Bài tập cơ chất lỏng.docx

8/18/2019 Bài tập cơ chất lỏng.docx

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PHẠM THANH HƯNG_40800884_LỚP:VP08CĐT

BÀI TẬP

I.Chương 7:

7.1. The propagation speed of small-amplitude surface waves in a region ofunform depth is given by

where h is depth of the un disturbed liquid and λ is wavelength. Using as acharacteristic length and !o as a characteristic velocity" obtain the dimension lessgroups that characteri#e the equation.

$i%i:

Ta &'t c(c s) v* th+ nguy,n sau:

λ¿=

λ

L ;h

¿=h

L ;c

¿= c

V o

Thay vo biu th+c v/n t)c truy0n sng" ta 234c:

C ¿2

V o2=

(

σL

( ρ V o2 ) 2 π

λ¿ +

gL

V o2 .2 π λ

¿

)tanh (

2π h¿

λ¿ )

5o 2" c(c nhm v* th+ nguy,n 26c tr3ng cho ph3ng tr8nh:

σL

( ρ V o2 )

; gL

V o2

7.9.

1


2/15


$i%i:

-;i l34ng 2Ω× V l gia t)c coriolic

-ch th+ nguy,n c(c 2;i l34ng?

@!ABT-1? @tABT-1? @CABDTE-F? @GpABD-F?

@HABT-1

-I) 2;i l34ng c th+ nguy,n:J

-I) 2;i l34ng c th+ nguy,n 2Kc l/p:L

-MhNn L 2;i l34ng c th+ nguy,n 2Kc l/p: "Gp"!

-I) 2;i l34ng v* th+ nguy,n :E

-M(c s) v* th+ nguy,n:

O1BΩL

V ?OEB ρ

ΔpV −2

L2 ?OLB

Vt

L

-


3/15


7.P.

$i%i: DBfQ!"R" CS


@DABD? @!ABT-1? @CABDTE-F? @RABE

-I) 2;i l34ng c th+ nguy,n:F


-MhNn L 2;i l34ng c th+ nguy,n 2Kc l/p: R"C"!

-I) 2;i l34ng v* th+ nguy,n :1-M(c s) v* th+ nguy,n:

OB F

Aa ρ

bV

c B F

L2a ( FT 2 L−4 )b ( LT −1 )c

= F

F b L

2a−4b+ cT

2b−c=[1 ]

{ b=12a−4 b+c=02b−c=0

→ {a=1b=1c=2

!/y" π = F

Aρ V 2

Iuy ra: F V 2

7..

$i%i:

GpBfQd"5"!"V" CS


3


4/15


@pABD-E? @!ABT-1? @CABDTE-F? @dAB@5AB ? @VABDT-E

-I) 2;i l34ng c th+ nguy,n:9


-MhNn L 2;i l34ng c th+ nguy,n 2Kc l/p: d"C"!

-I) 2;i l34ng v* th+ nguy,n :L


π 1= D

d

π 2= p

da ρ

bV

c B F L

−2

L

a

( FT 2

L

−4

)

b

( LT −1

)

c=

F L−2

F b L

a−4b+cT

2b−c=[1 ]

{ b=1a−4b+c=−22b−c=0

→ {a=0b=1c=2

π 2= p

ρV 2

π 3= μ

d

a

ρ

b

V

c B FT L

−2

L

a

( FT 2

L

−4

)

b

( L T −1

)

c=

F TL−2

F

b

L

a−4b+c

T

2b−c =[1 ]

{ b=1a−4b+c=−22b−c=1

→ {a=1b=1c=1

π 3= μ

ρdV

-Wiu th+c trX thnh:GOE BfQO1" OLS

7.1E.

4


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$i%i:ú=f ( ! " # ! μ ! ρ)


@ ú ABT-E? @YwABD-E? @CABDTE-F? @yAB ? @VABDT-E

-I) 2;i l34ng c th+ nguy,n:J


-MhNn L 2;i l34ng c th+ nguy,n 2Kc l/p: y"C" Yw

-I) 2;i l34ng v* th+ nguy,n :E


π 1=

ú ρa " #

b c →

L T −1

( FT 2 L−4 )a ( F L−2 )b Lc= LT

−1

F a+b L−4a−2b+ cT 2a= [1 ]

{ a+b=0

−4a−2b+c=12a=−1

→{a=

−12

b=1

2

c=0

π 1= ú

ρ−12 " #

1

2

= ú

uo

π 2=

μ

ρa" #

b

c →

FT L−2

( FT 2 L−4 )a

( F L−2 )b L

c=

FT L−2

F a+b

L−4a−2b+ c

T 2a= [1 ]

5


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{ a+b=1

−4a−2b+c=−22a=1

→ {a=

1

2

b=1

2

c=1

π 2= μ

ρ0.5

" #0.5

1=

$

uo

!Zi uo=( " # ρ )1

2

-Wiu th+c trX thnh:O1 BfQOES

ú

uo=f (

$

uo)

7.LP.

$i%i:• !8 2)i vZi chuyn 2Kng c[a m(y bay" t\ s) l]c n=ng v trNng l]c 2(ng ^ n,n

h_ s) l]c n=ng l 2(ng ^: F % ρ%V %

2 L%

2 =

F t

ρt V t 2 Lt

2

• Ta bi`n 2i biu th+c tr,n : p%

V %2 =

pt

V t 2

!8 m*i tr3ng chuyn 2Kng l nh3 nhau gia m* h8nh v th]c t` n,n CmBCt.p sut Bl]cdi_n t>ch n,n pDE.

7.F1.

6


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$i%i:!8 l]c c%n l 2(ng ^ so vZi l]c qu(n t>nh hn c(c l]c ^h(c n,n h_ s* l]c c%n lnh3 sau tr,n th]c t` v tr,n m* h8nh.

F d%

ρ%V %2 A%

= F dt

ρt V t 2 A t

F dt = F d%. ρt V t

2 At

ρ%V %2 A%

*i tr3ng th]c l ^h*ng ^h>: CtB1.EEJQ^gmLS*i tr3ng th nghi_m l n3Zc: CmB.1Q^gmLSRtRmBEJ!t!mBEJF

DdtBE1PQjS

7.FJ.

$i%i:• ]c ma s(t l 2(ng ^ nht so vZi l]c qu(n t>nh n,n ta dung m* h8nh 2kng

d;ng h_ s) enold.V % L%

&%=

V t Lt

& t

h*ng ^h>: tB1.LJE.1-JQmEsS j3Zc: mB1.7.1-9QmEsStmB1 !mB171QftsS• _ s) l]c c%n:

F d%

ρ%V %2 A%

= F dt

ρt V t 2 A t

7


8/15


F dt

F d%=

ρt V t 2 At

ρ%V %2 A%

=1.347

998.2 ×

2302

1712 ×10

2=0.244

7.9E.

$i%i:

3ng cong biu din l]c D theo v/n t)c c[a m* h8nh:

]c c[a v/t th/t:_ s) l]c c[a m* h8nh v v/t th/t l nh3 nhau:

C D= F

p

1

2 ρ A p V p

2=

F %

1

2 ρ A% V %

2

B F p= F %

V %2 ( A p A% )V p

2=0,022 5

2

0.152

V p2

Ta 234c b%ng gi( tr D p theo ! p:

8


9/15


k th:

II.Chương 8:P.1F

$i%i:Ta 2i dn v:DBL9 QpoundsS.gB19LLQ^gSnh t gia ra E 2u ^he. ]c n=ng c[a n3Zc l,n thnh ^he:

F = '. (

2

xBED


10/15


bS$radient (p sut:

{) p

) * =−2 '

( =−27.106(

pa

% ) ! & +ớ *>0

) p

) * =

2 '

( =27.106(

pa

% ) ! & +ớ *nh to(n trong bi gi%ng:,2 -= −112 μ ( ) p) * )h3

h=3

√6 μ

'

(

/6 ( μ%)

P.1J.

$i%i:• T>nh :

T ph3ng tr8nh c=n bzng l]c cho piton suy ra:

0 = pπ D

2

4 g =4.2(1g)

• T>nh l3u l34ng :p dng biu th+c 2{ 234c &=y d]ng trong l| thy`t ta 234c:Qa}}5S

,πD

= 112 μ

'- a3

,=94.106a

3

• T>nh a:!/n t)c 2i &u)ng c[a piton l vB1mmph~t

3u l34ng n3Zc tho(t ra ngoi: ,=π D

2

4 &=4,7.10−10(

%3

2 )

10


11/15


Iuy ra 2K rKng ^he hX: a=3

√ ,

94.106=1,7 ( μ%)

P.19.

$i%i: !/n t)c trung b8nh t;i b(n ^>nh r:

,=2 π3hú → ú= ,

2πh3

$radient (p sut:p dng c*ng th+c 2{ 234c thi`t l/p trong l| thuy`t:

,

- =

−112 μ ( ) p) * )h3

B) p

) 3 =−12 μ,

2 π h33

T8m pQrS p (3 )− p ( 4 )=∫

4

3 −12 μ,2π h

33

d3=6 μ,

π h3 ln (

4

3 )

pQ S B

B p (3 )=6 μ,

π h3 ln (

4

3 )

T>nh D: po= p ( 4o )=

6 μ,

π h3 ln (

4

4o)

11


12/15


F = po π 4o2+∫

40

4

2 π3 p (3 ) d3= po π 4o2+∫

40

46 μ,

π h3 .2π3 ln( 43 )d3

¿ po π 4o2−

12 μ, 42

h3 ∫

40

4

1

* ln ( * ) d*

F = po π 4o2+

12 μ, 42

h3 [ 14 (1− 4o

2

42 )+ 4o

2

2 42 ln( 4o 4 )]=3 μ, 4

2

h3 (1− 4o

2

42 )

P.EJ.

$i%i:QaS_ s) enolds:

e B !a B •a B .1J€9€EO€.J€1-91.F9€1-JB1.FQbSng sut ti`p:

" = μ ) u

) =

μV

a

QcSM*ng sut:]c ma s(t ̂ hong ^h> vZi 2u 2Nc: DBYR*men &o‚n: TBD. M*ng sut mt m(t:


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P.LE.

$i%i:ƒ't 1 phn t cht l„ng h8nh ^h)i hKp ch nh/t c di_n t>ch d&dy nh3 h8nh v….ch l]c t(c dng l,n phn t ta c:

∑ F *=(" *+ ) " *) d ) Ld*−" * Ld*− ρgL. d* . dTheo c*ng th+c b%o ton 2Kng l34ng bzng ph3ng ph(p th t>ch ^im so(t:

∑ F *= )

) t ∭

CV

❑

ρ⃗udV +∬C5

❑

ρ⃗ u⃗ 6d5=0

5o 2") " *

) = μ

)2

& *

) 2 = ρg

{& *= ρg2 μ 2+C 1 μ . +C 2" *= ρg+C 1M(c 2i0u ^i_n bi,n: {& * (0 )=7 o" * (h )=0 → {

C 2=7 oC 1=− ρgh

!/y" v/n t)c trong long cht l„ng: & *= ρg

2 μ

2− ρgh

μ +7 o

13


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P.LP.

$i%i:!8 a}} n,n dong ch%y &em nh3 chuyn dKng gia E tm ph†ng.ƒem ^ho%ng c(ch 2u vo v ra nh„ so vZi " ta c: BEO p dng c*ng th+c v0 l3u l34ng cht l„ng:

,

- =

7a

2 −

1

12 μ ( ) p) * )h3

,

b = 48a

2 −

1

12 μ ( Δ p

2π4 )a3

∆ p=24 μπ4

a3 (

48a

2 −

,

b )

M*ng sut ra:


15/15


P.FL.

$i%i:omen &o‚n do l]c ma s(t:

T =∫ 4

+

4

" .dA. 3=∫ 4

+

4 μ38

h (2 π 3 d 3 )3=2 πμ 8

h ∫

4+

4

33

d3=πμ8

2h ( 44− 4+

4)R

Ri

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