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
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)
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
(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
Loading page 4...
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
(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
Loading page 5...
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
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
Loading page 6...
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
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
Loading page 7...
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)
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)
Loading page 8...
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 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)
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 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)
Loading page 9...
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 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)
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 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)
Loading page 10...
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:
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:
Loading page 11...
''. 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
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
Loading page 12...
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./
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./
Loading page 13...
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
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
Loading page 14...
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
(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
Loading page 15...
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
(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
Loading page 16...
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:
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:
Loading page 17...
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
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
Loading page 18...
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
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
Loading page 19...
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
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
Loading page 20...
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)
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)
Loading page 21...
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
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
Loading page 22...
,]: 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
^duJOIVe 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)
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
^duJOIVe 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)
Loading page 23...
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*
;
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*
;
Loading page 24...
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=
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=
Loading page 25...
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
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
Loading page 26...
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'\, .
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'\, .
Loading page 27...
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;dPz Pt
TT
t, = if rr: (t.2o4xl.25s)(3oo):F53JE
'2 rl
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;dPz Pt
TT
t, = if rr: (t.2o4xl.25s)(3oo):F53JE
'2 rl
Loading page 28...
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.g7t
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)
Po' - Po' Po' Po'
- Po, p; p, :(o'9758)(o'ss82)Q7.l4xl)
wherefotM: = l.g7t
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)
Loading page 29...
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..
:'. -,:;,
';;:- -':- :..:.':.'
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..
:'. -,:;,
';;:- -':- :..:.':.'
Loading page 30...
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
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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Subject
Mechanical Engineering