Gas Dynamic lictures
Transcript of Gas Dynamic lictures
LECTURES
in
GASDYNAMICS
IN THE NAME OF ALLAH
FOR 3rd YEAR OF MECHANICAL POWER ENGINEERING
Prepared by
Assoc. Prof. Dr. Eng. Mostafa Nasre-meil, [email protected]
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)