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

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)

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

1.7
(a) dp = -pVdV
and dp: p t dp, or
Combining:
dp
, _L=_pVdV
pr
dp:-"p2vdv
dp4 =-tpVdVp
dP =-rpv2 dV
Vp
(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

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)flS
(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) un\rs
Evaluating the surface integral in Equation (6), we

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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 nfzdyS
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

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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.

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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.

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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:

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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,

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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

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vlz4.1
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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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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Solution Manual For Modern Compressible Flow: With Historical Perspective, 3rd Edition

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