exercícios resolvidos - brunetti - cap 5
TRANSCRIPT
Capítulo 5
Equação da Quantidade de Movimento para Regime Permanente
Neste capítulo admite-se ainda a hipótese de regime permanente para simplificar o raciocínio. O tratamento do regime variado, como já foi dito, será feito no Capítulo 10. O objetivo deste capítulo é mostrar como calcular a força resultante que um fluido aplica em superfícies com as quais está em contato. Essa resultante deve-se ao efeito normal, criado pelas pressões, e ao tangencial, provocado pelas tensões de cisalhamento. Pelo equacionamento utilizado, é possível verificar que a integral das forças normais e tangenciais reduz-se a uma solução bastante simplificada. Na solução dos problemas despreza-se o efeito do peso do fluido, que poderia ser obtido pelo produto do volume pelo seu peso específico. Esse cálculo poderia causar embaraços, no caso de volumes de figuras complexas; entretanto, será sempre um problema geométrico, que não tem nenhuma relação com os objetivos do capítulo. Exercício 5.1
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
Na escala efetiva p1 = 0, p2 = 0 e é dado do enunciado que v1 = 0.
N3,1324
35,0308,97,12
4D
vg
F
AvF:xSegundovQF
22
222
2s
222s2ms
x
x
=×π
××−=πγ
=
ρ−=→−=rr
kW99,11000
1464,37,12QHN
sm4,3
438,030
4D
vQ
m468,92
30g2
vH
HHHH
B
3222
2
222
B
p2B1 2,1
=×××=γ=
=×π
×=π
=
=×
==
+=+
Exercício 5.2
( ) ( ) ( )[ ]( )
( )[ ]
2,1p2
21
221
2,1p2
221
21
4
3
224
3
11
o1m
o11zS
o1m
o11zS
2o
1mo
11xS
o12m22
o11xS
Hzg2vvpHz
g2vp
g2v
sm5,7
108106
AQv;
sm3
1020106
AQv
60senvQ60senApF
60senvQ60senApF
v60cosvQ60cosApF
60cosvvQ1Ap60cosApF
++−
=γ
⇒++=γ
+
=××
===××
==
+=
−−−=
−+=
−+++−−=
−
−
−
−
( )N12660sen3106000.160sen1020106,63F
N285,760cos3106000.160cos1020106,63F
kPa6,63pm36,63120
35,7p
o3o43zS
o3o43xS
1
221
=××××+××××=
≅−×××+××××=
=⇒=++−
=γ
−−
−−
Exercício 5.3
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( ) ( ) ( )[ ]( )
( )
22222s
222222s
2323
2233322
322222s
23223322s
AvApF
v2vAvApF
v4v2080v
AA
vvAvAv
cosvvAvApF
vcosvAvcosAp1ApF
3,2x
3,2x
3,2x
3,2x
ρ−=
−ρ+=
=→==→=
θ−ρ+=
−θρ+θ+−−=
m5,7000.101050
2007,7h
pg2
vhHH
sm07,7
8400vv84000
1080v1000108010500
322
22
21
222
422
43
=×
+=→γ
+=→=
==→−=
×××−×××= −−
Exercício 5.4
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]
N6001020102010200F
N5601020102010180F
kPa180pm181110
10200Hzpp
Hzpp
skg201020000.1QQ
sm10
10201020
AQvvv
vQApF
v0Q0Ap1ApF
vQApF
0vQ1Ap0ApF
43zS
43xS
24
3
2,1p212
2,1p221
3m
4
3
21
1m11zS
1m2211zS
2m22xS
2m2211xS
=×+×××=
−=×−×××−=
=⇒=−−×
=−−γ
=γ
⇒++γ
=γ
=××=ρ=
=××
====
+=
−++−−=
−−=
−++−=
−
−
−
−
−
N820600560FFF 222
zS2xSS =+=+=
Exercício 5.5
REDUÇÃO
( )( ) Pa500.16123
21000000.84p
vv2
pg2
vvpp
sm1234vv4v
1530v
DD
vv4D
v4D
v
pg2
vpg2
vHH
222
22
211
22
21
12
212
2
1
2
2
112
22
2
21
1
2221
21
21
=−+=
−ρ
+=−
γ+=
=×=→=
⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=→
π=
π
γ+=
γ+→=
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( ) ( ) ( )[ ]( )
( ) N740.3123212415,0500.16
43,0000.84F
skg212
43,03000.1
4D
vQ
vvQApApF
vvQ1Ap1ApF
22s
221
1m
21m2211s
12m2211s
Rx
Rx
Rx
=−×+×π
×−×π
×=
=×π
××=π
ρ=
−+−=
−+++−−=
TURBINA
3
3m
TT
T233
T2
3T2
m
N000.1010000.1g
sm212,0
000.1212Q
Q
QNHQHN
Hppp
Hp
HHH
=×=ρ=γ
==ρ
=
γ=→γ=
γ−=→γ
=−γ
=−
( ) ( ) N242415,0800.2500.16AppF
Pa800.237,1000.10500.16p
m37,1212,0000.10
109,2H
232s
3
3T
Tx=
×π×−=−=
=×−=
=××
=
Exercício 5.6
( )[ ]
( ) ( )[ ]N792.810314,0000.10314,01018vQApF
vQ1ApF)bkW7,80107,25314,010N
m7,25120
5,21010
10218H
sm5,2
4,0314,04
DQ4v;
sm10
2,0314,04
DQ4v
zg2
vvppH
g2vp
Hz2
vp
QHN)a
41m11xS
1m11xS
34
22
4
4
T
222
2221
1
1
22
2121
T
222
T1
211
T
=××+××=+=
−+−−==×××=
=+−
+×−−
=
=×π×
=π
==×π×
=π
=
+−
+γ−
=⇒+γ
=−++γ
γ=
−
Exercício 5.7
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( ) ( )[ ]
( ) ( )[ ]
( ) ( )[ ]N120430F
vQvQApvQ1ApF)2
N180630F
vQvQApvQ1ApF)1sm4
1075
03,0AQv
sm6
1050
03,0AQv
sm03,0Q
skg30101003000.1AvQ
sm3
1090vv10000.1090.1
10100v000.11010010100090.1
AvApF
vQ1ApF
2y
2y
1y
1y
x
x
s
2m2m222m22s
s
1m1m111m11s
422
411
3
400m
020
420
430
2000s
1m00s
−=×−=
−=−−=++−=
=×=
=+=−+−−=
=×
==
=×
==
=
=×××=ρ=
==→+=
××−×××−=−
ρ−−=
+++−=
−
−
−
−−
Exercício 5.8
( ) ( ) ( )[ ]o1
o2m
o22
o11xS 30cosv60cosvQ60cosAp30cosApF +−+−+−=
( )
( )( )
N3401,2495,231F
N1,24930sen560sen101010000.130sen1020105,137F
N5,23130cos560cos101010000.130cos1020105,137F
kPa5,137pm75,131020
510Hg2
vvp
Hg2
vpg2
v
sm10
10101010
AQv;
sm5
10201010
AQv
)]30senv60senv(Q30senApF
)]30senv60senv(Q)60sen(Ap)30sen(Ap[F
30cosv60cosvQ30cosApF
22S
oo3o43yS
oo3o43xS
1
22
2,1p
21
221
2,1p
221
21
4
3
224
3
11
o1
o2m
o11yS
o1
o2m
o22
o11yS
o1
o2m
o11xS
=+=
=×+×××+××××=
−=×−×××+××××−=
=⇒=+−
=+−
=γ
+=γ
+
=××
===××
==
++=
−−+−+−−=
−+−=
−−
−−
−
−
−
−
Exercício 5.9
( )[ ]12m222111s vvQnApnApF rrrrr
−++−= ( ) ( )[ ]
( )
( ) N58958911782053,394
1,010150F
skg3,39Q
sm0393,0
41,05
4D
vQ
sm20
5105
DD
vv
vvQApF
vvQ1ApF
23
s
m322
11
22
2
112
21m11s
12m11s
x
x
x
=−=−+×π
××=
=→=×π
×=π
=
=⎟⎠⎞
⎜⎝⎛×=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
−+=
−+−−=
N7854
05,020000.1AvFF2
32
22sx
=×π
××=ρ==
Exercício 5.10
2h
hAghAgh2FF
AghAhF
Agh2Fgh2vAvF
2121dirxS
22dir
1xS1j2jxS
=⇒ρ=××ρ⇒=
ρ=γ=
××ρ=⇒=→ρ=
Exercício 5.11
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( ) ( ) ( )[ ]( ) ( )
( )( ) ( ) ( )[ ]
N3645sen55,220F
senvQsenvQsenApF
0senvQsenAp0ApF
N9,1445cos155,220F
sm55,2
1,0
10204
D
Q4vv
skg201020000.1QQ
cos1vQcosvvQcosApApF
vcosvQcosAp1ApF
os
2m2m22s
2m2211s
os
2
3
2j
21
3m
m21m2111s
12m2211s
y
y
y
x
x
x
−=××−=
θ−=θ−θ−=
−θ+θ+−=
=−××=
=×π
××=
π==
=××=ρ=
θ−=θ−+θ−=
−θ+θ+−−=
−
−
Exercício 5.12
( ) kW3,25107,02,27133,010QHNs
m133,04
15,052,7Q
m2,27205,730H
sm5,7
15,0000.1000.14
D
F4v
4DvF
g2v
zHg2
vHz
34TTT
32
2
T
22xS
2
222xS
22
1T
22
T1
=××××=ηγ=
=×π
×=
=−=
=×π×
×=
ρπ=⇒
πρ=
−=⇒=−
−
Exercício 5.13
N5201020106,2F
Pa106,22,11021036,1p
022,1p
ApF
45pistão
545p
HgOHp
pppistão
2
=×××=
×=×−××=
=×γ−×γ+
=
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( )
( ) ( ) ( )
( ) pistãoo
j
2
xs
o
j
oo21mxs
12ms
F60cos1AQF
60cos1AQQ60cos1Qv60cosvvQF
vvQF
=−ρ=
−ρ=−ρ=−=
−−=rrr
( ) ( ) sm233,0
60cos1000.110520520
60cos1
AFQ
3
o
4
ojpistão =
−×××
=−ρ
=−
Exercício 5.14
( )( )
( ) ( )( )( )
( )( )
sL10
sm01,0101010AvQ
sm10
60cos11060cos13010
60cos1A
60cos1Avv
60cos1Av60cos1Av
60cos1AvF
60cos1AvF
34
djdj
o
o
odj
oej
ejdj
odjd
2j
oeje
2j
odjd
2jdxS
oeje
2jexS
==××==
=+−
×=+
−=
+ρ=−ρ
+ρ=
−ρ=
−
Exercício 5.15
Adotando o eixo x na direção do jato do bocal:
sm2
1050000.1
30sen40A
30senGv
30senGFA
Fv
AvvQF
4
o
jj
os
j
sj
j2jjms
xx
x
=××
×=
ρ=
=→ρ
=
ρ==
−
Exercício 5.16
( ) ( )
( ) ( )
sm86,5
08,0000.18,1724vFF
D
F4v
4D
vF
Hg2
vHz
N8,17260cos14
1,063,6000.160cos14D
vF
sm63,68,2520Hzg2vH
g2v
z)a
221xS2xS
22
2xS2
222
22xS
2,0p
22
B0
o2
2o212
11xS
1,0p011,0p
21
0
=×π×
×=⇒=
ρπ=⇒
πρ=
+=+
=−××π
××=−π
ρ=
=−×=−=⇒+=
N376173250299FGFF
N1734
08,086,5000.14D
vF
N29960sen4
1,063,6000.160sen4D
vF)b
kW26,0107,0
62,00294,010QHN
sm0294,0
408,086,5
4D
vQ
m62,059,32086,5zH
g2v
H
2yS1ySsolo
22
222
22yS
o2
o212
11yS
34
B
BB
3222
2
2
02,0p
22
B
=−+=−+=
=×π
××=π
ρ=
−=××π
××−=π
ρ=
=×××
=η
γ=
=×π
×=π
=
=−+=−+=
−
Exercício 5.17
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
( ) ( )
( ) ( )
sm21120vuvvvu
sm20
60cos110000.1
20
60cos1A
Fu
N2030sen201010
11,030senGAv30senGAT
T60cos1Au60cos1uQF
jj
o4oj
s
o24
oofio
fioo
j2o
ms
x
apx
=+=+=→−=
=−×
=−ρ
=
=+××=+ε
μ=+τ=
=−ρ=−=
−
−−
Exercício 5.18
sL9,19
sm0199,0102094,9AvQ
sm94,9194,8vuv
sm94,8
1020000.1
1105,0
110330sen200
A
Av30Gsenu
Av30GsenAuF
34
jj
sj
4
32o
j
so
soj
2xS
==××==
=+=+=
=××
××
××+×=
ρε
μ+=
εμ+=ρ=
−
−
−−
Exercício 5.19
( )( )
sm7,705,1
604502nR2Rv
vcos1vvvAN
s
ssjjj
=×π×=π=ω=
θ−−ρ=
( )( ) kW115.41000.117,70170cos17,701001001,0000.1N o =××−−×××=
Exercício 5.20
N890.387307320072Fsm730
2,05,073
AQ
v
skg73172QQQ
skg723,02002,1AvQ
vQvQF
xS
22
2m2
3m1m2m
1111m
22m11mxS
−=×−×=
=×
=ρ
=
=+=+=
=××=ρ=
−=
Exercício 5.21
( )[ ]12m222111s vvQnApnApF rrrrr
−++−=
( )
m2,3208
g2vh
sm8
1096,1
107,15AQv
m1096,16,125
107,15000.1FQA
N6,1251081057,1F
Pa1057,1785,0102hp
ApFF
AQvQF
22
3
3
22
2323
s
22
34s
441p
ppps
2
22ms
x
x
x
x
===→=×
×==
×=××
=ρ
=
=×××=
×=××=γ=
==
ρ−=−=
−
−
−−
−
Exercício 5.22
Pa21602602,1
2v
pg2
vp
N35,0108,0602,1AvF22
si0
2s0
422sxS
=×
=ρ
=⇒=γ
−=×××−=ρ−= −
Exercício 5.23
sm155,53
101013020Hz
pg2v
Hg2
vz
pHHH
4
3
2,0p00
2
2,0p
22
00
2,0p20
=−+×
×=⎟⎟⎠
⎞⎜⎜⎝
⎛−+
γ=
+=+γ
→+=
( )[ ]12m222111s vvQnApnApF rrrrr−++−=
m115,04,13000.1
17674
v
F4D
4D
vAvF
sm4,135,51
10
1013020Hzp
g2v
N17674
1,015000.14D
vF
FAvvQF
222
s2
222
222
2s
4
3p0
02
22
222
2s
s2222ms
x
x
2,0
x
xx
=××π
×=
′πρ=′
′π′ρ=′′ρ=
=⎟⎟⎠
⎞⎜⎜⎝
⎛−+
××=⎟⎟
⎠
⎞⎜⎜⎝
⎛−′+
γ=′
=×π
××=π
ρ=
−=ρ−=−=′