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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
22
p
v
pv
pp
m
kg226,527395260
105RTp
)e
kJ627J688.62610956,9218TmcH)dkJ450J888.44910956,6618TmcI)c
K.kgJ6,661
393,16,921
k
cc)b
K.kgJ260
393,11393,16,921
k1kcR
1kkRc)a
=+×
×==ρ
==−××=Δ=Δ==−××=Δ=Δ
===
=−
×=−
=→−
=
Exercício 12.2
( )MJ21154,1UkTmcH)d
MJ151020460717568,47TmcU
kg568,47293287
2102RT
Vpm)c
C460K73325293T
TT
pp
mRTVpmRTVp
)b
K.kgJ717
14,1287
1kRc)a
K.kgJ287
29315.8R
p
6v
6
1
11
o2
1
2
1
2
11
22
v
=×=Δ=Δ=Δ=×−××=Δ=Δ
=×
××==
==×=⇒=→⎭⎬⎫
==
=−
=−
=
==
−
Exercício 12.3
( )
( )kgkJ3,93
s
m275.93201505,717Tcu
K.s
m5,71714,1
2871k
Rc
K.s
m28729315.8
MRR)b
C150K4234,0
27320
4,0
TT4,0
TT
)abs(kPa3714,0
103p4,0
ppV4,0V
VV
pp
)a
2
2v
2
2v
2
2
mol
o4,01k
12
1k
2
1
4,12k1
212
k
1
2
2
1
==−×=Δ=Δ
=−
=−
=
===
==+
==→=
==→=→=→⎟⎟⎠
⎞⎜⎜⎝
⎛=
−−
( )kgkJ6,130
000.11201505,004.1h
K.kgkJ5,004.1
14,12874,1
1kkRc
Tch)c
p
p
=×−×=Δ
=−
×=
−=
Δ=Δ
Exercício 12.4
K.kgJ562
500500ln3,461
573423ln872.1
pp
lnRTT
lncs1
2
1
2p −=×−×=−=Δ
Exercício 12.5
29,0342100
cv
sm3422932864,1kRTc
sm100
6,3360v
===Μ
=××==
==
Exercício 12.6
kPa9,5)abs(kPa1,94
6,02
14,11
120
21k1
pp
21k1
pp
14,14,1
21kk
2
0
1kk
20
−==
⎟⎠⎞
⎜⎝⎛ ×
−+
=
⎟⎠⎞
⎜⎝⎛ Μ
−+
=
⎟⎠⎞
⎜⎝⎛ Μ
−+=
−−
−
sm2093,3012874,16,0kRTv
mkg088,1
3,301287101,94
RTp
C3,28K3,3016,0
214,11
323
21k1
TT
3
3
o
22
0
=×××=Μ=
=××
==ρ
===−
+=
Μ−
+=
Exercício 12.7
( )
%45,01003,67
6,673,67êrro
6,6702,0000.13619,12v
m
kg19,129328710100
RTp
h2hg2ppg2v
ppg2
vívelIncompress
sm3,672932874,1196,0kRTMv
196,01100
72,10214,1
2M
1p
p1k
2MMk
1k1p
p
)abs(kPa72,10210072,2p
kPa72,2Pa272002,0000.136hp
1
3
3
HgHg
12121
21
4,114,1
k1k
01kk
20
0
Hg0
abs
=×−
=
=××=
=××
==ρ
γρ
=γ
γ=−
γ=→
γ=
γ+→
=××==
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−⎟
⎠⎞
⎜⎝⎛
−=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⇒⎟⎠⎞
⎜⎝⎛ −+=
=+=
==×=γ=
−
−
−
Exercício 12.8
mm970m97,011293287
4004,1214,1
000.13610100h
11RTv
k21kph1
ph
1RT1k
k2v
14,14,1
23
1kk
2
m
k1k
m2
==⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛+
××
×−×
=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛+
−γ
=⇒⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−⎟⎟
⎠
⎞⎜⎜⎝
⎛ γ+
−=
−
−−
%8,27100970
970700
mm700m7,0000.136120.95
gpp
h
Pa120.952
400189,12vpp
Hg
0
22
0
=×−
=ε
===ρ−
=
=×=ρ
=−
Exercício 12.9
)abs(Pa102,35,0000.13610hpp)a 45Hg12 ×=×−=γ−=
[ ] [ ][ ] [ ]
( )
%11310047,01111
1Q
QQêrro
QQ)c
skg518,0
5,0132,012
2932871047,095,0
AA
1
pp
12A
RTpCQ)b
47,032,0132,05,01
5,0132,015,332,0
5,05025
AA
e32,010
102,3pp
pp
1pp
AA
1
AA
1pp
11k
k
pp
m
invm
m
incmmmincm
2
5
2
1
2
1
2
21
Dm
429,12
2286,0714,0
1
25
4
1
2
1
2k2
1
22
1
2
2
1
2k1k
1
2
k1
1
2
=×⎟⎠
⎞⎜⎝
⎛−=
φ−=−=
−=⇒
φ=
=−−×
×××=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
φ=
=−××−−×−×
=φ
===×
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=φ
−
Exercício 12.10
⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
⎟⎟⎠
⎞⎜⎜⎝
⎛=φ
−
1
2k2
1
22
1
2
2
1
2k1k
1
2
k1
1
2
pp
1pp
AA
1
AA
1pp
11k
k
pp
kPa149101,5000.10200hpp 3Hg12 =××−=γ−= −
( )
( )
( )s
kg151,1444,01
745,0124
1,03662077
10200844,095,0Q
AA
1
pp
12A
RTpCQ
K.kgJ2077200.5
665,11665,1c
k1kR
1kkRc
844,0
745,01745,0444,01
444,01745,011665,1
665,1
745,0
444,01510
DD
AA
;745,0200149
pp
2
23
m
2
1
2
1
2
21
Dm
pp
665,12
2
2665,11665,1
665,11
22
1
2
1
2
1
2
=−−×
××π
×××
××=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
⎟⎟⎠
⎞⎜⎜⎝
⎛−
φ=
=×−
=−
=⇒−
=
=
−×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛×−
−×⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−×
−×=φ
=⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛===
−
Exercício 12.11
( )
( )
( )
( )
%4,51001310
12391310êrro)c
sm1310
5,115,1
405,12111015,1
0651.02v
ppp
k211pp2v)b
sm12391015,1
0651,02v
m
kg0651,05504189
105,1K.s
m4189532.14405,1
1405,1ck
1kRRTp
pp2v)a
5c
0
c0c0
0c
5c
3
50
2
2p
0
00
c0c
=×−
=
=⎟⎠
⎞⎜⎝
⎛ −×
+××−×=
−+−
ρ=
=×−=
=××
=ρ
=×−
=−
=→=ρ
−ρ
=
Exercício 12.12
s102,1477,591
440.7vv
st)c
m440.7502,147800.14s
s50296
800.14cRt
5,0296
2,147c
vsm2,147
6,3530v
2296
7,591cv
sm7,591
6,3130.2v
sm2962182874,1kRTc)b
211
222
111
=+
=+Δ
=
=×−=Δ
===
===Μ⇒==
===Μ⇒==
=××==
Exercício 12.13
sm8162952574,137,2kRTMv
37,225sen
1MM1
2sen
o
=×××==
==→=α
Exercício 12.14
3*
**
133,133,1
5
1kk
0*1kk
*0
**
0*2*0
mkg338,0
346462036.54
RTp
)abs(kPa54)abs(Pa036.54
2133,11
10
21k1
pp
21k1
pp
sm46134646233,1kRTv
K346
2133,11
403
21k1
TT
21k1
TT
=×
==ρ
==
⎟⎠⎞
⎜⎝⎛ −+
=
⎟⎠⎞
⎜⎝⎛ −+
=→⎟⎠⎞
⎜⎝⎛ −+=
=××==
=−
+=
−+
=→Μ−
+=
−−
−
Exercício 12.15
1.Tab1Ms →=
K5,4775738333,0T8333,0TT
)abs(MPa5283,015283,0p5283,0pp
s0
s
s0
s
=×=→=
=×=→=
skg338,0102438856,3Q
sm4385,4772874,11kRTMv
AvQm
kg856,35,477287
105283,0RTp
4m
sss
sssm
3
6
s
ss
=×××=
=×××==
ρ=
=××
==ρ
−
Exercício 12.16
21*
15
0
1
51atmHg1
cm1,2015340,1A340,1AA
8434,0102680.168
pp
)abs(Pa680.168505,0000.13610pphp
=×=⇒=→=×
=
=×+=⇒=γ−
Exercício 12.17
a) T0 = 373 K = 100 oC
bloqueadoestánão833,0102,1
10pp
)b5
5
0
s →=×
=
1.Tab8333,0pp
0
s →=
skg183,010196984,0AvQ
sm1963542874,152,0kRTMv
m
kg984,0354287
10RTp
3sssm
sss
3
5
s
ss
=××=ρ=
=×××==
=×
==ρ
−
c) A mesma
1.Tab3,0M)d →=
23m
5
cm6,151056,1073,1115
193,0v
QA
sm1153662874,13,0kRTMv
073,136628710127,1
RTp
=×=×
=ρ
=
=×××==
=××
==ρ
−
K3543739487,0T9487,0TT
52,0M
s0
s
s
=×=→=
=
)abs(Pa10127,1102,19395,0p9395,0pp
K3663739823,0T9823,0TT
55
0
0
×=××=→=
=×=→=
e) Ms=1 1.Tab→
⎪⎪⎩
⎪⎪⎨
⎧
=×=→=
×==→=
K3113738333,0T8333,0TT
)abs(Pa10893,15283,010p5283,0
pp
s0
s
55
00
s
skg396,0105,35312,1AvQ
s,m5,3533112874,11kRTMv
m
kg12,1311287
10RTp
3sssm
sss
3
5
s
ss
=××=ρ=
=×××==
=×
==ρ
−
Exercício 12.18
( ) mm64m064,0000.136
1029564,01h
9564,0pp
erpolandoint38,21026,1
103AA
ppp
1hphp)e
m1026,159,1102
59,1A
A)d
sm7,1367,2902874,14,0kRTv
59,1AA
4,0969,0
3007,290
TT
)c
)abs(Pa102p)b
)abs(Pa102pp
K300TT)a
5
0
A3
3
*A
Hg
00
A
AHg0
233
B*
2BB
*B
B
0
2
5A2M
52M0
10
==××−
=
⎩⎨⎧
=→→=×
×=
γ
⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⇒=γ−
×=×
==
=×××=Μ=
⎪⎩
⎪⎨⎧
=
=Μ→==
×=
×==
==
−
−
−−
Exercício 12.19 Fixando o sistema de referência no conduto, isto é, no avião: v1 = 180 m/s
→→===
=××==
1.Tab55,05,327
180cv
M
sm5,3272672874,1kRTc
1
11
1
1.Tab8,0M 2 →=
sm2542512874,18,0kRTMv 222 =×××==
Exercício 12.20
sm1,933452874,125,0kRTv
)abs(MPa55,0575,0957,0p957,0pp
K345349988,0T988,0TT
25,0
39,2009,137,2AA
AA
AA
009,1AA
)abs(MPa575,05913,034,0p5913,0
pp
K3498606,0300T8606,0
TT
9,03002874,1
312kRTv
222
20
2
20
2
2
*1
1
2*2
*1
00
1
00
1
1
11
=×××=Μ=
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=×=→=
=×=→=
=Μ
→=×=×=
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=
==→=
==→=
→=××
==Μ
Exercício 12.21
→== 21020
A
A*e
rpm4926022
103R2
vn
sm1032932874,13,0kRTMv
'e''
e'e =×
×π×=
π=→=××==
rpm36046022
755R2
vn
sm7552932874,12,2kRTMv
''e''''
e''e =×
×π×=
π=→=××==
K2839449,0267T9449,0
TT
)abs(kPa1288201,0100p8201,0
pp
00
1
00
1
==→=
==→=
K2518865,0283T8865,0TT
)abs(kPa80656,0128p656,0pp
20
2
20
2
=×=→=
=×=→=
2,2M
3,0M''e
'e
=
=
Exercício 12.22
22
ss
ms
3
5
s
ss
sss
s0
s
s
6
5
0
s
2G
G22
GG
mG
GGG
3
5
G
GG
G0
G
56G
0
G
G
cm413m0413,0794037,1
34v
QA
mkg037,1
33628710
RTp
)d
sm7943362874,116,2kRTv
K3366505173,0T5173,0TT
16,21,0
1010
pp
)c
m165,010147,24A4Dm10147,2
467394,334
vQ
A
sm4675422874,11kRTv)b
mkg394,3
5422871028,5
RTp
K5426508333,0T8333,0TT
)abs(Pa1028,5105283,0p5283,0pp
1)a
==×
=ρ
=
=×
==ρ
=×××=Μ=
⎪⎩
⎪⎨
⎧
=×=→=
=Μ==
=π××
=π
=⇒×=×
=ρ
=
=×××=Μ=
=××
==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=→=
×=×=→=
=Μ
−−
Exercício 12.23
→=
=
1M
K310T)a
G
0
8333,0TT
5283,0pp
0
G
0
G
=
=
.mesmoO)cK2583108333,0T08333T
)abs(kPa6,1067,2015283,0p5283,0p)bm
kg27,2310287107,201
RTp
)abs(kPa7,201)abs(kPa695.201p200.95p5283,0p200.95pp
200.95pp
7,0000.136phpp
0G
0G
3
3
0
00
0
00G0
G0
GHgG0
=×===×==
=××
==ρ
===−→=−
+=
×+=γ+=
Exercício 12.24
2s*
223
ss
ms
3
3
s
ss
xSsss
*s
s0
s
s
0
s
smxS
smemsmiiiS
cm3,17999,3
3,69999,3A
A)c
cm3,69m1093,6864.1348,0
5,4v
QA
mkg348,0
000.128710100
RTp
)b
N838818645,4Fsm1864000.12874,194,2kRTv
999,3AA
K000.1730.23665,0T3665,0TT
94,2
0294,0400.3
100pp
vQFvQvQvQnApF)a
===
=×=×
=ρ
=
=××
==ρ
=×−=⇒=×××=Μ=
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=
=×=→=
=Μ
==
−=
−=+−=
−
∑ rrrrr
Exercício 12.25
→== 76,11,0
176,0
A
A*3
)abs(MPa08,068,01164,0p1164,0pp
06,2M
)abs(MPa622,068,09143,0p9143,0pp
36,0M
''3
0
''3
''3
'3
0
'3
'3
=×=→=
=
=×=→=
=
skg1621,05,31021,5Q
sm5,3102402874,11kRTMv
mkg21,5
24028710359,0
RTp
)abs(MPa359,068,05283,0p5283,0pK2402888333,0T8333,0T
vAQ
m
GGG
3
6
G
GG
0G
0G
m
=××=
=×××==
=××
==ρ
=×===×==
ρ=
Exercício 12.26
⎪⎪⎪
⎩
⎪⎪⎪
⎨
⎧
=×=⇒=
=×=⇒=
=×=⇒=
→=Μ
=×=×
=ρ
=
=×××=Μ=
=××
==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=⇒=
=×=⇒=
→=Μ
−
2s*
s
s0
s
s0
s
s
222
GG
mG
GGG
3
6
G
GG
G0
G
G0
G
G
cm419251668,1A668,1AA
)abs(MPa0128,01,01278,0p1278,0pp
K1442605556,0T5556,0TT
2
cm251m1051,2295815,0
3,6v
QA
sm2952172874,11kRTv
mkg815,0
21728710053,0
RTp
K2172608333,0T8333,0TT
)abs(MPa053,01,05283,0p5283,0pp
1
Exercício 12.27
→== 089,1293319
A
A*s
)abs(MPa516,051,13417,0p3417,0pp
7358,0TT
34,1M
s0
s
0
s
s
=×=→=
=
=
Na realidade, existe também a solução subsônica, entretanto, com essa velocidade de saída, a temperatura seria um valor prático impossível.
N342.4110319000.2324,0AvFm
kg324,0544.528710516,0
RTp
AvF
K544.52874,134,1
000.2
kRM
vTkRTMv
42s
2ssS
3
6
s
ss
s2ssS
2
2
2s
2s
ssss
x
x
=×××=ρ=
=××
==ρ
ρ=
=××
==→=
−
Exercício 12.28
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
===⇒=
=
=Μ
→=
=××=××=
=⇒=×=×=
=××=××==
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=
=
=
→=Μ
⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪
⎨
⎧
=
=
===⇒=
=Μ
→=Μ
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=
=
=Μ
→=
)abs(kPa1,1218259,0100
8259,0p
p8259,0pp
9468,0TT
53,0
256,1AA
256,1176,11094,135,1
AA
A
A
AA
AA
)c
A075,1A93,0094,1176,11
A
AAA
AA
)b
46,2272,019298,07209,0
pp
p
p
pp
pp
Int)a
094,1A
A
7209,0pp
9108,0TT
7011,0
458,2pp
32,1TT
)abs(kPa1309298,0
1,1219298,0
pp9298,0
p
p
7011,0
5,1
176,1AA
6897,0TT
5,1
272,0pp
sy0
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
yOC
*y
y
y0
y
0
y
y
x
y
x
y
y0
x0x0
y0
y
x
*x
x
0
x
x
x0
x
K3179468,0300
9468,0T
T)e
)abs(kPa130p)d
s0
x0
===
=
K.kgJ8,20
458,2
32,1ln004.1
pp
TT
lncss)g
A6,213A184161,1Qsm1843002874,153,0kRTv
mkg161,1
30028710
RTp
AvQ)f
41,114,1
k1k
x
y
x
y
pxy
ssm
sss
3
5
s
ss
sssm
=×=
⎟⎠⎞
⎜⎝⎛
=−
=××=
=×××=Μ=
=×
==ρ
ρ=
−−
Exercício 12.29
→= 1278,0pp
x0
x
→= 2M x
→= 58,0M y
437,1213,1688,112
A
A
AA
A
A
A
A*y
y
x
*x
*x
s*y
s =××=××=
→= 437,1A
A*y
s
688,1A
A
5556,0TT
2M
*x
x
0
x
x
=
=
=
5,4p
p
687,1T
T
7209,0p
p
5774,0M
x
y
x
y
0
0
y
x
y
=
=
=
=
213,1A
A
7962,0p
p
9370,0T
T
*y
y
0
y
0
y
y
=
=
=
)abs(Pa1004,18650,0
109,0p8650,0pp
9594,0TT
46,0M
55
00
s
0
s
s
yy
×=×
=→=
=
=
m47,01036,1
1084,11028,8pph
)abs(Pa1028,81084,15,4p5,4p
)abs(Pa1084,11044,11278,0p1278,0p
php)b
K3139594,0300
9594,0T
T
)abs(kPa144)abs(Pa1044,17209,0
1004,17209,0
pp)a
5
44
Hg
xy
44xy
450x
yHgx
s0
550
0
x
yx
=×
×−×=
γ
−=
×=××==
×=××==
=γ=
===
=×=×
==
8650,0pp
x
7962,0pp
x
5283,0pp
x)c
y0
ss
y0
yyOC
x0
GG
=→
=→
=→
Exercício 12.30
⎩⎨⎧
=Μ ′′=Μ′
→=×
=
=Μ→=×
=
−
−
2,23,0
210
102AA
76,0756,010
1056,7pp
s3
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
1850,0pp
6175,0TT
76,1M
x0
x
0
x
x
=
=
=
447,3pp
502,1TT
8302,0p
p
6257,0M
x
y
x
y
x0
y0
y
=
=
=
=
→= 76,1M x
→= 6257,0M y
→=×
×=
−
−667,1
102,1
102
A
A3
3
*y
s
( )
( )
skg78,110217122,5AvQ
sm1715052874,138,0kRTMv
m
kg22,55052871056,7
Tp
)c
sm4224836257,032176,12874,1v
K4835209286,0T9286,0TK3215206175,0T6175,0T
TMTMkRkRTMkRTMvvv)b
3sssm
sss
3
5
ss
ss
0y
0x
yyxxyyxxyx
=×××=ρ=
=×××==
=××
=ρ
=ρ
=×−×××=Δ
=×===×==
−=−=−=Δ
−
Exercício 12.32 Ver o exercício 12.31 Exercício 12.33
233y*
y*y
y
0
y
0
y
m102,1166,1104,1
166,1A
A166,1A
A
7716,0pp
9286,0T
T
y
−−
×=×
==→=
=
=
)abs(Pa1035,89052,0
1052,79052,0p
p9052,0pp
K5209719,0505
9719,0T
T)a9719,0TT
38,0M
55
sy0
y0
s
s0
0
s
s
×=×
==→=
===→=
=
244
s*y*
y
s
y0
s
s
m107,33188,11040
188,1A
A188,1A
A
784,0pp
6,0M
−−
×=×
==→=
=
=
→== 9324,0
429400
TT
y0
s
→=×
×=
−
−128,1
107,33
1038
A
A4
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
660
60
00
y
x
xx
×=××=
×=
=−×
Exercício 12.34
⎪⎪⎪⎪⎪
⎩
⎪⎪⎪⎪⎪
⎨
⎧
=
=
=
=Μ
→=Μ
=×=×
=ρ
=
=×××=Μ=
=××
==ρ
⎪⎪⎩
⎪⎪⎨
⎧
=×=⇒=
=×=⇒=
→=Μ
−
783,3pp
562,1TT
7947,0p
p84,1
606,0)b
cm924m1024,9422818,2
110v
QA
sm4224442874,11kRTv
mkg818,2
44428710359
RTp
K4445338333,0T8333,0TT
)abs(kPa3596805283,0p5283,0pp
1)a
x
y
x
y
x0
y0
x
y
222
GG
mG
GGG
3
3
G
GG
G0
G
Gx0
G
G
7465,0p
p
9199,0TT
66,0M
y0
y
0
y
y
=
=
=
xx
y0y0
0
0p8876,0p8876,0
p
p=→=
2*ys*
y
ss
2y*y*
y
yy
2x*
x
x
x0
x
0
x
x
cm835.1154.159,1A59,1A59,1AA
4,0
cm154.1188,1
1371188,1A
A188,1A
A606,0)c
cm371.1484,1924A484,1AA
1537,0pp
5963,0TT
84,1
=×=⎪⎩
⎪⎨⎧
=⇒=→=Μ
===⎪⎩
⎪⎨⎧
⇒=→=Μ
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=×=⇒=
=
=
→=Μ
Exercício 12.35
9325,0400373
TT
0
1 ==
Na segunda situação *1
A
A não muda, mas o escoamento é supersônico.
188,1A
A*1 =
52,1M x =
CouK91274365TTT oxy =−=−=Δ
Exercício 12.36
24,1AA
857,0429,1225,1
mkg429,1
40028710164
RTp
)a
*x
0
3
3
0
00
=⎩⎨⎧→==
ρρ
=××
==ρ
188,1A
A6,0M
*1
1
=
=
K274400684,0T684,0T684,0TT
52,1M
0x0
x
x
=×==→=
=
K365274334,1T334,1T333,1T
T
6941,0M
xyx
y
y
=×==→=
=
sm3663332874,11kRTv
K3334008333,0T8333,0TT
1)c
skg29,410160517519,0AvQ
sm5172672874,158,1kRTv
mkg519,0
2672871040
RTp
)b
)abs(kPa401642423,0p2423,0p2423,0pp
K2674006670,0T6670,0T6670,0TT
58,1
24,1AA
GGG
G0
GG
4xxxm
xxx
3
3
x
xx
0x0
x
0x0
x
x
*x
=×××=Μ=
=×=⇒=⎩⎨⎧
→=Μ
=×××=ρ=
=×××=Μ=
=××
==ρ
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=×=×=⇒=
=×=×=⇒=
=Μ
→=
−
Exercício 12.37
1,0101
100pp
)a30
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
→==
=
666,6pp
059,2TT
521,042,2
5283,0p
p
pp
pp
x
y
x
y
y
x
x0
y0
x0
G
y0G
K2,1553005173,0T5173,0T5173,0TT
16,2M
0s0
s
s
=×==→=
=
K2832,155822,1T822,1T822,1T
T
5525,0M
xyx
y
y
=×==→=
=
sm1853162874,1521,0kRTv
K3163339487,0T9487,0TT
521,0
yyy
y0
yy
=×××=Μ=
=×=⇒=⎩⎨⎧
→=Μ
Exercício 12.39
2
máx
1
máx
2
máx
1
máxD
Lf
DLf
DL
fD
Lf
DL
fDLf ⎟
⎠
⎞⎜⎝
⎛+=⎟
⎠
⎞⎜⎝
⎛→⎟
⎠
⎞⎜⎝
⎛−⎟
⎠
⎞⎜⎝
⎛=
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 ?
01451,0D
Lf
)abs(Pa1086,8129,1
10129,1p
p129,1p
p
K6,284033,1
294033,1T
T033,1T
T
2
máx
45
2**2
2**2
=⎟⎠
⎞⎜⎝
⎛
×===→=
===→=
0145,401451,0075,0
1502,0D
Lf
1
máx =+×
=⎟⎠
⎞⎜⎝
⎛
)abs(Pa1031086,8389,3p389,3p
p
K7,3346,284176,1T176,1T
T32,0M
541*
1
1*1
1
×=××=→=
=×=→=
=
)abs(Pa1022,39315,0103p9315,0
pp
K3429799,0
7,334T9799,0TT
55
00
1
00
1
×=×
=→=
==→=
Exercício 12.40
)abs(MPa011,0259,027,0ppp
176,1TT
)abs(MPa259,00764,0389,3p389,3pp
32,0
224,4D
Lf
224,4025,0
2,13008,0299,5D
Lf
DL
fD
Lf
299,5D
Lf
)abs(MPa0764,0619,327,0
619,3p
p619,3pp
179,1TT
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 =
sm3342772874,11kRTMv 222 =×××==
Exercício 12.42
⎪⎩
⎪⎨⎧
=⎟⎟⎠
⎞⎜⎜⎝
⎛→=Μ
=⎪⎩
⎪⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛→=Μ
3050,0DLf
2
5222,0DLf
3
2H
máx2
1H
máx1
( )absMPa166,0843,014,0
843,0p
p843,0pp 1
00
11
1
===→=
( )absMPa124,034,1166,0
34,1
pp34,1
p
p11 0*
0*0
0===→=
( )
K2773338333,0T8333,0T8333,0TT
absMPa065,0124,05283,0p5283,0p5283,0pp
020
2
*02*
0
2
=×==→=
=×==→=
8,1012,0
1,02172,0f
D2172,0L
2172,03050,05222,0DLf
H
H
máx
=×
=×
=
=−=⎟⎟⎠
⎞⎜⎜⎝
⎛
Exercício 12.43
3M1 =
sm4204392874,11kRTMv *** =×××==
2M2 =
sm6852922874,12kRTMv 222 =×××==
m17,201,0
1,02172,0f
D2172,0L
2172,03050,0522,0D
LfD
LfD
Lf
2,1
2
máx
1
máx2,1
=×
==
=−=⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=
Exercício 12.44
K245
504,3pp
1775,15,288
1775,1T
T1775,1TT
31,0
8,4025,0
602,0D
Lf
*1
1**1
1
1
máx =
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
=
==⇒=
=Μ
→=×
=⎟⎟⎠
⎞⎜⎜⎝
⎛
( )
m22,501,0
1,05222,0f
D5222,0L5222,0D
Lf
absMPa46,02182,0
1,02182,0p
p2182,0p
p
K4394286,0188
4286,0T
T4286,0T
T
1máx1
máx
1**1
1**1
=×
==→=⎟⎟⎠
⎞⎜⎜⎝
⎛
===→=
===→=
( )
3050,0D
Lf
absMPa19,046,04083,0p4083,0p4083,0pp
K2924396667,0T6667,0T6667,0TT
2
máx
*2*
2
*2*
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛
=×==→=
=×==→=
skg219,0
4025,0314422,1
4DvQ
sm3142452874,11kRTv
mkg422,1
24528710100
RTp
22**
máx
***
3
3
*
**
=×π
××=π
ρ=
=×××=Μ=
=××
==ρ
Exercício 12.45 A leitura do termômetro é 600 K, uma vez que a temperatura de estagnação não se altera.
→== 22040
A
A*3
→= 2,2M x
→= 5471,0M y
→== 59,15,31
50
A
A*y
4
→= 4,0M 4
cm8504A4D 4
4 =π×
=π
=
059,108,0
502,0309,2DLf
DLf
DLf
4
máx
5
máx =×
−=−⎟⎟⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛
( )absMPa0468,05,009352,0p09352,0p09352,0pp
K3056005081,0T5081,0T5081,0TT
2,2M
xx
0x0
x
0x0
x
x
=×==→=
=×==→=
=
( )absMPa314,05,06281,0p6281,0p6281,0p
p
5471,0M
xyx
y00
0
0
y
=×==→=
=
23*y*
y
3 cm5,3127,1
4027,1
AA27,1
A
A===→=
( )absMPa281,0314,08956,0p8956,0p8956,0pp
4,0M
yy
040
4
4
=×==→=
=
( )
309,2D
Lf
absMPa197,059,1314,0
59,1
pp59,1
p
p
4
máx
0*0*
0
0 yy
=⎟⎟⎠
⎞⎜⎜⎝
⎛
===→=
→=⎟⎟⎠
⎞⎜⎜⎝
⎛059,1
DLf
5
máx
skg68,110501987,1AvQ
sm1985802874,141,0kRTv
mkg7,1
58028710283,0
RTp
4555m
555
3
6
s
55
=×××=ρ=
=×××=Μ=
=××
==ρ
−
MPa13,01,023,0p
)abs(MPa23,0108,0158,2p38,2p
p5,0M
ef5
5*5
5
=−=
=×=→=
=
