<1

Flow Direction

Continuous Wind Tunnel with diffuser

ThroatM=1

M>1

Test Section

Throat

M<1 M<1 M<1

Throat

Throat

M<1 M<1 M<1

Throat

Wind Tunnel Startup Sequence(A) Initial startup

M<1 M<1

(B) Frist throat sonic

M=1 ,A*1 M<1

Throat

M<1 M<1 M<1

Throat

Throat

M<1 M<1 M<1

Throat

(C) Shock in diverging section

M<1

(D) Shock in Test Section Entrance

M=1 ,A*1

M=1 , A*2

M>1

M=1 ,A*1

M>1

Throat

M<1 M>1 M<1

Throat

M<1 M>1 M<1

Throat

(E) Shock Swallowed بتلعتnأ الصدمة

(F) Shock-Free deceleration with variable-area diffuser throat.

M=1 ,A*1

M=1 , A*2

M>1

M=1 ,A*1

M>1

M>1

Throat

M<1 M>1 M<1

(G) Shock in diffuser ( fixed area ) throat.

M=1 ,A*1

M>1

End of Chapters 1,2, and 3.

CHAPTER 1 Fundamental Concepts of Compressible Fluid Dynamics

CHAPTER 2 Isentropic Flow (Simple Area change)

CHAPTER 3 Waves in Compressible Fluid Flow

(Stationary and Moving Normal Shock Wave)