<p>Quasi-1D Nozzle Review</p><p>Example: Want to find P,T,M, etc. given Po, Pe, and nozzle shape.</p><p>Quasi – Area is allowed to vary along x coordinate, but flow variables are functions of x only.</p><p>Start out with governing conservation equations:</p><p>Mass: r 蝌蝌 dV+ r U� ndS 0 Vt S</p><p>Momentum:</p><p>蝌蝌rUdV+ 蝌 r( U� n) UdS - + 蝌 pdS + 蝌 r fdV Fviscous t V S S V Energy:</p><p>抖 骣U2 骣 U 2 q r琪e+ dV + r 琪 e + U� ndS - � pU + ndS r dV r f U dV 抖蝌蝌 蝌 蝌 蝌 蝌 ( ) tV桫2 S 桫 2 S V t V</p><p>Assuming: 1. steady 2. inviscid 3. no body forces 4. 2D flow</p><p>Quasi-1D: Area is allowed to vary but flow variables are a function of x only</p><p>A, u,  A+dA,u+du,  +d Mass:</p><p>-ruA +( u + du)( r + d r )( A + dA) = 0 - ruA + ruA +rudA + r duA + d r uA +higher order terms = 0 divide by through ruA dA du dr + + = 0 A u r or d(r uA) = 0 Momentum:</p><p>2 骣PdA -ru A +( r + d r )( u + du)( u + du)( A + dA) = PA -( P + dP)( A + dA) + 2琪 桫 2 - ru2 A + ru2 A +ru2 dA + u 2 Ad r + r uAdu + r uAdu = PA - PA - PdA -AdP + PdA u( r udA+ uAd r + r Adu ) +ruAdu = - AdP dP= -r udu</p><p> dP dP dr dP = = -udu a2 = rd r r d r s dr u = - du r a2 Substituting equation into to get: dA du u + -du = 0 A u a2 which can be rearranged to get: 2 dA du骣 u dA du 2 +琪1 -2 = +( 1 -M ) = 0 A u桫 a A u dA du =(M 2 -1) A u which can be used to determine general flow behavior in a converging-diverging nozzle, as below:</p><p>M dA du < 1 < 0 > 0 dA <0 --> converging subsonic < 1 > 0 < 0 dA >0 --> diverging > 1 < 0 < 0 du <0 --> decelerating supersonic > 1 > 0 > 0 du >0 --> accelerating</p><p>Energy: We will not go through the derivation for the energy equation but, applying analysis as before will give:</p><p> dh+ udu =0 or cP dT = - udu</p><p>Total-Static-Mach Relations Isentropic Relations: g g P骣r 骣 T g -1 2=琪 2 = 琪 2 P1桫r 1 桫 T 1</p><p>Static-Total</p><p>From Energy Equation with uo=0 (total) u 2 cT= c T + 1 Po P 1 2 Rearranging T u2 o =1 + T2 cP T g R using cP = g -1 , T g -1 u2 o =1 + T2 g RT Inserting a= g RT and M= u a gives</p><p>T g -1 o =1 + M 2 T 2 g g -1 P 骣 g -1 2 o =琪1 + M P 桫 2 1 g -1 r 骣 g -1 2 o =琪1 + M r 桫 2 Given Eq.’s ,, and we can now:</p><p> Find any static property in an isentropic flow given Mach #, Po,To,o.  Use/control known total conditions to find mach # through nozzle</p><p>Area-Mach Relations From mass r*u * A * = r uA u* at A* M= = 1 � u * a * , giving a* 2 2 2 骣A骣r * 骣 a * 琪 * = 琪 琪 桫A桫r 桫 u or 2 2 *2 * 2 骣A骣r 骣ro 骣 a 琪 * = 琪 琪 琪 桫A桫ro 桫 r 桫 u using isentropic relations for the density terms g +1 2 g -1 骣A 1轾 2 骣 g - 1 2 琪* =2 犏 琪1 + M 桫A M 臌g +1 桫 2 Mach # is a function of this area ratio only. Must find A*.</p><p>Completely Subsonic Flow</p><p>Pe= P atm From isentropic relation 轾 g -1 2 骣P g M =犏 o -1 e 犏琪 g -1 桫Pe 臌犏 Can now determine A* and entire Mach # distribution P A o = 1 e * If: Then: Me =0, *  , A = 0 NO FLOW Pe A P o Ae As increases, Me increases, * decreases, A* increases Pe A Note that A/A* < 1 is not physically possible. That is, after 1st critical is reached, must have Amin = A*</p><p>Supersonic Flow Subsonic Flow ahead of throat. Follow supersonic A/A* branch after throat. * A= At</p><p>骣Ae Me = f 琪 桫A* g 1-g 骣 g -1 2 Pe= P o琪1 + M e 桫 2</p><p>Suppose we pick Po so that Pe = Patm</p><p> If we decrease Po, then Pe < Patm because Me is unchanged. Need weak oblique shocks to get a small pressure jump.  As Po decreases, need stronger oblique shocks until normal shock at exit, 2nd critical.  As Po decreases, shock moves up the nozzle. Eventually get to 1st critical.  Increasing Po from 3rd critical, Pe > Patm. Get Prandtl-Meyer expansion fan to get pressure decrease</p><p>Summary:</p><p>For our nozzle:</p><p>Po,3 rd 60 psig</p><p>Po,2 nd 20 psig</p><p>Po,1 st 8 psig</p>