Gas DynamicsCompressibility: It is measure of change in density that will be produced in fluid by specified change in pressure.In fluid flow, flow characteristics changesV p (Temperature changes are important incompressible flow)Th t d f th fl i hi h h i d it d t t i t t i. . V KE T The study of these flows in which change in density and temperature are important isknown as compressible fluid flow or gas dynamics (Compressibility effect areusually important only in gas flow)Fundamental Assumptions:1. Gas is treated as continuous medium. (Not true for rarefied flow where density isvery low e.g. flow over spacecraft flying at high altitudes.)2 No chemical change occur in flow field (No combustion or ionization of molecules) 2. No chemical change occur in flow field. (No combustion or ionization of molecules)3. Gas is perfect and obeys perfect gas law: p=RT (R for air = 287 J / kg K)4. Specific heats are constant (Not for high Temperature flow OR if T is high) i.e. gas is calorically perfect.Propulsion (by K. S. Bhati) 15. Gravitational, Magnetic and Electrical effects are negligible. (Not for MHD etc. flows)6. Overall effect of viscosity is small.Streamtube: Streamtube is defined by considering a closed curve drawn in flow. Series of streamlines will pass through this curve. Further downstream these streamlines can be joined by another curve. Since there can be NO flow normal to streamlines can be joined by another curve. Since there can be NO flow normal to streamline, streamlines passing through curve define walls called as streamtube.1- D flow: It is flow in which reference axis is chosen in such a way that velocity vectorh l t fl i t t d t has only one component e.g. flow in constant area duct. Strictly speaking, flow in variable area duct is NOT 1-D; but can be assumed 1-D if rate of change of area andcurvature are small enough such that one componentg pof velocity is dominant over other velocity components e.g. flow in C-D nozzle. Such flow are referredas Quasi-one dimensional flow.Conservation laws for 1- D steady flow:1. Continuity equation:Since there can be NO mass transfer across streamtube,Neglecting higher( )( )( )0m constAV d A dAV dVVAd AdV VdA == + + ++ +

Propulsion (by K. S. Bhati) 2Neglecting higher order terms00(1)VAd AdV VdAd dV dAV A + + =+ + =2.Momentum equation:Net force in x-direction due to pressure and frictional forces acting on control surface isPressure on curved surface,( )( )net curved projectedF pA p dp A dA p A dF= + + + [ ]0.5 ( )curvedp p p dp = + +Projected area of curved surface, ( )projectedA A dA A = + ( 0.5 ) F Adp pdA p dpdA dF = + + (Neglecting higher order terms)( 0.5 )netnetF Adp pdA p dpdA dFF Adp dF+ += (Neglecting higher order terms)By Newtons 2ndlaw,[ ]( )netF m V dV V = +

If effect of friction is neglected[ ](2) Adp dF AVdV =0 (3)dpVdVPropulsion (by K. S. Bhati) 3If effect of friction is neglected,0 (3)pVdV + =This is Euler equation for steady flow. (Bernoulli equation for incompressible flowcan be derived from this.)3. Steady Flow Energy equation (SFFE):2 21 21 2( )pV Vh q h w h c T + + = + + =We will restrict our attention to flows in which NO work is done. 1 2( )2 2pq2 2V VApplying this to control volume (shown in figure)2 21 21 2(4)2 2p pV Vc T q c T + + = +Applying this to control volume (shown in figure)2 2( )( )2 2p pV V dVc T dq c T dT++ + = + +For adiabatic flow2 2 (5)pcdT VdV dq + =Propulsion (by K. S. Bhati) 4For adiabatic flow,0(6)pcdT VdV + =4. Equation of State:l l l lp RT =Taking logarithm,Differentiating,ln ln ln ln(7)p T Rdp d dTp T= + += +5. Entropy consideration:Entropy places limitation on which flow process is physically possible and which is NOT.p T py p p p y y pSubtracting equation 3 from 5,pdpcdT Tds = (8)pdT dpds c RT p= Integrating,2 22 11 1ln ln(9)pT ps s c RT p = Propulsion (by K. S. Bhati) 52 1 2 21 11ln ln =1pps s T p Rcc T p = 12 1 2 1lns s T p = 1 21pc T p For isentropic flow s=0,2 2 2 11 1 1 2T p pT p p = = 2 21 1(10)pp = (11)pconst =This is true only if flow is also adiabatic (apart from being frictionless) i.e. for isentropic flow because in momentum equation frictionless flow was already considered.Propulsion (by K. S. Bhati) 6Example 1:Consider isothermal flow of air through variable area duct. At certain section velocity,pressure and temperature are given as 200 m/s, 25C and 120 kPa respectively. If velocity is decreasing at this section at rate of 30% per meter, find dp/dx, ds/dx & d/dx.Ans: By equation 5,dT VdV d TdAns: By equation 5,For isothermal T=0pcdT VdV dq Tdsds VdVd T d+ = ==It is given 20.30.3dx T dxdVV ds Vdx dx T= =From equation 3,dx dx Tdp dV pVdVVdx dx RT dx = = From equation 7, 1 d dpdx RTdx =This give dp/dx=-16.8 kPad/ =-0.196 kg/m3ds/dx=40 3 J/mKPropulsion (by K. S. Bhati) 7ds/dx=40.3 J/mK1-D Steady Isentropic Flow:Many flows in engineering practice can be adequately modeled by assuming them to be 1 D steady and isentropic 1-D, steady and isentropic.An isentropic flow is an adiabatic frictionless flow.Variable area flow:It is assumed that flow is quasi one dimensional and isentropic. We now first consider effect of change of area on isentropic flow considering a control volume.Recalling equation 1, 6, 7 and 11,0 (1)d dA dVA V + + =0(6)(7)pcdT VdVdp d dT + =(7)(11)pp Tpt= +Propulsion (by K. S. Bhati) 8(11)pconst=ln ln 0(12)pdp d =Using eq 12 in 7,(12)( 1) (13)ppdT d== Using eq 13 in 6,( 1)(13)( 1) 0pTdcT VdV= + =g q ,2( )( 1) 0pd V dVcT V + =But 2 2 22( 1)(14)( )pcT VV aMM a RTRcT = = = Hence21(1pRcTTd dVM= 5)Propulsion (by K. S. Bhati) 9Hence (1 MV = 5)Substituting this in equation 1,2( 1)(16)dA dVMA V= 2( ) ( )1= (17)1A VdV dAV M A Using equation 15 in this,221(18)1V M Ad M dAM A=Using equation 12 in this,22(19)1dp M dAp M A=Once dA/A is known, dV/V, d/ and dp/p can be calculated from above relations.Isentropic relations in terms of Mach no.:ppconst =By equation 11, Using equation of stateconstTconst==Propulsion (by K. S. Bhati) 10Using equation of state,1const =This gives isentropic relations:1122 2 2 2T p a 2 2 2 221 1 1 12 21 2(20)T p aT p aV V = = = By equation 4,Multiplying both side with 2/Cp1 21 22 21 22 22 2p pV VcT cTV V+ = + Propulsion (by K. S. Bhati) 111 21 21 22 2p pV VT TcT cT+ = + Using equation 14,22 121 22 ( 1) (21)2 ( 1)T MT M+ =+ Using equation 20,1 2212 122 ( 1)2 ( 1) (22)2 ( 1)T Mp MM+ + = 21 21212 1( )2 ( 1)2 ( 1)(23)p MM + + Stagnation conditions:2 121 2( ) (23)2 ( 1)M = + Stagnation conditions:Stagnation conditions are those that would exist if flow at any point in a fluid stream is isentropically brought to rest. (To define stagnation temperature it is only necessary thatflow is brought to rest adiabatically.). Even if flow is non-isentropic, stagnation condition at a point would be that if local flow is brought to rest isentropically. Substituting M2=0 in above equations,2011 (24)TM = +Propulsion (by K. S. Bhati) 12Likewise p0/p and 0/ can be obtained.1(24)2MT = +12011(25)2pM = + Critical conditions:Critical conditions are those that would exist if flow is isentropically brought to M2= 1.F 21 t 232 *( )22 ( 1)pM T From eq 21 to 23,2 *112 ( 1)(26)1M TT+ =+2 *1112 ( 1) (27)1M pp + = + 12 *112 ( 1) (28)1M + = + Substituting M1= 0 in above equation, 11 + 1* * * *1 12 2 2 2 T p a Propulsion (by K. S. Bhati) 131 10 0 0 02 2 2 2 1 1 1 1T p aT p a = = = = + + + + Pitot static tube:Pitot tube place in subsonic compressible flow will register stagnation pressure (In case of supersonic flow, shock wave stagnation pressure (In case of supersonic flow, shock waveproduce ahead of tube and hence flow is not isentropic).Since the disturbance produced by the tube is restricted to stagnation point region, pitot tube can be used to measure l it i ti 25 f ll velocity using equation 25 as follow:1201 11 1p p pM = + 121 12M + 212021 11 1 (29)2p pV MM = + For incompressible flow by Bernoullis equation:02 p pV =In order to determine velocity in incompressible flow only the difference between p0& p has to be measured and not their individual values. Propulsion (by K. S. Bhati) 14Example 2:Pitot tube indicate pressure of 134.3 kPa when place in air stream of T=303K and staticpressure 101kPa. Find flow velocity taking compressibility effect into account. pressure 101kPa. Find flow velocity taking compressibility effect into account.Ans: 31.16/pkg mRT = =Using this in equation 292 21M RTV = g q212011 12p pV RTV RTV = + ( )0.5211 24349 1 6993647 V + = ( )121 24349 1 69936.4724349 1 586.7VV+ = + =Propulsion (by K. S. Bhati) 150.155/ V m s =Example 3:Air flows through converging diverging duct with inlet area of 5 cm2and exit area of 3.8 cm2.At inlet section V, p and T are 100 m/s, 680 kPa & 333K. Find V,p, & T at exit. 3.8 cm .At inlet section V, p and T are 100 m/s, 680 kPa & 333K. Find V,p, & T at exit.Ans: By continuity, 12 2 1 12AV VA =222 1 12 2 12 2 1936.2/p A pM RT V kg msRT A RT = =2 2 122 22936.2 13404.3/p M Rkg ms cT= = =By isentropic relation,212TT Substituting this in above equation,22 113 3132828125p pTMT K c= = =Propulsion (by K. S. Bhati) 162 2 31 1 113282812.5 0.273MT K cM V RT = == =From equation 21,212 12 22 ( 1) 676 2 ( 1) 2 04MT T KM M+ = =+ +Substituting this above equation,2 232 322 ( 1) 2 0.4676M MM c + + = ( )2 322322 22 0.45 363.4MM M + + =( )2 26 2 22 2 2 26 4 2125 15 ( 5) 363.415 75 3634 125 0M M M MM M M M+ + + =From equation 21,6 4 22 2 2 2215 75 363.4 125 00.373M M M MM+ + + ==22 2 2329 135.6/T KV M RT m s == =From isentropic relation,Propulsion (by K. S. Bhati) 172 2 232 2652, 6.9/ p kPa kg m = =Example 4:At exit of rocket nozzle V & p are 3800 m/s & 100 kPa respectively. Nozzle exit has dia 30 cm. Temperature of gas in combustion area is 2400C. Find T at nozzle exit, pressure 30 cm. Temperature of gas in combustion area is 2400 C. Find T at nozzle exit, pressurein combustion area and thrust developed. Assume isentropic flow of gas which has molarmass 9 and specific heat ratio 1.3Ans: T01=2673 K, V2=3800 m/s, p2=100 kPa, R=8134/9=903.8 J/kgK, =1.3F ti 242( )From equation 24,201 22 2( 1)12T VT RT= +22 22673 0.3*380012*1.3*903.8 T T= +24.331830 2673T KT= 101 012 22673830p Tp T = = Propulsion (by K. S. Bhati) 18012215823 100* (0.15) 7 p kPaT pA kN kN == = =Effect of area on flow velocity:From equation 16,2( 1)dA dVM =Because A and V are positive,( 1) MA V= Area VelocityM < 1 (Subsonic)Increases Decreases DiffuserD I N l(Subsonic)Decreases Increases NozzleM > 1 (Super sonic)Increases Increases NozzleDecreases Decreases Diffuser2( 1)dA AMdV V= sonic)If M = 1, dA/dV = 0 i.e . A become maximum or minimum.1dV VdA V Propulsion (by K. S. Bhati) 191 dA VAdV RT V = Differentiating this w.r.t. V,2 221 11 ( 1) 0d A A V AA M + + + > Because second derivative is positive, Area will be minimum at M = 1.22 2 2 21 ( 1) 0 A MdV RT V V RT V = + = + = + > Effect of Area on Mach No.:Now we would like to examine how Mach No. changes when area changes.First of all, we know that, , ,Taking log and differentiating,V aM = (30)dV da dMV M= +By equation 13 & 15,2( 1)(31)V a MdT dVMT V = Propulsion (by K. S. Bhati) 20T Va RT =Speed of sound,12da dTa T= Taking log and differentiating,Using equation 31 in this212da dVMa V = Using equation 31 in this,2112dV dMM + = Substituting this in equation 30,2 V M Multiplying this with equation 16,2 211 ( 1) (32)2dA dMM MA M + = Propulsion (by K. S. Bhati) 21Area Mach No.Because A and M are positive,M < 1 (Subsonic)Increases Decreases DiffuserDecreases Increases NozzleM > 1 Increases Increases Nozzle M > 1 (Super sonic)Increases Increases NozzleDecreases Decreases DiffuserPropulsion (by K. S. Bhati) 22It follows that if a subsonic flow is to be accelerated to a supersonic velocity it must bepassed through convergent-divergent passage. The convergent portion accelerate flowto supersonic velocity. At minimum area section (throat) we have already said that M=1. to supersonic velocity. At minimum area section (throat) we have already said that M 1.In such nozzle, pressure decreases continuously throughout. If end pressure is NOT too low, flow will remain subsonic throughout as in Venturimeter i.e. supersonic flowwill not be generated in diverging section (will act as diffuser). In such case, it follow f ti 32 th t dM t b t th t (dA 0) i M h i from equation 32 that dM must be zero at throat (dA=0) i.e. M reaches maximum.*Mach no. criterion for incompressible flow:Change in velocity bring change in pressure, density and temperature, as can be seen f 3 15 d 31Propulsion (by K. S. Bhati) 23from eq 3, 15 and 31:Note: Star (*) marked sections can be skipped2(3)dp V dVp p V= Using a2=p/,2(33)dp dVMp V = 2 (15)d dVMV = At high Mach no change in pressure density & temperature are not negligible Hence2( 1)(31)dT dVMT V = At high Mach no. change in pressure, density & temperature are not negligible. Hence Mach no. is parameter that determine flow is compressible or not.By continuity equation,( )0Vx =0( ) . . 0xV VV V f xi ex x x + = = Should be very smallPropulsion (by K. S. Bhati) 24In terms of magnitude,VVx x

d dVV

Equation for speed of sound,2d dpadp dV =

By equation 3,2a Vdp VdV =

Hence flow is incompressible if,22VdV dVa V

22211VaM

Mach Waves:Consider a point source of disturbance moving with a uniform velocity u through gas210.3MM

Consider a point source of disturbance moving with a uniform velocity u through gas.Consider only series of waves emitted at time interval t. As body is moving, origin ofThese waves will be continuously changing.Propulsion (by K. S. Bhati) 25Let u < a (speed of sound). In this case, influence of disturbance will be observedeverywhere in space AND in sequence everywhere in space AND in sequence (i.e. 1stgenerated wave will be observed 1st).Let u > a i.e. point is moving with supersonic speed. In this case wave pattern is as follow.One can notice that here gas within the cone is alone aware of presence of particle.Vertex angle of this cone is Vertex angle of this cone istermed as Mach angle .1sina = =There will be jumps in values of flow variable when flow reaches conesinu M = =flow variable when flow reaches cone.Cone is therefore termed as conical Mach wave.This result is sometimes used in measurement of Mach no. of gas flow.Propulsion (by K. S. Bhati) 26*Speed of sound:Speed of sound is the rate of propagation of infinitesimal pressure pulse through a still fluid. It is a thermodynamic property of a fluid. fluid. It is a thermodynamic property of a fluid. Ratio of flow velocity to speed of sound is important dimensionless number known asMach number.VM < 1 ; Subsonic flowM ~ 1 ; Transonic flowDerivation of formula for speed of sound in fluid medium:VMa=;M > 1 ; Supersonic flowM > 5 ; Hypersonic flowDerivation of formula for speed of sound in fluid medium:Consider an infinitesimal pressure wave initiated by slight movement dV of piston to right.Wave will induce gas velocity dV behind it as it move through gas. Let the observer be traveling with (pressure pulse) wave front. By continuity eq.,( )( )( )( ) (32)a d a dVa a ad d dVdV d d = + = + ++A l i N t 2ndl l di ti( )(32)( )dV ad da dV a d = + = +Propulsion (by K. S. Bhati) 27Applying Newton 2ndlaw along ve x-direction,[ ]( ) ( )( )p p dp A m a dV maAdp Aaa dV a + = = Also 1(33)1dp a dVdT dpT p= ( 1) /(34)T pT pdT TdV a = = Putting expression for dv from eq 32 to eq 33,2( )1dp a d ddp da = + + Temperature gradients in wave are small.1(35)paddp = + Hence gas undergo isentropic processwhile passing through waveBy equation 11(35)spadpconst = Propulsion (by K. S. Bhati) 28By equation 11,i.e.ln lnpconstp const = =Differentiating,0dp dp =(36)spdp p pa RTd = = = *Momentum equation for 1-D Steady Isentropic Flow:Momentum equation is not used in above analysis because it will always give same result as that of energy equation. This is proved as follow:s gy q pBy equation 322 22 110 (37)2 2V V dp + =By equation 11,11/11/11 pp =Integral become,1 12 21/ 1/1 11/1 12 11 1 1 11p p dpp dp p p = = Propulsion (by K. S. Bhati) 291 1 1 1 121 21p p dp = 1 1 11 1221 p 221 2 1 21 1 11 11 1RT p a p dpp p = = Substituting this in equation 37122 22 21 22 1121 01a pV Vp + = 122 2 2 21 22 2 1 1211a paM aM + 0 =Propulsion (by K. S. Bhati) 3011 p 122 22 22 121 01a pM M + = Using equation 202 11 11 112 2a p 2 22 22 11 112 201 1p pM Mp p + = 12 222 112 21 1pM Mp + = + 212 122 ( 1)2 ( 1)p Mp M + = + This is same as equation 22. Hence it is not necessary to consider momentum equationin 1-D steady isentropic flow once you are considering energy equation and equation of1 22 ( 1) p M + Propulsion (by K. S. Bhati) 31in 1-D steady isentropic flow, once you are considering energy equation and equation ofstate because momentum equation was already used to derive isentropic relations.Normal Shock Waves:It has been seen experimentally that in supersonic flow, there may be region of sharp change; across which velocity decreases and pressure increases drastically This change; across which velocity decreases and pressure increases drastically. This extremely thin region is called shock wave.A shock wave is, in general, curved. However in practical situation many are straight.If flow is at right angle to straight wave, it is called normal shock wave; otherwise obliqueshock wave.Stationary shock wave is one which is not moving relative to coordinate system used.How shock wave form:Consider a piston cylinder arrangement in which gas is initially at rest. Piston is giveninitial velocity dV causing pressure pulse to move with speed of sound. Increase in d t t b hi d th i d i t d b ti 33 & 34 SiPropulsion (by K. S. Bhati) 32pressure and temperature behind the wave is depicted by equation 33 & 34. Since temperature increase across wave, speed of sound behind wave will be a+da.Now if moving piston (with velocity dV) is suddenly again given velocity increase dV, itsvelocity increases to 2dV causing another pressure pulse to move with speed a+da velocity increases to 2dV causing another pressure pulse to move with speed a+da relative to gas. Its velocity w.r.t. duct will be a+da+dV. Hence second wave is moving faster than first wave. If duct is long enough second wave will overtake first wave, however, it cannot pass through first wave because gas ahead of first wave is has zerovelocity and speed of sound is only a. Therefore, two waves will merge together. Ifpiston is given series of step increase in velocity, all pressure pulse will merge togetherforming a strong wave across which large change in pressure occur. This is calledshock wave Since back of wave is always trying to move faster than front wave willPropulsion (by K. S. Bhati) 33shock wave. Sinceback of waveis always trying to move faster than front, wave willremain thin. Pressure and temperature increase in flow direction of gas.Stationary Normal Shock Waves:Consider the control volume as shown in figure. (Even in variable area flow, area across shock wave i.e. C.V. is constant as shock wave is very thin). Applying conservation eqs: shock wave i.e. C.V. is constant as shock wave is very thin). Applying conservation eqs:Continuity1 1 2 2 (38) m V A VA = =

1 22 1VV =MomentumUsing eq 381 2 2 121 2( )( ) (39)p A pA m V Vp pp p VV V VV V = = =

Using eq 38,1 2 1 1 2 1 1 2 1121 2( )(39)( ) (40)p p VV V VV Vp pp p V V V V VV= == =Adding eq 39 & 401 2 2 2 2 1 2 1 222 2( )(40)1 1( ) (41)p p V V V V VVV V = = Propulsion (by K. S. Bhati) 342 22 1 1 21 2( )(41) V V p p = + By equation 4,Flow through control volume is adiabatic (because control surfaces lie in region where2 21 21 22 2p pV Vc T q c T + + = +Flow through control volume is adiabatic (because control surfaces lie in region wherethere are NO normal temperature gradients.01 01 02 p p pc T q c T c T + = =Therefore, stagnation temperature doesnt change across shock wave. Using gas law,2 21 1 2 2(42)V p V p R R 1 1 2 21 22 2 (42)2 1 2 12p pR Rp p + = + Equating this with eq 412 21 22 11 2212 1 1p pV Vp p = Multiplying by 2/p11 21 21 2 1 22 1 1( )1p pp p = + Propulsion (by K. S. Bhati) 352 21 121pp 2 21 11 1pp = + Rearranging,2 2 2 2 22 21 11 1p p pp p p = + 1 1 1 1 12 2 21 12 21 1 1p p pp p + = + 1 1 12 11 11 11p pp p + ++ + Using eq 381 2 1 2 22 1 1 2 211 1=(43)1 1p p V pV p p p + + = = + ++ + By gas law & eq 43,1 11 1 p p + + 21 p ++ 2 21 1T pT p=21 12 11(44)1ppp + = ++ Propulsion (by K. S. Bhati) 361 1p2 121pp + Ratio p2/p1is often termed as strength of shock wave (always>1) and equation 43 & 44as Rankine-Hugoniot normal shock wave relations.By equation 9, By equation 9,2 22 11 1ln lnpT ps s c RT p = 121 2 1 111pp s s p + + Using equation 441 2 1 12 121ln(45)11p s s pR p pp = + + Variation of (s2-s1)/R with p2/p1is plotted. Asp2/p1>1, s2>s1i.e. entropy across shock wave2p increase. Hence p2/p11 T2/T1>1 (Numerator>Denominator).Analyzing equation 43 we can say that / >1Propulsion (by K. S. Bhati) 37Analyzing equation 43 we can say that 2/1>1and V2/V111p pp p + > > 1 12 21 11 11 01 1p pp pp p + + +> 1 12 21 11 111 1p pp pp p + ++> + 1 121 1 2111p ppp V ++ 1 1 22 1 211111p VV pp = = > ++ 1p We have seen that stagnation temperature across shock wave doesnt change becauseflow across it is adiabatic. To see how stagnation pressure is changing consider asituation:Propulsion (by K. S. Bhati) 38situation:Example 5:Gas flow from larger reservoir and expandedisentropically up to Mach no M1, where shock wave isentropically up to Mach no M1, where shock wave occur. After the shock, flow is isentropically decelerated until velocity is again effectively zero insecond reservoir. A l i ti 9 b t i t i fi t i Applying equation 9 between a point in first reservoirand a point in second reservoir:02 02 02ln ln lnT p ps s c R RAs flow is isentropic before and after shock wave, s01=s1and s02=s2. Also because flow is adiabatic T =T02 0101 01 01ln ln lnps s c R RT p p = = is adiabatic, T02=T01.02 2 101lnp s sRR p= 02 2 101exp 1p s sp R = < Propulsion (by K. S. Bhati) 39Because entropy increase across shock wave, stagnation pressure decrease. If no shockis present, s2=s1i.e. p02=p01.Example 6:Normal shock wave occur at a point where p & T are 30 kPa & 243 K. If pressure ratio across shock wave is 2.7, find p, V & T downstream of shock wave. Also find change in across shock wave is 2.7, find p, V & T downstream of shock wave. Also find change in stagnation pressure.Ans: p1=30 kPa, T1=243K, p2/p1=2.7 hence p2=81 kPaF 43 & 442 21977 1366V TFrom eq 43 & 44,2 21 121.977, 1.366332 V TT K= ==From eq 4,221 01 122 ( )2 ( )pV c T TV c T T= = 2 02 201 12 ( )3.91pV c T TT TT T= =But02 201 02 0291 391T TT T TT T T= ==Propulsion (by K. S. Bhati) 400 2 102.91 3.91362 T T TT K= =Flow is still isentropic on either side of shock wave. Hence isentropic relation apply on both side independently i.e.,1p T 101 011 14.035121p Tp Tp kPa = = =01102 02121 1354p kPap T = = = 2 2021.354109.6 p Tp kPa= = =Also 02 012 2 21 01 111.4 2 ( ) 2*1006*(362 243)/pp p kPaV c T T m s = = = 2 2 22 02 21 12 ( ) 2*1006*(362 332)/489.3 / , 1.566ppV c T T m sV m s M= = = =Propulsion (by K. S. Bhati) 411 12489.3/ ,1.566245.V m s MV =27/ ,0.673 m s M =Normal Shock Wave Relations in terms of Mach Number:Applying conservation laws, Applying conservation laws,Continuity1 1 2 2m VA VAV a M = =

Momentum1 1 1 22 2 2 11 2 2 1(46)V a MV aMpA pA mV mV = = = 1 2 2 12 21 2 2 2 1 12 2pA pA mV mVp p V Va a = By equation 36,2 21 1 2 21 1 2 22 2a aV V + = + Dividing by a12/,2 22 21 1 1 2 2 21 12a aM Ma a + = + Propulsion (by K. S. Bhati) 42( ) ( )22 221 1 2 211 1aM Ma + = + 222 1 121 (47)1a Ma M += + 1 2 222 2 1211(48)1a Ma M MM M + += Equating this and equation 46,21 1 22 22 21 21 2( )12 21 1a M Ma aV V + + = +Using equation 36 in 42,1 22 22 22 21 21 1( 1) 2 ( 1) 2a aM M + = + g q ,Multiplying by ( 1)/a21 21 1222 1( 1) 2 ( 1) 22 ( 1)M Ma aa M + + + (49)Multiplying by (-1)/a12,2 11( )2 ( a= + 222 22 2(49)1)1 2 ( 1)MM M M + + Using equation 48Propulsion (by K. S. Bhati) 432 1 12 21 2 21 2 ( 1)1 2 ( 1)M M MM M M + += + + Using equation 48,2 2 2 4 22 1 1 22 2 2 4 2 1+2 2 ( 1)1+2 2 ( 1)M M M MM M M M + + = + + 1 2 2 12 4 2 2 2 4 2 4 2 22 2 1 2 1 2 1 21+2 2 ( 1)2 ( 1) 4 2 ( 1) 2 (M M M MM M MM MM MM + + + + + + +4 41 22 4 2 2 4 2 2 2 4 2 4 41)2 ( 1) 4 2 ( 1) 2 ( 1)MMM M MM MM MM MM = + + + + + 1 1 1 2 1 2 1 2 1 22 4 2 4 2 4 22 2 1 2 1 2212 ( 1) 4 2 ( 1) 2 ( 1)2 ( 1) 2 ( 1) 2 2 ( 1)M M MM MM MM MMM M MM MMM M = + + + + + + + += + 4 4 2 2 2 41 1 2 1 22 ( 1) 8 MM MM + +1( ) 1 1 2 1 24 4 2 2 2 2 2 2 22 1 1 2 2 1 2 12 2 2 2 2 2( )( 1)( ) 2 ( 1) 2 ( ) 2( ) 0( ) ( 1)( ) 2 2 0M M MM M M M MM M M M MM + + = + + = 2 1 2 1 1 2( ) ( 1)( ) 2 2 0 M M M M MM + + = 2 2 2 22 1 1 2 2 12 2 2( 1)( ) 2 2 0 M M MM M M + + = 2 2 21 2 1221222 ( 1) ( 1) 22 ( 1) (50)M M MMM = + + =Propulsion (by K. S. Bhati) 44221( )2 ( 1) M Substituting equation 50 in 49,222 ( 1) a M +2 121 1212 ( 1)2 ( 1)2 ( 1)2 ( 1)a MM aM + = + + 12 21 12 22 ( 1)2 ( 1) 2 ( 1)2 2 ( 1) ( 1) 2 ( 1)MM MM M + = + + 1 12 21 12 2 22 2 ( 1) ( 1) 2 ( 1)2 ( 1) 2 ( 1)M MM M + + + =2 2 21 12212 24 2( 1) 2( 1) ( 1)2 (M MMa T + + = = 211) 2 ( 1)(51)M + 1 1a T 2 2122 2 2 2 2 (51)( 1)Mp T a + Propulsion (by K. S. Bhati) 452 2 2 2 21 1 1 1 1p T a = = Using equation 47,2 22 1 12 21 12 ( 1) 1p M MM M + += =2 21 1 2212 22 ( 1) 112 ( 1)M p MM + ++ 2 21 12 21 11 2 ( 1)2 ( 1) 2 ( 1)M MM M + = + + 2 21 12 2 2 21 1 11 2 ( 1)2 1M MM M M + =+ + + 1 1 12 21 12 21 2 ( 1)(1 ) 1M MM M + =+ + +1 122 1(1 ) 12 (M Mp M + + + =1) (52)1Propulsion (by K. S. Bhati) 461p( )1 +2 2 21 1 1p Tp T=By gas law,2 221 11 22 21 12 ( 1) 2 ( 1)2 ( 1)1 ( 1)M MMM + =+ +Using eq 51 & 52,22 121 1( 1)(53)2 ( 1)MM +=+ Applying concept from example 2, flow is isentropic on either side of shock wave.02 02 2 2/ p p p p=01 01 1 1212 2/2 ( 1)p p p pM p + Using equation 25,2 221 12 ( 1)2 ( 1)M pM p += + Propulsion (by K. S. Bhati) 47Using equation 52,2 212 1212 ( 1) 2 ( 1)2 ( 1) 1M MM + = + + Using equation 50,2112 22 ( 1)2 ( 1)2 ( 1) 2 ( 1)MM M + + 2 202 1 1201 1( )2 ( 1) 2 ( 1)2 ( 1) 1p M Mp M =+ + 12 2 2 21 1 14 2( 1) 2( 1) ( 1) 2 ( 1) M M M + + 1 1 12 21 14 2( 1) 2( 1) ( 1) 2 ( 1)12 ( 1) 2 ( 1)M M MM M + + = + + 2 212 21 1( 1)2 ( 1) 2 ( 1)MM M += + 1212 ( 1)1M + 1 1( ) ( ) 1211202 1( 1) 2 1(54)p MM + = Propulsion (by K. S. Bhati) 48Equation 50 to 54 give property ratio across shock wave in term of upstream Mach no.1201 1(54)2 ( 1) 1 1Mp M = + + + From example 5,02 2 1ln p s sR= Using equation 54,011211ln( 1) 2 1RR ps s M = + 1122 1 1121( 1) 2 1ln(55)1 1 2 ( 1)s s MR MR M + = + + + This variation is plotted in figure. Since s2-s1>=0, it follows from figure that M1>=1. From eq 50, M21, T2~T1, p2~ p1, s2~s1i.e. in weak shock case flow is isentropic across the shock. T2T1, p2 p1, s2s1i.e. in weak shock case flow is isentropic across the shock.Using eq 20 & 36 in 46,1 11 11 2 2 2 2M a T T TM a T T T = = Using eq 21 in this,2 1 1 1 1122( 1)1 12 ( 1)M a T T TM M+ + g q ,This equation give value of downstream Mach no corresponding to upstream Mach no for a very weak1 122 22 ( 1) M M = + corresponding to upstream Mach no for a very weak shock wave (M2->1).Moving shock wave:The required results can be obtained from those that were derived above for stationary shock wave by noting velocities relative to coordinate systemPropulsion (by K. S. Bhati) 50velocities relative to coordinate system fixed to shock wave as shown in figure.V is induced velocity in still air.Pitot tube in supersonic flow:Shock wave interact with expansion wave, decaying rapidly to Mach wave. Thus pressure downstream of vicinity of nose of pitot tube i.e. p2is again equal to p1. downstream of vicinity of nose of pitot tube i.e. p2is again equal to p1.02 102 02 01p p pp p p= =Using eq 25 & 54:1 01 1121112 202 1( 1) 2 1 1p p pp M + 2 202 11 121 11( 1) 2 1 112 ( 1) 1 1 2p MM Mp M + = + + + + 21 1202 111( 1) 2 12 1 1p MMp + = + + Same can be obtained directly from Normal shock table.Propulsion (by K. S. Bhati) 51Operating characteristic of nozzles:Consider the flow of gas from a large reservoir through some duct. Because reservoir is large, stagnation conditions exist in it. Flow is analyzed by assuming it as quasi one large, stagnation conditions exist in it. Flow is analyzed by assuming it as quasi one dimensional and isentropic.By equation 4,211 02p pVcT cT + =Momentum equation lead to same result in isentropic flow (as discussed earlier), hence not used.2T Using isentropic relations (eq 18),211 001 12 1pTV cTT = 1 10 1 11 00 0 02 1 2 11pp p p RV cTp R p = = 10 12 p p Propulsion (by K. S. Bhati) 520 110 021(56)1p pVp = Using this eq in continuity:1 10 1 1 121p p pm AV A = =

0 1 1 0 10 0 0 011m AV Ap p = = 2 11 11 0 021 (57)1p pm A pp p =

This give mass flow rate in terms of pressure, area of a section and stagnation condition.Since mass flow rate is constant in steady state writing eq 57 for two different sections:0 01 p p Since mass flow rate is constant in steady state, writing eq 57 for two different sections:2 1 2 11 1 2 22 21 1p p p pA A 1 1 2 21 0 0 2 0 00 0 0 01 11 1p p p pA p A pp p p p = Propulsion (by K. S. Bhati) 53This gives,1111p 02 111 22(58)1pA pA pp = This relates pressure at any two section of duct to the areas of these section.201pp p yIt is convenient to choose conditions at some specific point for reference. Convenient point for this purpose is a point where M = 1. There may NOT actually be real point in flow where M =1, but conditions at such point are convenient to use as reference. Conditions at point where M=1 are known as critical conditions Conditions at point where M=1 are known as critical conditions.Taking section 1 as critical section andUsing *12(59)p eqs 56 to 58 become:0 (59)12 2pp= + Propulsion (by K. S. Bhati) 54*002 211 1pV = + Using this in eqs 56 to 58,* *02 (60)1pV a= =021( )12 2 2 + 1*0 012 2 211 1 1m A p + = + +

12( 1)*0 02 (61)1m A p+ = +

11221 1 1 *1111AAp p ++ = Propulsion (by K. S. Bhati) 550 01p pp p 1/22 1112 1(62)A p p + = We study the effect of change in upstream and downstream pressure on nature of flow d fl t th h l F thi f ll i diti d*0 0 (62)1 1 A p p = + + and mass flow rate through nozzle. For this purpose following conditions are assumed to be prevailing i.e. upstream chamber is kept at stagnation condition and downstream chamber conditions are varied. Pressure in downstream chamber is termed as back pressure.pConvergent Nozzle:Consider the following figure. If back pressure pb is initially equal to supply pressure p0, there will be no flow through nozzle As back pressure p is decreased subsonic flow there will be no flow through nozzle. As back pressure pbis decreased, subsonic flow starts. Till Me=1. Using equation 522Propulsion (by K. S. Bhati) 63equation 52,22 112 ( 1)11p Mp = +21212 ( 1) 12 2MM +111 M i.e. upstream conditions has to be supersonic if shock wave appear. For oblique shockthi diti i11sin 11iM this condition is11190 sinM For minimum shock wave angle , shock wave is mach wave as p2=p1. Now let us see how changes between these two limits From velocity diagram;1 21 2tan &tan( )N NL L = =Now let us see how changes between these two limits. From velocity diagram;1 21tan( )( ) Xl Using this and equation 93,Propulsion (by K. S. Bhati) 6412( )( )tanXlet = =tan tantan1 t tX =+21 tan tantan tan tan tan tan X X + = +2(1 )tan (1 tan )tan(1 )tantanX XX = +=22 21tan1 tan2cot ( sin 1)t (98)XM =+For limiting case of i.e. 90 and sin-1(1/M1), there is no turning of flow.12 21( )tan(98)2 ( cos ) M =+ +Looking at plot one can say that there is maximum angle through which gas can be turned at given M1. For flow over body involving greater angle than this, a detached shock occur which is curved in general. If a body involving a given turning angle, accelerate from a low to a high Mach number shock can be detached at low Mach andPropulsion (by K. S. Bhati) 65accelerate from a low to a high Mach number, shock can be detached at low Mach and become attached at higher Mach. If < maxthere are two possible values of at given M1. Solution giving larger is termed as strong shock solution. Example 7:Air is flowing over a wedge. Mach no, pressure and temperature in air stream are 3, 50 kPa and -20 C respectively. If wedge included angle is 4 which lead to the 50 kPa and20C respectively. If wedge included angle is 4 which lead to theformation of an oblique shock wave, find the Mach number, pressure and temperature inthe flow behind the shock wave.Ans: Solving equation 98 iteratively for M1=3 and =4,W bt i k h k l ti 22 54 We obtain weak shock solution =22.54.From normal shock table 1 1sin 1.15NM M = =MN2=0.8751, p2=68.82 kPa, T2=277.4 K2 2sin( )i ( ) 2752NM MM M = Reflection of oblique shock wave:2 2sin( ) 2.752NM M = =Reflection of oblique shock from plane wall is considered. An oblique shock is assumed to be generated from body that turns the flow through angle However flow adjacent to lowerPropulsion (by K. S. Bhati) 66through angle . However flow adjacent to lower flat wall must be parallel to wall. This is only possible if reflected wave is generated.For given M1and , determine M2and p2/p1from eq 50,52:2 22 212 ( 1) sini ( )MM + 2 2122 212 22 1( )sin ( )2 sin ( 1)2 sin ( 1)MMp M = FromM2and , determine M3and p3/p2.In real fluids, boundary layer exist on wall where 2 112 sin ( 1)1p Mp =+2,3p3p2 , y yvelocity start from zero. Hence flow is subsonic adjacent to wall and cannot sustain pressure discontinuities due to shock. This cause spreading out of pressure distribution.This interaction can also cause local separation bubble in boundary layer.Another point concerning reflection of shock wave is that for a given M1there is maximum angle through which shock wave can turn flow and this angle maxdecreases with Mach no. Therefore it is possible for a situation to arise that maxfor M2is less than angle required to bring flow parallel to wall. In such case Mach reflection occur near the wallPropulsion (by K. S. Bhati) 67Mach reflection occur near the wall.In above discussion it was assumed that shock was reflected off a flat wall. If wall is changing direction sharply at the point of shock impingement, flow sharply at the point of shock impingement, flow downstream of reflected shock must be parallel to wall downstream of reflection. Turning angle produced by reflected wave is 1- w.If fl t d ill If > If 1= w, no reflected wave will occur. If 1>w,expansion wave will be generated.Interaction of Shock wave:We have seen that oblique shock wave always decreases the mach no. and shock anglefor a given turning angle increases with decreasing mach no. Hence for a flow as shown in figure, oblique shock waves generated at each step converge into a single shock wavewhich is stronger than any of initial wave which is stronger than any of initial wave. Now the pressure and flow direction must be same for all streamlines downstream oflast wave. But two or more weaker wave cannot produce same change as single strong wave hence reflected shock must be generated. These reflected wave cannot equalize velocity density and entropyPropulsion (by K. S. Bhati) 68not equalize velocity, density and entropy. Hence sliplines exits, across which there isjump in these properties.Shock Polar:For given values of V1and a1we can plot all possible solutions for V2downstream of the shock. Following figure does this in velocity-component coordinates Vxand Vy, the shock. Following figure does this in velocity component coordinates Vxand Vy, with x parallel to V1. Such a plot is called a hodograph. The heavy dark line which looks like a fat airfoil is the locus, or shock polar, of all physically possible solutions for the given Ma1. The two dashed-line fishtails are solutions which increase V2; they are h i ll i ibl b th i l t th d l physically impossible because they violate the second law.Examining the shock polar we see that a given deflection line of small angle crosses the polar at two possible solution: p pthe strong shock, which greatly deceleratethe flow, and the weak shock, which causea much milder deceleration. The flow downstream of the strong shock is always downstream of the strong shock is always subsonic, while that of the weak shock is usually supersonic but occasionally sub-sonic if the deflection is large. Both types ofshock occur in practice. The weak shock is more prevalent, but the strong shock willoccur if there is a blockage or high-pressure condition downstream. Since the shock polar is only of finite size there is a maximum deflection which just grazes thePropulsion (by K. S. Bhati) 69polar is only of finite size, there is a maximum deflection max, which just grazes the upper edge of the polar curve. Expansion Waves:Prandtl Meyer Flow: Prandtl-Meyer Flow:Previously we have considered supersonic flow over a concave corner i.e. a corner involving positive angular change in flow direction. It was indicated that the flow over such a corner was associated with a oblique shock wave. Now consider the flow around a convex corner. To determine whether an oblique shock wave also occur in this case, it is assumed that it does occur. As there is no net force in direction parallel to shock wave,L =L and since V2 must be parallel to the downstream wall L1=L2and since V2 must be parallel to the downstream wall, geometrical considerations show that N2>N1(Draw lines along V1, V2& N1, and fit L1, L2to it). But N1& N2must be related by the normal shock wave relations, and N2cannot be greater than N1as expansive shock violate second law of thermodynamic. Therefore the flow over convex corner cannot take place through oblique shock.Considering a flow turned through differentially small angle d producing small change Considering a flow turned through differentially small angle d, producing small change in flow variables, it can be proved that221 ( )1 (99)2dM dMM = + Propulsion (by K. S. Bhati) 70221,,,,0MMdM d dV d dp d d d ds =Thus if flow around a corner consist of an infinite number differentially small angular changes it follows that for negative angular change waves diverge thus generating a region consisting of Mach waves and flow remain isentropic throughout. Such flows are region consisting of Mach waves and flow remain isentropic throughout. Such flows are called Prandtl-Meyer flows. As equation 99 apply locally at all points within the expansionfan, this is integrated to give relation between flow properties d.For initial condition =0 when M=1 give (dropping negative sign before -d since is ti f P dtl M fl t i t) negative for Prandtl-Meyer flow to exist):1 2 1 21 1tan ( 1) tan 1 M M + = The value for are listed in isentropic tables. To calculate the flow change produced by Prandtl Meyer expansion adopt the following procedure:tan ( 1) tan 11 1 M M +Prandtl-Meyer expansion adopt the following procedure:1. Find corresponding to M1. This is equivalent to assuming that the initial flow was generated by an expansion around hypothetical corner from a Mach no. of 1 (referenceMach no) to M1.2.2= 1+3. Find downstream Mach no. M2corresponding to this value of 2.4. Any other downstream property is obtained by noting that expansion is isentropic. 5 Boundaries of expansion wave is calculated by noting that they are Mach lines i ePropulsion (by K. S. Bhati) 715. Boundaries of expansion wave is calculated by noting that they are Mach lines i.e.1 111 1sin , sin = = R fl ti f i11 2,M MReflection of expansion wave:If 1is Prandtl-Meyer angle corresponding to initial flow condition, then 2= 1+ . Since flow in region 3 must again be parallel to wall, the reflected wave must also turn the flow through angangle of so that 3= 2+ = 1+ 2. Once this angle is determined, Mach no. in region 3 can be found. Region of interaction of incident and reflected waves is known as non simple region waves is known as non-simple region. Flow is isentropic throughout.Propulsion (by K. S. Bhati) 72Adiabatic flow in duct with friction:Effect of viscosity has been neglected in preceding discussion. This is often adequate assumption when dealing with flow through nozzle or short duct However for long duct assumption when dealing with flow through nozzle or short duct. However for long ducteffect of fluid friction at wall can be dominant. Friction cause pressure drop along flow direction. This cause density change in compressible flow and hence velocity change.Flow in constant area duct:Compressible adiabatic flow in constant area duct with frictional effect is known as Fanno flow. Applying momentum balance for small portion of duct: Applying momentum balance for small portion of duct:Net Force = Mass flow rate*Change in velocity( ) ( )wpA p dp A Pdx AVV dV V + = + Dividing by AV2:2 2wdp P dVdxV V A V =Using a2=p/2 21V V A VdV dp P Propulsion (by K. S. Bhati) 732 210(64)wdV dp PdxV M p V A + + =From continuity equation:T ki l d diff ti ti thi ( ) V const A const = =Taking log and differentiating this:0(65)d dVV+ =From energy equation:2VVcT const+Differentiating this:20 (66)pcT constdT VdV+ =From equation of state:0 (66)pc dT VdVRT+ =Taking log and differentiating this:p RTdp d dT=Propulsion (by K. S. Bhati) 74 (67)dp d dTp T= +We know that,VM =Taking log and differentiating this: RTdM dV dTNow our objective is to express dM in term of dx and dp,dT in term of dM (or dx). (68)2dM dV dTM V T= Multiplying equation 68 with V2:22V dTVdV RTM dMT = +Using this in equation 66:20V dTc dT RTM dM + + =Dividing by cpT:202pc dT RTM dMT + + =Propulsion (by K. S. Bhati) 75202p pdT R V dTM dMT c c T T+ + =Using cp=R/(-1):21( 1) 0dT dTMdM M+ +22( 1) 02( 1)(69)MdM MT TdT M dM+ + == Combining equation 65 and 67:2 (69)1 ( 1) / 2 T M Md dV dT + Using dV/V from equation 68:dp dV dTp V T= +Using dV/V from equation 68:2dp dM dTp M T= +Using dT/T from equation 69:21 ( 1)(70)pdp M dM + Propulsion (by K. S. Bhati) 762( )(70)1 ( 1) / 2pp M M= + Using dV/V from equation 68 in equation 64:1 dM dT dp P Using dT/T from equation 69:2 2102wdM dT dp PdxM T M p V A + + + =22 2 2( 1) / 2 101 ( 1) / 2wdM M dM dp PdxM M M M p V A + + =+ 2 2 21 101 ( 1) / 2wdM dp PdxM M M p V A + + =+ Using dp/p from equation 70:22 2 2 21 1 1 ( 1)0wdM M dM Pdx + + = 2 2 2 222 21 ( 1) / 2 1 ( 1) / 21 1M M M M M V AM dM + + + 20wPdx=Propulsion (by K. S. Bhati) 772 21 ( 1) / 2 M M M + 2V A Rearranging, 2 21 ( 1) / 2 M MdM P + Substituting this in equation 69 and 70:2 2( ) (71)1wdM PdxM M V A = 42 2( 1) (72)1wdT M PdxT M V A = 2 22 21 ( 1) (73)1wM Mdp Pdxp M V A + = Since wall shear stress, density and Mach no. are positive, it follow that M increases when M1. Thus friction causes Mach no. to tend toward 1.Once Mach no. of 1 is attained, change in downstream condition cannot effect upstreamp condition (information cannot flow more than speed of sound). Hence it follow thatfriction result in chocking of duct. We know that:1 ds dT dp = Propulsion (by K. S. Bhati) 78pc T p Using equation 72 and 73 in this:2 241 ( 1)( 1) 1M Mds M P + 2 2 21 ( 1)( 1) 11 1wpM Mds M Pdxc M M V Ad P + = + This show that entropy always increases irrespective of Mach no.22( 1)(74)wpds PM dxc V A = py y pDimensionless wall shear stress, also known as Fanning friction factor, is defined as:f 22(75)122w wffVV = =where fDis known as Darcy friction factor. In general,4Df f =Propulsion (by K. S. Bhati) 79(Re, / , )Hf Fn D M =where DHis hydraulic diameter based on circular cross section is defined as:4(76)HAD =and is measure of mean height of wall roughness.U i ti 75 d 76 i ti 71 (76)HDPUsing equation 75 and 76 in equation 71:2 221 ( 1) / 241 2M MdM fdxM M D + = 223 21 24 2(1 )1 ( 1) / 2HM M Df Mdx dMDM M = Integrating this:3 221 ( 1) / 2HMlDM M + 2123 204 2(1 )1 ( 1) / 2MlH Mf Mdx dMDM M = + 1lPropulsion (by K. S. Bhati) 80If is mean friction factor over the length l of duct,01f fdxl= Variation in friction factor along the duct is usually small and f can be assumed constant.2 24 2 1 1Mf M I t ti b ti l f ti( )13 24 2 1 1 21H Mf Ml dM pDM pM = = + Integrating by partial fraction,( )22 3 23 2111 M a b c Md eM M M pMM pM += + + +++Solving for constants a,b,c,d and e:( )11( 1) 0 1 ( 1)M M M pMM pMa p b e c d p p++= + = = = = +23 2( 1),0,1,( 1)4 2 1 1 ( 1)1MHa p b e c d p pf p p pMl dMD M M pM = + = = = = ++ += + ++122 212 1 1 1 1 1 2ln2 2 4 1H MD M M pMM pMM M M + + += + 22MdMpM + Propulsion (by K. S. Bhati) 811 2 12 2 4 1 M M M 1MpM + 2212 24 1 1 1 1 1 1ln ln 12 2MM fl MD M M M + + = + + + 11 2 22 21 22 21 ( 1) / 24 1 1 1 1lnHMD M M MM Mfl + + = + 2 22 21 22 1ln21 ( 1) / 2HlD M MM M + + In order to describe the result of this equation, it is convenient to select a reference valuefor M2. Since Mach no. always tend to 1 when friction is present, M2is chosen as 1.( )221 *2 2114 1 1ln (77)2 2 ( 1)HMf MlD M M + += ++ 2y p ,21 Hd*1( 1)TT M Likewise Integrating equation 69:d{ }212( 1)1 ( 1) / 2l l ( ) /T MT MdMT MT= + Propulsion (by K. S. Bhati) 82{ }2*ln ln 1 ( 1) / 2MTMT = + * 2( 1) / 2(78)1 ( 1) / 2TT M +=+*121 ( 1) / 21 ( 1)pT Md M dM + Integrating equation 70:1221 ( 1)1 ( 1) / 2pp Mdp M dMp M M+ = + 1 1* 21 ( 1)ln2 1 ( 1) / 2M Mp dM MdMp M M = ++ 21 ( 1) / 2 ln ln2 1 ( 1) / 2MM += ++ * 21 ( 1) / 2 (79)1 ( 1) / 2pp M M +=+ Propulsion (by K. S. Bhati) 83From definition of stagnation pressure (equation 24):1211 p p M + From this, for M=1:012p p M = + 1* *012p p+ = Dividing both the above equation:211 ( 1) / 2 p p M +10* *01 ( 1) / 2( 1) / 2p p Mp p + = + Using equation 79 in this:122( 1)0*1 1 ( 1) / 2(80)p M+ + = Propulsion (by K. S. Bhati) 84*0( )( 1) / 2 p M + Equation 77 to 80 are plotted for air:The Fanno line: The Fanno line:Fanno line shows flow process on T-s diagram.We know that,1(81)ds dT dp From equation 65 and 67: (81)ppc T p= qdp dV dTp V T= +and from equation 66:pc dTdV = Hence,2pdVVc dTdp dTV T== +Propulsion (by K. S. Bhati) 852p V TUsing this in equation 81:11pc Tds dT dT By SFEE for adiabatic flow,21ppc T T V = + 202p pVc T c T = +Hence,202 ( )pV c T T = 0112( )pds dT dT Tc T T T T = + 01 12pds dT dTc T T T = Propulsion (by K. S. Bhati) 86Integrating,1 12s T Tds dT dTc T T T = 1 1 101 120 12ln (82)p s T Tc T T TT T s s T 0 11 0 1ln(82)pc T T T = This give variation of entropy with temperature for given stagnation temperature andg py p g g parbitrary temperature T1. As entropy always increases, this show why Mach no. movetoward unity.Propulsion (by K. S. Bhati) 87Frictionless flow in duct with heat transfer:Effect of heat transfer is discussed here when viscosity effect is negligible. This is adequate assumption when large amount of heat is transferred such as in combustion adequate assumption when large amount of heat is transferred such as in combustion. It is also assumed that gas composition and hence specific heat ratio does not change.This may not be true in some combustion systems.Flow in constant area duct:Compressible frictionless flow in constant area duct with heat transfer, assuming constant momentum and mass flow but variable stagnation enthalpy is known as Rayleigh flow Rayleigh flow . Applying momentum balance for control volume shown in figure:Net Force = Mass flow rate*Change in velocity( ) ( ) pA p dp A AVV dV V + = + ( ) ( )(83)pA p dp A AVV dV Vdp VdV+ +From continuity equation:(83)pp p =Propulsion (by K. S. Bhati) 88 ( ) V const A const = =Taking log and differentiating this:0 (84)d dV + =From energy equation 5:20(84)V +Differentiating this:202p pVq c T c T = + =gTaking log of equation of state and differentiating it:0 (85)p pdq c dT VdV c dT = + =Taking log ofequation of state and differentiating it:dp d dTp T= +Using equation 83 and 84 in this:(86)VdV dV dTV T = +Propulsion (by K. S. Bhati) 89Since, p/=RT,( )p V TdT dV VdVT V RT= dV VdVdq c T VdV = + Using this in equation 85: 1pppdq c T VdVV RTcTc dV V dVV R+ = + 1ppV Rdq T VcdV V = 211( 1)pdV VV dVdq c Tc T V = ( )20( 1)1 = (87)pp pc T VdVdq c T M c dTV = Propulsion (by K. S. Bhati) 90VIf heat is added at M1 heat addition cause decrease in velocity. As dT0has same sign as dq, T0is maximum at M=1 as shown in below figure.Applying momentum equation between two sections of duct: pp y g qNet Force = rate of change of momentum:Using continuity equation AV = AV :1 2 1 1 2 1( ) ( ) p p A AV V V = Using continuity equation 1AV1= 2AV2:Since p/ =a2,2 21 1 1 2 2 2p V p V + = +2 21 1 2 222 1(1 ) (1 )1(88)p M p Mp M + = ++Propulsion (by K. S. Bhati) 912 121 2 (88)1pp M=+Using equation 24,2102 2 21 ( 1) / 2(89)p p M + By continuity and perfect gas law,02 2 2201 1 1( )(89)1 ( 1) / 2p pp p M= + 2 2 21 1 1T p VT p V=But,2 2 2 2 2V M a M TV M M T= =1 1 1 1 1222 2 22 (90)V M a M TT p M = 21 1 122 1 2 22( ) (91)T p MV M pV M = =Propulsion (by K. S. Bhati) 9221 2 1 1V M p Using equation 89 in 23,22 202 2 2 21 ( 1) / 2(92)T p M M + = As discussed earlier, T0is maximum at M2=1.2 201 1 1 1(92)1 ( 1) / 2 T p M M = + ( )2210max2201 11112(1 )1 ( 1) / 2MTT MM+=+ + Equation88 to 92 are plotted in following figure. Propulsion (by K. S. Bhati) 93Propulsion (by K. S. Bhati) 94