Introduction to Compressible Flow

1 1 Introduction to Compressible Flow0 DtD The density of a gas changes significantly along a streamlineCompressible FlowDefinition of Compressibility: the fractional change in volume of the fluid element per unit change in pressure ppppvdpp+dpp+dpp+dpp+dvv Compressible Number:2. Compressibility becomes important for High Speed Flows where M > M < Subsonic & incompressible <M < Subsonic & Compressible <M < transonic flow shock waves appear mixed subsonic and sonic flow regime <M < - Supersonic shock waves are present but NO subsonic flow M > Hypersonic Flow, shock waves and other flow changes are very strongsound of speed velocitylocal==cVM2 Compressible Flow3. Significant changes in velocity and pressure result in density variations throughout a flow field4. Large Temperature variations result in density a result we now have two new variables we must solve for:T & We need 2 new will solve: mass, linear momentum, energy and an equation of Effects of Compressibility on Flow1.

2 Choked Flow a flow rate in a duct is limited by the sonic condition 2. sound Wave/Pressure Waves rise and fall of pressure during the passage of an acoustic/ sound wave. The magnitude of the pressure change is very Shock Waves nearly discontinuous property changes in supersonic flow. (Explosions, high speed flight, gun firing, nuclear explosion)4. A pressure ratio of 2:1 will cause sonic flowApplications1. Nozzles and Diffusers and converging diverging nozzles2. Turbines, fans & pumps3. Throttles flow regulators, an obstruction in a duct that controls pressure One Dimensional Isentropic Flow Compressible pipe Control volume approach Steady, One-dimension, Uniform Flow Additional Thermodynamics Concepts are needed Restrict our analysis to ideal gasesThermodynamics Equation of State Ideal Gas LawRTp =Temperature is absolute and the specific volume is (volume per unit mass): 1=vK)J/( )J/(kmol8314air of massMolecular Constant Gas Universal = ===muMRRT hermodynamics Internal Energy & Enthalpy Internal Energy individual particle kinetic energy.

3 Summation of molecular vibrational and rotational energy. For an ideal gas Recall from our integral form of the Energy Equation for Enthalpyof an ideal gas:dvvudTTuudTv + =~~~()Tvuu,~~=dTcudv=~()Tuu~~=)(Thh=dTcd hp=pvuh+=~4 Thermodynamics Internal Energy & EnthalpydTcdhp=RdTuddhRTuhpvuh+=+=+=~~~R Tp= dTcudv=~Substituting:constRccRccRdTdTcdT cRdTuddhvpvpvp== +=+=+=~Thermodynamics Internal Energy & EnthalpyDefine the ratio of specific heats:constcckvp= Then,11 = =kRckkRcvpFor Air:cp= 1004 J/kg-Kk = 2ndLaw of Thermodynamics & Isentropic ProcessesCombining the 1stand 2ndLaws gives us Gibb s EquationrevTQds = We define entropy by: = = =212121pdpRTdTcdsdpdTcTdsdpdhTdspp dTcdhp=pRT= 15 The 2ndLaw of Thermodynamics & Isentropic ProcessesFor an Isentropic process: adiabatic and reversibleWe get the following power law relationship 121212lnlnppRTTcssp = kkkTTpp = = 1211212 Control Volume Analysis of a Finite Strength Pressure Wavec0=VTp TTpp + + + V Moving Wave of Frontal Area AThe Speed of sound (c) is the rate of propagation of a pressure wave of infinitesimalstrength through a still TTpp + + + VcV =Stationary WaveReference frame moving with wave()()()()()()()(A) 021 + = + += += ++ = = cVVccAVccAdAVcdAcdAnVCSrSteady State Continuity Equation(Solve for the induced velocity V):12 Control Volume Analysis of a Finite Strength Pressure WaveSmall Amplitude moderate frequency waves are isentropic and cVTp= TTpp + + + VcV =()()() ()(B) 12 VcpcVccAApppAVVmdAnVVFCSxx = = + = = &rSteady State Momentum Equation:(Find pand c)12 Now combine A & B and solve for the speed of sound .

4 0 oflimit in the 122 = + = + = pcppcconstpk= 6 Control Volume Analysis of a Finite Strength Pressure WaveCalculating the Speed of sound for an ideal gas:constpk= pkp= kRTpkc== kRTc=Typical Speeds of SoundFluidc (m/s)Gases:H21,294 Air340 Liquids:Water1,490 Ethyl Alcohol1,200 Data From White =vpcckFor Air:K)J/(kg287 =RExample 1: Speed of sound calculationDetermine the speed of sound in Argon (Ar) at 120 oC. MW = 40 kg/kmol:kRTc= =vpcckK)J/( )J/(kmol8314 = ==muMRR()() of a sound sourceand wave propagationV = 0V < cV > c Source moves to the right at a speed VMcV1sin== Zone of silenceV t3 c tV tV tMach cone7 Example 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer. Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard200 m M =3xExample 2: a needle nose projectile traveling at a speed of M=3 passes 200m above an observer.

5 Find the projectiles velocity and determine how far beyond the observer the projectile will first be heard()()() = ======= Steady Isentropic Flow Control Volume AnalysisApplications where the assumptions of steady, uniform, isentropic flow are reasonable:1. Exhaust gasses passing through the blades of a Diffuser near the front of a jet engine3. Nozzles on a rocket engine4. A broken natural gas line8 Steady Isentropic FlowpTh VdppdTTddhh++++ dVV+dx()()( )()dAdVddVAddAdVAdVdAVdVAdVdAAVVAdAAdVVd VAAVAVdAnVCS +++++++=+++=+ = = 222111 0rSteady State Continuity Equation:12 Steady Isentropic FlowpTh VdppdTTddhh++++ dVV+dx()()( )()dAdVddVAddAdVAdVdAVdVAdVdAAVVAdAAdVVd VAAVAVdAnVCS +++++++=+++=+ = = 222111 0rSteady State Continuity Equation:Only retain 1storder differential terms & divideBy VAVdVdAdA++= 012~ 0~ 0~ 0~ 0 Steady Isentropic FlowpTh VdppdTTddhh++++ dVV+dx()( )()12122122~~2pvupvuzzgVVmWQs+ ++ + = &&&Steady State Energy Equationwith 1 inlet & 1 exit:Neglecting potential energy and recalling: pvuh+=~1221222hhVVmWQs + = &&&()1221222 TTcVVmWQps + = &&&Assuming and ideal gas: 129 Steady Isentropic FlowpTh VdppdTTddhh++++ dVV+dxSteady State Energy Equationwith 1 inlet & 1 exit, neglecting potential energy & assuming Isentropic duct flow.

6 12122222hVhV+=+Assuming and ideal gas: 12122222 TcVTcVpp+=+1212221212 RTkkVRTkkV += +12 Stagnation ConditionsAssume the area A2is so big V2 ~ 0,thenohhVh=+=1212212 Stagnation enthalpyTcVTpo+=2212122222 TcVTcVpp+=+Similarly, as we adiabatically bring a fluid parcel to zero velocitythere is a corresponding increase in temperatureInsolatedwallsStagnation TemperatureStagnation Conditions maximum velocityTcVTpo+=22If the temperature, T is taken taken down to absolute zero, then (+) can be solved for the maximum velocity:(+)opTcV2max=No higher velocity is possible unless energy is added to the flow through heat transfer or shaft Conditions Mach number relations12112112222222+ =+ =+=+=MkcVkTTcTVTTTcVTopopoFor Ideal gases:111 = =kkRTTkkRTcppc2c1212+ =MkTToRecall, that the Mach number is defined as:cVM=Stagnation Conditions Isentropic pressure & density relationships11211121121121 + = = + = =kkookkkkooMkTTMkTTpp 1212+ =MkTToCritical Values: conditions when M = 1 +=12*kTTo1*12 +=kkokpp11*12 +=kok 21*12 +=kcco11 Critical Values: conditions when M = *= += *= += *= += kok For Air k = *= +=kccoIn all isentropic flow, all critical values are Values: conditions when M = 1 Critical Velocity: is the speed of sound c*2121**1212 += +===kkRTkckRTcVoo21*12 +=kccoExample 3: Stagnation ConditionsAir flows adiabatically through a duct.

7 At point 1the velocity is 240 m/s, with T1= 320K and p1= 170kPa. Compute(a) To(b) Po(c) ro(d) M(e) Vmax(f) V*12 Steady Isentropic Duct FlowpTh VdppdTTddhh++++ dVV+dxRecall, for Steady isentropic flow Continuity:VdVdAdA++= 0 For Compressible , isentropic flow the momentum equationis:VdVdpdVdp+=+= 202 Bernoulli s Equation!neglecting gravitySubstitute ( )into (*)( )(*)VdVdAdA = =+ =dpdVdpVdpdAdA 22112 Steady Isentropic Duct FlowpTh VdppdTTddhh++++ dVV+ =dpdVdpAdA 21 Recall that the speed of sound is: =pc2 = =22222111cVVdpcVdpAdA Substituting the Mach number: cVM=()221 MVdpAdA = Describes how the pressure behaves in nozzles and diffusers under various flow conditions12 Nozzle Flow Flow: M< 1 and dA< 0, then dP< 0: indicating a decrease in pressure in a converging Flow: M> 1 anddA< 0, then dP> 0: indicating an increase in pressure in a converging Flow: M< 1 and dA> 0, then dP> 0 : indicating an increase in pressure in a diverging Flow: M> 1dA> 0, then dP< 0 : indicating a decrease in pressure in a diverging channel.

8 ()221 MVdpAdA = PPPPPPPP13 Steady Isentropic Duct Flow Nozzles Diffusers and Converging Diverging Nozzles()221 MVdpAdA = Describes how the pressure behaves in nozzles and diffusers under various flow conditionsVdVdp+= 0 Recall, the momentum equation here is:VdVdp = Now substitute (**) into ( ) :()12 =MVdVAdA()12 =MVAdVdAOr,( )(**)Nozzle Flow Flow: M< 1 and dA< 0, then dV> 0: indicating an accelerating flow in a converging Flow: M> 1 anddA< 0, thendV< 0: indicating an decelerating flow in a converging Flow: M< 1 and dA> 0, thendV< 0 : indicating an decelerating flow in a diverging Flow: M> 1dA> 0, thendV> 0 : indicating an accelerating flow in a diverging channel.()12 =MVdVAdAConverging-Diverging NozzlesAminSubsonicSupersonicM = 1 AmaxSubsonicSupersonicM < 1 SubsonicSupersonicM > 1 Flow can not be sonic14 Choked Flow The maximum possible mass flow through a duct occurs when it s throat is at the sonic conditionConsider a converging Nozzle:VARTpVAm== &oooTp rpreceiverplenumeeVpMass Flow Rate (ideal gas):kRTVcVM==MARTkpAkRTMRTpm==&MARTkpm= &Choked FlowMARTkpm=&Mass Flow Rate (ideal gas):12121 + =kkoMkppRecall, the stagnation pressure and Temperature ratio and substitute:()kkooMkMARTkpm + +=1212211&1212+ =MkTTo()()1212*1121 + + +=kkkMkMAAThe critical area Ratio is:()kkookRTkApm + +=121*21&If the critical area (A*) is where M=1.