F Brunetti - Exercícios Resolvidos - Capitulo5

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


( ) ( ) ( )[ ]( )

( )

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+→=


( ) ( ) ( )[ ]( )

( ) 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


( ) ( )[ ]

( ) ( )[ ]

( ) ( )[ ]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


( ) ( ) ( )[ ]( ) ( )

( )( ) ( ) ( )[ ]

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+

=


( )

( ) ( ) ( )

( ) 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


( ) ( )

( ) ( )

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

→+=


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−=−=′

F Brunetti - Exercícios Resolvidos - Capitulo5

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