Chuong4 Cơ lưu chât
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Transcript of Chuong4 Cơ lưu chât
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CHNG 4: ONG LC HOC LU CHAT
I. Phng trnh vi phan chuyn ong cua lu chatII. Phng trnh nang lng
III. Tch phan phng trnh Euler
IV. Ptrnh Bernoulli cho dong chay lchat thc
V. ng dng phng trnh nng lng
VI. Phng trnh bien thien ong lngVII. ng dung phng trnh ong lng
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I. Phng trnh vi phan chuyn ong cua lu chat
1. Phng trnh Euler cho chuyen ong cua lu chat ly tng. Lu chat ly tng:=0 =0
khai niem ap suat:
Ngoai lc tac dung len phan t
tren phng x:
Lc khoi: Lc mat:
Phng trnh nh luat II Newton tren phng x cho phan t =>
Tng t:
hay
x
z
F
y
dx
dy
dz
p, 2
dx
x
pp
2
dx
x
pp
iip
1du
F grad pdt
. .x
dxdydz F
pdxdydz
x
1 yx x x zx x y z
udu u u upF u u u
dt x t x y z
1
1
y
y
zz
du pF
dt y
du pF
dt z
duF ma m
dt
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I. Phng trnh vi phan c.ong cua lu chat (tt)
2. Phng trnh Navier-Stokes cho chuyen ong cua lu chat thc
Phng trnh Navier-Stokes mieu ta dong chay cua cac chat lu trenc s bien thien ong lng trong nhng the tch vo cung nho cua chatlu n thuan ch la tong cua cac lc nht (nh lc ma sat), bien oi apsuat, trong lc, va cac lc khac tac ong len chat lu mot ng dungcua nh luat II Newton.
Lu chat thc:0 0 Ngoai lc tac dung len phan t tren
phng x:
Lc khoi:
Lc mat:
Viet phng trnh nh luat II Newton tren phng x cho phan t =>
1 yxx xx zxx
duF
dt x y z
yxxx zx dxdydzx y z
x
z
Fdx
dy
dz
dz
z
zx
zx
xx
zx
yx
dyy
yx
yx
dxx
xxxx
. . xdxdydz F
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Gia thiet Stokes:
vi
a ti phng trnh Navier-Stokes tren truc x:
Di dang vector:
oi vi lu chat khong nen c:
Lu y gia toc c tnh:
2 2 2
2 2 2
1 1
3
yx x x x x zx
udu u u u u upF
dt x x y z x x y z
21 1
3
duF grad p u u
dt
21du
F grad p udt
x y z
du u u u u uu u u u u
dt t x y z t
I. Phng trnh vi phan c.ong cua lu chat (tt)
1
3 xx yy zz p 23
ji lij ij ij
j i l
uu up
x x x
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II. Phng trnh nang lng
1. Phng trnh van tai nang lng
nh luat bao toan nang lng (L th nhat cua nhiet ong lc hoc):Toc o bien thien cua ong nang va noi nang bang tong cong c hoccua ngoai lc va cac dong nang lng khac tren 1 n v thi gian
e: noi nang (kh ly tng: ; chat long khong nen: ): dong nhiet rieng i vao qua be mat bao boc
nh luat truyen nhiet Fourier: Bien oi:
Thu c:
2
. .2
e
n n
V V S S
d ue dV F udV udS q dS
dt
eq
.
ij ij
n i j j i
S S Ve j j
n j j
S V V
udS u n dS u dV
q dS q dV T dV
q .grad T .T
2 1 1
.2
j ij j
j j i j
d ue F u u T
dt
Ve c T e cT
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II. Phng trnh nang lng (tt)
2. Phng trnh van tai ong nang
Phng trnh Navier di dang tensor:
Nhan phng trnh tren choui :
3. Phng trnh van tai noi nang Tr phng trnh van tai nang lng cho ptrnh van tai ong nang:
S dung gia thiet Stokes va cho lu chat khong nen c:
1 1j ijj j i
deT u
dt
1 yxx xx zxx
duF
dt x y z
1 ijii j
duF
dt
1 iji i j idu F udt
2
1 12
i ij iji j i j id u F u u u
dt
2
2
2
ji
j i
uudeT
dt x x
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III. Tch phan phng trnh euler
Phng trnh Euler dang Lambo-Gromeko:
Gia thiet:
= const
Phng trnh Euler dang Lambo-Gromeko thanh:
2 1
22
u ugrad u F grad p
t
F grad U
2
2 02
u p ugrad U u
t
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III. Tch phan phng trnh euler (tt)
1. Trng hp chuyen ong co the
Chuyen ong co the: va
Phng trnh Euler tren thanh:
Trong trng trong lc:U = - gz
(Tphan Lagrange)
oi vi chuyen ong on nh:
21
2
p uz C t
g t g
2
2
p uz C
g
u grad
0
2
0
2
p ugrad grad U
t
2
2
p uU C t
t
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III. Tch phan phng trnh euler (tt)
2. Trng hp lu chat chuyen ong on
nh, tphan doc ng dong Lay vi phan chieu dai ng
dong:
Nhan vo hng no vi pt. Euler:
Rut ra:
Trong trng trong lc: U = - gz
(Phng trnh Bernoulli)
ds
s
R
O
n
b
u
sd
nd
2
2 . 02
u p ugrad U u dst
2
02
p ud U
2
2
p uU C
2
2
p uz C
g
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III. Tch phan phng trnh euler (tt)
3. Trng hp lu chat chuyen ong on nh, tphan theo phng
vuong goc vi ng dong Phng trnh Euler trong he toa o t nhien:
Lay vi phan chieu dai ng phap tuyen vi ng dong: Nhan vo hng no vi pt. Euler:
Khi R :
Trong trng trong lc:U = - gz
(Tch phan Euler)
dn
n
pz C
2 22uu u pn grad U
t s R
2 22uu u pn dn grad U dn
t s R
2
n
u pdn d U
R
npU C
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III. Tch phan phng trnh euler (tt)
Yngha nang lng cua cac so hang tch phan Xet phng trnh Bernoulli. Qua trnh thiet lap qua cac bc:
Cac so hang:
Phng trnh Bernoulli la phng trnh bao toan nang lng
2
2 . 02
u p ugrad U u ds
t
2
02
p ud U
2
2p uU C
2
2
p uz C
g
Lc tren 1v Quang ng
klng lchat
Cong sinh ra t 1v klng lchat
Nang lng cua 1v trong lng lchat
Nang lng cua 1v klng lchat
va no khong thay oi trong cong
2
2
2
2
z p
u g
p uz
g
The nang cua 1v tlng lu chat (cot ap tnh)
ong nang cua 1v tlng lu chat (cot ap van toc)
Nang lng toan phan cua 1v tlng lu chat
(Cot ap toan phan)
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Xet oan dong chay on nh nam gia 2mcat t 1-1 va 2-2.
Xet 1 ng dong trong oan dong chay.Neu cho rang lu chat la ly tng, ptrnhBernoulli cho ng dong:
Phng trnh tren the hien tnh bao toan.Neu lu chat la thc th:
Bay gi xet 1 dong chay nguyen to. Nang lng cua no bien oi theo ptrnh:
Nh vay cho toan bo dong chay, nang lng cua no se bien oi theo ptrnh:
1
1
2
2
dQ
dQ
dQ
Q
2 2
1 1 2 21 2
2 2
p u p uz z
g g
2 2
1 1 2 21 2
2 2 f
p u p uz z h
g g
2 2
1 1 2 21 2
2 2 f
p u p uz dQ z dQ h dQg g
lchat)tlngv1cuanlngthatton:fh(
1 1 2 2
2 2
1 1 2 21 2
2 2 f
A A A A Q
p u p uz dQ dQ z dQ dQ h dQ
g g
IV. Phng trnh bernoulli cho dong chay lu chat thc
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IV. Phng trnh bernoulli cho dong chay lu chat thc
Thc hien cac tch phan:
Thay vao cho ket qua:
Ghi chu:
1. ieu kien ap dung pt Bernoulli cho dong chay: On nh, ; khong nen c=const; trong trng lc
Tai hai mcat ap dung pt, dong chay phai la bien oi cham. Trong oan dong chay gia 2 mcat, khong co nhap lu hoac tach lu.
2. Neu trong oan dong chay gia 2 mcat viet pt co turbine, may bm:
A
p pz dQ z Q
ieu kien : tai mcat t A dong chay la bien oi cham2 2
2 2A
u VdQ Q
g g
: he so hieu chnh ien nang,
31
1,05 1,10A
udA
A V
f f
Q
h dQ h Q
:fh ton that nang lng cua 1 v tlng lchat (ton that cot ap)
2 2
1 1 2 21 2
2 2 f
p V p Vz z h
g g
0t F g
f f T Bh h H H
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V. ng dng phng trnh nng lng
1. o lu tc (ong pitot)
Ap dung pt nang lng tren mot ng dong (Bernoulli) cho iem 1 va 2, vaxem lu chat ly tng
Ap dung tnh chat ap suat tnh hoc tuyet oi
Suy ra
Thay vao tren, ta co
2 2
1 1 2 21 2
2 2
1 1 2 2
1 2
2
1 2 12 1
2 2
2 2
1
2
k k
k k k
p V p Vz z
g g
p V p Vgz gz
V p pp p
1 1
2 2
k N k N
k M k M
p gz p gz
p gz p gz
2 1 2 1M N k M k N k kp p p p gz gz p p gh gh gh
2
1 1 12
k
k k
Vgh
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V. ng dng phng trnh nng lng
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V. ng dng phng trnh nng lng
1. o lu lng (ong Ventury) Ap dung pt nang lng gia hai mat
cat 1-1 va 2-2
C
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V. ng dng phng trnh nng lng
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V. ng dng phng trnh nng lng
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V. ng dng phng trnh nng lng
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V. ng dng phng trnh nng lng
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VI. Phng trnh bien thien ong lng
1. Phng trnh bien thien ong lng
Nguyen ly bien thien ong lng: toc o bien thien cua ong lngcua mot he vat chat bang vector tong ngoai lc tac dung len he.
Ap dung cho lu chat trong the tch kiem soat:
Bien oi:
oi vi dong chay on nh, ptrnh bien thien
ong lng la:
V
d
udV Rdt
V
S
uun
un.dS
n
n
V S
udV uu dS Rt
n
S
uu dS R
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VI. Phng trnh bien thien ong lng (tt)
2. Ptrnh bien thien lng cho dchay on nh cua lchat khong nen c
Xet the tch kiem soat la oan dong chay gia hai mcat 1-1 va 2-2Chia dien tch bao boc S=A1 +A2 + SnPtrnh bien thien ong lng thanh:
Tch phan th 3 bang khong con hai tchphan au c viet lai thanh:
Cac tch phan nay c thc hien:
Thay vao cho ket qua:
1 2 n
n n n
A A S
uu dS uu dS uu dS R
2
21
1
Sn
A2
A1
un=0
n
n
u
u
1 2A A
udQ udQ R
A
udQ V Q
2 2 1 1R Q V V
lng,hchnhhso:
2
1 1,02 1,05A
u dAA V
2 2 2 1 1 1R Q V Q V
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VII. ng dung phng trnh ong lng
1 1sin
NF Q V
2 21 22 12
N
q qF h h q
h h
1. Lc cua tia nc tac dung tren mot
tam phang nghieng goc , van toc
va lu lng en V1 va Q1. Xem
trong lng tia nc khong ang ke
Van toc am thanh 346 m/s (3 Mach=1035
m/s).
2. Lc cua dong chay tac dung len mot
tam chan co be rong bang 1 n v,
lu lng q va o sau h1, h2. Bo qua
ma sat ay.
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V du 1: Cho 1 voi co tiet dien A = 10 cm2, phun nc vi van toc v = 30m/svao tam phang at nam nghieng 1 goc=600 so vi phng ngang. Boqua ma sat, khong kh, hoi:
a) Neu tam phang ng yen (u = 0), lc F tac dung len tam phang, lu lngQ2, Q3.
b) Neu tam phang di chuyen (u = 10m/s), lc F tac dung len tam phang, phanlc N cua tam phang.
Giai:a) Lay the tch kiem soat nh hnh. Ngoai lc:
Trong lng nc trong TTKS
Phan lc cua tam phang
Phng trnh bien thien ong lng cho TTKS
Hay:
Chap nhan xap x:
G
'F
'2 2 2 3 3 3 1 1 1G F Q V Q V Q V
2 2 2 3 3 3 1 1 1(*)G F Q V Q V Q V
0 ( )
( 1,2,3)i
G G F
v v i
( )F u
V1 ,Q1
V2,Q2
V3,Q3
F
F
V d minh ha
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Chieu (*) len phng n:
- F = -.Q1.1.v1.sin => F=.A. v2
1.sin (**)Vi Q1 = v1.A ,1=1
Hay F=.A. v2.sin . The so F = 1000.10.10-4.302.sin600=779,4 N
Chieu (*) len phng :
0 =Q2 2v2 -Q3 3v3 -Q1 1v1cos
Suy ra: 0 = Q2 Q3 Q1cos (1)ptltc: Q1 = Q2 Q3 (2)
(1) va (2): Q2 = Q1(1+cos) / 2 ; Q3 = Q1(1-cos)/2
b) u = 10m/s
oi he quy chieu, xem tam phang ng yen, voi chuyen ong giat luivi van toc v1 = v-u. Suy ra: F =.A. (v-u)2.sin =346,4 N
Cong suat cua tam phang: N = F.u.sin =3000 W
Cong suat cua voi: Nv =Qv2/2g =.A. v3/2 = 13500 W
Hieu suat tam phang: = N/Nv= 22,22%
V d minh ha
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Cau hoi trac nghiem
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Cau hoi trac nghiem
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Cau hoi trac nghiem