Chapter 11

Compressible Flow Introduction z –variable density, and equation of state is important z Ideal gas equation of state—simple yet representative of actual gases at pressures and of interest z Energy equation is important, due to the significant variation of . 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 and 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= ρρδ−+− Vc δρδρδ V ρδ = cV δρ

-linear momentum conservation −ccAcρ +−()()()(δρδρ V + c − VA δ = pAp − +δ pA )

Q()()ρ +−= ρδρ δ cVAcA(continuity)

∴ −+−ccAcρ ()δρ V cA δ =− pA 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

= − wave )( cttr

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 .) 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. 718) Isentropic flow of an ideal gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.) Ratio A/A* AVρ ** ρρAV==*** A V or AV* ρ with VkRTV**== and Ma kRT

** ** AkRTρρ1 TT/ 0 →=* = ATρρkRT Ma Ma00 /T 1211kk++ 1 21 1 =+ ( )kk−−11 [1 Ma2 ] [ ]2 Makk++ 1 2 1 1 [(− 1) 2]Ma2

k +1 Ak11[(1)2]Ma+− 2 ⇒= []2(k − 1) Ak* Ma 1+− [( 1) 2]

Figure 11.10 (p. 639) The variation of area ratio with Mach number for isentropic flow of an ideal gas (k = 1.4, linear coordinate scales). Example 11.8 Air entering subsonically at standard atmospheric condition, A=0.1+x2 A 110.1+ x2 Ar=→==π 2 r()22 ( ) ππ

For the flow to be chocked

At throat, xA=→ 0* = 0.1 A 0.1+ xp2 T * =⇒, using (11.71) Ma⇒ , A 0.1 pT00 d

a c c 0.98 b

d 0.04 Example 11.9 Air entering supersonically at standard atmospheric condition, A=0.1+x2

For the flow to be chocked

At throat, xA=→ 0* = 0.1 A 0.1+ xp2 T * =⇒, using (11.71) Ma⇒ , A 0.1 pT00 d

c a

c 0.98

b d 0.04 Example 11.10 Ma=0.48 at throat, A=0.1+x2

The Ma at throat is 0.48 pTA Ma=⇒ 0.48 , , * pTA00 A 0.1 ==1.4 ⇒=A* 0.07 A**A

c

b subsonic-subsonic subsonic-subsonic(choked)

subsonic-supersonic(choked) supersonic-supersonic(choked)

Supersonic-subsonic (choked) Supersonic-supersonic (not-choked) 1. For a given (T0, p0), k and converging-diverging duct geometry, infinite number of isentropic subsonic to subsonic (not choked) and isentropic supersonic to supersonic (not choked) flow solutions exist. 2. For choked condition, the flow solutions are each unique.

ppppext ≥≤ΙΙor ext Ι,isentropic flow is possible

ppΙΙ≤≤ext pΙ, isentropic flow is not possible Non-isentropic choked flow

overexpanded

underexpanded

Oblique shock wave (3-D)

isentropic isentropic V11.5 Rocket engine start-up Entropy generation V11.6 Supersonic nozzle flow From Zucrow and Hoffman, Gas Dynamics 11.4.3 Constant Area Duct Flow z For constant area isentropic duct flow, the flow velocity, thus the fluid enthalpy and temperature are constant.

p0 pb

11.5 Nonisentropic flow of an ideal gas

z – adiabatic flow with friction. z – constant area duct flow with heat transfer but without friction 11.5.1 Adiabatic constant area duct flow with friction (Fanno flow) z Consider steady 1-D ideal gas constant area duct flow with friction

Continuity: mVA& = ρ =⇒=constρ V const Energy equation: 22 VV21− mh&[(21−+ h +g z21 − z)] =+ Q& W& 2 V 2 hhhhcTT+=, −= ( − ) 2 00p 0 V 2 TT+==0 const (stagnation temp.= const) 2c p 222 ()ρρVV ()T (11.75) ⇒+TTTTV22 =⇒+002 =, where ρ = const 22ccppρ p/R Eq. 11.75 allows us to calculate T for p in the Fanno flow. Tds equation (2nd law):

1 Tds=− dh() dp ρ dT dp ds=− c R p Tp

Tp22 ssc21−=p ln − R ln Tp11 Tp ss−=1 cp ln − R ln (11.76) Tp11

pTs111, , are considered reference values from the entrance.

From (11.75) and (11.76), the Fanno line for variation of p-T-s can be obtained. Ex 11.11 ExampleExample 11.1111.11 CompressibleCompressible FlowFlow withwith FrictionFriction ((FannoFanno Flow)Flow) z Air (k=1.4) enters [section (1)] an insulated, constant cross-sectional area duct with the following properties:

T0=284K

T1=286K

p1=99kPa(abs) For Fanno flow, determine corresponding value of fluid temperature and entropy change for various values of downstream pressures and plot the related Fanno line. ExampleExample 11.1111.11

To plot the Fanno line we use Eq. (75) and (76)

ρ T)V( 22 T + == ttanconsT (11.11.1) 22 0 P )R/p(c2 T p P1 lncss −=− lnR (11.11.2) T1 p1 k = 4.1 = /286.9 ⋅KkgJR

Rk From Eq. (14) c = /1004... ⋅== KkgJ (11.11.3) p k −1

p p1 (11.11.4) (1)+(69) V =ρ VRTkMa 11 =ρ= 11 kRTMa RT RT1 ExampleExample 11.1111.11

T1 286K T 1 == 993.0 = 2 (56) To 288K T −1k o + 2 Ma1 1 From Eq. (56) ⎛ ⎞ Ma1 ⎜ −= ⎟ = 2.002.0/1 ⎝ 399.0 ⎠

1kRT == /339... sm × 3 smPa )/339(2.01099 (11.11.4) ρV = = 2 ⋅smkg )/(8.81 ⋅ KKkgJ )286)(/6.289( For p= 48 kPa

22 (11.11.1) ρ )( TV T + 22 K =⇒== 7.278T288... K P Rpc )/(2 (11.11.2) T p 1 css P R ==−=− ⋅KkgJ )/(7.181...lnln T1 p1 ExampleExample 11.1111.11

For p=48kPa T=278.7K s-s1=181.7J/(kg‧K)

For p=41kPa T=275.6K s-s1=215.7J/(kg‧K)

For p=34kPa T=270.6K s-s1=251.0J/(kg‧K) Analysis of Fanno line Tds equation dp dp Tds=− dh =− dh RT ρ p For an ideal gas, dp Tds=− cP dT RT dh= c dT; p Q P dp dρ dT ddTρ pRT==+ρ or =−cdT RT() + P ρ T p ρ T ddVρ Continuity: ρV ==const, or − ρ V ddTρ dVdT →=Tds c dT − RT() + =c dT −− RT (+) PPρ TVT ds c 11dV →=P -(-R + ) dT T V dT T V 2 VdV dV c Energy eq.: P TT+ ==0 const → dT =−⇒ =− 2ccppdTV ds cc1 ∴ =+pp-(R ) dT TV 2 T (11.82) ds cc1 c For =→0 p =RcRcR(pp + ) → −==T dT TVV22 T p υ

VccRTkRT==(/)paυ a subsonic So, the Mach number at state a is 1.

ds <<0, VkRT →subsonic dT a ds >>0, VkRTa →supersonic supersonic dT Ma=1

supersonic

Since T0 is constant on the Fanno line, the temperature at point a is the critical temperature T*. -2nd law ds>0

subsonic flow supersonic flow Normal shock (acceleration) (deceleration) Summary of Fanno flow behavior

Friction drags acceleration by pressure drop Friction helps deceleration by pressure rise -To quantify the Fanno flow behavior, we need to combine relationship that represents the linear momentum law with the set of equations already derived.

pA11− pA 2 2−= Rx mV()2 − V 1

Rx →−−p12pVVVAAAmAVC =ρρ(21 − ), ( 1 = 2 = and & = =) A Q -Therefore, for the semi-infinitesimal control volumes τ π Ddx −−dp w =ρVdV A 8τ π D2 with fA==w , ρV 2 4

Vdx2 →− dp − fρρ = VdV 2D dp fρρ V22 dx d() V or + +=0 pp22 Dp dp fρρ V22 dx d() V ++=0 (11.88) pp22 Dp After some derivation, (11.88) becomes

1()(dV22 dMa) fk dx (1+−+=k Ma22 ) Ma 0 22V 22Ma D (1− Ma22 )dd (Ma ) x or = f 24 (11.96) {}1[(1)2]MaMa+−kkD -Integrate to the critical state

**22 Ma= 1 (1− Ma )dd (Ma ) l x = f 24 ∫∫Ma ⎡⎤l D ⎣⎦1[(1)2]Ma+−kk Ma

2* 11[(1Ma−−) kk++⎡⎤(1)2]Ma2 f (ll) 22+=ln ⎢⎥(11.98) kkMa 2 ⎣⎦1+− [(k 1) 2]Ma D -Note that the critical state does not have to exist, since for any two section in the Fanno flow,

f ( * − ) f ( * − ) f 2 − llll 1 ()−= . D D D ll 21

-Other fluid properties in the Fanno flow can also be derived, as summarized below. Summary of Property Relations for Fanno Flow Note: These equations correlate ratio of properties associated with different positions (a certain position and the choke position), between which friction loss exists. 2* 1 (1Ma−−) kk++1⎡⎤ [( 1) 2]Ma 2 f (ll) 22+=ln ⎢⎥ (11.98) kkkDMa 2⎣⎦ 1+− [( 1) 2]Ma Tk(1)2+ = , 2 (11.101) T* 1[(1)2]Ma+−k 1 Note: For p/p0, p*/p0*, T/T0 Vk⎡⎤[(+ 1) 2]Ma2 2 = , at the same position, ⎢⎥2 (11.103) V* ⎣⎦1[(1)2]Ma+−k isentropic relations in terms of Ma or Fig. D1 can be used. −1 ρ ⎛⎞V = ⎜⎟, (11.105) ρ**⎝⎠V 1 pTρ 1(1)2⎡⎤ k+ 2 == , ⎢⎥2 (11.107) pT***ρ Ma⎣⎦ 1+− [(k 1) 2]Ma same position different positions k +1 21(k − ) pp00pp*1⎡⎤⎛⎞ 2 2 ==⎢⎥⎜⎟()1[(1)2]Ma+−k (11.109) pppp00***Ma1⎣⎦⎝⎠ k+ Note: These curves p correlate ratio of 0 p * properties associated with 0 different positions (a certain position and the choke position), between which friction loss exists.

p Figure D2 (p. 719) p* Fanno flow of an ideal gas with k = 1.4. Example 11.12 Choked Fanno flow

Given pT00==101 kPa, 288 K What is the maximum flow rate at the duct? -For maximum flow rate, the flow must be choked at the exit. f ( * − ) f ( − ) × 202.0 1 = llll 12 = = 4.0 D D 1.0

TV11 p 1p0,1 Fig. D2 →=→= Ma1 0.63 ** 1.1, = 0.66, * = 1.7, * = 1.16 TV p p0 Tp11ρ 1 Fig. D1 with Ma1 =→= 0.63 0.93, = 0.76, = 0.83 Tp00,10ρ ,1 Since TC= =288 K 0 Or, perhaps more logical, * T 2 * TT10=×=×0.93 0.93 288 ==0.8333 ⇒=TT02 (0.8333) =⇐ 240 T Tk0 +1 = 268 K VRTk**==310m/s ( ⇐→ VV ) = 0.66 V* = 205m/s 21 VkRT11= Ma × 1 =×× 0.63 1.4 386.9 × 268 ρρ==→==pRT/ 1.23 0.83ρ 1.02 kg/m3 0,1 0,1 0,1 1 0,1 = 207m/s

mVA& = ρ11 =1.65 kg/s with friction, use Fig. D2 * p p1 1 * 1 pp20==××=,10.76 101 45kPa pp00,20,1==×= p 87kPa pp10,1 1.7 1.16 Example 11.13 Effect of duct length on choked Fanno flow

Pd = 45kPa

1m If the flow is choked.

* f (ll− 1 ) 0.02× 1 pV ==0.2, Fig. D2 →== Ma 0.7, 11 1.5, = 0.73, Dp0.1 1 **V

p11ρ Fig. D1 with Ma1 =→= 0.70 0.72, = 0.79 p00ρ ,1 * * p p1 1 pp20== p1 =×0.72 × 101 = 48.5kPa ( >=pd 45kPa) pp101 1.5 * ρρ10== 0.79,11 , VV 0.73

→=mAV& ρ111 =1.73 kg/s -For the same upstream stagnation state and downstream pressure, ll↑→mf&& ↓ or when is fixed: ↑→m ↓ Example 11.14 Unchoked Fanno flow 11.5.2 Frictionless constant-area duct flow with heat transfer (Rayleigh flow)

0 (frictionless flow) Momentum: p11AmVpAmVR+ 1=++ 2 2 2 x 22 ()ρρVV()RT or pp+=→+=const const ──(11.111) ρ p ρV 2 pp=+ ≠const, although there is no friction, why? 0 2

Continuity: ρVC= Tp Tds eq.: s −=scn1 pll − Rn ──(11.76) Tp11 Eqs. (11.111) and (11.76) can be used to construct a Rayleigh line with reference conditions. Example 11.15 Construction of a Rayleigh line for various downstream p2 (or T2), with entrance conditions T0, T1, p1

Given TTp011 , , ,→=ρ 11 , V and ρ V C can be calculated.

-Assume p2 (or T2), then from (11.111) T2 (or p2) can be obtained.

-From (11.76), s2 can be obtained. 2 ()ρVRT p +=const (11.111) p Tp ss−=1 Cnpll − Rn (11.76) Tp11 Discussion on the Rayleigh line ds -At point a on the Rayleigh line, = 0 dT -After some derivation, we have dsc V 1 =+p (11.115) dT T T[](/)(/) T V− V R ds For = 0, dT dsc V 1 =+p dT T T[](/)(/) T V− V R kR⎛⎞ T V ⎜⎟−+=V 0 kVR−1⎝⎠ kRT kV 2 ()kV−1 2 −+ =0 kk−−11 k − 1 VkRT2 =

VkRTaa=⇒= Ma a 1 Derivation of ds/dT dp=−ρ VdV ⎡ρVC= ⎢ dP ddVρ −= VdV ⎢ =− ρ ⎣⎢ ρ V pRT= ρ Since Tds equation dp dp dρ . dT Tds=− dh =+ ρ pTρ =+cdT VdV ρVdV dV dT p −=−+ or ρRT V T dscp V dV Vd11T =+ −=−+ dT T T dT RT V T dV c p V 1 TV dT =+ −= TTTVVR()//− VRdV At point bdTds , /= 0 dsc V 1 =+p dT T T[](/)(/) T V− V R dT 11 = ==0 ds ds c V −1 p +−[](/)(/)TV V R dT TT c V 1 p +→∞ TTTVVR[](/)(/)− TV ∴ =⇒VRTV2 = ⇒= RT VR V RT 1 Ma ==b = b ckkRT

-At point b, the flow is subsonic since k＞1. - Now, consider the energy equation,

22 ⎡⎤VV21− mh& ⎢⎥21−+ h +gz()21 − z = Qnet + W sheftnet ⎣⎦2 kR dh+= VdVδ q, dh = c dT = dT p k −1

cdTp += VdVδ q dT VdVδ q += TcTcTpp dV⎡⎤ V dT V2 δ q ⎢⎥+= VTdVkRTk⎣⎦⎢⎥/1()− cTp

2 −1 −1 dVδδ q⎡⎤ V dT ()kV−1 q⎡ V⎛⎞ T V 2 ⎤ ∴ =+⎢⎥ =−+−⎢ ⎜⎟()k 1Ma⎥ VcTTdVkRTpp⎣⎦ cTTVR⎣ ⎝⎠ ⎦

2 −1 δδqV⎡⎤2 q 22−1 =−+−=−⎢⎥ 1()kk 1 Ma⎣⎦⎡⎤ 1 Ma+−()k 1 Ma cTpp⎣⎦ RT cT δ q 1 = cT⎡⎤2 (11.121) p ⎣⎦1Ma−

Derivation for Property Relations for Rayleigh Flow -linear momentum 22 p +=+ρρVpaaa V where a is the reference state.

2 pVρ ρa 2 or +=+1 Va ppaa p a

2 ρρaa22kV a VVaa==== kVkRT, since aa pRTkRTaaaaρ

2 pVρ ρa 2 + =+11Vka =+ ppaa p a pV⎛⎞ρ 2 ⎜⎟1+=+ 1kVkRT , where = Ma ppa ⎝⎠ p 1+ k = 2 pka 1Ma+ (11.123) ρ VTTpT a ==Ma ⇒=Ma ρ VTTpTaaaaa

2 2 ⎡ ⎤ Tpρa T⎛⎞ p ()1Ma+ k ⎛⎞pk1+ =→=⎜⎟Ma =⎢ ⎥ = Tpρ T p1Ma+ k2 ⎜⎟Q 2 aa a⎝⎠ a ⎣ ⎦ ⎝⎠pka 1Ma+ ρ VT⎡(1Ma+ k ) ⎤ a ==Ma = Ma ⎢ 2 ⎥ (11.129) ρ VTaa⎣ 1Ma+ k⎦

- Due to heat transfer, T0 varies TTT T 00= a TTTT0,aaa0, ⎡⎤()1Ma+ k ⎡ 1 ⎤ =+⎡⎤1[(1)/2]Mak − 2 ⎣⎦⎢⎥2 ⎢ ⎥ ⎣⎦1Ma1(1)/2++−kk⎣ ⎦ 2()kk++− 1 Ma22() 1 [( 1) / 2]Ma = 2 2 (11.131) ⎣⎦⎡⎤1Ma+ k Summary of Property Relations for Rayleigh Flow

pk1+ = 2 (11.123) pka 1Ma+ 2 T ⎡⎤()1Ma+ k = ⎢⎥2 (11.128) Tka ⎣⎦1Ma+

ρa V ⎡⎤()1Ma+ k ==Ma ⎢⎥2 (11.129) ρ Vka ⎣⎦1Ma+ k ppp p ()1+ k ⎡⎛⎞2 ⎤ k−1 00==a 1 +− [(k 1) / 2]Ma 2 2 ⎢⎜⎟()⎥ (11.133) pppp0,aaa0, ()1Ma+ k ⎣⎝⎠k+1 ⎦

22 TTT T2()kk++− 1 Ma() 1 [( 1) / 2]Ma 00==a TTTT 2 2 (11.131) 0,aaa0, ⎣⎦⎡⎤1Ma+ k ρ V a , ρ Va

Figure D3 (p. 720) Rayleigh flow of an idea gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.) Effect of Ma and Heating/Cooling for Rayleigh Flow (Example 11.16)

Note: p0 is not constant, although frictionless 11.5.3 Normal Shock Waves -Normal shock waves involves: • deceleration from supersonic to subsonic (from Cengel and Cimbala, • a pressure rise Fluid Mechanics, 2006) • an increase of entropy

FIGURE 12–30 Schlieren image of a normal shock in a Laval nozzle. The Mach number in the nozzle just upstream (to the V11.6 Supersonic nozzle left) ofthe shock wave is about 1.3. flow Boundary layers distort the shape of the normal shock near the walls and lead to flow separation beneath the V11.7 Blast waves shock. Derivation for Normal Shock Waves z infinitesimal thin control volume surrounding the shock wave: friction and heat transfer negligible, A=C ─continuity: ρV = const ─linear momentum（friction negligible）:─same as Rayleigh line 2 ()ρVRT pV+=ρ 2 const or p + =const p ─energy: with Δzq=Δ=0, 0 V 2 hh+==const 2 0 For an ideal gas, 2 ()ρVT2 TT+==const 22 0 ─same as Fanno line 2/cpRp () Tp ─Tds relationship: s −=scn1 pll − Rn Tp11 Q: The irreversibility in shock wave is not from friction or heat transfer. What is it from? Since shock wave flows have the same energy eq. for Fanno flows and same momentum eq. for Rayleigh flows, thus for a given ρV, gas (R, k), and conditions at the inlet of the normal shock (Tx, px, sx), the conditions downstream of the shock (state y) will be on both a Fanno line and a Rayleigh line that pass through the inlet state (state x). Mom., mass, Tds eqs. - For Rayleigh line, Total energy, mass, Tds eqs. ppp yy= a px ppax py 11++kkpx Q ==22, ppaa1Ma++kkyx 1Ma 2 p y 1Ma+ k x ∴ = 2 (11.140) px 1Ma+ k y

- For Fanno line 1 * 2 ppyypp1(1)/2⎡ k+ ⎤ ==** and ⎢ 2⎥ pppxx pMa⎣ 1+− [( k 1) / 2]Ma ⎦ 1/2 p Ma⎡⎤ 1+− [(k 1) / 2]Ma2 →=y xx⎢⎥ (11.148) Total energy, mom., p Ma 2 xy⎣⎦⎢⎥1[(1)/2]Ma+−k y mass, Tds eqs. Combining (11.140) and (11.148) 1/2 ⎡⎤22 py 1[(1)/2]Ma+−kkx Maxx 1+ Ma →=⎢⎥22= pkx ⎣⎦⎢⎥1[(1)/2]Ma+− yy Ma 1+k Ma y 2 2 Max +− [2 /(k 1)] ⇒=Ma y 2 (11.149) [2kk /(− 1)]Max − 1 p 1Ma+ k 2 21kk− (11.149) into (11.140) ⇒=y x =Ma2 − 2 x (11.150) pkxy1Ma+ k++ 1 k 1

- For Fanno line, Eq. 11.101

Ty (1)/2kk++Tx (1)/2 *2== and *2 Tk1[(1)/2]Ma+− y Tk 1[(1)/2]Ma+− x * 2 TTyyT 1[(1)/2]Ma+−k x →=*2 = TTTx xy1[(1)/2]Ma+− k (11.144)

T [1+ [(kk−−− 1) / 2]Ma22 ]{[2 /(k 1)]Ma 1} (11.149) into (11.144) ⇒=y xx 22(11.151) Tkkxx{(+− 1) /[2( 1)]}Ma Summary of Property Relations across Shock Wave

If Max is known, property ratios across the shock can be known: 2 2 Max +− [2 /(k 1)] Ma y = 2 (11.149) [2kk /(− 1)]Max − 1

py 21kk2 − =−Ma x (11.150) pkx ++11 k

22 Ty [1+− [(kk 1) / 2]Maxx ]{[2 /(k− 1)]Ma − 1} = 22 (11.151) Tkkxx{(+− 1) /[2( 1)]}Ma

2 ρ y Vkxx(1)Ma+ == 2 ρxyVk(1)Ma2− x+ (11.154) kk kk+−1122kk−11− [Ma][1Ma]xx+ p0,y 22 = (11.156) p 21kk− 1 0,x [Ma]2 − k −1 kk+11x + Figure D4 (p. 721) Normal shock flow of an idea gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

Example 11.17 drop across a normal shock

ppyy0, For Max ↑⇒ ↑ and ↓ ppxx0,

p py 0,y 1 29 p p0,x x

0.5 10

0.06 1

12.5 5 135 Ma Ma x y

Across a normal shock, adverse pressure gradient occurs which can cause flow separation. Therefore, shock-boundary layer interactions are of great concern to designers of high speed flow device. Example 11.18 Supersonic flow pitot tube

Given: pTp0,yx=== 4114kPa, 0 555K, 82kPa

k k +1 2 k −1 ppp [Ma]x 0,yy== 0, 0,x 2 ⇐Rayleigh Pitot tube formula ppp 21kk− 1 xxx0, [Ma]2 − k −1 kk++11x 414kPa ==5 82kPa

Fig. D1→= Max 1.9

Vcxxxxx==Ma Ma kRT

Tx TT0,xy==⇒= 0, ,0.59327KTx T0,x

⇒=Vx 678 m/s

Note: Incompressible calculation for pitot tube would give the wrong result. Example 11.19 Normal shock in a converging-diverging duct p p (a) III == ? for normal shock at exit, (b) =? for shock at x 0.3 m pp00

x=-0.5 x=0.5

0.98 2.8

0.04 -shock at x=0.5m px From Ex 11.8: Max === 2.8 and 0.04 at x 0.5m p0,x

ppyy0, Fig. D4→= Max 2.8(at exit):= 9, = 0.38 ppxx0, ppp p yy==×==x 9 0.04 0.36 III p0,x ppxx0, p0,x p 0,y =−0.38 considerable energy loss p0,x

-shock at x=0.3m px From Ex 11.8: Max == 2.14, 0.1 p0,x

Fig. D4 with Max =→ 2.14 across the shock

ppyy0, == 5.2, May 0.56, = 0.66 ppxx0, (Conditions downstream the shock at x =0.3m)

Note: for isentropic A*=0.1 (a minimum) Consider the isentropic decelerating flow downstream the shock A Also, Fig. D4→= y 1.24 (Here, A * is used as dummy) A* 2 A2 0.1+ (0.5) ==2 1.842 Ay 0.1+ (0.3) AAA 22==×=y 1.24 1.842 2.28 AAA**y A →=A*0.152 = 2.28 Ap2 2 =→2.28, Fig. D1 (isentropic) Ma2 = 0.26, = 0.95 Ap* 0,y ppp →=220,y =×=0.95 0.66 0.63 ppp0,xyx 0, 0, pp Note: 0,yy===0.66, shock atxx 0.3m > 0, 0.36, shock at exit =0.5m pp0,xx0, Figure D1 (p. 718) Isentropic flow of an ideal gas with k = 1.4. (Graph provided by Dr. Bruce A. Reichert.)

11.7 Two-Dimensional Compressible Flow Supersonic - Flow acceleration across a Mach wave VV= tt12 VV> nn21

∴VV21> - Supersonic flows decelerate across compression Mach wave

from M. Van Dyke, An Album of Fluid Motion Small wedge angle-attached shock from M. Van Dyke, An Album of Fluid Motion

Large wedge angle-detached shock (from Cengel and Cimbala, Fluid Mechanics, 2006)

Figure 11.4 (p. 591) The schlieren visualization of flow (supersonic to subsonic) through a row of compressor airfoils. (Photograph provided by Dr. Hans Starken, Germany.) 3-D shock wave around a model space shuttle

from Gas Dynamics Lab, The Penn. State University, 2004