We need to do review for some thermodynamic relations: Equation of state P=ρ R T, h= u + pv = u + RT dh= du +R dt Cp dt=cv dt + R dt.

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Cmpressible Flw Cmpressible flw is the study f fluids flwing at speeds cmparable t the lcal speed f sund. his ccurs when fluid speeds are abut 30% r mre f the lcal acustic velcity. hen, the fluid density n lnger remains cnstant thrughut the flw field. his typically des nt ccur with fluids but can easily ccur in flwing gases. w imprtant and distinctive effects that ccur in cmpressible flws are () ching where the flw is limited by the snic cnditin that ccurs when the flw velcity becmes equal t the lcal acustic velcity and () shc waves that intrduce discntinuities in the fluid prperties and are highly irreversible. Since the density f the fluid is n lnger cnstant in cmpressible flws, there are nw fur dependent variables t be determined thrughut the flw field. hese are pressure, temperature, density, and flw velcity. w new variables, temperature and density, have been intrduced and tw additinal equatins are required fr a cmplete slutin. hese are the energy equatin and the fluid equatin f state. hese must be slved simultaneusly with the cntinuity and mmentum equatins t determine all the flw field variables. We need t d review fr sme thermdynamic relatins: Equatin f state P=ρ R, v u Cv respectively. v Cp h Specific heats in cnstant vlume and pressure p h= u + pv = u + R dh= du +R d Cp d=cv d + R d Cp = Cv + R

v p C C R,C R C v p he first law f thermdynamic: ds = du + p dv dv p du ds dv v R d C ds v v ln R ln C S v v ln ) C ( ln C S v ln C S Fr reversible, adiabatic prcess (Isentrpic), ΔS=0.0 and p p Stagnatin Cnditin Recall definitin f enthalpy which is the sum f internal energy u and flw energy P/ Fr high-speed flws, enthalpy and inetic energy are cmbined int stagnatin enthalpy h

Steady adiabatic flw thrugh duct with n shaft/electrical wr and n change in elevatin and ptential energy herefre, stagnatin enthalpy remains cnstant during steady-flw prcess If a fluid were brught t a cmplete stp ( = 0) herefre, h 0 represents the enthalpy f a fluid when it is brught t rest adiabatically. During a stagnatin prcess, inetic energy is cnverted t enthalpy. Prperties at this pint are called stagnatin prperties (which are identified by subscript ) - Stagnatin enthalpy is the same fr isentrpic and actual stagnatin states - ctual stagnatin pressure P,act is lwer than P due t increase in entrpy s as a result f fluid frictin. -Nnetheless, stagnatin prcesses are ften apprximated t be isentrpic, and isentrpic prperties are referred t as stagnatin prperties Fr an ideal gas, h = C p, which allws the h0 t be rewritten is the stagnatin temperature. It represents the temperature an ideal gas attains when it is brught t rest adiabatically. 3

/C p crrespnds t the temperature rise, and is called the dynamic temperature. Speed f sund Cnsider a duct with a mving pistn Creates a snic wave mving t the right Fluid t left f wave frnt experiences incremental change in prperties Fluid t right f wave frnt maintains riginal prperties Cnstruct C that enclses wave frnt and mves with it Cntinuity equatin (ass Balance) Energy Equatin (Energy balance) E in = E ut 4

Frm thermdynamic relatins: ds=dh-vdp ds=0 dh=dp/ρ Cmbining this with mass and energy equatins gives: Speed f sund fr Gases Frm ideal gas relatin (equatin f state p=ρr) Speed f sund fr Liquids dp E Bul mdulus f cmpressin d c E ach waves ach Number =/C Small disturbances created by a slender bdy in a supersnic flw will prpagate diagnally away as ach waves. hese cnsist f small isentrpic variatins in ρ,, 5

p, and h, and are lsely analgus t the water waves sent ut by a speedbat. ach waves appear statinary with respect t the bject generating them, but when viewed relative t the still air, they are in fact indistinguishable frm sund waves, and their nrmal-directin speed f prpagatin is equal t a, the speed f sund. he angle μ f a ach wave relative t the flw directin is called the ach angle. It can be determined by cnsidering the wave t be the superpsitin f many pulses emitted by the bdy, each ne prducing a disturbance circle (in -D) r sphere (in 3- D) which expands at the speed f sund a. t sme time interval t after the pulse is emitted, the radius f the circle will be at, while the bdy will travel a distance t. he ach angle is then seen t be tan ct t tan which can be defined at any pint in the flw. In the subsnic flw case where = /a < the expanding circles d nt calesce int a wave frnt, and the ach angle is nt defined. 6

ISENROPIC FLOW When the flw f an ideal gas is such that there is n heat transfer (i.e., adiabatic) r irreversible effects (e.g., frictin, etc.), the flw is isentrpic. he steady-flw energy equatin applied between tw pints in the flw field becmes Frm thermdynamic relatin and h=c p, h=c p 7

8 c c P P P =500 Pa, =400 K (Stagnatin cnditins) =0.469 =383K 3 / 3.9 (87)(383) 430,000 m g R P s m R / 84.487383.469 0 Example: ir is flwing isentrpically thrugh a duct is supplied frm a large supply tan in which the pressure is 500 Pa and temperature 400 K. What are the ach number, the temperature, density and fluid velcity v at a lcatin in this duct where the pressure is 430 Pa.4.4.4 430 500 0.469.4 400 P P

9 Critical (Snic ) Cnditin he values f the ideal gas prperties when the ach number is (i.e., snic flw) are nwn as the critical r snic prperties and are given by: c c P P Bth the critical (snic, a = ) and stagnatin values f prperties are useful in cmpressible flw analyses. Flw hrugh arying rea Duct Such flw ccurs thrugh nzzles, diffusers, and turbine blade passages, where flw quantities vary primarily in the flw directin. his flw can be apprximated as D isentrpic flw. Cntinuity Derived relatin (n image at right) is the differential frm f Bernulli s equatin (Euler equatin) and cmbining this with speed f sund gives: d C d d d C Substitute the result in cntinuity equatin: 0 d d d C

d d C d d Fr subsnic flw ( < ) d/d < 0 Fr supersnic flw ( > ) d/d > 0 Fr snic flw ( = ) d/d = 0 pplicatin: Cnverging r cnvergingdiverging nzzles are fund in many engineering applicatins Steam and gas turbines, aircraft and spacecraft prpulsin, industrial blast nzzles, trch nzzles Flw Cases in Cnverging Diverging Nzzle -: P > Pe > Pc Flw remains subsnic, and mass flw is less than fr ched flw. Diverging sectin acts as diffuser -: P e = Pc Snic flw achieved at thrat. Diverging sectin acts as diffuser. Subsnic flw at exit. Further decrease in Pb has n effect n flw in cnverging prtin f nzzle 3-:Pc > Pe > PE Fluid is accelerated t supersnic velcities in the diverging sectin as the pressure decreases. Hwever, acceleratin stps at lcatin f nrmal shc. Fluid decelerates and is subsnic at utlet. s Pe is decreased, shc appraches nzzle exit.

PE > Pe > 0 Flw in diverging sectin is supersnic with n shc frming in the nzzle. Withut shc, flw in nzzle can be treated as isentrpic. Flw Cases in Cnverging (runcated) Nzzle he highest velcity in a cnverging nzzle is limited t the snic velcity ( = ), which ccurs at the exit plane (thrat) f the nzzle ccelerating a fluid t supersnic velcities ( > ) requires a diverging flw sectin Frcing fluid thrugh a C-D nzzle des nt guarantee supersnic velcity, It requires prper bac (exit) pressure P e State : P b = P, there is n flw, and pressure is cnstant. State : P b < P0, pressure alng nzzle decreases. State 3: P b =P, flw at exit is snic, creating maximum flw rate called ched flw. State 4: P b < P b, there is n change in flw r pressure distributin in cmparisn t state 3 State 5: P b =0, same as state 4.

Find the Critical rea: m=ρ= ρ ) ( ) ( ) ( ) ( ) ( K he aximum ass Flw Rate If the snic cnditin des ccur in the duct, it will ccur at the duct minimum area. If the snic cnditin ccurs, the flw is said t be ched since the mass flw rate is maximum (m max ) Which is defined as the maximum mass flw rate the duct can accmmdate withut a mdificatin f the duct gemetry. max max R 0.686 P m.4 : having gases fr and R m

Ex: he reservir cnditins f air entering a cnverging-diverging nzzle are 00 Pa and 300 K. ach number at exit equals t 3.0 and the mass flw rate is.0 g/s. Determine: (a) thrat area (b) the exit area and (c) the air cnditins at exit sectin. m m max max 0.686 P R 0.686 (00,000 ) 87( 300 ) P t e=3.0 = 0.0048m e =4.34 e=0.08m ( K ) e ) ( e Frm tables at =3.0, (Pe/P)=0.07, (e/)=0.357 Pe=.7 Pa, e=07.k Ρ e =P e /(R e )=0.088 g/m 3 Shc waves Under the apprpriate cnditins, very thin, highly irreversible discntinuities can ccur in therwise isentrpic cmpressible flws. hese discntinuities are nwn as shc waves. Flw prcess thrugh the shc wave is highly irreversible and cannt be apprximated as being isentrpic. Shcs which ccur in a plane nrmal t the directin f flw are called nrmal shc waves, sme are inclined t the flw directin, and are called blique shcs. Nrmal Shc Wave Develping relatinships fr flw prperties befre and after the shc using cnservatin f mass, mmentum, and energy: 3

Cnservatin f mass Cnservatin f mmentum -------------() ------------() Cnservatin f energy C p C p -----------(3) Nw, frm equ(), P =P +ρ ρ -----(4) Equ() int (4) : P =P +ρ ρ -----(5) Substitute n ρ frm equ() int equ() P P -------(6) By substituting frm equ (5) int equ(6), we btain: 4

When using these equatins t relate cnditins upstream and dwnstream f a nrmal shc wave, eep the fllwing pints in mind:. Upstream ach numbers are always supersnic while dwnstream ach numbers are subsnic.. Stagnatin pressures and densities decrease as ne mves dwnstream acrss a nrmal shc wave while the stagnatin temperature remains cnstant (a cnsequence f the adiabatic flw cnditin). 3. Pressures increase greatly while temperature and density increase mderately acrss a shc wave in the dwnstream directin. 4. he critical/snic thrat area changes acrss a nrmal shc wave in the dwnstream directin and. 5. Shc waves are very irreversible causing the specific entrpy dwnstream f the shc wave t be greater than the specific entrpy upstream f the shc wave. EX: nrmal shc wave exists in a air flw with upatream =.0 and a pressure f 0 Pa and temperature f 5 C. Find the ach number, preeure, stagnatin pressure, temperature, stagnatin temperature and air velcity dwnstream f the shc wave Frm Shc wave table: =0.577, (P /P )=4.5, ( / )=.688 (P /P )=0.7 P=4.50=90 Pa =.688(73+5)=486 K = /C =0.577[.487486] / =55 m/s find P 0 and Frm Isentrpic table t =.0, (P /P )=0.8, ( / )=0.556 P =0/0.8 =56.5 Pa P =0.756.5=.6 Pa 5

=88/0.556=58 K = =58 K H.W. ir is supplied t the cnverging-diverging nzzle shwn here frm a large tan where P = Pa and = 400 K. nrmal shc wave in the diverging sectin f this nzzle frms at a pint where the upstream ach number is.4. he rati f the nzzle exit area t the thrat area is.6. Determine (a) the ach number dwnstream f the shc wave, (b) the ach number at the nzzle exit, and (c) the pressure and temperature at the nzzle exit. Fann and Rayleigh Lines Fann Line Cmbine cnservatin f mass and energy int a single equatin and plt n h-s diagram Fann Line : lcus f states that have the same value f h0 and mass flux Energy equatin: h =h+v / () cntinuity: d(ρ=cnst.) ρd+dρ=0..() Frm thermdynamic relatins: ds=dh-dp/ρ t the pint f maximum entrpy, ds=0: dh=dp/ρ d() dh =dh+vdv=0 -vdv=dp/ρ ρ frm equatin () ρ=-dρ/d -d=dpd/(-dρ) =dp/dρ=c Hence, the maximum entrpy ccurs at = Rayleigh Line Cmbine cnservatin f mass and mmentum int a single equatin and plt n h-s diagram :Rayleigh line 6

Pints f maximum entrpy crrespnd t a =. bve / belw this pint is subsnic / supersnic P+ρ =cnst (mmentum equatin) P+G=cnst dp+gd=0 dp+ρd=0 frm equ() d=-(/ρ) dρ dp=ρ (/ρ)dp =dp/dρ=c ls at =0, But nt necessarily the same value f stagnatin enthalpy here are pints where the Fann and Rayleigh lines intersect: pints where all 3 cnservatin equatins are satisfied Pint : befre the shc (supersnic) Pint : after the shc (subsnic) he larger a is befre the shc, the strnger the shc will be. Entrpy increases frm pint t pint : expected since flw thrugh the shc is adiabatic but irreversible Oblique shc and expansin waves ach waves can be either cmpressin waves (p >p) r expansin waves (p <p), but in either case their strength is by definitin very small ( p p <<p). bdy f finite thicness, hwever, will generate blique waves f finite strength, and nw we must distinguish between cmpressin and expansin types. he simplest bdy shape fr generating such waves is a cncave crner, which generates an blique shc (cmpressin), r a cnvex crner, which generates an expansin fan. he flw quantity changes acrss an blique shc are in the same directin as acrss a nrmal shc, and acrss an expansin fan they are in the ppsite directin. One imprtant difference is that p decreases acrss the shc, while the fan is isentrpic, s that it has n lss f ttal pressure, and hence p = p 7

t leading edge, flw is deflected thrugh an angle called the turning angle Result is a straight blique shc wave aligned at shc angle relative t the flw directin Due t the displacement thicness, is slightly greater than the wedge half-angle. Lie nrmal shcs, a decreases acrss the blique shc, and are nly pssible if upstream flw is supersnic Hwever, unlie nrmal shcs in which the dwnstream a is always subsnic, a f an blique shc can be subsnic, snic, r supersnic depending upn a and. ll equatins and shc tables fr nrmal shcs apply t blique shcs as well, prvided that we use nly the nrmal cmpnents f the ach number,n =,n/c=sinβ/c=sinβ,n=,n/c=sin(β-θ)/c=sin (β-θ) tanβ=,n/,t tan(β-θ)=,n/,t β But,t=,t (there is n pressure change in the tangential directin) Hence: tan,n tan, n,n Sin,n Sin Hence: tan Sin tan Sin β-θ Slving the abve relatin fr θ: 8

he blique shc chart abve reveals a number f imprtant features: -here is a maximum turning angle θmax fr any given upstream ach number. If the wall angle exceeds this, r θ>θmax, n blique shc is pssible. Instead, a detached shc frms ahead f the cncave crner. Such a detached shc is in fact the same as a bw shc discussed earlier. -If θ<θmax, tw distinct blique shcs with tw different β angles are physically pssible. he smaller β case is called a wea shc, and is the ne mst liely t ccur in a typical supersnic flw. he larger β case is called a strng shc, and is unliely t frm ver a straight-wall wedge. he strng shc has a subsnic flw behind it. 3-he strng-shc case in the limit θ 0 and θ 90, in the upper-left crner f the blique shc chart, crrespnds t the nrmal-shc case. 9

Prandtle-eyer Waves n expansin fan, smetimes als called a Prandtl- eyer expansin wave, can be cnsidered as a cntinuus sequence f infinitesimal ach expansin waves. understand the analysis clearly, we shall bac t explain ach cne r ach wave. Ct sinµ=ct/t=/ r µ=sin - (/) analyze this cntinuus change, we will nw cnsider the flw angle θ t be a flw field variable, lie r. crss each ach wave f the fan, the flw directin changes by d_, while the speed changes by d. Oblique-shc analysis dictates that nly the nrmal velcity cmpnent u can change acrss any wave, s that d must be entirely due t the nrmal-velcity change du.

Prandtle-eyer Functin he differential equatin () can be integrated if we first express in term f. C C ln C ln ln - Differentiatin the abve relatin: d d d d d Equatin () then becmes: d d () which can nw be integrated frm pint t any pint in the Prandtle-eyer wave d d ) ( ) ( (3) Where tan tan ) ( (4) θ: is the ttal turning f the crner Here ν() is called the Prandtle-eyer functin, and is shwn pltted fr =.4 Equatin (3) can be applied t any tw pints within an expansin fan, but the mst cmmn use is t relate the tw flw cnditins befre and after the fan. Reverting bac t ur previus ntatin where θ is the ttal turning f the crner, the relatin between θ and the upstream and dwnstream ach number is

Fan angle=μ -[μ -θ]

diabatic Flw ll the isentrpic flw relatins can be used thrugh the adiabatic flw, in any sectin, prvided that all variables f the relatin refer t the same sectin. P Ched hrat, P, P Isentrpic flw diabatic flw Find, and : 0.686 P m max R Hence m max R 0.686 P () m R max () 0.686 P divide () by () We gt: P = P Generally hrugh adiabatic flw: P =cnstant Ex: cnstant area adiabatic duct has the fllwing cnditins f air flw: t sectin, the pressure P=0.8 bar, =350K, air velcity=60 m/s. t sectin, ach number=0.5, Find P,, at sectin. 3

Sl. P = P R 60.4 87 350 0.46 P=80Pa =350K =60m/s =0.5 P P P =80(+0.(0.46) ) 3.5 =90.63 ( K ) ( ) / =.34.5 Hence: (/.5)90.63=(/.34)P P=80.3 Pa P P P=80.3/.863=67.7 = = (+0.(0.46) )=36.7K =345.4K =86.3m/s 4