Resolução do Capítulo 12 - Franco Brunetti

Prévia do material em texto

Capítulo 12 
 
ESCOAMENTO COMPRESSÍVEL 
 
 
Neste capítulo a Mecânica dos Fluidos funde-se com a Termodinâmica, devido à importância 
que os fenômenos térmicos adquirem. Por causa disso, a primeira parte do capítulo destina-se 
a uma compatibilização da nomenclatura e à introdução de conceitos que não haviam sido 
utilizados até este momento por estarem ligados aos efeitos térmicos. 
Nas aplicações é mais fácil trabalhar com energias por unidade de massa e não por unidade de 
peso, fazendo-se as devidas transformações. 
Esse assunto é extremamente vasto e complexo e o leitor que desejar um maior 
aprofundamento de seus conhecimentos deverá consultar livros dedicados apenas a ele. 
O objetivo deste capítulo consiste em alertar o leitor sobre as complicações advindas da 
variação da massa específica ao longo do escoamento e chamar a atenção para os fenômenos 
provocados por essa característica. Destacam-se ainda as mudanças de comportamento no 
escoamento supersônico, a existência de uma vazão em massa máxima nos condutos e o 
aparecimento da onda de choque. Todos esses fenômenos, abordados dentro de hipóteses 
simplificadoras, poderão orientar o leitor quando estiver lidando com algum problema prático 
sobre o assunto. 
 
Exercício 12.1 
 
( )
( )
( ) 3
5
2
2
2
p
v
p
v
pp
m
kg226,5
27395260
105
RT
p
)e
kJ627J688.62610956,9218TmcH)d
kJ450J888.44910956,6618TmcI)c
K.kg
J6,661
393,1
6,921
k
c
c)b
K.kg
J260
393,1
1393,16,921
k
1kcR
1k
kRc)a
=+×
×==ρ
==−××=Δ=Δ
==−××=Δ=Δ
===
=−×=−=→−=
 
 
Exercício 12.2 
 
( )
MJ21154,1UkTmcH)d
MJ151020460717568,47TmcU
kg568,47
293287
2102
RT
Vp
m)c
C460K733
2
5293T
T
T
p
p
mRTVp
mRTVp
)b
K.kg
J717
14,1
287
1k
Rc)a
K.kg
J287
29
315.8R
p
6
v
6
1
11
o
2
1
2
1
2
11
22
v
=×=Δ=Δ=Δ
=×−××=Δ=Δ
=×
××==
==×=⇒=→
⎭⎬
⎫
=
=
=−=−=
==
−
 
 
Exercício 12.3 
 
( )
( )
kg
kJ3,93
s
m275.93201505,717Tcu
K.s
m5,717
14,1
287
1k
Rc
K.s
m287
29
315.8
M
RR)b
C150K423
4,0
27320
4,0
T
T4,0
T
T
)abs(kPa371
4,0
103p
4,0
p
pV4,0V
V
V
p
p
)a
2
2
v
2
2
v
2
2
mol
o
4,01k
1
2
1k
2
1
4,12k
1
212
k
1
2
2
1
==−×=Δ=Δ
=−=−=
===
==+==→=
==→=→=→⎟⎟⎠
⎞
⎜⎜⎝
⎛=
−
−
 
 
( )
kg
kJ6,130
000.1
1201505,004.1h
K.kg
kJ5,004.1
14,1
2874,1
1k
kRc
Tch)c
p
p
=×−×=Δ
=−
×=−=
Δ=Δ
 
 
Exercício 12.4 
 
K.kg
J562
500
500ln3,461
573
423ln872.1
p
p
lnR
T
T
lncs
1
2
1
2
p −=×−×=−=Δ 
 
Exercício 12.5 
 
29,0
342
100
c
v
s
m3422932864,1kRTc
s
m100
6,3
360v
===Μ
=××==
==
 
 
Exercício 12.6 
 
kPa9,5)abs(kPa1,94
6,0
2
14,11
120
2
1k1
p
p
2
1k1
p
p
14,1
4,1
21k
k
2
0
1k
k
20
−==
⎟⎠
⎞⎜⎝
⎛ ×−+
=
⎟⎠
⎞⎜⎝
⎛ Μ−+
=
⎟⎠
⎞⎜⎝
⎛ Μ−+=
−−
−
 
 
s
m2093,3012874,16,0kRTv
m
kg088,1
3,301287
101,94
RT
p
C3,28K3,301
6,0
2
14,11
323
2
1k1
T
T
3
3
o
22
0
=×××=Μ=
=×
×==ρ
===−+
=
Μ−+
=
 
 
Exercício 12.7 
 
( )
%45,0100
3,67
6,673,67êrro
6,6702,0000.136
19,1
2v
m
kg19,1
293287
10100
RT
p
h2
hg2
ppg2v
pp
g2
v
ívelIncompress
s
m3,672932874,1196,0kRTMv
196,01
100
72,102
14,1
2M
1
p
p
1k
2MM
k
1k1
p
p
)abs(kPa72,10210072,2p
kPa72,2Pa272002,0000.136hp
1
3
3
Hg
Hg
121
21
2
1
4,1
14,1
k
1k
01k
k
20
0
Hg0
abs
=×−=
=××=
=×
×==ρ
γρ=γ
γ=−γ=→γ=γ+→
=××==
=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−⎟⎠
⎞⎜⎝
⎛
−=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−⎟⎟⎠
⎞
⎜⎜⎝
⎛
−=⇒⎟⎠
⎞⎜⎝
⎛ −+=
=+=
==×=γ=
−
−
−
 
 
Exercício 12.8 
 
mm970m97,011
293287
400
4,12
14,1
000.136
10100h
11
RT
v
k2
1kph1
p
h
1RT
1k
k2v
14,1
4,1
23
1k
k
2
m
k
1k
m2
==
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−⎟⎟⎠
⎞
⎜⎜⎝
⎛ +×××
−×=
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−⎟⎟⎠
⎞
⎜⎜⎝
⎛ +−γ=⇒⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−⎟⎟⎠
⎞
⎜⎜⎝
⎛ γ+−=
−
−
−
 
 
 
%8,27100
970
970700
mm700m7,0
000.136
120.95
g
pp
h
Pa120.95
2
400189,1
2
vpp
Hg
0
22
0
=×−=ε
===ρ
−=
=×=ρ=−
 
 
 
Exercício 12.9 
 
)abs(Pa102,35,0000.13610hpp)a 45Hg12 ×=×−=γ−= 
[ ] [ ][ ] [ ]
( )
%113100
47,0
1111
Q
Q
1
Q
QQ
êrro
Q
Q)c
s
kg518,0
5,01
32,012
293287
1047,095,0
A
A
1
p
p
12
A
RT
pCQ)b
47,0
32,0132,05,01
5,0132,015,332,0
5,0
50
25
A
A
e32,0
10
102,3
p
p
p
p
1
p
p
A
A
1
A
A
1
p
p
1
1k
k
p
p
m
invm
m
incmmm
incm
2
5
2
1
2
1
2
2
1
Dm
429,12
2286,0
714,0
1
2
5
4
1
2
1
2k
2
1
2
2
1
2
2
1
2k
1k
1
2
k
1
1
2
=×⎟⎠
⎞⎜⎝
⎛ −=φ−=−=
−=⇒φ=
=−
−×
×××=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
φ=
=−××−
−×−×=φ
===×=
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−−
⎟⎟⎠
⎞
⎜⎜⎝
⎛=φ
−
 
 
Exercício 12.10 
 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
⎟⎟⎠
⎞
⎜⎜⎝
⎛−−
⎟⎟⎠
⎞
⎜⎜⎝
⎛=φ
−
1
2k
2
1
2
2
1
2
2
1
2k
1k
1
2
k
1
1
2
p
p
1
p
p
A
A
1
A
A
1
p
p
1
1k
k
p
p
 
kPa149101,5000.10200hpp 3Hg12 =××−=γ−= − 
 
( )
( )
( )
s
kg151,1
444,01
745,012
4
1,0
3662077
10200844,095,0Q
A
A
1
p
p
12
A
RT
pCQ
K.kg
J2077200.5
665,1
1665,1c
k
1kR
1k
kRc
844,0
745,01745,0444,01
444,01745,01
1665,1
665,1
745,0
444,0
15
10
D
D
A
A
;745,0
200
149
p
p
2
23
m
2
1
2
1
2
2
1
Dm
pp
665,1
2
2
2665,1
1665,1
665,1
1
22
1
2
1
2
1
2
=−
−×××π××
×××=
⎟⎟⎠
⎞
⎜⎜⎝
⎛−
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
φ=
=×−=−=⇒−=
=
−×⎟⎟⎠
⎞
⎜⎜⎝
⎛
×−
−×⎟⎟⎠
⎞
⎜⎜⎝
⎛
−×−
×=φ
=⎟⎠
⎞⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛===
−
 
 
 
Exercício 12.11 
 
( )
( )
( )
( )
%4,5100
1310
12391310êrro)c
s
m1310
5,1
15,1
405,12
111015,1
0651.0
2v
p
pp
k2
11pp2v)b
s
m12391015,1
0651,0
2v
m
kg0651,0
5504189
105,1
K.s
m4189532.14
405,1
1405,1c
k
1kR
RT
p
pp2v)a
5
c
0
c0
c0
0
c
5
c
3
5
0
2
2
p
0
0
0
c0c
=×−=
=⎟⎠
⎞⎜⎝
⎛ −
×+××−×=
−+−ρ=
=×−=
=×
×=ρ
=×−=−=→=ρ
−ρ=
 
 
 
 
 
Exercício 12.12 
 
s10
2,1477,591
440.7
vv
st)c
m440.7502,147800.14s
s50
296
800.14
c
Rt
5,0
296
2,147
c
v
s
m2,147
6,3
530v
2
296
7,591
c
v
s
m7,591
6,3
130.2v
s
m2962182874,1kRTc)b
21
1
2
22
1
11
=+=+
Δ=
=×−=Δ
===
===Μ⇒==
===Μ⇒==
=××==
 
 
 
Exercício 12.13 
 
s
m8162952574,137,2kRTMv
37,2
25sen
1M
M
1
2
sen
o
=×××==
==→=α
 
 
Exercício 12.14 
 
3*
*
*
133,1
33,1
5
1k
k
0*1k
k
*
0
**
0*2
*
0
m
kg338,0
346462
036.54
RT
p
)abs(kPa54)abs(Pa036.54
2
133,11
10
2
1k1
p
p
2
1k1
p
p
s
m46134646233,1kRTv
K346
2
133,11
403
2
1k1
T
T
2
1k1
T
T
=×==ρ
==
⎟⎠
⎞⎜⎝⎛ −+
=
⎟⎠
⎞⎜⎝
⎛ −+
=→⎟⎠
⎞⎜⎝
⎛ −+=
=××==
=−+
=−+
=→Μ−+=
−−
−
 
 
Exercício 12.15 
 
 
1.Tab1Ms →= 
 
 
 
K5,4775738333,0T8333,0
T
T
)abs(MPa5283,015283,0p5283,0
p
p
s
0
s
s
0
s
=×=→=
=×=→=
 
s
kg338,0102438856,3Q
s
m4385,4772874,11kRTMv
AvQ
m
kg856,3
5,477287
105283,0
RT
p
4
m
sss
sssm
3
6
s
s
s
=×××=
=×××==
ρ=
=×
×==ρ
−
 
 
Exercício 12.16 
 
2
1*
1
5
0
1
5
1atmHg1
cm1,2015340,1A340,1
A
A
8434,0
102
680.168
p
p
)abs(Pa680.168505,0000.13610pphp
=×=⇒=→=×=
=×+=⇒=γ−
 
 
Exercício 12.17 
 
a) T0 = 373 K = 100 oC 
bloqueadoestánão833,0
102,1
10
p
p
)b
5
5
0
s →=
×
=
 
1.Tab8333,0
p
p
0
s →= 
 
 
s
kg183,010196984,0AvQ
s
m1963542874,152,0kRTMv
m
kg984,0
354287
10
RT
p
3
sssm
sss
3
5
s
s
s
=××=ρ=
=×××==
=×==ρ
−
 
c) A mesma 
 
1.Tab3,0M)d →= 
 
 
23m
5
cm6,151056,1
073,1115
193,0
v
Q
A
s
m1153662874,13,0kRTMv
073,1
366287
10127,1
RT
p
=×=×=ρ=
=×××==
=×
×==ρ
−
 
K3543739487,0T9487,0
T
T
52,0M
s
0
s
s
=×=→=
=
 
)abs(Pa10127,1102,19395,0p9395,0
p
p
K3663739823,0T9823,0
T
T
55
0
0
×=××=→=
=×=→=
 
e) Ms=1 1.Tab→
⎪⎪⎩
⎪⎪⎨
⎧
=×=→=
×==→=
K3113738333,0T8333,0
T
T
)abs(Pa10893,1
5283,0
10p5283,0
p
p
s
0
s
5
5
0
0
s
 
 
 
s
kg396,0105,35312,1AvQ
s
,m5,3533112874,11kRTMv
m
kg12,1
311287
10
RT
p
3
sssm
sss
3
5
s
s
s
=××=ρ=
=×××==
=×==ρ
−
 
 
Exercício 12.18 
 
( ) mm64m064,0
000.136
1029564,01h
9564,0
p
p
erpolandoint38,2
1026,1
103
A
A
p
p
p
1
hphp)e
m1026,1
59,1
102
59,1
A
A)d
s
m7,1367,2902874,14,0kRTv
59,1
A
A
4,0
969,0
300
7,290
T
T
)c
)abs(Pa102p)b
)abs(Pa102pp
K300TT)a
5
0
A
3
3
*
A
Hg
0
0
A
AHg0
23
3
B*
2BB
*
B
B
0
2
5
A2M
5
2M0
10
==××−=
⎩⎨
⎧ =→→=×
×=
γ
⎟⎟⎠
⎞
⎜⎜⎝
⎛ −
=⇒=γ−
×=×==
=×××=Μ=
⎪⎩
⎪⎨
⎧
=
=Μ
→==
×=
×==
==
−
−
−−
 
 
Exercício 12.19 
 
Fixando o sistema de referência no conduto, isto é, no avião: v1 = 180 m/s 
→→===
=××==
1.Tab55,0
5,327
180
c
v
M
s
m5,3272672874,1kRTc
1
1
1
1
 
 
 
 
 
1.Tab8,0M 2 →= 
 
 
 
s
m2542512874,18,0kRTMv 222 =×××== 
 
Exercício 12.20 
 
s
m1,933452874,125,0kRTv
)abs(MPa55,0575,0957,0p957,0
p
p
K345349988,0T988,0
T
T
25,0
39,2009,137,2
A
A
A
A
A
A
009,1
A
A
)abs(MPa575,0
5913,0
34,0p5913,0
p
p
K349
8606,0
300T8606,0
T
T
9,0
3002874,1
312
kRT
v
222
2
0
2
2
0
2
2
*
1
1
2
*
2
*
1
0
0
1
0
0
1
1
1
1
=×××=Μ=
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=×=→=
=×=→=
=Μ
→=×=×=
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
=
==→=
==→=
→=××==Μ
 
 
 
 
Exercício 12.21 
 
→== 2
10
20
A
A
*
e 
 
rpm49260
22
103
R2
v
n
s
m1032932874,13,0kRTMv
'
e''
e
'
e =××π×=π=→=××== 
rpm360460
22
755
R2
v
n
s
m7552932874,12,2kRTMv
''
e''''
e
''
e =××π×=π=→=××== 
K283
9449,0
267T9449,0
T
T
)abs(kPa128
8201,0
100p8201,0
p
p
0
0
1
0
0
1
==→=
==→=
 
K2518865,0283T8865,0
T
T
)abs(kPa80656,0128p656,0
p
p
2
0
2
2
0
2
=×=→=
=×=→=
 
2,2M
3,0M
''
e
'
e
=
=
 
 
 
 
 
 
 
Exercício 12.22 
 
22
ss
m
s
3
5
s
s
s
sss
s
0
s
s
6
5
0
s
2
G
G
22
GG
m
G
GGG
3
5
G
G
G
G
0
G
56
G
0
G
G
cm413m0413,0
794037,1
34
v
Q
A
m
kg037,1
336287
10
RT
p
)d
s
m7943362874,116,2kRTv
K3366505173,0T5173,0
T
T
16,2
1,0
10
10
p
p
)c
m165,010147,24
A4
Dm10147,2
467394,3
34
v
Q
A
s
m4675422874,11kRTv)b
m
kg394,3
542287
1028,5
RT
p
K5426508333,0T8333,0
T
T
)abs(Pa1028,5105283,0p5283,0
p
p
1)a
==×=ρ=
=×==ρ
=×××=Μ=
⎪⎩
⎪⎨
⎧
=×=→=
=Μ
==
=π
××=π=⇒×=×=ρ=
=×××=Μ=
=×
×==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=→=
×=×=→=
=Μ
−−
 
 
Exercício 12.23 
 
→=
=
1M
K310T)a
G
0
 
 
 
8333,0
T
T
5283,0
p
p
0
G
0
G
=
=
 
.mesmoO)c
K2583108333,0T08333T
)abs(kPa6,1067,2015283,0p5283,0p)b
m
kg27,2
310287
107,201
RT
p
)abs(kPa7,201)abs(kPa695.201p
200.95p5283,0p200.95pp
200.95pp
7,0000.136phpp
0G
0G
3
3
0
0
0
0
00G0
G0
GHgG0
=×==
=×==
=×
×==ρ
==
=−→=−
+=
×+=γ+=
 
 
Exercício 12.24 
 
2s*
223
ss
m
s
3
3
s
s
s
xSsss
*
s
s
0
s
s
0
s
smxS
smemsmiiiS
cm3,17
999,3
3,69
999,3
A
A)c
cm3,69m1093,6
864.1348,0
5,4
v
Q
A
m
kg348,0
000.1287
10100
RT
p
)b
N838818645,4F
s
m1864000.12874,194,2kRTv
999,3
A
A
K000.1730.23665,0T3665,0
T
T
94,2
0294,0
400.3
100
p
p
vQF
vQvQvQnApF)a
===
=×=×=ρ=
=×
×==ρ
=×−=⇒=×××=Μ=
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
=
=×=→=
=Μ
==
−=
−=+−=
−
∑ rrrrr
 
 
Exercício 12.25 
 
→== 76,1
1,0
176,0
A
A
*
3
 
 
 
 
 
)abs(MPa08,068,01164,0p1164,0
p
p
06,2M
)abs(MPa622,068,09143,0p9143,0
p
p
36,0M
''
3
0
''
3
''
3
'
3
0
'
3
'
3
=×=→=
=
=×=→=
=
s
kg1621,05,31021,5Q
s
m5,3102402874,11kRTMv
m
kg21,5
240287
10359,0
RT
p
)abs(MPa359,068,05283,0p5283,0p
K2402888333,0T8333,0T
vAQ
m
GGG
3
6
G
G
G
0G
0G
m
=××=
=×××==
=×
×==ρ
=×==
=×==
ρ=
 
 
 
 
 
 
Exercício 12.26 
 
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
=×=⇒=
=×=⇒=
=×=⇒=
→=Μ
=×=×=ρ=
=×××=Μ=
=×
×==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=⇒=
=×=⇒=
→=Μ
−
2
s*
s
s
0
s
s
0
s
s
222
GG
m
G
GGG
3
6
G
G
G
G
0
G
G
0
G
G
cm419251668,1A668,1
A
A
)abs(MPa0128,01,01278,0p1278,0
p
p
K1442605556,0T5556,0
T
T
2
cm251m1051,2
295815,0
3,6
v
Q
A
s
m2952172874,11kRTv
m
kg815,0
217287
10053,0
RT
p
K2172608333,0T8333,0
T
T
)abs(MPa053,01,05283,0p5283,0
p
p
1
 
 
 
Exercício 12.27 
 
→== 089,1
293
319
A
A
*
s
 
 
 
 
 
)abs(MPa516,051,13417,0p3417,0
p
p
7358,0
T
T
34,1M
s
0
s
0
s
s
=×=→=
=
=
 
 
Na realidade, existe também a solução subsônica, entretanto, com essa velocidade de 
saída, a temperatura seria um valor prático impossível. 
N342.4110319000.2324,0AvF
m
kg324,0
544.5287
10516,0
RT
p
AvF
K544.5
2874,134,1
000.2
kRM
v
TkRTMv
42
s
2
ssS
3
6
s
s
s
s
2
ssS
2
2
2
s
2
s
ssss
x
x
=×××=ρ=
=×
×==ρ
ρ=
=
××
==→=
−
 
 
 
 
 
Exercício 12.28 
 
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
===⇒=
=
=Μ
→=
=××=××=
=⇒=×=×=
=××=××==
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=
=
=
→=Μ
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
=
===⇒=
=Μ
→=Μ
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=
=
=Μ
→=
)abs(kPa1,121
8259,0
100
8259,0
p
p8259,0
p
p
9468,0
T
T
53,0
256,1
A
A
256,1
176,1
1094,135,1
A
A
A
A
A
A
A
A
)c
A075,1A93,0094,1
176,1
1
A
A
A
A
A
A
)b
46,2
272,0
19298,07209,0
p
p
p
p
p
p
p
p
Int)a
094,1
A
A
7209,0
p
p
9108,0
T
T
7011,0
458,2p
p
32,1
T
T
)abs(kPa130
9298,0
1,121
9298,0
p
p9298,0
p
p
7011,0
5,1
176,1
A
A
6897,0
T
T
5,1
272,0
p
p
s
y0
y0
s
0
s
s
*
y
s
x
*
x
*
y
y
*
x
s
*
y
s
*
x
*
y*
y
y
x
*
x
*
y
*
x
x
x0
x0
y0
y0
y
x
y
OC
*
y
y
y0
y
0
y
y
x
y
x
y
y0
x0
x0
y0
y
x
*
x
x
0
x
x
x0
x
 
K317
9468,0
300
9468,0
T
T)e
)abs(kPa130p)d
s
0
x0
===
=
 
 
K.kg
J8,20
458,2
32,1ln004.1
p
p
T
T
lncss)g
A6,213A184161,1Q
s
m1843002874,153,0kRTv
m
kg161,1
300287
10
RT
p
AvQ)f
41,1
14,1
k
1k
x
y
x
y
pxy
ssm
sss
3
5
s
s
s
sssm
=×=
⎟⎠
⎞⎜⎝
⎛
=−
=××=
=×××=Μ=
=×==ρ
ρ=
−−
 
Exercício 12.29 
 
→= 1278,0
p
p
x0
x 
 
 
 
 
 
 
 
→= 2M x 
 
 
 
 
 
 
 
 
 
→= 58,0M y 
 
 
 
 
437,1213,1
688,1
12
A
A
A
A
A
A
A
A
*
y
y
x
*
x
*
x
s
*
y
s =××=××= 
 
 
 
→= 437,1
A
A
*
y
s 
688,1
A
A
5556,0
T
T
2M
*
x
x
0
x
x
=
=
=
 
5,4
p
p
687,1
T
T
7209,0
p
p
5774,0M
x
y
x
y
0
0
y
x
y
=
=
=
=
 
213,1
A
A
7962,0
p
p
9370,0
T
T
*
y
y
0
y
0
y
y
=
=
=
 
)abs(Pa1004,1
8650,0
109,0p8650,0
p
p
9594,0
T
T
46,0M
5
5
0
0
s
0
s
s
y
y
×=×=→=
=
=
 
 
m47,0
1036,1
1084,11028,8pph
)abs(Pa1028,81084,15,4p5,4p
)abs(Pa1084,11044,11278,0p1278,0p
php)b
K313
9594,0
300
9594,0
T
T
)abs(kPa144)abs(Pa1044,1
7209,0
1004,1
7209,0
p
p)a
5
44
Hg
xy
44
xy
45
0x
yHgx
s
0
5
50
0
x
y
x
=
×
×−×=γ
−=
×=××==
×=××==
=γ=
===
=×=×==
 
8650,0
p
p
x
7962,0
p
p
x
5283,0
p
p
x)c
y0
s
s
y0
y
yOC
x0
G
G
=→
=→
=→
 
 
Exercício 12.30 
 
⎩⎨
⎧
=Μ ′′
=Μ′→=×=
=Μ→=×=
−
−
2,2
3,0
2
10
102
A
A
76,0756,0
10
1056,7
p
p
s
3
3
*
s
s6
5
0
s
 
Sim. Para ser totalmente subsônico Ms ≤ 0,3. Como Ms = 0,76, o escoamento passou para 
supersônico e posteriormente para subsônico através de uma onda de choque. 
 
Exercício 12.31 
 
→=×= −
−
4,1
10
104,1
A
A
)a
3
3
*
x
c 
 
 
 
 
 
 
 
 
 
 
 
 
1850,0
p
p
6175,0
T
T
76,1M
x0
x
0
x
x
=
=
=
 
447,3
p
p
502,1
T
T
8302,0
p
p
6257,0M
x
y
x
y
x0
y0
y
=
=
=
=
 →= 76,1M x
 
 
 
 
 
→= 6257,0M y 
 
 
 
 
 
 
 
→=
×
×= −
−
667,1
102,1
102
A
A
3
3
*
y
s 
 
 
 ( )
( )
s
kg78,110217122,5AvQ
s
m1715052874,138,0kRTMv
m
kg22,5
505287
1056,7
T
p
)c
s
m4224836257,032176,12874,1v
K4835209286,0T9286,0T
K3215206175,0T6175,0T
TMTMkRkRTMkRTMvvv)b
3
sssm
sss
3
5
ss
s
s
0y
0x
yyxxyyxxyx
=×××=ρ=
=×××==
=×
×=ρ=ρ
=×−×××=Δ
=×==
=×==
−=−=−=Δ
−
 
Exercício 12.32 
 
Ver o exercício 12.31 
 
 
 
Exercício 12.33 
 
 
 
 
 
 
 
 
 
 
23
3y*
y*
y
y
0
y
0
y
m102,1
166,1
104,1
166,1
A
A166,1
A
A
7716,0
p
p
9286,0
T
T
y
−− ×=×==→=
=
=
 
)abs(Pa1035,8
9052,0
1052,7
9052,0
p
p9052,0
p
p
K520
9719,0
505
9719,0
T
T)a9719,0
T
T
38,0M
5
5
s
y0
y0
s
s
0
0
s
s
×=×==→=
===→=
=
 
24
4
s*
y*
y
s
y0
s
s
m107,33
188,1
1040
188,1
A
A188,1
A
A
784,0
p
p
6,0M
−− ×=×==→=
=
=
 
→== 9324,0
429
400
T
T
y0
s
 
→=
×
×= −
−
128,1
107,33
1038
A
A
4
4
*
y
1 
 
 
 
 
000.136p5283,0p7465,0
)abs(Pa000.136pp
php
xy 00
Gy
yHgG
=−
=−
=γ+
 
→= 66,0M y 
 
 
( )absPa109,010013,18876,0p
)abs(Pa10013,1p
000.136p5283,0p8876,07465,0
66
0
6
0
00
y
x
xx
×=××=
×=
=−×
 
 
Exercício 12.34 
 
⎪⎪
⎪⎪
⎪
⎩
⎪⎪
⎪⎪
⎪
⎨
⎧
=
=
=
=Μ
→=Μ
=×=×=ρ=
=×××=Μ=
=×
×==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=⇒=
=×=⇒=
→=Μ
−
783,3
p
p
562,1
T
T
7947,0
p
p
84,1
606,0)b
cm924m1024,9
422818,2
110
v
Q
A
s
m4224442874,11kRTv
m
kg818,2
444287
10359
RT
p
K4445338333,0T8333,0
T
T
)abs(kPa3596805283,0p5283,0
p
p
1)a
x
y
x
y
x0
y0
x
y
222
GG
m
G
GGG
3
3
G
G
G
G
0
G
G
x0
G
G
 
7465,0
p
p
9199,0
T
T
66,0M
y0
y
0
y
y
=
=
=
 
x
x
y
0y0
0
0
p8876,0p8876,0
p
p
=→=
 
2*
ys*
y
s
s
2y*
y*
y
y
y
2
x*
x
x
x0
x
0
x
x
cm835.1154.159,1A59,1A59,1
A
A
4,0
cm154.1
188,1
1371
188,1
A
A188,1
A
A
606,0)c
cm371.1484,1924A484,1
A
A
1537,0
p
p
5963,0
T
T
84,1
=×=⎪⎩
⎪⎨
⎧ =⇒=→=Μ
===⎪⎩
⎪⎨
⎧ ⇒=→=Μ
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=×=⇒=
=
=
→=Μ
 
 
Exercício 12.35 
 
9325,0
400
373
T
T
0
1 == 
 
 
Na segunda situação 
*
1
A
A
 não muda, mas o escoamento é supersônico. 
 
 
188,1
A
A
*
1 = 
 
 
 
52,1M x = 
 
 
 
CouK91274365TTT oxy =−=−=Δ 
 
Exercício 12.36 
 
24,1
A
A
857,0
429,1
225,1
m
kg429,1
400287
10164
RT
p
)a
*
x
0
3
3
0
0
0
=⎩⎨
⎧→==ρ
ρ
=×
×==ρ
 
 
 
 
188,1
A
A
6,0M
*
1
1
=
=
 
K274400684,0T684,0T684,0
T
T
52,1M
0x
0
x
x
=×==→=
=
 
K365274334,1T334,1T333,1
T
T
6941,0M
xy
x
y
y
=×==→=
=
 
s
m3663332874,11kRTv
K3334008333,0T8333,0
T
T
1)c
s
kg29,410160517519,0AvQ
s
m5172672874,158,1kRTv
m
kg519,0
267287
1040
RT
p
)b
)abs(kPa401642423,0p2423,0p2423,0
p
p
K2674006670,0T6670,0T6670,0
T
T
58,1
24,1
A
A
GGG
G
0
G
G
4
xxxm
xxx
3
3
x
x
x
0x
0
x
0x
0
x
x
*
x
=×××=Μ=
=×=⇒=
⎩⎨
⎧→=Μ
=×××=ρ=
=×××=Μ=
=×
×==ρ
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=×=×=⇒=
=×=×=⇒=
=Μ
→=
−
 
 
 
Exercício 12.37 
 
1,0
101
100
p
p
)a
30
s =
×
= 
 
b) Se a onda de choque está na seção de saída, a montante tem-se a segunda solução 
isoentrópica, que corresponde à solução do item anterior. 
 
 
 
16,2M x = 
 
 
 
Exercício 12.38 
 
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=
=
=Μ
=Μ
→==
=
666,6
p
p
059,2
T
T
521,0
42,2
5283,0
p
p
p
p
pp
x
y
x
y
y
x
x0
y0
x0
G
y0G
 
 
 
K2,1553005173,0T5173,0T5173,0
T
T
16,2M
0s
0
s
s
=×==→=
=
 
K2832,155822,1T822,1T822,1
T
T
5525,0M
xy
x
y
y
=×==→=
=
 
s
m1853162874,1521,0kRTv
K3163339487,0T9487,0
T
T
521,0
yyy
y
0
y
y
=×××=Μ=
=×=⇒=
⎩⎨
⎧→=Μ
 
 
Exercício 12.39 
 
 
 
 
 
 
 
 
2
máx
1
máx
2
máx
1
máx
D
L
f
D
Lf
D
L
f
D
L
f
D
L
f
D
Lf ⎟⎠
⎞⎜⎝
⎛+=⎟⎠
⎞⎜⎝
⎛→⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛= 
 
 
 
 
 
M2 = 0,9 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
32,0M1 = 
 
 
 
 
 
 
 
L=15m (1) (2) 
p2 = 105 Pa(abs) 
T2 = 294 K 
M2 = 0,9
D = 7,5cm 
f = 0,02 
T0 ? 
 p0 ? 
01451,0
D
L
f
)abs(Pa1086,8
129,1
10
129,1
p
p129,1
p
p
K6,284033,1
294
033,1
T
T033,1
T
T
2
máx
4
5
2*
*
2
2*
*
2
=⎟⎠
⎞⎜⎝
⎛
×===→=
===→=
 
0145,401451,0
075,0
1502,0
D
L
f
1
máx =+×=⎟⎠
⎞⎜⎝
⎛ 
)abs(Pa1031086,8389,3p389,3
p
p
K7,3346,284176,1T176,1
T
T
32,0M
54
1*
1
1*
1
1
×=××=→=
=×=→=
=
 
)abs(Pa1022,3
9315,0
103p9315,0
p
p
K342
9799,0
7,334T9799,0
T
T
5
5
0
0
1
0
0
1
×=×=→=
==→=
 
Exercício 12.40 
 
)abs(MPa011,0259,027,0ppp
176,1
T
T
)abs(MPa259,00764,0389,3p389,3
p
p
32,0
224,4
D
L
f
224,4
025,0
2,13008,0299,5
D
L
f
D
L
f
D
L
f
299,5
D
L
f
)abs(MPa0764,0
619,3
27,0
619,3
p
p619,3
p
p
179,1
T
T
3,0
21
*
2
2*
2
2
H
máx
H
máx
1H
máx
2H
máx
H
máx
1*
*
1
*
1
1
=−=−=Δ
⎪⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
=
=×=⇒=
=Μ
→=⎟⎟⎠
⎞
⎜⎜⎝
⎛
=×−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=
===⇒=
=
→=Μ
 
 
Exercício 12.41 
 
5,0M1 = 
 
 
5,0M1 = 
 
 
 
 
1M2 = 
 
 
 
s
m3342772874,11kRTMv 222 =×××== 
 
Exercício 12.42 
 
⎪⎩
⎪⎨
⎧ =⎟⎟⎠
⎞
⎜⎜⎝
⎛→=Μ
=⎪⎩
⎪⎨⎧ ⎟⎟⎠
⎞
⎜⎜⎝
⎛→=Μ
3050,0
D
Lf
2
5222,0
D
Lf
3
2H
máx
2
1H
máx
1
 
 
( )absMPa166,0
843,0
14,0
843,0
p
p843,0
p
p 1
0
0
1
1
1
===→= 
( )absMPa124,0
34,1
166,0
34,1
p
p34,1
p
p
11 0*
0*
0
0 ===→= 
( )
K2773338333,0T8333,0T8333,0
T
T
absMPa065,0124,05283,0p5283,0p5283,0
p
p
02
0
2
*
02*
0
2
=×==→=
=×==→=
 
8,1
012,0
1,02172,0
f
D2172,0
L
2172,03050,05222,0
D
Lf
H
H
máx
=×=×=
=−=⎟⎟⎠
⎞
⎜⎜⎝
⎛
 
 
Exercício 12.43 
 
 
 
 
3M1 = 
 
 
 
 
s
m4204392874,11kRTMv *** =×××== 
 
 
 
 
 
2M2 = 
 
 
 
 
s
m6852922874,12kRTMv 222 =×××== 
 
m17,2
01,0
1,02172,0
f
D2172,0L
2172,03050,0522,0
D
Lf
D
Lf
D
Lf
2,1
2
máx
1
máx2,1
=×==
=−=⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
 
 
Exercício 12.44 
 
K245
504,3
p
p
1775,1
5,288
1775,1
T
T1775,1
T
T
31,0
8,4
025,0
602,0
D
Lf
*
1
1*
*
1
1
1
máx =
⎪⎪
⎪⎪
⎩
⎪⎪
⎪⎪
⎨
⎧
=
==⇒=
=Μ
→=×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
 
 
( )
m22,5
01,0
1,05222,0
f
D5222,0L5222,0
D
Lf
absMPa46,0
2182,0
1,0
2182,0
p
p2182,0
p
p
K439
4286,0
188
4286,0
T
T4286,0
T
T
1máx
1
máx
1*
*
1
1*
*
1
=×==→=⎟⎟⎠
⎞
⎜⎜⎝
⎛
===→=
===→=
 
( )
3050,0
D
Lf
absMPa19,046,04083,0p4083,0p4083,0
p
p
K2924396667,0T6667,0T6667,0
T
T
2
máx
*
2*
2
*
2*
2
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
=×==→=
=×==→=
 
s
kg219,0
4
025,0314422,1
4
DvQ
s
m3142452874,11kRTv
m
kg422,1
245287
10100
RT
p
22
**
máx
***
3
3
*
*
*
=×π××=πρ=
=×××=Μ=
=×
×==ρ
 
 
Exercício 12.45 
 
A leitura do termômetro é 600 K, uma vez que a temperatura de estagnação não se altera. 
 
 
 
→== 2
20
40
A
A
*
3 
 
 
 
 
 
 
→= 2,2M x 
 
 
 
→= 5471,0M y 
 
 
 
→== 59,1
5,31
50
A
A
*
y
4 
 
 
 
 
→= 4,0M 4 
 
 
 
cm8504
A4
D 44 =π
×=π= 
 
 
059,1
08,0
502,0309,2
D
Lf
D
Lf
D
Lf
4
máx
5
máx =×−=−⎟⎟⎠
⎞
⎜⎜⎝
⎛=⎟⎟⎠
⎞
⎜⎜⎝
⎛
 
 
( )absMPa0468,05,009352,0p09352,0p09352,0
p
p
K3056005081,0T5081,0T5081,0
T
T
2,2M
x
x
0x
0
x
0x
0
x
x
=×==→=
=×==→=
=
 
( )absMPa314,05,06281,0p6281,0p6281,0
p
p
5471,0M
xy
x
y
00
0
0
y
=×==→=
=
 
23*
y*
y
3 cm5,31
27,1
40
27,1
A
A27,1
A
A ===→= 
( )absMPa281,0314,08956,0p8956,0p8956,0
p
p
4,0M
y
y
04
0
4
4
=×==→=
=
 
( )
309,2
D
Lf
absMPa197,0
59,1
314,0
59,1
p
p59,1
p
p
4
máx
0*
0*
0
0 yy
=⎟⎟⎠
⎞
⎜⎜⎝
⎛
===→=
 
 
 
 
 
→=⎟⎟⎠
⎞
⎜⎜⎝
⎛
059,1
D
Lf
5
máx 
 
 
s
kg68,110501987,1AvQ
s
m1985802874,141,0kRTv
m
kg7,1
580287
10283,0
RT
p
4
555m
555
3
6
s
5
5
=×××=ρ=
=×××=Μ=
=×
×==ρ
−
 
MPa13,01,023,0p
)abs(MPa23,0108,0158,2p38,2
p
p
5,0M
ef5
5*
5
5
=−=
=×=→=
=