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Predavanja iz mehanike fluida M, Masinski fakultet.
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5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
1/10
CFD
CD
CD =CD(Re)
http://-/?-5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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n
Qi i= 1, 2, . . . , n
F(Q1, Q2, . . . , Qn) = 0,
F
Qi
m
F(1, 2, . . . , m) = 0.
p = n m Qi
Q Qj Qk Ql
i=Qxj Q
yjQ
zl Qi i= n 3.
x , y, z
[i] = [Qxj ] [Q
yj ] [Q
zl ] [Qi].
n
n
p
p
n
p
p
Qi
Qi = Qj = Qk =U Ql =D
=xUyDz
http://-/?-5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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= UD
=Re1
=xUyDz 1
= UD
= Re
x,y,z
SI
MLT
m kg M
L m L
t s T
A m2 L2
V m3 L3
U m/s LT1
a m/s2 LT2
F N MLT1
M Nm ML2T2
E
W J ML2T2
P W ML2T3
kg/m3 ML3
m kg/s MT1
V m3/s LT1
p
Pa ML1T2
Sij s1 T1
rad
rad/s T1
Pa s ML1T1
m2/s L2T1
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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p= f(U,,,L,D).
D [mm] L [m] V [m3/h] p [Pa] [kg/m3] [Pa s]
2.92 104
2.92 104
2.92 104
1 103
1 103
1 103
1.56 103
1.56 103
1.56 103
p
L
n = 6
F(p, U, , , L, D) = 0
p
=M L1T2
U
=LT1
=M L3
=M L1T1
L
=L
D
=L
{MLT}
m = n p = 6 3 = 3
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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pU
,,D
p=xyDz p, U=
abDc U, L = deDf L
p
M0L0T0 = (ML3)x (ML1T1)y Lz ML1T2
x + y+ 1 = 03x y+ z 1 = 0y 2 = 0
x= 1y= 2z= 2
,
p =D2 p
2
U
M0L0T0 = (ML3)a (ML1T1)b Lc LT1
a+ b= 03a b+ c 1 = 0b 1 = 0
a= 1b= 1c= 1
,
U= UD
=
UD
=Re
L
M0L0T0 = (ML3)e (ML1T1)f Lg L
e + f= 03e f+ g+ 1 = 0b= 0
e= 0f= 0
g= 1,
L= L
D
F1(p, U, L) = 0 F1D2 p
2
, Re, L
D = 0
D2 p
2 =F
Re,
L
D
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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D L V p 104 U 104Re 109p L/D
[mm] [m] [m3/h] [Pa] [kg/m3] [Pa s] m/s
2.92
2.471 3.732
2.92
4.942 17.78
2.92
8.236 56.46
10
1.765 0.8303
10 3.53 4.192
10
15.6
15.6
log p log Re
L/D
L/D
log p log Re
1e+06
1e+07
1e+08
1e+09
1e+10
1e+11
10000 100000
logRe
log
p
log
(p
/L
)
p Re p/L Re
L
D= 200
0.155Re1.75
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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D3 p
2L =f(Re)
p L
log
pL
= 1.75log Re 0.8097,
pL
= 0.155 Re1.75 D3 p
2L = 0.155 Re1.75
D3 p
2L = 0.155
UD
1.75
p U1.75
U
D
FDU2D2
=f
UD
w
U
D
wU2
=f
UD
,
L
D
P
V
D
P
3D5 =f
V
D3,D2
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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D
U
H/L
H L N
D
U =f
N,
H
L
Re 1
U
L
Cc c
FD/(0.5U2A)
d
FD = 3Ud Cc
FD = const U L; Cc= FDU L
= const;
Cc= 3
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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m
Um = U
5/21/2018 Dimenzionalna analiza, Mehanika fluida M.
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50km/h
350N
35 m
11 m/s
1 m
p0m= 0.2kPa