Chapter 11 Compressible Flow Introduction z Compressible flow –variable density, and equation of state is important z Ideal gas equation of state—simple yet representative of actual gases at pressures and temperatures of interest z Energy equation is important, due to the significant variation of temperature. 11.1 Ideal gas relationship = ρRTP For ideal gas, internal energy u=u(T) ∂udu constant pressure specific heat: c ==() υυ∂TdT T2 du=→−= cυυ dT u21 u c dT ∫T1 -For moderate changes in temperature: uucTT21− =−υ () 21 Enthalpy h=h(T) p hu=+ = uTRThT() + = () ρ ∂hdh constant pressure specific heat: cpp==() ∂TdT T2 dh=→−= cpp dT h21 h c dT ∫T1 -For moderate changes in temperature hhcTT21−=p () 21 − Since h=u+RT, dh=du+RdT or dh du =+→−=R ccR dT dT p υ cp Rk R kcc=∴ ( 1.4 for air) p == and υ ckkυ −11− Entropy 1st Tds equation Tds= du+ pd(1 /ρ ) p 1 Qh=+ u → dh = du + pd(1/ρ ) + dp ρρ 1 ∴Tds=− dh dp ---2nd Tds equation ρ du p dT R ds=+ d(1 /ρ ) = c + d(1 /ρ ) TTυ T(1 /ρ ) dh (1 /ρ ) cdT R =−dp =p −dp TT Tp For constant ccp , v : T21ρ ssc21−=υ ln + R ln( ) T12ρ Tp22 =−cRp ln ln( ) Tp11 For adiabatic and frictionless flow of any fluid ds= 0 or s21−= s 0 ← isentropic flow Comparison of isentropic and isothermal compression TTp21ρ 22 or cRcRυ ln( )+=−= ln( )p ln( ) ln( ) 0 TTp12ρ 11 Isentropic k process path, R TT2222ρ k−1 ρ k ln( )=→=R ln( ) ( ) ( ) Pvk = const kT−1 1111ρ T ρ kR TpTpk Pv = const ln(2222 )=→=R ln( ) ( )k−1 ( )k kT−1 1111 p T p Tpk ρ p ∴(222 )k−1 ==⇒= ( )k ( ) const, for isentropic flow k (11.25) Tp111ρ ρ 11.2 Mach number and speed of sound V Ma= , Vc--local flow velocity, --speed of sound c Sound generally consists of weak pressure pulses that move through air. Consider 1-D of infinitesimally thin weak pressure pulse moving at the speed of sound through a fluid at rest. fluid at rest Observer moving with control volume -conservation of mass ρAcAcV=()()ρδρ+− δ ρcc= ρρδδρδρδ−+− Vc V ρδ = cV δρ -linear momentum conservation −ccAcρ +−()()()(δρδρδ V + c − VA = pAp − +δ pA ) Q()()ρ +−=δρcVAcA δ ρ (continuity) ∴ −+−ccAcρ ()δρ V cA =− δ pA Isentropic process path −δVAcρδ=− pA → ρδδ V = pc/ Q ρδδρVc= (continuity) δ ppδδ p ∴cccδρ =→=⇒= 2 c δρδρ -Alternative derivation using conservation of energy instead of momentum equation δ pV⎛⎞2 +−=δδδ⎜⎟gz (loss) --from (5.103) ρ ⎝⎠2 δ (loss) 0 for frictionless flow; and δ (z) 0 δ pcV()− δδ22 c p ∴ +−=→0 ρδV = ρ 22 c δ p or = cVδ ρ δ p →=ρδδρVc = (from continuity: ρδδρVc=) c δ P δ p ∴ cc2 =⇒= δρδρ Assume the frictionless flow through the control volume is adiabatic, then the flow is isentropic. In the limit δ pp→∂ →0 ⎛⎞∂ p c = ⎜⎟ (11.34) ⎝⎠∂ ρ s - For isentropic flow of idea gas p = cρ k ⎛⎞∂pppkk−−11 ⎜⎟==ckρρk k ==⇒= k RTk c RTk (11.36) ⎝⎠∂ρρs ρ - More generally, use bulk modulus of elasticity dp⎛⎞∂ p Ev ==ρ ⎜⎟ d ρ / ρρ⎝⎠∂ s ∴ cE= v / ρ - For incompressible flow Ecv →∞ ∴ →∞ 11.3 Category of compressible flow - effect of compressibility on C of a sphere D Why CD increases with Ma? Can you explain physically? (from S.R. Turns, Thermal-Fluid Sciences) - Imagine the emission of weak pressure pulses from a point source r=() t− twave c where t → present time, twave → time wave emitted - stationary point source - moving point source V< C - source moving at V=c - source moving at V>c - Category of fluid flow 1. Incompressible flow Ma≤ 0.3 unrestricted, linear symmetrical and instantaneous pressure communication. 2. Compressible subsonic flow 0.3< Ma< 1.0 unrestricted, but noticeably asymmetrical pressure communication 3. Compressible supersonic flow Ma≥ 1 formation of Mach wave, pressure communication restricted to zone of action 4. transonic flow 0.9≤ Ma≤ 1.2 (modern aircraft) 5. hypersonic flow Ma≥ 5 (space shuttle) Example 11.4 Mach cone Mach cone from a rifle bullet from Gas Dynamics Lab, The Penn. State University, 2004 from M. Van Dyke, An Album of Fluid Motion Video: Mach cone of an airplane (Note the condensed cloud across the shock wave.) Ma=0.978 (Can you estimate the airplane speed?) 11.4 Isentropic flow of an idea gas - no heat transfer and frictionless 11.4.1 Effect of variation in flow cross-section area - conservation of mass mAV& ==ρ const. - Conservation of momentum for a inviscid and steady flow 1 0 dp++=ργ d() V2 dz 0 2 dp dV =− ρV 2 V Since mAVc& ==∴++=ρ , lnρ ln AVc ln ′ ddAdVρ differentiation →++=0 ρ AV dV dρ dA dp →−=+= ( ) (11.44) VAρ ρV 2 dA dp dρρρ dp d V 2 =−=(1 −⋅ ) A ρρVV22ρρdp dp V 2 =−(1 ) ρV 2 dp/ dρ (11.45) ⎛⎞∂p V Since c ==⎜⎟, and Ma ⎝⎠∂ρ s c dp 2 dA ∴ (1−= Ma ) (11.47) ρV 2 A dp dA1 dV −=− = (11.48) ρV 22A 1Ma− V diverging duct dp dA1 dV −=− = ρV 22A 1Ma− V converging duct Combining (11.44) and (11.48): ddAdAρρ1M ddAa2 += ⇒ = (11.49) ρρAA1Ma−−22A1Ma (11.49): For subsonic flow, density and area changes are in the same direction; for supersonic flow, density and area changes are in the opposite direction. dA A From (11.48): =−(1 − Ma2 ) dV V dA When Ma=→ 1 =0 ⇒The area associated with Ma=1 is either dV a minimum or a maximum. impossible Therefore the sonic conduction Ma=1 can be obtained in a converging-diverging duct at the minimum area location. For subsonic flow → converging diverging nozzle For supersonic flow →converging diverging diffuser throat throat subsonic supersonic subsonic Converging-diverging nozzle Converging-diverging diffuser What is the physical reason that supersonic flow develops in the diverging duct as long as sonic flow is reached at the throat? ΔP-induced flow ΔP-driven flow throat P0 Pe V avg Vavg Why can V only be accelerated to c in the converging duct, no matter how low the back pressure Pe is? 11.4.2 Converging–Diverging Duct Flow - For an isentropic flow p p0 kk==constant ρ ρ0 -streamwiseequation of motion for steady, frictionless flow dp V 2 +=dd( ) 0, γ z neglected ρ 2 1/k 2 ⎛⎞p p 1/kk 1/ p 0 dp V =→=0 ρρpp(/) +=d ()0 ⎜⎟Q kk 00 1/k ⎝⎠ρρ0 ρ 0 p 2 1/k kk−−112 kVp0 kk []0pp0 − −= k −12ρ0 kpVp 2 []0 −− = 0 k −12ρρ0 -For an idea gas p0 p ==RT0 , RT ρρ0 kR V 22V kR ∴ []TT−− = 0 or cTT() −− = 0 ( c = ) kk−1200pp 2Q − 1 V 2 ⇒−+hh()0 = where h is the stagnation enthalpy 0 2 0 kRT kRT V 2 k −1 0 =+ kRT=+ kRT V 2 kk−112− 0 2 T kV−−112 k 0 =+11M =+ a22 ()kRT= c TkRT22 T 1 = 2 (11.56) Tk0 1[(1)2]Ma+− p With = RT ρ ⎛⎞1 pTρρ00 ppp00k ==⇒= ⎜⎟Q kk () pT00ρρρρ⎝⎠ 0 p 11kk− pp− T p T p T ()kk=→ () =⇒ =() k−1 pp00 T 0 p 0 T 0 p 0 T 0 k p 1 k −1 ∴=[]2 (11.59) pk0 1[(1)2]Ma+− 1 using isentropic relation ρ 1 k −1 = []2 ρ0 1[(1)2]Ma+−k (11.60) T 1 = 2 Tk0 1[(1)2]Ma+− (11.56) k p 1 k −1 = []2 pk0 1[(1)2]Ma+− 1 ρ 1 k −1 = []2 ρ0 1[(1)2]Ma+−k T 1 = 2 Tk0 1[(1)2]Ma+− Figure D1 (p. 718) Isentropic flow of an ideal gas with k = 1.4. V 2 cT=+ cT pp0 2 Figure 11.7 The (T − s) diagram relating stagnation and static states. Figure 11.8 The T – s diagram for Venturi meter flow. -Any further decrease of the back pressure will not effect the flow in the converging portion of the duct. -At Ma=1 the information about pressure can not move upstream -Consider the choked flow where at the throat Ma=1, the state is called critical state p Ma =1, critical state (choked flow) p0 subsoinc 0.528 supersoinc Critical State: Set Ma=1 in (11.56), (11.59), (11.60) p * 2 k = ()k −1 pk0 + 1 For k=1.4 ⎛⎞p * ⎜⎟ = 0.528 p ⎝⎠0 k =1.4 * pk =1.4 = 0.528patm ** TT2 ⎛⎞ * =→ ⎜⎟ =0.833 or TTTk =1.4 =0.8330 = 0.833 atm Tk+1 T 00⎝⎠k=1.4 ρρ**pkT 212k + 1 ⎛⎞* ==0 ()()()kk−−11 = → ⎜⎟ =0.643 ρρTp* k++12k 1 00 ⎝⎠0k =1.4 Example 11.5 p0 =101 kPa 3 ρ0 =1.23 kg/m T0 = 288 K Find m& = (a)80 kPa, (b)40 kPa. Critical pressure pp*= 0.5280 =53.3 kPa (a) pa > p* ∴the throat is not choked k p 80 1 k −1 ==[ 2 ]→ Math = 0.587 pk0 101 1+− [( 1) 2]Ma 1 ρ 1 k −1 33 ==[ 2 ]ρρ0 1.23 kg/ m ⇒=1.04 kg/ m ρ0 1[(1)2]Ma+−k T 1 =⇒2 T =269 K Tk0 1[(1)2]Ma+− VkRTV=⇒Ma * =193 m/s mVA& ==ρ 0.0201 kg/s (b) pb=40 kPa < p*=53.3 the flow is choked at the throat Ma = 1 1 ρ 1 k −1 3 =⇒[]02 ρρ=.6340 =0.78 kg/m ρ0 1[(1)2]Ma+−k T 1 =⇒2 TK=240 Tk0 1[(1)2]Ma+− VkRTV=⇒Ma * =310 m/s mVA& ==ρ 0.0242 kg/s Figure D1 (p.