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Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition

Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition helps you tackle difficult exercises with expert guidance.

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    fnstructor's Solutions Manual to accompany MODERN COMPRESSIBLE FLOW Third Edition John D. Anderson, Jr. Curator for Aerodynamics National Air and Space Museum and Professor Erneritus University of Maryland

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    l.l .00812 1.2 e:#:sffiP:mr.g/m-l p RT (10X1.01 x 105) (1.38 x LO''11ZZO1 pr= p _(10)(1.01 x105) Rr pRT (11.0)(8314)(320) 87 x I0'"lm 0.,0345 kg- mole kg / n: 1.3 From the definition of enthalpy, h:e*pv:e*RT Eor a calorically perfect gas, this bebomes cpT:c"T+RT,or 6JF For a thermally perfect gzrs, EQ. (A1) is first differentiated dh: de + Rdt of, co dT: c" dT: Rdt of, (Al)

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    t.4 sz-sr :c, tn?-.^ f (a) R:1716 ft-lb/slug"R "r= h= W= 6oo6 ftJb/slug.R s2 - sr : (6006) tn (1.687) - (t716) tn 4.5 (b) R:287 joulelkg.K "r= h: W: too4.5 jouletkg"K sz - sr = (1004.5) tn (1.657) _287 n @.1) 52-sl = 59.9 ftJb/slue" s2-sl = 3'6 ioul / 1.5 Pr- Pr W:p pz 824.3 =-- r' RT, (1716X400) t.4 = 1800 (400/500) 0.4 0012 1.6 Volume of room : (20X15XB) = 2400 ff Standard sea level density :0.002377 slug/ft3 Mass of air : (0.002377)e400)= IS.ZO slud

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    1.7 (a) dp = -pVdV and dp: p t dp, or Combining: dp , _L=_pVdV pr dp:-"p2vdv dp 4 =-tpVdV p dP =-rpv2 dV V p (b) ,,: 1= - 1 7P (1'4[l^ol " 1d) :7'07 x 1o-6m2/N Y : r, p y2 + = - (7.07ii ro{y1r.23x10)2(0.01) + =FT?. ro=1 p (c) Here, & *iff belargerbytherati, (tooo\' p "[ rol y:(:8.7xrof (-*9' Comment: By increasing the velocity of a factor of 100, the fractional change in density is increased by factor of 104. This is just another indication of why high-speed flows must be dp: dP pr

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    CHAPTER 2 2-l Consider a two-dimensional body in a flow, as sketched in Figure A. A control volume is drawn around this body, as given in the dashed lines in Figure A. The control volume is bounded by: l. The upper and lower streamlines far above and below the body (ab and hi, respectively.) 2. Lines perpendicular to the flow velocity far ahead of and behind the body (ai and bh, respectively). . 3. A cut that surrounds and wraps the surface of the body (cdefg) The entire control volume is abcdefiria. The lyidth of the control volume in the z direction (perpendicular to the page) is unity. Stations I iand 2 arc inflow and outflow stations. respectively. Assume that the contour abhi is far enough from the body such that the pressure is inflow velocity ur is uniform across ai (as it would be in a freestream, or a test section of a wind tunnel.) The outflow velocity u2 is not uniform across bh, because the presence of the body has created u *uk at the outflow station. However, assume that both u1 and u2 8ro in the x direction: hence, u1 = coostdflt and u2: f(y). Consider the surface forces on the control volume shown in Figrre A. They stem from two contributions: l. The pressure distribution over the surface, abhi, -lJ out abhi

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    e_ _ 6 Figure A The sr:rface shear stress on ab and hi has been negfected. Also, note that in Figwe A.the cuts cd and fg are taken adjacent to each otheri hence any she6r stess or pressure distribution on one is ' equal and opposite to that on the other; i.e., the surface forces on cd and fg cancel each other. Also, note that the surface on def is the ewal and opposite reaction to the shear shess and i -* pressure distribution created by the flow over the surface of the body. To see this more clearly, examine Figrue B. On the left is shown the flow over the body. The moving fluid exerts pressure and shear stress distibutions over the body surface which create a resultant aerodynamic force per unit span R' on the body. In turn, by Newton's ttrird law, the body exerts equal and opposite pressure and shear stress distributions on the flow, i.e., on the part of the conftol surface bounded by def. Hence, the body'exerts a force -R on the contol surface, as shown on the ri of Figure B. With the above in mind, the total surface force on the entire control volume is Surface force: - lJ abhi

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    Morgover, this is the total force on the control volume shown in Figure A because the volumetric body force is negligible. Consider the integral form of the momentum equation as given by, Equation (2.1 l) in the text. The right-hand side of this equation is physically the foree on the fluid moving through the control volume. For the contol volume in Figure A, this force is simply the expression given by Equation (1). Hence, using Equation (2.11), with the right-hand side given by Equation (1), we have ovll dv+ 'V Q) fl S (pv'ds) v: - lJ n ns - n' abhi Flow exerts p and r on the surface of the body, giving a resultant aerodynamic force R Figure B Assuming steady flow, Equation (2) becomes R':-fl rou'ds)v-lJ nus a surface force on the section of .the control volume delthat equals -R (3)

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    Equation (3) is a vector equation. consider again the confrol volume in Figure A. Take the x component of Equation (3), nothing that the inflow and outflow velocities u1 and u2 dre in the x direction and the x component of R' is the aerodynamic drag per unit span D,: D':- # (pv'ds)'-lf (pds). S abhi (4) In Equation (4), the last term is the component of the pressure force in the x direction. [The expression G dS),. is the x component of the pressure force exerted on the elemental area dS of the control surface.] Recall that the boundaries of the control volume abhi are chosen far enough from the body such th"t p is constant along these boundaries. For a constant pressure. ll ro dS)* = o abhi because, looking along the x direction in Figure A, the pressure force on abhi pushing toward the ./ right exacfly balances the pressure force pushing toward the left. This is true no matter what the shape of abhi is, as long as p is constant along the surface. Therefore, substituting Equation (5) into (4), we obtain I tl D' : - (fl fnV.dS) u n\r s Evaluating the surface integral in Equation (6), we note from Figure A that: l. The sections ab, hi and def are steamlines of the flow. Since by definition V'is parallel to the streamlines and dS is perpendicular to the control surface, along these sections V and dS are perpendicular vectors, and hence V ' dS = 0. As a result. the contributions of ab, hi and def to the integral in Equation (6) are zero. (s) (6)

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    2' The cuts cd and fg are adjacent to each other. The mass flux out of one is identically the mass flux into the other. Hence, the contributions of cd and fg to the integral in Equation (6) cancel each other. As a result' the only contributions to the integral in Equation (6) come from sections ai and bh. These sections are oriented in the y direction. Also, the control volume has unit depth in the z direction (perpendicular to the page). Hence, for these sections, dS : dy(l). The integral in Equation (6) becomes fl rou'ds) u: - I, pru2rdy. JJ nfzdy S Note that the minus in front of the first tenn on the right-hand side of Equation (7) is due to v and ds being in opposite directions along ai (station I is an inflow boundary); in contrast, v and ds are in the same direction over hb (station 2 is anuoutflow boundary), and hence the second tenn has a positive sign. Before going fifther with Equation (7), consider the integral form of the continuity equation for steady flow. Applied to the control volume in Figure A, this becomes la 1b - J, prur dy + J n p2u2 dy: 0 or. J," p,u, at: Jrt pzwdy Multiplying Equation (8) by u1, which is a constant, we obtain J'" o'u"dv: IJ Pzwur dY Substituting Equation (9) into Equation (7), we have f rou'ds) u: - IJ pzvuray * J,' nizdy (7) (8) (e)

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    fl fon'ds)u=- IJ pzuz(ur-uz)dy S Substituting Equation (10) into Equation (6) yields o' : Jrt pzuz (ur - uz) dy (10) (l 1) (r2) Equation (l 1) is the desired result of this section; it expresses the drag of a body in terms of the known freestream velocity u1 and the flow-field properties p2 and u2, &croSS a vertical station downstream of the body. These downstearn properties can be measured in a wind tunnel, and the drag per unit span of the body D' can be obtained by evaluating the integrat in Equation (l l) numerically, using the measured data for p2 and u2 as a function of y. Examine Equation (11) more closely. The quantrty ur - uz is the velocity decrement at a given y location. That is, because of the drag on the body, there is a wake that trails downstream of the body. In this wake, there is a loss in flow velocif ur - uz. The quantiy mwis simply the mass flux; when multiplied by ot - u2, it gives the decrement in momentum. Therefore, the integral in Equation (11) is physically the decrement in momentum flow that exists across the -- W6ke, and from Equation (11), this #ake momentum decrement is equal to the drag on the body. For incompressible flow, p : constant and is known. For this case, Equation (11) becomes . D'=o JJ uz(ur-uz)dy Equation (12) is the answer to the questions posed at the beginning of this section. It shows how a measurement of the velocity distribution across the wake of a body can yield the drag. These velocity distributions are conventionally measwed with a Pitot rake:

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    ''. t.tl J.::'rit'- -!i,l:. l ;i l::11..-- , ,..::,r_..i l": .i UpTer f" ("\ 2.2 ia- I Lo,uet- Wa// f2 (r) Denote the pressure distributions on the upper andjower walls by pr(") and.p, (x) respectively. The walls are close enough to the modbl such that pu drd pz tr€ not necessarily equal to p-. Assume that faces ai and bh are far enough upstream and downstream of the model such that P: P- and v:0 and ai and bh. e I I I cl ?,, I I i iut tt. y-component of Eq. (2.11)inthe text: L:- # rpi at; v- lJ rnaSlv sabhi++ The fust integral = 0 over all surfaces, either because V' ds = 0 or because v = 0. Hence .bh L':- JJtnaSlv:-tJ Pudx- J n,d*1 abhi a i Minus sign because y-component is in downward direction. Note: In the above, the integrals over ia and bh cancel because p = p. on both faces. Hence L': hb c . 'f J p, o*- J puox i-a

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    CHAPTER 3 3.1 From Table A.1, for M: 0.7;pJp= l.3BTandTo/T: 1.09g. Hence. r^\ Po = p l?J : o.e (1.01 x to-5y11.387y : Note that P*/po can be obtained from Table A.l as the value of p/po for M: l. Hence, from Table A.l for M = 1; po/p* :1.893 and TolT* : 1.2. Thus. o* : n [&-) fa'.l :0.e (r.0r * ro)tr.s; 7)/rr8e3: - '\ p/ (p"/ r*:r t+) [+l :250 (t.os8)/1.2:ET&m \1./ \lol u*=.'ffi:^l@:bog2'A;d 1.26 x 10" N/m' (or 1.248 atrn x 10" N/m'(or 0.659 atm 3.2 po _ 1.5 x 106 _",., p 5x104 From Table 4.1' Fod @near interpolation) Also from Table A.1: Hence, T, T : 2.643 (linear interpolation) /r\ ro=r [+l :2oo(2.64T = ltt6E \ t./

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    3.3' a: JAT : rftr.4xr7r6x500) :1096ff/sec M : y la= 3000/1 09 6 = (2.7 4i, From Eq. (3.37) (M*;z: y+l 2.4 :3.6 3.4 From Table A.2,for Mr :3; pzlpr:10.33, pzlpr:3.857, po, /po,= 0.3283, and W4.4:.s2. rhus, pz : pr @zlp): (lxt.0t x tO511tO.:3) : 1.043 x 106 N/m2 (or 10.33 atm) Thus, // f.\ Pz= Pr I g I : (1)(l.ol x to)1t0.:3) : \p,/ . /^ \ _ Pz: pr | ,, | :1.23 (3.lsT=wqUffi \pr) 12: o:=#:#:F6m - pzR (4.744)(287) u: Jtar, = ,[(L4)Q87)Q66) = 554.8 m/sec w- a2M2= (554.9)(0.4752)= EOI.O rnAid From Table A.l, for Mr :3, Po, /p1 :36.73 a1d T", lTr:2.8. Hence: (P" \ t iu ,n. nnurn.Ao^\ po,: pl t ij fr: (1.01 x rc\ Q6.73X0.32s3) : h218 "' 10fr-/*-1 .i' '(-Lr)' *' -' M-:Tq

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    3.5 , 1.22 x 10' (a) pJp = ffii# :7.2r. From rable e.t: F<s3 , 7222 (b) pJp : ffi:3.413. From Table A.l; we find this would correspond to M- = 1.45. However, since this is supersonic, a normal shock sits in front of the Pitot tube. Hence, Po is now the total pressure behind a normal shock wave. Thus we have to use Table A.2. 1))) p^- /p, = t-z-!-'-:3.412. FromTableA.2' M;= l.jl ^ "? 2116 ' l3tg: n.85. FromTableA.2' M;=Til (c) Po, / Pr = .,OZO 3.6 For the shock compression, use Mr iIS isentropic compression, use pzlpt: (v1lv1)-7 : NORMAL SHOCK COMPRESSION " (pzpr)-t Mr pzlil =vzlYr Pzlqr a parameter, along with Table A.2. For the ISENTOPIC COMPRESSION vzlvr wlPr 1.0 t.2 1.5 t.7 2.0 2.5 3.0 I 1.34 1.86 2.20 2.67 J.JJ 3.86 1 0.745 0.537 0.455 0.375 0.30 0.259 1 l.5l 2.46 3.21 4.s0 7.13 10.3 I 0.7s 0.50 0.40 0.30 0.20 1 1.50 2.64 . 3.61 5.4 9.51

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    Note that: t (a) For a given decrease in volume, a shock wave yields a higher pressure and therefore many be construed to be more "effective" than an isentropic compression. (b) On the other hand, becarise the entropy increases across the shock, a shock wave is less efficient than an isentropic compression. To obtain a given pzlpt it requires more work via a shock. (c) Note that for vzlvr near unity, the shock and isentropic curves are essentially the same. This is a reflection that the entropy increase across weak shocks (Mr < 1.3) is negligible. 3.7 From Table A.1 fo M-::g, fJT =289.9;To: (289.8)(270):78246"4 This temperature is almost 8 times the surface temperature of the sun. Long before this temperature is evenreached, the air will dissociate adiorize,and y is no longer constant, nor

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    anywhere close to 1.4. The above answer is totally inaccurate! Since dissociation and ionization require energy, the actual temperature is considerably less, more like 11,000"K. The proper calculation of this answer must be made with techniques discussed in Chapter 14. 3.8 (a) From Table A.3 for M1:2.0: fr: 0.363,+: os2us,+: 0.7s34 From Table A.1 for Mr :2.0: ft: ,.* -T. Hence: (r*t/m\ ro* = I +- I t +l r, = f+^. ) (r.8X288) = 653.4oK ( Tq / \ T, / \0.793ry' / This is the total temperature at the exit, since for choked conditions at the exit, M2 = I and hence To, : To*, p2 : p*, etc. At the inlet, ( 1, : 1.8 Tr : (1.8X288):518.4"K. g: cp (t, -T", ): 1005 (653.4 - 581.4) : 1.357 x I pz=p*:p1/0.3636= #H= Wd Tz:T* =T110.5289: ^^ :F@ 0.5289 (b) From Table A.3, for M1 :0.2: prlp* : 2.273, TrlT* :0.2066, 1, /To* :0.7736 Frcim Table A.l, for M1:0.2:

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    1, :1.00g T, .n*_ft"-)fT",\ ( t \ ro* = t-r"J tiJ t' : Iorrru1 (l'008)(288): r672K 1, : 1.008 T1 : 1.008 (2SA;:290.3"K 9 : cp (f., - 1, ) = 1005 (1672 -290.3) = 1.389 x l0o ioul Pz : P* = p110.273 = I atm 2.273 Tz = T* : T1/o.2o6u : #= h39aB.. 3.9 I slug air + 0.06 slug fuel: 1.06 slug of mixture = 424loR 0.044.5 x los\ n: # :2.547.x107 ft-lb per slug of mixture c.= A -Q.4)(1716):6006 ftlb ' y-l 0.4 slugoR .n ." q -2.547 xl07 'or-to,-%- 6006 From Table A.1: At M1 : 0.2, 4, = Tr : 1.008 1, = 1.008 (1000): l008oR 1, = 4241 + To,:4241+ 1008 :5249"R FromTable A.3: For M1 :0.2, prlp* :2.273, TrlTo* :0.2066, T",flon = g,l

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    To* : T",10.1736 : 1008/0.1 736 : 5S06R l, [o* : 5249/5806 = 0.904 From Table A.3: For t", /To* :0.904, (by interpolation) N{r : 0594, pz/p* : 1.433,T21T* = 0.9895 n,=;?f pr:(1 *r(#) (10):tu0;d t, : -ftT 11 = (o.e8er, (#J (rooo) =F7sed 3.10 4, =.To* : 5806'R (from prob. 3.9) e:cp (T", - T",):6006 (5306- 1003)=2.88x lOiftlb/slug Let F : fueVair ratio by mass. o : lA (4'5 x lot) :2.88 x ro7 - (1+ FA) FA= 2.88 x 107 i 4.5 x 108 -2.88 x 10?) = 0.0684 0.03 (4.5 x 108) 3.11 q: : 1.311 x 107 ft lb per slug of mixture 1.311 x 10? 6006 :2182oR 1.03 q co 4, -1, = q, _ T", T* q* However, _ 2182 T* ^o 1, :.To* for choked flow

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    Also, H= l*r:l (sincetr4z:1) T22 To* : l2Tz: 1.2 (4800):5760"R. ,' L, - 2182 T" * 5760 :0.621 T", q* From Table A.3: For + : 0.6212, 3.12 From Table A.4: For M2 : 0.5, ry = \.069, pzlp* = 2.l3gand T2lT* : 1.143. 4L, * = 4L=, * * *= 1.069 + 4(0.q0-t(40) ! or.o, DDD From Table A.4: For 4L' * D = 41.069, m= 0.12i TrlT* : l.l96,prlp* : 8.675 n,= fr # ,,:(8.67s)(#3J (r arn) =Eoo ad 3.13 From Table A.4: = 2.63V, AgT TF For Mr :2.5.'iLtJr D :0.432, L= 0.2921. T' -P* 'T'F = 0.5333, po, / po * From Table A.l: For M1: 2:5,!"'- : 17.0g pr

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    4Lrn - T *flo=0.432*4ff =0.132 From Table A.4: For Y :0.132, M; = l-4q ,pzlp* :0.6125,T21T*:0.8314, po, / po *- r.169 * : ;i# p,: (0.6t2r, (#) (o s):m48 d ,,= h f t,=(0.8314)(#)(5zo;= Fro.FE po, = # H f n' = rr.rur>ffi(r7.0e)(0.s) :ltrs8;d 3.14 -L= 10. Also, p1 = 100 atm. Henc e,p2:10 ugn. If the exit is choked, the4p* =p2= Pz/ 10 atm. Thus, ptlp* :100/10: 10. From Table A.4, for p1lp* = 10, 4L, * I :55.83 D D=4inches:0.333ft. 55.83D a.l- = -= '4f , ss.83(0.333) = FgoE 4(0.00t 3.15 Starting with Eq. (3.95), repeated below: dp + pu du: - (ll2 p*, T and noting that ( pr'\ t' pu2: (rp) | \YP) a- (3.es)

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    Eq. (3.95) becomes dp *M du _ - A,I' flgl) p u 2 (D/ Fromp: pRT dp do dT ppT From pu: const dp:-du pu Co ining (A2) and (A3) dp_dT-du pTu Substitute (Aa) into (Al). r dr*(yMz_l)!* -tM' Tu2 From the adiabatic energy equation fr+ { =const 2 dh+udu:co dT*udu: A- dT*udu= 0 y-l -l-g.4 du I dr *14, $ =o y-I T /RTu y-l T u (Al) (42) (A3) (As) t: .f:.. .i of, dr '')M'jg T:-(T-r' u From the definition of Mach number

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    ,]: a2 M2: M2 yRT Thus, dr:M*1dT u M 2 T (A7) Substitute (.4.6) into (A7) du - dM - 1 rn-tt M2 iu u M 2" u ^du JOIVe IOr _. u +:# u*+(y-r)rvr't-' Substitute (,4.6) and (A8) into (A5) -(y-r) M' # U * +(y-r) att-,+ (TM, - t, g tl* :(y-r) rvr,t-, tM'z- |-Tr,/P+Mrl #tt * )<r-rlrl-'= ry (atr.) M--t) # u * +(y-r) M,r-' = - ry (9 or, finally + #(1 -M'?) # tt * j tr-r)rur't'' which is Eq. (3.96). (A8) g = cp (To,z - To,r)

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    AtM =2.5,from Table a.3, lU- = 0.7101. 'Tt€ Since cp To,t is the total enthalpy of the gas entering the duct, and q is given as 30%oof this total enthalpy, we have q =0.3 toI,, Hence k : (l.3xo.7ro r): 0.e23 T* Flom Table A.3, for + : O.g23,we have tot,, 0.3: l'' T* ;

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    vlz 4.1 I a1 :1ffi: {14rc(520t &r: 1118 fl/sec M1 = v1la1 : 335511118 : 3.0 Mo, : M1 sin P :3.0 sin (35)" = 1.72 From Table A.l. Pz/pr:3.285 Tzttr= 1.473 : Mn, :0.6355 ? P2 : 3.285 (2000) :@O tAlfr T2: 1.473 (5201:@ M,, : M1 cos p :3 cos (35o1: 2.457 w1:2.457 (1 I 18) = 2747 fVsec : w2 a2: JIRT2 = JQW =1357.3fflsec. u2 : Mo, a2= 0.6355 (1357.3):862.6 ff/sec Tan (F - 0): wlwz:862.612747 = 0.314 p-0=17.4 ,, , .. e = p - 17.4:35 -17.4=

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    vr= rli] **r' = @:Es79m;d 4.2 Mn= 2 From the O-B-M diagram: F :39.3o Mn, :Mr sin B =2 sin 39.3" = I.2668 From Table A.2: po, lpo, :FW 4.3 F1o- the O-B-M diagram, at M1 : 3, 0r* : 34.1o and p = 66o. Hence, the wedge can be no larger than a half-angle of 34.1e. : i Mn, = Mr sin F = (3.0) sin 66-o =2.74 From Table A.2: pzlpr:8.592 pr = 1.01 x 10s N/mz (standifd sea level pressure) pz= bp:8.592(1.01 x 10s1= 8.678 x lgs + pr m' This is the maximum pressure. Any higher preslure would conespond to a wedge half-angle > 0,or., the shock would be detached. 4.4 P' - 5. From Table A.2, Pr Mn, 2.2 sinP= : j'' : =, p=38.94o ' Mr 3.5 Yn, = 2.2, Mor: M1 sin p, Thus, From the O:B-M diagram, fo{ M1'.- 3,5

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    4.5 From the O-B-M diagram,lB :3S.1 Mn, : M1 sin B : (3.0) sin 3go = l.g5 From Table A.2, for Mn, = 1.85; pz/pr = 3.g26,T2/Tr : l.56gand M". : 0.6057 P2 : 3.826 p1 : 3.826 efi6)= 18096 h/f, Tz = 1.569 Tr : 1.569 (5te; =@ v,:ffi=ffi:lr-td 4.6 0 -_ __ : l5z=2oo M, = 9.6 Cr;,Q From the 0-p-M diagram, gt:34 Mo, : Mr sin 34o = (3.6X0.559)=2.01 From Table A.2: M,. :0.5752 M2= . y:"' :: = ' 0'5752 :2.378x2.38 - sn(p - 0) sin(34 - 20) F.q-* e-B-M diagram, for IvIz = 2.38and 02 : 20", p1-=, j'\, .

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    4.7 From the O-p-M diagram: For Mr = 2.g and 0 : 30o, 0r = I lo. Mn, Mr sin 30o : 2.9 (0.5) _ 1.4 From Table A.2: Mn,:1.4:pz/pr=2.12,T2/Tr= 1.255,po, /po, :0.95g2,Mn,: 0.7297 M- n.ra^1 1' Itdz: : ; :"' :. - -0"739-7 :2.272 sin(p _ 0) sin(30_ I l) From e-p-M diagram: For tr42 = 2.272and 02 : I l o, pz = 35.5o. For the reflected shock. designate the upstream normal Mach number as Mn, ,. Mn, ' : I\42 sin p2:2.272 sin (35.5") = 1.319 From Table A'2'for Mn- '= r.3r9;ptlpz:1.g66, Tzfiz= r.204, po,rpo, =0.97Sg,Mo.: 0.776 vr,:ffi=*ffi:lr.8, o, : 33pr : (1 .866)e.r2)(r) = F.e56;d Pz Pt TT t, = if rr: (t.2o4xl.25s)(3oo):F53JE '2 rl

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    From Table A.l for Mr = 2.g, p o,/p1= 27.14 Po' - Po' Po' Po' - Po, p; p, :(o'9758)(o'ss82)Q7.l4xl) wherefotM: = l.g7 t 0",= 30, = (6.398)(3.956) = 25.t a,,rr - p:- Clean enough within roundoff errors. Po, Es3-sud As a check, po, Po,/Ps can be calculated more directly from Table A.l = 6.398 4.8 (a) Ir rsr.s. Pr From Table A.2 for ]Wn, : 2.57: Po, = 0.4715*O Mo, = 0.5065. por From the O-p-M diagram, for Mr = 4 and,p = 40: 0 = 26.3o yr:-r,.-- 0.5065 - Sin(F - e) Si{A- 26,, = 2.139 From Table A.2,forMr =4.0, *,=0.13gg. From TableA.l, forMr:4.0, Hence, Po, = L f o, = (o.l38go(rsl.8Xl) Mn, = Mr sin F :4 Sin 40o :2.57 Po, (b)

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    From Table A.2, forNh: 2.139, no' : 0.6557 po. p", : + I 9nt : (0.6557x0.47r5xr5r.sxr) - Po, Po, Pr Note that the total pressure in case (b) is higher than case (a), indicating a more efficient shock compression to subsonic flow for case (b). The upstream total pressure is po,: (l5l.8xl) = 151.8 atrn. For case (a) the loss is total pressure = 151.8 -2r.07 atrn: 130.7 atut. For case (b) the loss in total presstre: 151.8 - 46.93 = 104.9 atm. Hence, case (b) experiences a smaller los! in po, ?nd therefore is more efficient. This is ,f generally true. Oblique shock inlets on jet engines are more effective devices than the older normal shock inlets. 4.9 From the O-B-M diagram, and Figure 4.22; For 0z= 20? and Mr : 3, g = 37 .8o. Mn,=M1 :sinP = (3 ) sin(37.8) = 1.839; P' - 3.783 Pr Mo, :0.607g; Mr: ,, Y,"', : o='90.t* - = = l.gg S:u;r(6-q Sin(37.s-20) For 0g : l5o and Mr :3, F:32.2 Mn,: Mr sin B: (3) sin(32.2): 1.60; b :2.82 Pr Mo, = 0.6684;M3: Mo' slun// - e) _ 0.6684 =2_26 Sin(322- It .:.. ', l'.. ,'j',.,..i.. :'. -,:;, ';;:- -':- :..:.':.'

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    For the upstream flow represented following calculations : by region 2, plot a press're deflection diagram from the 0t' p M"" =M2sinp Pt' Pz 9 = Pa' Pz o:02-oa, Pr Pz Pt 30 JJ.J 37.2 4t.6 46.7 53.5 For the upstrOam flow following calculations : I 1.09 1.20 1.32 t.45 1.60 M." =lrzl3sinp 3.783 4.6r 5.72 7.06 8.65 10.67 &- = &-&- o: oa_03 Pr Pr Pr 20 t6 l2 8 4 0 0 4 8 t2 t6 20 I 1.2r9 1.513 1.866 2.286 2.820 represeflted by region 3, plot a presswe deflection diagram from the 0+ p P+ Pz 26 29 JJ 36.8 4t.5 46.8 53.7 I 1.096 r.23 1.35 1.50 1.65 1.82 I r.23 1.598 t.96 2.458 3.01 3.698 2.82 3.47 4.51 5.53 6.93 8.49 10.43 -15 -l I -7 a -J I 5 9 0 4 8 t2 t6 20 24

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