ﻣﻜﺎﻧﻴﻚ ﺳﻴﺎﻻﺕ ﭘﻴﺸﺮﻓﺘﻪ ﻣﻬﻨﺪﺳﻲ ﺷﻴﻤﻲ
ﺗﺎﻟﻴﻒ ﻭ ﮔﺮﺩﺁﻭﺭﻱ:
ﺩﻛﺘﺮ ﻋﻠﻲ ﻣﺤﺒﻲ
ﺍﺳﺘﺎﺩ ﻣﻬﻨﺪﺳﻲ ﺷﻴﻤﻲ
)ﻋﻀﻮ ﻫﻴﺄﺕ ﻋﻠﻤﻲ ﺩﺍﻧﺸﮕﺎﻩ ﺷﻬﻴﺪ ﺑﺎﻫﻨﺮ ﻛﺮﻣﺎﻥ(
ﺣﻖ ﺗﻜﺜﻴﺮ ﻭ ﭼﺎپ ﻣﺨﺼﻮﺹ ﻣﻮﻟﻒ ﺍﺳﺖ.
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻳﺎﺩ ﺁﻭﺭﻱ :
ﺗﻌﺮﻳﻒ ﺳـﻴﺎﻝ :ﺳـﻴﺎﻝ ﺟﺴـﻤﻲ ﺍﺳـﺖ ﻛـﻪ ﺗﺤـﺖ ﺗـﺄﺛﻴﺮ ﺗـﻨﺶ ﺑﺮﺷـﻲ ) ﻓـﺮﻕ ﻧﻤـﻲ ﻛﻨـﺪ ﻛـﻪ ﺗـﻨﺶ ﺑﺮﺷـﻲ
ﻛﻮﭼﻚ ﺑﺎﺷﺪ ( ﺑﻪ ﻃﻮﺭ ﺩﺍﺋﻢ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺑﺪﻫﺪ.
ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ : 1
F0
ﺟﻬــﺖ ﻣﻄﺎﻟﻌــﻪ ﺣﺮﻛــﺖ ﺳــﻴﺎﻝ ﺍﺯ ﺭﻭﺵ ﺍﻭﻳﻠــﺮﻱ ﺍﺳــﺘﻔﺎﺩﻩ ﻣــﻲ ﻛﻨﻨــﺪ ﻛــﻪ ﺩﺭ ﻫــﺮ ﻧﻘﻄــﻪ ﺍﺯ ﻣﻴــﺪﺍﻥ ﺟﺮﻳــﺎﻥ
ﺳﺮﻋﺖ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﻣﻜﺎﻥ ﻭ ﺯﻣﺎﻥ ﻣﻲ ﺑﺎﺷﺪ ﻳﻌﻨﻲ:
)V = V (x, y, z, t
ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﻧﻤﺎﻳﺶ ﻛﺎﻣﻞ ﻫﺮ ﺧﺎﺻﻴﺖ ﺳﻴﺎﻝ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﻲ ﺑﺎﺷﺪ:
ﻭ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﺍﺋﻢ :
)η = η (x, y, z
ﻳﺎ
)η = η (x, y, z, t
dη
=0
dt
ﺑﻨﺎﺑﺮ ﺍﻳﻦ ﺩﺭ ﺟﺮﻳﺎﻥ ﺩﺍﺋﻢ ،ﺧﻮﺍﺹ ﺟﺮﻳﺎﻥ ﻣﻤﻜﻦ ﺍﺳﺖ ﺍﺯ ﻧﻘﻄﻪ ﺍﻱ ﺑﻪ ﻧﻘﻄﻪ ﺩﻳﮕﺮ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺗﻐﻴﻴﺮ ﻛﻨﺪ
ﻭﻟﻲ ﺁﻥ ﺧﻮﺍﺹ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ ﺑﺎ ﺯﻣﺎﻥ ﺛﺎﺑﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻳﻚ ،ﺩﻭ ،ﻭﺳﻪ ﺑﻌﺪﻱ :
ﺑﺴﺘﻪ ﺑﻪ ﺗﻌﺪﺍﺩ ﻣﺨﺘﺼﺎﺕ ﻣﻜﺎﻧﻲ ﻻﺯﻡ ﺑﺮﺍﻱ ﻣﺸﺨﺺ ﻧﻤﻮﺩﻥ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ،ﺟﺮﻳﺎﻥ ﺭﺍ ﻳﻚ – ﺩﻭ – ﻳﺎ ﺳﻪ ﺑﻌﺪﻱ
ﻣﻲ ﺧﻮﺍﻧﻨﺪ.
Velocity Field
۱
۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺧﻂ ﺟﺮﻳﺎﻥ : 1
F1
ﺧﻂ ﺟﺮﻳﺎﻥ ﺧﻄﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﭼﻨﺎﻥ ﺭﺳﻢ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺑﺮ ﺑﺮﺩﺍﺭﻫﺎﻱ ﺳﺮﻋﺖ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺍﺯ
ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﻣﻤﺎﺱ ﺑﺎﺷﺪ.
ﻣﺜﺎﻝ :1-1ﻣﻴﺪﺍﻥ ﺳﺮﻋﺘﻲ ﺑﻪ ﺻﻮﺭﺕ V = 2 yiˆ + ˆjﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﻭﺍﺣﺪ ﺳﺮﻋﺖ ft secﻭ yﺑﺮ
ﺣﺴﺐ ﻓﻮﺕ ﻣﻲ ﺑﺎﺷﺪ.
ﺍﻟﻒ :ﺑﻌﺪ ﺟﺮﻳﺎﻥ ﺭﺍ ﻣﺸﺨﺺ ﻛﻨﻴﺪ.
ﺏ :ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ v ، uﻭ wﺭﺍ ﺩﺭ ﻧﻘﻄﻪ) (x,y,z)=(1,2,0ﺣﺴﺎﺏ ﻛﻨﻴﺪ.
ﺝ :ﺷﻴﺐ ﺧﻂ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﺍﺯ ﻧﻘﻄﻪ ) (1,2,0ﻣﻲ ﮔﺬﺭﺩ ﺣﺴﺎﺏ ﻛﻨﻴﺪ.
ﺍﻟﻒ( ﭼﻮﻥ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺩﺍﺩﻩ ﺷﺪﻩ ﻓﻘﻂ ﺗﺎﺑﻊ ﻳﻜﻲ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﻣﻜﺎﻥ ﺍﺳﺖ) ،(yﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﻳﻚ ﺑﻌﺪﻱ
ﺍﺳﺖ.
Streamline
۲
۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺏ(
u = 2 y
V = uiˆ + vˆj + wkˆ
⇒ v = 1
ˆ
ˆ
=
2
+
V
y
i
j
w = 0
u = 4 ft
sec
x = 1
at (1,2,0 ) ⇒ y = 2 ⇒ v = 1 ft
sec
z = 0
w = 0
ﺝ(ﺑﻨﺎ ﺑـﻪ ﺗﻌﺮﻳـﻒ ﺧـﻂ ﺟﺮﻳـﺎﻥ ،ﺷـﻴﺐ ﺧـﻂ ﺟﺮﻳـﺎﻥ ﺩﺭ ﻧﻘﻄـﻪ ) (1,2,0ﭼﻨـﺎﻥ ﺍﺳـﺖ ﻛـﻪ ﺧـﻂ ﺟﺮﻳـﺎﻥ ﺑـﺮ
ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ ﺩﺭ ﺍﻳﻦ ﻧﻘﻄﻪ ﻣﻤﺎﺱ ﺍﺳﺖ.
v 1
=
u 4
= ) (1, 2 , 0
dy
dx
= ﺷﻴﺐ ﺧﻂ ﺟﺮﻳﺎﻥ
ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ :ﺑـﻪ ﺳـﻴﺎﻟﻲ ﮔﻔﺘـﻪ ﻣـﻲ ﺷـﻮﺩ ﻛـﻪ ﺩﺭ ﺁﻥ ﺗـﻨﺶ ﺑﺮﺷـﻲ ﻣﺴـﺘﻘﻴﻤﺎً ﺑـﺎ ﺷـﺪﺕ ﺗﻐﻴﻴـﺮ ﺷـﻜﻞ ﺁﻥ
ﻣﺘﻨﺎﺳــﺐ ﺑﺎﺷــﺪ ﺳــﻴﺎﻝ ﻧﻴﻮﺗــﻮﻧﻲ ﮔﻮﻳﻨــﺪ .ﺍﻛﺜــﺮ ﺳــﻴﻼﺗﻲ ﺍﺯ ﻗﺒﻴــﻞ ﺁﺏ ﻭ ﻫــﻮﺍ ﺗﺤــﺖ ﺷــﺮﺍﻳﻂ ﻋــﺎﺩﻱ ﻧﻴﻮﺗــﻮﻧﻲ
ﻫﺴﺘﻨﺪ.
ﺭﻓﺘﺎﺭ ﻳﻚ ﺍﻟﻤﺎﻥ ﺳﻴﺎﻝ ﺭﺍ ﺑﻴﻦ ﺩﻭ ﺻـﻔﺤﻪ ﺩﺭ ﺷـﻜﻞ ﺯﻳـﺮ ﻧﺸـﺎﻥ ﺩﺍﺩﻩ ﺷـﺪﻩ ﺍﺳـﺖ .ﺻـﻔﺤﻪ ﻓﻮﻗـﺎﻧﻲ ﺑـﺎ ﺳـﺮﻏﺖ
ﺛﺎﺑﺖ δuﺗﺤﺖ ﺗـﺄﺛﻴﺮ ﻧﻴـﺮﻭﻱ δ Fﺩﺭ ﺣـﺎﻝ ﺣﺮﻛـﺖ ﺍﺳـﺖ .ﺗـﻨﺶ ﺑﺮﺷـﻲ ﻭﺍﺭﺩﻩ ﺑـﺮ ﺍﻟﻤـﺎﻥ ﺳـﻴﺎﻝ ﺑـﻪ ﺻـﻮﺭﺕ
ﺯﻳﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
δFx dFx
=
δA →0 δA
dAy
y
τ yx = lim
y
۳
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻛﻪ δAyﺳﻄﺢ ﺍﻟﻤﺎﻥ ﺳﻴﺎﻝ ﺍﺳﺖ ﻛﻪ ﺑﺎ ﺻﻔﺤﻪ ﻓﻮﻗﺎﻧﻲ ﺩﺭ ﺗﻤﺎﺱ ﺍﺳﺖ.
ﺩﺭ ﺯﻣــﺎﻥ δ tﺍﻟﻤــﺎﻥ ﺳــﻴﺎﻝ ﺍﺯ ﺣﺎﻟــﺖ OPMNﺑــﻪ ﺣﺎﻟــﺖ OP ′M ′Nﺗﻐﻴﻴــﺮ ﺷــﻜﻞ ﺩﺍﺩﺍﻩ ﺍﺳــﺖ .ﺑﻨــﺎﺑﺮﺍﻳﻦ
ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻣﻲ ﺑﺎﺷﺪ:
δα dα
=
δ t dt
= δlimﺷﺪﺕ ﺗﻐﻴﻴﺮ ﺷﻜﻞ
t →0
ﺳﻴﺎﻝ ﻭﻗﺘﻲ ﻧﻴﻮﺗﻨﻲ ﺍﺳﺖ ﻛﻪ ﺗﻨﺶ ﺑﺮﺷﻲ ﻭﺍﺭﺩﻩ ﺑﺮ ﺁﻥ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺑﺎﺷﺪ:
δl
ﻭﻗﺘﻲ ﻛﻪ δαﻛﻮﭼﻚ ﺑﺎﺷﺪ ⇐
δy
dα
dt
∝ τ yx
δl = δu.δt
= tag δα ≈ δα
ﺩﺭ ﺷﺮﺍﻳﻂ ﺣﺪﻱ :
dα du
=
dt dy
δα δu
=
δt δy
⇒
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺮﺍﻱ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻨﻲ:
du
dy
∝ τ yx
ﻭﻳﺴﻜﻮﺯﻳﺘﻪ :
ﺿﺮﻳﺐ ﺗﻨﺎﺳﺐ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻓﻮﻕ ﺭﺍ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻣﻄﻠﻖ )ﻳﺎ ﺩﻳﻨﺎﻣﻴﻜﻲ( µﮔﻮﻳﻨﺪ.
du
dy
ﺑﻌﺪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ:
⇒ τ yx = µ
] µ : [Ft L2
] µ : [M Lt
cm. sec .
٤
cgs : Poise = gr
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻴﻨﻤﺎﺗﻴﻜﻲ ﺑﺎ ﻭﺍﺣﺪ : stoke
µ
ρ
2
sec .
ﺗﻘﺴﻴﻢ ﺑﻨﺪﻱ ﺣﺮﻛﺎﺕ ﺳﻴﺎﻝ:
=v
stoke = cm
ﻣﻜﺎﻧﻴﻚ ﺳﻴﺎﻻﺕ
ﺟﺴﻢ ﭘﻴﻮﺳﺘﻪ
ﻭﻳﺴﻜﻮﺯ
ﺩﺭﻫﻢ
ﻏﻴﺮ ﻭﻳﺴﻜﻮﺯ
ﺁﺭﺍﻡ
ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ
ﺗﺭﺍﮐﻡ ﭘﺫﻳﺭ
ﺗﺭﺍﮐﻡ ﻧﺎﭘﺫﻳﺭ
ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ
ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻭﻳﺴﻜﻮﺯ ﻭ ﻏﻴﺮ ﻭﻳﺴﻜﻮﺯ :
ﺧﺎﺻﻴﺖ ﺍﺻﻠﻲ ﺳﻴﺎﻝ ﻭﻳﺴﻜﻮﺯ :
ﺳـﻴﺎﻝ ﺩﺭ ﺗﻤــﺎﺱ ﻣﺴــﺘﻘﻴﻢ ﺑـﺎ ﺳــﻄﺢ ﻳــﻚ ﺟﺴـﻢ ﺟﺎﻣــﺪ ﺳــﺮﻋﺘﻲ ﺑﺮﺍﺑـﺮ ﺑــﺎ ﺟﺴــﻢ ﺟﺎﻣـﺪ ﺩﺍﺭﺩ .ﻳﻌﻨــﻲ ﺩﺭﺍﺛــﺮ
ﺗﻤﺎﺱ ﺳﻴﺎﻝ ﺑﺎ ﺟﺴﻢ ﺟﺎﻣﺪ ﺑﻪ ﺁﻥ ﻣﻲ ﭼﺴﺒﺪ .1
ﻻﻳﻪ ﻣﺮﺯﻱ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺩﺭ ﺭﻭﻱ ﻳﻚ ﺻﻔﺤﻪ ﻣﺴﻄﺢ :
F2
No–Slip Condition
٥
۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺟﺮﻳﺎﻥ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺩﻭ ﻧﺎﺣﻴﻪ ﺗﻘﺴﻴﻢ ﻧﻤﻮﺩ :
ﻻﻳﻪ ﻣﺮﺯﻱ ﻛﻪ ﺩﺭ ﺁﻥ ﺗﻨﺶ ﺑﺮﺷﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ) .ﻻﻳﻪ ﻧﺰﺩﻳﻚ ﺑﻪ ﺳﻄﺢ( ﻧﺎﺣﻴﻪ ﺧﺎﺭﺝ ﺍﺯ ﻻﻳﻪ ﻣﺮﺯﻱ ﻛﻪ ﺩﺭ ﺁﻥ ﺗﻨﺶ ﺑﺮﺷﻲ ﺻﻔﺮ ﺍﺳﺖ .ﺯﻳﺮﺍ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺻﻔﺮ ﺍﺳﺖ.ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺁﺭﺍﻡ ﻭ ﺩﺭﻫﻢ :
ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻟﺰﺝ ﺍﺯ ﻧﻘﻄﻪ ﻧﻈﺮ ﻣﺎﻛﺮﻭﺳﻜﻮﭘﻴﻚ ﺑﻪ ﺩﻭ ﺩﺳﺘﻪ ﺗﻘﺴﻴﻢ ﻣﻲ ﺷﻮﻧﺪ.
ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﻛﻪ ﺩﺭ ﺁﻥ ﺳﻴﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﻻﻳﻪ ﻫﺎﻳﻲ ﺩﺭ ﺭﻭﻱ ﻫﻢ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﻨﺪ ﻭ ﺍﺯ ﻧﻘﻄﻪ ﻧﻈﺮﻣﺎﻛﺮﻭﺳﻜﻮﭘﻴﻚ ﻫﻴﭻ ﮔﻮﻧﻪ ﺍﺧﺘﻼﻃﻲ 1ﺑﻴﻦ ﻻﻳﻪ ﻫﺎﻱ ﺳﻴﺎﻝ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ.
ﺟﺮﻳﺎﻥ ﺩﺭﻫﻢ ﺟﺮﻳﺎﻧﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﺳﻴﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﻻﻳﻪ ﻫﺎﻳﻲ ﺩﺭ ﺭﻭﻱ ﻫﻢ ﺣﺮﻛﺖ ﻧﻜﻨﻨﺪ ﻭ ﺍﺯ ﻧﻘﻄﻪ ﻧﻈﺮﻣﺎﻛﺮﻭﺳﻜﻮﭘﻴﻚ ﺍﺧﺘﻼﻁ ﺫﺭﺍﺕ ﺳﻴﺎﻝ ﺑﻴﻦ ﻻﻳﻪ ﻫﺎﻱ ﻣﺠﺎﻭﺭ ﺻﻮﺭﺕ ﮔﻴﺮﺩ.
F3
ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎ ﭘﺬﻳﺮ :
ﻭﻗﺘﻲ ρﺛﺎﺑﺖ ﺑﺎﺷﺪ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﺮﺍﻛﻢ ﻧﺎ ﭘﺬﻳﺮ ﮔﻮﻳﻨﺪ. ﻭﻗﺘﻲ ρﺗﻐﻴﻴﺮ ﻛﻨﺪ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﻧﺎﻣﻨﺪ.ﺟﺮﻳﺎﻥ ﻫﻮﺍ ﺩﺭ ﺳﺮﻋﺖ ﻫﺎﻱ ﻛﻢ ﺟﺮﻳﺎﻧﻲ ﺗﺮﺍﻛﻢ ﻧﺎ ﭘﺬﻳﺮ ﺍﺳﺖ ﻭﻟﻲ ﻭﻗﺘﻲ ﺳﺮﻋﺖ ﻫﻮﺍ ﺍﺯ ﻋﺪﺩ ﻣﺎﺥ 0.3ﺑﺰﺭﮔﺘﺮ ﺑﺎﺷﺪ
ρﺛﺎﺑﺖ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ ﻭ ﺟﺮﻳﺎﻥ ﻫﻮﺍ ﺟﺮﻳﺎﻧﻲ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﻣﻲ ﮔﺮﺩﺩ.
ﺳﺮﻋﺖ ﺳﻴﺎﻝ = v
ft
ﺩﺭ ﺷﺮﺍﻳﻂ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﺳﺮﻋﺖ ﺻﻮﺕ ﺩﺭ ﻫﻮﺍ ﺑﺮﺍﺑﺮ ﺑﺎ
sec
,
v
) ﺳﺮﻋﺖ ﺻﻮﺕ ﺩﺭ ﺳﻴﺎﻝ ( = c
c
=µ
1100ﺍﺳﺖ.
⟨100 mﺳﺮﻋﺖ ﻫﻮﺍ ⇒ M ⟨0.3
s
ﻣﻔﻬﻮﻡ ﭘﻴﻮﺳﺘﮕﻲ : 2ﺳﻴﺎﻝ ﺑﻪ ﻋﻨﻮﺍﻥ ﻳﻚ ﻣﺤﻴﻂ ﭘﻴﻮﺳﺘﻪ ﺍﺳﺖ.
ﻣﻔﻬﻮﻡ ﭘﻴﻮﺳﺘﮕﻲ ﭘﺎﻳﻪ ﺍﺻﻠﻲ ﻣﻜﺎﻧﻴﻚ ﺳﻴﺎﻻﺕ ﻛﻼﺳﻴﻚ ﻣﻲ ﺑﺎﺷﺪ.
ﻓــﺮﺽ ﭘﻴﻮﺳــﺘﮕﻲ ﺩﺭ ﻣﻄﺎﻟﻌــﻪ ﺳــﻴﺎﻻﺕ ﺗﺤــﺖ ﺷــﺮﺍﻳﻂ ﻃﺒﻴﻌــﻲ ﺻــﺎﺩﻕ ﻣــﻲ ﺑﺎﺷــﺪ .ﺍﻳــﻦ ﻓــﺮﺽ ﻭﻗﺘــﻲ ﻛــﻪ
ﻣﺘﻮﺳﻂ ﻓﺎﺻﻠﻪ ﺁﺯﺍﺩ ﺑـﻴﻦ ﻣﻮﻟﻜﻮﻟﻬـﺎﻱ ﺟﺴـﻢ ﺑـﺎ ﻛـﻮﭼﻜﺘﺮﻳﻦ ﺑﻌـﺪ ﻣﺸﺨﺼـﻪ ﻣﺴـﺄﻟﻪ ﻣـﻮﺭﺩ ﻧﻈـﺮ ﺩﺭ ﻳـﻚ ﺣـﺪ
ﺑﺎﺷﻨﺪ ﻧﻘﺾ ﻣﻲ ﺷﻮﺩ.
3
ﭘﻮﻳﺶ ﺁﺯﺍﺩ ﻣﻴﺎﻧﮕﻴﻦ ﻣﻮﻟﻜﻮﻟﻬﺎ :
* ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻓﺮﺽ ﭘﻴﻮﺳﺘﮕﻲ ﺧﻮﺍﺹ ﺳﻴﺎﻝ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ ﺗﻌﺮﻳﻒ ﻧﻤﻮﺩ.
4F
F5
۱
Mixing
Continuum
۳
Mean Free Path Of Molecules
۲
٦
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺗﻌﺮﻳﻒ ﺟﺮﻡ ﻣﺨﺼﻮﺹ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ:
ﺳﻴﺎﻝ ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣـﺎﺩﻩ ﺍﻱ ﺑـﻪ ﻃـﻮﺭ ﻧﺎﻣﺘﻨـﺎﻫﻲ ﻗﺎﺑـﻞ ﺗﻘﺴـﻴﻢ ،ﻳﻌﻨـﻲ ﻳـﻚ ﭘﻴﻮﺳـﺘﺎﺭ ،ﺩﺭ ﻧﻈـﺮ ﻣـﻲ ﮔﻴـﺮﻳﻢ ﻭ
ﺧﻮﺩﻣﺎﻥ ﺭﺍ ﺑﺎ ﺭﻓﺘﺎﺭ ﺗﻚ ﺗﻚ ﻣﻮﻟﻜﻮﻟﻬﺎ ﻣﺸﻐﻮﻝ ﻧﻤﻲ ﻛﻨﻴﻢ.
for all liquids and gases at atm pressure :
δV ′ = 10−9 mm3
δm
ρ = lim
δV →δV ′ δV
δv ′ﻛﻮﭼﻜﺘﺮﻳﻦ ﺟﺴﻤﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﻣﻲ ﺗﻮﺍﻥ ﺳﻴﺎﻝ ﺭﺍ ﭘﻴﻮﺳﺘﻪ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ.
ﺣﺠﻢ ﻫﺎﻱ ﻛﻮﭼﻜﺘﺮ ﺍﺯ ﺍﻳﻦ ﻣﻘﺪﺍﺭ ﻣﻨﺠﺮ ﺑﻪ ﺍﻳـﻦ ﻣﺴـﺄﻟﻪ ﻣـﻲ ﺷـﻮﺩ ﻛـﻪ ﺟـﺮﻡ ﺩﺭ ﻓﻀـﺎ ﺗﻮﺯﻳـﻊ ﺷـﺪﻩ ﺍﺳـﺖ ﺍﻣـﺎ
ﺩﺭ ﺫﺭﺍﺕ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻮﻟﻜﻮﻟﻬﺎ ،ﺍﺗﻢ ﻫﺎ ﻭ ﺍﻟﻜﺘﺮﻭﻥ ﻫﺎ ﻭ ...ﻣﺘﻤﺮﻛﺰ ﺷﺪﻩ ﺍﺳﺖ.
۷
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺩﺭ ﺣﺎﻟﺖ ﺣﺪﻱ ﺣﺠـﻢ ﺻـﻔﺮ ρ ،ﺑـﻲ ﻧﻬﺎﻳـﺖ )ﺣﺠﻤﺒﺨﺸـﻲ ﺍﺯ ﻫﺴـﺘﻪ( ﻭ ﻳـﺎ ﺑﺴـﻴﺎﺭ ﻛﻮﭼـﻚ ) ﺩﺭ ﺣﺠـﻢ ﺟـﺮﻡ
ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ ( ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ.
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺍﺯ ﻣﻴﺪﺍﻥ ﺟﺮﻳـﺎﻥ ﻣـﻲ ﺗـﻮﺍﻥ ﺟـﺮﻡ ﻣﺨﺼـﻮﺹ ﺭﺍ ﺑـﻪ ﺻـﻮﺭﺕ ﻓـﻮﻕ ﺗﻌﻴـﻴﻦ ﻧﻤـﻮﺩ .ﻳﻌﻨـﻲ
ﺟﺮﻡ ﻣﺨﺼﻮﺹ ﺭﺍ ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺖ:
) ρ = ρ ( x, y , z , t
ﻧﻜﺘﻪ :ﺩﺭ ﻧﺘﻴﺠﻪ ﻓﺮﺽ ﭘﻴﻮﺳﺘﺎﺭ ،ﻓـﺮﺽ ﻣـﻲ ﺷـﻮﺩ ﻫـﺮ ﺧﺎﺻـﻴﺖ ﺳـﻴﺎﻝ ﺩﺭ ﻫـﺮ ﻧﻘﻄـﻪ ﻱ ﻓﻀـﺎ ﻣﻘـﺪﺍﺭ ﻣﻌﻴﻨـﻲ
ﺩﺍﺭﺩ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺧﻮﺍﺹ ﺳﻴﺎﻝ ﺗﻮﺍﺑﻊ ﭘﻴﻮﺳﺘﻪ ﻱ ﻣﻜﺎﻥ ﻭ ﺯﻣﺎﻥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ.
ﻣﻌﺎﺩﻻﺕ ﺍﺻﻠﻲ ﺑﻪ ﻓﺮﻡ ﺍﻧﺘﮕﺮﺍﻝ ﺑﺮﺍﻱ ﺣﺠﻢ ﻛﻨﺘﺮﻝ :ﻣﻄﺎﻟﻌﻪ ﺣﺮﻛﺖ ﺳﻴﺎﻝ ﺭﺍ ﺑـﺎ ﺑـﻪ ﺩﺳـﺖ ﺁﻭﺭﺩﻥ ﻣﻌـﺎﺩﻻﺕ ﺍﺻـﻠﻲ ﺑـﻪ ﻓـﺮﻡ ﺍﻧﺘﮕـﺮﺍﻝ ﺟﻬـﺖ ﻛـﺎﺭﺑﺮﺩ ﺑـﺮﺍﻱ ﺣﺠـﻢ
ﻛﻨﺘﺮﻝ ﺁﻏﺎﺯ ﻣﻲ ﻛﻨﻴﻢ .ﭼﺮﺍ ﺣﺠﻢ ﻛﻨﺘـﺮﻝ؟ ﺯﻳـﺮﺍ ﺳـﻴﺎﻝ ﻣـﻲ ﺗﻮﺍﻧﻬـﺪ ﺑـﻪ ﻃـﻮﺭ ﭘﻴﻮﺳـﺘﻪ ﺗﻐﻴﻴـﺮ ﺷـﻜﻞ ﺑﺪﻫـﺪ ﻭ
ﺍﻏﻠﺐ ﺑﻲ ﻧﻬﺎﻳﺖ ﻣﺸﻜﻞ ﺍﺳـﺖ ﻛـﻪ ﺫﺭﺍﺕ ﺳـﻴﺎﻝ ﻣﺸـﺨﺺ ﻧﻤـﻮﺩ ﻭ ﺩﺭﻫـﺮ ﻟﺤﻈـﻪ ﺁﻧﻬـﺎ ﺭﺍ ﺩﻧﺒـﺎﻝ ﻧﻤـﻮﺩ .ﺩﻟﻴـﻞ
ﺩﻳﮕــﺮ ﺍﻳﻨﻜــﻪ ﻏﺎﻟﺒ ـﺎً ﺣﺮﻛــﺖ ﺳــﻴﺎﻝ ﻭ ﺭﻓﺘــﺎﺭ ﺁﻥ ﺩﺭ ﻳــﻚ ﻧﺎﺣﻴــﻪ ﻣﺸــﺨﺺ ﺍﺯ ﻓﻀــﺎ ﻣﻄــﺮﺡ ﺍﺳــﺖ ﻭ ﺣﺮﻛــﺖ
ﺫﺭﺍﺕ ﻣﺸﺨﺼﻲ ﺍﺯ ﺳﻴﺎﻝ ﻣﻮﺭﺩ ﻧﻈﺮ ﺍﺻﻼً ﻣﻄﺮﺡ ﻧﻴﺴﺖ.
ﻗﻮﺍﻧﻴﻦ ﺍﺻﻠﻲ ﺑﺮﺍﻱ ﻳﻚ ﺳﻴﺴﺘﻢ :
ﺑﻘﺎﻱ ﺟﺮﻡ :ﭼــﻮﻥ ﺳﻴﺴــﺘﻢ ﺑﻨــﺎ ﺑــﻪ ﺗﻌﺮﻳــﻒ ﻣﺠﻤﻮﻋــﻪ ﺍﻱ ﺍﺯ ﺫﺭﺍﺕ ﻣﺸــﺨﺺ ﻣــﺎﺩﻱ ﺍﺳــﺖ ،ﺑﻨــﺎﺑﺮﺍﻳﻦ ﺟــﺮﻡ ﺁﻥ ﺛﺎﺑــﺖ
ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﻳﻌﻨﻲ:
dµ
=0
dt sys
) → M = ∫mdm = ∫v ρdvﺟﺮﻡ ﺳﻴﺴﺘﻢ(
ﻗﺎﻧﻮﻥ ﺩﻭﻡ ﻧﻴﻮﺗﻦ :ﺑــﺮﺍﻱ ﺳﻴﺴــﺘﻤﻲ ﻛــﻪ ﻧﺴــﺒﺖ ﺑــﻪ ﻳــﻚ ﺩﺳــﺘﮕﺎﻩ ﻣﺨﺘﺼــﺎﺕ ﺍﻳﻨﺮﺳــﻲ )ﺳــﺎﻛﻦ( ﻣﺠﻤــﻮﻉ ﻧﻴﺮﻭﻫــﺎﻱ ﺧــﺎﺭﺟﻲ
ﻭﺍﺭﺩﻩ ﺑﺮ ﺳﻴﺴﺘﻢ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﻣﻤﻨﺘﻮﻡ ﺧﻄﻲ ﺳﻴﺴﺘﻢ ،ﻳﻌﻨﻲ:
d P
F = Fs + FB
dt sys
=F
ﻣﻤﻨﺘﻮﻡ ﺧﻄﻲ Pﺑﺮﺍﻱ ﻳﻚ ﺳﻴﺴﺘﻢ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ:
P sys = ∫ V dm = ∫ V ρdv
v
m
ﭼﮕﻮﻧﮕﻲ ﺍﺭﺗﺒﺎﻁ ﻣﺸﺘﻘﺎﺕ ﺳﻴﺴﺘﻢ ﺑﻪ ﺣﺠﻢ ﻣﻌﻴﺎﺭ:
ﺩﺭ ﺣﺎﻟــﺖ ﻛﻠــﻲ ﻫــﺮ ﺧﺎﺻــﻴﺖ ﻛﻤــﻲ )ﻣﻘــﺪﺍﺭﻱ( ﺳﻴﺴــﺘﻢ ﺭﺍ ﻣــﻲ ﺗــﻮﺍﻧﻴﻢ ﺑــﺎ ﻋﻼﻣــﺖ Nﻧﺸــﺎﻥ ﺑــﺪﻫﻴﻢ ﻭ
ﺧﺎﺻﻴﻴﺖ ﺷﺪﺗﻲ ﻣﺘﻨﺎﻇﺮ ﺑﺎ ﺁﻥ ﺭﺍ ﺑﺎ ηﻣﺸﺨﺺ ﻛﻨﻴﻢ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:
N sys = ∫ m ηdm = ∫ v ηρdv
۸
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺍﺻﻞ ﻣﻤﻨﺘﻮﻡ :
ﻣﺠﻤﻮﻉ ﺗﻤﺎﻡ ﮔﺸﺘﺎﻭﺭﻫﺎﻱ ﻣﺆﺛﺮ ﺑﺮ ﺳﻴﺴﺘﻢ = ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﻣﻤﻨﺘﻮﻡ ﺯﺍﻭﻳﻪ ﺍﻱ
sys
d H
dt
=T
: H sys = ∫mr × V dm = ∫vr × V ρdvﻣﻤﻨﺘﻮﻡ ﺯﺍﻭﻳﻪ ﺍﻱ ﺳﻴﺴﺘﻢ
: N = µ , η = 1ﺟﺮﻡ ﺳﻴﺴﺘﻢ
: N = P , η = Vﻣﻤﻨﺘﻮﻡ ﺧﻄﻲ ﺳﻴﺴﺘﻢ
: N = H , η = r × Vﻣﻤﻨﺘﻮﻡ ﺯﺍﻭﻳﻪ ﺍﻱ ﺳﻴﺴﺘﻢ
: N = E , η = eﺍﻧﺮژﻱ ﺳﻴﺴﺘﻢ
: N = S , η = sﺁﻧﺘﺮﻭﭘﻲ ﺳﻴﺴﺘﻢ
∂
dN
= ∫c.v.ηρdv + ∫c.s.ηρV ⋅ d A
dt sys ∂t
dN
ﻛﻪ ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﺧﺎﺻﻴﺖ ﻣﻘﺪﺍﺭﻱ Nﺩﺭ ﺳﻴﺴﺘﻢ ≡
dt sys
∂
ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﻫﺮ ﺧﺎﺻﻴﺖ ﻣﻘﺪﺍﺭﻱ Nﺩﺍﺧﻞ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ≡ ηρdv
∂t ∫c.v.
ﻛــﻪ ﺩﺭ ﺁﻥ ηﺧﺎﺻــﻴﺖ ﻣﻘــﺪﺍﺭﻱ ﻣﺘﻨــﺎﻇﺮ ﺑــﺎ Nﺍﺳــﺖ .ﻳﻌﻨــﻲ ﺧﺎﺻــﻴﺖ ﻛﻤــﻲ Nﺑــﺮ ﻭﺍﺣــﺪ ﺟــﺮﻡρdv .
ﺍﻟﻤﺎﻥ ﺟﺮﻡ ﺩﺍﺧﻞ ﺣﺠﻢ ﻛﻨﺘﺮﻝ.
ﻛﻞ ﺧﺎﺻﻴﺖ ﻣﻘﺪﺍﺭﻱ Nﺩﺍﺧﻞ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ﺍﺳﺖ ≡ ηρdv
∫
c.v.
ﺷﺪﺕ ﺧﺮﻭﺟﻲ ﺧﺎﻟﺺ ﺧﺎﺻﻴﺖ Nﺍﺯ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ≡ ∫ c.s. ηρV .d A
ﺳﺮﻋﺖ ﻧﺴﺒﺖ ﺑﻪ ﺳﻄﺢ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻣﻲ ﺷﻮﺩV ≡ .
ﺷﺪﺕ ﺧﺮﻭﺟﻲ ﺧﺎﺻﻴﺖ Nﺍﺯ ﺍﻟﻤﺎﻥ ﺳﻄﺢ dAﺣﺠﻢ ﻛﻨﺘﺮﻝ
ﺣﺠــﻢ ﻛﻨﺘــﺮﻝ ﺍﻳﻨﺮﺳــﻲ :ﺣﺠــﻢ ﻛﻨﺘﺮﻟــﻲ ﺍﺳــﺖ ﻛــﻪ ﻧﺴــﺒﺖ ﺑــﻪ ﻳــﻚ ﺩﺳــﺘﮕﺎﻩ ﻣﺨﺘﺼــﺎﺕ ﺍﻳﻨﺮﺳــﻲ )ﺳــﺎﻛﻦ(
ﺷﺘﺎﺏ ﻧﺪﺍﺷﺘﻪ ﺑﺎﺷﺪ.
≡ ηρV .d A
ﺁﻧﺎﻟﻴﺰ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺣﺮﻛﺖ ﺳﻴﺎﻝ:
ﺳﻴﺎﻝ ﺭﺍ ﻣﻲ ﺗـﻮﺍﻥ ﻳـﻚ ﺟﺴـﻢ ﭘﻴﻮﺳـﺘﻪ ﻓـﺮﺽ ﻧﻤـﻮﺩ ﻭ ﺩﺭ ﻧﺘﻴﺠـﻪ ﻣﻘـﺎﺩﻳﺮ ﺳـﺮﻋﺖ ،ﺟـﺮﻡ ﻣﺨﺼـﻮﺹ ﻭ ﻏﻴـﺮﻩ
ﺭﺍ ﺑﺮ ﺣﺴﺐ ﻣﺨﺘﺼﺎﺕ ﻣﻜﺎﻥ ﻭ ﺯﻣﺎﻥ ﺑﻴﺎﻥ ﻧﻤﻮﺩ ﻳﻌﻨﻲ:
ﻫﺮ ﺧﺎﺻﻴﺖ η = η (x, y, z, t) → η
V = V ( x, y, z , t ), ρ = ρ (x, y, z, t), etc...
ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ :
ˆV = uiˆ + vˆj + wk
۹
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻣﻌﺎﺩﻟﻪ ﺑﻘﺎﻱ ﺟﺮﻡ :
ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ﺭﺍ ﺑﺎ ﺯﻣﺎﻥ ﺛﺎﺑﺖ ﻓﺮﺽ ﻛﻨﻴﻢ ﺩﺍﺭﻳﻢ :
N = m ,η = 1
∂
⇒ ∫ c.v. ρdv + ∫ c.s. ρV .d A = 0
∂t
∂ρ
dv + ∫ c.s. ρV .d A = 0
∂t
∫ c .v .
divergence Theorem:
∫ c.s. f .d A = ∫ c.v. ∇. f dv
∂ρ
∂ρ
⇒ ∫ c.v.
dv + ∫ c.v. ∇.( ρV )dv =∫ c .v . ( + ∇.( ρV ))dv = 0
∂t
∂t
ﺍﻧﺘﮕــﺮﺍﻝ ﻓــﻮﻕ ﻣﺴــﺘﻘﻞ ﺍﺯ ﻣﺸﺨﺼــﺎﺕ ﺣﺠــﻢ ﻛﻨﺘــﺮﻝ ﻫﻤــﻮﺍﺭﻩ ﺑﺮﺍﺑــﺮ ﺻــﻔﺮ ﺍﺳــﺖ ﻭ ﺍﻳــﻦ ﻣﻄﻠــﺐ ﺩﺭ ﺻــﻮﺭﺗﻲ
ﺩﺭﺳﺖ ﺍﺳﺖ ﻛﻪ ﺗﺎﺑﻊ ﺯﻳﺮ ﺍﻧﺘﮕﺮﺍﻝ ﺻﻔﺮ ﺑﺎﺷﺪ ،ﻳﻌﻨﻲ:
∂ρ
+ ∇.( ρV ) = 0
∂t
)∂ρ ∂ ( ρu ) ∂ ( ρv) ∂ ( ρw
⇒
+
+
+
=0
dz
∂t
∂x
∂y
) ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﭘﻴﻮﺳﺘﮕﻲ (
ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺟﺮﻳﺎﻥ ﺩﺍﺋﻢ ﺑﺎﺷﺪ:
ﻭﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺑﺎﺷﺪ ﻳﻌﻨﻲ ﺛﺎﺑﺖ = ρ
∂ρ
= 0 ⇒ ∇.( ρ V) = 0
∂t
⇒ ∇.V = 0
ﺍﮔــﺮ ﺑﺨــﻮﺍﻫﻴﻢ ﻫﻤــﻴﻦ ﻣﺴــﺄﻟﻪ ﺭﺍ ﺍﺯ ﺭﻭﺵ ﺍﻟﻤــﺎﻥ ﮔﻴــﺮﻱ ﺣــﻞ ﻛﻨــﻴﻢ ﻳــﻚ ﺍﻟﻤــﺎﻥ ﻣﻜﻌــﺐ ﻣﺴــﺘﻄﻴﻞ ﺩﺭ ﻧﻈــﺮ
ﻣﻲ ﮔﻴﺮﻳﻢ :
∂
∫ c.v. ρdv + ∫ c.s. ρV .d A = 0
∂t
۱۰
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
Accumulation=Input –output
accumulation =
∂mc.v. ∂ρ
dxdydz
=
∂t
∂t
∂ρ
∂ρu
dxdydz = ρudydz − ( ρu +
dx)dydz
∂t
∂x
∂ρv
dy )dxdz
+ ρvdxdz − ( ρv +
∂y
∂ρw
dz )dxdy
+ ρwdxdy − ( ρw +
∂z
∂ρ
∂ ( ρu )
∂ ( ρv)
∂ ( ρw)
dxdydz = −
dxdydz −
dxdydz −
dxdydz
⇒
∂t
∂x
∂y
∂z
∂ρ ∂
∂
∂
⇒
+ ( ρu ) + ( ρy ) + ( ρw) = 0
∂t ∂x
∂y
∂z
⇒
: 1 ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ
F6
acc. =Input – output
∂ρ
∂
rdθdzdr = ( ρvr )rdθdz − ( ρvr + ( ρvr )dr )rdθdz
∂t
∂r
∂
+ ρvθ drdz − ( ρvθ +
( ρvθ )dθ )drdz
∂θ
∂
+ ρv z rdθdr − ( ρv z + ( ρv z )dz )rdθdr
∂z
∂ρ 1 ∂
∂
1 ∂
⇒
+
(rρvr ) +
( ρvθ ) + ( ρv z ) = 0
∂t r ∂r
∂z
r ∂θ
۱
Continuity Equation In Cylindrical Coordinat
۱۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﺍﭘﺮﺍﺗﻮﺭ ∇ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺎﺭﺗﺰﻳﻦ ﭼﻨﻴﻦ ﺍﺳﺖ :
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ :
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺮﻭﻱ :
∂
∂
∂
) ( ̂∇ = iˆ ( ) + ˆj ( ) + k
∂x
∂y
∂z
∂
∂ 1
∂ 1
∇. A = iˆr
(rAr ) +
) ( Aθ ) + ( Az
∂z
r ∂r
r ∂θ
∂
∂ 1
∂ 1
= ∇.ρV
(rρvr ) +
) ( ρvθ ) + ( ρv z
r ∂r
r ∂θ
∂z
x = r sin θ cosφ
y = r sinθ sinφ
z = r cosθ
∂ ∂ρ 1
∂
1
∂ 1
+ 2 (rρ 2 vr ) +
( ρvθ sin θ ) +
( ρvφ ) = 0
∂t r ∂r
r sinθ ∂θ
r sinθ ∂φ
∂ 1
1
∂
∂ 1
∇.ρV = 2 (r 2 ρvr ) +
( ρvθ sin θ ) +
) ( ρvφ
r ∂r
r sinθ ∂θ
r sinθ ∂φ
ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ :
∂ρ
+ ∇.( ρV ) = 0
∂t
ﺩﺭ ﻣﻮﺭﺩ ﻳﻚ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ )ﺛﺎﺑﺖ = ( ρﺩﺍﺭﻳﻢ :
∂u ∂v
+
=0
∂x ∂y
ﺍﮔﺮ ﺗﺎﺑﻊ ﭘﻴﻮﺳﺘﻪ ﻱ ) ψ ( x, y, tﺑﻪ ﻧﺎﻡ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺗﻌﺮﻳﻒ ﻛﻨﻴﻢ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ :
∂ψ
∂x
⇒ ∇.V = 0
, v=−
∂ψ
∂y
=u
ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﻳﻦ ﺗﺎﺑﻊ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﺻﺪﻕ ﻣﻲ ﻛﻨﺪ ﻳﻌﻨﻲ :
∂u ∂v ∂ 2ψ
∂ 2ψ
+
=
−
=0
∂x ∂y ∂x∂y ∂y∂x
ﺧﻄــﻮﻁ ﺟﺮﻳــﺎﻥ ﺩﺭ ﻫــﺮ ﻟﺤﻈــﻪ ﺑــﺮ ﺑــﺮﺩﺍﺭ ﺳــﺮﻋﺖ ﻣﻤــﺎﺱ ﻣــﻲ ﺑﺎﺷــﻨﺪ .ﺑﻨــﺎﺑﺮﺍﻳﻦ ﺍﮔــﺮ d rﺍﻟﻤــﺎﻥ ﻃــﻮﻝ ﺩﺭ
ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﺎﺷﺪ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ.
۱۲
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
V × d r = 0 = (iˆu + ˆjv) × (iˆdx + ˆjdy ) = kˆ(udy − vdx) = 0
ﻣﻌﺎﺩﻟﻪ ﺧﻂ ﺟﺮﻳﺎﻥ ﺩﺭ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ
ﺍﮔﺮ ﺑﻪ ﺟﺎﻱ uﻭ vﺑﺮ ﺣﺴﺐ ψﺟﺎﻧﺸﻴﻦ ﻛﻨﻴﻢ ﺩﺍﺭﻳﻢ:
⇒ udy − vdx = 0
∂ψ
∂ψ
dx +
dy = 0
∂x
∂y
ﺍﺯ ﺁﻧﺠــﺎ ﻛــﻪ ) ψ = ψ ( x, y, tﺩﺭ ﻟﺤﻈــﻪ ﻱ t0ﺩﺍﺭﻳــﻢ ) .ψ = ψ ( x, y, t0ﺩﺭ ﺍﻳــﻦ ﻟﺤﻈــﻪ ﺑــﺎ ﺗﺼــﻮﺭ ﺍﻳﻨﻜــﻪ
) ψ = ψ ( x, yﻣﻲ ﺗﻮﺍﻥ ﺗﻐﻴﻴﺮ ψﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻛﺮﺩ.
∂ψ
∂ψ
dx +
dy
∂x
∂y
⇒ dψ = 0 ⇒ ψ = constant along a streamline
= dψ
ﻳﻌﻨــﻲ ψﺩﺭ ﺭﻭﻱ ﺧــﻂ ﺟﺮﻳــﺎﻥ ﺛﺎﺑــﺖ ﺍﺳــﺖ .ﺑﻨــﺎﺑﺮﺍﻳﻦ ﺧﻄــﻮﻁ ﺛﺎﺑــﺖ = ψﺧﻄــﻮﻁ ﺟﺮﻳــﺎﻥ ﺭﺍ ﻧﺸــﺎﻥ ﻣــﻲ
ﺩﻫﻨﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎ ﺗﻮﺟﻪ ﺑـﻪ ﺗﻌﺮﻳـﻒ ﺧـﻂ ﺟﺮﻳـﺎﻥ ،ﺗﺸـﺨﻴﺺ ﻣـﻲ ﺩﻫـﻴﻢ ﻛـﻪ ﻫـﻴﭻ ﺟﺮﻳـﺎﻧﻲ ﺧـﻂ ﺟﺮﻳـﺎﻥ ﺭﺍ
ﻗﻄﻊ ﻧﻤﻲ ﻛﻨﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺑـﺮﺍﻱ ﺟﺮﻳـﺎﻥ ﺩﻭ ﺑﻌـﺪﻱ ﻭ ﺗـﺮﺍﻛﻢ ﻧﺎﭘـﺬﻳﺮ ﺩﺭ ﻫـﺮ ﻟﺤﻈـﻪ ﺍﺯ ﺯﻣـﺎﻥ ﺩﺑـﻲ ﺳـﻴﺎﻝ ﺑـﻴﻦ
ﺩﻭ ﺧﻂ ﺟﺮﻳﺎﻥ ﺩﺭ ﺗﻤﺎﻡ ﻣﻘﺎﻃﻊ ﻳﻜﺴﺎﻥ ﺍﺳﺖ.
ﺩﺑﻲ ﺩﺭ ﺭﻭﻱ ﺧﻂ ABﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ:
∂ψ
ﻭ ﭼﻮﻥ ﺛﺎﺑﺖ = xﺍﺳﺖ ﺩﺭ ﻧﺘﻴﺠﻪ dy :
∂y
∂ψ
dy
∂y
y2
y2
y1
y1
∫ = QAB = ∫ udy
= dψ
y 2 dψ
y2
∂ψ dψ
=
∫ = ⇒ QAB
⋅ dy = ∫ dψ =ψ 2 −ψ 1
y1 dy
y1
dy
∂y
ﺩﺑﻲ ﺩﺭ ﺭﻭﻱ ﺧﻂ BCﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
∂ψ
dx
∂x
۱۳
x2
x2
x1
x1
⇒
∫ QBC = ∫ vdx = −
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
∂ψ
ﻭ ﭼﻮﻥ ﺛﺎﺑﺖ = yﺍﺳﺖ ﺩﺭ ﻧﺘﻴﺠﻪ dx :
∂x
= dψ
ψ2
x2
ψ1
x1
⇒ QBC = − ∫ dψ = − ∫ dψ = ψ 2 − ψ 1
ﻳﻌﻨﻲ ﺩﺑﻲ ﺑﻴﻦ ﺩﻭ ﺧﻂ ﺟﺮﻳﺎﻥ ψ 1ﻭ ψ 2ﺩﺭ ﻫﺮ ﺣﺎﻝ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ:
Q = QAB = QBC = ψ 2 −ψ 1
ﻳﻌﻨــﻲ ﺩﺑــﻲ ﺣﺠﻤــﻲ ﺑــﻴﻦ ﻫــﺮ ﺩﻭ ﻧﻘﻄــﻪ ﺩﺭ ﻣﻴــﺪﺍﻥ ﺟﺮﻳــﺎﻥ ﺑﺮﺍﺑــﺮ ﺑــﺎ ﺗﻐﻴﻴــﺮ ﺩﺭ ﺗﻮﺍﺑــﻊ ﺟﺮﻳــﺎﻥ ﺑــﻴﻦ ﺁﻥ ﺩﻭ
ﻧﻘﻄﻪ ﺍﺳﺖ.
ﺟﻬﺖ:
Q1−2 = ∫12 (V .d A) = ∫12 dψ = ψ 2 −ψ 1
∂
∂
( ρu ) + ( ρv) = 0
∂x
∂y
∂ψ
∂ψ
= ρu
, ρv = −
∂y
∂x
Compressible flow:
•
m1→2 = ∫12 ρV .d A = ψ 2 −ψ 1
Incompressible flow in polar coordinate
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ) : (r ,θ
∂1
∂ 1
(rvr ) +
(vθ ) = 0 , vz = 0
rr
r ∂θ
∂v
∂
⇒ (rvr ) + θ = 0
∂r
∂θ
∂ ∂ψ
∂
∂ψ
( ⇒
)+
(−
)=0
∂r ∂θ
∂θ
∂r
∂ψ
1 ∂ψ
= ⇒ vr
, vθ = −
r ∂θ
∂r
Q1→2 = ψ 2 −ψ 1
Q1→2 = ∫∫ rdθdzvr
1 ∂ψ
r ∂θ
۱٤
2
= ∫ ∫ rdθdz
1
= ψ 2 −ψ 1
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
Incompressible Axisymmettic flow
∂
=0
∂θ
∂
∂ 1
(rvr ) + (v z ) = 0
⇒ ﭘﻴﻮﺳﺘﮕﻲ
r ∂r
∂z
= vθ
ﻛﻪ rﻭ zﻣﺘﻐﻴﺮﻫﺎﻱ ﻣﺴﺘﻘﻞ ﻫﺴﺘﻨﺪ.
∂
∂
(rvr ) + (v z ) = 0
∂r
∂z
∂
∂ψ
∂ ∂ψ
⇒ (−
( )+
)=0
∂r
∂z
∂z ∂r
1 ∂ψ
1 ∂ψ
⇒ vr = −
= , vz
r ∂z
r ∂r
) Q1→2 = 2π (ψ 2 −ψ 1
⇒
r2
Q1→2 = ∫ 2πrdr ⋅ u
r1
1 ∂ψ
⋅ rdr
r ∂r
) = 2π (ψ 2 −ψ 1
r2
∫ = 2π
r1
ﻣﺜــﺎﻝ :ﻣﻴــﺪﺍﻥ ﺳــﺮﻋﺖ ﺟﺮﻳــﺎﻥ ﺗــﺮﺍﻛﻢ ﻧﺎﭘــﺬﻳﺮ ﻭ ﭘﺎﻳــﺪﺍﺭ ﺑــﺎ V = Axiˆ − Ayˆjﺩﺍﺩﻩ ﺷــﺪﻩ ﺍﺳــﺖ ﻛــﻪ
A = 2s −1ﺍﺳﺖ .ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
∂ψ
∂ψ
, v=−
∂y
∂x
∂ψ
∂ψ
∫ = ⇒ψ
= u = Ax
)dy + f ( x) = Axy + f ( x
∂y
∂y
=u
) f(xﻳﻚ ﺗﺎﺑﻊ ﺍﺧﺘﻴﺎﺭﻱ ﺍﺳﺖ.
⇒ ψ = Axy + C
∂ψ
df
= − Ay −
df
v = −
= 0 ⇒ f ( x) = constant
∂x
⇒ dx
dx
v = − Ay
ﺧﻄــﻮﻁ ψﺛﺎﺑــﺖ ﻧﺸــﺎﻥ ﺩﻫﻨــﺪﻩ ﻱ ﺧﻄــﻮﻁ ﺟﺮﻳــﺎﻥ ﺩﺭ ﻣﻴــﺪﺍﻥ ﺟﺮﻳــﺎﻥ ﻫﺴــﺘﻨﺪ .ﺑــﺮﺍﻱ ﺗﺮﺳــﻴﻢ ،ﻣــﻲ ﺗــﻮﺍﻥ
ﺛﺎﺑﺖ Cﺭﺍ ﻫﺮ ﻣﻘـﺪﺍﺭ ﻣﻨﺎﺳـﺐ ﺍﻧﺘﺨـﺎﺏ ﻛـﺮﺩ .ﺛﺎﺑـﺖ Cﺭﺍ ﺻـﻔﺮ ﺍﻧﺘﺨـﺎﺏ ﻣـﻲ ﻛﻨـﻴﻢ ﺗـﺎ ﺧـﻂ ﺟﺮﻳـﺎﻧﻲ ﻛـﻪ ﺍﺯ
ﻣﺒﺪﺃ ﻣﻲ ﮔﺬﺭﺩ ﺑـﺎ ψ = ψ 1 = 0ﻧﺸـﺎﻥ ﺩﺍﺩﻩ ﺷـﻮﺩ .ﺩﺭ ﺍﻳـﻦ ﺻـﻮﺭﺕ ﻣﻘـﺪﺍﺭ ﻫـﺮ ﺧـﻂ ﺟﺮﻳـﺎﻥ ﺩﻳﮕـﺮ ،ﺟﺮﻳـﺎﻥ
ﺑﻴﻦ ﻣﺒﺪﺃ ﻭ ﺁﻥ ﺧﻂ ﺟﺮﻳﺎﻥ ﺭﺍ ﺍﺯ ﭼﭗ ﺑﻪ ﺭﺍﺳﺖ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ.
۱٥
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
m3
s
m
: ψ = 2 xyﺩﺭ ﺭﺑﻊ ﺍﻭﻝ
ﺩﺭ ﺭﺑﻊ ﺍﻭﻝ ،ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ u⟩ 0ﻭ v⟨0ﺟﺮﻳﺎﻥ ﺍﺯ ﭼﭗ ﺑﻪ ﺭﺍﺳﺖ ﻭ ﺑﻪ ﻃﺮﻑ ﭘﺎﻳﻴﻦ ﺍﺳﺖ.
m3
s
m
Q1 2 = ψ 2 −ψ 1 = 2
ﺩﺭ ﺭﺑﻊ ﺩﻭﻡ u ⟨0ﻭ v⟨0ﺟﺮﻳﺎﻥ ﺍﺯ ﺭﺍﺳﺖ ﺑﻪ ﭼﭗ ﻭ ﺑﻪ ﭘﺎﻳﻴﻦ ﺍﺳﺖ.
3
m
s
m
ﻋﻼﻣﺖ ﻣﻨﻔﻲ ﺑﺎ ﺟﺮﻳﺎﻥ ﺍﺯ ﺭﺍﺳﺖ ﺑﻪ ﭼﭗ ﺳﺎﺯﮔﺎﺭ ﺍﺳﺖ.
Q7 9 = ψ 9 −ψ 7 = −8 − (−4) = −4
ﻣﺜــﺎﻝ :ﻣﻴــﺪﺍﻥ ﺳــﺮﻋﺖ V = −3 yiˆ + 3xˆjﺩﺍﺩﻩ ﺷــﺪﻩ ﺍﺳــﺖ .ﺧــﺎﻧﻮﺍﺩﻩ ﺗــﺎﺑﻊ ﺟﺮﻳــﺎﻥ ψﻛــﻪ ﻣﻨﺠــﺮ ﺑــﻪ ﺍﻳــﻦ
ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﻣﻲ ﺷﻮﺩ ،ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ.
∂ψ
∂ψ
3
∫ = ⇒ϕ
)dy = ∫ −3 ydy = − y 2 + f ( x
∂y
∂y
2
∂ψ
v = −ﺑـﻪ ﺩﺳـﺖ
ﻛﻪ ) f(xﻳـﻚ ﺗـﺎﺑﻊ ﺍﺧﺘﻴـﺎﺭﻱ ﺍﺳـﺖ .ﺗـﺎﺑﻊ ) f(xﺭﺍ ﻣـﻲ ﺗـﻮﺍﻥ ﺑـﺎ ﺍﺳـﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻌﺎﺩﻟـﻪ
∂x
= u = −3 y
ﺁﻭﺭﺩ.
۱٦
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
∂ψ
∂ψ
)∂f ( x
)df ( x
= +3 x ⇒ 3 x = −
=−
=−
∂x
∂x
∂x
dx
)df(x
3
∫ = )⇒ f ( x
dx = ∫ −3 xdx = − x 2 + c
dx
2
3
3
⇒ψ = − x2 − y2 + c
2
2
v=−
Check:
∂ψ
∂
3
3
= (− x 2 − y 2 + c) = −3 y
∂y ∂y 2
2
∂ψ
∂
( =−
v=−
) = 3x
∂x
∂x
=u
ﺭﻭﺵ ﻻﮔﺮﺍﻧﮋ :ﺍﻳﻦ ﺭﻭﺵ ﻭﻗﺘـﻲ ﺑـﻪ ﻛـﺎﺭ ﻣـﻲ ﺭﻭﺩ ﻛـﻪ ﺩﻧﺒـﺎﻝ ﻧﻤـﻮﺩﻥ ﺫﺭﺍﺕ ﻣﺸـﺨﺺ ﺟـﺮﻡ ﺁﺳـﺎﻥ ﺑﺎﺷـﺪ .ﺑـﻪ
ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ ﺩﺭ ﻣﻜﺎﻧﻴﻚ ﺫﺭﻩ ﺍﻱ.
ﺑﺮﺩﺍﺭ ﻣﻜﺎﻥ ﺫﺭﻩ
) r = r (t
ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ ﺫﺭﻩ
V = dr
dt
2
a=d r 2
dt
ﺑﺮﺩﺍﺭ ﺷﺘﺎﺏ ﺫﺭﻩ
ﺭﻭﺵ ﺍﻭﻳﻠﺮ :ﺍﻳﻦ ﺭﻭﺵ ﻭﻗﺘـﻲ ﺑـﻪ ﻛـﺎﺭ ﻣـﻲ ﺭﻭﺩ ﻛـﻪ ﺩﻧﺒـﺎﻝ ﻧﻤـﻮﺩﻥ ﺫﺭﺍﺕ ﻣﺸـﺨﺺ ﺟـﺮﻡ ﺗﻘﺮﻳﺒـﺎً ﻛـﺎﺭﻱ ﻏﻴـﺮ
ﻣﻤﻜﻦ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﭼﻨﻴﻦ ﺷﺮﺍﻳﻄﻲ ﻣﻨﺎﺳﺐ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﺗﻮﺻﻴﻒ ﻣﻴﺪﺍﻥ ،ﻭ ﻳﺎ ﺍﻭﻳﻠﺮ ،ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ.
) V = V ( x, y , z , t
ﺩﺭ ﺍﻳﻦ ﺭﻭﺵ ﺧﻮﺍﺹ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ 1ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﺯﻣﺎﻥ ﻭ ﻣﻜﺎﻥ ﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ.
Dt
a = DV
F7
) ρ = ρ ( x, y , z , t
ﺷﺘﺎﺏ ﻳﻚ ﺫﺭﻩ ﺳﻴﺎﻝ:
ﺭﻭﺵ ﻻﮔﺮﺍﻧــﮋ :ﺑﺴــﻴﺎﺭ ﻣﺸــﻜﻞ ﺍﺳــﺖ ﻭ ﻋﻤــﻼً ﻏﻴــﺮ ﻣﻤﻜــﻦ ﻣــﻲ ﺑﺎﺷــﺪ ﭼــﻮﻥ ﺑــﺮﺍﻱ ﻫــﺮ ﺫﺭﻩ ﻣﻌﺎﺩﻟــﻪ ﺍﻱ
ﺟﺪﺍﮔﺎﻧﻪ ﻻﺯﻡ ﺍﺳﺖ ،ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﻧﻮﺷﺘﻦ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻫﺎ ﺑﺮﺍﻱ ﺑﺴﻴﺎﺭﻱ ﺍﺯ ﺫﺭﺍﺕ ﻣﺸﻜﻞ ﺁﻓﺮﻳﻦ ﺍﺳﺖ.
ﺷﺘﺎﺏ ﻳﻚ ﺫﺭﻩ ﺳﻴﺎﻝ ﺩﺭ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ :ﺭﻭﺵ ﺍﻭﻳﻠﺮﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺫﺭﻩ ﺍﻱ ﻛـﻪ ﺩﺭ ﻣﻴـﺪﺍﻥ ﺳـﺮﻋﺖ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ،ﺗﻮﺻـﻴﻒ ﻛﻠـﻲ ﺗـﺮﻱ ﺑـﺮﺍﻱ ﺷـﺘﺎﺏ ﻣـﻲ
ﺗــﻮﺍﻥ ﺑــﻪ ﺩﺳــﺖ ﺁﻭﺭﺩ .ﻣــﻲ ﺧــﻮﺍﻫﻴﻢ ﺑــﺎ ﺗﻮﺟــﻪ ﺑــﻪ ﻣﻴــﺪﺍﻥ ﺳــﺮﻋﺖ ) V = V ( x, y, z, tﺷــﺘﺎﺏ ﻳــﻚ ﺫﺭﻩ ﻱ
ﺳﻴﺎﻝ a p ،ﺭﺍ ﭘﻴﺪﺍ ﻣﻤﻨﻴﻢ.
Flow Field
۱۷
۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺫﺭﻩ ﺍﻱ ﺭﺍ ﺩﺭ ﻧﻈــﺮ ﺑﮕﻴﺮﻳــﺪ ﻛــﻪ ﺩﺭ ﻣﻴــﺪﺍﻥ ﺳــﺮﻋﺖ ﺣﺮﻛــﺖ ﻣــﻲ ﻛﻨــﺪ .ﺩﺭ ﺯﻣــﺎﻥ tﺫﺭﻩ ﺩﺭ ﻳــﻚ ﻣﻜــﺎﻥ ،x
z ،yﺍﺳﺖ ﻭ ﺳﺮﻋﺖ ﺁﻥ ﻣﺘﻨﺎﻇﺮ ﺍﺳﺖ ﺑﺎ ﺳﺮﻋﺖ ﺩﺭ ﺁﻥ ﻧﻘﻄﻪ ﺩﺭ ﺯﻣﺎﻥ : t
) V p = V ( x, y , z , t
t
ﺩﺭ ﺯﻣــﺎﻥ t + dtﺍﻳــﻦ ﺫﺭﻩ ﺑــﻪ ﻧﻘﻄــﻪ y + dy ، x + dxﻭ z + dzﺣﺮﻛــﺖ ﻛــﺮﺩﻩ ﻭ ﺳــﺮﻋﺖ ﺁﻥ ﭼﻨــﻴﻦ
ﺍﺳﺖ:
) = V ( x + dx, y + dy, z + dz , t + dt
t + dt
Vp
dV pﻳﻌﻨﻲ ﺗﻐﻴﻴﺮ ﺳﻌﺖ ﺫﺭﻩ ﺩﺭ ﺣﺮﻛﺖ ﺍﺯ ﻣﻜﺎﻥ rﺗﺎ r + d rﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ.
∂V
∂V
∂V
∂V
dx p +
dy p +
dz p +
dt
∂x
∂y
∂z
∂t
= dV p
ﺷﺘﺎﺏ ﻛﻞ ﺫﺭﻩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ:
∂V dx p ∂V dy p ∂V dz p ∂V
⋅
+
⋅
+
⋅
+
∂x dt
∂y dt
∂z dt
∂t
=
dV p
dt
= ap
ﺍﺯ ﻃﺮﻑ ﺩﻳﮕﺮ:
dz p
dt
=; w
dy p
dt
=; v
∂V
∂V
∂V ∂V
+v
+w
+
∂x
∂y
∂z ∂t
dx p
dt
=u
⇒ ap = u
ﺩﺭ ﻣﺤﺎﺳﺒﻪ ﺷـﺘﺎﺏ ﻳـﻚ ﺫﺭﻩ ﺳـﻴﺎﻝ ﺩﺭ ﻣﻴـﺪﺍﻥ ﺳـﺮﻋﺖ ﺍﺯ ﻳـﻚ ﻣﺸـﺘﻖ ﻭﻳـﮋﻩ ﺍﺳـﺘﻔﺎﺩﻩ ﻣـﻲ ﻛﻨـﻴﻢ ﻛـﻪ ﺁﻥ ﺭﺍ
ﺑﺎ ﻧﻤﺎﺩ DVﻧﺸﺎﻥ ﻣﻲ ﺩﻫﻴﻢ.
Dt
ﻣﺸﺘﻖ ﺍﺻﻠﻲ :
DV
→ substantial derivative
Dt
۱۸
DV
∂V
∂V
∂V ∂V
≡ ap = u
+v
+w
+
Dt
∂x
∂y
∂z ∂t
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺫﺭﻩ ﺳﻴﺎﻟﻲ ﻛﻪ ﺩﺭ ﻣﻴـﺪﺍﻥ ﺟﺮﻳـﺎﻥ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ﺑـﻪ ﺩﻭ ﺩﻟﻴـﻞ ﻣـﻲ ﺗﻮﺍﻧـﺪ ﺷـﺘﺎﺏ ﺩﺍﺷـﺘﻪ ﺑﺎﺷـﺪ .ﺫﺭﻩ ﺑـﻪ
ﺍﻳﻦ ﺩﻟﻴﻞ ﻛﻪ ﺑﻪ ﺩﺍﺧﻞ ﻳﻚ ﻧﺎﺣﻴﻪ ،ﻛـﻪ ﺳـﺮﻋﺖ ﺩﺭ ﺁﻧﺠـﺎ ﺑﻴﺸـﺘﺮ ﻳـﺎ ﻛﻤﺘـﺮ ﺍﺳـﺖ ﺟـﺎ ﺑـﻪ ﺟـﺎ ﻣـﻲ ﺷـﻮﺩ ﻣـﻲ
ﺗﻮﺍﻧﺪ ﺷﺘﺎﺏ ﺑﮕﻴﺮﺩ .ﺳﻴﺎﻝ ﺷﺘﺎﺏ ﺍﺿﺎﻓﻲ "ﻣﺤﻠﻲ" ﻧﻴﺰ ﺩﺍﺭﺩ .ﺯﻳﺮﺍ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺗﺎﺑﻌﻲ ﺍﺯ ﺯﻣﺎﻥ ﺍﺳﺖ.
DV
∂V
∂V
∂V ∂V
=u
+v
+w
+
Dt
∂x
∂y
∂z ∂t
= ap
ﺷﺘﺎﺏ ﺟﺎﺑﻪ ﺟﺎﻳﻲ 1ﻛﻪ ﺩﺭ ﺍﺛﺮ ﺗﻐﻴﻴﺮ ﻛﺮﺩﻥ ﻣﻜﺎﻥ ﺫﺭﻩ ﺩﺭ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺑﻪ ﻭﺟﻮﺩ ﻣﻲ ﺁﻳﺪ.
F8
ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺧﻮﺩ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺍﺳﺖ ﻛﻪ ﺷﺘﺎﺏ ﻣﺤﻠﻲ 2ﻧﺎﻡ ﺩﺍﺭﺩ
∂V
∂V
∂V
+v
+w
→
∂x
∂y
∂z
u
F9
∂V
→
∂t
substantial derivative ≡ derivative following the motion
) c( x,y,z,t
ﺷﺨﺼــﻲ ﺭﻭﻱ ﻳــﻚ ﭘــﻞ ﺍﻳﺴــﺘﺎﺩﻩ ﻭ ﺗﻌــﺪﺍﺩ ﻣــﺎﻫﻲ ﻛــﻪ ﺩﺭ ﺭﻭﺩﺧﺎﻧــﻪ ﺩﺭ ﻭﺍﺣــﺪ ﺯﻣــﺎﻥ ﺍﺯ ﺯﻳــﺮ ﭘــﻞ ﻋﺒــﻮﺭ ﻣــﻲ
ﻛﻨﻨﺪ ﺭﺍ ﺷﻤﺎﺭﺵ ﻛﻨﺪ.
x, y,z
∂c
∂t
→
ﺷﺨﺺ ﺳﻮﺍﺭ ﻳﻚ ﻗﺎﻳﻖ ﻣﻲ ﺑﺎﺷﺪ ﻛـﻪ ﻗـﺎﻳﻖ ﺑـﺎ ﺳـﺮﻋﺖ Vﺟﺮﻳـﺎﻥ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ﻭ ﺗﻌـﺪﺍﺩ ﻣـﺎﻫﻲ ﺭﺍ ﻣـﻲ
ﺷﻤﺎﺭﺩ.
Dc ∂c
∂c
∂c
∂c
=
+ u +υ + w
Dt ∂t
∂x
∂y
∂z
∂V
∂V
∂V
+v
+w
= (V ⋅ ∇)V
∂x
∂y
∂z
u
DV
∂V
= a = (V ⋅ ∇)V +
p
Dt
∂t
D
∂
∇⋅ = +V
Dt ∂t
∂u
Du
∂u
∂u ∂u
=u
= axp
+v +w +
∂x
Dt
∂y
∂z ∂t
∂v
Dv
∂v
∂v ∂v
=u +v +w +
= ayp
∂x
Dt
∂y
∂z ∂t
∂w
Dw
∂w
∂w ∂w
=u
= az p
+v
+w
+
∂x
Dt
∂y
∂z ∂t
⇒
Convective
Local Acceleration
۱۹
۱
۲
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﻣﺜﺎﻝ :ﺟﺮﻳﺎﻥ ﺗـﺮﺍﻛﻢ ﻧﺎﭘـﺬﻳﺮ ،ﺩﺍﺋـﻢ ﻭ ﻳـﻚ ﺑﻌـﺪﻱ ﺩﺍﺧـﻞ ﻛﺎﻧـﺎﻝ ﻧﺸـﺎﻥ ﺩﺍﺩﻩ ﺷـﺪﻩ ﺩﺭ ﺷـﻜﻞ ﺯﻳـﺮ ﺭﺍ ﺩﺭ ﻧﻈـﺮ
x
ﺑﮕﻴﺮﻳﺪ.ﻣﻴــﺪﺍﻥ ﺳــﺮﻋﺖ ﺑــﻪ ﺻــﻮﺭﺕ V = 10(1 + ) ˆj ft secﺩﺍﺩﻩ ﺷــﺪﻩ ﺍﺳــﺖ .ﻣﺆﻟﻔــﻪ xﺷــﺘﺎﺏ ﻳﻌﻨــﻲ
L
Du
Dt
،ﺑﺮﺍﻱ ﻳـﻚ ﺫﺭﻩ ﺳـﻴﺎﻝ ﺩﺭ ﺩﺍﺧـﻞ ﺍﻳـﻦ ﻣﻴـﺪﺍﻥ ﺳـﺮﻋﺖ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ﺭﺍ ﺑـﻪ ﺩﺳـﺖ ﺁﻭﺭﻳـﺪ .ﻣﺴـﻴﺮ
ﺫﺭﻩ ﺍﻱ ﻛــﻪ ﺍﺯ ﻣﺤــﻞ x = 0ﺩﺭ ﻟﺤﻈــﻪ t = 0ﻋﺒــﻮﺭ ﻣــﻲ ﻛﻨــﺪ ﺗــﺎﺑﻌﻲ ﺍﺯ ﺯﻣــﺎﻥ ﺍﺳــﺖ ﻳﻌﻨــﻲ ) . x p = f (t
ﺗﺎﺑﻊ ) f (tﺭﺍ ﺑﻪ ﺩﺳـﺖ ﺁﻭﺭﻳـﺪ ﻭ ﺳـﭙﺲ ﺑـﺎ ﺗﻮﺟـﻪ ﺑـﻪ ﺩﺍﺷـﺘﻦ ﻣﺴـﻴﺮ ﺑـﺎ ﺩﻭ ﺑـﺎﺭ ﻣﺸـﺘﻖ ﮔـﺮﻓﺘﻦ ﺍﺯﺁﻥ ﺭﺍﺑﻄـﻪ
ﺍﻱ ﺑﺮﺍﻱ ﻣﺆﻟﻔﻪ xﺷﺘﺎﺏ ﺫﺭﻩ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ.
∂V
∂V
∂V ∂V
DV
=u
+v
+w
+
∂x
∂y
∂z ∂t
Dt
∂u
∂u
∂u ∂u
Du
=u
+v +w +
∂x
∂y
∂z ∂t
Dt
x
) v = w = 0 , u = 10(1 +
L
∂u
Du
x 10 100
x
⇒
=u
= ⋅ ) = 10(1 +
) (1 +
∂x
Dt
L L
L
L
ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﺷـﺘﺎﺏ ﺫﺭﻩ ﺩﺭ ﻫـﺮ ﻧﻘﻄـﻪ ﺍﺯ ﻣـﺪﺍﻥ ﺳـﺮﻋﺖ ﺻـﺮﻓﺎً ﺑﺎﻳـﺪ ﻣﺨﺘﺼـﺎﺕ ﻧﻘﻄـﻪ ﻣـﻮﺭﺩ ﻧﻈـﺮ ﺭﺍ
ﺩﺭ ﺭﺍﺑﻄﻪ ﻓﻮﻕ ﺟﺎﻧﺸﻴﻦ ﻧﻤﻮﺩ.
ﺩﺭ ﻗﺴــﻤﺖ ﺩﻭﻡ ﺩﺭ ﻧﻈــﺮ ﺍﺳــﺖ ﻛــﻪ ﻳــﻚ ﺫﺭﻩ ﻣﺸــﺨﺺ ﺭﺍ ﺩﻧﺒــﺎﻝ ﻧﻤــﻮﺩ -ﻳﻌﻨــﻲ ﺫﺭﻩ ﺍﻱ ﻛــﻪ ﺍﺯ ﻣﺤــﻞ x = 0
ﺩﺭ ﺯﻣﺎﻥ t = 0ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ ﻣﻜﺎﻥ ﺁﻥ ﺭﺍ ﺩﺭ ﻫﺮ ﻟﺤﻈﻪ ﺩﺭ ﺩﺍﺧﻞ ﻛﺎﻧﺎﻝ ﻣﺸﺨﺺ ﻧﻤﻮﺩ.
) df (t
ﻣﺨﺘﺼــﺎﺕ ، xﻳﻌﻨــﻲ ،ﻣﺤــﻞ ﺍﻳــﻦ ﺫﺭﻩ ﺗــﺎﺑﻌﻲ ﺍﺯ ﺯﻣــﺎﻥ ﺧﻮﺍﻫــﺪ ﺑــﻮﺩ . x p = f (t ) .ﻫﻤﭽﻨــﻴﻦ
dt
= up
ﻧﻴــﺰ ﺗــﺎﺑﻌﻲ ﺍﺯ ﺯﻣــﺎﻥ ﺍﺳــﺖ .ﺳــﺮﻋﺖ ﺫﺭﻩ ﺩﺭ ﻫــﺮ ﻣﺤــﻞ ﺩﺭ ﻣﻴــﺪﺍﻥ ﺑﺮﺍﺑــﺮ ﺳــﺮﻋﺖ ﺩﺭ ﺁﻥ ﻧﻘﻄــﻪ ﺍﺯ ﻣﻴــﺪﺍﻥ
10 ftﻭ ﺩﺭ ﺯﻣــﺎﻥ t = tﻭ x = L
ﺍﺳــﺖ .ﻳﻌﻨــﻲ ﺳــﺮﻋﺖ ﺫﺭﻩ ﺩﺭ ﻫــﺮ ﻣﺤــﻞ x = 0ﻭ t = 0ﺑﺮﺍﺑــﺮ ﺑــﺎ
sec
ﺳﺮﻋﺖ ﺁﻥ ﺑﺮﺍﺑﺮ ﺑﺎ 20 ft secﺍﺳﺖ .ﻟﺬﺍ
۲۰
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
dx p
) df (t
x
) f (t
= 10(1 + ) = 10(1 +
)
dt
dt
L
L
) f (t
t
) df (t
) df (t
⇒
∫ ⇒ = 10dt
= ∫ 10dt
0
0
) 1 + f (t
) 1 + f (t
L
L
(
)
f
t
⇒ L ln(1 +
) = 10t
L
10 t
1 + f (t ) = e L
L
=
]− 1
L
= up
10 t
⇒ x p = f (t ) = L[e
d 2 f (t ) 100 10 t L
e
=
dt 2
L
=
d 2xp
dt 2
= ⇒ axp
ﺑﻨــﺎﺑﺮﺍﻳﻦ ﺩﻭ ﺭﻭﺵ ﻣﺨﺘﻠــﻒ ﺑــﺮﺍﻱ ﺑﻴــﺎﻥ ﺷــﺘﺎﺏ ﻳــﻚ ﺫﺭﻩ ﻣﺸــﺨﺺ ﺩﺭ ﺩﺳــﺖ ﺍﺳــﺖ ﻛــﻪ ﺍﺯ ﻣﺤــﻞ x = 0ﺩﺭ
ﺯﻣــﺎﻥ t = 0ﻋﺒــﻮﺭ ﻣــﻲ ﻛﻨــﺪ .ﺗﻮﺟــﻪ ﻛﻨﻴــﺪ ﮔﺮﭼــﻪ ﻣﻴــﺪﺍﻥ ﺟﺮﻳــﺎﻥ ﺩﺍﺋــﻢ ﺍﺳــﺖ ﻭﻟــﻲ ﻭﻗﺘــﻲ ﻛــﻪ ﻳــﻚ ﺫﺭﻩ
ﻣﺸﺨﺺ ﺭﺍ ﺩﻧﺒﺎﻝ ﻣﻲ ﻛﻨﻴﻢ ﻣﻜﺎﻥ ﻭ ﺷﺘﺎﺏ ﺗﻮﺍﺑـﻊ ﺯﻣـﺎﻥ ﻣـﻲ ﺑـﺎ ﺷـﻨﺪ .ﺍﻛﻨـﻮﻥ ﻧﺸـﺎﻥ ﻣـﻲ ﺩﻫـﻴﻢ ﻛـﻪ ﻫـﺮ ﺩﻭ
ﻋﺒﺎﺭﺕ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺑﺮﺍﻱ ﺷﺘﺎﺏ ﻳﻜﺴﺎﻥ ﻫﺴﺘﻨﺪ.
Du 100
x
=
) (1 +
Dt
L
L
ﺍﻟﻒ( ﺩﺭ ﻟﺤﻈﻪ t = 0ﻭ x p = 0ﺩﺭ ﻟﺤﻈﻪ t = 0ﺫﺭﻩ ﺩﺭ x = 0ﺍﺳﺖ
= axp
Du 100
0 100
=
= ) (1 +
Dt
L
L
L
ﺏ( ﺩﺭ x = Lﻭﻗﺘﻲ ﻛﻪ x p = Lﺍﺳﺖt = t 2 ،
= axp
100 10 t L
e
L
,
100
L
↔
Du 100
200
=
= )(1 + 1
Dt
L
L
= axp
= axp
)− 1
L
10t2
x p = L = L (e
10t2
=2
100 10t2 L 100
200
= ⇒ axp
=
= ×2
e
L
L
L
L
⇒e
ﺍﻳﻦ ﻣﺴﺄﻟﻪ ﺩﻭ ﺭﻭﺵ ﻣﺨﺘﻠﻒ ﺗﻮﺻﻴﻒ ﺣﺮﻛﺖ ﻳﻚ ﺫﺭﻩ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ.
The Equation of Motion
)(The differtial Equation of linear momentum
∂
dN
N
= ) sys ∫ c.v. ρηdv + ∫ c.s. ρηV ⋅ d A , η
∂t
dt
m
∂
) d (mV
= F = ∫ c.v. ρV dv + ∫ c.s. ρV V ⋅ d A
⇒ N = mV
∂t
dt
(
η =V
۲۱
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
•
•
∂
∫ c.v. ρV dv + ∑(mi Vi ) out − ∑(m i Vi ) in
∂t
∂
∂
∫ c.v. ρV du ≈ ( ρV )dxdydz
∂t
∂t
= ∑F
ﭼﻮﻥ ﺍﻟﻤﺎﻥ ﻛﻮﭼﻚ ﺍﺳﺖ ﺍﻧﺘﮕﺮﺍﻝ ﺗﺒﺪﻳﻞ ﺑﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﻲ ﺷﻮﺩ.
ﻗﺮﺍﺭﺩﺍﺩ ﻋﻼﻣﺖ ﺑﺮﺍﻱ ﺗﻨﺶ:
ﻳﻚ ﻣﺆﻟﻔـﻪ ﺗـﻨﺶ ﻣﺜﺒـﺖ ﺍﺳـﺖ ،ﻫﺮﮔـﺎﻩ ﺟﻬـﺖ ﻣﺆﻟﻔـﻪ ﺗـﻨﺶ ﻭ ﺻـﻔﺤﻪ ﺍﻱ ﻛـﻪ ﺍﻳـﻦ ﺗـﻨﺶ ﺑـﺮ ﺁﻥ ﻭﺍﺭﺩ ﻣـﻲ
ﺷﻮﺩ ﻫﺮ ﺩﻭ ﻣﺜﺒﺖ ﻳﺎ ﻫﺮ ﺩﻭ ﻣﻨﻔـﻲ ﺑﺎﺷـﻨﺪ .ﻣﺆﻟﻔـﻪ ﻫـﺎﻱ ﺗـﻨﺶ ﻣﻨﻔـﻲ ﻫﺴـﺘﻨﺪ ﻫـﺮ ﮔـﺎﻩ ﺟﻬـﺖ ﻣﺆﻟﻔـﻪ ﺗـﻨﺶ
ﻭ ﺻﻔﺤﻪ ﺍﻱ ﻛﻪ ﺗﻨﺶ ﺑﺮ ﺁﻥ ﺍﺛﺮ ﻣﻲ ﻛﻨﺪ ﺩﺍﺭﺍﻱ ﻋﻼﻣﺖ ﻣﺨﺎﻟﻒ ﺑﺎﺷﻨﺪ.
۲۲
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
ﺻـﻔﺤﻪ. ﻣﻨﻔـﻲ ﺍﺛـﺮ ﻣـﻲ ﻛﻨـﺪx ﻣﺜﺒـﺖ ﺩﺭ ﺟﻬـﺖτ yx ﺑﻨـﺎﺑﺮﺍﻳﻦ. ﻣﻨﻔـﻲ ﺍﺳـﺖy ﺻﻔﺤﻪ ﺑﺎﻻﻳﻲ ﻳﻚ ﺳـﻄﺢ
. ﻣﺜﺒﺖ ﺍﺛﺮ ﻣﻲ ﻛﻨﺪx ﻣﺜﺒﺖ ﺩﺭ ﺟﻬﺖτ yx ﺑﻨﺎﺑﺮﺍﻳﻦ. ﻣﺜﺒﺖ ﺍﺳﺖy ﭘﺎﻳﻴﻨﻲ ﻳﻚ ﺳﻄﺢ
∂
∂
( ρV )dxdydz + ρuV dydz + ( ρuV )dxdydz
∂t
∂x
∂
∂
+ ρvV dxdz + ( ρvV )dxdydz + ρwV dxdy + ( ρwV )dxdydz
∂y
∂z
∑F =
− ρuV dydz − ρvV dxdz − ρwV dxdy
∂
∂
∂
∂
⇒ ∑ F = dxdydz[ ( ρV ) + ( ρuV ) + ( ρvV ) + ( ρwV )]
∂t
∂x
∂y
∂z
∂
∂
∂
∂
( ρV ) + ( ρuV ) + ( ρvV ) + ( ρwV )
∂t
∂x
∂y
∂z
⇒V
∂ρ
∂V
∂ ( ρu )
∂V
∂ ( ρv)
∂V
∂ ( ρw)
∂V
+ρ
+V
+ ρu
+V
+ ρv
+V
+ ρw
∂t
∂t
∂x
∂x
∂y
∂y
∂z
∂z
= V[
∂ρ ∂ ( ρu ) ∂ ( ρv) ∂ ( ρw)
∂V
∂V
∂V
∂V
+
+
+
+u
+v
+w ]
] + ρ[
∂t
∂x
∂y
∂z
∂t
∂x
∂y
∂z
= V[
∂ρ
∂V
∂V
∂V
∂V
+ ∇ ⋅ ( ρV )] + ρ[
+u
+v
+w ]
∂t
∂t
∂x
∂y
∂z
↓↓
0= ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ
d V DV
=
dt
Dt
body forces
ΣF
surface forces
body forces : d F grav = ρ gdxdydz
if z ↑ + ⇒ g = − gkˆ
surface forces:
− P + τ xx
i
ﺑﺮ ﺭﻭﻱ ﺻﻔﺤﻪ ﻱ ﻋﻤﻮﺩ ﺑﺮ ﻣﺤﻮﺭj ← ﺗﻨﺶ ﺩﺭ ﺟﻬﺖσ ij = τ xy
τ xz
۲۳
τ yx
τ zx
− P + τ yy
τ zy
τ yz
− P + τ zz
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
⇒ dFx , surf = [
⇒
dFx ,surf
dv
dFy ,surf
dv
dFz ,surf
dv
∂
∂
∂
( σ xx ) + ( σ yx ) + ( σ zx )]dxdydz
∂x
∂y
∂z
=−
∂P ∂
∂
∂
+ (τ xx ) + (τ yx ) + (τ zx )
∂x ∂x
∂y
∂z
(1)
=−
∂P ∂
∂
∂
+ (τ xy ) + (τ yy ) + (τ zy )
∂y ∂x
∂y
∂z
(2)
=−
∂P ∂
∂
∂
+ (τ xz ) + (τ yz ) + (τ zz )
∂z ∂x
∂y
∂z
(3)
: ﻭ ﺟﻤﻊ ﺁﻧﻬﺎ ﺩﺍﺭﻳﻢk̂ ﻭĵ ﻭiˆ ( ﺑﻪ ﺗﺮﺗﻴﺐ ﺩﺭ3) ( ﻭ2 ) ( ﻭ1 ) ﺿﺮﺏ ﻣﻌﺎﺩﻻﺕ
۲٤
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
dF
dF
) surf = −∇P + (
) viscous
dv
dv
∂τ
∂τ
∂τ
∂τ
∂τ
∂τ
∂τ
∂τ
dF
∂τ
( ) viscous = iˆ( xx + yx + zx ) + ˆj ( xy + yy + zy ) + kˆ( xz + yz + zz )
dv
∂x
∂y
∂z
∂x
∂y
∂z
∂x
∂y
∂z
(
dF
) viscous = ∇ ⋅τ ij
⇒(
dv
∑F = ρ
τ xx τ yx τ zx
, τ ij = τ xy τ yy τ zy
τ τ τ
xz yz zz
τ ij → viscouse - stress tensor
DV
dxdydz
Dt
⇒ ρ gdxdydz − ∇P + ∇ ⋅τ ij = ρ
DV
dxdydz
Dt
for unit volume :
ρ g − ∇P + ∇ ⋅τ ij = ρ
DV
Dt
compact form
: ﺭﺍﺑﻄﻪ ﺑﻴﻦ ﺗﻨﺶ ﻭ ﺷﺪﺕ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺧﺎﻟﺺ
ﺭﻭﺍﺑـﻂ ﺑـﻴﻦ ﺗـﻨﺶ ﻭ ﺷـﺪﺕ ﺗﻐﻴﻴـﺮ ﺷـﻜﻞ ﺧـﺎﻟﺺ ﺑـﻪ ﺷـﺢ ﺯﻳـﺮ ﻣـﻲ،ﺑﺮﺍﻱ ﻳﻚ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ ﻭ ﺍﻳﺰﻭﺗﺮﻭﭘﻴﻚ
.ﺑﺎﺷﺪ
Isotropic fluid: A fluid whose properties are not dependent on the direction
along which they are measured.
∂u
∂u
+ λ∇ ⋅ V ⇒ τ xx = 2 µ
+ λ∇ ⋅ V
∂x
∂x
∂v
∂v
σ yy = − P + 2 µ + λ∇ ⋅ V ⇒ τ yy = 2 µ + λ∇ ⋅ V
∂y
∂y
∂w
∂w
σ zz = − P + 2 µ
+ λ∇ ⋅ V ⇒ τ zz = 2 µ
+ λ∇ ⋅ V
∂z
∂z
∂u ∂v ∂w
∇ ⋅V =
+
+
∂x ∂y ∂z
∂u ∂v
τ xy = τ yx = µ ( + )
∂y ∂x
∂w ∂u
τ xz = τ zx = µ ( + )
∂x ∂z
∂v ∂w
τ yz = τ zy = µ ( + )
∂z ∂y
σ xx = − P + 2µ
: ﻭ ﻣﺠﻤﻮﻉ ﺗﻨﺶ ﻫﺎﻱ ﻧﺮﻣﺎﻝ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ
σ xx + σ yy + σ zz = −3P(2 µ + 3λ )∇ ⋅ V
⋅ ∇ ﻭ ﺍﻳــﻦ ﺭﺍﺑﻄــﻪ ﻧﺸــﺎﻥ ﻣــﻲ ﺩﻫــﺪ ﻛــﻪ ﻣﻘــﺪﺍﺭ ﻓﺸــﺎﺭ ﺑﺮﺍﺑــﺮ ﺍﺳــﺖ ﺑــﺎV = 0 ﺑــﺮﺍﻱ ﺳــﻴﺎﻝ ﺗــﺮﺍﻛﻢ ﻧﺎﭘــﺬﻳﺮ
:ﻣﺘﻮﺳﻂ ﺳﻪ ﺗﻨﺶ ﻧﺮﻣﺎﻝ ﻳﻌﻨﻲ
۲٥
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
σ xx + σ yy + σ zz
3
P=−
2
λ=− µ
3
2
λ=+ µ
3
ﺑﺮﺍﻱ ﮔﺎﺯﻫﺎ
ﺑﺮﺍﻱ ﻣﺎﻳﻌﺎﺕ
ﻣﺜــﺎﻝ :ﻣﻘــﺪﺍﺭ ﺗــﻨﺶ ﻧﺮﻣــﺎﻝ σ xxﺭﺍ ﺩﺭ ﻧﻘﻄــﻪ ﺳــﻜﻮﻥ ﺷــﻜﻞ ﺯﻳــﺮ ﺣﺴــﺎﺏ ﻛﻨﻴــﺪ ﺩﺭ ﺻــﻮﺭﺗﻲ ﻛــﻪ ﻣﻘــﺪﺍﺭ
ﺳــﺮﻋﺖ ﺩﺭ ﺍﻣﺘــﺪﺍﺩ ﻣﺤــﻮﺭ xﻫــﺎ ﺑﺮﺍﺑــﺮ ﺑــﺎ
x2
u = 30 − 120ﺑﺎﺷــﺪ .ﻓــﺮﺽ ﻛﻨﻴــﺪ ﻛــﻪ ﺍﺯ ﺍﺛــﺮﺍﺕ ﻭﻳﺴــﻜﻮﺯﻳﺘﻪ
ﻣــﻲ ﺗــﻮﺍﻥ ﺩﺭ ﺟﺮﻳــﺎﻥ ﺁﺯﺍﺩ ﺗــﺎ ﻧﻘﻄــﻪ ﺳــﻜﻮﻥ ﺻــﺮﻑ ﻧﻈــﺮ ﻧﻤــﻮﺩ ﻭ ﺩﺭ ﺁﻧﺠــﺎ )ﺩﺭ ﺟﺮﻳــﺎﻥ ﺁﺯﺍﺩ( ﻓﺸــﺎﺭ ﺑﺮﺍﺑــﺮ ﺑــﺎ
5 psiﺍﺳﺖ .ﺁﺏ ﺑﺎ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ
lb f sec
ft 2
µ = 10 −5ﺍﺯ ﺭﻭﻱ ﺟﺴﻢ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ.
∂u
; ∇ ⋅V = 0
∂x
sec
σ xx = − P + 2µ
x → −∞ ⇒ u = 30 ⇒ V∞ = 30 ft
P∞ = 5 psi
ﺍﮔﺮ ﺍﺯ ﺍﺛـﺮﺍﺕ ﻭﻳﺴـﻜﻮﺯﻳﺘﻪ ﺩﺭ ﺟﺮﻳـﺎﻥ ﺁﺯﺍﺩ ﺗـﺎ ﻧﻘﻄـﻪ ﺳـﻜﻮﻥ ﺻـﺮﻑ ﻧﻈـﺮ ﻛﻨـﻴﻢ ﻣﻌﺎﺩﻟـﻪ ﺑﺮﻧـﻮﻟﻲ ﺭﺍ ﻣـﻲ ﺗـﻮﺍﻥ
ﻧﻮﺷﺖ :ﻧﻘﻄﻪ ﺳﻜﻮﻥ ﺩﺭ x = −2ﺍﺳﺖ.
2
P
∞V
P
2
∞+ ∞ = 0 ⇒ P0 = P∞ + 1 ρV
2
2
ρ ρ
lb
30 2
P0 = 5 × 144 +
× 1 ⋅ 94 = 1593 f 2
ft
2
240
240
)
( ⇒ σ xx = −1593 + 2 × 10 −5 ( 3 ) = −1593 + 2 × 10 −5
x
−8
lb
⇒ σ xx ≈ −1593 f 2
ft
ﻫﻤﺎﻧﻄﻮﺭ ﻛﻪ ﻣﻼﺣﻈﻪ ﻣـﻲ ﺷـﻮﺩ ﺗـﻨﺶ ﻫـﺎﻱ ﻧﺮﻣـﺎﻝ ﺑـﻪ ﺧـﻮﺑﻲ ﺑـﺎ ﻣﻨﻔـﻲ ﻓﺸـﺎﺭ ﺑﺮﺍﺑﺮﻧـﺪ .ﻭﻳﺴـﻜﻮﺯﻳﺘﻪ ﺍﻏﻠـﺐ
ﺳــﻴﺎﻻﺕ ﺑــﻪ ﻗــﺪﺭﻱ ﻛــﻢ ﺍﺳــﺖ ﻛــﻪ ﺟﻤﻠــﻪ ﻧﺎﺷــﻲ ﺍﺯ ﻭﻳﺴــﻜﻮﺯﻳﺘﻪ ﺩﺭ ﺭﺍﺑﻄــﻪ ﺗــﻨﺶ ﻫــﺎﻱ ﻧﺮﻣــﺎﻝ ﻧــﺎﭼﻴﺰ ﻣــﻲ
ﺑﺎﺷﺪ.
۲٦
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
DV
Dt
∂P ∂τ xx ∂τ yx ∂τ zx
Du
= ρg x −
+
+
+
∂x
∂x
∂y
∂y
Dt
∂P ∂
∂u 2
∂
∂u ∂v
Du
∂
∂w ∂u
= ρg x −
+ [ µ (2 ± ∇ ⋅ V )] + [ µ ( + )] + [ µ ( + )]
∂x ∂x
∂x 3
∂y
∂y ∂x
Dt
∂z
∂x ∂z
∂P ∂
∂
∂v 2
∂v ∂u
Dv
∂
∂w ∂v
= ρg y −
+ [ µ ( + )] + [ µ (2 ± ∇ ⋅ V )] + [ µ ( + )]
∂y ∂x
∂y
∂y 3
∂x ∂y
∂z
Dt
∂y ∂z
∂P ∂
∂w ∂u
Dw
∂
∂w ∂v
∂
∂w 2
= ρg z −
+ [µ (
+ )] + [ µ (
+ )] + [ µ (2
± ∇ ⋅ V )]
∂z ∂x
∂x ∂z
Dt
∂y
∂y ∂z
∂z
∂z 3
ρ g − ∇P + ∇ ⋅τ ij = ρ
⇒ρ
⇒ρ
⇒ρ
⇒ρ
ﻭµ = µ ( ρ ) ﻭP = P( ρ ) ﺍﻳــﻦ ﻣﻌــﺎﺩﻻﺕ ﺑــﻪ ﻫﻤــﺮﺍﻩ ﻣﻌﺎﺩﻟــﻪ ﭘﻴﻮﺳــﺘﮕﻲ ﻭ ﻣﻌﺎﺩﻟــﻪ ﻭ ﻣﻌﺎﺩﻟــﻪ ﺣﺎﻟــﺖ
ﺩﺍﻧﺴـﻴﺘﻪ ﻭ ﺳـﺮﻋﺖ ﻫـﺎﻱ ﻳـﻚ ﺳـﻴﺎﻝ ﺟـﺎﺭﻱ ﺩﻣـﺎ ﺛﺎﺑـﺖ ﺭﺍ،ﺷﺮﺍﻳﻂ ﻣـﺮﺯﻱ ﻭﺷـﺮﺍﻳﻂ ﺍﻭﻟﻴـﻪ ﻣـﻲ ﺗـﻮﺍﻥ ﻓﺸـﺎﺭ
.ﭘﻴﺪﺍ ﻛﺮﺩ
:ﺣﺎﻻﺕ ﺧﺎﺹ
i) for constant µ and constant ρ
⇒ ∇ ⋅V = 0
ρ
∂P
∂ 2u
∂ 2u
∂ 2v
Du
∂2w
∂ 2u
= ρg x −
+ 2µ 2 + µ 2 + µ
+µ
+µ 2
∂x
∂x
∂y
∂y∂x
Dt
∂z∂x
∂z
⇒ρ
∂P
∂ 2u ∂ 2u ∂ 2u
∂ ∂u ∂v ∂w
Du
= ρg x −
+ µ( 2 + 2 + 2 ) + µ
+
( +
)
∂x
∂x
∂y
∂z
∂x ∂x ∂y ∂z
Dt
⇒ρ
∂P
∂ 2u ∂ 2u ∂ 2u
Du
= ρg x −
+ µ( 2 + 2 + 2 )
∂x
Dt
∂x
∂y
∂z
ρ
∂P
∂ 2v ∂ 2v ∂ 2v
Dv
= ρg y −
+ µ( 2 + 2 + 2 )
∂y
Dt
∂x
∂y
∂z
∂P
∂2w ∂2w ∂2w
Dw
= ρg z −
+ µ( 2 + 2 + 2 )
ρ
∂z
Dt
∂x
∂y
∂z
⇒ρ
DV
= ρ g − ∇P + µ∇ 2 V
Dt
ii) for ∇ ⋅ τ ij = 0
⇒ρ
Navier-Stokes Equation
DV
= −∇P + ρ g
Dt
iii) acceleration terms in the Navier-Stokes equation are neglected.
ρ
DV
=0
Dt
⇒ ρ g − ∇P + µ∇ 2 V = 0
Stokes flow Equation or creeping flow
∂V
∂V
DV ∂V
∂V
+u
+v
+w
=
∂x
∂y
Dt ∂t
∂z
: ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺷﺒﻴﻪ ﻗﺎﻧﻮﻥ ﺩﻭﻡ ﻧﻴﻮﺗﻮﻥ ﺍﺳﺖ
۲۷
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
Sum of forces = acceleration × mass
ﻭ ﺑﻴﺎﻥ ﻣﻲ ﻛﻨﺪ ﻛﻪ ﺍﻟﻤـﺎﻥ ﺳـﻴﺎﻝ ﻛـﻪ ﺑـﺎ ﺳـﻴﺎﻝ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ﺑـﻪ ﺧـﺎﻃﺮ ﻧﻴﺮﻭﻫـﺎﻳﻲ ﻛـﻪ ﺑـﻪ ﺁﻥ ﺍﻋﻤـﺎﻝ
.ﻣﻲ ﺷﻮﺩ ﺷﺘﺎﺏ ﭘﻴﺪﺍ ﻣﻲ ﻛﻨﺪ
∂P ∂τ xx ∂τ yx ∂τ zx
∂u
∂u
∂u
∂u
+
+
+
= ρ( + u
+v
+w )
∂x
∂x
∂y
∂z
∂t
∂x
∂y
∂z
∂P ∂τ xy ∂τ yy ∂τ zy
∂v
∂v
∂v
∂v
+
+
+
= ρ( + u + v + w )
ρg y −
∂y
∂x
∂y
∂z
∂t
∂x
∂y
∂z
∂P ∂τ xz ∂τ yz ∂τ zz
∂w
∂w
∂w
∂w
+
+
+
= ρ(
+u
+v
+w )
ρg z −
∂z
∂x
∂y
∂z
∂t
∂x
∂y
∂z
ρg x −
Inviscid flow: Eulers Equation
τ ij = 0 : ﺑﺮﺍﻱ ﺳﻴﺎﻻﺕ ﺑﺪﻭﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ
Eulers Equation ⇒ ρ g − ∇P = ρ DV
Dt
ﺗﺒﺪﻳﻼﺕ ﺩﺍﺧﻠﻲ ﺍﻧﺮژﻱ ﺩﺭ ﻳﻚ ﺳﻴﺎﻝ ﺟﺎﺭﻱ-
The Equathin of Mechanical Energy
ρ
ρ
DV
= −∇P − [∇ ⋅τ ] + ρ g
Dt
[ ] ≡ vector
( ) ≡ scalar
: ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﺍﺳﻜﺎﻟﺮ ﺯﻳﺮ ﻣﻲ ﺭﺳﻴﻢ، ﺿﺮﺏ ﺍﺳﻜﺎﻟﺮ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢV ﺍﮔﺮ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍ ﺑﺎ
(
) (
) ( )) ( ← ﺍﺳﻜﺎﻟﺮ
D 1 2
v = − V ⋅ ∇P − V ⋅ [∇ ⋅ τ ] + ρ V ⋅ g
Dt 2
ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﻳﺖ ﺗﻐﻴﻴـﺮ ﺍﻧـﺮژﻱ ﺳـﻴﻨﺘﻴﻚ ﺑـﻪ ﺍﺯﺍﻱ ﺟـﺮﻡ ﺑـﺮﺍﻱ ﻳـﻚ ﺍﻟﻤـﺎﻥ ﺳـﻴﺎﻝ ﻛـﻪ ﺣﺮﻛـﺖ ﻣـﻲ ﻛﻨـﺪ ﺭﺍ
.ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ
ρ
∂ 1
∂ 1
∂ 1
D 1 2 ∂ 1
2
2
2
2
V = ρV + u ρV + v ρV + w ρV
∂x 2
∂y 2
∂z 2
Dt 2 ∂t 2
⇒ρ
1
D 1 2 ∂ 1
2
2
V = ρV + ∇ ⋅ ρV V
2
Dt 2 ∂t 2
∇ ⋅ PV = V ⋅ ∇P + P∇ ⋅ V
(∇ ⋅ [τ ⋅V ]) = (V ⋅ ∇ ⋅τ )+ (τ : ∇V )
double - dot product ↵
(V ⋅W ) = V ⋅W ...θ scaler product or dot product
(V ⋅V ) = V = V
vector product : [V × W ] = V ⋅ W sin θ
cross product : [V × V ] = 0
(τ : ∇V ) = ∑ ∑ τ ∂∂xv
2
2
i
ij
i
j
j
۲۸
ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ۱ ﻓﺼﻞ
∂ 1
1
ρV 2 + ∇ ⋅ ρV 2 V
∂t 2
2
= −∇ ⋅ PV + P∇ ⋅ V − ∇ ⋅ τ ⋅ V + τ : ∇V + ρ V ⋅ g
∂ 1
1
⇒ ρV 2 = − ∇ ⋅ ρV 2 V − ∇ ⋅ PV
∂t 2
2
rate of increase rate of addition rate of work done by
in kinetic energy kinetic energy by pressure of sorroundings
⇒
per unit volume
(
convectin per unit
volume
)
1
− τ : ∇V = µφv = µ ∑
2 i
i → x, y , z
j → x, y , z
ﻭ
∑
j
on fluid
( )
∂v ∂v j 2
− ∇ ⋅ V δ ij
i +
∂x j ∂xi 3
δ ij = 1 for i = j
δ ij = 0
2
for i ≠ j
.ﺍﻳﻦ ﺗﺮﻡ ﺟﺎﻳﻲ ﻛﻪ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺑﺎﻻﺳﺖ ﻣﺎﻧﻨﺪ ﺳﺮﻋﺖ ﻫﺎﻱ ﺑﺎﻻ ﻣﻬﻢ ﺍﺳﺖ
δ ij ≡ kronecker delta
۲۹
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
1-1ﺑﺮﺍﻱ ﺟﺮﻳﺎﻧﻲ ﺩﺭ ﺻﻔﺤﻪ ، xyﻣﻮﻟﻔﻪ xﺳﺮﻋﺖ ﺑﺎ ﺭﺍﺑﻄﻪ
ﺑﻴﺎﻥ ﺷﺪﻩ ﺍﺳﺖ،ﻛﻪ ﺩﺭ ﺁﻥ
) u = Ax( y − B
B = 2m ، A = 3m −1 s −1ﻭ yﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﻫﺴﺘﻨﺪ .ﻣﻮﻟﻔﻪ yﺭﺍ ﺑﻴﺎﺑﻴﺪ ،ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﻭ ﺗﺮﺍﻛﻢ
ﻧﺎﭘﺬﻳﺮ ﺑﺎﺷﺪ .ﺁﻳﺎ ﺍﻳﻦ ﻣﻮﻟﻔﻪ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻧﺎﭘﺎﻳﺎ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺻﺤﺖ ﺩﺍﺭﺩ؟ ﭼﻨﺪ ﻣﻮﻟﻔﻪ yﻣﻤﻜﻦ ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟
2-1ﻣﻮﻟﻔﻪ
y
ﺳﺮﻋﺖ ﺩﺭ ﻳﻚ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ،ﻭﺍﻗﻊ ﺩﺭ ﺻﻔﺤﻪ
xy
ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ، υ = − Bxy 3ﻛﻪ
ﺩﺭ ﺁﻥ B = 0.2m −3 s −1ﻭ xﻭ yﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﻫﺴﺘﻨﺪ.ﺳﺎﺩﻩ ﺗﺮﻳﻦ ﻣﻮﻟﻔﻪ xﺳﺮﻋﺖ ﺭﺍ ﺩﺭ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ،
ﻭ ﻣﻌﺎﺩﻟﻪ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﻴﺎﺑﻴﺪ.ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﮔﺬﺭﺍ ﺍﺯ ﻧﻘﺎﻁ ) (1,4ﻭ ) (2,4ﺭﺍ ﺭﺳﻢ ﻛﻨﻴﺪ.
3-1ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﻣﺠﻤﻮﻋﻪ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺯﻳﺮ ،ﻳﻚ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ؟
(a) Vr = U cosθ , Vθ = −U sin θ
, Vθ = k
2πr
2
(c) Vr = U cos θ 1 - a , Vθ = −U sin θ 1 + a
r
r
) (
) (
2πr
(b) Vr = − q
2
4-1ﻣﺠﻮﻋﻪ ﺗﻮﺍﺑﻊ ﺟﺮﻳﺎﻥ ψﺭﺍ ﺑﺮﺍﻱ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺯﻳﺮ ﺑﻴﺎﺑﻴﺪ.
∧ V = ( x + 2 y )i ∧ + ( x 2 − y ) j
5 -1ﻣﻮﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ ﺩﺭ ﻳﻚ ﺟﺮﻳﺎﻥ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ , u = 0 υ = − y 3 − 4 zﻭ
W = 3y 2 z
)ﺍﻟﻒ( ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﻳﻚ ﺑﻌﺪﻱ ﺍﺳﺖ ،ﺩﻭ ﺑﻌﺪﻱ ﺍﺳﺖ ﻳﺎ ﺳﻪ ﺑﻌﺪﻱ؟
)ﺏ( ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ؟
)ﺝ( ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﻴﺎﺑﻴﺪ.
6-1ﻣﻴﺪﺍﻥ
Ax
Ay
i+ 2
ﺳﺮﻋﺖ j
2
) (x + y
) (x + y 2
2
= Vﺭﺍ ﺩﺭ ﺻﻔﺤﻪ
xy
ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ A = 10 m s ،ﻭ xﻭ
2
yﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﻫﺴﺘﻨﺪ.ﺁﻳﺎ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ؟ﻋﺒﺎﺭﺕ ﺷﺘﺎﺏ ﺳﻴﺎﻝ ﺭﺍ ﺑﻴﺎﺑﻴﺪ .ﺳﺮﻋﺖ ﻭ ﺷﺘﺎﺏ ﺭﺍ ﺩﺭ
ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ ، xﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭ yﻭ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ y = xﺑﻴﺎﺑﻴﺪ.ﭼﻪ ﻧﺘﻴﺠﻪ ﺍﻱ ﻣﻲ ﺗﻮﺍﻥ ﮔﺮﻓﺖ؟
7 -1ﺟﺮﻳﺎﻥ ﻫﻮﺍ ﺑﺎ ﺳﺮﻋﺖ ﻛﻢ ﺭﺍ ﺑﻴﻦ ﺩﻳﺴﻚ ﻫﺎﻱ ﻣﻮﺍﺯﻱ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ.ﻓﺮﺽ ﻛﻨﻴﺪ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ
ﻏﻴﺮ ﭼﺴﺒﻨﺪﻩ ﺍﺳﺖ ،ﻭ ﺳﺮﻋﺖ ﻓﻘﻂ ﺑﻪ ﻃﻮﺭ ﺷﻌﺎﻋﻲ ﻭ ﺩﺭ ﻫﺮ ﻣﻘﻄﻊ ﺑﻪ ﻃﻮﺭ ﻳﻜﻨﻮﺍﺧﺖ ﺍﺳﺖ.ﺳﺮﻋﺖ ﺟﺮﻳﺎﻥ ﺩﺭ
R = 75mmﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ . V = 15 mﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﺭﺍ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻴﺎﻳﺪ .ﻭ ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ
ﻋﺒﺎﺭﺕ ﻛﻠﻲ ﺳﺮﻋﺖ ﺑﺮﺍﻱ
s
ri ≤ r ≤ R
ﭼﻨﻴﻦ ﺍﺳﺖ:
∧
V = V ( R )e r
r
۳۰
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
ﺷﺘﺎﺏ ﺫﺭﻩ ﺳﻴﺎﻝ ﺭﺍ ﺩﺭ r = riﻭ r = Rﺑﻴﺎﺑﻴﺪ.
8 -1ﻫﻮﺍ ﺍﺯ ﻃﺮﻳﻖ ﺳﻄﺢ ﻣﺘﺨﻠﺨﻠﻲ ﻭﺍﺭﺩ ﻓﻀﺎﻱ ﺑﺎﺭﻳﻚ ﺑﻴﻦ ﺩﻭ ﺻﻔﺤﻪ ﻣﻮﺍﺯﻱ ،ﺑﻪ ﻓﺎﺻﻠﻪ hﺍﺯ ﻳﻜﺪﻳﮕﺮ ،ﻣﻲ
ﺷﻮﺩ .ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻳﻚ ﺣﺠﻢ ﻛﻨﺘﺮﻝ ،ﻛﻪ ﺳﻄﺢ ﺧﺎﺭﺟﻲ ﺁﻥ ﺩﺭ ﻣﻜﺎﻥ xﺍﺳﺖ ،ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺳﺮﻋﺖ
ﻳﻜﻨﻮﺍﺧﺖ ﺩﺭ ﺟﻬﺖ xﻋﺒﺎﺭﺗﺴﺖ ﺍﺯ . u = υ 0 xﻣﻮﻟﻔﻪ ﺳﺮﻋﺖ ﺩﺭ ﺟﻬﺖ xﻭ ﺷﺘﺎﺏ ﻳﻚ ﺫﺭﻩ ﺳﻴﺎﻝ ﺩﺭ ﺑﻴﻦ
ﺻﻔﺤﺎﺕ ﺭﺍ ﺑﻴﺎﺑﻴﺪ.
h
9-1ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮﻱ ﺩﺭ ﻧﺎﺯﻟﻲ ﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ.ﻣﺴﺎﺣﺖ ﻧﺎﺯﻝ ﺑﺎ ﺭﺍﺑﻄﻪ ) A = A0 (1 − bxﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ ﻭ
ﺳﺮﻋﺖ ﻭﺭﻭﺩﻱ ﺩﺍﺭﺍﻱ ﺭﺍﺑﻄﻪ ) U = U 0 (1 − e − λtﺍﺳﺖ،ﻛﻪ ﺩﺭ ﺁﻥ ، b = 0.1m ، L = 5m ، A0 = 0.5m 2
λ = 0.2s −1ﻭ . U 0 = 5 m sﺷﺘﺎﺏ ﺭﺍ ﺩﺭ ﺧﻂ ﻣﺮﻛﺰﻱ ﺑﺮ ﺣﺴﺐ ﺯﻣﺎﻥ ﺑﻴﺎﺑﻴﺪ ﻭ ﺁﻥ ﺭﺍ ﺭﺳﻢ ﻛﻨﻴﺪ.
۳۱
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۱ﺍﺻﻮﻝ ﻣﻘﺪﻣﺎﺗﯽ ﻣﮑﺎﻧﻴﮏ ﺳﻴﺎﻻﺕ
10-1ﺟﺮﻳﺎﻥ ﻳﻚ ﺑﻌﺪﻱ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮﻱ ﺭﺍ ﺩﺭ ﻛﺎﻧﺎﻝ ﺩﺍﻳﺮﻩ ﺍﻱ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ .ﺳﺮﻋﺖ ﺩﺭ ﻣﻘﻄﻊ 1ﺩﺍﺭﺍﻱ
ﺭﺍﺑﻄﻪ U = U 0 + U 1 sin ωtﺍﺳﺖ،ﻛﻪ ﺩﺭ ﺁﻥ U = 2 m s ، U 0 = 20 m sﻭ . ω = 0.3 radﺍﺑﻌﺎﺩ ﻛﺎﻧﺎﻝ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ
s
R1 = 0.2m L = 1mﻭ . R2 = 0.1mﺷﺘﺎﺏ ﺭﺍ ﺩﺭ ﺧﺮﻭﺟﻲ ﻛﺎﻧﺎﻝ ﺑﻴﺎﺑﻴﺪ .ﺍﻳﻦ ﻧﺘﺎﻳﺞ ﺭﺍ ﺑﺮ ﺣﺴﺐ ﺯﻣﺎﻥ ﻭ ﺩﺭ ﻳﻚ
ﺳﻴﻜﻞ ﻛﺎﻣﻞ ﺭﺳﻢ ﻛﻨﻴﺪ .ﺩﺭ ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭ ،ﺷﺘﺎﺏ ﺩﺭ ﺧﺮﻭﺟﻲ ﻛﺎﻧﺎﻝ ﺭﺍ ﺑﺮﺍﻱ ﺣﺎﻟﺘﻲ ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻛﺎﻧﺎﻝ ﺩﺍﺭﺍﻱ
ﻣﺴﺎﺣﺖ ﺛﺎﺑﺖ ﺍﺳﺖ .ﺩﺭﺑﺎﺭﻩ ﺍﺧﺘﻼﻑ ﺍﻳﻦ ﻣﻨﺤﻨﻲ ﻫﺎ ﺗﻮﺿﻴﺢ ﺩﻫﻴﺪ.
۳۲
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﻧﺤﻮﻩ ﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﺗﻮﺯﻳﻊ ﻫﺎﻱ ﺳﺮﻋﺖ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺁﺭﺍﻡ ﺳﻴﺎﻻﺕ ﺭﺍ ﺩﺭ ﺳﻴﺴﺘﻢ ﻫﺎﻱ
ﺟﺮﻳﺎﻥ ﺳﺎﺩﻩ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﻴﻢ ﺩﺭ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﻫﺎ ﺍﺯ ﺗﻌﺮﻳﻒ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ،ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﻣﺮﺑﻮﻁ
ﺑﻪ ﺷﺎﺭﻫﺎﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻣﻮﻟﻜﻮﻟﻲ ﻭ ﻫﻤﺮﻓﺘﻲ ﻭ ﻣﻔﻬﻮﻡ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ ﻭﻗﺘﻲ
ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪ ﻣﻲ ﺗﻮﺍﻧﻴﻢ ﻛﻤﻴﺖ ﻫﺎﻱ ﺩﻳﮕﺮﻱ ﻣﺎﻧﻨﺪ ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ ،ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ،ﻳﺎ
ﺗﻨﺶ ﺑﺮﺷﻲ ﺩﺭ ﺳﻄﺢ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﻢ .
ﺭﻭﺵ ﻫﺎ ﻭ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺍﻳﻦ ﻓﺼﻞ ﻓﻘﻂ ﺩﺭ ﻣﻮﺭﺩ ﺟﺮﻳﺎﻥ ﺭﺍﺳﺖ ﺧﻂ ﭘﺎﻳﺎ ﺻﺎﺩﻕ ﺍﻧﺪ ﻣﻨﻈﻮﺭ ﺍﺯ "ﭘﺎﻳﺎ" ﺍﻳﻦ ﺍﺳﺖ
ﻛﻪ ﻓﺸﺎﺭ ،ﭼﮕﺎﻟﻲ ،ﻭ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺍﺯ ﺟﺮﻳﺎﻥ ﺑﺎ ﺯﻣﺎﻥ ﺗﻐﻴﻴﺮ ﻧﻤﻲ ﻛﻨﻨﺪ .
ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﻓﻘﻂ ﺑﺎ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺳﺮﻭﻛﺎﺭ ﺩﺍﺭﻳﻢ" .ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ" ﺟﺮﻳﺎﻥ ﻣﻨﻈﻤﻲ ﺍﺳﺖ ﻛﻪ ،ﻣﺜﻼً ﺩﺭ ﺟﺮﻳﺎﻥ
ﺩﺍﺧﻞ ﻟﻮﻟﻪ ،ﺑﺎ ﺳﺮﻋﺖ ﻫﺎﻳﻲ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﻱ ﻛﺎﻓﻲ ﭘﺎﻳﻴﻦ ،ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺩﺭ ﺁﻥ ﺫﺭﺍﺕ ﺭﻳﺰ ﺗﺰﺭﻳﻖ ﺷﺪﻩ ﺑﻪ
ﺩﺍﺧﻞ ﻟﻮﻟﻪ ﺩﺭ ﻃﻮﻝ ﺧﻄﻲ ﻧﺎﺯﻙ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﻨﺪ .ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺑﺎ "ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ" ﻛﻪ ﻛﺎﻣﻼً ﺁﺷﻔﺘﻪ ﻭ ﺑﻲ
ﻧﻈﻢ ﺍﺳﺖ ،ﺗﻔﺎﻭﺕ ﺁﺷﻜﺎﺭﻱ ﺩﺍﺭﺩ .ﺟﺮﻳﺎﻥ ﺍﺧﻴﺮ ﺩﺭ ﺳﺮﻋﺖ ﻫﺎﻱ ﺑﺎﻻ ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺩﺭ ﺁﻥ ﻫﺎ ﺫﺭﺍﺕ ﺑﻪ
ﺍﻃﺮﺍﻑ ﭘﺮﺍﻛﻨﺪﻩ ﻭ ﺩﺭ ﺗﻤﺎﻡ ﺳﻄﺢ ﻣﻘﻄﻊ ﻟﻮﻟﻪ ﭘﺎﺷﻴﺪﻩ ﻣﻲ ﺷﻮﻧﺪ.
1.2ﻣﻮﺍﺯﻧﻪ ﻱ ﻻﻳﻪ ﺍﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ
ﺑﺮﺍﻱ ﺣﻞ ﻣﺴﺌﻠﻪ ﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﺑﺨﺶ ﻫﺎﻱ 2.2ﺗﺎ 5.2ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﻧﺪ ،ﺑﺎﻳﺪ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﻱ ﺣﺮﻛﺖ ﺭﺍ
ﺭﻭﻱ "ﻻﻳﻪ ﻱ" ﻧﺎﺯﻛﻲ ﺍﺯ ﺳﻴﺎﻝ ﻧﻮﺷﺖ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﻱ ﺣﺮﻛﺖ ﭼﻨﻴﻦ ﺍﺳﺖ:
۳٤
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺍﻳﻦ ﺭﺍﺑﻄﻪ ،ﺑﻴﺎﻥ ﻣﺤﺪﻭﺩ ﻗﺎﻧﻮﻥ ﭘﺎﻳﺴﺘﮕﻲ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺳﺖ .ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﺍﺯ ﺍﻳﻦ ﺑﻴﺎﻥ ﻓﻘﻂ ﺩﺭ ﻣﻮﺭﺩ ﻳﻢ
ﻣﺆﻟﻔﻪ ﺍﺯ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ ،ﻛﻪ ﻣﺆﻟﻔﻪ ﺩﺭ ﺟﻬﺖ ﺟﺮﻳﺎﻥ ﺍﺳﺖ .ﺑﺮﺍﻱ ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ
ﺣﺮﻛﺖ ،ﺑﻪ ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﺷﺎﺭﻫﺎﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻫﻤﺮﻓﺘﻲ ﻭ ﺷﺎﺭﻫﺎﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻣﻮﻟﻜﻮﻟﻲ ﻧﻴﺎﺯ ﺩﺍﺭﻳﻢ ﻛﻪ
ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻣﻮﻟﻜﻮﻟﻲ ﺷﺎﻣﻞ ﻫﺮ ﺩﻭ ﺑﺨﺶ ﻓﺸﺎﺭﻱ ﻭ ﻭﻳﺴﻜﻮﺯ ﺍﺳﺖ .
ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺳﻴﺴﺘﻢ ﻫﺎﻳﻲ ﺑﻪ ﻛﺎﺭ ﻣﻲ ﺭﻭﺩﻛﻪ ﻳﻚ ﻣﺆﻟﻔﻪ ﻱ ﺳﺮﻋﺖ ﺩﺍﺭﻧﺪ ﻛﻪ ﺗﻨﻬﺎ ﺑﻪ
ﻳﻚ ﻣﺘﻐﻴﻴﺮ ﻣﻜﺎﻧﻲ ﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ ،ﺑﻪ ﻋﻼﻭﻩ ،ﺟﺮﻳﺎﻥ ﺑﺎﻳﺪ ﺭﺍﺳﺖ ﺧﻂ ﺑﺎﺷﺪ .
ﺭﻭﺵ ﻛﺎﺭ ﻣﺎ ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﺗﻌﺮﻳﻒ ﻭ ﺣﻞ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻭﻳﺴﻜﻮﺯ ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ﺍﺳﺖ :
• ﻣﺆﻟﻔﻪ ﻱ ﻏﻴﺮ ﺻﻔﺮ ﺳﺮﻋﺖ ﻭ ﻣﺘﻐﻴﺮ ﻣﻜﺎﻧﻲ ﺭﺍ ﻛﻪ ﺍﻳﻦ ﻣﺆﻟﻔﻪ ﺑﻪ ﺁﻥ ﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ ،ﺷﻨﺎﺳﺎﻳﻲ ﻛﻨﻴﺪ؛
• ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺘﻲ ﺑﻪ ﺻﻮﺭﺕ ﻣﻌﺎﺩﻟﻪ ) ، (1-1.2ﺭﻭﻱ ﻻﻳﻪ ﻱ ﻧﺎﺯﻛﻲ ﻋﻤﻮﺩ ﺑﺮ ﻣﺘﻐﻴﺮ ﻣﻜﺎﻧﻲ ﻣﺮﺑﻮﻁ
ﺑﻨﻮﻳﺴﻴﺪ؛
• ﺿﺨﺎﻣﺖ ﺍﻳﻦ ﻻﻳﻪ ﺭﺍ ﺑﻪ ﺳﻤﺖ ﺻﻔﺮ ﻣﻴﻞ ﺩﻫﻴﺪ ﻭ ﺍﺯ ﺗﻌﺮﻳﻒ ﻣﺸﺘﻖ ﺍﻭﻝ ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﻣﻌﺎﺩﻟﻪ ﻱ
ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﺘﻨﺎﻇﺮ ﺑﺮﺍﻱ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﻱ ﺣﺮﻛﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ؛
• ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺍﻧﺘﮕﺮﺍﻝ ﺑﮕﻴﺮﻳﺪ ﺗﺎ ﺗﻮﺯﻳﻊ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺣﺎﺻﻞ ﺷﻮﺩ؛
• ﻗﺎﻧﻮﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﻧﻴﻮﺗﻦ ﺭﺍ ﺑﻪ ﻛﺎﺭ ﺑﺒﺮﻳﺪ ﻭ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺑﺮﺍﻱ ﺳﺮﻋﺖ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ؛
• ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺍﻧﺘﮕﺮﺍﻝ ﺑﮕﻴﺮﻳﺪ ﺗﺎ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺣﺎﺻﻞ ﺷﻮﺩ؛
• ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ،ﻛﻤﻴﺖ ﻫﺎﻱ ﺩﻳﮕﺮ ،ﻣﺎﻧﻨﺪ ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ ،ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﻳﺎ ﻧﻴﺮﻭﻱ ﻭﺍﺭﺩ
ﺑﺮ ﺳﻄﻮﺡ ﺟﺎﻣﺪ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ .
ﺩﺭ ﺍﻧﺘﮕﺮﺍﻟﮕﻴﺮﻱ ﻫﺎﻱ ﺑﺎﻻ ،ﭼﻨﺪﻳﻦ ﺛﺎﺑﺖ ﺍﻧﺘﮕﺮﺍﻟﮕﻴﺮﻱ ﻇﺎﻫﺮ ﻣﻲ ﺷﻮﻧﺪ ﻛﻪ ﺑﺎﻳﺪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ "ﺷﺮﺍﻳﻂ
ﻣﺮﺯﻱ" – ﻳﻌﻨﻲ ﻋﺒﺎﺭﺕ ﻫﺎﻳﻲ ﺩﺭ ﻣﻮﺭﺩ ﺳﺮﻋﺖ ﻳﺎ ﺗﻨﺶ ﺩﺭ ﻣﺮﺯﻫﺎﻱ ﺳﻴﺴﺘﻢ – ﺁﻥ ﻫﺎ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﺮﺩ.
ﻣﺘﺪﺍﻭﻟﺘﺮﻳﻦ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ :
ﺍﻟﻒ( ﺩﺭ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﻫﺎﻱ ﺟﺎﻣﺪ -ﻣﺎﻳﻊ ،ﺳﺮﻋﺖ ﺳﻴﺎﻝ ﺑﺎ ﺳﺮﻋﺖ ﺣﺮﻛﺖ ﺳﻄﺢ ﺟﺎﻣﺪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ؛ ﺍﻳﻦ
ﻧﻜﺘﻪ ﻫﻢ ﺩﺭ ﻣﻮﺭﺩ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﻣﻤﺎﺳﻲ ﻭ ﻫﻢ ﺩﺭ ﻣﻮﺭﺩ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﻗﺎﺋﻢ ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ ،ﺻﺎﺩﻕ ﺍﺳﺖ.
ﺑﺮﺍﺑﺮﻱ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﻣﻤﺎ ﺳﻲ ﺭﺍ " ﺷﺮﻁ ﻋﺪﻡ ﻟﻐﺰﺵ" ﻣﻲ ﻧﺎﻣﻨﺪ .
۳٥
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺏ( ﺩﺭ ﺻﻔﺤﻪ ﻱ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﻣﺎﻳﻊ -ﻣﺎﻳﻊ ﺑﺎ xﺛﺎﺑﺖ ،ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺘﻲ ﻣﻤﺎﺳﻲ υ yﻭ υ zﺩﺭ
ﺳﺮﺍﺳﺮ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﭘﻴﻮﺳﺘﻪ ﺍﻧﺪ " ،ﺷﺮﻁ ﻋﺪﻡ ﻟﻐﺰﺵ" ،ﻫﻤﺎﻥ ﻃﻮﺭ ﻛﻪ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺗﺎﻧﺴﻮﺭ ﺗﻨﺶ
ﻣﻮﻟﻜﻮﻟﻲ ،ﻳﻌﻨﻲ p + τ xx ، τ xyﻭ τ xzﻧﻴﺰ ﭘﻴﻮﺳﺘﻪ ﺍﻧﺪ .
ﺝ( ﺩﺭ ﺻﻔﺤﻪ ﻱ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﻣﺎﻳﻊ -ﮔﺎﺯ ﺑﺎ xﺛﺎﺑﺖ ،ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺗﺎﻧﺴﻮﺭ ﺗﻨﺶ τ xyﻭ τ xzﺭﺍ ﺻﻔﺮ ﻣﻲ
ﮔﻴﺮﻳﻢ ،ﺑﻪ ﺷﺮﻁ ﺁﻥ ﻛﻪ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺩﺭ ﺳﻤﺖ ﮔﺎﺯ ﺧﻴﻠﻲ ﺑﺰﺭگ ﻧﺒﺎﺷﺪ .ﺍﻳﻦ ﻓﺮﺽ ﻣﻨﻄﻘﻲ ﺍﺳﺖ،
ﺯﻳﺮﺍ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﮔﺎﺯﻫﺎ ﺑﺴﻴﺎﺭ ﻛﻢ ﺗﺮ ﺍﺯ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﻣﺎﻳﻌﺎﺕ ﺍﺳﺖ .
ﺩﺭ ﻫﻤﻪ ﻱ ﺍﻳﻦ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻓﺮﺽ ﺑﺮ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻫﻴﭻ ﻣﺎﺩﻩ ﺍﻱ ﺍﺯ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﻋﺒﻮﺭ ﻧﻤﻲ ﻛﻨﺪ؛
ﻳﻌﻨﻲ ﺟﺬﺏ ﺳﻄﺤﻲ ،ﺟﺬﺏ ،ﺍﻧﺤﻼﻝ ،ﺗﺒﺨﻴﺮ ،ﺫﻭﺏ ،ﻳﺎ ﻭﺍﻛﻨﺶ ﺷﻴﻤﻴﺎﻳﻲ ﺩﺭ ﺳﻄﺢ ﺑﻴﻦ ﺩﻭ ﻓﺎﺯ ﺍﻧﺠﺎﻡ
ﻧﻤﻲ ﺷﻮﺩ.
∂
∫ c.v. ρV dv + ∫ c.s. ρV V ⋅ d A
∂t
↵0
= ∑F
•
•
∑ F = ∑ mi Vi − ∑ m i Vi
in
out
•
•
⇒ ∑ m i Vi − ∑ mi Vi + ∑ F = 0
out
in
rate of
rate of
sum of forces
momentum in − momentum out + acting in system = 0
Normal stress & shear stresses:
δ x = unit vector in the x direction
۳٦
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
Π ij = Pδ ij + τ ij where i and j may be x, y, or z
1
δ ij = kronecker delta =
0
i= j
i≠ j
Π ij = flux of j - momentum across a surface perpendicular to the i direction in the positive i direction
that is from the region of lesser xi to that of greater xi
τ yx ≡ flux of x - momentum in the positive y direction " flux" means " flow per unit area"
τ yx
:ﻧﻴﺮﻭﻳﻲ ﻛﻪ ﺳﻴﺎﻝ ﭘﺎﻳﻴﻨﻲ ﺑﻪ ﺳﻴﺎﻝ ﺑﺎﻻﻳﻲ ﺍﻋﻤﺎﻝ ﻣﻲ ﻛﻨﺪ
Π xy = τ xy
Π yz = τ yz ⇒ shear stresses
τ ≡ viscouse stress tensor
Π ≡ molecular stress tensor
viscous forces = ﻣﻮﻗﻌﻲ ﻛﻪ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺩﺍﺧﻞ ﺳﻴﺎﻝ ﺑﺎﺷﺪ
ﻧﻪ ﻣﻮﺍﺯﻱ ﻭ ﻧﻪ ﻋﻤﻮﺩ ﺑﺮ ﺳﻄﺢ ﻫﺴﺘﻨﺪ ﺑﻠﻜﻪ ﻳﻚ ﺯﺍﻭﻳﻪ ﺑﺎ ﺳﻄﺢ ﻣﻲτ x ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﻧﻴﺮﻭﻫﺎﻱ ﻭﻳﺴﻜﻮﺯ
.ﺳﺎﺯﻧﺪ
Velocity gradient as a “driving force” for momentum transport ← τ x → τ xx ,τ xy ,τ xz
.ﺍﻧﺘﻘﺎﻝ ﻣﻤﻨﺘﻮﻡ ﺍﺯ ﺟﺎﻳﻲ ﻛﻪ ﺳﺮﻋﺖ ﻛﻢ ﺍﺳﺖ ﻣﻨﺘﻘﻞ ﻣﻲ ﺷﻮﺩ
Π ij = force in the j direction on unit area perpendicular to the i direction where it is understood
the fluid in the region of lesser xi is exerting the force on the fluid of greater x i
۳۷
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
τ yx : force in the positive x direction on a plane perpendicular to the y direction, and that this
is the force exerted by the fluid in the region of the lessery on the fluid of greater y
Convective momentum transport :
ﻣﻤﻨﺘﻮﻡ ﺑﻪ ﻭﺳﻴﻠﻪ ﺣﺮﻛﺖ ﺑﺎﻟﻚ ﺳﻴﺎﻝ )ﺗﻮﺩﻩ ﺳﻴﺎﻝ( ﻧﻴﺰ. ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪΠ ij ﺍﻧﺘﻘﺎﻝ ﻣﻤﻨﺘﻮﻡ ﻣﻮﻟﻜﻮﻟﻲ ﺑﺎ
. ﮔﻮﻳﻨﺪconvective transport ﻣﻨﺘﻘﻞ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺑﻪ ﺁﻥ
∫ c . s . ρV V ⋅ d A
kg m m
m
d (mV )
⋅
⋅
⋅ m 2 = kg ⋅
=N=F=
3
2
m sec sec
sec
dt
.ﺍﻳﻦ ﻣﻘﺪﺍﺭ ﺳﻴﺎﻝ ﺍﺯ ﺳﻄﺢ ﻫﺎﺷﻮﺭ ﺯﺩﻩ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ
Q = vx × 1 = vx m
3
sec
:ﻣﻘﺪﺍﺭ ﻣﻤﻨﺘﻮﻣﻲ ﻛﻪ ﺍﻳﻦ ﻣﻘﺪﺍﺭ ﺳﻴﺎﻝ ﺑﺎ ﺧﻮﺩ ﺣﻤﻞ ﻣﻲ ﻛﻨﺪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ
ρV =
kg ⋅ m
sec = momentum V = ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ ﺳﻴﺎﻝ ﺩﺭ ﻣﺮﻛﺰ ﻣﻜﻌﺐ
m
m3
3
⇒ momentum flux across the shaded area
3
momentum momentum
= v x ρV m
×
=
sec
sec
m3
= ﻓﻼﻛﺲmomentum
ﺑﻪ ﺍﺯﺍﻱ ﻭﺍﺣﺪ ﺳﻄﺢ
2
m ⋅ sec
۳۸
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
. ﺭﻓﺘﻪ ﺍﺳﺖx ﺑﻪ ﻧﺎﺣﻴﻪ ﺑﺰﺭﮔﺘﺮx ﺍﻳﻦ ﻣﻘﺪﺍﺭ ﻓﻼﻛﺲ ﻣﻤﻨﺘﻮﻡ ﺍﺯ ﻧﺎﺣﻴﻪ ﻛﻤﺘﺮ
: ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ
. ﺳﻪ ﺑﺮﺩﺍﺭ ﻫﺴﺘﻨﺪ ﻛﻪ ﻫﺮ ﻛﺪﺍﻡ ﺳﻪ ﻣﺆﻟﻔﻪ ﺩﺍﺭﻧﺪρvz V ﻭρv y V ﻭρvx V
ρv x v y = convective flux of y - momentum across a surface perpendicular to the x direction.
τ xy = molecular fluxy of y - momentum across a surface perpendicular to the x direction.
۳۹
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ρV V = ∑ δ i ρvi V = ∑ δ i ρvi ∑ δ j v j = ∑∑ δ iδ j ρvi v j
i
i
j
i j
ﭼﻮﻥ ﻫﺮﻣﺆﻟﻔﻪ ρV Vﺩﻭ ﺍﻧﺪﻳﺲ ﺩﺍﺭﺩ ﻟﺬﺍ ρV Vﻳﻚ ﺗﺎﻧﺴﻮﺭ ﺍﺳﺖ ﻭ ﺑﻪ ﺁﻥ convective
momentum flux a tensorﮔﻮﻳﻨﺪ.
Total molecular momentum flux through a surface of orientation
] n= [n ⋅ Π ] = pn + [n ⋅τ
combined momentum = φ = Π + ρV V
φ xx = Π xx + ρv x v x = P + τ xx + ρv x v x
ﻓﺸﺎﺭ ﺑﻪ ﺳﻤﺖ ﺩﺍﺧﻞ ﻣﺜﺒﺖ = ⇒ φ = pδ + τ + ρV V P
φ xy = Π xy + ρv x v y = τ xy + ρv x v y
→ the combined flux of y - momentum across a surface perpendicular to the x direction
by molecular and convective mechanisms.
ﺩﺭ ﻣﻮﺭﺩ ﻋﻼﻣﺖ :ﺩﺭ ﻛﺘﺎﺏ ﺑﺮﺩ
τ yx ≡ force exerted by the fluid in the region of lessery on the fluid of greatery.
dv x
ﺩﺭ ﺑﻴﺸﺘﺮ ﻛﺘﺎﺏ ﻫﺎ ﺟﺎﻱ lesserﻭ greaterﻋﻮﺽ ﺷﺪﻩ ﻭ ﻟﺬﺍ
dy
τ yx = + µﻭ ﺩﺭ ﻛﺘﺎﺏ ﺑﺮﺩ ﺍﻳﻦ ﺗﻐﻴﻴﺮ
ﻋﻼﻣﺖ ﺑﺮﺍﻱ ﻫﻤﺎﻫﻨﮕﻲ ﺑﺎ ﻗﺎﻧﻮﻥ ﻓﻴﻚ ﻭ ﻓﻮﺭﻳﻪ ﺷﺪﻩ ﺍﺳﺖ.
2.2ﺟﺮﻳﺎﻥ ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ
1
F10
ﺟﺮﻳﺎﻥ ﻣﺎﻳﻌﻲ ﺍﺳﺖ ﻛﻪ ﺭﻭﻱ ﺻﻔﺤﻪ ﻱ ﺷﻴﺐ ﺩﺍﺭ ﺑﻪ ﻃﻮﻝ Lﻭ ﻋﺮﺽ Wﻣﻄﺎﺑﻖ ﺷﻜﻞ 1-2.2ﭘﺎﻳﻴﻦ ﻣﻲ ﺭﻭﺩ.
ﭼﻨﻴﻦ ﻓﻴﻠﻢ ﻫﺎﻳﻲ ﺩﺭ ﺍﺭﺗﺒﺎﻁ ﺑﺎ ﺑﺮﺝ ﻫﺎﻱ ﺟﺪﺍﺭﺧﻴﺲ ،ﺁﺯﻣﺎﻳﺶ ﻫﺎﻱ ﺗﺒﺨﻴﺮ ﻭ ﺟﺬﺏ ﮔﺎﺯ ،ﻭ ﭘﻮﺷﺶ ﻛﺎﺭﻱ ﻫﺎ
ﺑﺮﺭﺳﻲ ﺷﺪﻩ ﺍﻧﺪ .ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻭ ﭼﮕﺎﻟﻲ ﺳﻴﺎﻝ ﺭﺍ ﺛﺎﺑﺖ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ .
ﺗﻮﺻﻴﻒ ﻛﺎﻣﻞ ﺟﺮﻳﺎﻥ ﻣﺎﻳﻊ ،ﺑﻪ ﻋﻠﺖ ﺁﺷﻔﺘﮕﻲ ﺩﺭ ﻟﺒﻪ ﻫﺎﻱ ﺳﻴﺴﺘﻢ ) ( y = W ، y = 0 ، z = L ، z = 0ﺩﺷﻮﺍﺭ
ﺍﺳﺖ .ﺗﻮﺻﻴﻒ ﻛﺎﻓﻲ ﻏﺎﻟﺒﺎً ﺑﺎ ﭼﺸﻢ ﭘﻮﺷﻲ ﺍﺯ ﭼﻨﻴﻦ ﺁﺷﻔﺘﮕﻲ ﻫﺎﻳﻲ ،ﺑﻪ ﻭﻳﮋﻩ ﺍﮔﺮ Wﻭ Lﺩﺭ ﻣﻘﺎﻳﺴﻪ ﺑﺎ ﺿﺨﺎﻣﺖ
ﻓﻴﻠﻢ ، δ ،ﺑﺰﺭگ ﺑﺎﺷﻨﺪ ،ﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﺁﻫﻨﮓ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﻴﻦ ،ﺍﻧﺘﻈﺎﺭ ﻣﻲ ﺭﻭﺩ ﻛﻪ ﻧﻴﺮﻭﻫﺎﻱ ﻭﻳﺴﻜﻮﺯ ﺍﺯ
Flow Of A Falling Film
٤۰
۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺗﺪﺍﻭﻡ ﺷﺘﺎﺏ ﻣﺎﻳﻊ ﺭﻭ ﺑﻪ ﭘﺎﻳﻴﻦ ﺟﺪﺍﺭ ﺟﻠﻮﮔﻴﺮﻱ ﻛﻨﻨﺪ ،ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﻱ ﻛﻮﺗﺎﻫﻲ ﺩﺭ ﭘﺎﻳﻴﻦ ﺻﻔﺤﻪυ z ،
ﺍﺯ zﻣﺴﺘﻘﻞ ﺷﻮﺩ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﻣﺴﻠﻢ ﺩﺍﻧﺴﺘﻦ ﺍﻳﻦ ﻓﺮﺽ ﻫﺎ ﻣﻌﻘﻮﻝ ﺑﻪ ﻧﻈﺮ ﻣﻲ ﺭﺳﺪ .
ﻓﺮﺿﻴﺎﺕ:
ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ ﺑﺎ ρﻭ µﺛﺎﺑﺖ ؛
ﻃﻮﻝ ﻓﻴﻠﻢ = L؛ ﻋﺮﺽ ﻓﻴﻠﻢ = W؛ ﺿﺨﺎﻣﺖ ﻓﻴﻠﻢ = δ
ﺑﻌﺪ ﺍﺯ ﻣﺪﺗﻲ ﻛﻪ ﻓﻴﻠﻢ ﻣﺎﻳﻊ ﺣﺮﻛﺖ ﻛﺮﺩ vzﻣﺴﺘﻘﻞ ﺍﺯ zﻣﻲ ﺷﻮﺩ ﻭ ﻣﻲ ﺗﻮﺍﻥ ﻓﺮﺽ ﻛﺮﺩ ﻛﻪ:
) v z = vz ( xﻭ v x = 0ﻭ v y = 0ﻭ )P = P(x
ﭼﻮﻥ vzﺗﺎﺑﻊ xﺍﺳﺖ ﻋﻤﻮﺩ ﺑﻪ xﻳﻚ ﺷﻞ ﻧﺎﺯﻙ ﻳﺎ ﺍﻟﻤﺎﻥ ﻧﺎﺯﻙ ﺍﺯ ﺳﻴﺎﻝ ﻣﻲ ﮔﻴﺮﻳﻢ ﻭ ﺑﺮﺍﻱ ﺁﻥ ﻣﻮﺍﺯﻧﻪ
ﻣﻤﻨﺘﻮﻡ ﻣﻲ ﻧﻮﻳﺴﻴﻢ.
٤۱
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
1) Fully Developed: velocity not a function of flow direction ∂ (0) = 0
∂t
2) steady
(LW )(φxz ) x −(LW )(φxz ) x+ Dx + WDxφzz
z =0
− WDxφ zz
z=L
+ LWDxρ gasβ = 0
φ yz = ρv y v z + τ yz = 0
:∆ ﺩﺍﺭﻳﻢx → 0 ﺗﻘﺴﻴﻢ ﻭLWDx ﺍﮔﺮ ﻃﺮﻓﻴﻦ ﺭﺍ ﺑﺮ
lim
φ xz
x + Dx
x
∆x
∆x →0
⇒
−φ xz
∂φ xz φ zz
−
∂x
z =0
−
φ zz
−φ zz
L
z =0
−φ zz
z=L
L
z=L
= ρg cosβ
= ρg cosβ
generalization of Newton’slaw of viscosity
vx =0
φ xz = τ xz + ρvx vz ⇒ φ xz = τ xz
φ zz = P + τ zz + ρvz vz
( )
(
+
2
τ = − µ ∇V + ∇V + µ − κ ∇.V δ
3
]
∂v 2
⇒ τ zz = − µ 2 z + µ − κ ∇.V
∂z 3
∇ ⋅V = 0
( ) ﺑﺮﺍﻱ ﺳﻴﺎﻝ ﻏﻴﺮ ﻗﺎﺑﻞ ﺗﺮﺍﻛﻢ
⇒ τ zz = −2µ
∂v z
=0
∂z
( ﺍﺳﺖz ﻣﺴﺘﻘﻞ ﺍﺯvz ) ﭼﻮﻥ
∂v ∂v x
∂v x ∂v z
=0
+
= −µ z
∂x
∂x ∂z
∂z
τ xz = − µ
∇.V =
∂v x ∂v y ∂v z
+
+
∂x
∂y
∂z
∇.V = 0 ⇒
∂v z
=0⇒
∂z
ﺍﺳﺖz ﻣﺴﺘﻘﻞ ﺍﺯvz
∂v ∂v
∂v
τ yz = − µ z + y = 0 z = 0
∂y
∂y ∂z
∂v y
ﻭ
∂z
٤۲
=0
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
Dyadic product
∇V = ∑∑ δ iδ j
i
j
∂
vj
∂xi
δ = unit tensor with component δ ij
∇V = velocity gradient tensor with components
(∇V )
+
∂v j
∂xi
= transpose of the velocity gradient tensor with components
∂vi
∂
= ∑∑ δ iδ j vi
∂x j
xj
i
j
∇V
. = divergence of the velocity vector
ﺍﻓﺰﺍﻳﺶ ﺣﺠﻢ ﺣﺎﺻﻞ ﺍﺯ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺑﻪ ﺍﺯﺍﻱ ﻭﺍﺣﺪ ﺣﺠﻢ
ﺿﺮﻳﺐ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺣﺠﻤﻲ ﻣﺮﺑﻮﻁ ﺑﻪ ﺍﻧﺒﺴﺎﻁ ﺣﺠﻤﻲ
K = dilatational viscosity
K = 0 for monatomic gases at low density
∂v j
τ ij = − µ
∂xi
τ ij = τ ji
+
∂vi
∂x j
2
∂v
+ µ − K ∂v x + y + ∂v z
3
∂y
∂z
∂x
ﻭi and j
δ ij
i = 1,2,3
j = 1,2,3
ﻳﻜﻲ ﻫﺴﺘﻨﺪ ﻟﺬﺍ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺣﺬﻑ ﻣﻲz = L ﻭz = 0 ﺩﺭφ zz ﺍﺳﺖ ﻣﻘﺪﺍﺭvz = vz (x ) ﻭP = P(x ) ﭼﻮﻥ
.ﺷﻮﻧﺪ
⇒ φ zz = P + ρv z
⇒
∂τ xz
dτ
= ρ gasβ ⇒ xz = ρ gasβ ⇒ τ xz = (ρ gasβ )x + c1
∂x
dx
B.c.1 : at gas - liquid interface : at x = 0 τ xz = 0
⇒ 0 = 0 + c1 ⇒ c1 = 0
: ﻣﻄﺎﺑﻖ ﺷﻜﻞ
⇒ τ xz = (ρ gasβ )x
Newton’s law of viscosity : τ xz = − µ dvz
dx
٤۳
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻋﻼﻣﺖ ﻣﻨﻔﻲ :ﺩﺭ ﺟﻬﺖ ﺍﻓﺰﺍﻳﺶ xﺳﺮﻋﺖ ﻛﻢ ﻣﻲ ﺷﻮﺩ ﻭ dvzﻣﻨﻔﻲ ﺍﺳﺖ ﻟﺬﺍ ﺩﺭ ﻳﻚ ﻣﻨﻔﻲ ﺿﺮﺏ ﻣﻲ ﺷﻮﺩ
dx
ﻭ ﺍﻧﺘﻘﺎﻝ ﻣﻤﻨﺘﻮﻡ ﺍﺯ ﺳﺮﻋﺖ ﺑﻴﺸﺘﺮ ﺑﻪ ﺳﻤﺖ ﺳﺮﻋﺖ ﻛﻤﺘﺮ ﺍﺳﺖ.
ρ gasβ
ρ gasβ 2
dv z
= −
x ⇒ v = −
x + c2
dx
µ
µ
2
⇒
B.c.2 : no - slip boundary condition : at x = δ v z = 0
ρ gasβ 2
⇒ 0 = −( )δ 2 + c 2 ⇒ c 2 =
δ
2µ
2
ρgδ 2 cos β x
= ⇒ vz
1 − Parabolic Eq.
2µ
δ
ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ υ z ,maxﺁﺷﻜﺎﺭﺍ ﻫﻤﺎﻥ ﺳﺮﻋﺖ ﺩﺭ x = 0ﺍﺳﺖ؛ ﻳﻌﻨﻲ :
dv z
ρgδ 2 cos β
= = 0 ⇒ x = 0 ⇒ v z ,max
dx
2µ
.1ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ⟩ ⟨υ zﺩﺭ ﻣﻘﻄﻌﻲ ﺍﺯ ﻓﻴﻠﻢ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
٤٤
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
δ
w
∫ ∫ v dxdy = 1 v dA = 1 δ v (Wdx ) = 1 δ v dx
δ
∫A
∫ Wδ
∫δ
dxdy
∫ ∫
z
z
z
0
0
z
0
0
w
A
0
x 2 x ρgδ 2 cos β 2
∫0 1 − δ d δ = 3µ = 3 v z ,max
1
= v z
0
ρgδ 2 cos β
=
2µ
⇒ v z ,max = 1.5 v z
ﺍﻧﺘﮕﺮﺍﻝ ﺩﻭﮔﺎﻧﻪ ﺩﺭ ﻣﺨﺮﺝ ﻛﺴﺮ ﺳﻄﺮ ﻧﺨﺴﺖ ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﻓﻴﻠﻢ ﺍﺳﺖ .ﺍﻧﺘﮕﺮﺍﻝ ﺩﻭﮔﺎﻧﻪ ﺩﺭ ﺻﻮﺭﺕ ﻫﻤﻴﻦ ﻛﺴﺮ
ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺣﺠﻤﻲ ﺩﺭ ﺟﺰء ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺍﺯ ﻣﻘﻄﻊ υ z dxdyﺍﺳﺖ ﻛﻪ ﺭﻭﻱ ﺗﻤﺎﻡ ﻣﻘﻄﻊ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﻣﻲ
ﺷﻮﺩ .
.2ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ωﺍﺯﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﻳﺎ ﺍﺯ ﻃﺮﻳﻖ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺣﺎﺻﻞ ﻣﻴﺸﻮﺩ:
ρ 2 gWδ 3 cos β
3µ
= ρv z dxdg = ρWδ v z
δ
∫
w
0
∫=w
0
.3ﺿﺨﺎﻣﺖ ﻓﻴﻠﻢ δﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺮ ﺣﺴﺐ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﻳﺎ ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ،ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺍﺭﺍﺋﻪ ﺩﺍﺩ :
3µ v z
3µω
3µπ
=3 2
=3 2
ρg cos β
ρ gW cos β
ρ g cos β
=δ
π = ρδ vz = mass rate of flow per unit width of wall
.4ﻧﻴﺮﻭ ﺑﺮ ﻭﺍﺣﺪ ﺳﻄﺢ ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺭﻭﻱ ﺟﺰء ﺳﻄﺢ ﻋﻤـﻮﺩ ﺑـﺮ ﺍﻣﺘـﺪﺍﺩ xﺑﺮﺍﺑـﺮ ﺍﺳـﺖ ﺑـﺎ τ xz :ﻛـﻪ ﺩﺭ
ﻣﺤﺎﺳﺒﻪ x = δﺷﺪﻩ ﺑﺎﺷﺪ .ﺍﻳﻦ ﻧﻴﺮﻭﻳﻲ ﺍﺳﺖ ﻛﻪ ﺳﻴﺎﻝ ) ﻧﺎﺣﻴﻪ ﻱ xﻛﻤﺘﺮ( ﺑـﺮ ﺟـﺪﺍﺭ )ﻧﺎﺣﻴـﻪ ﻱ x
ﺑﻴﺸﺘﺮ ( ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ .ﻣﺆﻟﻔﻪ ﻱ zﻧﻴﺮﻭﻱ Fﻛﻪ ﺳﻴﺎﻝ ﺑﺮ ﺳﻄﺢ ﺟﺎﻣﺪ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ
ﺍﺯ ﺗﻨﺶ ﺑﺮﺷﻲ ﺭﻭﻱ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﺳﻴﺎﻝ ﺟﺎﻣﺪ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
dv
− µ z dydz
dx x=δ
∫ ∫ = )dydz
L W
0
0
∫ (τ
L W
xz x =δ
0
∫ = Fz
0
ρgδ cos β
= (LW )(− µ ) −
= ρgδLW cos β
µ
ﺍﻳﻦ ﻣﺆﻟﻔﻪ zﻭﺯﻥ ﺳﻴﺎﻝ ﺩﺭ ﻛﻞ ﻓﻴﻠﻢ ﺍﺳﺖ.
٤٥
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺸﺎﻫﺪﺍﺕ ﺗﺠﺮﺑﻲ ﻓﻴﻠﻢ ﻫﺎﻱ ﺭﻳﺰﺍﻥ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﻋﻤﻼً ﺳﻪ " ﺭژﻳﻢ ﺟﺮﻳﺎﻥ " ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻛﻪ ﻣﻲ ﺗﻮﺍﻥ ﺁﻥ ﻫﺎ
ﺭﺍ ﺑﺮ ﺣﺴﺐ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ، Re ،ﺟﺮﻳﺎﻥ ﺩﺳﺘﻪ ﺑﻨﺪﻱ ﻛﺮﺩ .ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺑﺮﺍﻱ ﻓﻴﻠﻢ ﻫﺎﻱ ﺭﻳﺰﺍﻥ ﺑﺎ ﺭﺍﺑﻄﻪ ﻱ ﺯﻳﺮ
ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ .
4 A 4δW
=
= 4δ
P
W
= DH
ρ v z DH
µ
ﻭ
4µ
µ
=
4δ v z ρ
µ
= Re
= ⇒ Re
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺳﻪ ﺭژﻳﻢ ﺟﺮﻳﺎﻥ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ :
ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺑﺎ ﻣﻮﺝ ﻫﺎﻱ ﻗﺎﺑﻞ ﭼﺸﻢ ﭘﻮﺷﻲ
Re < 20
ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺑﺎ ﻣﻮﺝ ﻫﺎﻱ ﺩﺭ ﺧﻮﺭ ﺍﻋﺘﻨﺎ
20 < Re < 1500
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ
Re > 1500
ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ ﺑﺎ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﻣﺘﻐﻴﺮ
1
F1
ﻣﺴﺌﻠﻪ ﻱ ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ ﺭﺍ ﺩﺭ ﺣﺎﻟﺖ ﻭﺍﺑﺴﺘﻪ ﺑﻮﺩﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺑﻪ ﻣﻜﺎﻥ ،ﻳﻌﻨﻲ
ﺩﻭﺑﺎﺭﻩ ﺣﻞ ﻛﻨﻴﺪ؛ ﺍﻳﻦ ﺣﺎﻟﺖ ﻫﻨﮕﺎﻣﻲ ﭘﺪﻳﺪ ﻣﻲ ﺁﻳﺪ ﻛﻪ ﻓﻴﻠﻢ ﻏﻴﺮﻫﻢ ﺩﻣﺎ ﺑﺎﺷﺪ ،ﻣﺎﻧﻨﺪ ﭼﮕﺎﻟﺶ ﺑﺨﺎﺭ ﺭﻭﻱ ﺟﺪﺍﺭ.
ﺩﺭ ﺍﻳﻦ ﺟﺎ µ 0ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﺳﻄﺢ ﻓﻴﻠﻢ ﻭ αﺛﺎﺑﺘﻲ ﺍﺳﺖ ﻛﻪ ﺳﺮﻋﺖ ﻛﺎﻫﺶ µﺑﺎ ﺍﻓﺰﺍﻳﺶ xﺭﺍ ﻧﺸﺎﻥ ﻣﻲ
ﺩﻫﺪ .ﺍﻳﻦ ﻧﻮﻉ ﺗﻐﻴﻴﺮﺍﺕ ﻣﻲ ﺗﻮﺍﻧﺪ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺎﻳﻊ ﭼﮕﺎﻟﻴﺪﻩ ﺑﻪ ﻃﺮﻑ ﭘﺎﻳﻴﻦ ﺟﺪﺍﺭ ،ﺑﺎ ﮔﺮﺍﺩﻳﺎﻥ ﺧﻄﻲ ﺩﻣﺎ ﺩﺭ ﻃﻮﻝ
ﻓﻴﻠﻢ ،ﭘﺪﻳﺪ ﺁﻳﺪ.
−α x / δ
µ = µ0 e
set upﻣﺴﺄﻟﻪ ﻫﻤﺎﻧﻨﺪ ﻗﺒﻞ ﺍﺳﺖ:
dv z
= ρgx cos β
dx
− µ 0 e −( x δ ) .
ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﻭ ﺍﻋﻤﺎﻝ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ:
Falling Film With Variable Viscosity
٤٦
۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
α 1 1 αx δ x
1
e α − α 2 − e αδ − α 2
ρgδ 2 cos β
µ0
= vz
ﺍﮔﺮ α = 0ﺑﺎﺷﺪ ﺑﻪ ﺣﺎﻟﺖ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺛﺎﺑﺖ ﺑﺮﻣﻲ ﮔﺮﺩﻳﻢ.
∞ if α = 0 ⇒ vz = ∞ −
ﺑﺮﺍﻱ ﺭﻓﻊ ﺍﺑﻬﺎﻡ eαﻭ eαx δﺭﺍﺑﺴﻂ ﻣﻲ ﺩﻫﻴﻢ.
1
α2 α3
1
+
+ ⋅ ⋅ ⋅ − 2
1 + α +
2
!2
!3
ρgδ cos β
α α
lim
=
2 2
3 3
α →0
µ
− 1 + αx + α x2 + α x3 + ⋅ ⋅ ⋅ x − 12
0
αδ α
δ
2!δ
3!δ
(v z )α =0
2
1 1
1 x3
ρgδ 2 cos β
1 x
lim + α + ⋅ ⋅ ⋅ −
α
+
+
⋅
⋅
⋅
=
2
3
α →0
2
3
2
3
µ
δ
δ
0
x 2
1 −
δ
α 1
2
2 2
e α − α 2 + α 3 − α 3
3.2ﺟﺮﻳﺎﻥ ﺩﺭ ﻟﻮﻟﻪ ﻣﺪﻭﺭ
ρgδ 2 cos β
=
2µ
ρgδ 2 cos β
µ
0
= ⇒ v z
0
1
F12
ﺟﺮﻳﺎﻥ ﺳﻴﺎﻻﺕ ﺩﺭ ﻟﻮﻟﻪ ﻫﺎﻱ ﻣﺪﻭﺭ ﻏﺎﻟﺒﺎ" ﺩﺭ ﻓﻴﺰﻳﻚ ،ﺷﻴﻤﻲ ،ﺯﻳﺴﺖ ﺷﻨﺎﺳﻲ ،ﻭ ﻣﻬﻨﺪﺳﻲ ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﺩ
.ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺳﻴﺎﻻﺕ ﺩﺭ ﻟﻮﻟﻪ ﻫﺎﻱ ﻣﺪﻭﺭ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻮﺍﺯﻧﻪ ً ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺗﻮﺻﻴﻒ ﺷﺪﻩ ﺩﺭ ﺑﺨﺶ
1.2ﺗﺤﻠﻴﻞ ﻛﺮﺩ .ﺗﻨﻬﺎ ﻭﻳﮋﮔﻲ ﺟﺪﻳﺪﻱ ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺟﺎ ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﺩ ،ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺍﺳﺖ
ﻛﻪ ﻣﺨﺘﺼﺎﺕ ﻃﺒﻴﻌﻲ ﺑﺮﺍﻱ ﺗﻮﺻﻴﻒ ﻣﻜﺎﻥ ﺩﺭ ﻟﻮﻟﻪ ﺍﻱ ﺑﺎ ﻣﻘﻄﻊ ﺩﺍﻳﺮﻫﺎﻱ ﺍﺳﺖ .
ﺣﺎﻝ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺣﺎﻟﺖ ﭘﺎﻳﺎﻱ ﺳﻴﺎﻟﻲ ﺑﺎ ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ ρﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ً µﺭﺍ ﺩﺭ ﻟﻮﻟﻪ ﺍﻱ ﻋﻤﻮﺩﻱ ﺑﻪ ﻃﻮﻝ L
ﻭ ﺷﻌﺎﻉ ﻣﻘﻄﻊ Rﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ .ﻣﺎﻳﻊ ﺗﺤﺖ ﺗﺎًﺛﻴﺮ ﺍﺧﺘﻼﻑ ﻓﺸﺎﺭ ﻭ ﮔﺮﺍﻧﺶ ﺑﻪ ﺳﻤﺖ ﭘﺎﻳﻴﻦ ﺟﺎﺭﻱ ﻣﻲ
ﺷﻮﺩ ،ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﺩﺭ ﺷﻜﻞ 1-3.2ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ .ﻃﻮﻝ ﻟﻮﻟﻪ ﺭﺍ ﻧﺴﺒﺖ ﺑﻪ ﺷﻌﺎﻉ ﻣﻘﻄﻊ ﺁﻥ ،ﺑﺴﻴﺎﺭ
ﺯﻳﺎﺩ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ ،ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ "ﺁﺛﺎﺭ ﺍﻧﺘﻬﺎﻳﻲ " ﺩﺭ ﺑﺨﺶ ﻋﻤﺪﻩ ﺍﻱ ﺍﺯ ﻃﻮﻝ ﻟﻮﻟﻪ ﺑﻲ ﺍﻫﻤﻴﺖ ﺷﻮﻧﺪ ؛ ﺑﻪ
ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ﻣﻲ ﺗﻮﺍﻧﻴﻢ ﺍﺯ ﺍﻳﻦ ﻧﻜﺘﻪ ﭼﺸﻢ ﭘﻮﺷﻲ ﻛﻨﻴﻢ ﻛﻪ ﺩﺭ ﻭﺭﻭﺩﻱ ﻭ ﺧﺮﻭﺟﻲ ﻟﻮﻟﻪ ،ﺟﺮﻳﺎﻥ ﺍﻟﺰﺍﻣﺎً ﺑﺎ ﺟﺪﺍﺭ
ﻟﻮﻟﻪ ﻣﻮﺍﺯﻱ ﺍﺳﺖ.
Flow Through A Circular Tube
٤۷
۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ:
) P = P(z
,
vθ = 0
,
vr = 0
,
) v z = v z (r
ﺩﺭ ﻧﺘﻴﺠﻪ ﻳﻚ ﺍﻟﻤﺎﻥ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺑﻪ ﺿﺨﺎﻣﺖ ∆rﻭ ﻃﻮﻝ Lﺍﻧﺘﺨﺎﺏ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ﻋﻤﻮﺩ ﺑﺮ ﺟﻬﺖ rﻣﻲ ﺑﺎﺷﺪ:
ﺁﻫﻨﮓ ﻭﺭﻭﺩﻱ ﻣﺆﻟﻔﻪ ﻱ zﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺳﻄﺢ ﺣﻠﻘﻮﻱ ﺩﺭ : z = 0
z =0
) (2πr∆r )(φ zz
ﺁﻫﻨﮓ ﺧﺮﻭﺟﻲ ﻣﺆﻟﻔﻪ ﻱ zﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺳﻄﺢ ﺣﻠﻘﻮﻱ ﺩﺭ : z = L
z=L
) (2πr∆r )(φ zz
ﺁﻫﻨﮓ ﻭﺭﻭﺩﻱ ﻣﺆﻟﻔﻪ ﻱ zﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺳﻄﺢ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺩﺭ : r
(2πrL)(φ rz ) r = (2πrLφ rz ) r
ﺁﻫﻨﮓ ﺧﺮﻭﺟﻲ ﻣﺆﻟﻔﻪ ﻱ zﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺳﻄﺢ ﺍﺳﺘﻮﻧﻪ ﺍﻱ ﺩﺭ : r + ∆r
(2π (r + ∆r ) L)(φ rz ) r + ∆r = (2πrLφ rz ) r + ∆r
ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺶ ﻭﺍﺭﺩ ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺑﺮ ﭘﻮﺳﺘﻪ ﻱ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ :
(2πr∆rL) ρg
ﺍﻛﻨﻮﻥ ﺍﻳﻦ ﺳﻬﻢ ﻫﺎ ﺭﺍ ﺩﺭ ﻣﺆﺍﺯﻧﻪ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﻴﻢ :
٤۸
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
(2πrLφrz ) r − (2πrLφrz ) r +∆r + (2πr∆r )(φzz ) z=0 − (2πr∆r )(φzz ) z=L + (2πr∆rL )ρg = 0
: ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ،∆ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﻢr → 0 ﺗﻘﺴﻴﻢ ﻛﻨﻴﻢ ﻭ ﺣﺪ ﻋﺒﺎﺭﺕ ﺭﺍ ﻭﻗﺘﻲ2πL∆r ﻭﻗﺘﻲ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﺑﺮ
(rφrz ) r +∆r −(rφrz ) r
lim
∆r →0
∆r
⇒
φ
∂
(rφrz ) = zz
∂r
z =0
φ zz
=
−φ zz
L
z=L
z =0
−φ zz
L
z=L
+ ρg r
+ ρg r
φrz = τ rz + ρvr v z
φ zz = P + τ zz + ρv z v z
ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ
Appendix B.۱ (Bird):
∂v
∂v
τ rz = − µ r + z
∂r
∂z
٤۹
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
) (
∂v z 2
+ µ − k ∇.V
∂z 3
∂v z
∂z
τ zz = − µ 2
∂v z
∂r
ﻭ
τ zz = −2 µ
vr =0
⇒ τ rz = − µ
∇ .V =0
∂v z
+ ρvr v z
∂r
∂v z
+ ρv z v z
∂z
⇒ φrz = − µ
φ zz = P + τ zz + ρv z v z = P − 2µ
ﺳﭙﺲ ﻓﺮﺿﻲ ﺭﺍ ﻛﻪ ﺩﺭ ﺍﺑﺘﺪﺍﻱ ﻣﺴﺌﻠﻪ ﺍﺧﺘﻴﺎﺭ ﻛﺮﺩﻳﻢ ﻣﻨﻈﻮﺭ ﻣﻲ ﻛﻨﻴﻢ؛ﺍﻳﻦ ﻓﺮﺽ ﻛﻪ
) P = P(z
,
vθ = 0
,
vr = 0
,
) v z = v z (r
ﺣﺎﻝ ﺍﻳﻦ ﺳﺎﺩﻩ ﺳﺎﺯﻱ ﻫﺎ ﺭﻭ ﺍﻧﺠﺎﻡ ﻣﻲ ﺩﻫﻴﻢ :
(1
(2
(3
vr = 0 ⇒ ρvr vz = 0
ﺗﺮﻡ ρvz vzﺩﺭ ﺍﺑﺘﺪﺍ ﻭ ﺍﻧﺘﻬﺎﻱ ﻟﻮﻟﻪ ﻳﻜﻲ ﺍﺳﺖ.
⇒ ) vz = vz (r
∂v z
= 0 ⇒ φ zz = P
∂z
d
P − PL
(P − ρg 0 ) − (PL − ρgL ) P0 − PL
⇒ (rz rz ) = 0
+ ρg r = 0
r =
r
dr
L
L
L
P = P − ρgz
⇒ ) v z = v z (r
ﻛﻤﻴﺖ ﻣﺸﺨﺺ ﺷﺪﻩ ﺑﺎ Pﺭﺍ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﻣﻲ ﻧﺎﻣﻨﺪ .ﺑﻪ ﻃﻮﺭ ﻛﻠﻲ ﺍﻳﻦ ﻛﻤﻴﺖ ﺑﺎ ﺭﺍﺑﻄﻪ ﻱ P = P + ρgh
ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ ﻛﻪ hﻓﺎﺻﻠﻪ " ﺑﻪ ﻃﺮﻑ ﺑﺎﻻ" ﺍﺳﺖ؛ﻳﻌﻨﻲ ﺩﺭ ﺟﻬﺘﻲ ﻣﺨﺎﻟﻒ ﺑﺎ ﺟﻬﺖ ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺶ ﺍﺯ ﺻﻔﺤﻪ ﻱ
ﻣﺮﺟﻊ ﺍﻧﺘﺨﺎﺑﻲ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺭ ﺍﻳﻦ ﻣﺴﺄﻟﻪ . h = − z :
ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻗﺒﻞ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ :
c
P0 − PL
r + 1
r
2L
τ rz =
B.C.1 : at r = 0 τ rz = finite
ﺛﺎﺑﺖ c1ﺑﺎﻳﺪ ﺻﻔﺮ ﺑﺎﺷﺪ ﺩﺭ ﻏﻴﺮ ﺍﻳﻦ ﺻﻮﺭﺕ ﺩﺭ r = 0ﻣﺤﻮﺭ ﻟﻮﻟﻪ ﻣﻘﺪﺍﺭ ﻓﻼﻛﺲ ﻣﻤﻨﺘﻮﻡ ﺑﻲ ﻧﻬﺎﻳﺖ ﺧﻮﺍﻫﺪ ﺷﺪ.
٥۰
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
P −P
⇒ τ rz = 0 L r
2L
dv
dv P − P
τ rz = − µ z ⇒ − µ z = 0 L r
dr
dr 2 L
P −P
dv
⇒ z = − 0 L r
dr
2 µL
P −P
⇒ vz = − 0 L r 2 + c2
4µL
B.C.2 : at r = R ⇒ v z = 0 ⇒ c2 =
Parabolic ⇒ vz
(P0 − PL )R 2
4 µL
2
(
P0 − PL )R 2 r
1−
=
4 µL
R
: ﻭ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯr = 0 ﻫﻨﮕﺎﻣﻲ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﺩ ﻛﻪυ z ,max ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ.1
dv z
=0 ⇒ r =0
dr
(P − PL )R 2
⇒ v z ,max = 0
4 µL
v z ,max ,
٥۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
.۲ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ υ z ﺍﺯ ﺗﻘﺴﻴﻢ ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺣﺠﻤﯽ ﮐﻞ ﺑﺮ ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ ﺣﺎﺻﻞ ﻣﯽ ﺷﻮﺩ:
1
1 R
) ∫ A v z dθ = ∫ v z (2πrdr
A
A 0
2π R
⇒ ∫ ∫ v z rdrdθ
v z = 0 2π 0 R
∫ ∫ rdrdθ
= v z
0
1
v z ,max ⇒ v z ,max = 2 v z
2
=
0
(P − PL )R 2
= 0
8µL
vz
.3ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ωﺣﺎﺻﻞ ﺿﺮﺏ ﻣﺴﺎﺣﺖ ﻣﻘﻄﻊ πR 2ﭼﮕﺎﻟﻲ ، ρﻭﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ υ z
ﺍﺳﺖ:
Hagen - Poiseuille equation
π (P0 − PL )R 4 ρ
= ω = ρ v z πR
8µL
2
.4ﻣﺆﻟﻔﻪ ﻱ zﻧﻴﺮﻭﻳﻲ ﻛﻪ ﺳﻴﺎﻝ ﺑﺮ ﺳﻄﺢ ﺧﻴﺲ ﻟﻮﻟﻪ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ ، Fzﺩﺭﺳﺖ ﺑﺮﺍﺑﺮ ﺍﻧﺘﮕﺮﺍﻝ ﺗﻨﺶ ﺑﺮﺷﻲ τ rz
ﺭﻭﻱ ﺳﻄﺢ ﺧﻴﺲ ﺷﺪﻩ ﺍﺳﺖ :
dv
Fz = (2πRL ) − µ z = πR 2 (P0 − PL ) = πR 2 (P0 − PL ) + πR 2 Lρg
dr r = R
ﺍﻳﻦ ﻧﺘﻴﺠﻪ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﻧﻴﺮﻭﻱ ﻭﻳﺴﻜﻮﺯ Fzﺑﺎ ﻧﻴﺮﻭﻱ ﻓﺸﺎﺭﻱ ﺧﺎﻟﺺ ﻭ ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺶ ﺧﻨﺜﻲ ﻣﻲ ﺷﻮﺩ .
ﺍﻳﻦ ﺩﻗﻴﻘﺎً ﻫﻤﺎﻥ ﻧﺘﻴﺠﻪ ﺍﻱ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﻧﻴﺮﻭ ﺭﻭﻱ ﺳﻴﺎﻝ ﺩﺍﺧﻞ ﻟﻮﻟﻪ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻣﺪ.
ﺣﺎﻝ ﻫﻤﻪ ﻱ ﻓﺮﺽ ﻫﺎﻱ ﺍﺧﺘﻴﺎﺭ ﺷﺪﻩ ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﻣﻌﺎﺩﻟﻪ ﻱ ﻫﻴﮕﻦ – ﭘﻮﺋﺎﺯﻭﻱ ﺭﺍﺟﻤﻊ ﺑﻨﺪﻱ ﻣﻲ
ﻛﻨﻴﻢ.
ﺍﻟﻒ( ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺍﺳﺖ؛ ﻳﻌﻨﻲ Reﺑﺎﻳﺪ ﺍﺯ ﺣﺪﻭﺩ 2100ﻛﻮﭼﻚ ﺗﺮ ﺑﺎﺷﺪ.
ﺏ( ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ ﺍﺳﺖ )ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ(.
ﺝ( ﺟﺮﻳﺎﻥ "ﭘﺎﻳﺎ " ﺍﺳﺖ )ﻳﻌﻨﻲ ﺑﺎ ﺯﻣﺎﻥ ﺗﻐﻴﻴﺮ ﻧﻤﻲ ﻛﻨﺪ(.
ﺩ( ﺳﻴﺎﻝ ﻧﻴﻮﺗﻨﻲ ﺍﺳﺖ.
٥۲
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻩ( ﺍﺯ ﺁﺛﺎﺭ ﺍﻧﺘﻬﺎﻳﻲ ﭼﺸﻢ ﭘﻮﺷﻲ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﻭﺍﻗﻊ "ﻃﻮﻝ ﻭﺭﻭﺩﻱ" ﭘﺲ ﺍﺯ ﻣﺪﺧﻞ ﻟﻮﻟﻪ ،ﺑﺎﻳﺪ ﺍﺯ ﻣﺮﺗﺒﻪ ﻱ
Le = 0.035D Reﺑﺎﺷﺪ ﺗﺎ ﺗﻮﺯﻳﻊ ﺳﻬﻤﻮﻱ ﺍﻳﺠﺎﺩ ﺷﻮﺩ .ﺍﮔﺮ ﻗﻄﻌﻪ ﻱ ﻣﻮﺭﺩ ﻧﻈﺮ ﺍﺯ ﻟﻮﻟﻪ ﺷﺎﻣﻞ ﻧﺎﺣﻴﻪ ﻱ ﻭﺭﻭﺩﻱ
ﺑﺎﺷﺪ ،ﺑﺎﻳﺪ ﺗﺼﺤﻴﺤﻲ ﺭﺍ ﺩﺭ ﺁﻥ ﺍﻋﻤﺎﻝ ﻛﺮﺩ ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ، Le Lﺗﺼﺤﻴﺢ ﺟﺰﺋﻲ ﺍﺧﺘﻼﻑ ﻓﺸﺎﺭ ﻳﺎ ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ
ﺟﺮﻣﻲ ﻫﺮﮔﺰ ﺍﺯ L Leﺑﺰﺭﮔﺘﺮ ﻧﺨﻮﺍﻫﺪ ﺷﺪ.
ﻭ( ﺳﻴﺎﻝ ﻣﺎﻧﻨﺪ ﭘﻴﻮﺳﺘﺎﺭ ﺭﻓﺘﺎﺭ ﻣﻲ ﻛﻨﺪ – ﺍﻳﻦ ﻓﺮﺽ ﻣﻌﺘﺒﺮ ﺍﺳﺖ – ﻣﮕﺮ ﺑﺮﺍﻱ ﮔﺎﺯ ﻫﺎﻱ ﺑﺴﻴﺎﺭ ﺭﻗﻴﻖ ﻳﺎ ﻟﻮﻟﻪ ﻫﺎﻱ
ﻣﻮﻳﻴﻦ ﺑﺴﻴﺎﺭ ﺑﺎﺭﻳﻚ ،ﻛﻪ ﺩﺭ ﺁﻥ ﻫﺎ ﻣﺴﺎﻓﺖ ﺁﺯﺍﺩ ﻣﻴﺎﻧﮕﻴﻦ ﻣﻮﻟﻜﻮﻟﻲ ﺑﺎ ﻗﻄﺮ ﻟﻮﻟﻪ ﻗﺎﺑﻞ ﻣﻘﺎﻳﺴﻪ ﺍﺳﺖ )ﻧﺎﺣﻴﻪ ﻱ
ﺟﺮﻳﺎﻥ ﻟﻐﺰﺷﻲ ( ،ﻳﺎ ﺍﺯ ﻗﻄﺮ ﻟﻮﻟﻪ ﺑﺴﻴﺎﺭ ﺑﺰﺭﮔﺘﺮ ﺍﺳﺖ ) ﺭژﻳﻢ "ﺟﺮﻳﺎﻥ ﻛﻨﻮﺳﻦ" ﻳﺎ " ﺟﺮﻳﺎﻥ ﻣﻮﻟﻜﻮﻟﻲ ﺁﺯﺍﺩ"(
ﺯ( ﺭﻭﻱ ﺟﺪﺍﺭ ﻟﻮﻟﻪ ﻟﻐﺰﺵ ﺭﺥ ﻧﻤﻲ ﺩﻫﺪ ،ﺑﻨﺎﺑﺮﺍﻳﻦ ﺷﺮﻁ ﻣﺮﺯﻱ 2ﻣﻌﺘﺒﺮ ﺍﺳﺖ؛ ﺍﻳﻦ ﻓﺮﺿﻲ ﻋﺎﻟﻲ ﺑﺮﺍﻱ ﺳﻴﺎﻻﺕ
ﺧﺎﻟﺺ ﺗﺤﺖ ﺷﺮﺍﻳﻂ ﻣﻔﺮﻭﺽ ﺩﺭ )ﻭ( ﺍﺳﺖ.
ﻣﺜﺎﻝ :ﻋﺒﺎﺭﺗﻲ ﺑﺮﺍﻱ ﺑﻴﺎﻥ ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ωﺩﺭ ﮔﺎﺯ ﺍﻳﺪﻩ ﺁﻝ ﺑﺎ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺩﺭ ﻟﻮﻟﻪ ﻱ ﻣﺪﻭﺭ ﻃﻮﻳﻞ ﺑﻪ ﺩﺳﺖ
ﺁﻭﺭﻳﺪ .ﺟﺮﻳﺎﻥ ﻫﻢ ﺩﻣﺎ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ .ﻓﺮﺽ ﻛﻨﻴﺪ ﺗﻐﻴﻴﺮ ﻓﺸﺎﺭ ﺩﺭ ﻟﻮﻟﻪ ﭼﻨﺪﺍﻥ ﺯﻳﺎﺩ ﻧﻴﺴﺖ؛ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﻣﻲ
ﺗﻮﺍﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺭﺍ ﺩﺭ ﻫﻤﻪ ﺟﺎ ﺛﺎﺑﺖ ﮔﺮﻓﺖ.
ﺣﻞ :ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﺗﻘﺮﻳﺒﻲ ،ﺑﺎ ﻓﺮﺽ ﺍﻳﻦ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻱ ﻫﻴﮕﻦ – ﭘﻮﺋﺎﺯﻭﻱ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ
ﻃﻮﻝ ﻛﻮﭼﻚ dzﺍﺯ ﻟﻮﻟﻪ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ ،ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ﺣﻞ ﻛﺮﺩ :
π (P0 − PL )R 4 ρ
πR 4 ρ dP
= ⇒ω
−
8µL
8µ dz
=ω
P0
P
ρ
ideal gas law ⇒ ρ = 0 P
P0
ρ0
πR 4 ρ 0
dP
− P
dz
8µ P0
PdP
PL
∫
P0
L
πR 4 ρ 0
πR 4 ρ 0
PdP ⇒ ∫ ωdz = −
0
8µ P0
8µ P0
)
=
ρ
= ⇒ω
⇒ ωdz = −
(
πR 4 ρ 0 2
2
2
2
= ⇒ω
) P0 − PL , P0 − PL = (P0 + PL )(P0 − PL
16µL P0
(P0 + PL ) → P
avg
2
π (P − P )R 4 ⋅ ρ avg
) πR 4 ρ 0 (P0 + PL
(P0 − PL ) = 0 L
⋅
8µL P0
2
8µL
٥۳
= ⇒ω
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
ﺭﺍﺑﻄﻪ ﺍﻱ. ﻳﻚ ﺳﻴﺎﻝ ﺑﻴﻨﮕﻬﺎﻡ ﺍﺯ ﻳﻚ ﻟﻮﻟﻪ ﻋﻤﻮﺩﻱ ﺗﺤﺖ ﮔﺮﺍﺩﻳﺎﻥ ﻓﺸﺎﺭ ﻭ ﻧﻴﺮﻭﻱ ﺟﺎﺫﺑﻪ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ:ﻣﺜﺎﻝ
1
. ﺩﺭ ﻟﻮﻟﻪ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ، ﺩﺑﻲ ﺣﺠﻤﻲ ﺳﻴﺎﻝ،Q ﺑﺮﺍﻱ
F13
τ rz = τ 0 − µ 0
dv z
dr
momentum flux distribution :
P0 − PL
r
2L
τ rz =
dv z P0 − PL
=
r
dr 2 L
P − PL 2 τ 0
r +
r + c2
⇒ v z = − 0
µ0
4µ0 L
⇒ τ 0 − µ0
B.C : at r = R ⇒ v z = 0
P − PL 2 τ 0
R +
R + c2
⇒ 0 = − 0
L
4
µ
µ
0
0
P − PL 2 τ 0
R −
R
⇒ c2 = 0
µ0
4µ0 L
⇒ vz =
(P0 − PL )R 2 1 − r 2 − τ 0 R 1 − r
4µ0 L
R
µ 0
R
r ≥ r0
r0 = radius of the plug - flow region
P0 − PL
r0
2L
τ0 =
: ﻗﺮﺍﺭ ﺩﻫﻴﻢ ﺩﺍﺭﻳﻢr = r0 ﺍﮔﺮ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ
۱
Bingham Flow In A Circular Tube
٥٤
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
vz =
(P0 − PL )R 2 1 − r0
4µ0 L
R
R
r0
0
0
R
Q = ∫ v z 2πrdr = 2π ∫ v z .rdr + 2π ∫ v z .rdr
r0
4
π (P0 − PL )R 4 4 τ 0 1 τ 0
1 − +
Q=
8µ 0 L
3 τ R 3 τ R
τR =
(P0 − PL )R
2L
τR τ0
↓ Buckingham - Reiner Equation
if τ 0 = 0 ⇒ Hagen - Poiseuille Equation
τ R τ 0 ﺑﺮﺍﻱ ﺣﺮﻛﺖ ﺳﻴﺎﻝ ﺑﺎﻳﺪ
[ ]→0
at τ R = τ 0 ⇒ Q = 0
٥٥
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
4.2ﺟﺮﻳﺎﻥ ﺩﺭ ﺣﻠﻘﻪ
۱
F14
ﻳﻚ ﺳﻴﺎﻝ ﻏﻴﺮ ﻗﺎﺑﻞ ﺗﺮﺍﻛﻢ ﺩﺭ ﺣﺎﻟﺖ ﭘﺎﻳﺪﺍﺭ ﺑﻴﻦ ﻓﻀﺎﻱ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻫﻢ ﻣﺤﻮﺭ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ ﻛﻪ ﺷﻌﺎﻉ ﺍﺳﺘﻮﺍﻧﻪ
ﺩﺍﺧﻠﻲ kRﻭ ﺷﻌﺎﻉ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ Rﺍﺳﺖ .ﭘﺮﻭﻓﻴﻞ ﺳﺮﻋﺖ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
ﺑﺎ ﮔﺮﻓﺘﻦ ﻳﻚ ﺍﻟﻤﺎﻥ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﻫﻤﺎﻧﻨﺪ ﻣﺜﺎﻝ ﻟﻮﻟﻪ ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﺯﻳﺮ ﻣﻲ ﺭﺳﻴﻢ:
d
(rτ rz ) = P0 − PL r , P = P + ρgz
dr
L
ﺩﺭ ﺍﻳﻨﺠﺎ zﻭ hﻫﻢ ﺟﻬﺖ ﻫﺴﺘﻨﺪ .ﻓﺸﺎﺭ ﻭ ﻧﻴﺮﻭﻱ ﺟﺎﺫﺑﻪ ﺑﺮ ﺧﻼﻑ ﻫﻢ ﻋﻤﻞ ﻣﻲ ﻛﻨﻨﺪ.
c
r + 1
r
P − PL
⇒ τ rz = 0
2L
ﻣﻘﺪﺍﺭ τ rzﺭﺍ ﺩﺭ r = kRﻭ r = Rﻧﻤﻲ ﺩﺍﻧﻴﻢ ﻭﻟﻲ ﻣﻲ ﺩﺍﻧﻴﻢ ﻛﻪ ﺳﺮﻋﺖ ﺩﺭ ﻳﻚ r = λRﻣﻘﺪﺍﺭ ﺁﻥ ﻣﺎﻛﺰﻳﻤﻢ ﻭ
τ rz = 0ﺍﺳﺖ.
Flow Through An Annulus
٥٦
۱
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
at r = λR ⇒ τ rz = 0
(P − PL ) λR 2
c
P − PL
⇒0= 0
λR + 1 ⇒ c1 = − 0
λR
2L
2L
(
⇒ τ rz =
)
(P0 − PL )R r − λ2 R
R
2L
r
. ﺭﺍ ﻣﻲ ﺩﺍﻧﻴﻢλ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻣﺎ ﻣﻔﻬﻮﻡ ﻓﻴﺰﻳﻜﻲλ ﺑﺎc1 ﻋﻠﺖ ﻋﻮﺽ ﻛﺮﺩﻥ. ﻣﺠﻬﻮﻝ ﺍﺳﺖλ ﻛﻪ
τ rz = − µ
⇒
dv z
dr
(P − PL )R r − λ2 R
dv z
=− 0
dr
2 µL R
r
⇒ vz
2
(
P0 − PL )R 2 r
r
=−
− 2λ2 Ln
+c
R
4 µL
R
B.C.1 : ⇒ at r = kR
vz = 0
B.C.2 : ⇒ at r = R
vz = 0
2
(P0 − PL )R 2 2
R − 2λ2 Lnk + c2 = 0
−
4 µL
2
− (P0 − PL )R (1 + c ) = 0
2
4 µL
(
2λ2 =
c2 = −1 ,
⇒ τ rz =
vz =
)
1− k 2
Ln(1 k )
(P0 − PL )R r −
1 − k R
R 2 Ln(1 k ) r
2L
2
(P0 − PL )R 2 1 − r 2 +
4 µL
R
1 − k 2 r
Ln(1 k ) Ln R
k → 0 ⇒ flow is in circular tubes.
: ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ.1
v z ,max = v z
r =λR
=
(P0 − PL )R 2 1 −
4 µL
1 − k 2
1 − k 2
−
1
Ln
2 Ln(1 k )
2 Ln(1 k )
٥۷
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
.2ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺍﺯ ﺭﺍﺑﻄﻪ ﻱ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
1− k 4 1− k 2
−
2
Ln(1 k )
1− k
R
2
2π
∫ ∫ v rdrdθ = (P − P )R
= π
8µL
∫ ∫ rdrdθ
L
0
z
kR
R
2
kR
0
vz
0
.3ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
π (P0 − PL )R 4 ρ
(1 − k 2 )
(
1 − k 4 )−
Ln(1 k )
8µL
2
= ω = πR 2 (1 − k 2 )ρ v z
.4ﻧﻴﺮﻭﻳﻲ ﻛﻪ ﺳﻴﺎﻝ ﺑﺮ ﺳﻄﻮﺡ ﺟﺎﻣﺪ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ ﺍﺯ ﺟﻤﻊ ﺯﺩﻥ ﻧﻴﺮﻭﻫﺎﻱ ﻭﺍﺭﺩ ﺑﺮ ﺍﺳﺘﻮﺍﻧﻪ ﻫﺎﻱ ﺩﺍﺧﻠﻲ ﻭ
ﺧﺎﺭﺟﻲ ،ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ،ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
) − ΡL
)(2πRL ) ≡ πR (1 − κ )(Ρ
2
2
0
rz r = R
)(2πkRL) + (τ
r = kR
Fz = (− τ rz
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺩﺭ ﺑﺎﻻ ﻓﻘﻂ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﻣﻌﺘﺒﺮﻧﺪ .ﮔﺬﺍﺭ ﺁﺭﺍﻡ – ﻣﺘﻼﻃﻢ ﺩﺭ ﻫﻤﺴﺎﻳﮕﻲ
Re = 2000ﺭﻭﻱ ﻣﻲ ﺩﻫﺪ ﻭ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ :
2 R(1 − k ) v z ρ
µ
5.2ﺟﺮﻳﺎﻥ ﺩﻭ ﺳﻴﺎﻝ ﺍﻣﺘﺰﺍﺝ ﻧﺎﭘﺬﻳﺮ ﻣﺠﺎﻭﺭ ﻫﻢ
= Re
1
F15
ﺗﺎ ﺍﻳﻨﺠﺎ ﻭﺿﻌﻴﺖ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﺑﺎ ﻣﺮﺯﻫﺎﻱ ﺟﺎﻣﺪ-ﺳﻴﺎﻝ ﻭ ﻣﺎﻳﻊ-ﮔﺎﺯ ﺭﺍ ﺑﺮﺭﺳﻲ ﻛﺮﺩﻳﻢ .ﺍﻛﻨﻮﻥ ﻧﻤﻮﻧﻪ ﺍﻱ ﺍﺯ ﻳﻚ
ﻣﺴﺌﻠﻪ ﻱ ﺟﺮﻳﺎﻥ ﺑﺎ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﻣﺎﻳﻊ – ﻣﺎﻳﻊ ﺭﺍ ﻣﻄﺮﺡ ﻣﻲ ﻛﻨﻴﻢ .
Flow Of Two Adjacent Immiscible Fluids
٥۸
۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺗﺎﺛﻴﺮ ﺩﻭ ﻣﺎﻳﻊ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﺍﻣﺘﺰﺍﺝ ﻧﺎﭘﺬﻳﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ Zﺩﺭ ﻳﻚ ﺷﻜﺎﻑ ﺍﻓﻘﻲ ﻧﺎﺯﻙ ﺑﻪ ﻃﻮﻝ Lﻭ ﻋﺮﺽ ، Wﺗﺤﺖ
ﺗﺎﺛﻴﺮ ﮔﺮﺍﺩﻳﺎﻥ ﻓﺸﺎﺭ ﺍﻓﻘﻲ ( p0 − pl ) Lﺟﺮﻳﺎﻥ ﺩﺍﺭﻧﺪ .ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﻃﻮﺭﻱ ﺗﻨﻈﻴﻢ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﻧﻴﻤﻲ ﺍﺯ
ﺷﻜﺎﻑ ﺑﺎ ﺳﻴﺎﻝ ) 1ﻛﻪ ﻓﺎﺯ ﭼﮕﺎﻝ ﺗﺮ ﺍﺳﺖ( ﻭ ﻧﻴﻢ ﺩﻳﮕﺮ ﺁﻥ ﺑﺎ ﺳﻴﺎﻝ ) 2ﻛﻪ ﻓﺎﺯ ﻛﻢ ﭼﮕﺎﻝ ﺗﺮ ﺍﺳﺖ( ﭘﺮ ﺷﻮﺩ.
ﺟﺮﻳﺎﻥ ﺍﻳﻦ ﺳﻴﺎﻝ ﻫﺎ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﺍﻱ ﺁﻫﺴﺘﻪ ﺍﺳﺖ ﻛﻪ ﻫﻴﭻ ﻧﺎﭘﺎﻳﺪﺍﺭﻱ ﺭﺥ ﻧﺨﻮﺍﻫﺪ ﺩﺍﺩ؛ ﻳﻌﻨﻲ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﺩﻗﻴﻘﺎً
ﺻﻔﺤﻪ ﺍﻱ ﻣﻲ ﻣﺎﻧﺪ .ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺗﻮﺯﻳﻊ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﻢ.
ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ :
=0
x + ∆x
+ LWφ xz x − LWφ xz
z=L
−φ zz
L
z =0
z=L
− W∆xφ zz
z =0
−φ xz x φ zz
=
∆x
x + ∆x
∆x→0
z=L
−φ zz
L
٥۹
z =0
W∆xφ zz
φ xz
lim
∂φ xz φ zz
=
∂x
⇒
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
φ xz = τ xz + ρv x v z = τ xz
ρv x v z → 0
φ zz = P + τ zz + ρv z v z = P − 2µ
∂v z
+ ρv z v z
∂z
∂v z
→0
∂z
∂τ xz (P + ρv z v z )0 − (P + ρv z v z )L
=
∂x
L
(ρvz vz )0 =(ρvz vz )L
⇒
⇒
dτ xz P0 − PL
=
dx
L
: ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺑﺮﺍﻱ ﺩﻭ ﻧﺎﺣﻴﻪ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ. ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ2 ﻭ1 ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺑﺮﺍﻱ ﻫﺮ ﺩﻭ ﻓﺎﺯ
P − PL
⇒ τ xz = 0
x + C1
L
P0 − PL
I
x + C1
L
P − PL
II
= 0
x + C1
L
⇒
τ xz I =
⇒
τ xz II
ﺩﺭ ﺳﺮﺍﺳﺮ ﻓﺼﻞ ﻣﺸﺘﺮﻙτ xz ﻣﺜﻼً ﭘﻴﻮﺳﺘﮕﻲ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ،ﺑﻼﻓﺎﺻﻠﻪ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﻳﻜﻲ ﺍﺯ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ
: ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ،ﺳﻴﺎﻝ – ﺳﻴﺎﻝ
B.C.1: at x = 0 τ xz =τ xz
I
II
⇒ C1 = C1 = C1
I
⇒
⇒
II
τ xz I = C1I
τ xz II = C1II
− µI
dv z
P − PL
= 0
x + C1
dx L
− µ II
dv z
P − PL
= 0
x + C1
dx
L
I
vz = −
(P0 − PL )x 2 − C1
vz = −
(P0 − PL )x 2 −
I
⇒
II
II
2 µ I .L
2 µ .L
II
µI
C1
µ
II
x + C2
I
x + C2
II
٦۰
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
.ﻣﺠﻬﻮﻝ ﻫﺴﺘﻨﺪ ﻛﻪ ﺑﻪ ﺳﻪ ﺷﺮﻁ ﻣﺮﺯﻱ ﻧﻴﺎﺯ ﺩﺍﺭﻳﻢ
B.C.2 :
at x = 0
vz = vz
B.C.3 :
at x = −b
vz = 0
B.C.4 :
at x = +b
vz = 0
B.C.2 ⇒ C2 = C2
I
I
II
ﻭ
C2
I
ﻭC1 ﺳﻪ ﺛﺎﺑﺖ
II
I
II
II
B.C.3 ⇒ -
(P0 − PL )b 2 + C1b + C I = 0
B.C.4 ⇒ -
(P0 − PL )b 2 + C1b + C II = 0
⇒ C1 = −
C2
2 µ I .L
2 µ II .L
µI
2
µ II
2
(P0 − PL )b . µ I − µ II
µ I + µ II
2
I
(P − P )b 2µ = C II
I
C2 = + 0 I L
2
2 µ .L µ I + µ II
2L
:ﻟﺬﺍ ﻓﻼﻛﺲ ﻣﻤﻨﺘﻮﻡ ﻭ ﭘﺮﻭﻓﻴﻞ ﺳﺮﻋﺖ ﺑﻪ ﻓﺮﻡ ﺫﻳﻞ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ
τ xz =
vx
I
vz
II
(P0 − PL )b x − 1 µ I − µ II
b
L
2 µ I + µ II
(
P0 − PL )b 2
=
2
µ I − µ II x x
+ I
−
II
µ + µ b b
(P − PL )b 2
= 0
µ I − µ II x x
+ I
−
II
µ + µ b b
2 µ I .L
2 µ .L
II
2µ I
I
II
µ + µ
2 µ II
I
II
µ + µ
2
ﺁﻧﮕﺎﻩ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺳﻬﻤﻮﻱ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﻭ ﺍﻳﻦ ﻫﻤﺎﻥ ﺗﻮﺯﻳﻌﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﻣﻮﺭﺩ،ﺍﮔﺮ ﻫﺮ ﺩﻭ ﺳﺮﻋﺖ ﺑﺮﺍﺑﺮ ﺑﺎﺷﻨﺪ
ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺩﺭ ﻫﺮ ﻻﻳﻪ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺩﺳﺖ.ﺳﻴﺎﻝ ﺧﺎﻟﺺ ﺟﺎﺭﻱ ﺑﻴﻦ ﺻﻔﺤﻪ ﻫﺎﻱ ﻣﻮﺍﺯﻱ ﺍﻧﺘﻈﺎﺭ ﻣﻲ ﺭﻭﺩ
:ﺁﻭﺭﺩ ﻭ ﻧﺘﺎﻳﺞ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ
v z =
(P0 − PL )b 2
1 0 I
=
v
dx
z
b ∫−b
12 µ I .L
v z =
(P0 − PL )b 2
1 b II
=
v
dx
z
b ∫0
12 µ II .L
I
II
7 µ I + µ II
I
II
µ +µ
µ I + 7 µ II
I
II
µ +µ
٦۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺛﺎﺑﺖ ﻛﻨﻴﺪ:
x 1 µ I − µ II
=
: plane of zero shear stress
b 2 µ I + µ II
ﺁﻳﺎ ﺍﻣﻜﺎﻥ ﺩﺍﺭﺩ؟
6.2ﺟﺮﻳﺎﻥ ﺧﺰﺷﻲ ﺣﻮﻝ ﻳﻚ ﻛﺮﻩ
1
F16
ﺩﺭ ﺑﺨﺶ ﻫﺎﻱ ﮔﺬﺷﺘﻪ ﺟﻨﺪﻳﻦ ﻣﺴﺌﻠﻪ ﻱ ﻣﻘﺪﻣﺎﺗﻲ ﺟﺮﻳﺎﻥ ﻭﻳﺴﻜﻮﺯ ﺭﺍ ﺣﻞ ﻛﺮﺩﻳﻢ .ﺩﺭ ﻫﻤﻪ ﻱ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﻫﺎ ﺑﺎ
ﺟﺮﻳﺎﻥ ﺭﺍﺳﺖ ﺧﻄﻲ ﺳﺮﻭ ﻛﺎﺭ ﺩﺍﺷﺘﻴﻢ ﻛﻪ ﻓﻘﻂ ﻳﻚ ﻣﺆﻟﻔﻪ ﺳﺮﻋﺖ ﻏﻴﺮ ﺻﻔﺮ ﺩﺍﺷﺘﻨﺪ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺍﻃﺮﺍﻑ ﻳﻚ
ﻛﺮﻩ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﻣﻨﺤﻨﻲ ﺷﻜﻞ ﻫﺴﺘﻨﺪ ﻭ ﻣﺎ ﺩﻭ ﻣﺆﻟﻔﻪ ﺳﺮﻋﺖ ﻏﻴﺮ ﺻﻔﺮ ﺩﺍﺭﻳﻢ ) (vr , vθﻟﺬﺍ ﺍﻳﻦ ﻣﺴﺄﻟﻪ ﺑﺎ
ﺗﻜﻨﻴﻚ ﻫﺎﻱ ﺍﻳﻦ ﺑﺨﺶ ﻗﺎﺑﻞ ﺣﻞ ﻧﻴﺴﺖ .ﺍﻳﻦ ﻣﺴﺄﻟﻪ ﺑﻌﺪﺍً ﺣﻞ ﻣﻲ ﺷﻮﺩ ﻭ ﺩﺭ ﺍﻳﻨﺠﺎ ﻣﺎ ﻓﻘﻂ ﺍﺯ ﻧﺘﺎﻳﺞ ﺣﻞ ﺍﺳﺘﻔﺎﺩﻩ
ﻣﻲ ﻛﻨﻴﻢ.ﻣﺴﺌﻠﻪ ﺍﻱ ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺟﺎ ﺑﺮﺭﺳﻲ ﻣﻲ ﻛﻨﻴﻢ ﺑﺎ "ﺟﺮﻳﺎﻥ ﺧﺰﺷﻲ" ﺳﺮﻭﻛﺎﺭ ﺩﺍﺭﺩ ﻛﻪ ﺟﺮﻳﺎﻧﻲ ﺑﺴﻴﺎﺭ ﺁﻫﺴﺘﻪ
ﺍﺳﺖ .ﺍﻳﻦ ﻧﻮﻉ ﺟﺮﻳﺎﻥ ﺭﺍ "ﺟﺮﻳﺎﻥ ﺍﺳﺘﻮﻛﺲ" ﻧﻴﺰ ﻣﻲ ﻧﺎﻣﻨﺪ.
ﺳﻴﺎﻝ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮﻱ ﺭﺍ ﺩﺭ ﺍﻃﺮﺍﻑ ﻛﺮﻩ ﺍﻱ ﺑﻪ ﺷﻌﺎﻉ Rﻭ ﻗﻄﺮ Dﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ .ﺍﻳﻦ ﺳﻴﺎﻝ ،ﺑﺎ
ﭼﮕﺎﻟﻲ ρﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ، µﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩﻱ zﻭ ﺑﻪ ﻃﺮﻑ ﺑﺎﻻ ،ﺑﺎ ﺳﺮﻋﺖ ﻳﻜﻨﻮﺍﺧﺖ ∞ vﺑﻪ ﻛﺮﻩ ﻱ ﺛﺎﺑﺖ
ρv∞ D
ﻧﺰﺩﻳﻚ ﻣﻲ ﺷﻮﺩ ﺩﺭ ﺍﻳﻦ ﻣﺴﺌﻠﻪ "ﺟﺮﻳﺎﻥ ﺧﺰﺷﻲ " ﺑﻪ ﻣﻌﻨﺎﻱ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻋﺪﺩ
µ
= Reﻛﻤﺘﺮ ﺍﺯ 0.1
ﺍﺳﺖ .ﻣﺸﺨﺼﻪ ﻱ ﺍﻳﻦ ﺭژﻳﻢ ﺟﺮﻳﺎﻥ ﻧﺒﻮﺩ ﺟﺮﻳﺎﻥ ﮔﺮﺩﺍﺑﻲ ﺩﺭ ﭘﺎﻳﻴﻦ ﺩﺳﺖ ﻛﺮﻩ ﺍﺳﺖ.
Creeping Flow Around A Sphere
٦۲
۱
ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ۲ ﻓﺼﻞ
(ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻭ ﻓﺸﺎﺭ ﺑﻪ ﻓﺮﻡ ﺫﻳﻞ ﺍﺳﺖ )ﺑﻌﺪﺍً ﺍﺛﺒﺎﺕ ﻣﻲ ﺷﻮﺩ
3 R 1 R 3
vr = v∞ 1 − + cos θ
2 r 2 r
3
3R 1R
vθ = v∞ − 1 + + sin θ
4 r 4 r
vφ = 0
3 µv∞ R
P = P0 − ρgz −
cos θ
2 R r
2
vr = 0
at r = R ⇒
vθ = 0
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﻌﺎﺩﻻﺕ ﺑﺎﻻ
at r → ∞ ⇒ vr = v∞ cos θ , vθ = −v∞ sin θ
v z = vr cos θ − vθ sin θ = v∞ cos 2 θ + v∞ sin 2 θ
↓
Eq. A.6 − 33
(
)
= v∞ sin 2 θ + cos 2 θ = v∞
. ﺍﺳﺖv∞ ﺳﺮﻋﺖz ﻳﻌﻨﻲ ﺩﺭﺩﻭﺭﺩﺳﺖ ﺩﺭ ﺟﻬﺖ
at r → ∞ ⇒ v x = v y = 0
Eq. A.6 − 31 : v x = vr sin θ cos φ + vθ . cos θ cos φ − sin φvφ
⇒ v x = v∞ cos θ sin θ cos φ − v∞ sin θ cos θ cos φ = 0
v y = (sin θ sin φ )vr + (cos θ sin φ )vθ + cos φvφ = 0
Eq. A.6.32↵
٦۳
vφ → 0
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺆﻟﻔﻪ ﻫﺎﻱ τﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺮﻭﻱ:
) (
∂vr 2
+ µ − k ∇.V
∂r 3
τ rr = − µ 2
) (
1 ∂vθ vr 2
+ + µ − k ∇.V
r 3
r ∂θ
τ θθ = − µ 2
) (
1 ∂vφ vr + vθ cot θ 2
+ µ − k ∇.V
+
r
3
r sin θ ∂φ
τ φφ = − µ 2
∂ vθ 1 ∂vr
+
∂r r r ∂r
τ rθ = τ θr = − µ r
for incompresssible flow
∇V
. =0
1 ∂vθ vr
∂vr
+
, τ θθ = − µ 2
∂r
r
r ∂θ
v + v cot θ
τ φφ = −2µ . r θ
r
3 1
∂vr
= v∞ − R. − 2 + R 3 − 3r −4 cos θ
∂r
2 r 2
⇒ τ rr = −2 µ
(
)
2
4
3µv∞ R R
= ⇒ τ rr
− + cos θ
R r r
τ rr = −2τ θθ = −2τ φφ
3 µv∞ R
= = τ θr
sin θ
2 R r
4
τ rθ
ﺗﻨﺶ ﻫﺎﻱ ﻧﺮﻣﺎﻝ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﺴﺄﻟﻪ ﺑﻪ ﺟﺰ ﺩﺭ r = Rﻏﻴﺮ ﺻﻔﺮ ﻫﺴﺘﻨﺪ.
ﺣﺎﻝ ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﻧﻴﺮﻭﻳﻲ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﻢ ﻛﻪ ﺳﻴﺎﻝ ﺟﺎﺭﻱ ﺭﻭﻱ ﻛﺮﻩ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ .ﺑﻪ ﻋﻠﺖ ﺗﻘﺎﺭﻥ ﺣﻮﻝ ﻣﺤﻮﺭ
zﻧﻴﺮﻭ ﺑﺮﺍﻳﻨﺪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺧﻮﺍﻫﺪ ﺑﻮﺩ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﻧﻴﺮﻭﻱ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻣﺆﻟﻔﻪ ﻫﺎﻱ z
ﻧﻴﺮﻭﻫﺎﻱ ﻗﺎﺋﻢ ﻭ ﻣﻤﺎﺳﻲ ﺭﻭﻱ ﺳﻄﺢ ﻛﺮﻩ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ.
٦٤
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻧﻴﺮﻭﻱ ﻗﺎﺋﻢ
ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﻭﺍﻗﻊ ﺑﺮ ﺳﻄﺢ ﻛﺮﻩ ،ﺳﻴﺎﻝ ﻧﻴﺮﻭﻳﻲ ﺑﺮ ﻭﺍﺣﺪ ﺳﻄﺢ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ ﻛﻪ ﺑﺮ ﺳﻄﺢ ﻋﻤﻮﺩ ﺍﺳﺖ ﻭ ﻣﻘﺪﺍﺭ ﺁﻥ
ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ . − (P + τ rr )r =R :ﭼﻮﻥ ﺳﻴﺎﻝ ﺩﺭ ﻧﺎﺣﻴﻪ ﻱ rﺑﻴﺸﺘﺮ ﺍﺳﺖ ﻭ ﻛﺮﻩ ﺩﺭ ﻧﺎﺣﻴﻪ rﻛﻤﺘﺮ ،ﺑﺎﻳﺪ ﺑﺮ ﺍﺳﺎﺱ
ﻗﺮﺍﺭﺩﺍﺩ ﻋﻼﻣﺖ ﻣﻨﻬﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﻢ.
1
ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ
∂vr
∂r
τ rr = −2 µ
∂vr
>0
∂r
ﺩﺭ ﻧﺘﻴﺠﻪ 2µ ∂vrﻣﺜﺒﺖ ﺍﺳﺖ ﭘﺲ τ rrﺩﺭ ﻣﻌﺎﺩﻟﻪ 1ﺩﺭ ﻳﻚ ﻣﻨﻔﻲ ﺿﺮﺏ ﻣﻲ ﺷﻮﺩ .ﻭ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ
∂r
σ rr = − P + τ rrﺍﺳﺖ ﻛﻪ
− (P + τ rr ) r = R cosθ
∂v
τ rr = 2 µ r
∂r
ﻟﺬﺍ ﻣﺆﻟﻔﻪ zﺍﻳﻦ ﻧﻴﺮﻭﻱ ﺭﻭﻱ ﺳﻄﺢ ﻛﺮﻩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
٦٥
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﻧﻴﺮﻭﻱ ﻋﻤﻮﺩﻱ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﻢ ﻣﺆﻟﻔﻪ ﻓﺸﺎﺭ ﻗﺒﻠﻲ ﺭﺍ ﺩﺭ ﺳﻄﺢ ﺍﻟﻤﺎﻥ ﺿﺮﺏ ﻭ ﺭﻭﻱ ﺳﻄﺢ ﻛﺮﻩ ﺍﻧﺘﮕﺮﺍﻝ
ﻣﻲ ﮔﻴﺮﻳﻢ.
0 ≤θ ≤π
0 ≤ φ ≤ 2π
θ :0 →π
φ : 0 → 2π
cos θ .R 2 sin θdθdφ
r =R
∞≤ 0≤r
π
) ∫ − (P + τ
rr
2π
0
∫ = Fn
0
ﺑﺮﺍﻱ ﻫﺮ ﻧﻮﻉ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ ﻭ ﻏﻴﺮ ﻗﺎﺑﻞ ﺗﺮﺍﻛﻢ )ﻣﻌﺎﺩﻟﻪ ﻗﺒﻞ( ﺗﻨﺶ ﻫﺎﻱ ﻧﺮﻣﺎﻝ ﺩﺭ ﻣﺮﺯﻫﺎﻱ ﺟﺎﻣﺪ
ﺳﻴﺎﻝ ﺻﻔﺮ ﻫﺴﺘﻨﺪ) .ﺍﺛﺒﺎﺕ ﻛﻨﻴﺪ(
=0
r =R
τ rr
∞3 µv
cos θ
2 R
2π π
∞3 µv
⇒ Fn = ∫ ∫ − P0 + ρgR cos θ +
cos θ cos θ .R 2 sin θdθdφ
0
0
2 R
4
∞⇒ Fn = πR 3 ρg + 2πµRv
3
P r = R = P0 − ρgR cos θ −
↓
↓
form drag
buoyant force
The force in the direction of flow exerted by the fluid on the solid is called drag.
There is no net drag in potential flow
٦٦
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻧﻴﺮﻭﻱ ﻣﻤﺎﺳﻲ
ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺭﻭﻱ ﺳﻄﺢ ﺟﺎﻣﺪ ﻳﻚ ﺗﻨﺶ ﺑﺮﺷﻲ ﻧﻴﺰ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻛﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﻤﺎﺳﻲ ﻋﻤﻞ ﻣﻲ ﻛﻨﺪ .ﻧﻴﺮﻭﻳﻲ ﻛﻪ
ﺳﻴﺎﻝ )ﻧﺎﺣﻴﻪ ﻱ zﺑﻴﺸﺘﺮ( ﺩﺭ ﺍﻣﺘﺪﺍﺩ − θﺑﺮ ﻭﺍﺣﺪ ﺳﻄﺢ ﺟﺎﻣﺪ )ﻧﺎﺣﻴﻪ ﻱ rﻛﻤﺘﺮ( ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﺪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
. (+ τ rθ ) r = Rﻣﺆﻟﻔﻪ ﻱ zﺍﻳﻦ ﻧﻴﺮﻭ ﺏ ﻭﺍﺣﺪ ﺳﻄﺢ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ (+ τ rθ ) r = R sin θ :ﺣﺎﻝ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﺭﺍ ﺩﺭ ﺟﺰء
ﺳﻄﺢ R 2 sin θdθdφﺿﺮﺏ ﻣﻲ ﻛﻨﻴﻢ ﻭ ﺭﻭﻱ ﻛﻞ ﺳﻄﺢ ﻛﺮﻩ ﺍﻧﺘﮕﺮﺍﻝ ﻣﻲ ﮔﻴﺮﻳﻢ.
٦۷
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﺩﺭ ﻧﺘﻴﺠﻪ ﻧﻴﺮﻭﻱ ﺑﺮﺍﻳﻨﺪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
sin θ )R 2 sin θdθdφ
r =R
π
∫ (τ θ
r
2π
0
∫ = Ft
0
∞3 µv
sin θ
2 R
2π π 3 µv
∞
⇒ Ft = ∫ ∫
sin θ . sin θ R 2 sin θdθdφ
0
0
2 R
⇒ Ft = 4πµRv∞ = friction drag = wall drag
=
r =R
τ rθ
4
∞F = Fn + Ft = πR 3 ρg + 2πµRv∞ + 4πµRv
3
↓
↓
↓
friction
form
drag
buogant
force
drag
4
∞⇒ F = πR 3 ρg + 6πµRv
3
4
∞⇒ F = Fb + FK = πR 3 ρg + 6πµRv
3
↓
↓
kinetic
force
buogant
force
ﺟﻤﻠﻪ ﻱ ﺍﻭﻝ ﻧﻴﺮﻭﻱ ﺷﻨﺎﻭﺭﻱ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺳﻴﺎﻝ ﺩﺭ ﺣﺎﻝ ﺳﻜﻮﻥ ﺣﻀﻮﺭ ﺩﺍﺭﺩ؛ ﺍﻳﻦ ﻧﻴﺮﻭ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﺟﺮﻡ
ﺳﻴﺎﻝ ﺟﺎﺑﻪ ﺟﺎ ﺷﺪﻩ ،ﺿﺮﺏ ﺩﺭ ﺷﺘﺎﺏ ﮔﺮﺍﻧﺸﻲ .ﺟﻤﻠﻪ ﻱ ﺩﻭﻡ ،ﻧﻴﺮﻭﻱ ﺟﻨﺒﺸﻲ ،ﺍﺯ ﺣﺮﻛﺖ ﺳﻴﺎﻝ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ
.ﺭﺍﺑﻄﻪ ﻱ ﺯﻳﺮ ﻗﺎﻧﻮﻥ ﺍﺳﺘﻮﻛﺲ ﻧﺎﻣﻴﺪﻩ ﻣﻲ ﺷﻮﺩ :
FK = 6πµRv∞ Re < 0.1
ﻣﺜﺎﻝ :ﺭﺍﺑﻄﻪ ﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ ﻛﻪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻥ ﺑﻪ ﻭﺳﻴﻠﻪ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺳﺮﻋﺖ ﺣﺪﻱ vt 1ﻳﻚ ﻛﺮﻩ ﻛﻮﭼﻚ
ﺑﻪ ﺷﻌﺎﻉ Rﺩﺭ ﻳﻚ ﺳﻴﺎﻝ ،ﺑﺘﻮﺍﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻴﺎﻝ ﺭﺍ ﭘﻴﺪﺍ ﻛﺮﺩ.
F17
ﺩﺭ ﺳﺮﻋﺖ ﺣﺪ ﺩﺍﺭﻳﻢ :
∑ F = ma = 0
mg − Fb − FD = 0
← ρ sﺩﺍﻧﺴﻴﺘﻪ ﺟﺎﻣﺪ
← ρﺩﺍﻧﺴﻴﺘﻪ ﺳﻴﺎﻝ
Terminal Velocity
٦۸
۱
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
⇒ mg = Fb + FD
4 3
4
πR ρ s g = πR 3 ρg + 6πµRvt
3
3
≤ 0.1
Dvt ρ
µ
٦۹
= Re
for
2 2 (ρ s − ρ )g
R
q
vt
=⇒µ
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
1.2ﺟﺮﻳﺎﻥ ﺭﺍ ﺩﺭ ﺷﻜﺎﻑ ﺑﺎﺭﻳﻚ )ﺷﻜﻞ1.2ﺭﺍ ﺑﺒﻴﻨﻴﺪ(
)ﺍﻟﻒ( ﺳﻴﺎﻟﻲ ﻧﻴﻮﺗﻮﻧﻲ ﺩﺭ ﺷﻜﺎﻑ ﺑﺎﺭﻳﻚ ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺗﻮﺳﻂ ﺩﻭ ﺟﺪﺍﺭ ﻣﻮﺍﺯﻱ ﺑﻪ ﻓﺎﺻﻠﻪ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ
ﻣﻲ ﺩﺍﻧﻴﻢ ﻛﻪ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﺍﺛﺎﺭ ﻟﺒﻪ ﺍﻫﻤﻴﺖ ﻧﺪﺍﺭﻧﺪ .ﻣﻮﺍﺯﻧﻪ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﺯﻳﺮ
ﺭﺍ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭ ﺳﺮﻋﺖ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ:
P0 − PL
x
L
τ xz =
( p0 − p L )B 2 1
2
x
−
B
2 µL
= υz
)ﺏ( ﻧﺴﺒﺖ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺑﻪ ﺳﺮﻋﺖ ﻣﺎﻛﺰﻳﻤﻢ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﭼﻪ ﻗﺪﺭ ﺍﺳﺖ؟
)ﺝ( ﻫﻤﺘﺎﻱ ﺷﻜﺎﻓﻲ ﻣﻌﺎﺩﻟﻪ ﻫﻴﮕﻦ-ﭘﻮﺋﺎﺯﻭﻱ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ
)ﺩ( ﻧﻤﻮﺩﺍﺭﻱ ﺑﺎ ﻣﻌﻨﺎ ﺑﺮﺍﻱ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﺍﻳﻦ ﻧﻜﺘﻪ ﺗﺮﺳﻴﻢ ﻛﻨﻴﺪ ﻛﻪ ﻫﺮﮔﺎﻩ ، B = Wﺗﺤﻠﻴﻞ ﺑﺎﻻ ﻗﺎﺑﻞ ﻛﺎﺭﺑﺮﺩ ﻧﻴﺴﺖ.
)ﻩ( ﭼﮕﻮﻧﻪ ﻣﻲ ﺗﻮﺍﻥ ﻧﺘﻴﺠﻪ ﻗﺴﻤﺖ )ﺏ( ﺭﺍ ﺍﺯ ﻧﺘﺎﻳﺞ ﺑﺨﺶ 5.2ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﺩ؟
2.2ﺟﺮﻳﺎﻥ ﻓﻴﻠﻢ ﺭﻭﻱ ﺳﻄﺢ ﺧﺎﺭﺟﻲ ﻟﻮﻟﻪ ﻣﺪﻭﺭ)ﺷﻜﻞ 2.2ﺭﺍ ﺑﺒﻴﻨﻴﺪ( .ﺩﺭ ﺍﺯﻣﺎﻳﺶ ﺟﺬﺏ ﮔﺎﺯ ،ﺳﻴﺎﻟﻲ
ﻭﻳﺴﻜﻮﺯ ﺩﺭ ﻟﻮﻟﻪ ﻱ ﻣﺪﻭﺭ ﻛﻮﭼﻚ ﺑﺎﻻ ﻣﻲ ﺭﻭﺩ ﻭ ﺳﭙﺲ ﺍﺯ ﺭﻭﻱ ﺳﻄﺢ ﺧﺎﺭﺟﻲ ﺁﻥ ﭘﺎﻳﻴﻦ ﻣﻲ ﺍﻳﺪ .ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ
ﺣﺮﻛﺖ ﺭﺍ ﺭﻭﻱ ﭘﻮﺳﺘﻪ ﺍﻱ ﺩﺭ ﻓﻴﻠﻢ ،ﺑﻪ ﺿﺨﺎﻣﺖ ∆rﻣﻄﺎﺑﻖ ﺷﻜﻞ 2.2ﺑﻨﻮﻳﺴﻴﺪ .ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻛﻪ ﭘﻴﻜﺎﻥ ﻫﺎﻱ
"ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭﺭﻭﺩﻱ" ﻭ" ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺧﺮﻭﺟﻲ" ﻫﻤﻴﺸﻪ ﺩﺭ ﺟﻬﺖ ﻣﺜﺒﺖ ﻣﺤﻮﺭ ﻣﺨﺘﺼﺎﺕ ﺍﻧﺪ ،ﻫﺮ ﭼﻨﺪ ﺩﺭ
ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺳﻄﻮﺡ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺩﺭ ﺟﻬﺖ ﻣﻨﻔﻲ rﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ.
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺴﺌﻠﻪ ﻫﺎ
)ﺍﻟﻒ( ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ) ﺑﺎ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺁﺛﺎﺭ ﺍﻧﺘﻬﺎﻳﻲ( ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
2
ρgR 2 r
r
2
= υz
1 − + 2a ln
4 µ R
R
)ﺏ( ﻋﺒﺎﺭﺗﻲ ﺑﺮﺍﻱ ﺑﻴﺎﻥ ﺍﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ﺩﺭ ﻓﻴﻠﻢ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ.
)ﺝ(ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻫﺮﮔﺎﻩ ﺿﺨﺎﻣﺖ ﻓﻴﻠﻢ ﺧﻴﻠﻲ ﻛﻢ ﺑﺎﺷﺪ ،ﻧﺘﻴﺠﻪ ﻗﺴﻤﺖ )ﺏ( ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﻣﺮﺑﻮﻁ ﺑﻪ ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ
ﺟﺮﻣﻲ ﺩﺭ ﺑﺨﺶ 2.2ﺳﺎﺩﻩ ﻣﻲ ﺷﻮﺩ.
3.2ﺟﺮﻳﺎﻥ ﺣﻠﻘﻪ ﺍﻱ ﺩﺭ ﺣﺎﻟﺘﻲ ﻛﻪ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﺩﺍﺧﻠﻲ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﺤﻮﺭﻱ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ)ﺷﻜﻞ 3.2ﺭﺍ
ﺑﺒﻴﻨﻴﺪ( .ﻣﻴﻠﻪ ﺍﻱ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺑﻪ ﺷﻌﺎﻉ kRﺑﺎ ﺳﺮﻋﺖ υ z = υ 0ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺣﻔﺮﻩ ﺍﻱ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺑﻪ ﺷﻌﺎﻉ Rﻣﻄﺎﺑﻖ
ﺷﻜﻞ ﺩﺭ ﺣﺮﻛﺖ ﺍﺳﺖ .ﻓﺸﺎﺭ ﺩﺭ ﺩﻭ ﺳﺮ ﺣﻔﺮﻩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ،ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﺳﻴﺎﻝ ﺩﺭ ﻧﺎﺣﻴﻪ ﻱ ﺣﻠﻘﻪ ﺍﻱ ،ﺻﺮﻓﺎ ﺑﻪ
ﺳﺒﺐ ﺣﺮﻛﺖ ﻣﻴﻠﻪ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ.
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺴﺌﻠﻪ ﻫﺎ
)ﺍﻟﻒ( ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﻧﺎﺣﻴﻪ ﻱ ﺣﻠﻘﻮﻱ ﺑﺎﺭﻳﻚ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ.
)ﺏ( ﺍﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ﺩﺭ ﻧﺎﺣﻴﻪ ﻱ ﺣﻠﻘﻮﻱ ﺑﻴﺎﺑﻴﺪ.
)ﺝ( ﻧﻴﺮﻭﻱ ﻭﻳﺴﻜﻮﺯ ﻭﺍﺭﺩ ﺑﺮ ﻣﻴﻠﻪ ﺩﺭ ﻃﻮﻝ Lﺭﺍ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ.
)ﺩ( ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻧﺘﻴﺠﻪ ﻗﺴﻤﺖ)ﺝ( ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﻓﺮﻣﻮﻝ" ﺷﻜﺎﻑ ﺗﺤﺖ" ﺿﺮﺏ ﺩﺭ "ﺗﺼﺤﻴﺢ ﺍﻧﺤﻨﺎ"
ﻧﻮﺷﺖ.ﻣﺴﺌﻠﻪ ﻫﺎﻳﻲ ﺍﺯ ﺍﻳﻦ ﻧﻮﻉ ﺩﺭ ﺑﺮﺭﺳﻲ ﻋﻤﻠﻜﺮﺩ ﻗﺎﻟﺐ ﻫﺎﻱ ﭘﻮﺷﺶ ﺳﻴﻢ ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﻧﺪ.
4.2ﭘﺪﻳﺪﻩ ﻫﺎﻱ ﭼﮕﺎﻟﻲ ﭘﺎﻳﻴﻦ ﺩﺭ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﺩﺭ ﻟﻮﻟﻪ)ﺷﻜﻞ .(4.2ﺑﺎ ﻛﺎﻫﺶ ﻓﺸﺎﺭ ﺩﺭ ﺳﻴﺴﺘﻢ
ﺑﺮﺭﺳﻲ ﺷﺪﻩ ﺩﺭ ﻣﺜﺎﻝ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﺩﺭ ﻟﻮﻟﻪ ﻱ ﻣﺪﻭﺭ ﺍﻓﻘﻲ ) ،ﺩﺭ ﺑﺨﺶ (3.2ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻫﺎ ﺍﻧﺤﺮﺍﻑ ﺣﺎﺻﻞ
ﻣﻲ ﺷﻮﺩ .ﺭﻓﺘﺎﺭ ﮔﺎﺯ ﭼﻨﺎﻥ ﺍﺳﺖ ﻛﻪ ﮔﻮﻳﻲ ﺭﻭﻱ ﺟﺪﺍﺭ ﻟﻮﻟﻪ ﻣﻲ ﻟﻐﺰﺩ .ﻣﺮﺳﻮﻡ ﺍﺳﺖ ﻛﻪ ﺷﺮﻁ ﻣﺮﺯﻱ ﻣﺘﺪﺍﻭﻝ "ﻋﺪﻡ
ﻟﻐﺰﺵ" ﺭﺍ ﺑﻨﺎ ﺑﻪ ﺁﻥ ﺩﺭ ﺟﺪﺍﺭ ﻟﻮﻟﻪ υ Z = 0ﺑﺎ ﺷﺮﻁ ﺯﻳﺮ ﺟﺎﻳﮕﺰﻳﻦ ﻛﻨﻨﺪ :
dυ z
, r=R
dr
υ Z = −ς
ﻛﻪ ﺩﺭ ﺍﻥ ςﺿﺮﻳﺐ ﻟﻐﺰﺵ ﺍﺳﺖ .ﻧﺘﻴﺠﻪ ﺑﻪ ﺩﺳﺖ ﺍﻣﺪﻩ ﺩﺭ ﻣﺜﺎﻝ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ ﺩﺭ ﻟﻮﻟﻪ ﻱ ﻣﺪﻭﺭ ﺍﻓﻘﻲ ) ،ﺩﺭ
ﺑﺨﺶ (3.2ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻱ ﺑﺎﻻ ﺑﻪ ﻣﻨﺰﻟﻪ ﺷﺮﻁ ﻣﺮﺯﻱ ،ﺩﻭﺑﺎﺭﻩ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ .ﺑﻪ ﻋﻼﻭﻩ ﺍﺯ ﺍﻳﻦ ﻧﻜﺘﻪ ﻱ
ﺗﺠﺮﺑﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ ﻛﻪ ﺿﺮﻳﺐ ﻟﻐﺰﺵ ﺑﺎ ﻋﻜﺲ ﻓﺸﺎﺭ ﺗﻐﻴﻴﺮ ﻣﻲ ﻛﻨﺪ ﻭ ﺩﺍﺭﻳﻢ ς = ς 0 Pﻛﻪ ﺩﺭ ﺁﻥ ς 0ﻣﻘﺪﺍﺭﻱ
ﺛﺎﺑﺖ ﺍﺳﺖ .ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺍﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ :
π ( p0 − p L )R 4 ρ ave
4ζ 0
1 +
8µL
RPave
1
ﻛﻪ ﺩﺭ ﺁﻥ ) Pavg = 2 ( p0 + p Lﻭ ρ avgﭼﮕﺎﻟﻲ ﻣﺘﻮﺳﻂ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ ﺩﺭ pavgﺍﺳﺖ.
ﻭﻗﺘﻲ ﻓﺸﺎﺭ ﺑﻴﺸﺘﺮ ﻛﺎﻫﺶ ﭘﻴﺪﺍ ﻛﻨﺪ ،ﺭژﻳﻢ ﺟﺮﻳﺎﻧﻲ ﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺩﺭ ﺍﻥ ﻣﺴﺎﻓﺖ ﺁﺯﺍﺩ ﻣﻴﺎﻧﮕﻴﻦ ﻣﻮﻟﻜﻮﻝ
ﻫﺎﻱ ﮔﺎﺯ ﻧﺴﺒﺖ ﺑﻪ ﺷﻌﺎﻉ ﻟﻮﻟﻪ ﺑﺰﺭگ ﺍﺳﺖ)ﺟﺮﻳﺎﻥ ﻛﻨﻮﺳﻦ(.ﺩﺭ ﺍﻳﻦ ﺭژﻳﻢ:
=w
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺴﺌﻠﻪ ﻫﺎ
2m 4 3 P0 − PL
πR
πkT 3
L
=w
ﻛﻪ ﺩﺭ ﺁﻥ mﺟﺮﻡ ﻣﻮﻟﻜﻮﻟﻲ ﻭ kﺛﺎﺑﺖ ﺑﻮﻟﺘﺰﻣﻦ ﺍﺳﺖ .ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﻧﺘﻴﺠﻪ ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﻫﻤﻪ
ﻱ ﺑﺮﺧﻮﺭﺩ ﻫﺎﻱ ﻣﻮﻟﻜﻮﻝ ﻫﺎ ﺑﺎ ﺳﻄﻮﺡ ﺟﺎﻣﺪ ﺍﺯ ﻧﻮﻉ ﭘﺨﺸﻲ ﺍﺳﺖ ﻧﻪ ﺑﺎﺯﺗﺎﺑﻲ.ﻧﺘﺎﻳﺞ ﺩﺭ ﺷﻜﻞ ﺧﻼﺻﻪ ﺷﺪﻩ ﺍﺳﺖ.
5.2ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﺩﺭ ﺷﺒﻜﻪ ﺍﻱ ﺍﺯ ﻟﻮﻟﻪ ﻫﺎ)ﺷﻜﻞ .(5.2ﺳﻴﺎﻟﻲ ﺩﺭ ﺷﺒﻜﻪ ﻱ ﻟﻮﻟﻪ ﻛﺸﻲ ﺍﺯ ﻧﻘﻄﻪ ﻱ Aﺗﺎ
ﻧﻘﻄﻪ ﻱ Bﻣﻄﺎﺑﻖ ﺷﻜﻞ ،ﺟﺮﻳﺎﻥ ﺍﺭﺍﻡ ﺩﺍﺭﺩ .ﻋﺒﺎﺭﺗﻲ ﺑﺮﺍﻱ ﺍﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ωﺳﻴﺎﻝ ﻭﺭﻭﺩﻱ ﺩﺭ ) Aﻳﺎ
ﺧﺮﻭﺟﻲ ﺩﺭ ،( Bﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ﺍﻓﺖ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﻱ PA − PBﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ .ﺍﺯ ﺍﺧﺘﻼﻝ ﺩﺭ ﺍﺗﺼﺎﻻﺕ
ﻣﺨﺘﻠﻒ ﺷﺒﻜﻪ ﭼﺸﻢ ﭘﻮﺷﻲ ﻛﻨﻴﺪ.
3π ( PA − PB ) R 4 ρ
20uL
=ω
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
ﻣﺴﺌﻠﻪ ﻫﺎ
6.2ﻋﻤﻠﻜﺮﺩ ﻏﺒﺎﺭﮔﻴﺮ ﺍﻟﻜﺘﺮﻭﺳﺘﺎﺗﻴﻜﻲ)ﺷﻜﻞ.(6.2
)ﺍﻟﻒ( ﻳﻚ ﺩﺳﺘﮕﺎﻩ ﻏﺒﺎﺭﮔﻴﺮ ﺍﺯ ﺻﻔﺤﻪ ﻫﺎﻳﻲ ﺑﺎ ﺑﺎﺭ ﻣﺨﺎﻟﻒ ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺟﺮﻳﺎﻥ ﮔﺎﺯ ﺣﺎﻭﻱ ﻏﺒﺎﺭ ﺍﺯ ﺑﻴﻦ
ﺍﻥ ﻫﺎ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ .ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﻣﻌﻴﺎﺭﻱ ﺑﺮﺍﻱ ﻃﻮﻝ ﻣﻴﻨﻴﻤﻢ ﻏﺒﺎﺭﮔﻴﺮ ﺑﺮ ﺣﺴﺐ ﺑﺎﺭ ﺫﺭﻩ eﺷﺪﺕ ﻣﻴﺪﺍﻥ
ﺍﻟﻜﺘﺮﻳﻜﻲ εﺍﺧﺘﻼﻑ ﻓﺸﺎﺭ ) ( p0 − p1ﺟﺮﻡ ﺫﺭﻩ mﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﮔﺎﺯ µﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ .ﻳﻌﻨﻲ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ ﻛﻪ
ﺑﻪ ﺍﺯﺍﻱ ﻛﺪﺍﻡ ﻃﻮﻝ Lﻛﻮﭼﻜﺘﺮﻳﻦ ﺫﺭﻩ ) ﺑﻪ ﺟﺮﻡ (mﺩﺭﺳﺖ ﻗﺒﻞ ﺍﺯ ﺍﻥ ﻛﻪ ﻓﺮﺻﺘﻲ ﺑﺮﺍﻱ ﺧﺎﺭﺝ ﺷﺪﻥ ﺍﺯ ﻣﺠﺮﺍ ﭘﻴﺪﺍ
ﻛﻨﺪ ،ﺑﻪ ﻛﻒ ﺍﻥ ﻣﻲ ﺭﺳﺪ .ﻓﺮﺽ ﻛﻨﻴﺪ ﺟﺮﻳﺎﻥ ﺑﻴﻦ ﺩﻭ ﺻﻔﺤﻪ ﺍﺭﺍﻡ ﺍﺳﺖ ،ﻫﻢ ﭼﻨﻴﻦ ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ ﺳﺮﻋﺖ ﺫﺭﻩ
ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺑﺎ ﺳﺮﻋﺖ ﺳﻴﺎﻝ ﺩﺭ ﺍﻳﻦ ﺍﻣﺘﺪﺍﺩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ .ﺑﻪ ﻋﻼﻭﻩ ﻓﺮﺽ ﻛﻨﻴﺪ ﭘﺴﺎﻱ ﺍﺳﺘﻮﻛﺲ ﺭﻭﻱ ﻛﺮﻩ ﻭ ﻧﻴﺰ
ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺶ ﻭﺍﺭﺩ ﺑﺮ ﻛﺮﻩ ،ﺩﺭﻫﻨﮕﺎﻡ ﺍﻓﺰﺍﻳﺶ ﺳﺮﻋﺖ ﺍﻥ ﺩﺭﺍﻣﺘﺪﺍﺩ ﻣﻨﻔﻲ xﻗﺎﺑﻞ ﭼﺸﻢ ﭘﻮﺷﻲ ﺍﺳﺖ.
)ﺏ( ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﺭﺍ ﺑﺎ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺷﺘﺎﺏ ﺩﺭ ﺍﻣﺘﺪﺍﺩ xﺍﻣﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﭘﺴﺎﻱ ﺍﺳﺘﻮﻛﺲ ﺩﻭﺑﺎﺭﻩ ﺣﻞ ﻛﻨﻴﺪ.
)ﺝ( ﺳﻮﺩﻣﻨﺪﻱ ﺟﻮﺍﺏ ﻫﺎﻱ) ﺍﻟﻒ( ﻭ)ﺏ( ﺭﺍ ﺑﺎﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ﺫﺭﺍﺕ ﺍﺋﺮﻭﺳﻞ ﭘﺎﻳﺪﺍﺭ ﻗﻄﺮﻱ ﺩﺭ ﺣﺪﻭﺩ 1ﺗﺎ 10
ﻣﻴﻜﺮﻭﻥ ﻭ ﭼﮕﺎﻟﻲ ﺣﺪﻭﺩ
cm 3
1 gﺩﺍﺭﻧﺪ ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﺪ.
7.2ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻨﺞ ﺑﺎ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﺍﻓﺘﺎﻥ ﺍﺯ ﻳﻚ ﻃﺮﻑ)ﺷﻜﻞ 7.2ﺭﺍ ﺑﺒﻴﻨﻴﺪ( .ﺍﺳﺘﻮﺍﻧﻪ ﻋﻤﻮﺩﻱ ﺑﻠﻨﺪ )ﺑﻪ
ﺷﻌﺎﻉ ( Rﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﻫﺮ ﺩﻭ ﺳﺮ ﺍﻥ ﺩﺭﭘﻮﺵ ﺩﺍﺭﺩ ﻭ ﻳﻚ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﻛﻮﭼﻚ ﺗﻮﭘﺮ)ﺑﻪ ﺷﻌﺎﻉ (kRﺩﺭ
ﺩﺍﺧﻞ ﺍﻥ ﺍﺳﺖ .ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﻛﻮﭼﻚ ﭘﺮﻩ ﻫﺎﻳﻲ ﺩﺍﺭﺩ ﻛﻪ ﺳﺒﺐ ﻣﻲ ﺷﻮﻧﺪ ﻫﻤﻮﺍﺭﻩ ﺑﺎ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﺑﺰﺭگ ﻫﻢ ﻣﺤﻮﺭ
ﺑﻤﺎﻧﺪ .ﻣﻲ ﺗﻮﺍﻥ ﺍﻫﻨﮓ ﺳﻘﻮﻁ ﺍﺳﺘﻮﺍﻧﻪ ﻛﻮﭼﻚ ﺩﺭ ﻇﺮﻑ ﺭﺍ ﻭﻗﺘﻲ ﺑﺎ ﺳﻴﺎﻟﻲ ﭘﺮ ﺷﺪﻩ ﺍﺳﺖ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻛﺮﺩ .ﺭﺍﺑﻄﻪ
ﺍﻱ ﺑﻴﺎﺑﻴﺪ ﻛﻪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻴﺎﻝ ﺑﺮ ﺣﺴﺐ ﺳﺮﻋﺖ ﺣﺪﻱ υ 0ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﻛﻮﭼﻚ ﻭ ﻛﻤﻴﺖ ﻫﺎﻱ ﻫﻨﺪﺳﻲ ﻣﺨﺘﻠﻒ
ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺭﻭﻱ ﺷﻜﻞ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻥ ﺑﻪ ﺩﺳﺖ ﺍﻳﺪ.
۷٤
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
)ﺍﻟﻒ( ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺷﻜﺎﻑ ﺣﻠﻘﻮﻱ ﺍﺯ ﺭﺍﺑﻄﻪ ﻱ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺍﻳﺪ :
(1 − ξ 2 ) − (1 + k 2 )ln 1ξ
υz
=−
υ0
(1 − k 2 ) − (1 + k 2 )ln 1k
) (
ξ = rR
)ﺏ( ﻣﻮﺍﺯﻧﻪ ﻱ ﻧﻴﺮﻭ ﺭﻭﻱ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﺗﻮﭘﺮ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ:
(ρ 0 − ρ )g (kR )2 ln 1 − 1 − k 2
k 1 + k 2
2υ 0
=µ
ﻛﻪ ﺩﺭ ﺍﻥ ρﻭ ρ 0ﺑﻪ ﺗﺮﺗﻴﺐ ﭼﮕﺎﻟﻲ ﺳﻴﺎﻝ ﻭ ﭼﮕﺎﻟﻲ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﺗﻮﭘﺮ ﺍﺳﺖ.
)ﺝ( ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻭﻗﺘﻲ ﻋﺮﺽ ﺷﻜﺎﻑ ﻛﻢ ﺑﺎﺷﺪ ،ﻧﺘﻴﺠﻪ)ﺏ( ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺮ ﺣﺴﺐ ﺗﻮﺍﻥ ﻫﺎﻱ ε = 1 − kﺑﺴﻂ
ﺩﺍﺩ ﺗﺎ ﻧﺘﻴﺠﻪ ﺯﻳﺮ ﺣﺎﺻﻞ ﺷﻮﺩ :
(ρ 0 − ρ )gR 2ε 3 1 − 1 ε − 13 ε 2 + ....
20
2
6υ 0
=µ
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۲ﻣﻮﺍﺯﻧﻪ ی ﻻﻳﻪ ﺍی ﺍﻧﺪﺍﺯﻩ ﺣﺮﮐﺖ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ
8.2ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ ﺭﻭﻱ ﺳﻄﺢ ﻣﺨﺮﻭﻃﻲ)ﺷﻜﻞ 8.2ﺭﺍ ﺑﺒﻴﻨﻴﺪ(.ﺳﻴﺎﻟﻲ ﺩﺭ ﻟﻮﻟﻪ ﻱ ﻣﺪﻭﺭ ﺑﻪ ﻃﺮﻑ ﺑﺎﻻ ﺩﺭ ﺟﺮﻳﺎﻥ
ﺍﺳﺖ ﻭ ﺳﭙﺲ ﺭﻭﻱ ﺳﻄﺢ ﻣﺨﺮﻭﻃﻲ ﭘﺎﻳﻴﻦ ﻣﻲ ﺍﻳﺪ .ﻣﻄﻠﻮﺏ ﺍﺳﺖ ﺗﻌﻴﻴﻦ ﺿﺨﺎﻣﺖ ﻓﻴﻠﻢ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ
ﻓﺎﺻﻠﻪ ﻱ sﺑﻪ ﻃﺮﻑ ﻗﺎﻋﺪﻩ ﻣﺨﺮﻭﻁ.
)ﺍﻟﻒ( ﻓﺮﺽ ﻛﻨﻴﺪ ﻧﺘﺎﻳﺞ ﺑﺨﺶ 2.2ﺩﺭ ﻫﺮ ﻧﺎﺣﻴﻪ ﻱ ﻛﻮﭼﻚ ﺍﺯ ﺳﻄﺢ ﻣﺨﺮﻭﻁ ﺗﻘﺮﻳﺒﺎ ﺻﺎﺩﻕ ﺍﺳﺖ .ﻧﺸﺎﻥ ﺩﻫﻴﺪ
ﻛﻪ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ ﺭﻭﻱ ﺣﻠﻘﻪ ﺍﻱ ﺍﺯ ﻣﺎﻳﻊ ﻣﻮﺟﻮﺩ ﺑﻴﻦ sﻭ s + ∆sﺑﻪ ﻧﺘﻴﺠﻪ ﻱ ﺯﻳﺮ ﻣﻨﺘﻬﻲ ﻣﻲ ﺷﻮﺩ :
) (
d
(sδ ⟨υ ⟩ ) = 0 or d sδ 3 = 0
ds
ds
)ﺏ( ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺍﻧﺘﻜﺮﺍﻝ ﺑﮕﻴﺮﻳﺪ ﻭ ﺛﺎﺑﺖ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺭﺍ ﺑﺎ ﺑﺮﺍﺑﺮ ﻧﻬﺎﺩﻥ ﺍﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ωﺑﻪ ﻃﺮﻑ
ﺑﺎﻻﻱ ﻟﻮﻟﻪ ﻱ ﻭﺳﻄﻲ ﻭ ﻫﻤﻴﻦ ﺍﻫﻨﮓ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺭﻭ ﺑﻪ ﭘﺎﻳﻴﻦ ﺭﻭﻱ ﺳﻄﺢ ﻣﺨﺮﻭﻃﻲ ﺩﺭ s = Lﻣﺤﺎﺳﺒﻪ ﻛﻨﻴﺪ.
ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺯﺍ ﺑﺮﺍﻱ ﺿﺨﺎﻣﺖ ﻓﻴﻠﻢ ﺑﻪ ﺩﺳﺖ ﺍﻭﺭﻳﺪ:
3µw
L
πρ gL sin 2 β s
2
δ =3
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
۷۷
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﺩﺭ ﻓﺼﻞ ،2ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺭﻭﺵ ﻣﻮﺍﺯﻧﻪ ﻱ ﻻﻳﻪ ﺍﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ،ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺭﺍ ﺑﺮﺍﻱ ﭼﻨﺪ ﺳﻴﺴﺘﻢ ﺟﺮﻳﺎﻥ
ﺳﺎﺩﻩ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻳﻢ .ﺳﭙﺲ ﺍﺯ ﺗﻮﺯﻳﻊ ﻫﺎﻱ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺑﺮﺍﻱ ﻳﺎ ﻓﺘﻦ ﻛﻤﻴﺖ ﻫﺎﻱ ﺩﻳﮕﺮ) ،ﻣﺎﻧﻨﺪ ﺳﺮﻋﺖ
ﻣﺘﻮﺳﻂ ﻭ ﻧﻴﺮﻭﻱ ﭘﺴﺎ( ﻭ ﺍﺯ ﺭﻫﻴﺎﻓﺖ ﻣﻮﺍﺯﻧﻪ ﻱ ﻻﻳﻪ ﺑﺮﺍﻱ ﺁﺷﻨﺎ ﻛﺮﺩﻥ ﻣﺒﺘﺪﻳﺎﻥ ﺑﺎ ﻧﻤﺎﺩ ﮔﺬﺍﺭﻱ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ
ﺣﺮﻛﺖ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪ .ﺍﮔﺮ ﭼﻪ ﺩﺭ ﻓﺼﻞ 2ﺻﺮﺍﺣﺘﺎً ﺍﺯ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ ﺳﺨﻨﻲ ﺑﻪ ﻣﻴﺎﻥ ﻧﻴﺎﻣﺪ،ﺍﻣﺎ ﺑﻪ ﻃﻮﺭ ﺿﻤﻨﻲ ﺍﺯ
ﺍﻳﻦ ﺍﻳﺪﻩ ﻧﻴﺰ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩﻳﻢ.
ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﻻﻳﻪ ﺑﺮﺍﻱ ﻫﺮ ﻣﺴﺌﻠﻪ ﺍﻱ ﻛﻪ ﺑﺎ ﺁﻥ ﺭﻭﺑﻪ ﺭﻭ ﻣﻲ ﺷﻮﻳﻢ ،ﻛﺎﺭ ﺩﺷﻮﺍﺭ ﻭ ﺧﺴﺘﻪ ﻛﻨﻨﺪﻩ ﺍﻱ ﺍﺳﺖ .ﺁﻥ
ﭼﻪ ﻧﻴﺎﺯ ﺩﺍﺭﻳﻢ ﻳﻚ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ ﻛﻠﻲ ﻭ ﻳﻚ ﻣﻮﺍﺯﻧﻪ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻛﻠﻲ ﺍﺳﺖ ﻛﻪ ﺑﺘﻮﺍﻥ ﺩﺭ ﻫﺮ ﻣﺴﺌﻠﻪ ﺍﻱ،ﺍﺯ
ﺟﻤﻠﻪ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺷﺎﻣﻞ ﺣﺮﻛﺖ ﻏﻴﺮ ﺭﺍﺳﺖ ﺧﻂ ،ﺍﺯ ﺁﻥ ﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ.ﻣﺤﻮﺭ ﺍﺻﻠﻲ ﺑﺤﺚ ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ﻫﻤﻴﻦ
ﻣﻮﺿﻮﻉ ﺍﺳﺖ .ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﻛﻪ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﻢ ،ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ )ﺑﺮﺍﻱ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ( ﻭ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ
)ﺑﺮﺍﻱ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ( ﻧﺎﻡ ﺩﺍﺭﻧﺪ .ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻫﺎ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﻧﻘﻄﻪ ﻱ ﺷﺮﻭﻉ ﺑﺮﺍﻱ ﻣﻄﺎﻟﻌﻪ ﻱ
ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺷﺎﻣﻞ ﺟﺮﻳﺎﻥ ﻫﻢ ﺩﻣﺎﻱ ﺳﻴﺎﻝ ﺧﺎﻟﺺ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ.
ﺩﺭ ﺑﺨﺶ ،1.3ﺑﺎ ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ ﺭﻭﻱ ﺟﺰء ﺣﺠﻤﻲ ﻛﻮﭼﻚ ،ﻛﻪ ﺳﻴﺎﻝ ﺍﺯ ﺁﻥ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ ،ﻣﻌﺎﺩﻟﻪ ﻱ
ﭘﻴﻮﺳﺘﮕﻲ ﺭﺍ ﺑﻪ ﺩﺳﺖ ِﻣﻲ ﺁﻭﺭﻳﻢ .ﺳﭙﺲ ﺍﻧﺪﺍﺯﻩ ﻱ ﺍﻳﻦ ﺟﺰء ﺭﺍ ﺑﻪ ﺳﻤﺖ ﺻﻔﺮ ﻣﻴﻞ ﻣﻲ ﺩﻫﻴﻢ ) ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺑﺎ
ﺳﻴﺎﻝ ﺑﻪ ﺻﻮﺭﺕ ﭘﻴﻮﺳﺘﺎﺭ ﺑﺮﺧﻮﺭﺩ ﻣﻲ ﻛﻨﻴﻢ ( ،ﻭ ﻣﻌﺎﺩﻟﻪ ﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺟﺰﺋﻲ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ.
ﺩﺭ ﺑﺨﺶ ،2.3ﺑﺎ ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺭﻭﻱ ﺟﺰء ﺣﺠﻤﻲ ﻛﻮﭼﻚ ،ﻭ ﻣﻴﻞ ﺩﺍﺩﻥ ﺣﺠﻢ ﺍﻳﻦ ﺟﺰء ﺑﻪ
ﺳﻤﺖ ﻣﻘﺪﺍﺭﻱ ﺑﻲ ﻧﻬﺎﻳﺖ ﻛﻮﭼﻚ ،ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ .ﺩﺭ ﺍﻳﻦ ﺟﺎ ﻧﻴﺰ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﻱ
ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺟﺰﺋﻲ ﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ .ﺑﺎ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ﻭ ﻛﻤﻚ ﮔﺮﻓﺘﻦ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ ﻣﻲ ﺗﻮﺍﻥ
ﻫﻤﻪ ﻱ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﻓﺼﻞ 2ﻭ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺑﺴﻴﺎﺭ ﭘﻴﭽﻴﺪﻩ ﺗﺮ ﺭﺍ ﺣﻞ ﻛﺮﺩ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ﻣﻌﺎﺩﻟﻪ ﺍﻱ
ﺑﺴﻴﺎﺭ ﻣﻬﻢ ﺩﺭ ﭘﺪﻳﺪﻩ ﻫﺎﻱ ﺍﻧﺘﻘﺎﻝ ﺍﺳﺖ.
1.3ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ
ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺑﺎ ﻧﻮﺷﺘﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ ﺭﻭﻱ ﺟﺰء ﺣﺠﻢ ∆x∆y∆zﺛﺎﺑﺖ ﺩﺭ ﻓﻀﺎ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ ﻛﻪ ﺳﻴﺎﻝ ﺩﺭ ﺁﻥ
ﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ :
ﺣﺎﻝ ﺑﺎﻳﺪ ﺍﻳﻦ ﺑﻴﺎﻥ ﺳﺎﺩﻩ ﻱ ﻓﻴﺰﻳﻜﻲ ﺭﺍ ﺑﻪ ﺯﺑﺎﻥ ﺭﻳﺎﺿﻲ ﺗﺮﺟﻤﻪ ﻛﻨﻴﻢ.
۷۸
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﻛﺎﺭ ﺭﺍ ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺩﻭ ﻭﺟﻪ ﺳﺎﻳﻪ ﺧﻮﺭﺩﻩ ﺁﻏﺎﺯ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ﺑﺮ ﻣﺤﻮﺭ xﻋﻤﻮﺩﻧﺪ .ﺁﻫﻨﮓ ﻭﺭﻭﺩ ﺟﺮﻡ ﺑﻪ ﺟﺰء
ﺣﺠﻢ ،ﺍﺯ ﻃﺮﻳﻖ ﻭﺟﻪ ﺳﺎﻳﻪ ﺧﻮﺭﺩﻩ ﺩﺭ xﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ، (ρυ x ) x ∆y∆z :ﻭ ﺁﻫﻨﮓ ﺧﺮﻭﺝ ﺟﺮﻡ ﺍﺯ ﻭﺟﻪ ﺳﺎﻳﻪ
ﺧﻮﺭﺩﻩ ﺩﺭ x + ∆xﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ . (ρυ x ) x + ∆x ∆y∆z :ﺑﺮﺍﻱ ﺩﻭ ﻭﺟﻪ ﺩﻳﮕﺮ ﻫﻢ ﻣﻲ ﺗﻮﺍﻥ ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﻣﺸﺎﺑﻬﻲ
ﻧﻮﺷﺖ .ﺁﻫﻨﮓ ﺍﻓﺰﺍﻳﺶ ﺟﺮﻡ ﺩﺭ ﺩﺍﺧﻞ ﺟﺰء ﺣﺠﻢ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ . ∆x∆y∆z (∂ρ ∂t ) :ﺑﻨﺎﺑﺮﺍﻳﻦ ﻣﻮﺍﺯﻧﻪ ﻱ ﺟﺮﻡ
ﭼﻨﻴﻦ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ :
∂ρ
] = ∆y∆z [(ρυ x ) x −(ρυ x ) x + ∆x ]+ ∆z∆x (ρυ y ) y −(ρυ y ) y + ∆y + ∆x∆y[(ρυ z ) z −(ρυ z ) z + ∆z
∆x∆y∆z
]
[
∂t
ﺑﺎ ﺗﻘﺴﻴﻢ ﻛﺮﺩﻥ ﻣﻌﺎﺩﻟﻪ ﺑﺮ ∆x∆y∆zﻭ ﮔﺮﻓﺘﻦ ﺣﺪ ﻭﻗﺘﻲ ، ∆z ، ∆y ، ∆xﺑﻪ ﺳﻤﺖ ﺻﻔﺮ ﻣﻴﻞ ﻣﻲ ﻛﻨﻨﺪ ،ﻭ ﺁﻥ ﮔﺎﻩ
ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻌﺮﻳﻒ ﻫﺎﻱ ﻣﺸﺘﻖ ﻫﺎ ﻱ ﺟﺰﺋﻲ ،ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ :
∂
∂
∂
∂ρ
= − ρυ x + ρυ y + ρυ z
∂z
∂y
∂t
∂x
ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ ﺍﺳﺖ ،ﻛﻪ ﺁﻫﻨﮓ ﺯﻣﺎﻧﻲ ﺗﻐﻴﻴﺮ ﭼﮕﺎﻟﻲ ﺳﻴﺎﻝ ﺩﺭ ﻧﻘﻄﻪ ﺍﻱ ﺛﺎﺑﺖ ﺩﺭ ﻓﻀﺎ ﺭﺍ ﺗﻮﺳﻴﻒ ﻣﻲ
ﻛﻨﺪ .ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻤﺎﺩ ﮔﺬﺍﺭﻱ ﺑﺮﺩﺍﺭﻱ ،ﻣﻲ ﺗﻮﺍﻥ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺖ:
∂ρ
) = −(∇.ρV
∂t
ﺩﺭ ﺍﻳﻨﺠﺎ ) (∇.ρVﺭﺍ "ﺩﻳﻮﺭژﺍﻧﺲ " ρVﻣﻲ ﻧﺎﻣﻨﺪ.ﺑﺮﺩﺍﺭ ρVﺷﺎﺭ ﺟﺮﻡ ﺍﺳﺖ ﻭ ﺩﻳﻮﺭژﺍﻧﺲ ﺁﻥ ﻣﻔﻬﻮﻡ ﺳﺎﺩﻩ ﺍﻱ
ﺩﺍﺭﺩ؛ ﺁﻫﻨﮓ ﺧﺎﻟﺺ ﺷﺎﺭ ﺟﺮﻡ ﺧﺮﻭﺟﻲ ﺩﺭ ﻭﺍﺣﺪ ﺣﺠﻢ.
ﺻﻮﺭﺕ ﺧﺎﺹ ﺑﺴﻴﺎﺭ ﻣﻬﻤﻲ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ ﻣﺮﺑﻮﻁ ﺑﻪ ﺳﻴﺎﻟﻲ ﺑﺎ ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ ﺍﺳﺖ:
)ﺳﻴﺎﻝ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ(
(∇V
. )= 0
ﺍﻟﺒﺘﻪ ﻫﻴﭻ ﺳﻴﺎﻟﻲ ﻭﺍﻗﻌﺎً ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻧﻴﺴﺖ ،ﺍﻣﺎ ﻏﺎﻟﺒﺎً ﺩﺭ ﻛﺎﺭﺑﺮﺩ ﻫﺎﻱ ﻣﻬﻨﺪﺳﻲ ﻭ ﺯﻳﺴﺖ ﺷﻨﺎﺳﻲ ،ﻓﺮﺽ ﺛﺎﺑﺖ
ﺑﻮﺩﻥ ﭼﮕﺎﻟﻲ ﺑﻪ ﺳﺎﺩﻩ ﺳﺎﺯﻱ ﭼﺸﻢ ﮔﻴﺮ ﻭ ﺧﻄﺎﻱ ﺑﺴﻴﺎﺭ ﻛﻢ ﻣﻨﺘﻬﻲ ﻣﻲ ﺷﻮﺩ.
۷۹
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
2.3ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ
ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ،ﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺭﺍ ﺭﻭﻱ ﺟﺰء ﺣﺠﻢ ∆x∆y∆zﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ
ﻣﻲ ﻧﻮﻳﺴﻴﻢ :
ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻱ ) (1-2.3ﺑﺴﻂ ﻣﻌﺎﺩﻟﻪ ﻱ ) (1-1.2ﺑﻪ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺣﺎﻟﺖ ﻧﺎﭘﺎﻳﺎﺳﺖ .ﺑﻨﺎﻳﺮﺍﻳﻦ ﺗﻘﺮﻳﺒﺎً
ﻫﻤﺎﻥ ﺷﻴﻮﻩ ﻱ ﻓﺼﻞ 2ﺭﺍ ﭘﻴﺶ ﻣﻲ ﮔﻴﺮﻳﻢ .ﺍﻣﺎ ﻋﻼﻭﻩ ﺑﺮ ﮔﻨﺠﺎﻧﺪﻥ ﺟﻤﻠﻪ ﻱ ﺣﺎﻟﺖ ﻧﺎﭘﺎﻳﺎ ،ﺑﺎﻳﺪ ﺑﻪ ﺳﻴﺎﻝ ﺍﻣﻜﺎﻥ
ﺑﺪﻫﻴﻢ ﻛﻪ ﺍﺯ ﻫﺮ ﺷﺶ ﻭﺟﻪ ﺟﺰء ﺣﺠﻢ ﻋﺒﻮﺭ ﻛﻨﺪ .ﺑﻪ ﺧﺎﻃﺮ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﺪ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻱ ) (1-2.3ﻣﻌﺎﺩﻟﻪ ﺍﻱ
ﺑﺮﺩﺍﺭﻱ ﺍﺳﺖ ﻭ ﻣﺆﻟﻔﻪ ﻫﺎﻳﻲ ﺩﺭ ﻫﺮ ﺳﻪ ﺍﻣﺘﺪﺍﺩ z, y, xﺩﺍﺭﺩ .ﻣﺆﻟﻔﻪ ﻱ xﻫﺮ ﺟﻤﻠﻪ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻱ ) (1-2.3ﺭﺍ ﺑﻪ
ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ؛ ﻣﺆﻟﻔﻪ ﻫﺎﻱ z, yﺭﺍ ﻣﻴﺘﻮﺍﻥ ﺑﻪ ﺷﻴﻮﻩ ﻱ ﻣﺸﺎﺑﻪ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ .
ﺍﺑﺘﺪﺍ ﺁﻫﻨﮓ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻭﺭﻭﺩﻱ ﻭ ﺧﺮﻭﺟﻲ ﺍﺯ ﺟﺰء ﺣﺠﻢ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﻜﻞ
1-2.3ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ .ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺯ ﻃﺮﻳﻖ ﺩﻭ ﻣﻜﺎﻧﻴﺴﻢ ﺑﻪ ﺟﺰء ﺣﺠﻢ ∆x∆y∆zﻭﺍﺭﺩ ﻳﺎ ﺍﺯ ﺁﻥ ﺧﺎﺭﺝ
ﻣﻲ ﺷﻮﺩ :ﺍﻧﺘﻘﺎﻝ ﻫﻤﺮﻓﺘﻲ ﻭ ﺍﻧﺘﻘﺎﻝ ﻣﻮﻟﻜﻮﻟﻲ
ﺁﻫﻨﮓ ﻭﺭﻭﺩ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺯ ﻃﺮﻳﻖ ﻭﺟﻪ ﺳﺎﻳﻪ ﺧﻮﺭﺩﻩ ﺩﺭ xﺍﺯ ﻃﺮﻳﻖ ﻫﻤﻪ ﻱ ﻣﻜﺎﻧﻴﺴﻢ ﻫﺎ -ﻫﻢ
ﻫﻤﺮﻓﺘﻲ ﻭ ﻫﻢ ﻣﻮﻟﻜﻮﻟﻲ -ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ (φ xx ) x ∆y∆z :ﻭ ﺁﻫﻨﮓ ﺧﺮﻭﺝ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺯ ﻭﺟﻪ ﺳﺎﻳﻪ ﺧﻮﺭﺩﻩ ﺩﺭ
x + ∆xﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ . (φ xx ) x+ ∆x ∆y∆z :ﺁﻫﻨﮓ ﻫﺎﻱ ﻭﺭﻭﺩ ﻭ ﺧﺮﻭﺝ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺯ ﺍﻳﻦ ﻭﺟﻮﻩ ﺩﺭ
yﻭ ، y + ∆yﺑﻪ ﺗﺮﺗﻴﺐ ،ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ (φ yx ) y ∆z∆x :ﻭ . (φ yx ) y + ∆y ∆z∆xﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺁﻫﻨﮓ ﻫﺎﻱ ﻭﺭﻭﺩ
ﻭ ﺧﺮﻭﺝ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺯ ﻭﺟﻮﻩ zﻭ ، z + ∆zﺑﻪ ﺗﺮﺗﻴﺐ ،ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ (φ zx ) z ∆x∆y :ﻭ
. (φ zx ) z + ∆z ∆x∆yﻭﻗﺘﻲ ﺍﻳﻦ ﺳﻬﻢ ﻫﺎ ﺭﺍ ﺑﺎ ﻫﻢ ﺟﻤﻊ ﻛﻨﻴﻢ ،ﺁﻫﻨﮓ ﺧﺎﻟﺺ ﺍﻓﺰﺍﻳﺶ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ،
ﺭﻭﻱ ﻫﺮ ﺳﻪ ﺟﻔﺖ ﻭﺟﻪ ،ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
۸۰
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
)(2-2.3
)
−φ zx z + ∆z
zx z
) + ∆x∆y(φ
y + ∆y
− φ yz
yx y
) + ∆z∆x(φ
x + ∆x
− φ xx
x
∆y∆z (φ xx
ﻧﻴﺮﻭﻱ ﺧﺎﺭﺟﻲ)ﻣﻌﻤﻮﻻً ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺸﻲ( ﻧﻴﺰ ﺑﺮ ﺳﻴﺎﻝ ﺩﺍﺧﻞ ﺟﺰء ﺣﺠﻢ ﻭﺍﺭﺩ ﻣﻲ ﺷﻮﺩ .ﻣﻮﻟﻔﻪ ﻱ xﺍﻳﻦ ﻧﻴﺮﻭ
ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :
ρg x ∆x∆y∆z
) (3-2.3
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ) (2-2.3ﻭ) (3-2.3ﻣﺆﻟﻔﻪ ﻫﺎﻱ xﺳﻪ ﺟﻤﻠﻪ ﻱ ﺳﻤﺖ ﺭﺍﺳﺖ ﻣﻌﺎﺩﻟﻪ ﻱ ) (1-2.3ﺭﺍ ﺑﻪ ﺩﺳﺖ ﻣﻲ
ﺩﻫﻨﺪ .ﻣﺠﻤﻮﻉ ﺍﻳﻦ ﺳﻪ ﺟﻤﻠﻪ ﺭﺍ ﺑﺎﻳﺪ ﺑﺎ ﺁﻫﻨﮓ ﺍﻓﺰﺍﻳﺶ ﻣﺆﻟﻔﻪ ﻱ xﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﺩﺍﺧﻞ ﺟﺰء ﺣﺠﻢ ﻳﻌﻨﻲ
∆x∆y∆z∂(ρυ x ) / ∂tﺑﺮﺍﺑﺮ ﮔﺮﻓﺖ .ﭘﺲ ﺍﺯ ﺍﻧﺠﺎﻡ ﺍﻳﻦ ﻛﺎﺭ ،ﻣﺆﻟﻔﻪ ﻱ xﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺑﻪ ﺩﺳﺖ ﻣﻲ
ﺁﻳﺪ .ﻫﺮﮔﺎﻩ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﺑﺮ ∆x∆y∆zﺗﻘﺴﻴﻢ ﻛﻨﻴﻢ ﻭ ﻭﻗﺘﻲ ∆y ، ∆xﻭ ∆zﺑﻪ ﺳﻤﺖ ﺻﻔﺮ ﻣﻴﻞ ﻣﻲ ﻛﻨﻨﺪ ﺣﺪ
ﺑﮕﻴﺮﻳﻢ،ﻣﻌﺎﺩﻟﻪ ﻱ ﺯﻳﺮ ﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ :
∂
∂
∂
∂
ρυ x = − φ xx + φ yx + φ zx + ρg x
∂t
∂y
∂z
∂x
) (4-2.3
ﺩﺭ ﺍﻳﻦ ﺟﺎ ﺍﺯ ﺗﻌﺮﻳﻒ ﻣﺸﺘﻖ ﺟﺰﺋﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩﻫﺎﻳﻢ .ﺑﺮﺍﻱ ﻣﺆﻟﻔﻪ ﻫﺎﻱ yﻭ zﻣﻮﺍﺯﻧﻪ ﻱ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻧﻴﺰ ﻣﻲ
ﺗﻮﺍﻥ ﻣﻌﺎﺩﻻﺕ ﻣﺸﺎ ﺑﻬﻲ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ:
∂
(ρv y ) = − ∂ φ xy + ∂ φ yy + ∂ φ zy + ρg y
∂t
∂y
∂z
∂x
) 5-2.3ﻭ (6-2.3
∂
(ρvz ) = − ∂ φxz + ∂ φ yz + ∂ φzz + ρg z
∂t
∂y
∂z
∂x
ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﺎﺩﮔﺬﺍﺭﻱ ﺑﺮﺩﺍﺭ -ﺗﺎ ﻧﺴﻮﺭ ،ﺍﻳﻦ ﺳﻪ ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺖ :
i = x, y , z
) (7-2.3
∂
ρυi = −[∇.φ ]i + ρg i
∂t
ﻳﻌﻨﻲ ﺑﺎ ﺗﺒﺪﻳﻞ ﻣﺘﻮﺍﻟﻲ iﺑﻪ z, y, xﻣﻲ ﺗﻮﺍﻥ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ) 5،4-2.3ﺗﺎ (6ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ .ﻛﻤﻴﺖ ﻫﺎﻱ ρυ i
ﻣﺆﻭﻟﻔﻪ ﻫﺎﻱ ﺩﻛﺎﺭﺗﻲ ﺑﺮﺩﺍﺭ ρVﻫﺴﺘﻨﺪ ﻛﻪ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺩﺭ ﻭﺍﺣﺪ ﺣﺠﻢ ﺩﺭ ﻧﻘﻄﻪ ﺍﻱ ﺍﺯ ﺳﻴﺎﻝ.
ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ،ﻛﻤﻴﺖ ﻫﺎﻱ ρg iﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺑﺮﺩﺍﺭ ρgﻫﺴﺘﻨﺪ ﻛﻪ ﻧﻴﺮﻭﻱ ﺧﺎﺭﺟﻲ ﺑﺮ ﻭﺍﺣﺪ ﺣﺠﻢ ﺍﺳﺖ.
ﺟﻤﻠﻪ ﻱ ، − [∇.φ ]iﻣﺆﻟﻔﻪ ﻱ iﺍﻡ ﺑﺮﺩﺍﺭ ] − [∇.φﺍﺳﺖ.
۸۱
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﻭﻗﺘﻲ ﻣﺆﻟﻔﻪ iﺍﻡ ﻣﻌﺎﺩﻟﻪ 7-2.3ﺭﺍ ﺩﺭ ﺑﺮﺩﺍﺭ ﻳﻜﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ iﺍﻡ ﺿﺮﺏ ﻛﻨﻴﻢ ﻭ ﺳﻪ ﻣﺆﻟﻔﻪ ﺭﺍ ﺑﺎ ﻫﻢ ﺟﻤﻊ ﻛﻨﻴﻢ
ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ :
∂
ρV = −[∇.φ ] + ρg
∂t
)(8-2.3
ﻛﻪ ﻋﺒﺎﺭﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﻗﺎﻧﻮﻥ ﭘﺎﻳﺴﺘﮕﻲ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺍﺳﺖ .ﺍﻳﻦ ﺗﺮﺟﻤﻪ ﻱ ﻣﻌﺎﺩﻟﻪ ﻱ) (1-2.3ﺑﻪ ﻧﻤﺎﺩ ﻫﺎﻱ
ﻣﻜﺎﻧﻴﻜﻲ ﺍﺳﺖ.
ﺗﺎﻧﺴﻮﺭ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻣﺮﻛﺐ φﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﺠﻤﻮﻉ ﺗﺎﻧﺴﻮﺭ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻫﻤﺮﻓﺘﻲ ρVVﻭ ﺗﺎﻧﺴﻮﺭ
ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﻣﻮﻟﻜﻮﻟﻲ πﻭ ﺍﻳﻦ ﻛﻪ ﻛﻤﻴﺖ ﺍﺧﻴﺮ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﻣﺠﻤﻮﻉ pδﻭ τﻧﻮﺷﺖ .ﻫﻨﮕﺎﻣﻲ
ﻛﻪ ﻋﺒﺎﺭﺕ φ = ρVV + pδ + τﺭﺍ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻱ ) (8-2.3ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻴﻢ ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ
ﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ :
∂
ρV = −[∇.ρVV ] − ∇p − [∇.τ ] + ρg
∂t
ﺩﺭ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ∇pﺑﺮﺩﺍﺭﻱ ﻣﻮﺳﻮﻡ ﺑﻪ " ﮔﺮﺍﺩﻳﺎﻥ )ﺍﺳﻜﺎﻟﺮ( " pﺍﺳﺖ ﻛﻪ ﮔﺎﻫﻲ ﺑﻪ ﺻﻮﺭﺕ " " grad pﻧﻮﺷﺘﻪ
ﻣﻲ ﺷﻮﺩ.ﻧﻤﺎﺩ ] [∇.τﺑﺮﺩﺍﺭﻱ ﻣﻮﺳﻮﻡ ﺑﻪ "ﺩﻳﻮﺭژﺍﻧﺲ )ﺗﺎﻧﺴﻮﺭ( " τﺍﺳﺖ ﻭ ] [∇.ρVVﺑﺮﺩﺍﺭﻱ ﺑﻪ ﻧﺎﻡ "ﺩﻳﻮﺭژﺍﻧﺲ
)ﺿﺮﺏ ﺩﻭ ﺗﺎﻳﻲ( " ρVVﺍﺳﺖ.
ﺭﻭﺵ ﺩﻳﮕﺮ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﺳﻪ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍ ﺑﻪ ﻓﺮﻡ ﻓﺸﺮﺩﻩ ﺫﻳﻞ ﺑﻨﻮﻳﺴﻴﻢ:
ρg
+
] ∇P − [∇.τ
۸۲
−
]
[
− ∇.ρV V
φ = ρV V + ρδ + τ
∂
= ρV
⇒
∂t
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
) [∇.ρVV ] = [ρV .∇V ] + V (∇.ρV
∂ρ
) = −(∇.ρV
∂ρ
∂t
⇒ [∇.ρVV ] = [ρV .∇V ] − V
∂t
v1w1 v1w 2 v1w3
vw = v 2 w1 v 2 w 2 v 2 w 3
v3 w1 v3 w 2 v3 w 3
dyadic product vw :
]
[
∂
∂ρ
⋅⋅⋅
ρV = − ρV .∇V + V
∂t
∂t
[
]
∂V
∂ρ
∂ρ
+V
+ ρV .∇V − V
⋅⋅⋅ =
∂t
∂t
∂t
∂ D
∇= + V .
Dt ∂t
]
3.3
[
⇒ρ
DV ∂V
=
+ V .∇V
Dt
∂t
DV
∂V
=ρ
+ ρV .∇V
Dt
∂t
⇒ρ
DV
= −∇P − [∇.τ ] + ρ g
Dt
⇒ρ
ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻐﻴﻴﺮ ﺑﺮﺍﻱ ﺣﻞ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺟﺮﻳﺎﻥ
ﺑﺮﺍﻱ ﺗﻮﺻﻴﻒ ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ ﺩﺭ ﺩﻣﺎﻱ ﺛﺎﺑﺖ ،ﺑﻪ ﻃﻮﺭ ﻛﻠﻲ ﺑﻪ ﺍﻃﻼﻋﺎﺕ ﺯﻳﺮ ﻧﻴﺎﺯ ﺩﺍﺭﻳﻢ :
(1ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ
(2ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ
(3ﻣﺆﻟﻔﻪ ﻫﺎﻱ τ
) p = p (ρ
(4ﻣﻌﺎﺩﻟﻪ ﺣﺎﻟﺖ
(5ﻣﻌﺎﺩﻻﺕ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ) µ = µ (ρ ) ، κ = κ (ρ
ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻫﺎ ،ﻫﻤﺮﺍﻩ ﺑﺎ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻭ ﺍﻭﻟﻴﻪ ﻱ ﻻﺯﻡ ،ﺑﻪ ﻃﻮﺭ ﻛﺎﻣﻞ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ،ﭼﮕﺎﻟﻲ ،ﻭ ﺳﺮﻋﺖ ﺩﺭ ﺳﻴﺎﻝ ﺭﺍ
ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺩﻫﻨﺪ .ﺁﻥ ﻫﺎ ﺑﻪ ﻧﺪﺭﺕ ﺩﺭ ﺷﻜﻞ ﻛﺎﻣﻞ ﺧﻮﺩ ﺑﺮﺍﻱ ﺣﻞ ﻣﺴﺌﻠﻪ ﻫﺎﻱ ﺩﻳﻨﺎﻣﻴﻚ ﺳﻴﺎﻝ ﺑﻪ ﻛﺎﺭ ﻣﻲ
ﺭﻭﻧﺪ .ﻣﻌﻤﻮﻻً ﺩﺭ ﺍﻳﻦ ﻓﺼﻞ ،ﺑﺮﺍﻱ ﺳﻬﻮﻟﺖ ﻛﺎﺭ ،ﺍﺯ ﺻﻮﺭﺕ ﻫﺎﻱ ﻣﺤﺪﻭﺩ ﺷﺪﻩ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ.
ﺍﮔﺮ ﺑﺘﻮﺍﻥ ﭼﮕﺎﻟﻲ ﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺭﺍ ﺛﺎﺑﺖ ﻓﺮﺽ ﻛﺮﺩ ،ﺁﻧﮕﺎﻩ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺯﻳﺮ:
ﻣﻌﺎﺩﻟﻪ ﻱ) ( 4-1.3ﻭ ﺟﺪﻭﻝ ﺏ4.
ﻣﻌﺎﺩﻟﻪ ﻱ ﭘﻴﻮﺳﺘﮕﻲ
ﻣﻌﺎﺩﻟﻪ ﻱ ﻧﺎﻭﻳﻪ-ﺍﺳﺘﻮﻛﺲ ﻣﻌﺎﺩﻟﻪ ﻱ) (6-5.3ﻭ ﺟﺪﻭﻝ ﻫﺎﻱ ﺏ ،5،6.ﻭ 7
ﻫﻤﺮﺍﻩ ﺑﺎ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻭ ﻣﺮﺯﻱ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ .ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻳﻦ ﻫﺎ ﻣﻲ ﺗﻮﺍﻥ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﻭ ﺳﺮﻋﺖ ﺭﺍ ﺑﻪ ﺩﺳﺖ
ﺁﻭﺭﺩ.
۸۳
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﻣﺜﺎﻝ :ﺩﺭ ﻳﻚ ﻟﻮﻟﻪ ﺑﻠﻨﺪ ﺩﺭ ﺣﺎﻟﺖ ﭘﺎﻳﺪﺍﺭ ﻳﻚ ﺳﻴﺎﻝ ﺩﺭ ﺣﺎﻟﺖ ﺁﺭﺍﻡ ﺟﺎﺭﻱ ﺍﺳﺖ ﺑﺎ ﺗﻮﺟﻪ ﻣﻌﺎﺩﻻﺕ ﭘﻴﻮﺳﺘﮕﻲ ﻭ
ﺣﺮﻛﺖ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
ﻓﺮﺿﻴﺎﺕ ρ , µ (1 :ﺛﺎﺑﺖ
vz (2ﻓﻘﻂ ﺗﺎﺑﻌﻲ ﺍﺯ r
symmetry (3
ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ :
∂ ∂ρ 1
(ρrvr ) + 1 ∂ (ρvθ ) + ∂ (ρvz ) = 0
+
∂t r ∂r
∂z
r ∂θ
↵0
↵0
↵0
s.s.
vr = 0
vθ = 0
∂ 2vz
=0
∂z 2
ﻣﻌﺎﺩﻟﻪ ﻱ zﺣﺮﻛﺖ :
⇒
∂v z
=0
∂z
⇒
∂v v ∂v
∂ρ
∂v
∂v
ρ z + vr z + θ z + v z z = −
r ∂θ
∂r
∂z
∂z
∂t
1 ∂ ∂v z 1 ∂ 2v z ∂ 2v z
+ µ
+ 2
r
+ 2
2
r
r
r
r
∂
∂
∂
∂z
θ
∂ρ
1 ∂ ∂v z
+µ
r
r ∂r ∂r
∂z
0=−
∂ρ
=0
∂r
ρﻓﻘﻂ ﺗﺎﺑﻊ zﺍﺳﺖ.
⇒
∂ρ
θ - equation of motion ⇒ −
=0
∂θ
∂ρ
1 ∂ ∂v z
⇒
= µ
r
r ∂r ∂r
∂z
ﻓﻘﻂ ﺗﺎﺑﻊ z
ﻓﻘﻂ ﺗﺎﺑﻊ r
r - equation of motion ⇒ −
dρ
1 d dv z
r
= = c0
r dr dr
dz
= c0 z + c1
c 2
= c0 ⇒ v z = 0 r + c2 Lnr + c3
4µ
ﭼﻬﺎﺭ ﺛﺎﺑﺖ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ :
۸٤
⇒µ
dρ
= c0 ⇒ ρ
dz
1 d dv z
µ
r
r dr dr
ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ۳ﻓﺼﻞ
B.C.1 : at
z=0
B.C.2 : at
z=L
ρ = ρ0
ρ = ρL
B.C.3 : at r = R
vz = 0
B.C.4 : at r = 0
v z = finite
ρ 0 = 0 + c1 ⇒ c1 = ρ 0
ρ L = c0 L + c1 ⇒ c0 =
⇒ ρ = ρ0 −
0=
ρ0 − ρ L
L
ρ L − ρ0
L
⋅z
c0 2
R + c2 LnR + c3
4µ
dv z
dr
r =0
c3 = −
vz =
=0⇒
dv z c0
c
r+ 2
=
dr 2 µ
r
⇒ c2 = 0 ,
c0 2
R
4µ
c0 2 c0 2 c0 R 2
r −
R =
4µ
4µ
4µ
⇒ vz = −
⇒ vz =
r2
R 2 − 1
ρ0 − ρ L R 2 r 2
⋅
−1
L
4 µ R 2
(ρ 0 − ρ L )R 2 1 − r 2
4 µL
R
ﻓﻴﻠﻢ ﺭﻳﺰﺍﻥ ﺑﺎ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﻣﺘﻐﻴﺮ:ﻣﺜﺎﻝ
. ﻛﻨﻴﺪset up ﻣﺴﺄﻟﻪ ﺭﺍτ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺑﺮ ﺣﺴﺐ
vx = v y = 0 ,
τ xz = τ zx = − µ
:ﻓﺮﺿﻴﺎﺕ
( ﺣﺎﻟﺖ ﭘﺎﻳﺎ1
( ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ2
v z = v z ( x ) (3
: ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯτ ﺗﻨﻬﺎ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﻏﻴﺮ ﺻﻔﺮ
dv z
dx
۸٥
ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ۳ﻓﺼﻞ
: ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯτ ﻣﮋﻟﻔﻪ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺑﺮ ﺣﺴﺐ
∂v x
∂v
∂v
∂v
∂P
+ v x x + v y x + v z x = −
∂x
∂y
∂z
∂x
∂t
∂τ
∂τ
∂τ
x ﻣﺆﻟﻔﻪ: − xx + yx + zx + ρg x
∂y
∂z
∂x
∂P
⇒0=−
+ ρg sin β
∂x
ρ
y
⇒ ﻣﺆﻟﻔﻪ0 = −
:
∂P
∂y
∂v
∂v
∂P
∂v z
+ vx z + v y z = −
∂x
∂z
∂z
∂t
∂τ yz ∂τ zz
∂τ
+ ρg z
+
z ﻣﺆﻟﻔﻪ: − xz +
∂y
∂z
∂x
∂P dτ xz
⇒0=−
−
+ ρg cos β
∂z
dx
ρ
0=−
∂P
+ ρg sin β ⇒ P = ρg sin β .x + f ( y, z )
∂x
∂P
= 0 ⇒ f ( y, z )
∂y
at x = 0
. ﺍﺳﺖz ﻓﻘﻂ ﺗﺎﺑﻊ
P = Patm ⇒ Patm = 0 + f (z )
⇒ f (z ) = Patm
⇒ P = ρg sin β .x + Patm
⇒
∂P
=0
∂z
۸٦
.
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
d
(τ xz ) = ρg cos β
dx
-ﺑﻘﻴﻪ ﺣﻞ ﻫﻤﺎﻧﻨﺪ ﻗﺒﻞ ﺍﺳﺖ) .ﺩﺭ ﻓﺼﻞ (2
⇒
ﻣﺜﺎﻝ :ﻳﻚ ﺳﻴﺎﻝ ﻏﻴﺮ ﻗﺎﺑﻞ ﺗﺮﺍﻛﻢ ﺑﻪ ﻓﺮﻡ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻫﻢ ﻣﺤﻮﺭ ﻗﺮﺍﺭ ﺩﺍﺭﺩ ﻛﻪ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ
ﺳﺎﻛﻦ ﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ ﺑﺎ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ Ω 0ﻣﻲ ﭼﺮﺧﺪ .ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻭ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺑﺮﺷﻲ ﺭﺍ ﺑﺮﺍﻱ ﺳﻴﺎﻝ
ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
ﺩﺭ ﺑﺨﺶ ﺣﻠﻘﻮﻱ ﻣﻮﺭﺩ ﺑﺮﺭﺳﻲ ،ﺳﻴﺎﻝ ﺣﺮﻛﺖ ﺩﺍﻳﺮﻩ ﺍﻱ ﺩﺍﺭﺩ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺭ ﺣﺎﻟﺖ ﭘﺎﻳﺪﺍﺭ ﻭ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺳﻴﺎﻝ ﺩﺭ
ﻳﻚ ﻣﺴﻴﺮ ﺩﺍﻳﺮﻩ ﺍﻱ ﻣﻲ ﭼﺮﺧﺪ.
ﻣﻲ ﺗﻮﺍﻥ ﻓﺮﺽ ﻛﺮﺩ :
) , P = P(r , z
vθ = vθ (r ) , vr = 0 , v z = 0
pﺑﻪ ﻋﻠﺖ ﮔﺮﺍﻧﺶ ﺑﻪ zﻭ ﺑﻪ ﻋﻠﺖ ﻧﻴﺮﻭﻱ ﮔﺮﻳﺰﺍﺯﻣﺮﻛﺰ ﺑﻪ rﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ.
ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ :
∂ ∂ρ 1
(ρrvr ) + 1 ∂ (ρvθ ) + ∂ (ρvz ) = 0
+
∂t r ∂r
r ∂θ
∂z
∂v
) → θ = 0 ⇒ vθ = vθ (r
∂θ
۸۷
ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ۳ﻓﺼﻞ
: r ﻣﺆﻟﻔﻪ
∂vr
ρ
∂t
2
∂vr vθ ∂vr vθ
∂v
∂P
+
−
+ v z r = −
∂r
∂z
r
∂r
r ∂θ
+ vr
2
2
∂ 1 ∂
(rvr ) + 12 ∂ v2r − 22 ∂vθ + ∂ v2r + ρg r
+ µ
r ∂θ
∂z
r ∂θ
∂r r ∂r
. ﺍﻳﺠﺎﺩ ﻱ ﺷﻮﺩr ﻧﻴﺮﻭﻱ ﺩﺭ ﺟﻬﺖθ ﺑﻪ ﻋﻠﺖ ﺣﺮﻛﺖ ﺳﻴﺎﻝ ﺩﺭ ﺟﻬﺖ
v
∂P
⇒ −ρ θ =−
r
∂r
2
Centrifugal force
: θ ﻣﺆﻟﻔﻪ
∂v v ∂v v v
1 ∂P
∂v
∂v
ρ θ + vr θ + θ θ + r θ + v z z = −
∂r
r ∂θ
r
∂z
r ∂θ
∂t
2
2
∂ 1 ∂
(rvθ ) + 12 ∂ v2θ + 22 ∂vr + ∂ v2θ + ρgθ
+ µ
r ∂θ ∂z
r ∂θ
∂r r ∂r
⇒0=
d 1 d
(rvθ )
dr r dr
: z ﻣﺆﻟﻔﻪ
∂v v ∂v
∂v
∂P
∂v z
+ vr z + θ z + v z z = −
∂r
r ∂θ
∂z
∂z
∂t
ρ
1 ∂ ∂v z 1 ∂ 2 v z ∂ 2 v z
+ µ
+ 2 + ρg z
r
+ 2
2
∂z
r ∂r ∂r r ∂θ
∂P
⇒0=−
+ ρg
∂z
ﻭ ﻣﻌﺎﺩﻟﻪ ﺳﻮﻡ ﺍﺛﺮ ﮔﺮﺍﻧﺶ ﺭﺍ ﺑﺮ.ﻣﻌﺎﺩﻟﻪ ﺍﻭﻝ ﺑﻪ ﻣﺎ ﻣﻲ ﮔﻮﻳﺪ ﻛﻪ ﻧﻴﺮﻭﻱ ﮔﺮﻳﺰﺍﺯﻣﺮﻛﺰ ﭼﻪ ﺍﺛﺮﻱ ﺭﻭﻱ ﻓﺸﺎﺭﺩﺍﺭﺩ
. ﻭ ﻣﻌﺎﺩﻟﻪ ﺩﻭﻡ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ،ﻓﺸﺎﺭ )ﺍﺛﺮ ﻫﻴﺪﺭﻭﺳﺘﺎﺗﻴﻜﻲ( ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ
d 1 d
(rvθ ) = 0 ⇒ 1 d (rvθ ) = c1
dr r dr
r dr
d
1
⇒ (rvθ ) = c1r ⇒ rvθ = c1r 2 + c2
dr
2
1
c
⇒ vθ = c1r + 2
2
r
۸۸
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ،ﻫﻤﺎﻥ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻣﺮﺑﻮﻁ ﺑﻪ ﺳﻴﺎﻟﻲ ﺍﺳﺖ ﻛﻪ ﺭﻭﻱ ﺩﻭ ﺳﻄﺢ ﺍﺳﺘﻮﺍﻧﻪ ﻧﻤﻲ ﻟﻐﺰﺩ.
vθ = 0
r = kR
B.C.1 : at
vθ = Ω 0 R
r=R
B.C.2 : at
c
1
c1 (kR ) + 2
kR
2
c
1
B.C.2 ⇒ Ω 0 R = c1 R + 2
R
2
r kR
−
kR r
⇒ vθ = Ω 0 R
1
−k
k
= B.C.1 ⇒ 0
ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻣﻲ ﺗﻮﺍﻥ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺟﺪﻭﻝ ﺏ 1.ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ :
∂ vθ 1 ∂vr
+
∂r r r ∂θ
τ rθ = − µ r
∂ vθ
∂r r
⇒ τ rθ = − µr
r kR
−
Ω 0 R
∂
kR r
= − µr
∂r
1
r0 − k
k
⇒ τ rθ
2
1 k
= −2 µΩ 0 R 2 2
2
r 1 − k
ﮔﺸﺘﺎﻭﺭ ﻭﺍﺭﺩ ﺑﺮ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ ﺍﺯ ﺿﺮﺏ ﺷﺎﺭ ﺍﻧﺪﺍﺯﻩ ﺣﺮﻛﺖ ﺭﻭﺑﻪ ﺩﺍﺧﻞ ) ، (− τ rθﺳﻄﺢ ﺍﺳﺘﻮﺍﻧﻪ ،ﻭ ﺑﺎﺯﻭﻱ ﺍﻫﺮﻡ،
ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ،ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺩﺭ ﺍﻳﻦ ﺳﻴﺴﺘﻢ:
k2
Tz = (− τ rθ )r =kR ⋅ 2πkRL ⋅ kR = 4πµΩ 0 R 2 L
2
1− k
2
(Re )crit . = Ω 0 R ρ
µ
ﻣﻲ ﺗﻮﺍﻥ ﺗﻌﻴﻴﻦ ﻛﺮﺩ ﻛﻪ ﭼﻪ ﻣﻮﻗﻊ ﺟﺮﻳﺎﻥ ﺑﻪ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ ﻭ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺑﻪ ﻃﻮﺭ ﻗﻮﻱ ﻭﺍﺑﺴﺘﻪ ﺑﻪ
R − kR
)ﺷﻌﺎﻉ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ() /ﺿﺨﺎﻣﺖ ﺁﻧﻮﻟﻮﺱ( =
R
ﻳﺎ ) (1 − kﺍﺳﺖ.
ﺑﺮﺍﻱ ﺯﻣﺎﻧﻲ ﻛﻪ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ ﺳﺎﻛﻦ ﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ ﻣﻲ ﭼﺮﺧﺪ .ﻣﻴﻨﻴﻤﻢ ﻣﻘﺪﺍﺭ ﺭﻳﻨﻮﻟﺪﺯ ﺑﺤﺮﺍﻧﻲ ﺩﺭ 50000
ﺑﺮﺍﻱ 1 − kﺑﺮﺍﺑﺮ ﺑﺎ 0.05ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ.
۸۹
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﺍﮔﺮ ﺳﻴﻠﻨﺪﺭ ﺧﺎﺭﺟﻲ ﺳﺎﻛﻦ ﻭ ﺳﻴﻠﻨﺪﺭ ﺩﺍﺧﻠﻲ ﺑﺎ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ Ωiﺑﭽﺮﺧﺪ ،ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺑﻪ ﻓﺮﻡ ﺫﻳﻞ ﺑﻪ
ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ :
R r
−
vθ = Ω i kR r R
1
−k
k
ﻭ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ:
41.3
(1 − k ) 2
3
41.3µ
, k ≈1
2
3
) ρR 2 (1 − k
≈
ρ
2
(Re )crit = Ωi kR
µ
41.3µ
=
2
3
) ρR 2 k (1 − k
≈ ⇒ Ω i ,crit
ﻣﺜﺎﻝ :ﻳﻚ ﺳﻴﺎﻝ ﺑﺎ ﺩﺍﻧﺴﻴﺘﻪ ﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺛﺎﺑﺖ ﺩﺭ ﻳﻚ ﻇﺮﻑ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺑﻪ ﺷﻌﺎﻉ Rﻣﺎﻧﻨﺪ ﺷﻜﻞ ﻗﺮﺍﺭ ﺩﺍﺭﺩ.
ﻇﺮﻑ ﺑﺎ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ Ωﺣﻮﻝ ﻣﺤﻮﺭﺵ ﻣﻲ ﭼﺮﺧﺪ .ﻣﺤﻮﺭ ﺍﺳﺘﻮﺍﻧﻪ ﻋﻤﻮﺩﻱ ﺍﺳﺖ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ
g r = gθ = 0ﻭ . g z = − gﺑﺎ ﻓﺮﺽ ﺣﺎﻟﺖ ﭘﺎﻳﺪﺍﺭ ﺷﻜﻞ ﺳﻄﺢ ﺁﺯﺍﺩ ﻣﺎﻳﻊ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
۹۰
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﻣﻨﺎﺳﺐ ﺍﺳﺖ.ﺩﺭ ﺣﺎﻟﺖ ﭘﺎﻳﺎ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ υ rﻭ υ zﻫﺮ ﺩﻭ
ﺻﻔﺮﻧﺪ ﻭ υθﻓﻘﻂ ﺑﻪ rﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ .ﺑﻪ ﻋﻼﻭﻩ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ ، Pﺑﻪ ﻋﻠﺖ ﻧﻴﺮﻭﻱ ﮔﺮﺍﻧﺶ ،ﺑﻪ zﻭ ﺑﻪ ﻋﻠﺖ
ﻧﻴﺮﻭﻱ ﮔﺮﻳﺰﺍﺯﻣﺮﻛﺰ ،ﺑﻪ rﻭﺍﺑﺴﺘﻪ ﺍﺳﺖ،ﺍﻣﺎ ﺑﻪ θﻭﺍﺑﺴﺘﻪ ﻧﻴﺴﺖ.
vz = 0
vr = 0
) vθ = vθ (r
) P = P(r , z
ﺑﺎ ﺍﻳﻦ ﻓﺮﺽ ﻫﺎ ،ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ ، 0 = 0ﻭ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺩﺍﺭﻳﻢ :
v
∂P
فلؤم r -
= :ρ θه
r
∂r
∂ ∂ 1
(rvθ )
فلؤم θ −
: 0=µ ه
∂r r ∂r
∂P
فلؤم z -
: 0=−ه
− ρg
∂z
cr c
⇒ vθ = 1 + 2
2
r
2
ﺍﺯ ﻣﺆﻟﻔﻪ θﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺍﻧﺘﮕﺮﺍﻝ ﻣﻲ ﮔﻴﺮﻳﻢ ﻭ υθﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ.ﻛﻪ ﺩﺭ ﺁﻥ c1ﻭ c2ﺛﺎﺑﺖ ﻫﺎﻱ ﺍﻧﺘﮕﺮﺍﻝ
ﮔﻴﺮﻱ ﺍﻧﺪ .ﭼﻮﻥ υθﻧﻤﻲ ﺗﻮﺍﻧﺪ ﺩﺭ r = 0ﻧﺎﻣﺘﻨﺎﻫﻲ ﺷﻮﺩ ،ﺛﺎﺑﺖ c2ﺑﺎﻳﺪ ﺻﻔﺮ ﺑﺎﺷﺪ.
ﻣﻲ ﺗﻮﺍﻧﻴﻢ ﻓﺮﺽ ﻛﻨﻴﻢ ﻛﻪ ﻣﺎﻳﻊ ﺑﻪ ﺻﻮﺭﺕ ﺟﺴﻢ ﺻﻠﺐ ﻣﻲ ﭼﺮﺧﺪ :
=0
∂ vθ
∂r r
τ rθ = − µr
1
⇒ vθ = c1r , at r = R ⇒ vθ = RΩ
2
1
⇒ RΩ = c1 R ⇒ c1 = 2Ω
2
1
⇒ vθ = × 2Ω × r ⇒ vθ = Ωr
2
ﺍﻳﻦ ﺑﻴﺎﻥ ﻣﻲ ﺩﺍﺭﺩ ﻛﻪ ﻫﺮ ﺟﺰ ﺳﻴﺎﻝ ﺩﺭ ﺣﺎﻝ ﭼﺮﺧﺶ ﻣﺎﻧﻨﺪ ﻳﻚ ﺟﺴﻢ ﺟﺎﻣﺪ ﻋﻤﻞ ﻣﻲ ﻛﻨﺪ.
∂P
2
∂r = ρΩ r
⇒
هفلؤم z -
∂P = − ρg
∂z
∂P
∂P
dz
dr +
= P = P(r , z ) ⇒ dP
∂z
∂r
هفلؤم r -
1
∫
P = − ρgz + ρΩ 2 r 2 + c
→⇒ dP = ρΩ 2 rdr − ρgdz
2
۹۱
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
⇒ P = Patm
r =0
z = z1
,
at
⇒ Patm = − ρgz1 + 0 + c ⇒ c = Patm + ρgz1
r2
r2
ρΩ 2
2
ρΩ 2
2
⇒ P − Patm = − ρg (z − z1 ) +
if P = Patm ⇒ 0 = − ρg ( z − z1 ) +
ﻣﻌﺎﺩﻟﻪ ﺳﻄﺢ ﺁﺯﺍﺩ ﻣﺎﻳﻊ )ﺳﻬﻤﻲ(
2
r
Ω2
⇒ z − z1 =
2g
ﻣﺜﺎﻝ :ﻛﺮﻩ ﻱ ﺗﻮﭘﺮﻱ ﺑﻪ ﺷﻌﺎﻉ Rﺑﺎ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ Ωﺩﺭ ﺣﺠﻢ ﺯﻳﺎﺩﻱ ﺍﺯ ﻳﻚ ﺳﻴﺎﻝ ﺳﺎﻛﻦ ﺑﻪ ﺁﻫﺴﺘﮕﻲ ﻣﻲ
ﭼﺮﺧﺪ)ﺷﻜﻞ ﺭﺍ ﺑﺒﻴﻨﻴﺪ( ﻋﺒﺎﺭﺕ ﻫﺎﻳﻲ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﻭ ﺳﺮﻋﺖ ﺩﺭ ﺳﻴﺎﻝ ﻭ ﮔﺸﺘﺎﻭﺭ Tzﻻﺯﻡ ﺑﺮﺍﻱ ﺗﺪﺍﻭﻡ ﺣﺮﻛﺖ
ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ .ﻓﺮﺽ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﻛﺮﻩ ﺑﻪ ﺍﻧﺪﺍﺯﻩ ﻛﺎﻓﻲ ﺁﻫﺴﺘﻪ ﻣﻲ ﭼﺮﺧﺪ ﻭ ﻣﻴﺘﻮﺍﻥ ﺍﺯ ﺣﺎﻟﺖ ﻣﺮﺑﻮﻁ ﺑﻪ ﺟﺮﻳﺎﻥ
ﺧﺰﺷﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ .ﺍﻳﻦ ﻣﺜﺎﻝ ﻧﺤﻮﻩ ﻱ ﻧﻮﺷﺘﻦ ﻭ ﺣﻞ ﻣﺴﺌﻠﻪ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺮﻭﻱ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ.
ﻓﺮﺿﻴﺎﺕ:
(1ﺟﺮﻳﺎﻥ ﺧﺰﺷﻲ ﭘﺎﻳﺎ
) V = δ φ vφ (r ,θ ) , vr = 0 , vθ = 0(2ﺻﻮﺭﺕ ﻛﻠﻲ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ(
) P = P(r ,θ )(3ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ(
(4ﺟﺮﻳﺎﻥ ﺣﻮﻝ ﻣﺤﻮﺭ zﻣﺘﻘﺎﺭﻥ ﺍﺳﺖ ﻟﺬﺍ ﻭﺍﺑﺴﺘﮕﻲ ﺑﻪ ﺯﺍﻭﻳﻪ φﻭﺟﻮﺩ ﻧﺨﻮﺍﻫﺪ ﺩﺍﺷﺖ.
ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ :
∂ ∂ρ 1
1
∂
(ρvθ sin θ ) + 1 ∂ (ρvφ ) = 0
+ 2
ρr 2 v r +
∂t r ∂r
r sin θ ∂θ
r sin θ ∂φ
)
۹۲
(
ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ۳ﻓﺼﻞ
⇒
∂vφ
∂φ
. ﺍﺳﺖ، ﻣﺘﻘﺎﺭﻥz ﻧﺴﺒﺖ ﺑﻪ ﻣﺤﻮﺭ
= 0 ⇒ vφ = vφ (r , θ )
: ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﻱ ﺣﺮﻛﺖ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺧﺰﺷﻲ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ
: r ﻣﺆﻟﻔﻪ
2
2
∂vr
vφ
∂vr vθ ∂vr
∂vr vθ + vφ
∂P
+ vr
+
+
⋅
−
=−
ρ
∂t
∂r
∂r
r ∂θ r sin θ ∂φ
r
1 ∂2
∂
∂vr
∂ 2 vr
1
1
+ µ 2 2 r 2 vr + 2
+
sin
θ
∂θ r 2 sin 2 θ ∂φ 2
r sin θ ∂θ
r ∂r
∂P
⇒0=−
vφ = 0
creeping flow
∂r
(
)
: θ ﻣﺆﻟﻔﻪ
2
∂vθ
vφ ∂vθ vr vθ vφ cot θ
∂vθ vθ ∂vθ
+ vr
+
+
+
−
ρ
∂t
r
r
∂
∂
r
∂
r
r
sin
θ
θ
φ
1 ∂ 2 ∂vθ 1 ∂ 1
∂ 2 vφ 2 ∂vr
∂
1 ∂P
1
2 cos θ ∂vφ
v
+ µ 2
⋅
(
)
+
+ 2
− 2 2
sin
θ
r
+ 2
θ
2
2
2
r ∂θ
r ∂θ r sin θ ∂φ
r sin θ ∂φ
r ∂r ∂r r ∂θ sin θ ∂θ
1 ∂P
⇒0=−
r ∂θ
=−
: φ ﻣﺆﻟﻔﻪ
∂vφ vθ ∂vφ
vφ ∂vφ vφ v r vθ vφ
∂vφ
1 ∂P
+ vr
+
+
+
+
cot θ = −
∂r
r ∂θ r sin θ ∂φ
r
r
r sin θ ∂φ
∂t
ρ
1 ∂ 2 ∂vφ
r
+ µ 2
r ∂r ∂r
0=
∂ 2 vφ
1 ∂ 1 ∂
1
2 ∂v r
2 cos θ ∂vθ
+ 2
(vφ sin θ ) + 2 2
+ 2
+ 2
+ ρg φ
2
r sin θ ∂φ r sin 2 θ ∂φ
r sin θ ∂φ
r ∂θ sin θ ∂θ
1 ∂ 2 ∂vφ 1 ∂ 1 ∂
r
+ 2
(vφ sin θ )
2
r ∂r ∂r r ∂θ sin θ ∂θ
(1)
: ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺧﻼﺻﻪ ﻛﺮﺩ
۹۳
ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ۳ﻓﺼﻞ
B.C.1 :
at
r = R , vr = 0 , vθ = 0 , vφ = (R sin θ )Ω
B.C.2 :
at
r →∞
vr → 0
B.C.3 :
at
r →∞
P → P0
, vθ → 0 , vφ → 0
. ﻭ ﺩﻭﺭ ﺍﺯ ﻛﺮﻩ ﺍﺳﺖz = 0 ﻓﺸﺎﺭ ﺳﻴﺎﻝ ﺩﺭ ﻧﻘﻄﻪP0 ﻭP = P + ρgz ﻛﻪ ﺩﺭ ﺁﻥ
: ﺍﺳﺖ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢvφ
vφ = f (r )sin θ
at
= vφ (r , θ )
( ﭼﻮﻥ1) ﺑﺮﺍﻱ ﺣﻞ ﻣﻌﺎﺩﻟﻪ
(2)
.ﻛﻪ ﺍﻳﻦ ﻓﻘﻂ ﻳﻚ ﻓﺮﺽ ﺍﺳﺖ ﻭﻟﻲ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺭﺍ ﺍﺭﺿﺎ ﻣﻲ ﻛﻨﺪ
r = R ⇒ vφ =
f (R )sin θ
RΩ
∂vφ
:( ﺩﺍﺭﻳﻢ1) ( ﺩﺭ ﻣﻌﺎﺩﻟﻪ2) ﺑﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻣﻌﺎﺩﻟﻪ
∂
∂
vφ sin θ ) =
f (r )sin 2 θ = 2 f (r )sin θ ⋅ cos θ
(
∂r
∂θ
∂θ
∂ 1 ∂
(vφ sin θ ) = ∂ 1 ⋅ 2 f sin θ cosθ = 2 f (− sin θ )
∂θ sin θ ∂θ
∂θ sin θ
d df
⇒ r2
sin θ − 2 f sin θ = 0
dr dr
⇒
= f ′ sin θ
(
,
)
d 2 df
r
−2f =0
dr dr
2
d3y
dy
2 d y
ax
+
+ bx + cy = 0
3
2
dx
dx
dx
n3
n1
n2
y = c1 x + c2 x + c3 x
x3
n(n − 1)(n − 2 ) + an(n − 1) + bn + c = 0
2
d 2 df
df
2 d f
r
r
r
=
+
2
dr dr
dr
dr 2
⇒ 2rf ′ + r 2 f ′′ − 2 f = 0 ⇒ r 2
d2 f
df
+ 2r
−2f =0
2
dr
dr
df
d2 f
n −1
f =r ⇒
,
= nr
= n(n − 1)r n − 2
2
dr
dr
n−2
n −1
2
− 2r n = 0
⇒ n(n − 1)r ⋅ r + 2r nr
n
(
n(n − 1)r n + 2nr n − 2r n = 0
)
⇒ n 2 − n + 2n − 2 = 0 ⇒ n 2 + n − 2 = 0
−1+ 3
n=
=1
− 1 ± 1 + 8
2
⇒n=
2
n = − 1 − 3 = −2
2
۹٤
Euler Eq.
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
c2
r2
f = c1 r n1 + c 2 r n2 = c1 r +
c
⇒ vφ (r , θ ) = c1 r + 22 sin θ
r
c
RΩ sin θ = c1 R + 22 sin θ
R
0 = c1 × ∞ + 0 ⇒ c1 = 0
⇒ c2 = R 3Ω
2
R
⇒ vφ (r , θ ) = ΩR sin θ
r
ﮔﺸﺘﺎﻭﺭ ﻻﺯﻡ ﺑﺮﺍﻱ ﺗﺪﺍﻭﻡ ﭼﺮﺧﺶ ﻛﺮﻩ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻣﻲ ﻛﻨﻴﻢ .ﺍﻳﻦ ﮔﺸﺘﺎﻭﺭ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﻧﻴﺮﻭﻱ ﻣﻤﺎﺳﻲ
ﻭﺍﺭﺩ ﺑﺮ ﺳﻴﺎﻝ ﺗﻮﺳﻂ ﺟﺰء ﺳﻄﺢ ﺟﺎﻣﺪ ،ﻛﻪ ﺭﻭﻱ ﺗﻤﺎﻡ ﺳﻄﺢ ﻛﺮﻩ ﻣﺤﺎﺳﺒﻪ ﺷﻮﺩ ،ﺿﺮﺏ ﺩﺭ ﺑﺎﺯﻭﻱ ﺍﻫﺮﻡ ﺑﺮﺍﻱ ﺁﻥ
ﺟﺰء :
, dA = R 2 sin θdθdφ
R sin
θ
∫ (τ φ )dA ⋅
وزاب
⋅ (R sin θ )R 2 sin θdθdφ
)
π
r
r =R
0
π
) ∫ (τ φ
r
0
2π
2π
∫ = Tz
0
∫ = ⇒ Tz
0
∂v
1
∂ v
⋅ r + r φ
∂r r
r sin θ ∂φ
τ rφ = − µ
d v
1
= ΩR 3 sin θ 3 ⇒ φ = ΩR 3 sin θ − 3r −4
r
r dr r
1
d v
⇒ φ = −3ΩR 3 sin θ ⋅ 4
dr r
r
(
vφ
1
1
⇒ τ rφ = − µ − 3ΩR 3 sin θ ⋅ 4 ⋅ r = 3µΩR 3 sin θ ⋅ 3
r
r
⇒ τ rφ r = R = 3µΩ sin θ
2
3
3
3
∫0 (3µΩ sin θ )(R sin θ )(R sin θdθdφ ) = 6πµΩR ∫0 sin θdθ = 8πµΩR
π
π
2π
∫ = Tz
0
ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ،ﺍﻧﺤﺮﺍﻑ ﺍﺯ"ﺟﺮﻳﺎﻥ ﺍﻭﻟﻴﻪ " ﻣﻌﺎﺩﻟﻪ ﻱ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ )ﺩﺭ ﺑﺎﻻ ﻣﺤﺎﺳﺒﻪ ﺷﺪﻩ ﺍﺳﺖ(.
ﺭﺥ ﻣﻲ ﺩﻫﺪ .ﺑﻪ ﻋﻠﺖ ﺁﺛﺎﺭ ﻧﻴﺮﻭﻱ ﮔﺮﻳﺰﺍﺯﻣﺮﻛﺰ،ﺳﻴﺎﻝ ﺑﻪ ﻃﺮﻑ ﻗﻄﺐ ﻫﺎﻱ ﻛﺮﻩ ﻛﺸﻴﺪﻩ ﻣﻲ ﺷﻮﺩ ﻭ ﺩﺭ ﺍﺳﺘﻮﺍ ﺑﻪ
ﻃﺮﻑ ﺑﻴﺮﻭﻥ ﺭﺍﻧﺪﻩ ﻣﻲ ﺷﻮﺩ .ﺑﺮﺍﻱ ﺗﻮﺻﻴﻒ ﺍﻳﻦ "ﺟﺮﻳﺎﻥ ﺛﺎﻧﻮی " ﺑﺎﻳﺪ ﺟﻤﻠﻪ ﻱ V ⋅ ∇Vﺭﺍ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻱ
ﺣﺮﻛﺖ ﺑﮕﻨﺠﺎﻧﻴﻢ.
4.3ﺗﺤﻠﻴﻞ ﺍﺑﻌﺎﺩﻱ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻐﻴﻴﺮ
ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ ﺩﺭ ﻣﻮﺭﺩ ﺟﺮﻳﺎﻥ ﮔﺬﺭﺍ ﺍﺯ ﺳﻴﺴﺘﻤﻲ ﻛﻪ ﻧﻤﻲ ﺗﻮﺍﻥ ﺑﺎ ﺣﻞ ﺗﺤﻠﻴﻠﻲ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻐﻴﻴﺮ ﺁﻥ ﺭﺍ ﺗﺤﻠﻴﻞ
ﻛﺮﺩ ،ﺩﺍﺩﻩ ﻫﺎﻱ ﺗﺠﺮﺑﻲ ﺩﺍﺭﻳﻢ ﻳﺎ ﺍﺯ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﻋﻜﺲ ﮔﺮﻓﺘﻪ ﺍﻳﻢ .ﻣﺜﺎﻟﻲ ﺍﺯ ﺍﻳﻦ ﻧﻮﻉ ﺳﻴﺴﺘﻢ ،ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﮔﺬﺭﺍ ﺍﺯ
۹٥
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
ﺟﺮﻳﺎﻥ ﺳﻨﺞ ﺭﻭﺯﻧﻪ ﺍﻱ ﺗﻌﺒﻴﻪ ﺷﺪﻩ ﺩﺭ ﻳﻚ ﻟﻮﻟﻪ ﺍﺳﺖ )ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺳﻨﺞ ﺍﺯ ﻗﺮﺻﻲ ﺗﺸﻜﻴﻞ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺭﻭﺯﻧﻪ ﺍﻱ
ﺩﺭ ﻣﺮﻛﺰ ﺁﻥ ﺍﻳﺠﺎﺩ ﺷﺪﻩ ﺍﺳﺖ؛ ﺩﻳﺴﻚ ﺭﺍ ﺩﺭ ﻟﻮﻟﻪ ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻨﺪ ﻭ ﺩﺭ ﺑﺎﻻ ﺩﺳﺖ ﻭ ﭘﺎﻳﻴﻦ ﺩﺳﺖ ﺁﻥ ﻭﺳﺎﻳﻞ
ﺣﺴﮕﺮ ﻓﺸﺎﺭ ﻧﺼﺐ ﻣﻲ ﻛﻨﻨﺪ( .ﺣﺎﻝ ﻓﺮﺽ ﻛﻨﻴﺪ ﻛﻪ ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺳﻴﺴﺘﻢ ﺁﺯﻣﺎﻳﺸﻲ ﺭﺍ ﺑﺰﺭﮔﺘﺮ )ﻳﺎ ﻛﻮﭼﻜﺘﺮ(
ﻛﻨﻴﻢ ،ﺗﺎ ﺳﻴﺴﺘﻢ ﺟﺪﻳﺪﻱ ﺑﺴﺎﺯﻳﻢ ﻛﻪ ﺩﻗﻴﻘﺎً ﻫﻤﺎﻥ ﺍﻟﮕﻮﻱ ﺟﺮﻳﺎﻥ ﺩﺭ ﺁﻥ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﺩ )ﺍﻟﺒﺘﻪ ﺑﺎ ﻣﻘﻴﺎﺱ ﺑﺰﺭﮔﺘﺮ
)ﻳﺎ ﻛﻮﭼﻜﺘﺮ(( .ﻗﺒﻞ ﺍﺯ ﻫﺮ ﭼﻴﺰ ،ﺑﻪ ﺗﺸﺎﺑﻪ ﻫﻨﺪﺳﻲ ﻧﻴﺎﺯ ﺩﺍﺭﻳﻢ :ﻳﻌﻨﻲ ﺑﺎﻳﺪ ﻧﺴﺒﺖ ﻫﺎﻱ ﻫﻤﻪ ﺍﺑﻌﺎﺩ ﻟﻮﻟﻪ ﻭ ﺻﻔﺤﻪ
ﺭﻭﺯﻧﻪ ﺩﺍﺭ ﺩﺭ ﺳﻴﺴﺘﻢ ﺍﺻﻠﻲ ﻭ ﺳﻴﺴﺘﻢ ﺑﺰﺭگ)ﻳﺎ ﻛﻮﭼﻚ( ﺷﺪﻩ ﻣﺴﺎﻭﻱ ﺑﺎﺷﺪ .ﺑﻪ ﻋﻼﻭﻩ ﺑﺎﻳﺪ ﺗﺸﺎﺑﻪ ﺩﻳﻨﺎﻣﻴﻜﻲ
ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ :ﻳﻌﻨﻲ ﮔﺮﻭﻩ ﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ )ﻣﺎﻧﻨﺪ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ( ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺑﺎﻳﺪ
ﻫﻤﺎﻧﻨﺪ ﺑﺎﺷﻨﺪ .ﺑﺮﺍﻱ ﺩﺭﻙ ﻫﺮ ﭼﻪ ﺑﻬﺘﺮ ﺗﺸﺎﺑﻪ ﺩﻳﻨﺎﻣﻴﻜﻲ ﺑﺎﻳﺪ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻐﻴﻴﺮ ﺭﺍ ،ﻫﻤﺮﺍﻩ ﺑﺎ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻭ
ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ،ﺑﻪ ﺻﻮﺭﺕ ﺑﺪﻭﻥ ﺑﻌﺪ ،ﻧﻮﺷﺖ.
ﺑﺮﺍﻱ ﺳﺎﺩﻩ ﺗﺮ ﺷﺪﻥ ﻣﻮﺿﻮﻉ ،ﺑﺤﺚ ﺭﺍ ﺑﻪ ﺳﻴﺎﻝ ﻫﺎﻱ ﺑﺎ ﭼﮕﺎﻟﻲ ﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﺛﺎﺑﺖ ﻣﺤﺪﻭﺩ ﻣﻲ ﻛﻨﻴﻢ :
∇ ⋅V = 0
D
ρ V = −∇P + µ∇ 2 V
Dt
ﺩﺭ ﺍﻏﻠﺐ ﺳﻴﺴﺘﻢ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ "ﻋﺎﻣﻞ ﻫﺎﻱ ﻣﻘﻴﺎﺱ" ﺯﻳﺮ ﺭﺍ ﺷﻨﺎﺳﺎﻳﻲ ﻛﺮﺩ :ﻃﻮﻝ ﻣﺸﺨﺼﻪ ، l0ﺳﺮﻋﺖ
ﻣﺸﺨﺼﻪ ، υ 0ﻭ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﻣﺸﺨﺼﻪ ) P0 = p0 + rgh0ﻣﺜﻼً ﺍﻳﻦ ﻋﺎﻣﻞ ﻫﺎ ﻣﻲ ﺗﻮﺍﻧﻨﺪ ﻗﻄﺮ ﻟﻮﻟﻪ ،ﺳﺮﻋﺖ
ﺟﺮﻳﺎﻥ ﻣﺘﻮﺳﻂ ،ﻭ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﺩﺭ ﺧﺮﻭﺟﻲ ﻟﻮﻟﻪ ﺑﺎﺷﻨﺪ (،ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ،ﻣﻲ ﺗﻮﺍﻥ ﻣﺘﻐﻴﺮ ﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﻭ
ﻋﻤﻠﮕﺮ ﻫﺎﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺭﺍ ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻛﺮﺩ.
ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜﺎﻝ ﺑﺮﺍﻱ ﻳﻚ ﻟﻮﻟﻪ :
fube
l0 = D
v0 = vavg
P0 = Pexit of
ﺳﭙﺲ ﻣﺘﻐﻴﺮﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﺭﺍ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ.
P − P0
µv0 l0
x
vt
y
z
= *, y
= *, z
, t* = 0
l0
l0
l0
l0
= *x
P − P0
V
= *, P
v0
ρv0 2
= *V
= ** P
or
ﺩﻭ ﮔﺰﻳﻨﻪ ﺑﺮﺍﻱ ﻓﺸﺎﺭ ﺑﺪﻭﻥ ﺑﻌﺪ ﭘﻴﺸﻨﻬﺎﺩ ﻛﺮﺩﻩ ﺍﻳﻢ؛ ﮔﺰﻳﻨﻪ ﺍﻭﻝ ﺑﺮﺍﻱ ﺍﻋﺪﺍﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺑﺰﺭگ ،ﻭ ﮔﺰﻳﻨﻪ ﺩﻭﻡ ﺑﺮﺍﻱ
ﺍﻋﺪﺍﺩ ﺭﻳﻨﻮﻟﺪﺯ ﻛﻮﭼﻚ ﻣﻨﺎﺳﺐ ﺍﺳﺖ.
۹٦
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
∂
∂
∂
* +δy * +δz
*
∂x
∂y
∂z
∂2
2
* ∂z
+
∂2
2
* ∂y
+
∇ * = l0 ∇ = δ x
∂2
2
*∂x
2
∇ * = l0 ∇ 2
2
D l0 D
=
Dt * v0 Dt
∂v x ∂v y ∂v z
+
+
∂x ∂y
∂z
= ∇ ⋅V = 0 ⇒ ∇ ⋅V
∂v
∂v
v
∂x
* ⋅ v x = x ⇒ x* = x
∂x
∂x ∂x
v0
*
*
*
∂v x
∂x* 1
1 ∂v
,
=
= ⋅ x
∂x l0
∂x
v0 ∂x
*
∂v x
∂v
l ∂v
v ∂v
*= 0 x ⇒ x = 0 x
*
∂x
∂x l0 ∂x
v0 ∂x
*
*
⇒
*
v ∂v
∂v z v0 ∂v z
⇒
= 0 y* ,
=
∂y l0 ∂y
* ∂z l0 ∂z
*
∂v y
*
*
*
v0 ∂v x ∂v y ∂v z v0
= ∇ ⋅V
* + * + * = ∇ ⋅V
*
l0 ∂x
∂y
∂z l0
⇒ ∇ ⋅V = 0 ⇒ ∇ ⋅V * = 0
D
ρ V = −∇P + µ∇ 2 V
Dt
2
l0 v0 ρ
µ
= Re
v D
1
*∇
* ρ 0 * V *v0 = − ∇* ρv0 2 P* + µ ⋅ 2 v0 V
l0 Dt
l0
l0
2
D
1
* ⋅ ∇* ⋅ V
1 ⇒ * V * = −∇* P* +
Dt
Re
ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﺍﺯ ** Pﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﻢ ﺩﺍﺭﻳﻢ:
inertial forces
viscous forces
1
* 1 *2
* D
V =−
⋅ ∇ * P ** +
∇ V
*
Re
Re
Dt
= Re
⇒2
ﺍﻧﺪﺍﺯﻩ ﺍﻳﻦ ﮔﺮﻭﻩ ﺑﺪﻭﻥ ﺑﻌﺪ ﻧﺸﺎﻧﻪ ﺍﻱ ﺍﺯ ﺍﻫﻤﻴﺖ ﻧﺴﺒﻲ ﻧﻴﺮﻭ ﻫﺎﻱ ﻟﺨﺘﻲ ﻭ ﻭﻳﺴﻜﻮﺯ ﺩﺭ ﺳﻴﺴﺘﻢ ﺳﻴﺎﻝ ﺍﺳﺖ.
* D
V = −∇* P* Euler Eq.
*
Dt
1
* 1 *2
2) Re 1 ⇒ 2 ⇒ 0 = −
⋅ ∇* P** +
∇ ⋅ V creeping flow
Re
Re
⇒ 1) Re → ∞ ⇒ 1
ﮔﺮﻭﻩ ﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﺩﻳﮕﺮﻱ ﺩﺭ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﻧﺪ؛ ﺩﻭ ﻣﻮﺭﺩ ﺍﺯ ﺍﻳﻦ ﮔﺮﻭﻩ ﻫﺎ ﻛﻪ ﺩﺭ
ﻣﺴﺌﻠﻪ ﻫﺎﻳﻲ ﻣﻄﺮﺡ ﻣﻲ ﺷﻮﻧﺪ ﻛﻪ ﺑﺎ ﻓﺼﻞ ﻣﺸﺘﺮﻙ ﺳﻴﺎﻝ -ﺳﻴﺎﻝ ﺳﺮﻭﻛﺎﺭ ﺩﺍﺭﻧﺪ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ :
۹۷
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
2
v
Fr = 0 = Froude number
l0 g
= Weber number
σ
l0 v0 ρ
2
= We
ﻣﺜﺎﻝ :ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺟﺮﻳﺎﻥ ﻳﻚ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻮﻧﻲ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺭﺍ ﻛﻪ ﺍﺯ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﻣﺪﻭﺭ ﻣﻲ ﮔﺬﺭﺩ ،ﺑﻪ ﺻﻮﺭﺕ
ﺗﺠﺮﺑﻲ ﺑﺮﺭﺳﻲ ﻛﻨﻴﻢ .ﻫﺪﻑ ﺍﻃﻼﻉ ﺍﺯ ﻧﺤﻮﻩ ﻱ ﻭﺍﺑﺴﺘﮕﻲ ﺍﻟﮕﻮﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻭ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﺑﻪ ﻗﻄﺮ ﺍﺳﺘﻮﺍﻧﻪ ،ﻃﻮﻝ،
ﺳﺮﻋﺖ ﻧﺰﺩﻳﻚ ﺷﺪﻥ ،ﭼﮕﺎﻟﻲ ،ﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻴﺎﻝ ﺍﺳﺖ.ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺍﻳﻦ ﺑﺮﺭﺳﻲ ﺭﺍ ﭼﮕﻮﻧﻪ ﺑﺎﻳﺪ ﺍﻧﺠﺎﻡ ﺩﺍﺩ ﺗﺎ
ﺗﻌﺪﺍﺩ ﺁﺯﻣﺎﻳﺶ ﻫﺎﻱ ﻻﺯﻡ ﺑﻪ ﺣﺪﺍﻗﻞ ﻣﻤﻜﻦ ﺑﺮﺳﺪ.
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﻛﻪ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﻮﺻﻴﻒ ﻣﻲ ﻛﻨﻨﺪ ،ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﻭ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ :
(∇ ⋅V ) = 0
D
V = −∇P + µ∇ 2 V
Dt
ﺷﺮﻁ ﺍﻭﻟﻴﻪ ﺑﺮﺍﻱ t = 0ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :
1
∞L ⇒ V = δ x v
2
z
or if
2
4
ρ
if x 2 + y 2 D
ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺑﻪ ﺍﺯﺍﻱ t ≥ 0ﻭ zﻫﺮ ﭼﻪ ﺑﺎﺷﺪ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ :
∞V → δ x v
V =0
∞P → P
∞ → B.C.1 : as x 2 + y 2 + z 2
2
B.C.2 : if x 2 + y 2 ≤ D
and z ≤ L
4
2
B.C.3 : as x → −∞ at y = 0
ﺣﺎﻝ ﻣﺴﺌﻠﻪ ﺭﺍ ﺑﺮ ﺣﺴﺐ ﻣﺘﻐﻴﺮ ﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﺷﺪﻩ ﺑﺎ ﻃﻮﻝ ﻣﺸﺨﺼﻪ ، Dﺳﺮﻋﺖ ∞ υﻭﻓﺸﺎﺭ ﺍﺻﻼﻉ ﺷﺪﻩ ∞p
ﺑﺎﺯﻧﻮﻳﺴﻲ ﻣﻲ ﻛﻨﻴﻢ .ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻐﻴﻴﺮ ﺑﺪﻭﻥ ﺑﻌﺪ ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ :
]
[
)
(
* ∂V
* 1 *2
∇ ⋅V = 0 ,
+ V * ⋅ ∇ * V * = −∇ * P * +
∇ V
*
Re
∂t
ρv D
∞ = Re
µ
ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻭ ﻣﺮﺯﻱ ﻣﺘﻨﺎﻇﺮ ﻋﺒﺎﺭﺕ ﺍﻧﺪ ﺍﺯ :
۹۸
*
*
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
V* = δx
1
) (L 2
2
V* →δx
V* =0
P* → 0
or if z *
4
2
2
2
if x * + y * 1
2
I.C.
2
∞ → * B.C.1 as x * + y * + z
1
) (L 2
2
≤ * and z
4
2
2
B.C.2 if x * + y * ≤ 1
B.C.3 as x * → −∞ at y = 0
ﺍﮔﺮ ﺑﺘﻮﺍﻧﻴﻢ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﺗﻐﻴﻴﺮ ﺭﺍ ،ﻫﻤﺮﺍﻩ ﺑﺎ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ﺣﻞ ﻛﻨﻴﻢ ،ﺟﻮﺍﺏ ﻫﺎﻱ ﺁﻥ ﻫﺎ ﺑﺎﻳﺪ
ﺻﻮﺭﺕ ﺯﻳﺮ ﺭﺍ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ :
(
)
) = P (x , y , z , t , Re, L D
V * = V * x* , y * , z * , t * , Re, L D
*
*
*
*
*
*
P
ﻳﻌﻨﻲ ﺳﺮﻋﺖ ﺑﺪﻭﻥ ﺑﻌﺪ ﻭ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﺑﺪﻭﻥ ﺑﻌﺪ ﻣﻲ ﺗﻮﺍﻧﻨﺪ ﻓﻘﻂ ﺑﻪ ﭘﺎﺭﺍﻣﺘﺮ ﻫﺎﻱ Reﻭ L Dﻭ ﻣﺘﻐﻴﺮ
ﻫﺎﻱ ﻣﺴﺘﻘﻞ ﺑﺪﻭﻥ ﺑﻌﺪ * z * ، y * ، xﻭ * tﻭﺍﺑﺴﺘﻪ ﺑﺎﺷﻨﺪ.
ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺗﺤﻠﻴﻞ ﺍﺑﻌﺎﺩﻱ ﻣﺴﺌﻠﻪ ﻛﺎﻣﻞ ﻣﻲ ﺷﻮﺩ .ﻣﺴﺌﻠﻪ ﺟﺮﻳﺎﻥ ﺭﺍ ﺣﻞ ﻧﻜﺮﺩﻩ ﺍﻳﻢ ،ﺍﻣﺎ ﺩﺭ ﻣﻮﺭﺩ ﻣﺠﻤﻮﻋﻪ
ﻣﻨﺎﺳﺒﻲ ﺍﺯ ﻣﺘﻐﻴﺮﻫﺎﻱ ﺑﺪﻭﻥ ﺑﻌﺪ ،ﺑﺮﺍﻱ ﺑﻴﺎﻥ ﺩﻭﺑﺎﺭﻩ ﻣﺴﺌﻠﻪ ،ﺗﺼﻤﻴﻢ ﮔﺮﻓﺘﻪ ﺍﻳﻢ ﻭ ﺑﻪ ﺻﻮﺭﺕ ﺟﻮﺍﺏ ﺩﺳﺖ ﻳﺎﻓﺘﻪ
ﺍﻳﻢ .ﺍﻳﻦ ﺗﺤﻠﻴﻞ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴﻢ ﺍﻟﮕﻮﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻋﺒﻮﺭﻱ ﺍﺯ ﻛﻨﺎﺭ ﺍﺳﺘﻮﺍﻧﻪ ﺭﺍ ﻓﻬﺮﺳﺖ ﻛﻨﻴﻢ،
ﻛﺎﻓﻲ ﺍﺳﺖ ﺁﻧﻬﺎ ﺭﺍ ﺑﻪ ﺍﺯﺍﻱ ﺭﺷﺘﻪ ﺍﻱ ﺍﺯ ﺍﻋﺪﺍﺩ ﺭﻳﻨﻮﻟﺪﺯ Re = Dυ ∞ ρ µﻭ ﻣﻘﺎﺩﻳﺮ L Dﺛﺒﺖ ﻛﻨﻴﻢ )ﻣﺜﻼً ﺑﻪ
ﺻﻮﺭﺕ ﻋﻜﺲ (؛ ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺗﺤﻘﻴﻘﺎﺕ ﺟﺪﺍﮔﺎﻧﻪ ﺩﺭ ﻣﻮﺭﺩ ﻧﻘﺶ ، ρ ، υ ∞ ، D ، Lﻭ µﺿﺮﻭﺕ ﻧﺪﺍﺭﺩ .ﺑﺎ ﺍﻳﻦ ﻧﻮﻉ
ﺳﺎﺩﻩ ﺳﺎﺯﻱ ﻣﻘﺪﺍﺭ ﺯﻳﺎﺩﻱ ﻭﻗﺖ ﻭ ﻫﺰﻳﻨﻪ ﺻﺮﻓﻪ ﺟﻮﻳﻲ ﻣﻲ ﺷﻮﺩ.
ﺑﺮﺍﻱ ﺗﺸﺎﺑﻪ ﻳﻚ ﻣﺪﻝ ﺁﺯﻣﺎﻳﺸﮕﺎﻫﻲ ﻭ ﻳﻚ ﻣﺪﻝ ﻭﺍﻗﻌﻲ ﻣﺜﺎﻝ ﺑﺎﻻ:
geometric similarity
dynamic similarity
۹۹
) ( D ) = (L D
II
I
⇒ 1) L
2)(Re )I = (Re )II
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
1.3ﺗﻌﻴﻴﻦ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻨﺞ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﭼﺮﺧﺎﻥ .ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ
ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻨﺞ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﭼﺮﺧﺎﻥ ،ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﻣﺤﻠﻮﻟﻬﺎﻱ ﺳﺎﻛﺎﺭﻭﺯ ﺑﺎ ﻏﻠﻈﺖ ﻭﺯﻧﻲ ﺣﺪﻭﺩ 0 0 60ﺭﺍ ﺩﺭ
ﺩﻣﺎﻳﻲ ﺣﺪﻭﺩ 20 cﺗﻌﻴﻴﻦ ﻛﻨﻴﻢ .ﺍﻳﻦ ﺍﺳﺒﺎﺏ ﻳﻚ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ ﺑﻪ ﻗﻄﺮ 4cmﺩﺍﺭﺩ ﻛﻪ ﺗﻮﺳﻂ ﺍﺳﺘﻮﺍﻧﻪ ﻫﻢ
ﻣﺮﻛﺰﻱ ﺑﻪ ﻗﻄﺮ 4.5cmﺍﺣﺎﻃﻪ ﺷﺪﻩ ﺍﺳﺖ .ﻃﻮﻝ 4cm ، Lﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻣﺤﻠﻮﻝ ﺳﺎﻛﺎﺭﻭﺯ 0 0 60ﺩﺭ ﺩﻣﺎﻱ 20 c
ﺩﺭ ﺣﺪﻭﺩ 57cpﻭ ﭼﮕﺎﻟﻲ ﺁﻥ ﺣﺪﻭﺩ
cm 3
1.29 gﺍﺳﺖ.
ﺑﺮ ﺍﺳﺎﺱ ﺗﺠﺮﺑﻪ ﻱ ﻗﺒﻠﻲ ﻣﻤﻜﻦ ﺍﺳﺖ ﺁﺛﺎﺭ ﺍﻧﺘﻬﺎﻳﻲ ﻣﻬﻢ ﺑﺎﺷﻨﺪ ،ﻭ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺗﺼﻤﻴﻢ ﻣﻲ ﮔﻴﺮﻳﻢ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺳﻨﺞ
ﺭﺍ ،ﺑﺎ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﭼﻨﺪ ﻣﺤﻠﻮﻝ ﺷﻨﺎﺧﺘﻪ ﺷﺪﻩ ،ﻛﻪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺁﻧﻬﺎ ﺩﺭ ﺣﺪﻭﺩ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻣﺤﻠﻮﻟﻬﺎﻱ ﻣﺠﻬﻮﻝ
ﺍﺳﺖ ،ﻛﺎﻟﻴﺒﺮﻩ ﻛﻨﻴﻢ.ﻣﻘﺪﺍﺭﻱ ﻣﻌﻘﻮﻝ ﺑﺮﺍﻱ ﮔﺸﺘﺎﻭﺭﻱ ﻛﻪ ﺑﺎﻳﺪ ﺩﺭ ﻫﻨﮕﺎﻡ ﻛﺎﻟﻴﺒﺮﻩ ﺳﺎﺯﻱ ﻭﺍﺭﺩ ﻛﺮﺩ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ.
ﻫﺮﮔﺎﻩ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻫﺎﻱ ﮔﺸﺘﺎﻭﺭ ﺩﺭ ﺣﺪﻭﺩ 100 dyne cmﻗﺎﺑﻞ ﺍﻋﺘﻤﺎﺩ ﺑﺎﺷﻨﺪ ﻭ ﺑﺘﻮﺍﻥ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ﺭﺍ ﺑﺎ
ﺩﻗﺖ 0.5
0
0
ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻛﺮﺩ ،ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ﺑﺮﺁﻳﻨﺪ ﭼﻘﺪﺭ ﺧﻮﺍﻫﺪ ﺑﻮﺩ؟
2.3ﺟﺮﻳﺎﻥ ﺑﻴﻦ ﺍﺳﺘﻮﺍﻧﻪ ﻫﺎﻱ ﻫﻢ ﻣﺤﻮﺭ ﻛﺮﻩ ﻫﺎﻱ ﻫﻢ ﻣﺮﻛﺰ.
ﺍﻟﻒ( ﻓﻀﺎﻱ ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻫﻢ ﻣﺤﻮﺭ ﺑﺎ ﺳﻴﺎﻟﻲ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺩﺭ ﺩﻣﺎﻱ ﺛﺎﺑﺖ ﭘﺮ ﻣﻲ ﺷﻮﺩ .ﺷﻌﺎﻉ ﻫﺎﻱ ﺳﻄﻮﺡ ﺗﺮ
ﺷﺪﻩ ﻱ ﺩﺍﺧﻠﻲ ﻭ ﺧﺎﺭﺟﻲ ،ﺑﻪ ﺗﺮﺗﻴﺐ KR ،ﻭ Rﻭ ﺳﺮﻋﺖ ﻫﺎﻱ ﺯﺍﻭﻳﻪ ﺍﻱ ﭼﺮﺧﺶ ﺍﺳﺘﻮﺍﻧﻪ ﻫﺎﻱ ﺩﺍﺧﻠﻲ ﻭ ﺧﺎﺭﺟﻲ
Ω iﻭ Ω oﻫﺴﺘﻨﺪ .ﻣﻄﻠﻮﺑﺴﺖ ﺗﻌﻴﻴﻦ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﺳﻴﺎﻝ ﻭﮔﺸﺘﺎﻭﺭﻫﺎﻳﻲ ﻛﻪ ﺳﻴﺎﻝ ﺑﺮ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻭﺍﺭﺩ ﻣﻲ
ﻛﻨﺪ ﺗﺎ ﺣﺮﻛﺖ ﺑﺮﻗﺮﺍﺭ ﺑﻤﺎﻧﺪ.
ﺏ(ﻗﺴﻤﺖ )ﺍﻟﻒ( ﺭﺍ ﺑﺮﺍﻱ ﺩﻭ ﻛﺮﻩ ﻫﻢ ﻣﺮﻛﺰ ﺗﻜﺮﺍﺭ ﻛﻨﻴﺪ.
3.3ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺩﺭ ﻣﺠﺮﺍﻱ ﻣﺮﺑﻌﻲ.
)ﺍﻟﻒ( ﻣﺠﺮﺍﻱ ﻣﺴﺘﻘﻴﻤﻲ ﺩﺭ ﺍﻣﺘﺪﺍﺩ zﺑﻪ ﻃﻮﻝ Lﺑﺎ ﻣﻘﻄﻊ ﻣﺮﺑﻌﻲ ﻣﻔﺮﻭﺽ ﺍﺳﺖ ﻛﻪ ﻣﺮﺯﻫﺎﻱ ﺁﻥ ﺭﺍ ﺧﻄﻮﻁ
x = ± Bﻭ y = ± Bﺗﺸﻜﻴﻞ ﻣﻲ ﺩﻫﺪ .ﻫﻤﻜﺎﺭﻱ ﺑﻪ ﺷﻤﺎ ﮔﻔﺘﻪ ﺍﺳﺖ ﻛﻪ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ
ﻣﻲ ﺁﻳﺪ:
)(1-3.3
2
2
(
P0 − PL )B 2 x y
=
1−
1−
B
B
4 µL
υz
ﭼﻮﻥ ﺍﻳﻦ ﻫﻤﻜﺎﺭ ﮔﺎﻫﻲ ﺩﺭ ﮔﺬﺷﺘﻪ ﺗﻮﺻﻴﻪ ﻫﺎﻱ ﻧﺎﺩﺭﺳﺘﻲ ﻛﺮﺩﻩ ﺍﺳﺖ ،ﺧﻮﺩ ﺭﺍ ﻣﻮﻇﻒ ﻣﻲ ﺩﺍﻧﻴﺪ ﻛﻪ ﻧﺘﻴﺠﻪ ﺭﺍ
ﺑﺮﺭﺳﻲ ﻛﻨﻴﺪ .ﺁﻳﺎ ﺍﻳﻦ ﻧﺘﻴﺠﻪ ﺩﺭ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻭ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﺮﺑﻮﻃﻪ ﺻﺪﻕ ﻣﻲ ﻛﻨﺪ؟
)ﺏ( ﻃﺒﻖ ﻣﻘﺎﻟﻪ ﺑﺮﻛﺮ ،ﺁﻫﻨﮓ ﺟﺮﻳﺎﻥ ﺟﺮﻣﻲ ﺩﺭ ﻣﺠﺮﺍﻱ ﻣﺮﺑﻌﻲ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ:
0.563( p0 − p L )B 4 ρ
µL
ﺿﺮﻳﺐ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﺭﺍ ﺑﺎ ﺿﺮﻳﺒﻲ ﻛﻪ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ) (1-3.3ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ ،ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﺪ.
=w
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ۳ﻣﻌﺎﺩﻟﻪ ﻫﺎی ﺗﻐﻴﻴﺮ ﺑﺮﺍی ﺳﻴﺴﺘﻢ ﻫﺎی ﻫﻢ ﺩﻣﺎ
4.3ﺟﺮﻳﺎﻥ ﺷﻌﺎﻋﻲ ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﻫﻢ ﻣﺤﻮﺭ.
ﺳﻴﺎﻟﻲ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺭﺍ ﺩﺭ ﺩﻣﺎﻱ ﺛﺎﺑﺖ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﺷﻌﺎﻋﻲ ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﻣﺘﺨﻠﺨﻞ ﺑﺎ
ﺷﻌﺎﻉ ﻫﺎﻱ ﺩﺍﺧﻠﻲ ﻭ ﺧﺎﺭﺟﻲ KRﻭ Rﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ.
)ﺍﻟﻒ( ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﺑﻪ υ r = c rﻛﻪ ﻣﻘﺪﺍﺭ cﺩﺭ ﺁﻥ ﺛﺎﺑﺖ ﺍﺳﺖ.
)ﺏ( ﻣﻮﻟﻔﻪ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺭﺍ ﺳﺎﺩﻩ ﻛﻨﻴﺪ ﺗﺎ ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﺯﻳﺮ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﺍﺻﻼﺡ ﺷﺪﻩ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ:
dυ
dP
dP
dP
= − ρυ r r ,
=0 ,
=0
dr
dr
dθ
dz
)ﺝ( ﺍﺯ ﻋﺒﺎﺭﺕ dp drﺍﻧﺘﮕﺮﺍﻝ ﺑﮕﻴﺮﻳﺪ ﺗﺎ ﻧﺘﻴﺠﻪ ﺯﻳﺮ ﺣﺎﺻﻞ ﺷﻮﺩ:
)ﺩ(ﻫﻤﻪ ﻣﻮﻟﻔﻪ ﻫﺎﻱ ﻏﻴﺮ ﺻﻔﺮ τﺭﺍ ﺑﺮﺍﻱ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺑﻨﻮﻳﺴﻴﺪ.
)ﻩ(ﻣﺴﺌﻠﻪ ﻛﺮﻩ ﻫﺎﻱ ﻫﻢ ﻣﺮﻛﺰ ﺭﺍ ﺗﻜﺮﺍﺭ ﻛﻨﻴﺪ.
R 2
1
2
P (r ) − P ( R ) = ρ [υ r (R )] 1 −
2
r
5.3ﺷﻜﻞ ﺳﻄﺢ ﺁﺯﺍﺩ ﺩﺭ ﺟﺮﻳﺎﻥ ﺣﻠﻘﻮﻱ ﻣﻤﺎﺳﻲ.
)ﺍﻟﻒ( ﻣﺎﻳﻌﻲ ﺩﺭ ﻓﻀﺎﻱ ﺣﻠﻘﻮﻱ ﺑﻴﻦ ﺩﻭ ﺍﺳﺘﻮﺍﻧﻪ ﻱ ﻋﻤﻮﺩﻱ ﺑﻪ ﺷﻌﺎﻉ ﻫﺎﻱ KRﻭ Rﻭﺍﻗﻊ ﺍﺳﺖ ﻭ ﺍﺯ ﺑﺎﻻ ﺑﺎ ﺟﻮ
ﺗﻤﺎﺱ ﺩﺍﺭﺩ.ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﻭﻗﺘﻲ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ ﺑﺎ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ Ω iﻣﻲ ﭼﺮﺧﺪ ﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ ﺛﺎﺑﺖ
ﺍﺳﺖ ،ﺳﻄﺢ ﺁﺯﺍﺩ ﻣﺎﻳﻊ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺍﺳﺖ:
2
ξ − 2 + 4 ln ξ − ξ 2
1 k 2 RΩ i
= ZR − Z
2 g 1 − k 2
) (ξ − 2 − 1) + 4k − 2 ln ξ − k − 4 (ξ 2 − 1
1 k 2 RΩ o
= ZR − Z
2 g 1 − k 2
)
(
ﻛﻪ ﺩﺭ ﺁﻥ Z Rﺍﺭﺗﻔﺎﻉ ﻣﺎﻳﻊ ﺩﺭ ﺟﺪﺍﺭ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ ξ = r Rﺍﺳﺖ.
)ﺏ( ﻗﺴﻤﺖ )ﺍﻟﻒ( ﺭﺍ ﺗﻜﺮﺍﺭ ﻛﻨﻴﺪ ،ﺍﻣﺎ ﺍﻳﻦ ﺑﺎﺭ ﻓﺮﺽ ﻛﻨﻴﺪ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺍﺧﻠﻲ ﺛﺎﺑﺖ ﺍﺳﺖ ﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺧﺎﺭﺟﻲ ﺑﺎ ﺳﺮﻋﺖ
ﺯﺍﻭﻳﻪ ﺍﻱ Ω oﻣﻲ ﭼﺮﺧﺪ .ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺷﻜﻞ ﺳﻄﺢ ﻣﺎﻳﻊ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ:
)
(
)ﺝ( ﻧﻤﻮﺩﺍﺭﻱ ﺑﺮﺍﻱ ﻣﻘﺎﻳﺴﻪ ﺷﻜﻞ ﺳﻄﺢ ﻣﺎﻳﻊ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺗﺮﺳﻴﻢ ﻛﻨﻴﺪ.
۱۰۱
2
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
۱۰۲
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
1-4ﺣﺮﻛﺖ ﻳﻚ ﺟﺰء ﺳﻴﺎﻝ )ﺳﻴﻨﻤﺎﺗﻴﻚ(
ﺣﺮﻛﺖ )ﺳﻴﻨﻤﺎﺗﻴﻚ( ﻳﻚ ﺟﺰءﺳﻴﺎﻝ ﺭﺍ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ .ﺑﺮﺍﻱ ﺳﻬﻮﻟﺖ ﻳﻚ ﺟﺰ ﺳﻴﺎﻝ ﺑﺎ ﭼﮕـﺎﻟﻲ
ﺛﺎﺑﺖ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ.
ﻭﻗﺘﻲ ﺟﺰ ﺑﻲ ﻧﻬﺎﻳﺖ ﻛﻮﭼﻚ ﺑﻪ ﺟﺮﻡ dmﺩﺭ ﻳﻚ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ،ﺧﻴﻠﻲ ﭼﻴﺰﻫﺎ ﻣﻲ ﺗﻮﺍﻧﺪ ﺑﺮﺍﻱ
ﺁﻥ ﺭﻭﻱ ﺩﻫﺪ :
ﺣﺮﻛﺖ ﺍﻧﺘﻘﺎﻟﻲ
1
F18
ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺟﺰ ﺳﻴﺎﻝ ﺣﺎﻟﺖ ﺧﻮﺩﺵ ﺭﺍ ﺣﻔﻆ ﻣﻲ ﻛﻨﺪ ﻭ ﺗﻐﻴﻴﺮ ﻣﻜﺎﻥ ﺧﻄﻲ ﻣﻲ ﺩﻫﺪ.
- Translation
۱۰۳
۱
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺣﺮﻛﺖ ﭼﺮﺧﺸﻲ
1
F19
ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺟﺰء ﺳﻴﺎﻝ ﺣﻮﻝ ﻳﻜﻲ ﺍﺯ ﻣﺤﻮﺭﻫﺎﻱ ﻣﺨﺘﺼﺎﺕ )ﻳﺎ ﺣﻮﻝ ﻫﺮ ﺳﻪ( ﻣﻲ ﭼﺮﺧﺪ ﻭﻟﻲ ﺷـﻜﻞ ﺧـﻮﺩﺵ ﺭﺍ
ﺣﻔﻆ ﻣﻲ ﻛﻨﺪ ﻟﺬﺍ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺁﻥ ﺻﻔﺮ ﺍﺳﺖ.
ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺧﻄﻲ
2
F20
ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺻﻔﺤﺎﺕ ﺟﺰء ﺳﻴﺎﻝ ﻛﻪ ﺩﺭ ﺍﺑﺘﺪﺍ ﺑﻪ ﻃﻮﺭ ﻋﻤﻮﺩﻱ ﻫﺴﺘﻨﺪ ﺑﻪ ﻃﻮﺭ ﻋﻤﻮﺩﻱ ﻣﻲ ﻣﺎﻧﻨـﺪ .ﺷـﻜﻞ ﺟـﺰء
ﺳﻴﺎﻝ ﺗﻐﻴﻴﺮ ﻣﻲ ﻛﻨﺪ ﺑﺪﻭﻥ ﺍﻳﻨﻜﻪ ﻭﺿﻌﻴﺖ ﺁﻥ ﺗﻐﻴﻴﺮ ﻛﻨﺪ.
- Rotation
- Linear Deformation
۱۰٤
۱
۲
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ
1
F21
ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺑﺎﻋﺚ ﻭﺍﭘﻴﭽﺶ ﺟﺰء ﺳﻴﺎﻝ ﻣﻲ ﺷﻮﺩ ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﺻﻔﺤﺎﺕ ﺩﺭ ﺍﺑﺘﺪﺍ ﻋﻤﻮﺩﻱ،
ﺩﻳﮕﺮ ﻋﻤﻮﺩﻱ ﻧﻤﻲ ﻣﺎﻧﻨﺪ .ﺑﻪ ﻃﻮﺭ ﻛﻠﻲ ﻳﻚ ﺟﺰء ﺳﻴﺎﻝ ﺿﻤﻦ ﺣﺮﻛﺖ ﻣﻲ ﺗﻮﺍﻧﺪ ﺗﺤﺖ ﺗﺄﺛﻴﺮ ﺗﺮﻛﻴﺒﻲ ﺍﺯ ﺍﻧﺘﻘﺎﻝ,
ﭼﺮﺧﺶ ,ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺧﻄﻲ ﻭ ﺯﺍﻭﻳﻪ ﺍﻱ ﻗﺮﺍﺭ ﮔﻴﺮﺩ.
ﺩﺭ ﺍﺷﻜﺎﻝ ﺑﺎﻻ ،ﭼﻬﺮ ﻣﺆﻟﻔﻪ ﺣﺮﻛﺖ ﺳﻴﺎﻝ ﺑﺮﺍﻱ ﺣﺮﻛﺖ ﺩﺭ ﺻﻔﺤﻪ xyﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻛﻠﻲ
ﺳﻪ ﺑﻌﺪﻱ ،ﺣﺮﻛﺖ ﻫﺎﻱ ﻣﺸﺎﺑﻬﻲ ﺑﺮﺍﻱ ﺫﺭﻩ ﺩﺭ ﺻﻔﺤﻪ ﻫﺎﻱ yzﻭ xzﺭﺳﻢ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﺍﻧﺘﻘﺎﻝ ﻳﺎ ﭼﺮﺧﺶ ﺧﺎﻟﺺ،
ﺟﺰء ﺳﻴﺎﻝ ﺷﻜﻞ ﺧﻮﺩ ﺭﺍ ﺣﻔﻆ ﻣﻲ ﻛﻨﺪ؛ ﻫﻴﭻ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﺍﻧﺘﻘﺎﻝ ﺧﺎﻟﺺ ﻳﺎ ﭼﺮﺧﺶ
ﺧﺎﻟﺺ ﻫﻴﭻ ﺗﻨﺶ ﺑﺮﺷﻲ ﺭﺍ ﺑﻪ ﻭﺟﻮﺩ ﻧﻤﻲ ﺁﻭﺭﺩ).ﺩﺭ ﻳﻚ ﺳﻴﺎﻝ ﻧﻴﻮﺗﻨﻲ ﺗﻨﺶ ﺑﺮﺷﻲ ﺑﺎ ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ
ﺍﻱ ﺑﻪ ﻃﻮﺭ ﻣﺴﺘﻘﻴﻢ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ(.
1-1-4ﭼﺮﺧﺶ ﺳﻴﺎﻝ
2
F2
ﭼﺮﺧﺶ) ( ωﺫﺭﻩ ﺳﻴﺎﻝ ]ﺣﻮﻝ ﻳﻚ ﻣﺤﻮﺭ[ ﺑﻪ ﺻﻮﺭﺕ ﻣﺘﻮﺳﻂ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ﺩﻭ ﭘﺎﺭﻩ ﺧﻂ ﻋﻤﻮﺩ ﺑﺮﻫﻢ ﺍﺯ ﺫﺭﻩ
ﻛﻪ ﺩﺭ ﺻﻔﺤﻪ ﻋﻤﻮﺩ ﺑﺮ ﺁﻥ ﻣﺤﻮﺭ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ،ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ .ﭼﺮﺧﺶ ﻛﻤﻴﺘﻲ ﺑﺮﺩﺍﺭﻱ ﺍﺳﺖ .ﺫﺭﻩ ﺍﻱ ﻛﻪ ﺩﺭ
ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺳﻪ ﺑﻌﺪﻱ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ ﻣﻲ ﺗﻮﺍﻧﺪ ﺣﻮﻝ ﻫﺮ ﺳﻪ ﻣﺤﻮﺭ ﻣﺨﺘﺼﺎﺕ ﺑﭽﺮﺧﺪ.ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ،ﺩﺭﺣﺎﻟﺖ
ﻛﻠﻲ،
ω = iˆω + ˆjω + kˆω
z
ﻛﻪ ﺩﺭ ﺁﻥ ω xﭼﺮﺧﺶ ﺣﻮﻝ ﻣﺤﻮﺭ ω y ، xﭼﺮﺧﺶ ﺣﻮﻝ
ﻣﺜﺒﺖ ﭼﺮﺧﺶ ﺑﺎ ﻗﺎﻋﺪﻩ ﺩﺳﺖ ﺭﺍﺳﺖ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ.
ﻣﺤﻮﺭ y
y
x
ﻭ ω zﭼﺮﺧﺶ ﺣﻮﻝ ﻣﺤﻮﺭ zﺍﺳﺖ .ﺳﻮﻱ
۱
- Angular Deformation
۲- Angular Velocity
۱۰٥
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺑﺮﺍﻱ ﻳﺎﻓﺘﻦ ﻋﺒﺎﺭﺗﻲ ﺭﻳﺎﺿﻲ ﺑﺮﺍﻱ ﭼﺮﺧﺶ ﺳﻴﺎﻝ ،ﺣﺮﻛﺖ ﻳﻚ ﺟﺰء ﺳﻴﺎﻝ ﺭﺍ ﺩﺭ ﺻﻔﺤﻪ xyﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ.
ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﻱ ﺩﺍﺧﻞ ﺟﺮﻳﺎﻥ ﺑﺎ ) u (x, yﻭ ) v(x, yﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ.ﭼﺮﺧﺶ ﻳﻚ ﺫﺭﻩ
ﺳﻴﺎﻝ ﺩﺭ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺩﺭ ﺷﻜﻞ 2-4ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ.
ﻓﻘﻂ ﺍﮔﺮ ﺳﺮﻋﺖ ﺩﺭ ﻧﻘﺎﻁ aﻭ bﺑﺎ ﺳﺮﻋﺖ ﺩﺭ oﺗﻔﺎﻭﺕ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ،ﺩﻭ ﺧﻂ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ oaﻭ obﺩﺭ ﺑﺎﺯﻩ
ﺯﻣﺎﻧﻲ ∆tﺗﺎ ﻣﻜﺎﻥ ﻫﺎﻱ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﻣﻲ ﭼﺮﺧﻨﺪ.
ﺍﺑﺘﺪﺍ ﭼﺮﺧﺶ ﺧﻂ oaﺑﻪ ﻃﻮﻝ ∆xﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ .ﭼﺮﺧﺶ ﺍﻳﻦ ﺧﻂ ﺑﺮ ﺍﺛﺮ ﺗﻐﻴﻴﺮﺍﺕ ﻣﺆﻟﻔﻪ yﺳـﺮﻋﺖ
ﺍﺳﺖ .ﺍﮔﺮ ﻣﺆﻟﻔﻪ yﺳﺮﻋﺖ ﺩﺭ ﻧﻘﻄﻪ vo ، oﺑﺎﺷﺪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺑﺴﻂ ﺗﻴﻠﻮﺭ ،ﻣﺆﻟﻔـﻪ yﺳـﺮﻋﺖ ﺩﺭ ﻧﻘﻄـﻪ aﻣـﻲ
ﺗﻮﺍﻥ ﭼﻨﻴﻦ ﻧﻮﺷﺖ:
∂v
∆x
∂x
v a = vo +
ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ﺧﻂ oaﭼﻨﻴﻦ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ،
∆η
∆α
= lim
= lim ∆x
∆t → 0 ∆t
∆t
ωoa
∆t → 0
ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ،
∂v
∆x∆t
∂x
= ∆η
∂υ ∆x∆t
∂υ
∂x ∆x
=
= lim
∆t
∂x
∆t →0
ﭼﺮﺧﺶ ﺧﻂ obﺑﻪ ﻃﻮﻝ ∆yﺍﺯ ﺗﻐﻴﻴﺮﺍﺕ ﻣﺆﻟﻔﻪ xﺳﺮﻋﺖ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ.
۱۰٦
ωoa
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺍﮔﺮ ﻣﺆﻟﻔﻪ xﺳﺮﻋﺖ ﺩﺭ ﻧﻘﻄﻪ ﻱ oﺑﺎ uﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﻮﺩ ،ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺑﺴﻂ ﺳﺮﻱ ﺗﻴﻠﻮﺭ ،ﻣﺆﻟﻔﻪ ﺳﺮﻋﺖ ﺭﺍ ﺩﺭ
ﻧﻘﻄﻪ bﻣﻲ ﺗﻮﺍﻥ ﭼﻨﻴﻦ ﻧﻮﺷﺖ :
∂u
∆y
∂y
∂u
∆y∆t
∂y
∆ζ = −
ub = u o +
∆ζ
∆β
∆y
ωob = lim
= lim
∆t →0 ∆t
∆t →0
∆t
∂u ∆y∆t
⋅ −
∂u
∂y ∆y
=−
= ⇒ ωob
∆t
∂y
,
)ﻋﻼﻣﺖ ﻣﻨﻔﻲ ﺭﺍ ﻭﺍﺭﺩ ﻣﻲ ﻛﻨﻴﻢ ﺗﺎ ﺑﺮﺍﻱ ωobﻣﻘﺪﺍﺭ ﻣﺜﺒﺘﻲ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ .ﻃﺒـﻖ ﻗـﺮﺍﺭﺩﺍﺩ ﻋﻼﻣـﺖ ﻣـﺎ ،ﭼـﺮﺧﺶ
ﭘﺎﺩﺳﺎﻋﺘﮕﺮﺩ ﻋﻼﻣﺖ ﻣﺜﺒﺖ ﺩﺍﺭﺩ(.
ﭼﺮﺧﺶ ﺟﺰء ﺳﻴﺎﻝ ﺣﻮﻝ ﻣﺤﻮﺭ zﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪ ﺍﻱ ﻣﺘﻮﺳﻂ ﺩﻭ ﭘﺎﺭﻩ ﺧﻂ oaﻭ obﻛﻪ ﺩﺭ
ﺻﻔﺤﻪ xyﻗﺮﺍﺭ ﺩﺍﺭﺩ .ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ،
1
(ωoa + ωob ) = 1 ∂v − ∂u
2
2 ∂x ∂y
= ωz
ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﭼﺮﺧﺶ ﺯﻭﺝ ﺧﻂ ﻫﺎﻱ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ ﻛﻪ ﺩﺭ ﺻﻔﺤﺎﺕ yzﻭ xzﻓﺮﺍﺭﺩﺍﺭﻧﺪ ،ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ
ﻛﻪ :
1 ∂u ∂ω
ωy = −
2 ∂z ∂x
,
ﺑﻨﺎﺑﺮﺍﻳﻦ،
ˆj ∂u − ∂ω + kˆ ∂v − ∂u
∂z ∂x ∂x ∂y
1 ∂ω ∂v
ω x = −
2 ∂y ∂z
1 ∂ω ∂v
⇒ ω = iˆω x + ˆjω y + kˆω z = iˆ
− +
2 ∂y ∂z
ˆj
ˆi
ˆk
1
∂
= ∇ ×V
∂z 2
w
1
∇ ×V
2
ˆk
∂
∂y
v
= curl V = ∇ × V ⇒ ω
vz
۱۰۷
∂ 1
2 ∂x
u
=
ωz
ˆj
vy
ˆi
V ×W = vx
ωy
ωx
]
[
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺗﺤﺖ ﭼﻪ ﺷﺮﺍﻳﻄﻲ ﻣﻲ ﺗﻮﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺩﺍﺷﺖ؟ ﺫﺭﻩ ﺳﻴﺎﻟﻲ ﻛﻪ ،ﺑﺪﻭﻥ ﭼﺮﺧﺶ ،ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻧﻲ
ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ ﺗﺤﺖ ﺗﺎﺛﻴﺮ ﻧﻴﺮﻭﻱ ﺣﺠﻤﻲ ﻳﺎ ﻧﻴﺮﻭﻫﺎﻱ ﺳﻄﺤﻲ ﻋﻤﻮﺩﻱ)ﻓﺸﺎﺭﻱ( ﻧﻤﻲ ﺗﻮﺍﻧﺪ ﭼﺮﺧﺶ ﺩﺍﺷﺘﻪ
ﺑﺎﺷﺪ .ﺑﺮﺍﻱ ﺍﻳﺠﺎﺩ ﭼﺮﺧﺶ ﺩﺭ ﺫﺭﻩ ﺳﻴﺎﻝ،ﻛﻪ ﺍﺑﺘﺪﺍ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ ،ﺑﺎﻳﺪ ﺗﻨﺶ ﺑﺮﺷﻲ ﺑﺮ ﺳﻄﺢ ﺫﺭﻩ ﺍﻋﻤﺎﻝ ﺷﻮﺩ.
ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ ﺗﻨﺶ ﺑﺮﺷﻲ ﺑﺎ ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﻣﺘﻨﺎﺳﺐ ﺍﺳﺖ،ﺫﺭﻩ ﺍﻱ ﻛﻪ ﺩﺭ ﺍﺑﺘﺪﺍﻱ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ
ﺑﺪﻭﻥ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺍﺯ ﺧﻮﺩ ﭼﺮﺧﺶ ﻧﺸﺎﻥ ﻧﻤﻲ ﺩﻫﺪ .ﺗﻨﺶ ﺑﺮﺷﻲ ﺍﺯ ﻃﺮﻳﻖ ﭼﺴﺒﻨﺪﮔﻲ ﺑﻪ ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ
ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺍﺭﺗﺒﺎﻁ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ .ﻭﺟﻮﺩ ﻧﻴﺮﻭﻫﺎﻱ ﭼﺴﺒﻨﺪﻩ ﺑﻪ ﺍﻳﻦ ﻣﻌﻨﻲ ﺍﺳﺖ ﻛﻪ ﺟﺮﻳﺎﻥ ﭼﺮﺧﺸﻲ ﺍﺳﺖ.
ﺑﺮﺍﻱ ﻧﺎﺣﻴﻪ ﻫﺎﻳﻲ ﺍﺯ ﻳﻚ ﺟﺮﻳﺎﻥ ﻛﻪ ﺩﺭ ﺁﻧﺠﺎﻫﺎ ﻧﻴﺮﻭﻫﺎﻱ ﭼﺴﺒﻨﺪﮔﻲ ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﻫﺴﺘﻨﺪ ،ﺣﺎﻟﺖ ﺑﻲ ﭼﺮﺧﺸﻲ
ﻓﺮﺽ ﺻﺤﻴﺤﻲ ﺍﺳﺖ).ﭼﻨﻴﻦ ﻧﺎﺣﻴﻪ ﺍﻱ ﺩﺭ ﺧﺎﺭﺝ ﻻﻳﻪ ﻣﺮﺯﻱ ﺩﺭ ﺟﺮﻳﺎﻥ ﺭﻭﻱ ﻳﻚ ﺳﻄﺢ ﺻﻠﺐ ﻭﺟﻮﺩ ﺩﺍﺭﺩ(.ﺑﺎ
ﺗﻌﺮﻳﻒ ﻛﻤﻴﺘﻲ ﺑﻪ ﻧﺎﻡ ﺑﺮﺩﺍﺭ ﮔﺮﺩﺍﺑﻲ ، ξ ،ﻛﻪ ﺩﻭ ﺑﺎﺑﺮ ﺑﺮﺩﺍﺭ ﭼﺮﺧﺶ ﺍﺳﺖ،ﺿﺮﻳﺐ 1ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺣﺬﻑ ﻛﺮﺩ،
2
ξ ≡ 2ω = ∇ × V
ﮔﺮﺩﺍﺑﻲ ﻳﻚ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﺑﺮﺍﻱ ﭼﺮﺧﺶ ﺫﺭﻩ ﺍﻱ ﺍﺯ ﺳﻴﺎﻝ ﺑﻪ ﻫﻨﮕﺎﻡ ﺣﺮﻛﺖ ﺁﻥ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺍﺳﺖ .ﺩﺭ
ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ،ﮔﺮﺩﺍﺑﻲ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ
1 ∂vz ∂vθ
∂v ∂v 1 ∂ (rvθ ) 1 ∂vr
−
curl V = ∇ × V = eˆr
−
+ eˆθ r − z + kˆ
r ∂θ
∂z ∂r r ∂r
r ∂θ ∂z
if curl V = 0 ⇒ irrotational flow
2-1-4ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺳﻴﺎﻝ
ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺳﻴﺎﻝ ﺷﺎﻣﻞ ﺗﻐﻴﻴﺮ ﺯﺍﻭﻳﻪ ﺑﻴﻦ ﺩﻭ ﺧﻂ ﻋﻤﻮﺩ ﺑﺮﻫﻢ ﺩﺍﺧﻞ ﺟﺰء ﺳﻴﺎﻝ ﺍﺳﺖ .ﺑﺎ ﻣﺮﺍﺟﻌﻪ ﺑﻪ
ﺷﻜﻞ ،3-4ﻣﻲ ﺑﻴﻨﻴﻢ ﻛﻪ ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺟﺰء ﺳﻴﺎﻝ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﺁﻫﻨﮓ ﻛﺎﻫﺶ ﺯﺍﻭﻳﻪ ﺑﻴﻦ ﺧﻂ
ﻫﺎﻱ oaﻭ obﻛﻪ ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ،
dγ dα dβ
=
+
dt
dt dt
۱۰۸
−
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
∂v ∆x∆t
∆η
∆α
∂v
∂x ∆x
∆
x
= lim
= lim
= lim
=
∆t →0 ∆t
∆t →0 ∆t
∆t →0
∆t
∂x
∂u ∆y∆t
∆ξ
∂y ∆y
∆β
∂u
∆y
= lim
= lim
= lim
=
∆t →0 ∆t
∆t →0 ∆t
∆t →0
∆t
∂y
ﺩﺭ ﻧﺘﻴﺠﻪ ،ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺩﺭ ﺻﻔﺤﻪ
xy
dα
dt
dβ
dt
ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :
dγ dα dβ ∂v ∂u
=
+
= +
dt
dt dt ∂x ∂y
−
ﺗﻨﺶ ﺑﺮﺷﻲ ﺍﺯ ﻃﺮﻳﻖ ﭼﺴﺒﻨﺪﮔﻲ ﺳﻴﺎﻝ ﺑﻪ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﺍﺭﺗﺒﺎﻁ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﺟﺮﻳﺎﻥ ﭼﺴﺒﻨﺪﻩ ﺍﻱ
)ﻛﻪ ﺩﺭ ﺁﻥ ﺷﻴﺐ ﻫﺎﻱ ﺳﺮﻋﺖ ﻭﺟﻮﺩ ﺩﺍﺭﺩ( ﺑﺴﻴﺎﺭ ﺑﻌﻴﺪ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺳﺮﺗﺎﺳﺮ ﺟﺮﻳﺎﻥ ∂υ ∂x ،ﻣﺴﺎﻭﻱ ﻭ ﻣﺘﻀﺎﺩ
∂y
∂u
ﺑﺎﺷﺪ .ﻭﺟﻮﺩ ﻧﻴﺮﻭﻫﺎﻱ ﭼﺴﺒﻨﺪﻩ ﻳﻌﻨﻲ ﺍﻳﻦ ﻛﻪ ﺟﺮﻳﺎﻥ ﭼﺮﺧﺸﻲ ﺍﺳﺖ.
ﺗﻌﺮﻳﻒ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺑﻪ ﻗﺮﺍﺭ ﺯﻳﺮ ﺍﺳﺖ:
)(6-4
shear stress
rate of shear strain
= viscosity
dr
dt
,
∝τ
∂u ∂v
⇒ τ xy = τ yx = µ +
∂y ∂x
∂u
1 ∂v
∂u
1 ∂v
ω z = −
2 ∂x ∂y
•
ε xy → shear - strain rate
2-4ﺩﻳﻨﺎﻣﻴﻚ ﺟﺮﻳﺎﻥ ﻏﻴﺮ ﻭﻳﺴﻜﻮﺯ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
•
ε xy = +
2 ∂x ∂y
ﻣﻴﺪﺍﻥ ﺗﻨﺶ ﺩﺭ ﻳﻚ ﺳﻴﺎﻝ ﺍﻳﺪﻩ ﺁﻝ )ﻏﻴﺮ ﻭﻳﺴﻜﻮﺯ( ﺩﺭ ﺳﻴﺎﻝ ﺍﻳﺪﻩ ﺁﻝ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻫﻤﻮﺍﺭﻩ ﺻﻔﺮ ﺍﺳﺖ .ﻳﻌﻨﻲ
µ = 0ﺑﻨﺎﺑﺮﺍﻳﻦ ﻛﻠﻴﻪ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺗﻨﺶ ﺑﺮﺷﻲ ﺩﺭ ﺳﻴﺎﻝ ﺍﻳﺪﻩ ﺁﻝ ﺑﺮﺍﺑﺮ ﺻﻔﺮﻧﺪ ﻭ ﻓﻘﻂ ﺗﻨﺶ ﻫﺎﻱ ﻋﻤﻮﺩﻱ ﻏﻴﺮ
ﺻﻔﺮﻧﺪ .ﺍﻛﻨﻮﻥ ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﺗﻨﺶ ﻋﻤﻮﺩﻱ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ ﺩﺭ ﺗﻤﺎﻡ ﺟﻬﺎﺕ ﻳﻜﺴﺎﻥ ﺍﺳﺖ .ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ
ﺗﻨﺶ ﻋﻤﻮﺩﻱ ﺩﺭ ﻳﻚ ﻧﻘﻄﻪ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺎ ﻛﻤﻴﺖ ﻋﺪﺩﻱ ﻧﺸﺎﻥ ﺩﺍﺩ.
۱۰۹
ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ٤ ﻓﺼﻞ
d F = ρ adv
dxdydz
dxdydz
=ρ
az
2
2
sin αds = dy ﭼﻮﻥas π − α = dy ds ⇒ dy = sin αds
2
dz
dz
⇒ −σ zz + σ nn = ρg
+ ρa z
2
2
dz → 0 ⇒ σ zz = σ nn
⇒ dFz = −σ zz ⋅ dydx + σ nn sin αdsdx − ρg
(
)
dFy = −σ yy dzdx + σ nn cos αdsdx = ρ
:y ﺩﺭ ﺟﻬﺖ
dxdydz
ay
2
cos αds = dz
⇒ as dy → 0 ⇒ σ yy = σ nn
ﺩﺭ ﻧﺘﻴﺠﻪ ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﺗﻨﺶ ﻧﺮﻣﺎﻝ ﺩﺭ ﺳﻴﺎﻝ ﺍﻳﺪﻩ ﺍﻝ ﺩﺭ ﻫﻤﻪ ﺟﻬﺎﺕ ﺑﺎ ﻫﻢ ﺑﺮﺍﺑـﺮ ﺍﻧـﺪ ﻭ ﻧﻴـﺰ ﺑـﺎ ﻣﻨﻔـﻲ ﻓﺸـﺎﺭ
.ﺗﺮﻣﻮﺩﻳﻨﺎﻣﻴﻜﻲ ﻣﺴﺎﻭﻱ ﻣﻲ ﺑﺎﺷﺪ
⇒ σ xx = σ yy = σ nn = − P
۱۱۰
ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ٤ ﻓﺼﻞ
: ﻣﻌﺎﺩﻟﻪ ﻣﻤﻨﺘﻮﻡ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﺪﻭﻥ ﺍﺻﻄﻜﺎﻙ3-4
g x = − g cos α x
g y = − g cos α y
g z = − g cos α z
dz
,
dx
∂z
, g y = −g
∂y
∂z
g z = −g
,
∂z
, cosα x =
g x = −g
∂z
∂x
: ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ
DV
Dt
∂P
Du
=ρ
ρg x −
∂x
Dt
∂P
Dv
=ρ
ρg y −
∂y
Dt
∂P
Dw
ρg z −
=ρ
∂z
Dt
ρ g − ∇P = ρ
: ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩﻱ ﻭ ﺭﻭ ﺑﻪ ﺑﺎﻻ ﺑﺎﺷﺪ ﺩﺭ ﺁﻥ ﺻﻮﺭﺕ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪz ﺍﮔﺮ ﻣﺤﻮﺭ
,gy = 0
,
g x = 0 , gz = −g
ρ g = − ρg k
ρg x , ρg y , ρg z , g z = − g
ρ g
⇒ ρ g = − ρg∇z
( )
⇒ ρ g = ρ − gkˆ = − ρg∇z
ρ g = ρg xiˆ + ρg y ˆj + ρg z kˆ = − ρg
⇒ − ρg∇z − ∇P = ρ
∂z ˆ
∂z ˆ
∂z
i − ρg
j − ρg kˆ = − ρg∇z
∂x
∂y
∂z
( )
DV
∂v
= ρ + V ⋅∇ V
Dt
∂t
DV
− ρgkˆ − ∇P = ρ
Dt
۱۱۱
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮﺩﺭﻣﻲ ﺁﻳﺪ:
1 ∂P
∂v
∂v v ∂v
∂v v
gr −
= a r = r + vr r + θ r + v z r − θ
∂t
∂r r ∂θ
∂z
r
ρ ∂r
∂v
∂v v ∂v
∂v v v
1 ∂P
gθ −
= aθ = θ + vr θ + θ θ + v z θ − r θ
∂t
∂r r ∂θ
∂z
r
ρr ∂θ
1 ∂P
∂v
∂v v ∂v
∂v
gz −
= a z = z + vr z + θ z + v z z
∂t
∂r r ∂θ
∂z
ρ ∂z
2
ﺍﮔﺮ ﻣﺤﻮﺭ zﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩﻱ ﺭﻭ ﺑﻪ ﺑﺎﻻ ﺑﺎﺷﺪ:
g r = gθ = 0
gz = −g
4-4ﻣﻌﺎﺩﻻﺕ ﺍﻭﻳﻠﺮ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺧﻂ ﺟﺮﻳﺎﻥ :
ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ،ﻛﻪ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﻱ ﺩﺍﺧﻞ ﺟﺮﻳﺎﻥ ﺑﻪ ﻃﻮﺭ ﻣﻤﺎﺱ ﺑﺮ ﺑﺮﺩﺍﺭﻫﺎﻱ ﺳﺮﻋﺖ ﺭﺳﻢ ﻣﻲ ﺷﻮﺩ ،ﻧﻤﺎﻳﺶ
ﺗﺮﺳﻴﻤﻲ ﻣﻨﺎﺳﺒﻲ ﺭﺍ ﺑﻪ ﻭﺟﻮﺩ ﻣﻲ ﺁﻭﺭﺩ .ﺩﺭ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﺫﺭﻩ ﺳﻴﺎﻟﻲ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ
ﺯﻳﺮﺍ ،ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﺍﺳﺖ ،ﺧﻄﻮﻁ ﻣﺴﻴﺮ ﻭ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺑﺮ ﻫﻢ ﻣﻨﻄﺒﻖ ﻫﺴﺘﻨﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ،ﺑﺮﺍﻱ ﺗﻮﺻﻴﻒ
ﺣﺮﻛﺖ ﺫﺭﻩ ﻱ ﺳﻴﺎﻝ ﺩﺭ ﺟﺮﻳﺎﻧﻲ ﭘﺎﻳﺎ ،ﻓﺎﺻﻠﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﺨﺘﺼﻪ ﻣﻨﺎﺳﺒﻲ ﺑﺮﺍﻱ ﻧﻮﺷﺘﻦ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ
ﺣﺮﻛﺖ ﺍﺳﺖ .ﺍﺯ "ﻣﺨﺘﺼﺎﺕ ﺧﻂ ﺟﺮﻳﺎﻧﻲ" ﺑﺮﺍﻱ ﺗﻮﺻﻴﻒ ﺟﺮﻳﺎﻥ ﻧﺎﭘﺎﻳﺎ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ .ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ
ﻧﺎﭘﺎﻳﺎ ﻧﻤﺎﻳﺸﻲ ﺗﺮﺳﻴﻤﻲ ﺭﺍ ﺑﺮﺍﻱ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﻟﺤﻈﻪ ﺍﻱ ﻣﻲ ﺩﻫﺪ.
ﺑﺮﺍﻱ ﺳﻬﻮﻟﺖ ،ﺟﺮﻳﺎﻥ ﺩﺭ ﺻﻔﺤﻪ yzﺭﺍ،ﺷﻜﻞ ،1-4ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ .ﻣﻲ ﺧﻮﺍﻫﻴﻢ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺣﺮﻛﺖ ﺭﺍ ﺑﺮ
ﺣﺴﺐ ﻣﺨﺘﺼﻪ ، sﻓﺎﺻﻠﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ،ﻭ ﻣﺨﺘﺼﻪ ، nﻋﻤﻮﺩ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ،ﺑﻨﻮﻳﺴﻴﻢ .ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ ﺑﺮﺩﺍﺭ
ﺳﺮﻋﺖ ﺑﺎﻳﺪ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﻤﺎﺱ ﺑﺎﺷﺪ ،ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺑﺎ ) V = V ( s, tﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ .ﻓﺸﺎﺭ ﺩﺭ ﻣﺮﻛﺰ ﺟﺰء
ﺳﻴﺎﻝ p ،ﻣﻲ ﺑﺎﺷﺪ.ﺍﮔﺮ ﻗﺎﻧﻮﻥ ﺩﻭﻡ ﻧﻴﻮﺗﻦ ﺭﺍ ﺩﺭ ﺟﻬﺖ ﺧﻂ ﺟﺮﻳﺎﻥ ) (sﺑﺮﺍﻱ ﺟﺰء ﺳﻴﺎﻝ ﺑﺎ ﺣﺠﻢ ds dn dxﺑﻪ
ﻛﺎﺭ ﺑﺒﺮﻳﻢ ،ﺩﺭ ﺁﻥ ﺻﻮﺭﺕ ﺑﺎ ﺻﺮﻑ ﻧﻈﺮ ﻛﺮﺩﻥ ﺍﺯ ﻧﻴﺮﻭﻫﺎﻱ ﭼﺴﺒﻨﺪﻩ ﺩﺍﺭﻳﻢ،
۱۱۲
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
∂P ds
∂P ds
⋅ dndx − ρg cos θdndxds = ρas dndxds
⋅ dndx − P +
P−
∂s 2
∂s 2
∂P
− ρg cos θ = ρa s
∂s
∂z
= cos θ
∂s
DVs ∂Vs
∂V
= Vs = Vs (s, t ) ⇒ a s
=
+ Vs s
Dt
∂t
∂s
∂V
∂z ∂V
1 ∂P
+V
= −g
⇒−
,
ρ ∂s
∂s
∂s ∂t
⇒−
V = Vs
ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻫﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ) Vs = Vs (s, tﻭ a sﺷﺘﺎﺏ ﻛﻠﻲ ﺫﺭﻩ ﻱ ﺳﻴﺎﻝ ﺩﺭ ﺟﻬﺖ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﻲ ﺑﺎﺷﺪ ﻭ
ﺳﺮﻋﺖ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﻤﺎﺱ ﺍﺳﺖ،ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺯﻳﺮ ﻧﻤﺎﺩ sﺩﺭ Vsﺯﺍﻳﺪ ﺍﺳﺖ ﻭ ﻣﻲ ﺗﻮﺍﻥ ﺁﻥ ﺭﺍ ﺣﺬﻑ ﻛﺮﺩ.
ﺩﺭ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﺑﺎ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﻧﻴﺮﻭﻫﺎﻱ ﺣﺠﻤﻲ ،ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺟﻬﺖ ﺧﻂ ﺟﺮﻳﺎﻧﻲ ﭼﻨﻴﻦ ﻣﻲ ﺷﻮﺩ :
ﻭ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﻛﺎﻫﺶ ﺳﺮﻋﺖ ﺑﺎ ﺍﻓﺰﺍﻳﺶ ﻓﺸﺎﺭ ﻫﻤﺮﺍﻩ ﺍﺳﺖ ﻭ ﺑﺮﻋﻜﺲ.
۱۱۳
1 ∂P
∂V
= −V
∂s
ρ ∂s
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺑﺮﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻥ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺟـﺰ ﺳـﻴﺎﻝ ﺍﺯ ﻗـﺎﻧﻮﻥ ﺩﻭﻡ ﻧﻴـﻮﺗﻦ ﺩﺭ
ﺟﻬﺖ nﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻛﻨﻴﻢ.ﺑﺎ ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﻧﻴﺮﻭﻫﺎﻱ ﺍﺻﻄﻜﺎﻙ ،ﺩﺍﺭﻳﻢ،
∂P dn
∂P dn
P − ⋅ dsdx − P + ⋅ dsdx − ρg cos βdxdnds = ρan dxdnds
∂n 2
∂n 2
anﺷﺘﺎﺏ ﺫﺭﻩ ﺳﻴﺎﻝ ﺩﺭ ﺟﻬﺖ nﻭ βﺯﺍﻭﻳﻪ ﺑﻴﻦ ﺟﻬﺖ nﻭ ﻋﻤﻮﺩ ﺍﺳﺖ.
∂z
∂n
= cos β
ﻭ
∂P
− ρg cos β = ρan
∂n
⇒−
ﺷﺘﺎﺏ ﻋﻤﻮﺩﻱ ﺟﺰ ﺳﻴﺎﻝ ﺑﻪ ﻃﺮﻑ ﻣﺮﻛﺰ ﺍﻧﺤﻨﺎ ﺧﻂ ﺟﺮﻳﺎﻥ ﻭﺩﺭ ﺟﻬﺖ nﻣﻨﻔﻲ ﺍﺳﺖ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺷﺘﺎﺏ ﻣﺮﻛﺰﻱ
ﺩﺭ ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﺷﻜﻞ 4-4ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﭼﻨﻴﻦ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ :
V2
R
⇒ an = −
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ R ،ﺷﻌﺎﻉ ﺍﻧﺤﻨﺎﻱ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﺳﺖ .ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ
ﭼﻨﻴﻦ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ :
∂P
∂z
V2
+ ρg
=ρ
∂n
∂n
R
⇒
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﺩﺭ ﺻﻔﺤﻪ ﺍﻓﻘﻲ ،ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ﭼﻨﻴﻦ ﻣﻲ ﺷﻮﺩ :
1 ∂P V 2
=
ρ ∂n
R
ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﺩﺭ ﺟﻬﺖ ﺧﺎﺭﺝ ﻣﺮﻛﺰ ﺍﻧﺤﻨﺎﻱ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ،ﻓﺸﺎﺭ ﺍﻓﺰﺍﻳﺶ ﻣﻲ ﻳﺎﺑﺪ .ﻫﺮ ﺟﺎ ﺧﻄـﻮﻁ ﺟﺮﻳـﺎﻥ
ﻣﺴﺘﻘﻴﻢ ﺍﺳﺖ،ﺷﻌﺎﻉ
ﺍﻧﺤﻨﺎﻱ R
ﺑﻲ ﻧﻬﺎﻳﺖ ﺍﺳﺖ ﻭ ﻫﻴﭻ ﺗﻐﻴﻴﺮ ﻓﺸﺎﺭﻱ ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﻧﺪﺍﺭﺩ.
5-4ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ-ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ :
1-5-4ﺍﺳﺘﻨﺘﺎﺝ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﺧﻂ ﺟﺮﻳﺎﻧﻲ
ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ :
∂z
∂V
1 ∂P
−g
=V
∂s
∂s
ρ ∂s
ﺍﮔﺮ ﺫﺭﻩ ﺳﻴﺎﻝ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﻪ ﻓﺎﺻﻠﻪ dsﺣﺮﻛﺖ ﻛﻨﺪ ،ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ،
)ﺗﻐﻴﻴﺮ ﻓﺸﺎﺭ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
)ﺗﻐﻴﻴﺮ ﺍﺭﺗﻔﺎﻉ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
)ﺗﻐﻴﻴﺮ ﺳﺮﻋﺖ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﭘﺲ ﺍﺯ ﺿﺮﺏ ﻛﺮﺩﻥ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ، dsﻣﻲ ﺗﻮﺍﻧﻴﻢ ﺑﻨﻮﻳﺴﻴﻢ،
۱۱٤
−
∂P
ds = dP
∂s
∂z
ds = dz
∂s
∂V
ds = dV
∂s
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
dP
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
− gdz = VdV
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
+ gdz + VdV = 0
−
ρ
dP
ρ
ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻣﻲ ﺩﻫﺪ :
2
V
= constant
2
+ gz +
dP
ρ
∫
ﻗﺒﻞ ﺍﺯ ﺍﻳﻨﻜﻪ ﺑﺘﻮﺍﻥ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ ،ﺑﺎﻳﺪ ﺭﺍﺑﻄﻪ ﺑﻴﻦ ﻓﺸﺎﺭ pﻭ ﭼﮕﺎﻟﻲ ρﺭﺍ ﻣﺸﺨﺺ ﻛﻨﻴﻢ .ﺑﺮﺍﻱ ﺣﺎﻟﺖ
ﺧﺎﺹ ﭼﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ، ρ = const. ،ﻣﻌﺎﺩﻟﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺩﺭ ﻣﻲ ﺁﻳﺪ :
V2
= constant
2
+ gz +
P
ρ
ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎ :
(1ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ
(2ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
(3ﺟﺮﻳﺎﻥ ﺑﻲ ﺍﺻﻄﻜﺎﻙ
(4ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ
ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﻛﺎﺭﺍ ﻭ ﻣﻔﻴﺪ ﺍﺳﺖ ﺯﻳﺮﺍ ﺗﻐﻴﻴﺮﺍﺕ ﻓﺸﺎﺭ ﺭﺍ ﺑﻪ ﺗﻐﻴﻴﺮﺍﺕ ﺳﺮﻋﺖ ﻭ ﺍﺭﺗﻔﺎﻉ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ
ﺟﺮﻳﺎﻥ ﺑﻪ ﻫﻢ ﺍﺭﺗﺒﺎﻁ ﻣﻲ ﺩﻫﺪ .ﻭﻟﻲ ،ﻓﻘﻂ ﻭﻗﺘﻲ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﺎ ﭼﻬﺎﺭ ﻣﺤﺪﻭﺩﻳﺖ ﮔﻔﺘﻪ ﺷﺪﻩ ﺑﻪ ﻛﺎﺭ ﻣﻲ
ﺭﻭﺩ ،ﻧﺘﺎﻳﺞ ﺁﻥ ﺻﺤﻴﺢ ﻭ ﻗﺎﺑﻞ ﻗﺒﻮﻝ ﺍﺳﺖ .ﻫﺮ ﻭﻗﺖ ﻣﻲ ﺧﻮﺍﻫﻴﺪ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ ،ﺑﻪ
ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺁﻥ ﺗﻮﺟﻪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﺪ).ﺑﻪ ﻃﻮﺭ ﻛﻠﻲ ،ﺛﺎﺑﺖ ﺑﺮﻧﻮﻟﻲ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ
ﻣﺨﺘﻠﻒ ﺩﺍﺭﺍﻱ ﻣﻘﺎﺩﻳﺮ ﻣﺨﺘﻠﻒ ﺍﺳﺖ(.
2-5-4ﺍﺳﺘﻨﺘﺎﺝ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﺩﻛﺎﺭﺗﻲ
V
ﺷﻜﻞ ﺑﺮﺩﺍﺭﻱ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺭﺍ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﻛﺮﺩ .ﺍﮔﺮ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺩﺭ
ﻣﺨﺘﺼﺎﺕ ﺩﻛﺎﺭﺗﻲ z , y, xﺑﻴﺎﻥ ﺷﻮﺩ ،ﺑﻬﺘﺮ ﺍﺳﺖ ﺍﺯ ﻧﻤﺎﺩ ﺑﺮﺩﺍﺭﻱ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ .ﺍﺳﺘﻨﺘﺎﺝ ﺭﺍ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ
2
ﻣﺤﺪﻭﺩ ﻣﻲ ﻛﻨﻴﻢ؛ ﺍﺯ ﺍﻳﻦ ﺭﻭ ﻧﺘﻴﺠﻪ ﻧﻬﺎﻳﻲ ﻋﻤﻠﻴﺎﺕ ،ﻣﻌﺎﺩﻟﻪ ∫ dP + gz + V = constantﺧﻮﺍﻫﺪ ﺑﻮﺩ.
2
ρ
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺭﺍ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺩﻛﺎﺭﺗﻲ ﻣﻲ ﺗﻮﺍﻥ ﭼﻨﻴﻦ ﺑﻴﺎﻥ ﻛﺮﺩ :
) (
DV
∂V
∂V
∂V
= ˆ∇P − gk
=u
+v
'+ w
= V ⋅∇ V
Dt
∂x
∂y
∂z
ρ
1
−
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺑﺎ ﺭﺍﺑﻄﻪ ) V = V (x, y, zﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ .ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺧﻄﻮﻃﻲ ﻫﺴﺘﻨﺪ ﻛﻪ
ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﻣﻤﺎﺱ ﺑﺮ ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺭﺳﻢ ﻣﻲ ﺷﻮﻧﺪ .ﺣﺮﻛﺖ ﻳﻚ ﺫﺭﻩ ﺩﺭ ﺍﻣﺘﺪﺍﺩﻳﻚ
۱۱٥
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﻣﺸﺨﺺ ﻣﻲ ﺷﻮﺩ .ﺩﺭ ﺑﺎﺯﻩ ﺯﻣﺎﻧﻲ ، dtﺫﺭﻩ ﺩﺍﺭﺍﻱ ﺑﺮﺩﺍﺭ ﺟﺎﺑﺠﺎﻳﻲ dsﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ
ﺟﺮﻳﺎﻥ ﺍﺳﺖ.
ﺍﮔﺮ ﺿﺮﺏ ﻧﻘﻄﻪ ﺍﻱ ﺟﻤﻠﻪ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍ ﺑﺎ ﺟﺎﺑﺠﺎﻳﻲ dsﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﺤﺎﺳﺒﻪ ﻛﻨﻴﻢ ﻳﻚ ﻣﻌﺎﺩﻟﻪ
ﺍﺳﻜﺎﻟﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ ﻛﻪ ﻓﺸﺎﺭ ، pﺳﺮﻋﺖ Vﻭ ﺍﺭﺗﻔﺎﻉ zﺭﺍ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺑﻪ ﻫﻢ ﺍﺭﺗﺒﺎﻁ ﻣﻲ
ﺩﻫﺪ.ﺑﺎ ﻣﺤﺎﺳﺒﻪ ﻱ ﺿﺮﺏ ﻧﻘﻄﻪ ﺍﻱ dsﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺩﺍﺭﻳﻢ :
) (
∇P ⋅ d s − gkˆ ⋅ d s = V ⋅ ∇ V ⋅ d s
1
ρ
−
ˆd s = dxiˆ + dyˆj + dzk
[
]
1 ˆ ∂P ˆ ∂P ˆ ∂P
+ j
ˆ+ k ⋅ dxiˆ + dyˆj + dzk
i
∂y
∂z
ρ
ρ ∂x
∂P
1 ∂P
∂P
1
= − ⋅ dx + ⋅ dy + ⋅ dz = − dP
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
∂y
ρ ∂x
ρ
∂z
∇P ⋅ d s = −
]
)ﺩﺭﺍﻣﺘﺪﺍﺩ ( s
d
ds
[
1
−
= ∇⋅ s
− gkˆ ⋅ d s = − gkˆ ⋅ dxiˆ + dyˆj + dzkˆ = − gdz
) (V ⋅ ∇)V = 12 ∇(V ⋅V )− V × (∇ ×V
(V ⋅ ∇)V ⋅ d s = 12 ∇(V ⋅V )− V × (∇ ×V ) ⋅ d s
})
) (
( {
1
= ∇ V ⋅V ⋅ d s − V × ∇ × V ⋅ d s
2
)] (u ⋅ [v × w]) = (v ⋅ [w × u
ﺟﻤﻠﻪ ﺁﺧﺮ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺻﻔﺮ ﺍﺳﺖ .ﺯﻳﺮﺍ Vﺑﻪ ﻣﻮﺍﺯﺍﺕ d sﺍﺳﺖ.ﺩﺭ ﻧﺘﻴﺠﻪ،
]
[
∇ ×V ⋅ V × d s
(V ⋅ ∇)V ⋅ d s = 12 ∇(V ⋅V )⋅ d s = 12 ∇(V )⋅ d s
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
2
]
[
1 ˆ ∂V 2 ˆ ∂V 2 ˆ ∂V 2
= i
+j
+k
ˆ⋅ dxiˆ + dyˆj + dzk
2 ∂x
∂y
∂z
∂V 2
∂V 2 1
1 ∂V 2
=
⋅ dx +
dy +
dz = d V 2
∂y
∂z 2
2 ∂x
1
) V ⋅ ∇ V ⋅ d s = d (V 2
2
) (
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
) (
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
ﺑﺎ ﺟﺎﻳﮕﺬﺍﺭﻱ ﺍﻳﻦ ﺳﻪ ﺟﻤﻠﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺩﺍﺭﻳﻢ،
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
) (
1
+ d V 2 + gdz = 0
ρ 2
۱۱٦
dP
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺩﺍﺭﻳﻢ،
)ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
2
V
+ gz = constant
2
+
dP
ρ
∫
ﺍﮔﺮ ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ ﺑﺎﺷﺪ ،ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺭﺍ ﻣﻲ ﻳﺎﺑﻴﻢ،
P
1
+ V 2 + gz = constant
ρ 2
ﻧﻜﺘﻪ ( ﻓﺸﺎﺭ ﺗﺮﻣﻮﺩﻳﻨﺎﻣﻴﻜﻲ= ﻓﺸﺎﺭ ﺍﺳﺘﺎﺗﻴﻜﻲ
=P
ﻓﺸﺎﺭ ﺗﺮﻣﻮﺩﻳﻨﺎﻣﻴﻜﻲ ﺗﻮﺳﻂ ﻳﻚ ﺭﺍﺑﻄﻪ ﺗﺮﻣﻮﺩﻳﻨﺎﻣﻴﻜﻲ )ﻣﻌﺎﺩﻟﻪ ﺣﺎﻟﺖ( ﺑﺮ ﺣﺴﺐ ﺣﺠﻢ ﻭ ﺩﻣﺎ ﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ.
σ xx + σ yy + σ zz
3
P=−
ﻓﺸﺎﺭ ﺍﺳﺘﺎﺗﻴﻜﻲ :ﻓﺸﺎﺭﻱ ﻛﻪ ﺫﺭﻩ ﻣﺘﺤﺮﻙ ﺳﻴﺎﻝ ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ) .ﻟﺬﺍ ﻳﻚ ﻧﺎﻡ ﺑﻲ ﻣﺴﻤﻲ ﺍﺳﺖ!(
6-4ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ
ﺑﺮﺍﻱ ﻳﻚ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﻏﻴﺮ ﭼﺴﺒﻨﺪﻩ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﻧﺘﮕﺮﺍﻝ ﮔﺮﻓﺘﻴﻢ ﻭ
ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻳﻢ،
P
1
+ V 2 + gz = constant
ρ 2
ﻣﻌﺎﺩﻟﻪ ﺭﺍ ﺑﻴﻦ ﻫﺮ ﺩﻭ ﻧﻘﻄﻪ ﻱ ﺩﻟﺨﻮﺍﻩ ﺭﻭﻱ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ.ﺑﻪ ﻃﻮﺭ ﻛﻠﻲ ،ﻣﻘﺪﺍﺭ ﺛﺎﺑﺖ ﺍﺯ
ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺗﺎ ﺧﻂ ﺟﺮﻳﺎﻥ ﺩﻳﮕﺮ ﺗﻐﻴﻴﺮ ﻣﻲ ﻛﻨﺪ..
ﺍﮔﺮ ،ﻋﻼﻭﻩ ﺑﺮ ﺍﻳﻨﻜﻪ ﺟﺮﻳﺎﻥ ﻏﻴﺮ ﭼﺴﺒﻨﺪﻩ ،ﭘﺎﻳﺎ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﻧﻴﺰ ﺑﺎﺷﺪ )
ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﻃﻮﺭﻱ ﺑﺎﺷﺪ ﻛﻪ ،( 2ω = ∇ × V = 0ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺭﺍ ﺑﻴﻦ ﻫﺮ ﺩﻭ ﻧﻘﻄﻪ
ﺩﻟﺨﻮﺍﻩ ﺩﺭ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ.ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﻣﻘﺪﺍﺭ ﺛﺎﺑﺖ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺑﺮﺍﻱ ﺗﻤﺎﻡ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ
ﻣﻘﺪﺍﺭﻱ ﺛﺎﺑﺖ ﺍﺳﺖ ..ﺑﺮﺍﻱ ﺍﺛﺒﺎﺕ ﺍﻳﻦ ﻣﻮﺿﻮﻉ ،ﺑﺎ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺑﻪ ﺷﻜﻞ ﺑﺮﺩﺍﺭﻱ ﺷﺮﻭﻉ ﻣﻲ ﻛﻨﻴﻢ،
) (
) (V ⋅ ∇)V = 12 ∇(V ⋅V )− V × (∇ ×V
∇P − gkˆ = V ⋅ ∇ V
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ،ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ ، (∇ × V ) = 0ﺩﺍﺭﻳﻢ :
ﻭ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺭﺍ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﻣﻲ ﺗﻮﺍﻥ ﭼﻨﻴﻦ ﻧﻮﺷﺖ :
1
ρ
) (V ⋅ ∇)V = 12 ∇(V ⋅V
) (
) (
1
1
∇P − gkˆ = ∇ V ⋅ V = ∇ V 2
ρ
2
2
۱۱۷
−
1
−
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺩﺭ ﺑﺎﺯﻩ ﺯﻣﺎﻧﻲ dtﻳﻚ ﺫﺭﻩ ﺳﻴﺎﻝ ﺍﺯ ﺑﺮﺩﺍﺭ ﻣﻜﺎﻥ rﺗﺎ ﺑﺮﺩﺍﺭ ﻣﻜﺎﻥ r + d rﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ .ﺗﻐﻴﻴـﺮ ﻣﻜـﺎﻥ d r
ﺗﻐﻴﻴﺮ ﻣﻜﺎﻥ ﺑﻲ ﻧﻬﺎﻳﺖ ﻛﻮﭼﻚ ﺍﺧﺘﻴﺎﺭﻱ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺩﻟﺨﻮﺍﻩ ﺍﺳﺖ .ﺍﮔﺮ ﺿﺮﺏ ﺩﺍﺧﻠـﻲ ˆ d r = dxiˆ + dyˆj + dzkﺭﺍ
ﺑﺎ ﻫﺮ ﻳﻚ ﺍﺯ ﺟﻤﻠﻪ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺩﺍﺭﻳﻢ :
1
1
− ∇P ⋅ d r − gkˆ ⋅ d r = ∇(V 2 )⋅ d r
2
ρ
1
dP
⇒−
) − gdz = d (V 2
2
ρ
dP 1
⇒
+ d (V 2 ) + gdz = 0
ρ 2
ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺩﺍﺭﻳﻢ،
V2
+ gz = constant
2
+
dP
ρ
∫
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ، ρ = const. ،ﻭ
P
1
+ V 2 + gz = constant
ρ 2
ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ d rﻳﻚ ﺗﻐﻴﻴﺮ ﻣﻜﺎﻥ ﺍﺧﺘﻴﺎﺭﻱ ﺑﻮﺩ ﺑﺮﺍﻱ ﻳﻚ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ،ﺗـﺮﺍﻛﻢ ﻧﺎﭘـﺬﻳﺮ ﻭ ﻏﻴـﺮ ﭼﺴـﺒﻨﺪﻩ ﻛـﻪ ﺑـﻲ
ﭼﺮﺧﺶ ﻧﻴﺰ ﺑﺎﺷﺪ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺑﻴﻦ ﻫﺮ ﺩﻭ ﻧﻘﻄﻪ ﺩﻟﺨﻮﺍﻩ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺻـﺤﺖ ﺩﺍﺭﺩ .ﺩﺭ ﻧﺘﻴﺠـﻪ ﺛﺎﺑـﺖ ﻣﻌﺎﺩﻟـﻪ
ﺑﺮﻧﻮﻟﻲ ﺑﺮﺍﻱ ﺗﻤﺎﻡ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﻣﻘﺪﺍﺭﻱ ﺛﺎﺑﺖ ﺍﺳﺖ.
7-4ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﻧﺎﭘﺎﻳﺎ -ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ
ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺭﺍ ﻧﺒﺎﻳﺪ ﺑﻪ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﻣﺤﺪﻭﺩ ﻛﺮﺩ .ﻫﺪﻑ ﺍﻳﻦ ﻗﺴﻤﺖ ﻳﺎﻓﺘﻦ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺑﺮﺍﻱ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ
ﻧﺎﭘﺎﻳﺎ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﻭ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﻛﺎﺭﺑﺮﺩ ﺁﻥ ﺑﺎ ﻣﺜﺎﻝ ﺍﺳﺖ.
ﻣﻌﺎﺩﻟﻪ ﺗﻜﺎﻧﻪ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﺍﺻﻄﻜﺎﻙ ﭼﻨﻴﻦ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺑﺎﺷﺪ :
DV
1
= ∇P − g∇z
ρ
Dt
−
ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺑﺮﺩﺍﺭﻱ ﺍﺳﺖ.ﺍﮔﺮ ﻃﺮﻓﻴﻦ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍ ﺩﺭ d sﺿﺮﺏ ﺩﺍﺧﻠﻲ ﻛﻨﻴﻢ ﻛـﻪ ﺩﺭ ﺁﻥ d sﺟـﺰ
ﻓﺎﺻﻠﻪ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﺳﺖ ،ﺑﻪ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺍﺳﻜﺎﻟﺮ ﺗﺒﺪﻳﻞ ﻣﻲ ﺷﻮﺩ.ﺍﺯ ﺍﻳﻦ ﺭﻭ،
DVs
∂V
∂V
DV
1
ds = Vs s ds + s ds
= ∇P ⋅ d s − g∇z ⋅ d s
= ⋅ds
Dt
Dt
ρ
∂s
∂t
−
ﺑﺎ ﺑﺮﺭﺳﻲ ﺟﻤﻠﻪ ﺍﻱ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ،ﺗﻮﺟﻪ ﻣﻲ ﻛﻨﻴﻢ ﻛﻪ :
)ﺗﻐﻴﻴﺮ ﻓﺸﺎﺭ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
∇P ⋅ d s = dP
)ﺗﻐﻴﻴﺮ zﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
∇z ⋅ d s = dz
)ﺗﻐﻴﻴﺮ Vsﺩﺭ ﺍﻣﺘﺪﺍﺩ ( s
۱۱۸
∂Vs
ds = dVs
∂s
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﺑﺎ ﺟﺎﻳﮕﺬﺍﺭﻱ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ،ﺩﺍﺭﻳﻢ،
ﺑﺎ ﺍﻧﺘﮕﺮﺍﻝ ﮔﻴﺮﻱ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﺯ ﻧﻘﻄﻪ 1ﺗﺎ 2ﺩﺍﺭﻳﻢ،
∂Vs
ds
∂t
− gdz = Vs dVs +
2 ∂V
V − V1
s
+ 2
∫ + g (z 2 − z1 ) +
ds = 0
1
2
∂t
ρ
2
2
dP
dP
ρ
2
∫
1
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ،ﭼﮕﺎﻟﻲ ﺛﺎﺑﺖ ﺍﺳﺖ .ﺑﺮﺍﻱ ﺍﻳﻦ ﺣﺎﻟﺖ ﺧﺎﺹ ﻣﻌﺎﺩﻟﻪ ﭼﻨﻴﻦ ﻣﻲ ﺷﻮﺩ :
2 ∂V
P1 V1
P V
s
+
∫ + gz1 = 2 + 2 + gz 2 +
ds
1
∂t
2
ρ 2
ρ
2
2
ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎ :
(1ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
(2ﺟﺮﻳﺎﻥ ﺑﻲ ﺍﺻﻄﻜﺎﻙ
(3ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ
∂V
ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﻱ ﺟﻤﻠﻪ ﺍﻧﺘﮕﺮﺍﻝ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ،ﺗﻐﻴﻴﺮ ∂tSﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ sﺑﺎﻳﺪ ﻣﻌﻠﻮﻡ ﺑﺎﺷﺪ؛ sﻓﺎﺻﻠﻪ
∂V
ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﻧﻘﻄﻪ 1ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻣﻲ ﺷﻮﺩ) ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ∂tS = 0ﺍﺳﺖ(.ﻣﻌﺎﺩﻟﻪ
ﺑﺎﻻ ﺭﺍ ﺑﺮﺍﻱ ﻫﺮ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﻣﺤﺪﻭﺩﻳﺖ ﻫﺎﻱ ﺁﻥ ﺑﺎ ﺷﺮﺍﻳﻂ ﻓﻴﺰﻳﻜﻲ ﺳﺎﺯﮔﺎﺭ ﻫﺴﺘﻨﺪ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ.
ﻣﺜﺎﻝ :1-4ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﻧﺎﭘﺎﻳﺪﺍﺭ ﻟﻮﻟﻪ ﻃﻮﻳﻠﻲ ﺑﻪ ﻣﺨﺰﻧﻲ ﻣﺘﺼﻞ ﺷﺪﻩ ﻛﻪ ﺩﺭ ﺍﺑﺘﺪﺍ ﺗﺎ ﻋﻤﻖ 3mﺁﺏ ﺩﺍﺭﺩ .ﻟﻮﻟﻪ ﺑﻪ
ﻗﻄﺮ 150mmﻭ ﺑﻪ ﻃﻮﻝ 6mﺍﺳﺖ .ﺑﻪ ﻋﻨﻮﺍﻥ ﺗﻘﺮﻳﺐ ﺍﻭﻝ ،ﺍﺯ ﺍﺻﻄﻜﺎﻙ ﺻـﺮﻑ ﻧﻈـﺮ ﻛﻨﻴـﺪ .ﭘـﺲ ﺍﺯ ﺑﺮﺩﺍﺷـﺘﻦ
ﺩﺭﭘﻮﺵ ﺍﺯ ﺳﺮ ﺁﺯﺍﺩ ﻟﻮﻟﻪ ،ﺳﺮﻋﺖ ﺟﺮﻳﺎﻥ ﺧﺮﻭﺟﻲ ﺍﺯ ﻟﻮﻟﻪ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ ﺯﻣﺎﻥ ﺑﻴﺎﺑﻴﺪ.
۱۱۹
−
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
ﻓﺮﺿﻴﺎﺕ:
(1ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ
(2ﺟﺮﻳﺎﻥ ﺑﺪﻭﻥ ﺍﺻﻄﻜﺎﻙ
(3ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺧﻂ ﺟﺮﻳﺎﻥ 1ﺗﺎ 2
P1 = P2 = Patm (4
2
V1 ≅ 0 (5
z 2 = 0 (6
(7ﺛﺎﺑﺖ = z1 = 3
(8ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺳﺮﻋﺖ ﺩﺭ ﻣﺨﺰﻥ ،ﺑﻪ ﺟﺰ ﺑﺮﺍﻱ ﻧﺎﺣﻴﻪ ﻛﻮﭼﻚ ﻧﺰﺩﻳﻚ ﺑﻪ ﻭﺭﻭﺩﻱ ﻟﻮﻟﻪ
2 ∂V
V
P V
s
ds
∫ + 1 + gz1 = 2 + 2 + gz 2 +
1
ρ 2
ρ
∂t
2
2
2
P1
2 ∂V
V
s
∫ gz1 = 2 +
ds
1
∂t
2
2 ∂V
L ∂V
s
ds
≈
∫1 ∂t
∫0 ∂ts ds
2
ﻓﺮﺽ : 8
ﺩﺭ ﻟﻮﻟﻪ ﻫﻤﻪ ﺟﺎ Vs = V2ﺍﺳﺖ.
L ∂V
∂Vs
dV
2
∫ = ds
ds = L 2
0 ∂t
0 ∂t
dt
2
V
dV
dV2
dt
⇒ ⇒ gz1 = 2 + L 2
=
2
dt
2
2L
2 gz1 − V2
L
z1 = h
∫⇒
V t
t dt
dV2
1
= tanh −1 2
∫=
⇒
2
2 gh 2 L
0 2 gz − V
0 2L
2 gz1
1
2
V2
t
⇒
= tanh
⋅ 2 gh
2 gh
2L
V2
∫
,
s
2 gh = 2 × 9.81 × 3 = 7.67 m
t
2 gh = 0.639t
2L
) ⇒ V2 = 7.67 tanh (0.639t
۱۲۰
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
θ
1-4ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺩﺭ ﻳﻚ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ψ = −Ur sin θ + q πﻋﺒـﺎﺭﺕ ﻣﻴـﺪﺍﻥ
2
ﺳﺮﻋﺖ ،ﻭﻧﻘﻄﻪ ﻳﺎ ﻧﻘﺎﻁ ﺭﻛﻮﺩ ﺭﺍ ﻛﻪ ﺩﺭ ﺁﻥ V = 0ﺑﻴﺎﺑﻴﺪ ﻭﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺩﺭ ﺁﻧﺠﺎ ψ = 0؟
2-4ﻣﻮﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ ﺩﺭ ﻳﻚ ﺟﺮﻳﺎﻥ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ υ = − y 3 − 4 zﻭ . w = 3 y 2 z
ﺍﻟﻒ -ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﻳﻚ ﺑﻌﺪﻱ ﺍﺳﺖ ،ﺩﻭﺑﻌﺪﻱ ﻳﺎ ﺳﻪ ﺑﻌﺪﻱ؟
ﺏ -ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﺗﺮﺍﻛﻢ ﭘﺬﻳﺮ؟
ﺝ -ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
3-4ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻧﻲ ﺩﺍﺭﺍﻱ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ )
ﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ؟ﺁﻳﺎ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ؟
(
2π x 2 + y 2
ψ = − Aﻭ A = cteﺁﻳﺎ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﻭ ﺗﺮﺍﻛﻢ
4-4ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ∧ V = Axyi ∧ + By 2 jﺭﺍ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ ،ﻛـﻪ ﺩﺭ ﺁﻥ A = 4m −1 s −1ﻭ B = −2m −1 s −1
ﻣﺨﺘﺼــﺎﺕ ﺑــﺮ ﺣﺴــﺐ ﻣﺘــﺮ ﻫﺴــﺘﻨﺪ.ﭼﺮﺧﺶ ﺳــﻴﺎﻝ ﻭ ﮔــﺮﺩﺵ ﭘﻴﺮﺍﻣــﻮﻥ ﻣﻨﺤﻨــﻲ ﻣﺤــﺪﻭﺩ ﺷــﺪﻩ ﺑــﺎ ﺧﻄــﻮﻁ
y = o, x = 1, y = 1ﻭ x = 0ﺭﺍ ﺑﻴﺎﺑﻴﺪ.ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﻌﻴﻴﻦ ﻭ ﭼﻨﺪ ﺧﻂ ﺟﺮﻳﺎﻥ ﺭﺍ ﺩﺭ ﺭﺑﻊ ﺍﻭﻝ ﺭﺳﻢ ﻛﻨﻴﺪ؟
5-4ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻧﻲ ﺩﺍﺭﺍﻱ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ψ = x 2 − y 2ﺍﺳﺖ.ﻣﻴﺪﺍﻥ ﺳـﺮﻋﺖ ﺭﺍ ﺑﻴﺎﺑﻴـﺪ .ﻧﺸـﺎﻥ ﺩﻫﻴـﺪ ﻛـﻪ ﺍﻳـﻦ
ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ .ﭼﻨﺪ ﺧﻂ ﺟﺮﻳﺎﻥ ﺭﺍ ﺭﺳﻢ ﻛﻨﻴﺪ ﻭ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺭﺍ ﺩﺭ ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭ ﻧﺸﺎﻥ ﺩﻫﻴﺪ؟
6-4ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺩﺭ ﻧﺰﺩﻳﻚ ﻣﺮﻛﺰ ﻳﻚ ﮔﺮﺩﺑﺎﺩ ﺑﺎ ﺭﺍﺑﻄﻪ ﺗﻘﺮﻳﺒﻲ ﺯﻳﺮ ﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ :
ﺁﻳﺎ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ؟ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
∧ −q
k
∧
eθ
er +
=V
2πr
2πr
7-4ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ،ﺟﺮﻳﺎﻥ ﻭﻳﺴﻜﻮﺯﻱ ﺩﺭ ﻓﻀﺎﻱ ﺑﻴﻦ ﺩﻭ ﺻﻔﺤﻪ ﺑﺰﺭگ ﻣﻮﺍﺯﻱ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ .ﻣﻴﺪﺍﻥ
mm
ﺳﺮﻋﺖ ﺩﺭ ﺍﻳﻦ ﻓﻀﺎ ﺑﺎ ﺭﺍﺑﻄﻪ ∧ V = U y iﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ،ﻛﻪ ﺩﺭ ﺁﻥ
s
h
، t = 0ﭘﺎﺭﻩ ﺧﻂ ﻫﺎﻱ acﻭ bdﻧﺸﺎﻥ ﮔﺬﺍﺭﻱ ﺷﺪﻩ ﺍﻧﺪ .ﻣﻜﺎﻥ ﻧﻘﺎﻁ ﻧﺸﺎﻥ ﮔﺬﺍﺭﻱ ﺷﺪﻩ ﺭﺍ ﺩﺭ t = 1.5sﺑﻴﺎﺑﻴﺪ ﻭ
U = 4ﻭ . h = 4mmﺩﺭ
ﺍﻳﻦ ﻧﻘﺎﻁ ﺭﺍ ﺭﺳﻢ ﻛﻨﻴﺪ .ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﺯﺍﻭﻳﻪ ﺍﻱ ﻭ ﺁﻫﻨﮓ ﭼﺮﺧﺶ ﻳﻚ ﺫﺭﻩ ﺳﻴﺎﻝ ﺭﺍ ﺩﺭ ﺍﻳﻦ ﻣﻴﺪﺍﻥ
ﺳﺮﻋﺖ ﺑﻴﺎﺑﻴﺪ؟
۱۲۱
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
8-4ﻣﻨﺤﻨﻲ ﺳﺮﻋﺖ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻛﺎﻣﻼ ﻓﺮﺍﮔﻴﺮ ﺩﺭ ﻳﻚ ﻟﻮﻟﻪ ﺩﺍﻳﺮﻩ ﺍﻱ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :
r 2
Vz = Vmax 1 −
R
ﺁﻫﻨﮓ ﺗﻐﻴﻴﺮ ﺷﻜﻞ ﻫﺎﻱ ﺧﻄﻲ ﻭ ﺯﺍﻭﻳﻪ ﺍﻱ ﺭﺍ ﺩﺭ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ,ﻭ ﻋﺒﺎﺭﺕ ﺑﺮﺩﺍﺭ ﮔﺮﺩﺍﺑﻲ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
9-4ﺩﺭ ﻣﺠﺮﺍﻱ ﺷﻜﻞ ﺯﻳﺮ ﻫﻮﺍ ﺍﺯ ﭼﭗ ﺑﻪ ﺭﺍﺳﺖ ﺟﺮﻳﺎﻥ ﺩﺍﺭﺩ .ﺩﺭ ﻛﺪﺍﻡ ﻗﺴﻤﺖ ﺟﺮﻳﺎﻥ ﻣﻲ ﺗﻮﺍﻥ ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ
ﺭﺍ ﺍﻳﺪﻩ ﺍﻝ ﻭ ﻏﻴﺮﭼﺮﺧﺸﻲ ﻓﺮﺽ ﻛﺮﺩ؟ﭼﺮﺍ؟ﻓﺸﺎﺭ ﺩﺭ ﭼﻪ ﻧﻘﻄﻪ ﺍﻱ ﻣﻴﻨﻴﻤﻢ ﺍﺳﺖ؟
10-4ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺭﺍ ﺑﺮﺍﻱ ﺭﻭ ﺣﺎﻟﺖ ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﻭ ﺩﺭﻫﻢ ﺩﺭ ﻣﻘﻄﻊ 1-1ﺭﺳﻢ ﻛﻨﻴﺪ؟
11-4ﺩﺭ ﺟﺮﻳﺎﻥ ﻏﻴﺮ ﭼﺮﺧﺸﻲ ﺁﺏ ﺑﺎ ﺩﻣﺎ 15c ﺍﻃﺮﺍﻑ ﺍﺳﺘﻮﺍﻧﻪ ,ﺍﮔﺮ ﺳﺮﻋﺖ ﻭ ﻓﺸﺎﺭ ﺩﺭ ﻓﺎﺻﻠﻪ ﺑﻴﻨﻬﺎﻳﺖ ﺩﻭﺭ ﺍﺯ
ﺟﺴﻢ ﺑﻪ ﺗﺮﺗﻴﺐ 20 m sﻭ ﻓﺸﺎﺭ ﺍﺗﻤﺴﻔﺮ ﺑﺎﺷﻨﺪ ,ﺁﻳﺎ ﺩﺭ ﻧﻘﻄﻪ ﺍﻱ ﻛﻪ ﺳﺮﻋﺖ ﺑﻪ ﺑﻴﺸﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺧﻮﺩ ﻣﻲ ﺭﺳﺪ ,
ﺍﻣﻜﺎﻥ ﺗﺒﺨﻴﺮ )ﻛﺎﻭﻳﺘﺎﺳﻴﻮﻥ( ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟
۱۲۲
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٤ﻣﻘﺪﻣﻪ ﺍی ﺑﺮ ﺗﺤﻠﻴﻞ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﯽ ﺣﺮﮐﺖ ﺳﻴﺎﻝ
12-4ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺳﻴﺎﻝ ﻏﻴﺮ ﻗﺎﺑﻞ ﺗﺮﺍﻛﻢ ﻭ ﺑﺪﻭﻥ ﺍﺻﻄﻜﺎﻙ ﺑﻪ ﺻﻮﺭﺕ ψ = x 2 + y 2ﻣﻲ ﺑﺎﺷﺪ ﺁﻳﺎ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ
ﻛﻤﻚ ﺭﺍﺑﻄﻪ ﺑﺮﻧﻮﻟﻲ ﺍﺧﺘﻼﻑ ﻓﺸﺎﺭ ﺭﺍ ﻣﺎ ﺑﻴﻦ ﺩﻭ ﻧﻘﻄﻪ ) (x, y ) = (1,1ﻭ ) (x, y ) = (1,2ﻣﺤﺎﺳﺒﻪ ﻧﻤﻮﺩ؟
13-4ﻣﻴﺪﺍﻥ ﺳﺮﻋﺘﻲ ﺑﻪ ﺻﻮﺭﺕ ∧ V = 2 yi ∧ + jﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﻭﺍﺣﺪ ﺳﺮﻋﺖ ﺑﺮﺣﺴﺐ ﻓﻮﺕ ﺑﺮ
ﺛﺎﻧﻴﻪ ﻣﻲ ﺑﺎﺷﺪ ﺷﻴﺐ ﺧﻂ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﺍﺯ ﻧﻘﻄﻪ ) (1,2,0ﻣﻴﮕﺬﺭﺩ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ؟
∧
∧
∧
14-4ﻳﻚ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻪ ﺻﻮﺭﺕ ﺭﺍﺑﻄﻪ ﻣﻘﺎﺑﻞ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ V = axi + byj + czkﻣﻌﺎﺩﻟﻪ ﺧﻄﻮﻁ
ﺟﺮﻳﺎﻥ ﺩﺭ ﻟﺤﻈﻪ t = 0ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ؟
15-4ﻳﻚ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺩﻭ ﺑﻌﺪﻱ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺗﻮﺳﻂ ﺭﻭﺍﺑﻂ ﺯﻳﺮ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ
2
2
u = 2 x + yﻭ V = yf ( x) + x 2ﻣﻘﺪﺍﺭ ﺗﺎﺑﻊ ) f (xﺑﺮﺍﻱ ﺁﻧﻜﻪ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺍﺯ ﻧﻈﺮ ﻓﻴﺰﻳﻜﻲ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ
ﺑﺎﺷﺪ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ؟
16-4ﺩﺭ ﺟﺮﻳﺎﻥ 2ﺑﻌﺪﻱ ﺍﮔﺮ
u = 2 xy
ﺑﺎﺷﺪ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ؟
2
V = x + 1
۱۲۳
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
۱۲٤
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
1-5ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺭﺍﺑﻄﻪ ﺍﻱ ﺑﻪ ﻧﺎﻡ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ φﺭﺍ ﺑﺮﺍﻱ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺑﻲ ﭼﺮﺧﺸﻲ ﻣﻲ ﺗﻮﺍﻧﻴﻢ ﻓﺮﻣﻮﻝ ﺑﻨﺪﻱ ﻛﻨﻴﻢ .ﺑﺮﺍﻱ ﺍﻧﺠﺎﻡ
ﺍﻳﻦ ﻛﺎﺭ ،ﺑﺎﻳﺪ ﺍﺯ ﺍﺗﺤﺎﺩ ﺑﺮﺩﺍﺭﻱ ﺍﺻﻠﻲ ﺯﻳﺮ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﻢ،
curl (grad φ ) = ∇ × ∇φ = 0
ﺍﺗﺤﺎﺩ ﺑﺎﻻ ﻭﻗﺘﻲ ﺻﺤﺖ ﺩﺍﺭﺩ ﻛﻪ φﺗﺎﺑﻊ ﺍﺳﻜﺎﻟﺮ ) ﺍﺯ ﻣﺨﺘﺼﺎﺕ ﻓﻀﺎﻳﻲ ﻭ ﺯﻣﺎﻧﻲ ( ﺑﺎﺷﺪ ﻭ ﺩﺍﺭﺍﻱ ﻣﺸﺘﻖ ﻫﺎﻱ ﺍﻭﻝ
ﻭ ﺩﻭﻡ ﭘﻴﻮﺳﺘﻪ ﺑﺎﺷﺪ.
ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺑﺮﺍﻱ ﻳﻚ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﻛﻪ ﺩﺭ ﺁﻥ ، ∇ × V = 0ﺗﺎﺑﻊ ﺍﺳﻜﺎﻟﺮ φﺑﺎﻳﺪ ﻭﺟﻮﺩ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ﺑﻪ ﻃﻮﺭﻱ
ﻛﻪ ﮔﺮﺍﺩﻳﺎﻥ φﻣﺴﺎﻭﻱ ﺑﺎ ﺑﺮﺩﺍﺭ ﺳﺮﻋﺖ Vﺑﺎﺷﺪ .ﺑﺮﺍﻱ ﺍﻳﻨﻜﻪ ﺟﻬﺖ ﻣﺜﺒﺖ ،ﺟﺮﻳﺎﻥ ﺩﺭ ﺟﻬﺖ ﻛﺎﻫﺶ φﺑﺎﺷﺪ )
ﻣﺸﺎﺑﻪ ﺑﺎ ﺟﻬﺖ ﻣﺜﺒﺖ ﺍﻧﺘﻘﺎﻝ ﮔﺮﻣﺎ ﻛﻪ ﺩﺭ ﺟﻬﺖ ﻛﺎﻫﺶ ﺩﻣﺎ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ( φ ،ﺭﺍ ﻃﻮﺭﻱ ﺗﻌﺮﻳﻒ ﻣﻲ ﻛﻨﻴﻢ
ﻛﻪ:
V ≡ −∇φ
ﺍﺯ ﺍﻳﻦ ﺭﻭ
∂φ
∂z
w=−
,
∂φ
∂y
v=−
∂φ
∂x
,
u=−
ﺑﺎ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﻛﻪ ﺑﻪ ﺍﻳﻦ ﻃﺮﻳﻖ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ ،ﺷﺮﻁ ﺑﻲ ﭼﺮﺧﺶ ،ﺑﻪ ﻃﻮﺭ ﺍﺗﺤﺎﺩﻱ ﺑﺮﻗﺮﺍﺭ ﻣﻲ ﺷﻮﺩ.
ﻧﻜﺘﻪ( ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ،ﺟﺮﻳﺎﻧﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﺫﺭﺍﺕ ﺳﻴﺎﻝ ﻛﻪ ﺩﺭ ﺁﻥ ﺫﺭﺍﺕ ﺳﻴﺎﻝ ﻛﻪ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ
ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﻨﺪ ﭼﺮﺧﺶ ﻧﺪﺍﺭﻧﺪ ،ﺑﺮﺍﻱ ∇ × V = 0 ، ω = 0ﺍﺳﺖ.
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺩﺍﺭﻳﻢ،
∂φ
∂z
Vz = −
,
1 ∂φ
r ∂θ
Vθ = −
,
∂φ
∂r
Vr = −
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ φﻓﻘﻂ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﻭﺟﻮﺩ ﺩﺍﺭﺩ .ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ، ψﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﺭﺍ ﺑﺮﺍﻱ
ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺑﺮﻗﺮﺍﺭ ﻣﻲ ﻛﻨﺪ؛ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺗﺤﺖ ﺗﺎﺛﻴﺮ ﻣﺤﺪﻭﺩﻳﺖ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﻗﺮﺍﺭ ﻧﺪﺍﺭﺩ.
ﺩﺭ ﻧﺎﺣﻴﻪ ﻫﺎﻳﻲ ﺍﺯ ﻳﻚ ﺟﺮﻳﺎﻥ ﻛﻪ ﺩﺭ ﺁﻧﺠﺎ ﻧﻴﺮﻭﻫﺎﻱ ﭼﺴﺒﻨﺪﻩ ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﻫﺴﺘﻨﺪ ،ﺑﻲ ﭼﺮﺧﺸﻲ ﻓﺮﺿﻲ
ﺻﺤﻴﺢ ﺍﺳﺖ ).ﻣﺜﻼ ﻳﻚ ﭼﻨﻴﻦ ﻧﺎﺣﻴﻪ ﺩﺭ ﺧﺎﺭﺝ ﻻﻳﻪ ﻣﺮﺯﻱ ﺩﺭ ﺟﺮﻳﺎﻥ ﺭﻭﻱ ﺳﻄﺤﻲ ﺻﻠﺐ ﻭﺟﻮﺩ ﺩﺍﺭﺩ( .ﻧﻈﺮﻳﻪ
ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺸﻲ ﺑﻪ ﺻﻮﺭﺕ ﺳﻴﺎﻝ ﺍﻳﺪﻩ ﺁﻝ ﻣﻮﻫﻮﻣﻲ ،ﻛﻪ ﭼﺴﺒﻨﺪﮔﻲ ﺁﻥ ﺻﻔﺮ ﺍﺳﺖ) ( µ = 0ﺑﻴﺎﻥ ﻣﻲ ﺷﻮﺩ
ﺍﺯ ﺁﻧﺠﺎ ﻛﻪ ﺩﺭ ﺟﺮﻳﺎﻧﻲ ﺑﻲ ﭼﺮﺧﺸﻲ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺭﺍ ﺑﺎ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ φﻣﻲ ﺗﻮﺍﻥ ﺗﻌﺮﻳﻒ ﻛﺮﺩ ،ﺍﻳﻦ ﻧﻈﺮﻳﻪ ﺭﺍ
ﺍﻏﻠﺐ ﻧﻈﺮﻳﻪ ﺟﺮﻳﺎﻥ ﭘﺘﺎﻧﺴﻴﻞ ﻣﻲ ﮔﻮﻳﻨﺪ.
ﻧﻜﺘﻪ( φﺑﺮ ﺧﻼﻑ ψﺳﻪ ﺑﻌﺪﻱ ﻭ ﺳﻪ ﻣﺆﻟﻔﻪ ﺩﺍﺭﺩ.
۱۲٥
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
2-5ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ،ﺑﻲ ﭼﺮﺧﺶ ﻭ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ؛ ﻣﻌﺎﺩﻟﻪ
ﻻﭘﻼﺱ
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﻭﺑﻌﺪﻱ ،ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﺑﻲ ﭼﺮﺧﺶ ﻋﺒﺎﺭﺕ ﻫﺎﻱ ﻣﺆﻟﻔﻪ ﻫﺎﻱ ﺳﺮﻋﺖ uﻭ ، vﺭﺍ ﺑﺮ ﺣﺴﺐ ﺗﺎﺑﻊ
ﺟﺮﻳﺎﻥ ψﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ φﭼﻨﻴﻦ ﺩﺍﺭﻳﻢ
∂φ
∂x
∂φ
v=−
∂y
∂ψ
∂y
∂ψ
v=−
∂x
=u
ﺑﺎﺟﺎﻳﮕﺬﺍﺭﻱ uﻭ vﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺩﺭ ﺷﺮﻁ ﺑﻲ ﭼﺮﺧﺸﻲ،
u=−
∂v ∂u
−
=0
∂x ∂y
⇒ω =0
ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ،
)(1-5
∂ 2ψ ∂ 2ψ
+ 2 = ∇ 2ψ = 0
2
∂x
∂y
ﺑﺎ ﺟﺎﻳﮕﺬﺍﺭﻱ uﻭ vﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ،
ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻭﺭﻳﻢ
∂u ∂v
+
=0
∂x ∂y
)(2-5
∂ 2φ ∂ 2φ
+ 2 = ∇ 2φ = 0
2
∂x ∂y
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ) (1-5ﻭ) (2-5ﺷﻜﻞ ﻫﺎﻱ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﻫﺴﺘﻨﺪ .ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺍﻛﺜﺮﻋﻠﻮﻡ
ﻓﻴﺰﻳﻜﻲ ﻭ ﻣﻬﻨﺪﺳﻲ ﻭﺍﺭﺩ ﻣﻲ ﺷﻮﺩ .ﻫﺮ ﺗﺎﺑﻊ ψﻭ φﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﺑﺮﻗﺮﺍﺭ ﻛﻨﺪ ﻳﻚ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺩﻭ
ﺑﻌﺪﻱ ،ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﭼﺮﺧﺸﻲ ﻣﻤﻜﻦ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ.
ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ψﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﺛﺎﺑﺖ ﺍﺳﺖ .ﺑﺮﺍﻱ
ﺛﺎﺑﺖ =
،ψ
=0
dψﻭ
∂ψ
∂ψ
dx +
dy = 0
∂y
∂x
= dψ
ﺷﻴﺐ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ – ﺧﻂ ψﺛﺎﺑﺖ -ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ :
∂ψ ∂x
dy
−v v
=−
=
=−
∂ψ ∂y
dx ψ
u u
۱۲٦
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻳﻚ ﺧﻂ φﺛﺎﺑﺖ dφ = 0 ،ﻭ
∂φ
∂φ
dx +
dy = 0
∂y
∂x
= dφ
ﺩﺭ ﻧﺘﻴﺠﻪ ،ﺷﻴﺐ ﻳﻚ ﺧﻂ ﭘﺘﺎﻧﺴﻴﻞ – ﺧﻂ φﺛﺎﺑﺖ -ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ،
∂φ ∂x
dy
u
=−
=−
∂φ ∂y
dx φ
v
ﺷﺮﻁ ﺭﻳﺎﺿﻲ ﺗﻌﺎﻣﺪ
1
dy
⇒ =−
(dy dx )ψ
dx φ
ﺷﻴﺐ ﻳﻚ ﺧﻂ ψﺛﺎﺑﺖ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﻣﻨﻔﻲ ﻣﻌﻜﻮﺱ ﺷﻴﺐ ﺧﻂ φﺛﺎﺑﺖ ﺩﺭ ﺁﻥ ﻧﻘﻄﻪ.
ﺩﺭ ﻧﺘﻴﺠﻪ ﺭﻭﺍﺑﻂ ﻓﻮﻕ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﻨﺪ ﻛﻪ ﺧﻄﻮﻁ ﺛﺎﺑﺖ = φﺑﺮ ﺛﺎﺑﺖ = ψﺩﺭ ﺟﺮﻳﺎﻥ ﻏﻴﺮ ﭼﺮﺧﺸﻲ ﺩﻭ ﺑﻌﺪﻱ
ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺑﺮ ﻫﻢ ﻋﻤﻮﺩﻧﺪ.
3-5ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺳﺎﺩﻩ ﺻﻔﺤﻪ ﺍﻱ
ﺑﺎ ﺗﺮﻛﻴﺐ ﻧﻘﺶ ﻫﺎﻱ ﺳﺎﺩﻩ ﺟﺮﻳﺎﻥ ،ﺍﻧﻮﺍﻉ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ .ﺗﻮﺍﺑﻊ ψﻭ φﺑﺮﺍﻱ
ﭘﻨﺞ ﺟﺮﻳﺎﻥ ﺳﺎﺩﻩ ﺩﻭ ﺑﻌﺪﻱ -ﻳﻜﻨﻮﺍﺧﺖ ،ﺳﻜﻮﻥ ،ﭼﺸﻤﻪ ﻭ ﭼﺎﻩ ،ﮔﺮﺩﺍﺏ ﻭ ﺩﻭ ﻗﻄﺒﻲ – ﺑﺮﺭﺳﻲ ﻣﻲ ﻛﻨﻴﻢ .ﺗﻮﺍﺑﻊ
ψﻭ φﻫﺮ ﺟﺮﻳﺎﻥ ﺳﺎﺩﻩ ﺭﺍ ﺍﺯ ﺭﻭﻱ ﺗﻮﺍﺑﻊ ψﻭ φﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ.
-1ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ
ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺘﻲ ﺑﺎ ﺳﺮﻋﺖ ﺛﺎﺑﺖ ﻭ ﻣﻮﺍﺯﻱ ﺑﺎ ﻣﺤﻮﺭ xﻣﻌﺎﺩﻟﻪ ﭘﻴﻮﺳﺘﮕﻲ ﻭ ﺷﺮﻁ ﺑﻲ ﭼﺮﺧﺸﻲ ﺭﺍ ﺑﻪ
ﻃﻮﺭ ﺍﺗﺤﺎﺩﻱ ﺑﺮﻗﺮﺍﺭ ﻣﻲ ﻛﻨﺪ.
v=0
۱۲۷
ﻭ ∞u = U
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
∂v
∂u
=0 ﻭ
=0
∂x
∂y
∂v
∂u
=0 ﻭ
=0
∂y
∂x
∂v ∂u
⇒
−
=0
()ﺑﻲ ﭼﺮﺧﺸﻲ
∂x ∂y
∂u
∂v
∂u ∂v
=0 ,
=0 ⇒
+
=0
()ﭘﻴﻮﺳﺘﮕﻲ
∂x
∂y
∂x ∂y
∂ψ
∂φ
=−
= const
u = U∞ =
∂y
∂x
∂ψ
∂φ
=−
v=0=−
∂x
∂y
∂ψ
∂ψ
∂ψ
=0 ,
= U∞ ⇒
= U ∞ ⇒ ψ = U ∞ ⋅ y + c1
∂x
∂y
∂y
at y = 0
∂φ
∂y
=0
ψ = 0 ⇒ c1 = 0 ⇒ ψ = U ∞ y
,
−
ψ = U ∞ r sin θ
∂φ
dφ
= U∞ ⇒
= −U ∞ ⇒ φ = −U ∞ ⋅ x + c2
∂x
dx
at x = 0 , φ = 0 ⇒ c2 = 0 ⇒ φ = −U ∞ ⋅ x
dy
dx
ψ =c
=
dy
dx
φ =c
=−
()ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻗﻄﺒﻲ
φ = −U ∞ r cos θ
v
0
=
=0
u U∞
u −U∞
=
= −∞
v
0
: ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ
۱۲۸
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
ψ = ψ ( x, y )
∂ψ
∂ψ
dψ =
dx +
dy
∂x
∂y
⇒ dψ = Adx + Bdy
⇒ ψ = Ax + By + C
at x = y = 0 , ψ = 0 ⇒ C = 0
ψ = Ax + By
∂ψ
∂ψ
=B , v=−
= −A
∂y
∂x
∂ψ
∂ψ
dψ =
dx +
dy = Adx + Bdy
∂x
∂y
u=
⇒ ψ = Ax + By
∂φ
∂φ
=B , v=−
= −A
u=−
∂x
∂y
∂φ
∂φ
dφ =
dx +
dy
∂x
∂y
⇒ φ = − Bx + Ay
dy
v −A
=
ψ =c =
B
dx
u
dy
u
B
B
=−
=+
φ =c = −
dx
v
−A
A
V = Biˆ − Aˆj
۱۲۹
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
1
-2ﺟﺮﻳﺎﻥ ﺳﻜﻮﻥ
ﺗﺎﺑﻊ ψﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ) (∇ 2ψ = 0ﺻﺪﻕ ﻣﻲ ﻧﻤﺎﻳﺪ ﻟﺬﺍ ﻣﻲ ﺗﻮﺍﻧﺪ ﺗﺎﺑﻊ ﻳﻚ ﺟﺮﻳﺎﻥ ﭘﺘﺎﻧﺴﻴﻞ
ﺑﺎﺷﺪ.
ψ = Axy ⇒ u = Ax , v = − Ay
∂φ
1
= Ax ⇒ φ = − Ax 2 + C1
u=−
F23
∂x
2
∂φ
1
= − Ay ⇒ φ = Ay 2 + C2
v=−
∂y
2
∂φ
∂φ
= dφ
dx +
dy
∂x
∂y
A 2
y − x2
= ⇒φ
2
)
-3ﺟﺮﻳﺎﻥ ﭼﺸﻤﻪ ﺍﺯ ﻣﺒﺪﺃ
(
2
F24
ﭼﺸﻤﻪ ﺳﺎﺩﻩ ﻳﻚ ﻧﻘﺶ ﺟﺮﻳﺎﻥ ﺩﺭ ﺻﻔﺤﻪ xyﺍﺳﺖ ﻛﻪ ﺩﺭ ﺁﻥ ﺟﺮﻳﺎﻥ ﺍﺯ ﻣﺤﻮﺭ zﺑﻪ ﻃﻮﺭ ﺷﻌﺎﻋﻲ ﺑﻪ ﻃﺮﻑ ﺧﺎﺭﺝ
ﺍﺳﺖ ﻭ ﺩﺭ ﺗﻤﺎﻡ ﺟﻬﺖ ﻫﺎ ﺑﻪ ﻃﻮﺭ ﻣﺤﻮﺭﻱ ﺍﺳﺖ .ﻗﺪﺭﺕ qﭼﺸﻤﻪ ﺁﻫﻨﮓ ﺷﺎﺭﺵ ﺣﺠﻢ ﺑﺮﺍﻱ ﻋﻤﻖ ﻭﺍﺣﺪ ﺍﺳﺖ .ﺩﺭ
ﺷﻌﺎﻉ ﺩﻟﺨﻮﺍﻩ ، rﺍﺯ ﻳﻚ ﭼﺸﻤﻪ ،ﺳﺮﻋﺖ ﻣﻤﺎﺳﻲ Vθﺻﻔﺮ ﺍﺳﺖ؛ ﺳﺮﻋﺖ ﺷﻌﺎﻋﻲ Vrﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ﺁﻫﻨﮓ
ﺷﺎﺭﺵ ﺣﺠﻢ ﺩﺭ ﻋﻤﻖ ﻭﺍﺣﺪ ، q ،ﺗﻘﺴﻴﻢ ﺑﺮ ﻣﺴﺎﺣﺖ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﻋﻤﻖ ﻭﺍﺣﺪ . 2πr ،ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﺑﺮﺍﻱ ﻳﻚ ﭼﺸﻤﻪ
q
2πr
=
. Vr
ﺩﺭ ﻳﻚ ﭼﺎﻩ ﺳﺎﺩﻩ ،ﺟﺮﻳﺎﻥ ﺷﻌﺎﻋﻲ ﺑﻪ ﻃﺮﻑ ﺩﺍﺧﻞ ﺍﺳﺖ ،ﭼﺎﻩ ﻳﻚ ﭼﺸﻤﻪ ﻣﻨﻔﻲ ﺍﺳﺖ .ﺗﻮﺍﺑﻊ ψﻭ φﻳﻚ ﭼﺎﻩ،
ﻣﻨﻔﻲ ﺗﻮﺍﺑﻊ ψﻭ φﻳﻚ ﺟﺮﻳﺎﻥ ﭼﺸﻤﻪ ﻫﺴﺘﻨﺪ .
Stagnation Flow
Source
۱۳۰
۱
۲
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﻣﺒﺪﺍ ﭼﺎﻟﻪ ﻳﺎ ﭼﺸﻤﻪ ﻳﻚ ﻧﻘﻄﻪ ﺗﻜﻴﻦ ﺍﺳﺖ ﺯﻳﺮﺍ ﻭﻗﺘﻲ ﺷﻌﺎﻉ ﺑﻪ ﺻﻔﺮ ﻧﺰﺩﻳﻚ ﻣﻲ ﺷﻮﺩ ،ﺳﺮﻋﺖ ﺷﻌﺎﻋﻲ ﺑﻪ ﺑﻲ
ﻧﻬﺎﻳﺖ ﻧﺰﺩﻳﻚ ﻣﻲ ﺷﻮﺩ ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﺩﺭ ﺣﺎﻟﻲ ﻛﻪ ﺟﺮﻳﺎﻥ ﻭﺍﻗﻌﻲ ﻣﻲ ﺗﻮﺍﻧﺪ ﺷﺒﻴﻪ ﭼﺸﻤﻪ ﻳﺎ ﭼﺎﻩ ﺑﺮﺍﻱ ﺑﻌﻀﻲ ﻣﻘﺎﺩﻳﺮ
rﺑﺎﺷﺪ ،ﭼﺸﻤﻪ ﻭ ﭼﺎﻩ ﻫﻴﭻ ﺗﻔﺴﻴﺮ ﻓﻴﺰﻳﻜﻲ ﺩﻗﻴﻘﻲ ﻧﺪﺍﺭﻧﺪ .ﺍﺭﺯﺵ ﺍﻭﻟﻴﻪ ﻣﻔﻬﻮﻡ ﭼﺸﻤﻪ ﻭ ﭼﺎﻩ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻭﻗﺘﻲ
ﺑﺎ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺳﺎﺩﻩ ﺩﻳﮕﺮ ﺗﺮﻛﻴﺐ ﻣﻲ ﺷﻮﻧﺪ ،ﻧﻘﺶ ﻫﺎﻱ ﺟﺮﻳﺎﻧﻲ ﺭﺍ ﺑﻪ ﻭﺟﻮﺩ ﻣﻲ ﺁﻭﺭﻧﺪ ﻛﻪ ﺑﻪ ﻃﻮﺭ ﻛﺎﻣﻞ ﺟﺮﻳﺎﻥ
ﻫﺎﻱ ﻭﺍﻗﻌﻲ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﻨﺪ.
ﺩﺭ ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺗﺎﺑﻊ ψﺑﻪ ﻓﺮﻡ ﺭﻭﺑﻪ ﺭﻭ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ.
ψ = Aθ
A
1 ∂ψ A
, vθ = 0 , vr = +
=
r
r ∂θ
r
= vr
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺎﺭﺗﺰﻳﻦ ﺗﺎﺑﻊ ψﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻣﻲ ﺁﻳﺪ:
y
x
ψ = A tan −1
u
⇒ u = vr cos θ
vr
= cos θ
A
= cos θ
r
= ⇒u
Ax
Ay
, v = vr sin θ = 2
2
x +y
x + y2
= ⇒u
x Ax
=
r r2
⋅
A
2
x +y
2
2
ﺩﺑﻲ ﺧﺮﻭﺟﻲ ﺍﺯ ﻧﻘﻄﻪ :o
۱۳۱
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
2π
2π
0
0
strength = q = ∫ vr (rdθ × 1) = ∫ vr rdθ
2π
A
q
= ⋅ rdθ = 2πA ⇒ A
0
r
2π
q
q
= ⇒ vr
= ⇒ψ
θ
2πr
2π
∂φ
q
q dr
vr = −
=
⇒ dφ = −
∂r 2πr
2π r
q
⇒φ = −
ln r + c1
2π
⇒ Singular point r = 0
q
−q
or
ln x 2 + y 2
φ = − ln r
=φ
2π
2π
∫=⇒q
ﺟﺮﻳﺎﻥ ﭼﺎﻩ ﺑﻪ ﻃﺮﻑ ﻣﺒﺪﺃ :1
F25
vθ = 0
,
=φ
,
q
ln r
2π
-4ﮔﺮﺩﺍﺏ ﭘﺘﺎﻧﺴﻴﻞ
q
2πr
q
ψ =− θ
2π
vr = −
2
F26
ﻧﻘﺶ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﺩﺭ ﺁﻥ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ،ﺩﺍﻳﺮﻩ ﻫﺎﻱ ﻫﻢ ﻣﺮﻛﺰ ﻫﺴﺘﻨﺪ ﻳﻚ ﮔﺮﺩﺍﺏ ﺍﺳﺖ .ﺩﺭ ﻳﻚ ﮔﺮﺩﺍﺏ ﺁﺯﺍﺩ )ﺑﻲ
ﭼﺮﺧﺶ( ﺫﺭﺍﺕ ﺳﻴﺎﻝ ﺑﻪ ﻫﻨﮕﺎﻡ ﺣﺮﻛﺖ ﭘﻴﺮﺍﻣﻮﻥ ﻣﺮﻛﺰ ﮔﺮﺩﺍﺏ ﻧﻤﻲ ﭼﺮﺧﻨﺪ.ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﺩﺭ ﻳﻚ ﮔﺮﺩﺍﺏ ﺑﻲ
ﭼﺮﺧﺶ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﻭ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ،ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ
ﻣﻴﺎﻥ ﺩﻭ ﻧﻘﻄﻪ ﺩﻟﺨﻮﺍﻩ ﺩﺭ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺻﺤﺖ ﺩﺍﺭﺩ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﺭ ﻳﻚ ﺻﻔﺤﻪ ﺍﻓﻘﻲ
dP = −Vθ dVθ
1
ρ
- Sink
- Irrotational Vortex
۱۳۲
۱
۲
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﻣﺆﻟﻔﻪ ﻣﻌﺎﺩﻟﻪ ﺍﻭﻳﻠﺮ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﻋﻤﻮﺩ ﺑﺮ ﺧﻂ ﺟﺮﻳﺎﻥ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ :
2
1 dP Vθ
=
ρ dr
r
ﺑﺎ ﺗﺮﻛﻴﺐ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻫﺎ ﺩﺍﺭﻳﻢ،
2
Vθ
dr
ρ
r
⇒ Vθ dr + rdVθ = 0 ⇒ d (rVθ ) = 0
= = −Vθ dVθ
⇒ vr = 0
dP
⇒
⇒ rVθ = constant
ψ = Aθ
⇒
φ = − A ln r
− ∂ψ
∂r
φ = − Aθ
A
r
ψ = − A ln r
= vθ
= vθ
ﺧﻄﻮﻁ ﭘﺘﺎﻧﺴﻴﻞ :ﺷﻌﺎﻋﻲ ،ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ :ﺩﺍﻳﺮﻩ
ﭘﻴﺮﺍﻣﻮﻥ ﻫﺮ ﻣﻨﺤﻨﻲ ﺑﺴﺘﻪ : 1
ﮔﺮﺩﺵ 2ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ:
F27
F28
Γ = ∫V ⋅ d s
c
ﺍﻧﺘﮕﺮﺍﻝ ﺧﻄﻲ ﻣﺆﻟﻔﻪ ﻣﻤﺎﺳﻲ ﺳﺮﻋﺖ ﺣﻮﻝ ﻳﻚ ﻣﻨﺤﻨﻲ ﺑﺴﺘﻪ ﺛﺎﺑﺖ ﺩﺭ ﺟﺮﻳﺎﻥ ﻛﻪ ﺩﺭ ﺁﻥ d sﻳﻚ ﺑـﺮﺩﺍﺭ ﺟﺰﺋـﻲ
ﺑﻪ ﻃﻮﻝ dsﻭ ﻣﻤﺎﺱ ﺑﺮ ﻣﻨﺤﻨﻲ ﺍﺳﺖ ﻭ ﻣﺴﻴﺮ ﺩﺭ ﺧﻼﻑ ﺟﻬﺖ ﻋﻘﺮﺑﻪ ﻫﺎﻱ ﺳﺎﻋﺖ ﺳﻮﻱ ﻣﺜﺒﺖ ﻣﻲ ﺑﺎﺷﺪ.
Γ = ∫ V cos α ds = ∫ V ⋅ d s = − ∫ ∇φ ⋅ ds = ∫ dφ = φ f − φi
c
c
c
c
Flow Along A Closed Curve
Circulation
۱۳۳
۱
۲
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
2π
strength = Γ = ∫ vθ ds = ∫ vθ .rdθ = 2πA ⇒ A =
0
c
Singular point r = 0 ⇒ vθ =
Γ
2πr
,
ψ =-
Γ
Ay
u= 2
ln x 2 + y 2
,
x + y2
2π
1 ∂φ
dφ
Γ
Γ
vθ = −
=
⇒
=−
r ∂θ 2πr
dθ
2π
Γ
⇒
φ =− θ
⇒
φ = −Aθ
2π
ψ =−
θ = 2π
⇒
:ﺭﻭﺵ ﺩﻳﮕﺮ
Γ
2π
Γ
ln r
2π
,
v=−
Ax
x + y2
2
Γ = 2πA
.( ﺟﺮﻳﺎﻥ ﺑﻪ ﻫﺮ ﺩﻟﻴﻞ ﺍﺯ ﻗﺒﻞ ﺩﺍﺭﺍﻱ ﭼﺮﺧﺶ ﺑﺎﺷﺪ1
(ω ≠ 0) ( ﺩﺭ ﭼﻪ ﻣﻮﻗﻊ ﺫﺭﺍﺕ ﻳﻚ ﺟﺮﻳﺎﻥ ﻣﻲ ﭼﺮﺧﻨﺪ؟2
ﻓﻘﻂ ﺗﺤﺖ ﺗﺄﺛﻴﺮ ﮔﺸﺘﺎﻭﺭ ﻧﺎﺷﻲ ﺍﺯ ﺗﻨﺶ ﻫﺎﻱ ﺑﺮﺷﻲ ﻣﻲ ﺗﻮﺍﻧﻨﺪ،( ﺍﮔﺮ ﺫﺭﺍﺕ ﺩﺍﺭﺍﻱ ﭼﺮﺧﺶ ﺍﻭﻟﻴﻪ ﻧﺒﺎﺷﻨﺪ3
.ﺑﭽﺮﺧﻨﺪ
1 ∂V z ∂Vθ
∂V ∂V
∇ × V = eˆr
−
+ eˆθ r − z
∂z
∂r
∂z
r ∂θ
ˆ 1 ∂
(rVθ ) − 1 ∂ (rVr )
+ k
r ∂θ
r ∂r
: rθ ﭼﺮﺧﺶ ﺻﻠﺐ ﮔﻮﻧﻪ ﺩﺭ ﺻﻔﺤﻪ-
Vθ = rω
Vr = 0
11 ∂
(rVθ ) − 1 ∂ (rVr )
r ∂θ
2 r ∂r
11 ∂
(rVθ ) = 1 ⋅ 1 ∂ (r × rω ) = 1 (2ωr ) = ω
⇒ ωz =
2 r ∂r
2 r ∂r
2r
ωz =
. ﻳﻚ ﻛﻤﻴﺖ ﺛﺎﺑﺖ ﺍﺳﺖω z
: ( ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ )ﮔﺮﺩﺍﺏ-
c
, Vr = 0
r
1 1 ∂
(rVθ ) = 1 ⋅ 1 ∂ r × c = 0
ωz = ⋅
r
2 r ∂r
2 r ∂r
Vθ =
۱۳٤
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺩﺭ ﻣﺮﻛﺰ :
ﺑﺮﺍﻱ sourceﻭ : sink
∞ = ∇ ×V
،
Γ=0
ﻣﺎﻧﻨﺪ ﺧﺎﻟﻲ ﺷﺪﻥ ﺁﺏ ﻛﻒ ﺣﻤﺎﻡ ﻳﺎ ﺣﻤﺎﻡstrength ....
φ = − A ln r = − A ln rf + A ln ri → 0
r f = ri
-5ﺟﺮﻳﺎﻥ ﺩﻭ ﻗﻄﺒﻲ :1
F29
ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺍﺯ ﺗﺮﻛﻴﺐ ﻳﻚ ﭼﺸﻤﻪ ﻭ ﻳﻚ ﭼﺎﻩ ﺑﺎ ﻗﺪﺭﺕ ﻫﺎﻱ ﻣﺴﺎﻭﻱ ﺑﻪ ﻃﻮﺭ ﺭﻳﺎﺿﻲ ﺍﻳﺠﺎﺩ ﻣﻲ ﺷﻮﺩ،ﺑﻪ ﻋﺒﺎﺭﺗﻲ،
ﺑﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻳﻚ ﭼﺎﻩ ﻭ ﭼﺸﻤﻪ ﺑﻪ ﻓﺎﺻﻠﻪ εﺍﺯ ﻣﺒﺪﺃ ﻣﺨﺘﺼﺎﺕ ﺩﺭ ﺭﻭﻱ ﻣﺤﻮﺭ xﻭﻗﺘﻲ ﻓﺎﺻﻠﻪ εﺑﻪ ﺳﻤﺖ ﺻـﻔﺮ
ﻣﻴﻞ ﻧﻤﺎﻳﺪ ﺟﺮﻳﺎﻥ ﺟﺪﻳﺪﻱ ﺑﻪ ﻧﺎﻡ ﺩﺍﺑﻠﺖ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ ﻳﻌﻨﻲ:
Doublet
۱۳٥
۱
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
y
y
− A tan −1
x+ε
x −ε
∂f
f (x + ε , y ) − f (x − ε , y )
= lim
ε →0 ⇒
∂x ε →0
ε
ψ = A tan −1
ψ = 2εA
: ﺩﺍﺭﻳﻢf ﺩﺭ ﻣﻮﺭﺩ ﻫﺮ ﺗﺎﺑﻊ
ﻟﺬﺍ
y
∂
−1 y
A tan
= 2εA− 2
2
x
∂x
x +y
ﺑـﻪε → 0 ⇒ A → ∞ ﻧﺸـﺎﻥ ﻣـﻲ ﺩﻫـﻴﻢ ﻳﻌﻨـﻲµ = 2εA ﺭﺍ ﺛﺎﺑﺖ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ ﻭ ﺁﻥ ﺭﺍ ﺑـﺎεA ﻣﻘﺪﺍﺭ
. = ﺛﺎﺑﺖ ﻟﺬﺍ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻳﻚ ﺩﺍﺑﻠﺖ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺍﺳﺖµ = 2εA ﻗﺴﻤﻲ ﻛﻪ
ψ=
− µy
x2 + y2
ψ = Aθ
q
source :
⇒ ψ = µθ , φ = −
ln r = − µ ln r
2π
A = µ
q = 2πµ = strength
ψ = − µθ
sink
φ = µ ln r
۱۳٦
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
source : (a,0 )
sink : (− a,0 )
source
2πµ = strength
sink
φ = − µ ln r1 + µ ln r2
: P ﺑﺮﺍﻱ ﻫﺮ ﺩﻭ ﺩﺭ ﻧﻘﻄﻪ
a 2
a
r1 = r + a − 2ar cos θ = r 1 + − 2 cos θ
r
r
2
2
2
2
a 2
a
r2 = r + a + 2ar cos θ = r 1 + + 2 cos θ
r
r
2
2
a
a
µ
µ
a
a
φ = − ln r1 2 − ln r2 2 = − ln r 2 + ln 1 + − 2 cos θ − ln r 2 − ln 1 + + 2 cos θ
2
2
r
r
r
r
x
x
2
2
2
2
(
)
ln(1 + x ) = x −
x2 x3 x4
+
−
+ ...
2
3
4
2
a 2
a 2
1
a
a
− 2 cos θ − − 2 cos θ + 1 [
µ
2 r
3
r
r
φ = − r
2
1
1
2
3
+ [ ] + [ ]
3
2
]
3
a 2
a
... − + 2 cos θ
r
r
cos θ a 2 cos θ a 4 cos θ 4 a 2 cos3 θ
⇒ φ = 2 aµ
+
−
−
+ ...
3 r
r
r r
r r
r
ﺍﮔﺮ2aµ = µ new , a → 0
2
: ﻣﺜﺒﺖx ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺑﺮﺍﻱ ﻳﻚ ﺩﺍﺑﻠﺖ ﺩﻭ ﺑﻌﺪﻱ ﺩﺭ ﻣﺒﺪﺃ ﺑﺎ ﺟﻬﺖ
⇒φ =
µ new ⋅ cos θ
r
۱۳۷
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
∂φ
1 ∂ψ
1 ∂φ ∂ψ
=−
=
vθ = −
,
∂r
r ∂θ
r ∂θ ∂r
∂ψ
µ cos θ
∂ψ µ new
⇒
= − new
= 2 sin θ
,
∂θ
r
r
∂r
µ new cos θ
∂ψ
⇒
=−
⇒
r
∂θ
∂ψ µ new
= 2 sin θ
r
∂r
µ cos θ
µ
dψ = − new
dθ + new
sin θ ⋅ dr
r
r2
µ sin θ 2 µ new
sin θ + C
⇒ ψ = − new
−
r
r
2 µ sin θ
stream function for doublet
⇒ ψ = − new
r
µ x
−µ y
cartesian coord. ⇒ φ = 2 new 2 , ψ = 2 new 2
x +y
x +y
vr = −
x2 + y2 =
⇒ x2 +
µx
µx
⇒ x2 + y2 −
=0
φ
φ
µ 2 µx
µ2
2
y
−
+
=
4φ 2 φ
4φ 2
2
µ
µ2
⇒ x − + y 2 = 2
2φ
4φ
µy
µ
⇒ x + y +
ψ =− 2
2
x +y
2ψ
2
2
µ2
=
4ψ 2
۱۳۸
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺧﻄﻮﻁ φﺛﺎﺑﺖ ﺩﺍﻳﺮﻩ ﻫﺎﻳﻲ ﻫﺴﺘﻨﺪ ﻛﻪ ﺍﺯ ﻣﺒﺪﺃ ﻣﺨﺘﺼﺎﺕ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﻨﺪ ﻭ ﻣﺮﻛﺰ ﺁﻧﻬﺎ ﺭﻭﻱ ﻣﺤﻮﺭ xﻫﺎ ﺍﺳﺖ ﻭ
ﺧﻄﻮﻁ ψﺛﺎﺑﺖ ﺩﺍﻳﺮﻩ ﻫﺎﻳﻲ ﻫﺴﺘﻨﺪ ﻛﻪ ﺍﺯ ﻣﺒﺪﺃ ﻣﺨﺘﺼﺎﺕ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﻨﺪ ﻭ ﻣﺮﻛﺰ ﺁﻧﻬﺎ ﺭﻭﻱ ﻣﺤﻮﺭ yﻫﺎ ﺍﺳﺖ.
Origin is a singular point where the velocity goes to infinity
Matab:
)[x, y ] = meshgrid(− 1 : 0.02 : 1
λ =1
Psc = − y. x. ^ 2 + y.^ 2
)contour ( x, y, psc,100
Axis square
Grid off
Axis off
Solving flow problems using a stream function
ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻗﻄﺒﻲ ﺩﺍﺭﻳﻢ:
µ sin θ
µ sin θ
)
2
r2
2 µxy
(x
vθ = −
,
v=−
,
+ y2
µ cos θ
φ =−
r
2
ﻳﺎ
)
ψ =−
r
µ cos θ
vr = −
r2
µ x2 − y2
u=−
2
x2 + y2
µx
φ =− 2
x + y2
)
(
(
1
4-5ﺗﺮﻛﻴﺐ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺳﺎﺩﻩ ﺻﻔﺤﻪ ﺍﻱ :
F30
φﻭ ψﻫﺮ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ ﺑﺮﻗﺮﺍﺭ ﻣـﻲ ﻛﻨﻨـﺪ .ﺍﺯ ﺁﻧﺠـﺎ
ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﭘﺎﺭﻩ ﺍﻱ ﻫﻤﮕﻦ ،ﺧﻄﻲ ﺍﺳﺖ ،ﺣﻞ ﻫﺎ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺮ ﻫﻢ ﻧﻬﺎﺩ )ﺑـﺎ ﻫـﻢ
ﺟﻤﻊ ﻛﺮﺩ( ﻭ ﻧﻘﺶ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﭘﻴﭽﻴﺪﻩ ﺗﺮ ﻭ ﺟﺎﻟـﺐ ﺗـﺮ ﺭﺍ ﺑـﻪ ﻭﺟـﻮﺩ ﺁﻭﺭﺩ .ﺍﺯ ﺍﻳـﻦ ﺭﻭ ﺍﮔـﺮ ψ 1ﻭ ψ 2ﻣﻌﺎﺩﻟـﻪ
ﻻﭘﻼﺱ ﺭﺍ ﺑﺮﻗﺮﺍﺭ ﻛﻨﻨﺪ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ψ 3 = ψ 1 + ψ 2ﻧﻴﺰ ﻣﻌﺎﺩﻟﻪ ﻻﭘﻼﺱ ﺭﺍ ﺑﺮﻗـﺮﺍﺭ ﻣـﻲ ﻛﻨﻨـﺪ .ﺟﺮﻳـﺎﻥ ﻫـﺎﻱ
ﺳﺎﺩﻩ ﺻﻔﺤﻪ ﺍﻱ ﻣﺼﺎﻟﺢ ﺳﺎﺧﺘﻤﺎﻧﻲ ﺩﺭ ﺍﻳﻦ ﻓﺮﺁﻳﻨﺪ ﺗﺮﻛﻴﺐ ﻫﺴﺘﻨﺪ.
ﻣﺜﺎﻝ :ﺟﺮﻳﺎﻥ ﺭﻭﻱ ﻳﻚ ﺍﺳﺘﻮﺍﻧﻪ -ﺗﺮﻛﻴﺐ ﺩﻭ ﻗﻄﺒﻲ ﻭ ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ
ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ،ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻭ ﺑﻲ ﭼﺮﺧﺶ ،ﺗﺮﻛﻴـﺐ ﺩﻭ ﻗﻄﺒـﻲ ﻭ ﺟﺮﻳـﺎﻥ ﻳﻜﻨﻮﺍﺧـﺖ ،ﺟﺮﻳـﺎﻥ ﭘﻴﺮﺍﻣـﻮﻥ
ﺍﺳﺘﻮﺍﻧﻪ ﺩﻭﺍﺭ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ .ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺍﻳﻦ ﻧﻘـﺶ ﺟﺮﻳـﺎﻥ ﺭﺍ ﺑـﻪ ﺩﺳـﺖ ﺁﻭﺭﻳـﺪ .ﻣﻴـﺪﺍﻥ
ﺳﺮﻋﺖ ﺭﺍ ﺑﻴﺎﺑﻴﺪ ،ﻣﻜﺎﻥ ﻧﻘﺎﻁ ﺭﻛﻮﺩ ﻭ ﺳﻄﺢ ﺍﺳﺘﻮﺍﻧﻪ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ ﻭ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﺳﻄﺤﻲ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ.
Super Position
۱۳۹
۱
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
µ sin θ
r
− Uy = −Ur sin θ +
µ cos θ
r
µ sin θ
r
ψ = ψ d + ψ uf = +
φ = φd + φuf = −Ur cos θ −
ﺩﺭ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﻫﺮ ﺧﻂ ﺟﺮﻳﺎﻧﻲ ﺭﺍ ﻟﺒﻪ ﻱ ﻳﻚ ﺟﺴﻢ ﺩﻭ ﺑﻌﺪﻱ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ.
ﺩﺭ ﺗﻜﻨﻴﻚ ﺑﺮ ﻫﻢ ﻧﻬﺶ ﺧﻂ ﺟﺮﻳﺎﻧﻲ ﺭﺍ ﺑﺎﻳﺪ ﭘﻴﺪﺍ ﻛﺮﺩ ﻛﻪ ﻣﺴﺎﺣﺘﻲ ﺭﺍ ﻛﻪ ﺷﻜﻞ ﺁﻥ ﺍﺯ ﻧﻈﺮ ﻋﻤﻠـﻲ ﻣﺴـﻠﻢ ﺑﺎﺷـﺪ،
ﺍﺣﺎﻃﻪ ﻛﻨﺪ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ،ﻧﻘﺶ ﺧﻂ ﺟﺮﻳﺎﻥ ﻭﺍﻗﻊ ﺩﺭ ﺧﺎﺭﺝ ﺍﻳﻦ ﻧﺎﺣﻴﻪ ،ﺟﺮﻳﺎﻥ ﭘﻴﺮﺍﻣﻮﻥ ﺍﻳﻦ ﺟﺴـﻢ ﺭﺍ ﺗﺸـﻜﻴﻞ
ﻣﻲ ﺩﻫﺪ.
)ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻗﻄﺒﻲ(
µ sin θ
r
ψ = −Ur sin θ +
ﺩﺍﻳﺮﻩ ﺑﻪ ﺷﻌﺎﻉ µﻗﺴﻤﺘﻲ ﺍﺯ ﺧﻂ ﺟﺮﻳﺎﻥ ﺍﺳﺖ
r
θ =π
.
ﺩﺭ ﻧﺘﻴﺠﻪ ﺧﻂ ﺟﺮﻳﺎﻥ ⇐ ψ = 0
sinθ = 0 ⇒ θ = 0
µ
for ψ = 0 ⇒ sin θ Ur − = 0 ⇒
µ
µ
r
= Ur − = 0 ⇒ r
r
U
۱٤۰
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
θ =0
µ
ﻧﻘﺎﻁ
AﻭB
ﻧﻘﺎﻁ ﺳﻜﻮﻥ ﻫﺴﺘﻨﺪ ﻛﻪ ﺩﺭ ﺁﻥ ﻫﺎ ﺳﺮﻋﺖ ﺻﻔﺮ ﻣﻲ ﺑﺎﺷﺪ.
=r
θ =π
B
µ
,
=r
A
U
U
∂φ
µ cos θ
= +U cos θ −
vr = −
∂r
r2
µ sin θ
1 ∂φ
= −U sin θ −
vθ = −
r ∂θ
r2
⇒ VA = VB = 0
ﻧﻘﺎﻁ ﺳﻜﻮﻥ ﻣﻌﻤﻮﻻً ﺩﺭ ﻣﻜﺎﻥ ﻫﺎﻳﻲ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﻛﻪ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺩﺭ ﺁﻧﺠﺎﻫﺎ ﺍﺯ ﻫﻢ ﺑﺎﺯ ﻣﻲ ﺷﻮﻧﺪ.
ﻧﺎﺣﻴﻪ ﻱ ﺩﺍﻳﺮﻩ ﺍﻱ ﺍﺣﺎﻃﻪ ﺷﺪﻩ ﺗﻮﺳﻂ ﻗﺴﻤﺘﻲ ﺍﺯ ﺧﻂ ﺟﺮﻳﺎﻥ ψ = 0ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺻﻠﺐ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ
ﻧﻈﺮ ﮔﺮﻓﺖ ﻛﻪ ﺩﺭ ﺟﺮﻳﺎﻥ ﺑﺪﻭﻥ ﺍﺻﻄﻜﺎﻛﻲ ﻛﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﻱ ﺯﻳﺎﺩﻱ ﺍﺯ ﺍﺳﺘﻮﺍﻧﻪ ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﻣﺤﻮﺭ ﺍﺳﺘﻮﺍﻧﻪ ﺑﻪ
ﻃﻮﺭ ﻳﻜﻨﻮﺍﺧﺖ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ ،ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﺷﺪﻩ ﺑﺎﺷﺪ .ﺁﻧﻬﺎﻳﻲ ﻛﻪ ﺩﺭ ﺧﺎﺭﺝ ﺍﺯ ﺩﺍﻳﺮﻩ ﻫﺴﺘﻨﺪ ﻧﻘﺶ ﺟﺮﻳﺎﻧﻲ ﺭﺍ ﻣـﻲ
ﺳﺎﺯﻧﺪ ﺩﺭ ﺣﺎﻟﻲ ﻛﻪ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺩﺍﺧﻞ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﻧﺎﺩﻳﺪﻩ ﮔﺮﻓﺖ.
V = vr eˆr + vθ eˆθ
Vr = 0
Vθ = 0
: V = 0 ⇒ ﻧﻘﺎﻁ ﺳﻜﻮﻥ
۱٤۱
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﭼﻮﻥ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﺭﺍ ﻣﻴﺎﻥ ﻫﺮ ﺩﻭ ﻧﻘﻄﻪ ﺩﻟﺨﻮﺍﻩ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﻛﺎﺭ ﺑﺮﺩ.
ﻣﻌﺎﺩﻟﻪ ﺑﺮﻧﻮﻟﻲ ﻣﻴﺎﻥ ﻧﻘﻄﻪ ﺩﻭﺭﺩﺳﺖ ﻭ ﻳﻚ ﻧﻘﻄﻪ ﺭﻭﻱ ﺳﻄﺢ ﺍﺳﺘﻮﺍﻧﻪ
U2
P V2
+ gz = +
+ gz
2
ρ
ρ 2
1
⇒ P − P∞ = ρ U 2 − V 2
2
+
ﺳﺮﻋﺖ ﺩﺭ ﺩﻭﺭﺩﺳﺖ → U 2
)
ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺳﻄﺢ r = aﺩﺍﺭﻳﻢ :
(
vr = 0
µ
⇒
µ
vθ = − 2 − U sin θ
a
∞P
= at r = a
r
ﺯﻳﺮﺍ µ = Ua 2 :
µ
⇒ V r =a = vr + vθ = − 2 − U sin 2 θ = 4U 2 sin 2 θ
a
2
2
)
)
(
2
(
1
1
ρ U 2 − 4U 2 sin 2 θ = ρU 2 1 − 4 sin 2 θ
2
2
2
= ∞⇒ P − P
⇒ Vmax = 2U
∞P − P
= 1 − 4 sin 2 θ
1
ρU 2
2
at θ = 0 , π
at θ = π , 3π
2
2
= ⇒ Pmax
at θ = 0 , π
or
)(stagnation point
2
⇒V = 0
1
ρU 2
2
2
= −3ρ U
(1 − 4 sin θ )cosθdθ =0
2
⇒ Pmin
2
, 3π
2
(P − P∞ ) cosθ (adθ ) = ρaU ∫0
2
2π
2
at θ = π
2π
∫ = Drag
0
ﺍﻳﻦ ﺗﻨﺎﻗﺾ ﺭﺍ ﭘﺎﺭﺍﺩﻭﻛﺲ ﺩﺍﻻﻣﺒﺮ )(d’Alembert’ s paradoxﮔﻮﻳﻨﺪ ﭼﻮﻥ ﺍﺯ ﺁﺛـﺎﺭ ﻭﻳﺴـﻜﻮﺯﻳﺘﻪ ﺩﺭ ﺳﺮﺗﺎﺳـﺮ
ﻛﻞ ﺟﺮﻳﺎﻥ ﺻﺮﻑ ﻧﻈﺮ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﻣﻲ ﺩﺍﻧﻴﻢ ﻛﻪ ﺍﺛﺮ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺩﺭ ﻳﻚ ﻧﺎﺣﻴﻪ ﻱ ﻧـﺎﺯﻙ ﻣﺠـﺎﻭﺭ ﺑـﺎ ﻣـﺮﺯ ﺑـﺮﺍﻱ
ﺍﺭﺯﻳﺎﺑﻲ ﻧﻴﺮﻭﻱ ﺩﺭﺍگ ﺍﻫﻤﻴﺖ ﺩﺍﺭﺩ.
Conformal mapping :
z = x + iy
i = −1
z → complex variable if :
) f ( z ) = f ( x + iy ) = f1 ( x, y ) + if 2 ( x, y
df
ﻳﻚ
dz
ﻭﺍﺣﺪ ﺩﺭ ﻫﺮ ﻧﻘﻄﻪ ﺩﺍﺭﺩ
→ f (z ) → analytic function
۱٤۲
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
∂ 2 f1 ∂ 2 f1
∂2 f2 ∂2 f2
⇒ 2 + 2 =0=
+ 2
∂y
∂x
∂y
∂x 2
∂ψ
∂ψ
,
vx = −
vy = +
∂y
∂x
∂φ
∂φ
,
vx = −
vy = −
∂x
∂y
∂φ ∂ψ
∂φ
∂ψ
⇒
=
=−
,
∂x ∂y
∂y
∂x
then
∂ 2φ ∂ 2ψ
∂ 2 = ∂ ∂
x y
x
⇒ 2
2
∂ φ = − ∂ ψ
∂y 2
∂x∂y
⇒
Cauchy - Riemann Equations
∂ 2φ ∂ 2φ
+
= ∇ 2φ = 0
∂x 2 ∂y 2
∂ 2ψ
∂ 2φ
=
−
∂x 2
∂x∂y
∂ 2ψ ∂ 2ψ
⇒
+ 2 = ∇ 2ψ = 0
2
2
2
∂x
∂y
∂ ψ = ∂ φ
2
∂x
∂x∂y
⇒ f ( z ) = φ ( x , y ) + iψ ( x , y )
∂ψ
∂ψ ∂φ
df ∂φ
=
+i
=+
−
i = −v x + iv y
∂x
∂y ∂y
dz ∂x
Poolar coordinat :
z = x + iy = reiθ = r cos θ + ir sin θ
r = x 2 + y 2 , θ = tan -1 y
x
R2
ﻛﻪ ﻳﻚ ﺳﻴﺎﻝ ﺑﺎR ﺟﺮﻳﺎﻥ ﭘﺘﺎﻧﺴﻴﻠﻲ ﺭﺍ ﺍﻃﺮﺍﻑ ﻳﻚ ﺳﻴﻠﻨﺪﺭ ﺑﻪ ﺷﻌﺎﻉf (z ) = v∞ z + ﻧﺸﺎﻥ ﺩﻫﻴﺪ: ﻣﺜﺎﻝ
z
. ﺍﺯ ﺭﻭﻱ ﺁﻥ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ ﺭﺍ ﺷﺮﺡ ﻣﻲ ﺩﻫﺪ ﻭ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻭ ﻓﺸﺎﺭ ﺭﺍ ﺑﻴﺎﺑﻴﺪv∞ ﺳﺮﻋﺖ
f ( z ) = φ ( x , y ) + iψ ( x , y )
R2
R 2 (x − iy )
= v ∞ x + iy + 2
f (z ) = v ∞ x + iy +
x + iy
x + y 2
2
2
R
R
+ iv ∞ y 1 − 2
⇒ f (z ) = v∞ 1 + 2
x + y2
x + y2
φ ( x, y )
ψ ( x, y )
2
R
⇒ ψ ( x, y ) = v∞ y1 − 2
2
+
x
y
1
⇒ Ψ ( X , Y ) = Y 1 − 2
2
X +Y
,
Ψ=
ψ
, X =x , Y= y
R
R
v∞ R
۱٤۳
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
z = x + iy = re iθ
,
z 2 = r 2 e 2 iθ
R2
df
= v∞ 1 − 2
dz
z
⇒
R2
df
= v∞ 1 − 2 e −2iθ
dz
r
R2
= v∞ 1 − 2 (cos 2θ − i sin 2θ )
r
2
2
R
df
R
= v∞ 1 − 2 cos 2θ + i 2 sin 2θ
⇒
dz
r
r
⇒
R2
R2
= v∞ 1 − 2 cos 2θ + i 2 sin 2θ = −v x + iv y
r
r
R2
v x = −v∞ 1 − 2 cos 2θ
r
R2
v y = v∞ 2 sin 2θ
r
at r = R on the cylinder surface :
[
]
v 2 = v x + v y = v∞ (1 − cos 2θ ) + (sin 2θ ) = 4v∞ sin 2 θ
2
(
2
2
)
1
ρ v x 2 + v y 2 + P = constant
2
1
1
2
⇒ ρv 2 + P = ρv∞ + P∞
2
2
1
2
⇒ P − P∞ = ρv∞ 1 − 4 sin 2 θ
2
(
2
2
2
)
۱٤٤
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺯﻣﺎﻧﻲ ﻛﻪ ﺑﻴﺶ ﺍﺯ ﻳﻚ ﻣﻮﻟﻔﻪ ﺳﺮﻋﺖ ﺳﻴﺎﻝ ﻏﻴﺮ ﺻﻔﺮ ﺩﺍﺷﺘﻪ ﺩﺍﺭﻳﻢ ،ﺣﻞ ﻣﻌﺎﺩﻻﺕ ﻛﺎﻣﻞ ﻧﺎﻭﻳﺮ ﺍﺳـﺘﻮﻙ ﺩﺭ ﺩﻭ ﻳـﺎ
ﺳﻪ ﺑﻌﺪ ﻣﺸﻜﻞ ﺍﺳﺖ.ﻳﻚ ﺭﺍﻩ ﻣﻨﺎﺳﺐ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻓﺸﺎﺭ ﺭﺍ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﺣﺬﻑ ﻛﻨﻴﻢ ،ﺑﺮﺍﻱ ﺣﺬﻑ ﻓﺸﺎﺭ ﺍﺯ
ﻣﻌﺎﺩﻟﻪ ﺣﺮﻛﺖ ﻛﺮﻝ ﻣﻲ ﮔﻴﺮﻳﻢ.
𝐷
𝑉 �⃗ = −∇𝑃 + 𝜇∇2
⃗�
𝑉
𝑡𝐷
𝜌
𝜕
𝑉[�⃗ +
𝑉∇ �⃗ .
𝑉 �⃗ ]) = −∇𝑃 + 𝜇∇2
⃗�
𝑉 (𝜌⇒
𝑡𝜕
𝑉∇�⃗ .
𝑉�∇ �⃗ ]=1
𝑉 �⃗ .
𝑉[�⃗ �+
𝑉 × ∇[ × ⃗�
]] ⃗�
𝑉[
2
𝜕
�⃗ + [𝑉.
𝑉∇ ⃗���
𝑉 �⃗ ]) =curl(- 1 ∇𝑃) + 𝜗 curl(∇2
) ⃗�
𝑉 (curl
𝑡𝜕
𝜌
𝑉∇�⃗ .
𝑉�∇ �⃗ ]= curl�1
𝑉 �⃗ .
𝑉[�⃗ �� + curl+
𝑉 × ∇[ × ⃗�
]] ⃗�
𝑉[curl
2
𝑉[ × ∇[ �⃗ � +
𝑉 × ∇[ × ⃗�
𝑉 × ∇[ �⃗ ]]]=𝜗∇2
] ⃗�
𝑉 × ∇�
𝜕
𝑡𝜕
Equation of charge for the vorticity
ﺳﭙﺲ ﻣﻌﺎﺩﻟﻪ ﺑﺮ ﺣﺴﺐ 𝞧ﺑﻴﺎﻥ ﺧﻮﺍﻫﺪ ﺷﺪ ﻛﻪ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺩﺭﺟﻪ ﭼﻬﺎﺭ ﺑﺪﺳﺖ ﺧﻮﺍﻫﺪ ﺁﻣﺪ.
𝜕
𝑦𝜕
�
𝛹𝜕
𝑥𝜕
𝜕
𝑥𝜕
𝜕
𝑧𝜕
𝛹𝜕 � 𝑘 � +
0
−
𝑦𝜕
𝜕
𝑥𝜕
𝛹𝜕
𝑦𝜕
−
�𝑗�−
)
𝜕
⎤𝜕
𝜕
𝑦𝜕 � 𝑖 = ⎥
𝑧𝜕
⎥𝑧𝜕
𝛹𝜕
0
⎦⎥ 0
𝑥𝜕
𝛹𝜕2𝛹 𝜕2
𝜕𝑦 2
+
𝜕𝑥 2
𝑘
(𝑘 ) +
𝛹 ) = +𝑘∇2
𝑗
𝜕
𝑦𝜕
𝛹𝜕
𝑥𝜕
𝛹 𝜕2
𝑖
𝜕 ⎡
𝑉 × ∇ = ⃗�
𝑥𝜕 ⎢ = ⃗�
𝑉curl
⎢
𝛹𝜕 ⎢−
𝑦𝜕 ⎣
𝑧𝜕𝑦𝜕
)− 𝑗(0+
𝜕𝑦 2
𝜕𝑥 2
𝛹 𝜕2 𝛹 𝜕2
+
𝛹 𝜕2
�⃗ = 𝑖(0 +
𝑉 × ∇⇒
𝑧𝜕𝑥𝜕
(𝑘 � +
𝛹𝜕
𝑦𝜕
𝑗+
𝛹𝜕
𝑥𝜕
𝜕
𝑖= �−
𝑧𝜕
)𝛹 �⃗ � =+𝑘 𝜕 (∇2
𝑉 × ∇�
𝑡𝜕
𝜕
𝑡𝜕
𝑉 × ∇[ × ⃗�
𝑉[ = ]] ⃗�
𝑉 �⃗ ×(+𝑘∇2 𝛹)] ,
𝛹𝜕 𝑗 �⃗ =- 𝑖 𝜕𝛹 +
𝑉[
𝑦𝜕
𝑥𝜕
𝛹𝜕
𝛹𝜕
𝑘
)̂𝚥0 �= 𝜕𝑥 . ∇2 𝛹𝚤̂ − 𝜕𝑦 . ∇2 𝛹(−
𝑘
⎤
⎥0
⎥
⎦⎥ 0
۱٤٥
𝑗
𝜕
𝑦𝜕
𝛹 . ∇2
𝛹𝜕
𝑦𝜕
𝛹 ∇2
𝑗
𝜕
𝑦𝜕
0
𝑖
𝜕
�=
𝑥𝜕
0
𝑖
𝜕 ⎡
𝑉 × ∇[ × ⃗�
𝑥𝜕 ⎢ = ]]] ⃗�
𝑉[ × ∇[
⎢
𝛹 ⎢𝜕𝛹 . ∇2
𝑥𝜕 ⎣
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
)𝛹 (∇2
𝜕
.
�𝛹 . ∇2
𝛹𝜕
𝑦𝜕 𝑥𝜕
𝛹𝜕
𝑥𝜕
�
𝑘 . ∇2 𝛹 −
)𝛹 (∇2 𝛹) = +𝑘𝜗∇2 (∇2
𝜕
𝜕
𝑦𝜕
𝑘 . ∇2 𝛹� −
𝛹 𝜕2
𝛹𝜕
𝑦𝜕 𝑥𝜕
𝑔𝜕
𝑦𝜕
𝑥𝜕
𝑦𝜕
�
𝑔𝜕
)𝛹 (∇2
𝜕
.
𝛹𝜕
𝑥𝜕 𝑦𝜕
𝑥𝜕
𝑔𝜕�=)
(∇2 𝛹) −
𝜕
.
𝛹𝜕
𝑦𝜕 𝑥𝜕
)𝑔𝑓,
𝜕
.
)𝑦𝜕 (𝑥,
𝜕
.
𝑦𝜕 𝑥𝜕
𝛹𝜕
(∇2 𝛹) +
𝜕
𝛹𝜕
.
)𝛹
𝛹𝜕
𝑥𝜕 𝑦𝜕
( 𝛹 = 𝜗∇2
=�
𝑥𝜕
𝑥𝜕 𝑦𝜕
𝜕
𝑘 (∇2 𝛹) +
)𝛹 (∇2 𝛹)= 𝜗∇2 (∇2
𝑓𝜕
𝑦𝜕
𝑘 (∇2 𝛹) −
𝑦𝜕𝑥𝜕
.
𝛹𝜕
�
𝜕
𝑘 = 𝑖(0) − 𝑗(0) +
𝛹𝜕
𝑦𝜕
(∇2
𝜕
𝑘 .∇2 𝛹 +
.
𝛹𝜕
)𝛹 𝜕(𝛹,∇2
)𝑦𝜕(𝑥,
𝑦𝜕
ﺑﻘﻴﻪ ﻣﻌﺎﺩﻻﺕ ﺩﺭ ﻣﺨﺘﺼﺎﺕ ﻛﺮﻭﻱ ﻭ ﺍﺳﺘﻮﺍﻧﻪ ﺍﻱ ﺩﺭ ﺟﺪﻭﻝ 4.2 -1ﻛﺘﺎﺏ ﺑﺮﺩ ﻣﻲ ﺑﺎﺷﻨﺪ.
𝑦𝜕𝑥𝜕
𝑘 (∇2 𝛹) −
𝑥𝜕 𝑦𝜕
𝜕
𝛹 𝜕2
)𝛹
𝛹𝜕
𝑥𝜕
𝜕
𝑘=
𝑘⇒
𝑡𝜕
𝜕
⇒ (∇2 𝛹) −
𝑡𝜕
𝜕
⇒ (∇2 𝛹) +
𝑡𝜕
(∇2
𝜕 �=
𝑥𝜕
)𝛹 𝜕(𝛹,∇2
)𝑦𝜕(𝑥,
ﻣﺜﺎﻝ :ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺟﺪﻭﻝ 4.2 -1ﻛﺘﺎﺏ ﺑﺮﺩ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺭﺍ ﺑﺮﺍﻱ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ،ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻳﻚ ﺳﻴﺎﻝ
ﻧﻴﻮﺗﻨﻲ ﺍﻃﺮﺍﻑ ﻳﻚ ﻛﺮﻩ ﺳﺎﻛﻦ ﺑﻪ ﺷﻌﺎﻉ Rﺩﺭ set up ، Re<<1ﻛﻨﻴﺪ .ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻭ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﺳﻴﺎﻝ
1
ﺍﻃﺮﺍﻑ ﻛﺮﻩ ﺭﺍ ﻧﻴﺰ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
F31
ﻓﺮﺿﻴﺎﺕ :
(1ﺟﺮﻳﺎﻥ ﭘﺎﻳﺪﺍﺭ axiymmetric flow (3 creeping flow(2
𝛹 sin 𝜃) = 𝜗𝐸 4
𝛹𝜕 1
𝜃𝜕 𝑟
cos 𝜃 −
𝛹𝜕
(
𝛹 2𝐸 2
𝑟𝜕 𝜃 𝑟 2 sin2
𝛹𝜕
−
1
From table 4.2-1 Bird
)𝛹 𝜕(𝛹,∇2
)𝜃𝑟 2 sin 𝜃 𝜕(𝑟,
, 𝑣𝜃 =+
𝑟𝜕 𝜃 𝑟 sin
=0
⃗�
𝑉𝐷
𝑡𝐷
𝛹𝜕
1
𝑡𝜕
𝑣𝑟 =-
𝜃𝜕 𝜃 𝑟 2 sin
𝜌 ⇒Creeping flow
⇒ 𝐸 2 (𝐸 2 𝛹) = 0
2
1
(𝐸 2 𝛹) +
𝜕
⇒ 𝐸4 𝛹 = 0
𝜕 𝜃 𝜕 2 sin2
𝜕 1
⇒� 2+ 2
�
�� 𝛹 = 0
𝑟𝜕
𝜃𝜕 𝜃 𝑟 𝜕𝜃 sin
Creeping Flow Around A Sphere
۱٤٦
۱
ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ٥ ﻓﺼﻞ
B.C.1: at
B.C.2: at
B.C.3:
r=R
𝑣𝑟 = −
r=R
𝑣𝜃 =+
1
1
𝑟 2 sin 𝜃
1
𝜕𝛹
𝑟 sin 𝜃 𝜕𝑟
𝜕𝛹
𝜕𝜃
=0
=0
r→ ∞𝛹 → − 𝑣∞ . 𝑟 2 . sin2 𝜃
2
r→ ∞𝑣𝑧→ 𝑣∞
𝑣𝑧= 𝑣𝑟 cos 𝜃 − 𝑣𝜃 sin 𝜃
⇒−
⇒−
1
𝜕𝛹
𝑟 2 sin 𝜃 𝜕𝜃
1
𝜕𝛹
𝑟 2 sin 𝜃 𝜕𝜃
1
cos 𝜃 −
cos 𝜃 −
1
1 𝜕𝛹
𝑟 𝜕𝑟
𝜕𝛹
𝛹 = − 𝑣∞ . 𝑟 2 . sin2 𝜃⇒
2
𝜕𝛹
𝜕𝜃
=−𝑣∞ . 𝑟 2 cos 𝜃.sin 𝜃
1
⇒−
𝑟 2 sin 𝜃
𝑟 𝜕𝑟
𝑟
1 𝜕𝛹
1
𝜕𝛹
𝑟 sin 𝜃 𝜕𝑟
𝜕𝜃
sin 𝜃 → 𝑣∞
r→ ∞
1
− 𝑣∞ . 𝑟 2 (2. cos 𝜃.sin 𝜃)
2
cos 𝜃( − 𝑣∞ . 𝑟 2 cos 𝜃.sin 𝜃) = −𝑣∞ . sin2 𝜃. 𝑟
=− (−𝑣∞ . sin2 𝜃. 𝑟) = +𝑣∞ sin2 𝜃
𝑣𝑧= 𝑣𝑟 cos 𝜃 − 𝑣𝜃 sin 𝜃 =𝑣∞ (sin2 𝜃 + cos 𝜃 2 ) = 𝑣∞
1
r→ ∞ ⇒ 𝛹 → − 𝑣∞ . 𝑟 2 . sin2 𝜃
2
𝛹(𝑟, 𝜃) = 𝑓(𝑟). sin2 𝜃
𝐸 4 𝛹 = 𝐸 2 (𝐸 2 𝛹)
𝐸2𝛹 =
𝜕 2 𝛹 sin 𝜃 𝜕
1 𝜕𝛹
+ 2
�
�
2
∂𝑟
𝑟 𝜕𝜃 sin 𝜃 𝜕𝜃
𝜕𝛹 𝜕𝑓 2
𝜕2
𝜕2
𝑑2
2
=
sin 𝜃 ⇒
=
sin
𝜃
=
sin2 𝜃
𝜕𝑟
𝜕𝑟
𝜕𝑟 2 𝜕𝑟 2
𝑑𝑟 2
𝜕𝛹
𝜕𝜃
=2 𝑓(𝑟) cos 𝜃 . sin 𝜃
1 𝜕𝛹
= 2 𝑓(𝑟) cos 𝜃
sin 𝜃 𝜕𝜃
𝜕
1 𝜕𝛹
�
� = −2𝑓(𝑟) sin 𝜃
𝜕𝜃 sin 𝜃 𝜕𝜃
𝑑2 𝑓 2
sin 𝜃
𝑑 2 𝑓 2𝑓
𝐸 𝛹 = 2 sin 𝜃 + 2 (−2𝑓(𝑟) sin 𝜃) = � 2 − 2 � sin2 𝜃
𝑑𝑟
𝑟
𝑑𝑟
𝑟
2
۱٤۷
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
𝑓𝑑 2 𝑓 2
𝑓𝑑 2 𝑓 2
2
𝛹) = � 2 − 2 � sin 𝜃 . � 2 − 2 � sin2 𝜃 = 0
𝑟𝑑
𝑟
𝑟𝑑
𝑟
2 (𝐸 2
4
𝐸=𝛹 𝐸
𝑑2
2
𝑑2
2
𝑓𝑑2 𝑓 2
𝑔 = � ⇒ � 2 − 2� � 2 − 2� 𝑓 = 0 , � 2 − 2
𝑟
𝑟
𝑟
𝑟𝑑
𝑟𝑑
𝑟𝑑
𝑔𝑑2 𝑔 2
𝑓𝑑𝑔 𝑑3 𝑓 2 𝑑𝑓 4
⇒ � 2 − 2� = 0
=
−
+
𝑟𝑑
𝑟
𝑑𝑟 𝑑𝑟 3 𝑟 2 𝑑𝑟 𝑟 3
𝑓𝑑2 𝑔 𝑑 4 𝑓 2 𝑑2 𝑓 4 𝑑𝑓 4 𝑑𝑓 12
=
−
+
+
− 4
𝑟𝑑 𝑑𝑟 2 𝑑𝑟 4 𝑟 2 𝑑𝑟 2 𝑟 3 𝑑𝑟 𝑟 3
𝑟
𝑓𝑑 2 𝑔 2𝑔 𝑑4 𝑓 4 𝑑 2 𝑓 8 𝑑𝑓 8
−
=
−
+
−
=0
𝑑𝑟 2 𝑟 2 𝑑𝑟 4 𝑟 2 𝑑𝑟 2 𝑟 3 𝑑𝑟 𝑟 4
𝑓 𝑑2
= 𝑐𝑛(𝑛 − 1)𝑟 𝑛−2 ,
2
𝑟𝑑
𝑓 𝑑4
= 𝑐𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)𝑟 𝑛−4
𝑑𝑟 4
⇒
Fourth-order Euler Eq:
𝑓𝑑
= 𝑐𝑛𝑟 𝑛−1 ,
⇒ 𝑟𝑐 = )𝑟(𝑓
𝑟𝑑
𝑛
𝑓 𝑑3
= 𝑐𝑛(𝑛 − 1)(𝑛 − 2)𝑟 𝑛−3 ,
𝑑𝑟 3
⇒ 𝑐𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3)𝑟 𝑛−4 − 4𝑐𝑛(𝑛 − 1)𝑟 𝑛−4 + 8𝑐𝑛𝑟 𝑛−4 − 8𝑐𝑟 𝑛−4 = 0
⇒ 𝑛(𝑛 − 1)(𝑛 − 2)(𝑛 − 3) − 4𝑛(𝑛 − 1) + 8𝑛 = 0
𝑛 = 1 , 𝑛 = 2 , 𝑛 = 4 , 𝑛 = −1
𝐹(𝑟) = 𝐶1 𝑟 −1 + 𝐶2 𝑟 + 𝐶3 𝑟 2 + 𝐶4 𝑟 4
ﺑﺮﺍﻱ ﺍﺭﺿﺎﻱ ﺷﺮﻁ ﻣﺮﺯﻱ ﺳﻮﻡ ﺑﺎﻳﺪ:
1
𝐶3 = − 𝑉∞ 𝐶4 = 0
2
𝐶1
1
+ 𝐶2 𝑟 − × (𝑉∞ 𝑟 2 ))𝑠𝑖𝑛𝜃 2
𝑟
2
( = )𝜃 → Ψ(𝑟,
1
𝜕Ψ
𝐶2
𝐶1
.
=
𝑉(
−
2
−
2
)𝑐𝑜𝑠ө
∞
(𝑟 2 𝑠𝑖𝑛ө) 𝜕ө
𝑟
𝑟3
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺩﻭ ﺷﺮﻁ ﻣﺮﺯﻱ ﺍﻭﻝ:
1
𝜕Ψ
𝐶2 𝐶1
.
= (−𝑉∞ + − 3 )𝑠𝑖𝑛ө
𝑟𝜕 )(𝑟𝑠𝑖𝑛ө
𝑟 𝑟
𝑉𝑟 = −
𝑉𝜃 = +
1
3
𝑟 ∞𝑉 = 𝐶1 = − 𝑉∞ 𝑅3 𝐶2
4
4
۱٤۸
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
𝑅 3
𝑅 1
𝑉𝑟 = 𝑉∞ (1 − ( ) + ( )3 )𝑐𝑜𝑠ө
𝑟 2
𝑟 2
𝑅 3
𝑅 1
𝑉𝜃 = −𝑉∞ (1 − ( ) − ( )3 )𝑠𝑖𝑛ө
𝑟 4
𝑟 4
ﺑﺮﺍﻱ ﺑﺪﺳﺖ ﺍﻭﺭﺩﻥ ﺗﻮﺯﻳﻊ ﻓﺸﺎﺭ ﻣﻌﺎﺩﻻﺕ ﻓﻮﻕ ﺭﺍ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻱ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻙ ﻛﺮﻭﻱ ﺩﺭ ﺟﻬﺖ ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻴﻢ ﻭ
ﺩﺭ ﻧﺘﻴﺠﻪ:
𝜕Ƥ
𝑅 ∞𝑉𝜇
𝜃𝑠𝑜𝑐 = ( 2 )( )3
𝑟𝜕
𝑅
𝑟
𝜕Ƥ
𝜇𝑉∞ 𝑅 2
(=
)( ) sinθ
𝜕ө
𝑟 𝑅
𝑇𝑎𝑏𝑙𝑒 𝐵. 7
)θﻭ𝑟( Ƥ = Ƥ
𝜕Ƥ
𝜕Ƥ
). 𝑑𝑟 + ( ). 𝑑ө
𝑟𝜕
𝜕ө
𝑉∞ 𝑅 3
𝜇𝑉∞ 𝑅 2
( )( ) 𝑐𝑜𝑠ө. 𝑑𝑟 +
)( ) 𝑠𝑖𝑛ө. 𝑑ө
2
𝑟 𝑅
𝑟 𝑅
(=𝑑Ƥ
→ 𝑑 Ƥ = (µ
1 −2
𝜇𝑉∞ 𝑅 2
∞𝑉𝜇
3
𝑐 ) � � (−𝑐𝑜𝑠ө) +
( → Ƥ = ( 2 )𝑐𝑜𝑠ө. 𝑅 . (− 𝑟 ) +
𝑅
2
𝑅
𝑟
1
∞𝑉µ
𝑅
𝜇𝑉∞ 𝑅 2
( Ƥ = − × ( 2 ). 𝑅. ( )2 𝑐𝑜𝑠ө −
𝑐 )( ) 𝑐𝑜𝑠ө +
2
𝑅
𝑟
𝑟 𝑅
→
𝑧𝑔𝜌 Ƥ = Ƥ +
𝑧=0
𝑡𝑎
Ƥ = Ƥ0
𝑅 ∞𝑉𝜇 3
( →Ƥ=−
𝑐 )( )𝑐𝑜𝑠ө +
𝑟 𝑅 2
∞
→
𝑟 𝑡𝐴
3 𝜇𝑉∞ 𝑅 2
( Ƥ = −𝜌𝑔𝑧 −
𝑐 )( ) 𝑐𝑜𝑠ө +
𝑟 𝑅 2
→ 𝐶 = Ƥ0
𝑐 Ƥ0 = 0 − 0 +
3 𝜇𝑉∞ 𝑅 2
( Ƥ = Ƥ0 − 𝜌𝑔𝑧 −
)( ) 𝑐𝑜𝑠ө
𝑟 𝑅 2
ﻗﺒﻼ 𝑘𝑓ﻧﻴﺮﻭﻱ ﺩﺭﺍگ ﺭﺍ ﺭﻭﻱ ﻛﺮﻩ ﭘﻴﺪﺍ ﻛﺮﺩﻳﻢ.
۱٤۹
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺭﻭﺷﻲ ﺩﻳﮕﺮ:
𝑑𝑖𝑢𝑙𝑝 𝑒= 𝑓𝑜𝑟𝑐𝑒 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦) = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑣𝑖𝑠𝑐𝑜𝑤 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎﻛﺎﺭﻱ ﻛﻪ ﺭﻭﻱ ﻛﺮﻩ ﺑﻪ
ﻭﺳﻴﻠﻪ ﺳﻴﺎﻝ ﺍﻧﺠﺎﻡ ﻣﻲ ﮔﻴﺮﺩ
0 ≤ 𝛷 ≤ 2ᴨ
0 ≤ ө ≤ −ᴨ
∞
ᴨ
2ᴨ
𝛷𝑑𝑓𝑘 𝑣∞ = � � � (− (𝜁: 𝛻𝑣⃗)𝑟 2 𝑑𝑟𝑠𝑖𝑛ө𝑑ө
𝑅
0
0
𝑠𝑠𝑜𝑙 𝑛𝑜𝑖𝑡𝑐𝑖𝑟𝐹 = 𝑣𝑑)𝑣∇ 𝐸𝑣 = − � (𝜁:
𝑣
ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻱ ﺑﺮﻧﻮﻟﻲ ) 𝑠𝑠𝑜𝑙 𝑟𝑜𝑛𝑖𝑚 (𝑚𝑎𝑗𝑜𝑟 𝑙𝑜𝑠𝑠 +ﻓﺼﻞ 𝑏𝑖𝑟𝑑7ﺗﺒﺪﻳﻞ ﺍﻧﺮﺯﻱ ﻣﻜﺎﻧﻴﻜﻲ ﺑﻪ ﮔﺮﻣﺎ
∂vr 2
1 ∂vθ vr 2
→ Fk . v∞ = 2πµ � � [2 � � + 2 � .
� +
∂r
r ∂θ
r
0
R
∞
π
vr vθ cot θ 2
∂ vθ
1 ∂vr 2
) + (r � � +
) ] × r 2 drsinөdөdΦ
+( +
r
r
∂r r
r ∂θ
𝑅𝑒˂0.1
!→ Fk = 6πµv∞ R lengthy
𝑠𝑡𝑜𝑘𝑠′𝑠 𝑙𝑜𝑤.
− (𝜁: 𝛻𝑣) = µ𝜑𝑣 → (𝐵𝑖𝑟𝑑)𝐵. 7
𝑠𝑡𝑛𝑒𝑖𝑑𝑎𝑟𝑔 𝑦𝑡𝑖𝑠𝑜𝑐𝑠𝑖𝑣 𝑒𝑔𝑟𝑎𝐿
Lubrication rapid extrusion high − speed flight
ﻓﺮﺍﻳﻨﺪﻱ ﻛﻪ ﻳﻚ ﻣﺎﺩﻩ ﺟﺎﻣﺪ ﻧﻴﻤﻪ ﻧﺮﻡ ﺳﺮﺩ ﻳﺎ ﺩﺍﻍ ﺑﺎ ﺍﻋﻤﺎﻝ ﻧﻴﺮﻭ ﺗﺒﺪﻳﻞ ﺑﻪ ﻗﻄﻌﺎﺗﻲ ﺑﺎﺍﺷﻜﺎﻝ ﺩﻟﺨﻮﺍﻩ ﻣﻲ ﺷﻮﺩ.
۱٥۰
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
1-5ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺳﻴﺎﻟﻲ ﺑﺎ ﭼﮕﺎﻟﻲ 1500 kg3ﺑﺎ ∧ V = ( Ax − By )ti ∧ − ( Ay + Bx )tjﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ.ﻛﻪ ﺩﺭ ﺁﻥ
x ، A = 1s −2 , B = 2 s −2ﻭ y
m
ﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﻭ ﺑﺮ tﺣﺴﺐ ﺛﺎﻧﻴﻪ ﺍﺳﺖ ﻧﻴﺮﻭﻫﺎﻱ ﺣﺠﻤﻲ ﻗﺎﺑﻞ ﺻﺮﻓﻨﻈﺮ ﻫﺴﺘﻨﺪ.
∇Pﺭﺍ ﺩﺭ ﻧﻘﻄﻪ ) (x, y ) = (1,2ﺩﺭ ﻟﺤﻈﻪ t = 1sﺣﺴﺎﺏ ﻛﻨﻴﺪ؟
∧
2-5ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺩﺭ ﮔﺮﺩﺍﺑﻲ ﺁﺯﺍﺩ)ﺑﻲ ﭼﺮﺧﺶ( ﺑﺮﺍﻱ r > 0ﺑﺎ
2πr
2
ﺣﺴﺐ ﻣﺘﺮ ﺍﺳﺖ .ﺳﻴﺎﻝ ﺭﺍ ﺑﻲ ﺍﺻﻄﻜﺎﻙ ﺑﺎ ρ = 1000 kg3ﻭ k = 20π mﺩﺭ
s
m
ﺷﻌﺎﻋﻲ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺗﺎﺑﻌﻲ ﺍﺯ rﺑﻴﺎﻥ ﻛﻨﻴﺪ ﻭ ﺗﻐﻴﻴﺮ ﻓﺸﺎﺭ ﺑﻴﻦ r1 = 1mﻭ r2 = 2m
θ
3-5ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﻳﻚ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﻋﺒﺎﺭﺕ ﺍﺳﺖ
V = keﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺍﻥ rﺑﺮ
ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ ﮔﺮﺍﺩﻳﺎﻥ ﻓﺸﺎﺭ
ﺑﻴﺎﺑﻴﺪ؟
ﺍﺯ ∧ V = ( Ax − By )ti ∧ − ( Ay + Bx )tj
ﻛﻪ ﺩﺭ ﺁﻥ
x ، A = 1s −2 , B = 2s −2ﻭ yﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﻭ tﺑﺮ ﺣﺴﺐ ﺛﺎﻧﻴﻪ ﺍﺳﺖ ﺁﻳﺎ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺗﺮﺍﻛﻢ ﻧﺎ ﭘﺬﻳﺮ ﺍﺳﺖ؟ ﺁﻳﺎ
ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﭘﺎﻳﺎ ﺍﺳﺖ ﻳﺎ ﻧﺎﭘﺎﻳﺎ؟ ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ ﻭ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
4-5ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻳﻚ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ ψ = Ayx 2 − By 3ﻛﻪ ﺩﺭ ﺁﻥ
ﻭﻣﺨﺘﺼﺎﺕ ﺑﺮ ﺣﺴﺐ ﻣﺘﺮ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ ﻣﻲ ﺷﻮﺩ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
1
A = 1m −1 s −1ﻭ A = m −1 s −1
3
5-5ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻧﻲ ﺑﺎ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ψ = x 2 − y 2ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﻣﺘﻨﺎﻇﺮ ﺭﺍ ﺑﻴﺎﺑﻴﺪ ﻧﺸﺎﻥ
ﺩﻫﻴﺪ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻲ ﭼﺮﺧﺶ ﺍﺳﺖ ﻭ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
6-5ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺍﺏ ﺑﺎ ﺭﺍﺑﻄﻪ ∧ V = Ax 2 y 2 i ∧ − Bxy 3 jﺭﺍ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ ﻛﻪ ﺩﺭ ﺍﻥ A = 32ﻭ
m s
ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺭﺍ ﺑﻴﺎﺑﻴﺪ , ωﺳﻴﺎﻝ ﺭﺍ ﺗﻌﻴﻴﻦ ﻛﻨﻴﺪ .ﺑﺎ ﺻﺮﻓﻨﻈﺮ ﺍﺯ ﮔﺮﺍﻧﺶ ﺍﻳﺎ ﻣﻴﺘﻮﺍﻥ ﺍﺧﺘﻼﻑ ﻓﺸﺎﺭ ﺑﻴﻦ
) (1,1,1ﺭﺍ ﺣﺴﺎﺏ ﻛﺮﺩ؟ ﺍﮔﺮ ﻣﻴﺘﻮﺍﻧﻴﺪ ﺍﻧﺮﺍ ﺣﺴﺎﺏ ﻛﻨﻴﺪ ﻭ ﺍﮔﺮ ﻧﻪ ﺗﻮﺿﻴﺢ ﺩﻫﻴﺪ ﭼﺮﺍ؟
2
m2s
)(0,0,0
=B
ﻭ
7-5ﺑﺎ ﺗﺮﻛﻴﺐ ﭼﺎﻫﻲ ﺑﺎ ﻗﺪﺭﺕ q = 2800 mﻭ k = 5600 mﮔﺮﺩﺍﺑﻲ ﺍﺯﺍﺩ ﺑﺎ ﻗﺪﺭﺕ ﻣﺪﻝ ﺗﻘﺮﻳﺒﻲ ﻳﻚ ﮔﺮﺩﺑﺎﺩ
2
2
s
s
ﺑﺪﺳﺖ ﻣﻲ ﺍﻳﺪ .ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺭﺍ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻴﺎﺑﻴﺪ.ﺷﻌﺎﻋﻲ ﺭﺍ ﻛﻪ ﺑﻴﺶ ﺗﺮ ﺍﺯﺍﻥ
ﺑﺘﻮﺍﻥ ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺖ ﺗﺨﻤﻴﻦ ﺑﺰﻧﻴﺪ .ﻓﺸﺎﺭ ﭘﻴﻤﺎﻧﻪ ﺍﻱ ﺭﺍ ﺩﺭ ﺍﻥ ﺷﻌﺎﻉ ﺑﻴﺎﺑﻴﺪ؟
8-5ﻳﻚ ﭼﺸﻤﻪ ﻭ ﻳﻚ ﭼﺎﻩ ﺑﺎ ﻗﺪﺭﺕ ﻣﺴﺎﻭﻱ ﺭﻭﻱ ﻣﺤﻮﺭ xﺑﻪ ﺗﺮﺗﻴﺐ ﺩﺭ x = −aﻭ x = aﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ .ﺗﺎﺑﻊ
ﺟﺮﻳﺎﻥ ﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﺭﺍ ﺑﺮﺍﻱ ﺍﻳﻦ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺑﻴﺎﺑﻴﺪ .ﻧﻘﺎﻃﻲ ﺭﺍ ﺭﻭﻱ ﻣﺤﻮﺭ ﭘﻴﺪﺍ ﻛﻨﻴﺪ ﻛﻪ ﺍﺯ ﺑﻴﻦ ﺍﻧﻬﺎ
ﻧﺼﻒ ﺍﻫﻨﮓ ﺷﺎﺭﺵ ﻛﻞ ﺣﺠﻢ ﻣﻲ ﮔﺬﺭﺩ؟
3
9-5ﻳﻚ ﭼﺸﻤﻪ ﻭ ﻳﻚ ﭼﺎﻩ ﺑﺎ ﻗﺪﺭﺕ ﻣﺴﺎﻭﻱ q = 3π mﺭﻭﻱ ﻣﺤﻮﺭ xﺑﻪ ﺗﺮﺗﻴـﺐ ﺩﺭ x = −aﻭ x = aﻗـﺮﺍﺭ
s
m
ﺩﺍﺭﻧﺪﻳﻚ ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ ﺑﺎ ﺳﺮﻋﺖ
s
U = 20ﺩﺭ ﺟﻬﺖ xﻣﺜﺒﺖ ﺍﻓﺰﻭﺩﻩ ﻣﻲ ﺷﻮﺩﺗﺎ ﺟﺮﻳـﺎﻥ ﻋﺒـﻮﺭﻱ ﺍﺯ ﺭﻭﻱ
۱٥۱
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺟﺴﻢ ﺭﺍﻧﻜﻴﻦ ﺑﺪﺳﺖ ﺍﻳﺪ . .ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﻭ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ ﻭ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺟﺮﻳﺎﻥ ﺗﺮﻛﻴﺒﻲ ﺭﺍ ﺑﻴﺎﺑﻴﺪ .ﻣﻘﺪﺍﺭ ﺛﺎﺑﺖ
ψﻣﺮﺑﻮﻁ ﺑﻪ ﺧﻂ ﺟﺮﻳﺎﻥ ﺭﻛﻮﺩ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ .ﺍﮔﺮ a = 0.3mﻧﻘﺎﻁ ﺭﻛﻮﺩ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟
10-5ﺍﮔﺮ ﺁﺏ ﺍﺯ ﺳﻮﺭﺍﺧﻲ ﺩﺭ ﻛﻒ ﻳﻚ ﻣﺨﺰﻥ ﺑﺰﺭگ ﺧﺎﺭﺝ ﺷﻮﺩ ﺷﻜﻞ ﺧﻄﻮﻁ ﺟﺮﻳﺎﻥ ﺑﻪ ﭼﻪ ﺷﻜﻞ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ψ = Γ ln rﺍﺳﺖ؟ ﺑﺎ ﻧﻮﺷﺘﻦ ﺭﺍﺑﻄﻪ ﺑﺮﻧﻮﻟﻲ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﺑﻴﻦ ﻧﻘﺎﻁ ﻣﺸﺨﺺ ﺷﺪﻩ
2π
Γ2
= Z sﻣﻲ ﺑﺎﺷﺪ .ﺷﻜﻞ ﺳﻄﺢ ﺁﺯﺍﺩ
ﺩﺭ ﺷﻜﻞ ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻣﻌﺎﺩﻟﻪ ﺳﻄﺢ ﺁﺯﺍﺩ ﺩﺭ ﻗﺴﻤﺖ ﻏﻴﺮﭼﺮﺧﺸﻲ
8 gπ 2 r 2
ﺩﺭ ﻗﺴﻤﺖ ﭼﺮﺧﺸﻲ ﺑﻪ ﭼﻪ ﺻﻮﺭﺕ ﺍﺳﺖ؟
11-5ﻳﻚ ﮔﺮﺩﺍﺑﻪ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺷﻜﻞ ﻳﻚ ﺟﺮﻳﺎﻥ ﻭﺭﺗﻜﺲ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﻣﺪﻝ ﻛﺮﺩ ﺩﺭ ﺁﻥ:
ﻣﺸﺨﺺ ﻛﻨﻴﺪ ﻛﻪ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﺩﺭ ﺩﺍﺧﻞ ﻭ ﺧﺎﺭﺝ ﺷﻌﺎﻉ Rﭼﺮﺧﺸﻲ ﻳﺎ ﻏﻴﺮ ﭼﺮﺧﺸﻲ ﺍﺳﺖ؟ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ
ﻣﻌﺎﺩﻟﻪ ﻣﻤﻨﺘﻮﻡ ﺩﺭ ﺟﻬﺖ rﺑﺎ ﺍﻳﻦ ﻓﺮﺽ ﻛﻪ ∞ = Pﺩﺭ ∞ = rﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ .ﻣﻘﺪﺍﺭ ﻭ ﻣﻜﺎﻥ ﻛﻤﺘﺮﻳﻦ ﻓﺸﺎﺭ ﺭﺍ
ﺑﻴﺎﺑﻴﺪ؟
12-5ﭘﺎﻳﻪ ﭘﻠﻲ ﺑﺎ ﺑﻪ ﺑﺮﺩﻥ ﺷﻜﻠﻲ)ﺧﻂ ﺟﺮﻳﺎﻥ ﻣﻴﺎﻧﻲ (ψ = 0ﺑﺮﺍﻱ ﺟﺮﻳﺎﻧﻲ ﻛﻪ ﺗﺮﻛﻴﺒﻲ ﺍﺯ ﻳﻚ ﭼﺸﻤﻪ ﺑﻪ
ﺷﺪﺕ Qﺩﺭ ﺟﻬﺖ ﻣﺤﻮ Uﺩﺭ ﺟﻬﺖ ﻣﻨﻔﻲ ﻣﺤﻮﺭ xﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ.
۱٥۲
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
ﺍﻟﻒ -ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﻛﻪ ﺟﺮﻳﺎﻥ ﺭﻭﺩﺧﺎﻧﻪ ﺳﺮﻋﺖ ﺭﻭﻱ Uﺑﻪ ﻃﺮﻑ ﭘﺎﻳﻪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ﺗﻐﻴﻴﺮﺍﺕ ﻓﺸﺎﺭ ﺩﺭ ﺭﻭﻱ ﻣﺤﻴﻂ
ﭘﺎﻳﻪ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ؟
2
sin θ
2 sin 2θ
1
=P
ρU
−
2
θ
θ
ﺏ -ﺗﻮﺿﻴﺢ ﺩﻫﻴﺪ ﺩﺭ ﭼﻪ ﺳﺮﻋﺘﻲ ﺍﺯ ﺟﺮﻳﺎﻥ ,ﺍﻣﻜﺎﻥ ﻛﺎﻭﻳﺘﺎﺳﻴﻮﻥ ﺭﻭﻱ ﭘﺎﻳﻪ ﭘﻞ ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟
13-5ﺗﺮﻛﻴﺐ ﻳﻚ ﺟﺮﻳﺎﻥ ﺁﺯﺍﺩ ﺑﺎ ﺳﺮﻋﺖ , 10 m sﻳﻚ ﺩﻭﺑﻠﺖ ﺑﻪ ﻗﺪﺭﺕ 40ﻭ ﻳﻚ ﻭﺭﺗﻜﺲ ﻏﻴﺮﭼﺮﺧﺸﻲ ﺑﺎ
ﺳﻴﺮﻛﻮﻻﺳﻴﻮﻥ 200 m 2 sﺭﺍ ﻛﻪ ﺗﺸﻜﻴﻞ ﺟﺮﻳﺎﻥ ﺍﻳﺪﻩ ﺍﻝ ﺍﻃﺮﺍﻑ ﺍﺳﺘﻮﺍﻧﻪ ﺑﺎ ﺳﻴﺮﻛﻮﻻﺳﻴﻮﻥ ﻣﻲ ﺩﻫﺪ ﺩﺭ ﻧﻈﺮ
ﺑﮕﻴﺮﻳﺪ .ﺍﮔﺮ ﺩﺭ ) (x, p ) = (∞, oﺑﺎﺷﺪ ﻭ ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ ﻭﺟﻮﺩ ﺩﺍﺷﺖ ﺑﺎﺷﺪ ﻧﻘﺎﻁ ﺳﻜﻮﻥ,ﻧﻘﺎﻁ ﻣﺎﻛﺰﻳﻤﻢ ﻭ
ﻣﻴﻨﻴﻤﻢ ﺳﺮﻋﺖ ﻭ ﻓﺸﺎﺭ ﺭﻭﻱ ﺳﻄﺢ ﺳﻴﻠﻨﺪﺭ ﺑﺪﺳﺖ ﺑﻴﺎﻭﺭﻳﺪ؟
14-5ﺑﺎ ﺗﺮﻛﻴﺐ ﺩﻭ ﻭﺭﺗﻜﺲ ﺁﺯﺍﺩ ﺩﺭ ﺧﻼﻑ ﻳﻜﺪﻳﮕﺮ ﻭ ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺯﻳﺮ ,ﺧﺎﻧﻮﺍﺩﻩ ﺍﻱ ﺍﺯ ﺑﻴﻀﻲ
ﻫﺎﻱ ﻛﺸﻴﺪﻩ ﻣﻮﺳﻮﻡ ﺑﻪ ﺑﻴﻀﻲ ﻛﻠﻮﻳﻦ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ .ﻧﺸﺎﻥ ﺩﻫﻴﺪ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺍﻳﻦ ﻣﺠﻤﻮﻋﻪ ﺍﺯ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﺪﺳﺖ
ﻣﻲ ﺁﻳﺪ ﻭ ﺗﺎﺑﻊ ﭘﺘﺎﻧﺴﻴﻞ ﺭﺍ ﺑﺪﺳﺖ ﺁﻭﺭﻳﺪ.
x 2 + ( y + a) 2
x 2 + ( y − a) 2
۱٥۳
1
2
ψ = U ∞ y − k ln
ﻣﺴﺌﻠﻪ ﻫﺎ
ﻓﺼﻞ ٥ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ
۰-۰
15-5ﺍﺳﺘﻮﺍﻧﻪ ﺑﺴﻴﺎﺭ ﻃﻮﻻﻧﻲ ﺑﺎ ﺳﺮﻋﺖ ﺩﻭ ﻣﺘﺮﻳﺂﺏ ﺳﺎﻛﻦ ﺣﺮﻛﺖ ﻣﻲ ﻛﻨﺪ .ﺍﮔﺮ ﺳـﺮﻋﺖ 10ﻭ 20ﺑﺎﺷـﺪ ,ﺩﺭ ﺩﻭ
ﺣﺎﻟﺖ ﺟﺮﻳﺎﻥ ﺍﻳﺪﻩ ﺍﻝ ﻭ ﺣﻘﻴﻘﻲ ﻛﻤﺘﺮﻳﻦ ﻓﺸﺎﺭ ﺍﺳﺘﻮﺍﻧﻪ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ.
16-5ﺍﮔﺮ ﻳﻚ ﭼﺸﻤﻪ ﺑﺎ ﺷﺪﺕ 0.2 mﻭ ﻳﻚ ﮔﺮﺩﺍﺏ ﺑﺎ ﺷﺪﺕ
3
s
ﺳﺮﻋﺖ ﻭ ﺗﺎﺑﻊ ﺟﺮﻳﺎﻥ ﺑﺮﺍﻱ ﻣﻨﺒﻊ ﻭ ﮔﺮﺩﺍﺏ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ؟
17-5ﻳﻚ ﭼﺸﻤﻪ ﺩﻭ ﺑﻌﺪﻱ ﺑﺎ ﻗﺪﺭﺕ
m2
m.s
3
1 mﺩﺭ ﻣﺒﺪﺍً ﻗـﺮﺍﺭ ﺩﺍﺷـﺘﻪ ﺑﺎﺷـﺪ ﭘﺘﺎﻧﺴـﻴﻞ
s
9.42ﻭﺍﻗﻊ ﺩﺭ ﻣﺒﺪﺍً ﻣﺨﺘﺼﺎﺕ ﺩﺭ ﻣﺴـﻴﺮ ﻳـﻚ ﺟﺮﻳـﺎﻥ ﻳﻜﻨﻮﺍﺧـﺖ ﺑـﺎ
ﻣﺤﻮﺭ xﻫﺎ ﺯﺍﻭﻳﻪ 45ﻣﻲ ﺳﺎﺯﺩ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﺳﺖ ﺍﮔﺮ ﺳﺮﻋﺖ ﺍﻳﻦ ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺖ
ﻱ ﺳﻜﻮﻥ ﺗﺎ ﻣﺒﺪﺍً ﻣﺨﺘﺼﺎﺕ ﭼﻨﺪ ﺳﺎﻧﺘﻲ ﻣﺘﺮ ﺍﺳﺖ؟
m
s
U = 3ﺑﺎﺷﺪ ﻓﺎﺻـﻠﻪ ﻧﻘﻄـﻪ
2
18-5ﻳﻚ ﺩﺍﺑﻠﺖ ﺑﺎ ﭘﺘﺎﻧﺴﻴﻞ ﺳﺮﻋﺖ φ = 90 cos θ m ﺩﺭ ﻣﺒﺪﺍً ﻣﺨﺘﺼـﺎﺕ ﻗـﺮﺍﺭ ﺩﺍﺭﺩ) rﺑـﺮ ﺣﺴـﺐ ﻣﺘـﺮ(
s
r
ﺟﺮﻳﺎﻥ ﻳﻜﻨﻮﺍﺧﺘﻲ ﺑﺎ ﺳﺮﻋﺖ ﺩﻩ ﻣﺘﺮ ﺑﺮ ﺛﺎﻧﻴﻪ ﺍﺯ ﺭﻭﻱ ﺁﻥ ﻣﻲ ﮔﺬﺭﺩ ﺑﺎ ﻓـﺮ ﺽ ﺟﺮﻳـﺎﻥ ﺍﻳـﺪﻩ ﺁﻝ ﻓﺎﺻـﻠﻪ ﺩﻭ ﻧﻘﻄـﻪ
ﺳﻜﻮﻥ ﺣﺎﺻﻞ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﭼﻨﺪ ﻣﺘﺮ ﺍﺳﺖ؟
۱٥٤
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
۱٥٥
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
٦ ﻓﺼﻞ
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ
32
1
F
ﺟﺮﻳﺎﻥ ﺳـﻴﺎﻝ، ﺍﺗﻤﺴﻔﺮ، ﺭﻭﺩﺧﺎﻧﻪ ﻫﺎ،ﺍﻏﻠﺐ ﺟﺮﻳﺎﻧﻬﺎﻱ ﻣﻬﻢ ﻃﺒﻴﻌﻲ ﻭﺻﻨﻌﺘﻲ ﻣﺘﻼﻃﻢ ﻫﺴﺘﻨﺪ)ﺟﺮﻳﺎﻥ ﺍﻗﻴﺎﻧﻮﺱ ﻫﺎ
(... ﺍﺗﺎﻕ ﺍﺣﺘﺮﺍﻕ ﻭ ﺩﺭ ﺍﻃﺮﺍﻑ ﺍﺗﻮﻣﺒﻴﻞ ﻭ،ﺩﺭ ﺩﺍﺧﻞ ﻣﺒﺪﻟﻬﺎﻱ ﺣﺮﺍﺭﺗﻲ
. ﻭ ﺭﻳﺎﺿﻲ ﺁﻥ ﻏﻴﺮ ﺧﻄﻲ ﺍﺳﺖ2 ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﺑﻪ ﻃﻮﺭ ﻣﺪﻝ ﻭﻗﻮﻳﺎ ﺷﺎﻧﺴﻲ ﺑﺎ ﺯﻣﺎﻥ ﺗﻐﻴﻴﺮ ﻣﻲ ﻛﻨﺪ
F3
: ﭼﻨﺪ ﺗﻌﺮﻳﻒ ﺍﺯ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ
1) Turbulence is irregular and seemingly random.
2) Turbulence is highly diffusive.rapid mixing significantly increases momentum,heat,and mass transfer.
3) Turbulence is rotational and three-dimensional motion.
4) Turbulence is asso ciated with high levels of vorlicity fluctuation.
5) Turbulence is highly dissipative.it needs a source of energy to be maintained.
6) Turbulence is a continuum phenomenon.the smallest scale of turbulence is
much larger than the molecular scales in most engineering applications.
7) Turbulence is a manifestation of flow and not of fluid,origin of Turbulence:
Turbulence is associated with high Reynolds number.its origion is rooted in the
instability of shear flows.
Turbulence is also generated in buoyancy driven flows.
: ﻣﻌﺎﺩﻻﺕ ﺣﺎﻛﻢ ﺑﺮ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ
�⃗ =0
𝛻.𝑉
1
𝜕𝑡
𝜌
�⃗
𝜕𝑣
ﺟﺮﻳﺎﻥ ﺭﺍ ﺗﺮﺍﻛﻢ ﻧﺎﭘﺬﻳﺮ ﻓﺮﺽ ﻣﻲ ﻛﻨﻴﻢ
+ (𝑣⃗. ∇)𝑣⃗ = − × (∇Ƥ) + 𝑣∇2 𝑣⃗
No single pair of random functions v(x,y,z)and p(x,y,z,t)→
۱
۲
Turbulent
Randomly Time-Dependent
۱٥٦
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺑﺮﺍﻱ ﺣﻞ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻻﺯﻡ ﺍﺳﺖ ﻛﻪ ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻭ ﺷﺮﺍﻳﻂ ﺍﻭﻟﻴﻪ ﻣﺸﺨﺺ ﺑﺎﺷﺪ .ﺣﻞ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺩﺭﺣﺎﻟﺖ ﻛﻠـﻲ
ﻣﻤﻜﻦ ﻧﻴﺴﺖ ﻭ ﺍﮔﺮ ﻣﻤﻜﻦ ﻫﻢ ﺑﺎﺷﺪ ﺍﻃﻼﻋﺎﺕ ﻟﺤﻈﻪ ﺍﻱ ﺩﺭ ﻣﻮﺭﺩ ﺳﺮﻋﺖ ﻓﺸﺎﺭ ﺗﻨﺶ ﺑﺮﺷﻲ ﻭ ﻏﻴﺮﻩ ﻣـﻲ ﺩﺍﺩ ﻛـﻪ
ﭼﻨﺪﺍﻥ ﺟﻬﺖ ﻣﻮﺍﺭﺩ ﻣﻬﻨﺪﺳﻲ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻲ ﺑﺎﺷﺪ.ﺑﻠﻜﻪ ﻣﻘﺎﺩﻳﺮ ﻣﺘﻮﺳﻂ ﻫﺴـﺘﻨﺪ ﮔـﻪ ﺩﺭ ﻣﻬﻨﺪﺳـﻲ ﻛـﺎﺭﺑﺮﺩ
ﺩﺍﺭﻧﺪ.ﻟﺬﺍ ﻣﻘﺎﺩﻳﺮ ﻓﻮﻕ ﺭﺍ ﺑﻪ ﻗﺴﻤﺖ ﻣﺘﻮﺳﻂ ﻭﻗﺴﻤﺖ ﻧﻮﺳﺎﻧﻲ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺗﻘﺴﻴﻢ ﻣﻲ ﻛﻨﻴﻢ.
ﺑﻨﺎ ﺑﺮﺍﻳﻦ ﻣﻲ ﺗﻮﺍﻥ ﻧﻮﺷﺖ:
𝑤 = 𝑊 𝑈 = 𝑢� + 𝑢′ , 𝑉 = 𝑣̅ + 𝑣 ′ ,
� + 𝑤′
𝑃 = 𝑝̅ + 𝑝′
𝑈=instantaneous velocity
velocity
�=time_mean
𝑈
𝑈′=fluctuating velocity
۱٥۷
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺑﺮﺍﻱ ﺍﻧﺪﺍﺯﻩ ﮔﻴﺮﻱ 𝑈′ﺩﺭ ﺟﺮﻳﺎﻥ ﺗﻮﺭ ﺑﺎﻟﻨﺖ ﺍﺯ ﻭﺳﻴﻠﻪ ﻱ ﺑﻨﺎﻡ ﺑﺎﺩ ﺳﻨﺞ ﺳﻴﻢ ﺩﺍﻍ 1ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﺷﻮﺩ.ﺍﺳـﺎﺱ ﺍﻳـﻦ
ﺩﺳﺘﮕﺎﻩ ﻳﻚ ﺳﻴﻢ ﺧﻴﻠﻲ ﻧﺎﺯﻙ ﺣﺪﻭﺩﺍ" ﺑﻪ ﻗﻄﺮ10-3mmﻣﻲ ﺑﺎﺷﺪ ﻛﻪ ﺑﻄﻮﺭ ﻛﺸﻴﺪﻩ ﺩﺭ ﻣﺴﻴﺮ ﺟﺮﻳﺎﻥ ﻗـﺮﺍﺭ ﻣـﻲ
ﮔﻴﺮﺩ ﻭ ﺑﺼﻮﺭﺕ ﺍﻟﻜﺘﺮﻳﻜﻲ ﺗﺎﻳﻚ ﺩﻣﺎﻳﻲ ﻛﻤﻲ ﺑﺎﻻ ﺗﺮ ﺍﺯ ﺳﻴﺎﻝ ﮔﺮﻡ ﻣﻴﺸﻮﺩ
F34
ﺳﻴﺎﻝ ﺍﺯ ﺭﻭﻱ ﺁﻥ ﻋﺒﻮﺭ ﻣﻲ ﻛﻨﺪ ﻭ ﺁﻥ ﺭﺍ ﺳﺮﺩ ﻣﻲ ﻛﻨﺪ ﻛﻪ ﻣﻘﺪﺍﺭ ﺳﺮﺩ ﺷﺪﻥ ﻭﺍﺑﺴﺘﻪ ﺑﻪ ﺳﺮﻋﺖ ﻟﺤﻈـﻪ ﺍﻱ ﺳـﻴﺎﻝ
ﺩﺍﺭﺩ .ﺧﺮﻭﺟﻲ ﺍﺯ ﺩﺳﺘﮕﺎﻩ hot-wireﺭﻭﻱ ﻳﻚ oscilloscopeﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ ﻣﺎﻧﻨﺪ ﺷﻜﻞ ﺯﻳﺮ:
�
𝑈ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻮﺳﻴﻠﻪ ﻳﻚ ﻟﻮﻟﻪ ﭘﻴﺘﻮﺕ ۲ﺍﻧﺪﺍﺯﻩ ﮔﺮﻓﺖ ﺩﺭ ﻧﺘﻴﺠﻪ :
F35
�𝑢 𝑈 ′ = 𝑢 −
ﺗﻌﺮﻳﻒ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺯﻣﺎﻧﯽ : ۳
F36
𝑡𝑑)𝑧 � (x,y,z)=( 1 )∫𝑇 𝑢(𝑥, 𝑦,
𝑈
𝑇 0
Tﻳﻚ ﺯﻣﺎﻥ ﺑﺰﺭگ ﻣﻲ ﺑﺎﺷﺪ ﻧﺴﺒﺖ ﺑﻪ ﻃﻮﻻ ﻧﻲ ﺗﺮﻳﻦ ﭘﺮﻳﻮﺩ ﻧﻮﺳﺎﻧﺎﺕ .ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺁﺏ ﻭﮔﺎﺯ ﺗﻮﺭ
ﺑﺎﻟﻨﺖT~5secﺑﻪ ﺍﻧﺪﺍﺯﻩ ﻛﺎﻓﻲ ﺧﻮﺏ ﻣﻲ ﺑﺎﺷﺪ.
ﭘﺮﻳﻮﺩ :ﺑﻪ ﻣﺪﺕ ﺯﻣﺎﻧﻲ ﻛﻪ ﻃﻮﻝ ﻣﻲ ﻛﺸﺪ ﺗﺎ ﻧﻮﺳﺎﻧﮕﺮ ﻳﻚ ﻧﻮﺳﺎﻥ ﺍﻧﺠﺎﻡ ﺩﻫﺪ.
ﻓﺮﻛﺎﻧﺲ :ﺗﻌﺪﺍﺩ ﻧﻮﺳﺎﻥ ﻫﺎﻳﻲ ﻛﻪ ﻳﻚ ﻧﻮﺳﺎﻧﮕﺮ ﺩﺭ ﻣﺪﺕ ﻳﻚ ﺛﺎﻧﻴﻪ ﺍﻧﺠﺎﻡ ﻣﻲ ﺩﻫﺪ ﻓﺮﻛﺎﻧﺲ ﮔﻮﻳﻨﺪ.
1
𝑇
𝑇 1
→ � (𝑢 − ū)𝑑𝑡 = 𝑢� − 𝑢� = 0
𝑇 0
=𝑓
=�
𝑈
Hot Wire Anemometer
Pitot Tube
۳
Time – Mean Velocity
۱
۲
۱٥۸
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝑢�′ = 0
1 𝑇 2
2
����
𝑢′ = � 𝑢′ 𝑑𝑡 ≠ 0
𝑇 0
ﻣﻘﺪﺍﺭ�
𝑈ﺭﺍ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻥ ﺍﺯ ﻃﺮﻳﻖ ﻣﺘﻮﺳﻂ ﻛﻞ ۱ﺑﺪﺳﺖ ﺁﻭﺭﺩ.
F37
ﺷﺪﺕ ﺗﻼﻃﻢ : 2ﻣﻌﻴﺎﺭﻱ ﺍﺯ ﺍﻧﺪﺍﺯﻩ ﻱ ﻧﻮﺳﺎﻧﺎﺕ ﺗﻼﻃﻤﻲ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ.
𝑛
)𝑡( 𝑖𝑢 � =< 𝑢 >= ∑𝑖=1
𝑈
𝑛
F38
> �𝑢 <����۲ /
=�𝑢′ﺷﺪﺕ ﺗﻼﻃﻢ
ﺷﺪﺕ ﺗﻮﺭ ﺑﺎﻟﻨﺖ 1ﺗﺎ 10ﺩﺭ ﺻﺪ ﺩﺭ ﻗﺴﻤﺖ ﺍﺻﻠﻲ ﺟﺮﻳﺎﻥ ﺗﻮﺭ ﺑﺎﻟﻨﺖ ﻭ 25ﺩﺭﺻﺪ ﻳﺎ ﺑﻴﺸﺘﺮﺩﺭ ﻧﺰﺩﻳﻚ ﺩﻳﻮﺍﺭﻩ ﺍﺳﺖ.
ﭘﺲ ﺍﺯ ﺗﻌﺮﻳﻒ ﻛﻤﻴﺖ ﻫﺎﻱ ﻫﻤﻮﺍﺭﻩ ﺷﺪﻩ ﻱ ﺯﻣﺎﻧﻲ 3ﻭ ﺷﺮﺡ ﺑﻌﻀﻲ ﺍﺯ ﺧﻮﺍﺹ ﻛﻤﻴﺖ ﻫﺎﻱ ﻧﻮﺳﺎﻧﻲ ،ﺍﻛﻨﻮﻥ ﻣﻲ ﺗـﻮﺍﻧﻴﻢ
F39
ﺑﻪ ﺳﺮﺍﻍ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﻫﻤﻮﺍﺭﻩ ﺷﺪﻩ ﻱ ﺯﻣﺎﻧﻲ ﺗﻐﻴﻴﺮ ﺑﺮﻭﻳﻢ.ﺑﺎ ﻫﺪﻑ ﻫﺮ ﭼﻪ ﺳﺎﺩﻩ ﺗـﺮ ﻛـﺮﺩﻥ ﻣﺮﺍﺣـﻞ ﭘـﻲ ﺭﻳـﺰﻱ ﺍﻳـﻦ
ﻣﻌﺎﺩﻻﺕ ،ﺩﺭ ﺍﻳﻦ ﺟﺎ ﻓﻘﻂ ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺳﻴﺎﻟﻲ ﺑﺎ ﭼﮕﺎﻟﻲ ﻭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺛﺎﺑﺖ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﻲ ﮔﻴﺮﻳﻢ .ﻛﺎﺭ ﺭﺍ ﺑﺎ ﻧﻮﺷـﺘﻦ
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﭘﻴﻮﺳﺘﮕﻲ ﻭ ﺣﺮﻛﺖ ﻭ ﺟﺎﻳﮕﺰﻳﻦ ﻛﺮﺩﻥ ﻫﺎ ﺁﻏﺎﺯ ﻣﻲ ﻛﻨﻴﻢ :
Reyuolds time averaging:
= if a is a constant value
�𝑢 = �𝑢
𝑎 = 𝑎� )۱
2) �𝑎 = 𝑎�,
�������
𝑎 )3
�𝑎 + 𝑏=𝑏�+
�������
̅𝑣𝑢 + 𝑣 =𝑢� +
� ′𝛹 = 0
������)4
𝛷 = 𝛹𝛷′
Ψ=mean qualitity
����=𝑢′
� . 𝑢�=0
u′ū
𝛹�
𝛷� +
𝛹́𝛷� 𝛹́ +
́𝛹́𝛷� +
����� )5
𝛷== 𝛹𝛷
Ensemble Average
Intensity Of Turbulence
۳
Time – Mean Velocity
۱
۲
۱٥۹
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻓﺼﻞ ٦
�����
�����
�����
�����
�����
𝛹�
𝛷� +
𝛷� 𝛹́ +
𝛹́
𝛷� +
𝛷= ́𝛹 ́
𝛹�
𝛷� +
́𝛹 ́
𝛷
����́𝑢����+𝑢�𝑣̅ +
𝑣𝑢
����́𝑢𝑣́ ,
𝑣́ ≠ 0
���2 =𝑢�2 +𝑢́���2
𝑢
́�
��� ́
)𝛷� +
��� �
𝛷𝜕 �
�������� ���
𝛷𝜕
𝛷(𝜕
𝛷𝜕
𝛷𝜕 ́𝛷𝜕
=
= + = +
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
�
𝛷𝜕
𝑥𝜕
=
𝑥𝜕
���
𝛷𝜕
𝑥𝜕
)6
→
�𝑢𝜕 �
𝑢𝜕
=
𝑥𝜕 𝑥𝜕
���������
���������
)𝑡𝜕𝛷(𝑥,𝑡) 𝜕𝛷(𝑥,
=
𝑡𝜕
𝑡𝜕
)7
���������
)� (x,t
𝛷=)𝑡 𝛷(𝑥,
=0
ﭘﻴﻮﺳﺘﮕﯽ : ۱
)𝑡�������� 𝜕𝑢�(𝑥,
)𝑡𝜕𝑢(𝑥,
𝑡𝜕
=
𝑡𝜕
)𝑡�������� 𝜕𝛷�(𝑥,
)𝑡𝜕𝛷(𝑥,
,
=
𝑡𝜕
𝑡𝜕
→
)𝑡�������� 𝜕𝑢�(𝑥,
)𝑡𝜕𝑢(𝑥,
=
𝑡𝜕
𝑡𝜕
F40
𝜕
𝜕
𝜕
𝑤𝜕 𝑣𝜕 𝑢𝜕
(𝑢� + 𝑢́ ) +
𝑤( (𝑣̅ + 𝑣́ ) +
+
+
→=0
� + 𝑤́) = 0
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑧𝜕 𝑦𝜕 𝑥𝜕
́𝑤𝜕 �
𝑤𝜕 ́𝑣𝜕 ̅𝑣𝜕 ́𝑢𝜕 �𝑢𝜕
+
+
+
+
+
=0
𝑧𝜕
𝑧𝜕 𝑦𝜕 𝑦𝜕 𝑥𝜕 𝑥𝜕
)(1
������������������������������������
�
�
𝑢𝜕
́𝑢𝜕
�𝑣𝜕
́𝑣𝜕
𝑤𝜕
́𝑤𝜕
=0
𝑧𝜕
+
𝑧𝜕
+
𝑦𝜕
+
𝑦𝜕
+
𝑥𝜕
+
𝑥𝜕
̅𝑣𝜕 ����
���� ����
����
́𝑣𝜕 ����
𝑤𝜕
���� �
́𝑤𝜕
𝑢𝜕
́𝑢𝜕 �
+
+
+
+
+
=0
𝑧𝜕
𝑧𝜕 𝑦𝜕 𝑦𝜕 𝑥𝜕 𝑥𝜕
�
𝑤𝜕 ̅𝑣𝜕 �𝑢𝜕
+
+
=0
𝑧𝜕 𝑦𝜕 𝑥𝜕
ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻱ) (1ﺩﺍﺭﻳﻢ :
Continuity
۱٦۰
۱
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻓﺼﻞ ٦
́𝑤𝜕 ́𝑣𝜕 ́𝑢𝜕
+
+
=0
𝑧𝜕 𝑦𝜕 𝑥𝜕
ﺣﺮﮐﺖ : ۱
F41
𝑢 = 𝜇∇2
ﺿﺮﺏ uﺩﺭ ﭘﻴﻮﺳﺘﮕﻲ :
𝑝𝜕
𝑥𝜕
)+
𝑢𝜕
𝑧𝜕
𝑤+
𝑢𝜕
𝑦𝜕
𝑣+
𝑢𝜕
𝑥𝜕
𝑢+
𝑢𝜕
𝑡𝜕
(𝜌
𝑣𝜕
𝑤𝜕
𝑢𝜕
𝑢+
𝑢+
=0
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑢
𝑢𝜕
𝑢𝜕
𝑣𝜕
𝑢𝜕
𝑝𝜕 𝜕𝑤 1
𝑢𝜕
𝑢+ 2
𝑣+
𝑢+
𝑤+
𝑢+
+
𝑢 = 𝑣∇2
𝑥𝜕
𝑦𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕 𝜌 𝑧𝜕
𝑡𝜕
𝑝𝜕 𝜕𝑢 𝜕𝑢2 𝜕(𝑢𝑣) 𝜕(𝑢𝑤) 1
+
+
+
+
𝑢 = 𝑣∇2
→
𝑦𝜕
𝑧𝜕
𝑥𝜕 𝜌
𝑡𝜕
𝑥𝜕
ﺣﺎﻝ ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﻣﺘﻮﺳﻂ ﺯﻣﺎﻧﯽ ﻣﯽ ﮔﻴﺮﻳﻢ :
𝑣𝑢(𝜕 ���2
𝑢𝜕 �𝑢𝜕
𝑤𝑢(𝜕 )����
̅𝑝𝜕 ����) 1
+
+
+
+
�𝑢 = 𝑣∇2
𝑡𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕 𝜌
𝑤�𝑢(𝜕 ́𝑣
������(𝜕 )�
̅𝑝𝜕 𝑢́𝑤́ ) 1
����́𝑢(𝜕 ) ̅𝑣�𝑢(𝜕 𝜕𝑢� 𝜕𝑢�2 𝜕𝑢́���2
+
+
+
+
+
+
+
�𝑢 ≠ v∇2
→
𝑡𝜕
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕 𝜌
𝑧𝜕
ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻱ ﺑﺎﻻ ﺭﺍ ﺍﺯ ﻫﻢ ﻛﻢ ﻣﻲ ﻛﻨﻴﻢ:
𝑤𝜕 ̅𝑣𝜕 �𝑢𝜕
�
�𝑢𝜕
̅𝑣𝜕
𝑤𝜕
�
+
+
�𝑢 → = 0
�𝑢 +
�𝑢 +
=0
𝑧𝜕 𝑦𝜕 𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝜕𝑢� 𝜕𝑢�2 𝜕𝑢́ 2
̅𝑣𝜕
����́𝑢(𝜕 �𝑢𝜕
) ́𝑣
𝑤𝜕
�
̅𝑝𝜕 𝜕𝑢� 𝜕(𝑢́ 𝑤́ ) 1
�𝑢𝜕
→
+
+
�𝑢 +
̅𝑣 +
+
�𝑢 +
𝑤+
�
+
+
�𝑢 −
𝑡𝜕
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑦𝜕
𝑦𝜕
𝑧𝜕
𝑧𝜕
𝑧𝜕
𝑥𝜕 𝜌
𝑥𝜕
̅𝑣𝜕
𝑤𝜕
�
�𝑢 −
�𝑢 −
�𝑢 = 𝑣∇2
𝑦𝜕
𝑧𝜕
ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ :
�𝑢𝜕
�𝑢𝜕
̅𝑣𝜕
̅𝑝𝜕 𝜕𝑢� 1
) ́𝑤 ́𝑢(𝜕 ) ́𝑣 ́𝑢(𝜕 𝜕𝑢́���2
2
→
�𝑢 +
̅𝑣 +
𝑤+
�
+
= 𝑣∇ 𝑢� −
−
−
𝑡𝜕
𝑥𝜕
𝑦𝜕
𝑥𝜕 𝜌 𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
Momentum
۱٦۱
۱
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
̅𝑣𝜕
̅𝑣𝜕
̅𝑣𝜕
̅𝑝𝜕 𝜕𝑣̅ 1
����́𝑢𝜕
����́𝑣𝜕 𝑣́ 𝜕𝑣́���2
́𝑤
�𝑢 +
̅𝑣 +
𝑤+
�
+
= 𝑣∇2 𝑣̅ −
−
−
𝑥𝜕
𝑦𝜕
𝑦𝜕 𝜌 𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑡𝜕
����
𝑤𝜕
�
𝑤𝜕
�
𝑤𝜕
̅𝑝𝜕 � 1
����́𝑢𝜕
����́𝑣𝜕 ́𝑤
𝑤𝜕 ́𝑤
́2
𝑤𝜕
�
�𝑢 +
̅𝑣 +
𝑤+
�
+
𝑤 = 𝑣∇2
�−
−
−
𝑥𝜕
𝑦𝜕
𝑧𝜕 𝜌 𝑧𝜕
𝑡𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
ﺍﮔﺮ 𝑖𝑢 = 𝑖�𝑢 ﻧﺸﺎﻥ ﺩﻫﻴﻢ ﺳﻪ ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﺭﺍﺑﻔﺮﻡ ﺧﻼﺻﻪ ﺫﻳﻞ ﻣﻲ ﻧﻮﻳﺴﻴﻢ.
������́𝑢𝜕
𝚥 ́𝑢 𝚤
𝑗𝑥𝜕
𝜌−
𝑧 𝑗 = 𝑥, 𝑦,
𝑖𝑢 𝜕2
𝑗𝑥𝜕 𝑗
𝑝𝜕
𝑥𝜕 𝜇 � = − 𝜕𝑥 +
𝑠𝑒𝑠𝑠𝑒𝑟𝑡𝑠 𝑠𝑑𝑙𝑜𝑛𝑦𝑒𝑅
𝑖
𝑖𝑢𝜕
𝑗𝑥𝜕
𝑗𝑢 +
𝑖𝑢𝜕
𝑡𝜕
�𝜌
𝑖𝑢𝜕
=0
𝑖𝑥𝜕
𝑧 𝑖 = 𝑥, 𝑦,
𝑇
�����́𝑢𝜌𝜁𝑖𝑗 𝑇 = −
𝑖𝑗𝜁 = 𝚥 ́𝑢 𝚤
ﺍﻳﻦ ﻓﻼﻛﺲ ﻫﺎﻱ ﺍﺿﺎﻓﻲ ﻛﻪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻣﻤﻨﺘﻮﻡ ﻇﺎﻫﺮ ﺷـﺪﻩ ﺍﻧـﺪ ﻳـﻚ ﻧـﻮﻉ ﺍﻧﺘﻘـﺎﻝ ﻣﻤﻨﺘـﻮﻡ ﺑﺨـﺎﻃﺮ ﻧﻮﺳـﺎﻧﺎﺕ
ﺗﻮﺭﺑﺎﻟﻨﺘﻲ ﺍﺳﺖ.
𝑗𝑢𝜕
𝜕
𝑢𝜕
�����
) ́𝚥𝑢 𝚤́
𝑢𝜌 � = 𝜕𝑥 (−𝑝𝛿𝑖𝑗 + 𝜇 �𝜕𝑥 𝑖 + 𝜕𝑥 � −
𝑖
𝑗
𝑗
𝑗𝑢𝜕 𝑖𝑣𝜕
+
𝑟𝑜𝑠𝑛𝑒𝑡 𝑡𝑠𝑠𝑒𝑟𝑡𝑠 𝑠𝑢𝑜𝑐𝑠𝑖𝑣𝑛𝑎𝑒𝑚 ≡ �
𝑖𝑥𝜕 𝑗𝑥𝜕
ﻭ ﺩﺍﺭﻳﻢ symmentric :ﻳﻚ ﺗﺎﻧﺴﻮﺭ turbulent stress
𝑖𝑢𝜕
𝑗𝑥𝜕
𝑗𝑢 +
𝑖𝑢𝜕
𝑡𝜕
�𝜌 →
� 𝜇 𝜁𝑖𝑗 = −𝜌𝛿𝑖𝑗 +
�����́ρu
ı ú ȷ = turbulent stress tensor
����́𝑢𝜌𝜁𝑥𝑦 𝑇 = −
́𝑣
𝑟𝑖𝑑_𝑥 𝑒= 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑢𝑥 𝑜𝑓 𝑦𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖 𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑡ℎ
����́𝑢𝜌 𝑣́ −
́𝑤
����́𝑢𝜌 −𝜌𝑢́���2 −
𝑇
����́𝑢𝜌𝜁 = � −
����́𝑣𝜌 𝑣́ − 𝜌𝑣́���2 −
́𝑤
����
����́𝑢𝜌−
����́𝑣𝜌 𝑤́ −
𝑤𝜌 𝑤́ −
́2
ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﺩﺭ ﺻﻮﺭﺗﻴﻜﻪ ﺑﺨﻮﺍﻫﻴﻢ ﺣﻞ ﺗﺤﻠﻴﻠﻲ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﺍﻣﻜـﺎﻥ ﭘـﺬﻳﺮ ﻧﻴﺴـﺖ .ﺯﻳـﺮﺍ ﭼﻬـﺎﺭ ﻣﻌﺎﺩﻟـﻪ ﺩﺍﺭﻳـﻢ
)ﭘﻴﻮﺳﺘﮕﻲ ﻭ ﺳﻪ ﻣﻮﻟﻔﻪ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻛﺲ(ﻭ ﺩﻩ ﻣﺠﻬﻮﻝ )ﺳﻪ ﻣﻮﻟﻔﻪ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ,ﻓﺸﺎﺭ ﻣﺘﻮﺳﻂ ﻭﺷـﺶ ﻣﻮﻟﻔـﻪ
ﻣﺨﺘﻠﻒ ﺗﺎﻧﻮﺭﺗﺶ ﺭﻳﻨﻮﻟﺪﺯ(ﺑﻨﺎﺑﺮﺍﻳﻦ ﻣﺠﻬﻮﻻﺕ ﺍﺯ ﺗﻌﺪﺍﺩ ﻣﻌﺎﺩﻻﺕ ﺑﻴﺸﺘﺮ ﻫﺴﺘﻨﺪ.
ﻃﺮﻳﻘﻪ ﻱ ﺑﺮ ﺧﻮﺭﺩ ﺑﺎ ﻣﺴﺎﻳﻞ ﻣﺘﻼﻃﻢ :
۱٦۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻳﻜﻲ ﺍﺯ ﺭﻭﺷﻬﺎﻳﻲ ﻛﻪ ﺟﻬﺖ ﺑﺪﺳﺖ ﺍﻭﺭﺩﻥ ﺣﻞ ﺗﺤﻠﻴﻠﻲ ﻣﺴﺎﻳﻞ ﻣﺘﻼﻃﻢ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺪﻟﻬﺎﻱ ﺗﻴﻮﺭﻳـﻚ
ﭘﺪﻳﺪﻩ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﺍﺳﺖ.
ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻱ ﮔﺮﺩﺍﺑﻲ ﺑﻮﺳﻴﻨﺴﻚ : 1
F42
�𝑢𝜕
�𝑢𝜕
�𝑢𝜕
����� →
𝜇 = 𝑦𝑥𝜁
𝑇𝜇 +
𝑦𝜕
𝑦𝜕
𝑦𝜕
�𝑢𝜕
𝑦𝜕
����́𝑢𝜌−
𝑇 𝜇 = ́𝑣
���� →
) 𝑇 𝜇 𝜁𝑥𝑦 = (𝜇 +
ﺑﺎ ﺍﻳﻦ ﻛﺎﺭ ́−ρ𝑢́ vﺗﺒﺪﻳﻞ ﺑﻪ ﻣﺠﻬﻮﻝ ﺩﻳﮕﺮ ﻱ 𝑇 𝜇 ﻣﻲ ﮔﺮﺩﺩ ﻛﻪ ﺧﺎﺻﻴﺖ ﺟﺮﻳﺎﻥ ﺍﺳﺖ ﻧﻪ ﺧﺎﺻﻴﺖ ﺳـﻴﺎﻝ ﻭﻳـﻚ
ﺧﺎﺻﻴﺖ ﺗﺮﻣﻮ ﺩﻳﻨﺎﻣﻴﻜﻲ ﻣﺎﻧﻨﺪ μﻧﻴﺴﺖ.ﺩﺭ ﻣﻮﺭﺩ ﺟﺮﻳﺎﻧﻬﺎﻱ ﻣﺨﺘﻠﻒ ﻣﻘﺎﺩﻳﺮ ﺗﻔﺎﻭﺕ ﺑﺮﺍﻱ 𝑇 µﭘﻴﺸﻨﻬﺎﺩ ﺷـﺪﻩ ﻳـﺎ
ﻳﻚ ﻣﻘﺪﺍﺭ ﻋﻤﻮﻣﻲ ﺑﺮﺍﻱ 𝑇 µﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ.
ﺗﺌﻮﺭﻱ ﻃﻮﻝ ﺍﺧﺘﻼﻁ ﭘﺮﺍﻧﺘﻞ :2
F43
���������ﻳﻚ ﺭﻭﺵ ﺩﻗﻴﻖ ﺗﺮ ﺗﻮﺳﻂ ﭘﺮﺍﻧﺘﻞ ﺍﺭﺍﻳﻪ ﺷﺪﻩ ﺍﺳﺖ.
ﺟﻬﺖ ﻣﺤﺎﺳﺒﻪ ﺗﺮﻡ́ −ρu ́v
ﺍﮔﺮ ﻳﻚ ﺗﻜﻪ ﺳﻴﺎﻝ ﺍﺯ 𝑦0 + lmﺑﻪ 𝑦0ﺑﻴﺎﻳﺪ ﻳﻚ ﻧﻮﺳﺎﻥ ﻣﺜﺒﺖ ﺭﻭﻱ ) u� (𝑦0ﺍﻋﻤﺎﻝ ﻣـﻲ ﻛﻨـﺪ ﻭ ﺩﺑـﻲ ﺟﺮﻣـﻲ
ﻟﺤﻈﻪ ﺍﻱ ﮔﺬﺭﻧﺪﻩ ﺍﺯ ﺻﻔﺤﻪ ﺩﺭ 𝑦 = 𝑦0ﺑﺮﺍﺑﺮ ﺑﺎ ́𝑣−ρﺍﺳـﺖ ﻭ ﺳـﺮﻋﺖ ﻣﺤـﻮﺭﻱ ﺁﻥ ́𝑢 u� (𝑦0 ) +ﻣـﻲ ﺑﺎﺷـﺪ
ﺑﻄﻮﺭﻳﻜﻪ ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﺻﻔﺤﻪ 𝑦 = 𝑦0ﻣﻘﺪﺍﺭ ﻫﻤﻨﺘﻮﻡ ́ −ρ𝑢́ vﻣﻨﺘﻘﻞ ﻣـﻲ ﺷـﻮﺩ ﺍﮔـﺮ ﺍﻳـﻦ ﺗﻜـﻪ ﺳـﻴﺎﻝ ﺍﺯ
ﺻﻔﺤﻪ ﺯﻳﺮﻱ ﺑﺎﻻ ﺑﻴﺎﻳﺪ ﻫﻤﻴﻦ ﻣﻘﺪﺍﺭ ﻫﻤﻨﺘﻮﻡ ﻣﻨﺘﻘﻞ ﻣﻲ ﺷﻮﺩ ﺑﻄﻮﺭﻳﻜﻪ ﻣﻘﺪﺍﺭ ﺧﺎﻟﺺ ﺍﻧﺘﻘﺎﻝ ﻫﻤﻨﺘـﻮﻡ ﻣﺤـﻮﺭﻱ
ﺩﺭ ﺟﻬﺖ ﻋﻤﻮﺩ ﺑﺮ ﺣﺮﻛﺖ ﺑﻮﺳﻴﻠﻪ ﻧﻮﺳﺎﻧﺎﺕ ﺳﺮﻋﺖ ﺑﺮﺍﺑﺮ ﺑﺎ �́���
𝑣𝑢−ρﺍﺳﺖ.
)Eddy Vicosity ( Boussinesζ۱۸۷۷
Mixing Length
۱٦۳
۱
۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
lmﺷﺒﻴﻪ ﺑﻪ )λﻣﺴﺎﻓﺖ ﺁﺯﺍﺩ ﻣﻴﺎﻧﮕﻴﻦ (1ﺑﺮﺍﻱ ﻣﻮﻟﻜﻮﻝ ﻫﺎﺳﺖ ﻛﻪ ﺩﺭ ﺍﻳﻨﺠﺎ lmﻓﺎﺻﻠﻪ ﺑﻴﻦ ﺩﻭ ﺗﻜﻪ ﺳﻴﺎﻝ 2ﺍﺳـﺖ
.ﻳﻚ ﺗﻜﻪ ﺳﻴﺎﻝ ﻛﻪ ﺍﺯ ﻻﻳﻪ ﺑﺎﻻﻳﻲ ﺑﻪ ﻻﻳﻪ ﭘﺎﻳﻴﻨﻲ ﻭﺍﺭﺩ ﻣﻲ ﺷﻮﺩ ﻳﻚ ﻧﻮﺳـﺎﻧﻲ ﺩﺭ ﻣﻘـﺪﺍﺭ ﺳـﺮﻋﺖ ﺩﺭ 𝑦0ﺍﻳﺠـﺎﺩ
ﻣﻴﻜﻨﺪ.
F45
F4
) 𝑢́ = 𝑢�(𝑦0 + 𝑙𝑚 ) − 𝑢�(𝑦0
�𝑢𝜕
|) − 𝑢�(𝑦0
𝑦𝜕
�
𝑢𝜕
�𝑢1́ | ≈ 𝑙𝑚 � | with 𝑣̅ (𝑦0 ) < 0
𝑦𝜕
ﺍﮔﺮ ﺗﻜﻪ ﺳﻴﺎﻝ ﺍﺯ ﻻﻳﻪ ﺯﻳﺮﻱ ﺑﻪ ﻻﻳﻪ 𝑦 = 𝑦0ﺑﻴﺎﺑﻴﺪ،
�𝑢𝜕
�𝑢𝜕
)� = +𝑙𝑚 � � 𝑤𝑖𝑡ℎ 𝑣̅ > 0
𝑦𝜕
𝑦𝜕
𝑚𝑙 𝑢́ ≈ |𝑢�(𝑦0 +
𝑚𝑙 |𝑢́ | = |𝑢�(𝑦0 ) − 𝑢�(𝑦0 − 𝑙𝑚 )| = �𝑢�(𝑦0 ) − (𝑢�(𝑦0 ) −
�𝑢𝑑
�𝑢𝑑
�𝑢𝑑
1
� � 𝑚𝑙 = �� � 𝑚𝑙 �𝑢́� � = �𝑙𝑛 � � +
𝑦𝑑
𝑦𝑑
𝑦𝑑
2
Up word
down word
ﭘﺮﺍﻧﺘﻞ ﻓﺮﺽ ﻛﺮﺩ :
�𝑢𝑑
�
𝑦𝑑
� 𝑚𝑙 �𝑣́ ̅ � = 𝑐1 �𝑢́� � = 𝑐1
𝑇
𝑦𝑥𝜁
����́𝑢= −ρ
𝑦 𝑓𝑜 𝑡𝑛𝑢𝑙𝑓 𝑡𝑛𝑒𝑙𝑢𝑏𝑟𝑢𝑡 = ́𝑣
𝑟𝑖𝑑𝑒 𝑒− 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑐𝑟𝑜𝑠𝑠 𝑎 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑎𝑟 𝑡𝑜 𝑡ℎ
ﺍﺯ ﺭﻭﻱ ﺷﻜﻞ ﺩﻳﺪﻩ ﻣﻲ ﺷﻮﺩﻛﻪ ﻭﻗﺘﻲ ﻛﻪ ́𝑣ﻣﺜﺒﺖ ﺍﺳﺖ ﻳﻚ ﺳﻴﺎﻝ ﺑﻪ ﻧﺎﺣﻴﻪ ﺍﻱ ﻣﻲ ﺭﻭﺩ ﻛﻪ ﻣﻘﺪﺍﺭ �𝑢 ﺑﺎﻻﺗﺮﻱ
����ﻣﻘﺪﺍﺭﻱ ﻣﻨﻔﻲ ﻭ
ﺩﺍﺭﺩ ﻟﺬﺍ ﺩﺍﺭﻳﻢ ́𝑣 ́𝑢−ﻭ ﻭﻗﺘﻲ ﻛﻪ ́𝑣ﻣﻨﻔﻲ ﺍﺳﺖ ́𝑢ﻣﺜﺒﺖ ﺍﺳﺖ ﻭ ﺩﺍﺭﻳﻢ ́𝑣 ́𝑢−ﻭ ﻣﺘﻮﺳﻂ ﺍﻥ ́𝑣 ́𝑢
ﻏﻴﺮ ﺻﻔﺮ ﺍﺳﺖ.
0 < 𝑐2 ≤ 1
ﭼﻮﻥ ﻣﻘﺪﺍﺭ 𝑚𝑙 ﺭﺍ ﻧﻤﻲ ﺩﺍﻧﻴﻢ −𝑐3ﺭﺍ ﺟﺰء 𝑚𝑙ﮔﺮﻓﺘﻪ ﺍﻳﻢ .
����
� ̅ ́𝑣�� �́𝑢� 𝑢́ 𝑣́ = −𝑐2
𝑑𝑢� 2
�𝑢𝑑
𝑇
2
𝑚𝑙𝜌 =
( )2
𝑦𝑥𝜁 → �
𝑦𝑑
𝑦𝑑
2
����́𝑢
𝑚𝑙 𝑣́ = −𝑐3
�
Mean Free Path
Clump
۱٦٤
۱
۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻭ ﭼﻮﻥ ﺗﻨﺶ ﺑﺎﻳﺪ ﺑﺎ ﻋﻼﻣﺖ
�
𝑢𝑑
𝑦𝑑
ﺗﻐﻴﻴﺮ ﻋﻼﻣﺖ ﺩﻫﺪ ﺑﻬﺘﺮ ﺍﺳﺖ ﺑﻨﻮﻳﺴﻴﻢ :
𝑠𝑠𝑜𝑝𝑟𝑜𝑑𝑡𝑡𝑠 𝑚𝑖𝑥𝑖𝑛𝑔 − 𝑙𝑒𝑛𝑔𝑡ℎ ℎ𝑦𝑝𝑜𝑡ℎ
�
𝑢𝑑
2
𝑚𝑙𝜌 = 𝑇 𝜇 →
� �
𝑦𝑑
�𝑢𝑑 �𝑢𝑑
�
𝑦𝑑 𝑦𝑑
�𝑢𝑑
𝑦𝑑
𝑇
2
𝑦𝑥𝜁
𝑚𝑙𝜌 =
�
𝑡
|2 |1
𝑦𝑥𝜁 →
𝑚𝑙𝜌 =
�
𝑢𝑑
𝑦𝑑
𝑇
𝑦𝑥𝜁
𝑇𝜇 =
����́𝑢𝜌 ﻭ ﻳـﺎ 𝑇 𝜇 ﻣﺠﻬـﻮﻝ ﺟﺪﻳـﺪ 𝑚𝑙 ﺭﺍ ﻭﺍﺭﺩ ﻣـﻲ ﻧﻤﺎﻳـﺪ
ﺍﻳﻦ ﺗﺌﻮﺭﻱ ﺍﻃﻼﻋﺎﺕ ﺟﺪﻳﺪ ﺍﺭﺍﺋﻪ ﻧﻤﻲ ﻛﻨﺪ ﺑﻠﻜﻪ ﺑﺠﺎﻱ ́𝑣
ﻛﻪ 𝑚𝑙 ﺑﺎﺯ ﻫﻢ ﺗﺎﺑﻌﻲ ﺍﺯ ﺣﺎﻟﺖ ﺟﺮﻳﺎﻥ ﺍﺳﺖ ﻧﻪ ﺳﻴﺎﻝ ﻭ 𝑚𝑙 ﺗﺎﺑﻌﻲ ﺍﺯ ﻣﻜﺎﻥ ﺍﺳﺖ.ﻃﺮﻳﻖ ﻃﻮﻝ ﺍﺧﺘﻼﻁ ﺩﺭ ﺍﻳﻦ ﺍﺳﺖ
ﻛﻪ ﺑﻄﻮﺭ ﻣﻨﻄﻘﻲ ﻣﻲ ﺗﻮﺍﻥ ﺍﻧﺘﻈﺎﺭ ﺩﺍﺷﺖ ﻛﻪ ﺩﺭ ﺑﻌﻀﻲ ﺍﺯ ﺟﺮﻳﺎﻧﻬﺎﻱ ﺩﺭ 𝑚𝑙ﺗﺎﺑﻊ yﺑﺎﺷـﺪ .ﭘﺮﺍﻧﺘـﻞ ﺭﻭﺍﺑـﻂ ﺫﻳـﻞ ﺭﺍ
ﺑﺮﺍﻱ 𝑚𝑙ﭘﻴﺸﻨﻬﺎﺩ ﻛﺮﺩ .
) ﻓﺎﺻﻠﻪ ﺍﺯ ﺟﺪﺍﺭ ∶ 𝑙𝑚 = 𝑘1 𝑦 , (𝑦 = 2ﺗﻼﻃﻢ ﺩﺭ ﺟﺪﺍﺭ
)ﻋﺮﺽ ﻣﻨﻄﻘﻪ ﺍﺧﺘﻼﻁ ∶ 𝑙𝑚 = 𝑘2 𝑏 , (𝑏 = 4ﺗﻼﻃﻢ ﺁﺯﺍﺩ
1
3
ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻫﻤﻮﺍﺭﻩ ﺷﺪﻩ ﻱ ﺯﻣﺎﻧﻲ ﺩﺭ ﻧﺰﺩﻳﻜﻲ ﺟﺪﺍﺭ : 5
F50
ﺑﺮﺍﻱ ﻳﻚ ﺻﻔﺤﻪ ﺻﺎﻑ ﺟﺮﻳﺎﻥ ﻧﺰﺩﻳﻚ ﺻﻔﺤﻪ ﺭﺍﻣﻲ ﺗﻮﺍﻥ ﭼﻨﻴﻦ ﺗﻘﺴﻴﻢ ﺑﻨﺪﻱ ﻛﺮﺩ.
Wall Turbulence
Distance From Wall
۳
Free Turbulence
٤
With Of Mixing Zone
٥
The Time – Smoothed Velocity Profile Near A Wall
۱
۲
۱٦٥
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
(1ﺯﻳﺮ ﻻﻳﻪ ﻭﻳﺴﻜﻮﺯ : 1ﺩﺭ ﻓﺎﺻﻠﻪ ﺑﺴﻴﺎﺭ ﻛﻢ ﺍﺯ ﺟﺪﺍﺭ ،ﻛﻪ ﺩﺭ ﺁﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻧﻘﺶ ﻣﻬﻤﻲ ﺩﺍﺭﺩ؛
(۲ﻻﻳﻪ ﻱ ﻣﻴﺎﻧﮕﻴﺮ : ۲ﻛﻪ ﺩﺭ ﺁﻥ ﮔﺬﺍﺭ ﺑﻴﻦ ﺯﻳﺮ ﻻﻳﻪ ﻫﺎﻱ ﻭﻳﺴﻜﻮﺯ ﻭ ﻟﺨﺖ ﺭﺥ ﻣﻲ ﺩﻫﺪ؛) ﻣﻨﻄﻘﻪ ﺍﻧﺘﻘﺎﻟﻲ
ﺑﻴﻦ ﻻﻳﻪ 1ﻭ 3ﺍﺳﺖ(.
(۳ﺯﻳﺮ ﻻﻳﻪ ﻟﺨﺖ : 3ﺩﺭ ﺁﻏﺎﺯ ﺟﺮﻳﺎﻥ ﺗﻼﻃﻤﻲ ﺍﺻﻠﻲ ،ﻛﻪ ﺩﺭ ﺁﻥ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻧﻘﺶ ﻓﺮﻋﻲ ﺩﺍﺭﺩ؛
(4ﺟﺮﻳﺎﻥ ﺗﻼﻃﻤﻲ ﺍﺻﻠﻲ : 4ﻛﻪ ﺩﺭ ﺁﻥ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻫﻤﻮﺍﺭﻩ ﺷﺪﻩ ﻱ ﺯﻣﺎﻧﻲ ﺗﻘﺮﻳﺒﺎً ﺗﺨﺖ ﺍﺳﺖ ﻭ
ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺍﻫﻤﻴﺘﻲ ﻧﺪﺍﺭﺩ.
F51
F52
F53
F54
ﺩﺭ ﺯﻳﺮ ﻻﻳﻪ ﻟﺨﺖ ،ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﺑﻮﺳﻴﻠﻪ ﻳﻚ y) lengthscaleﻓﺎﺻﻠﻪ ﺍﺯ ﺩﻳﻮﺍﺭ ( ﻭ ﻳﻚ velocity scale
ﻣﺪﻝ ﻣﻲ ﺷﻮﺩ.
𝑤𝜁
𝑦=0
ﺩﺭ ﺍﻳﻦ ﻧﺎﺣﻴﻪ :
ﻭ ﺗﻨﺶ ﺑﺮﺷﻲ ﺗﻘﺮﻳﺒﺎ 𝑤𝜁ﺍﺳﺖ.
𝜌
�=U=shear velocity
𝑙𝑙𝑎𝑤 𝑒𝜁𝑤 = 𝑡𝑖𝑚𝑒 𝑠𝑚𝑜𝑜𝑡ℎ𝑒𝑑 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ
𝑡𝑤𝑎𝑙𝑦𝑟𝑜𝑐 𝑛𝑎𝑚𝑟𝑎𝑘 𝑛𝑜𝑣 = 𝑘 = 0.4
𝑙𝑚 = 𝑘𝑦 ,
Vicoussub Layer
Buffer Layer
۳
Inerfial Sublayer
٤
Main Turbulent Stream
۱
۲
۱٦٦
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝑢𝜕 �
�
𝑢𝜕
𝑇
2
𝑦𝑥𝜁
𝑚𝑙= ρ
� �
𝑦𝜕 𝑦𝜕
∗𝑢 �𝑢𝑑 𝜕𝑢� 2
→ )
=
𝑦𝜕
𝑦𝑘 𝑦𝑑
( → 𝜁𝑤 = 𝜌𝑘 2 𝑦 2
𝑦𝑑 ∗𝑢
∗𝑢
𝑐 → 𝑢� = 𝑙𝑛𝑦 +
𝑦 𝑘
𝑘
1
𝑦𝑢
30 < 𝑦 + < 300
ﻣﺜﻞ ﻋﺪﺩ feﺑﺮ ﻣﺒﻨﺎﻱ y𝑏~5.5
𝑣
𝑐 = 𝑙𝑛𝑦 +
𝑟
�
𝑢
𝑢
1
= �𝑢𝑑 →
= → 𝑢+
= → 𝑢+ = 𝑙𝑛𝑦 + + 𝑏 , 𝑦 +
𝑟
ﻭﺧﻴﻠﻲ ﻧﺰﺩﻳﻚ ﺑﻪ ﺩﻳﻮﺍﺭ ﺩﺭ ﺯﻳﺮ ﻻﻳﻪ ﻭﻳﺴﻜﻮﺯ ,ﻧﻮﺳﺎﻧﺎﺕ ﺗﻮﺭﺑﺎﻟﻨﺖ ﻛﻮﭼﻚ ﻫﺴﺘﻨﺪ ﻭ ﻣﻲ ﺗﻮﺍﻥ ﻧﻮﺷﺖ
�𝑢𝑑 𝜇 𝑤𝜁 �𝑢𝑑
�𝑢𝑑
→
=
𝑣 = → 𝑢2
𝜌
𝑦𝑑
𝑦𝑑 𝜌
𝑦𝑑
𝑓𝑜𝑟 0 < 𝑦 + ≤ 5
→ 𝑢+ = 𝑦 +
�
𝑢
𝜁𝑤 = µ
=1
𝑑𝑢+
𝑑𝑦 +
→
) 𝑑( +
𝑑𝑢+
�𝑢𝑑
= 1 → 𝑢𝑢+𝑦 = 1 → + = 1
𝑢
𝑦𝑑
𝑦𝑑
) (𝑑
𝑦
𝑟𝑒𝑦𝑎𝑙𝑏𝑢𝑠
𝑟𝑎𝑛𝑖𝑛𝑎𝑙
۱٦۷
0 < 𝑦+ < 5
�𝑢
= 𝑦+
𝑢+
)1
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝑡𝑟𝑎𝑛𝑠𝑖𝑙𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛 5 < 𝑦 + < 30
𝑟𝑒𝑦𝑎𝑙 𝑡𝑛𝑒𝑙𝑢𝑏𝑟𝑢𝑡
ﺑﺮﺍﻱ ﻟﻮﻟﻪ:
+
𝑢
=
4.2
−
𝑦𝑛𝑙5.7
+ 5.1(𝑙𝑛𝑦 + )2 − 0.7(𝑙𝑛𝑦 + )3
𝑢+
𝑢
1
=
) ln(9.78𝑦 +
+
0.418
𝑢
30 < 𝑦 + < 1000
0 < 𝑦+ < 5
5 < 𝑦 + < 30
𝑦 + > 30
𝑤𝜁
𝜌
� = , 𝑢+
)2
1 𝑦+ 3
( 𝑢+ = 𝑦 + �1 −
�)
4 14.5
𝑢+ = 5 ln(𝑦 + + 0.205) − 3.27
𝑢+ = 2.5𝑙𝑛𝑦 + + 5.5
𝜌 𝑦𝑢+
= 𝑦,
𝜇
�𝑢
𝑢 = +
𝑢
+
+
ﻣﺜﺎﻝ :ﺩﺭ ﻳﻚ ﻟﻮﻟﻪ ﺁﺏ ﺑﻠﻨﺪ ﻭ ﺻﺎﻑ ﺁﺏ ﺑﻪ ﻃﻮﺭ ﭘﺎﻳﺪﺍﺭ ﺑﺮﺍﻱ ﺷﺮﺍﻳﻂ ﺫﻳﻞ ﺩﺭ ﺟﺮﻳﺎﻥ ﺍﺳﺖ.
R=۷.٦۲cm
𝑅
ﺩﺭ = 𝑦ﻣﻘﺪﺍﺭ
2
𝑡𝜇
𝜇
ﺭﺍﻣﺤﺎﺳﺒﻪ ﻛﻨﻴﺪ؟
𝜌 = ۱۰۰۰kg/m۳
𝑤𝜁=wall shear stress=۰.۱٦pa
V=۱.۰۲*۱۰-۷ m۲/s
𝑟𝑌 =𝑅−
Equation of motion:
𝑑 Ƥ0 − Ƥ𝐿 1
)𝑣(
)𝑡(
𝑧𝑟̅𝜏 = 𝑧𝑟̅𝜏 (𝑟𝜏̅𝑟𝑧 ) ,
−
𝑧𝑟̅𝜏 +
𝐿
𝑟𝑑 𝑟
, 𝑎𝑡 𝑟 = 0 𝜏̅𝑟𝑧 = 0
𝑟 𝑅) 𝐿(Ƥ0 − Ƥ
.
𝐿2
𝑅
= 𝑧𝑟̅𝜏 ⇒
𝑤𝜏 = 𝑧𝑟̅𝜏 ⇒ 𝑅 = 𝑟 𝑡𝑎
𝑅) 𝐿(Ƥ0 − Ƥ
𝐿2
𝑦𝑦 =𝑅−𝑟 ⇒𝑟 =𝑅−
۱٦۸
=0
,
𝑟
𝑅
𝑦𝑅−
𝑦
� � = 𝜏𝑤 �1 −
𝑅
𝑅
= 𝑤𝜏
𝑤𝜏 = 𝑧𝑟̅𝜏
� 𝑤𝜏 = 𝑧𝑟̅𝜏
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝑧̅𝑣𝑑
𝑧̅𝑣𝑑
𝑧̅𝑣𝑑
𝑡𝜇 −
) 𝑡𝜇 = (𝜇 +
𝑟𝑑
𝑟𝑑
𝑦𝑑
𝑦
𝜇𝜏̅𝑟𝑧 = −
𝜇𝑡 1
𝑧𝑟̅𝜏
�𝑅 1 𝜏𝑤 �1 −
⇒
= .
−1= .
−1
𝜇
𝑦𝑑𝜇 𝑑𝑣̅𝑧 ⁄
𝑦𝑑𝜇 𝑑𝑣̅𝑧 ⁄
𝑦
� �1 −
𝑡𝜇
= + 𝑅 +−1
𝜇
𝑦𝑑𝑑𝑣 ⁄
⇒
= , 𝑣+
𝜌 ∗𝑢𝑦
= 𝑦
𝜇
1
𝑡𝜇
⇒ = 0.0052
= 2 − 1 = 95
𝜇
0.0052
𝑑𝑣 +
�
𝑑𝑦 +
𝑧̅𝑣
,
∗𝑢
𝜌 𝑅� �𝜏 ⁄𝜌 .
𝑅
𝑤
⇒ 𝑦+ = 2
= 485
2
𝜇
𝑦 + =485
+
=𝑦
ﺍﻳﻦ ﺗﺎﻳﻴﺪ ﻣﻲ ﻛﻨﺪ ﻛﻪ ﺩﺭ ﻓﺎﺻﻠﻪ ﺩﻭﺭ ﺍﺯ ﺩﻳﻮﺍﺭ ﺍﻧﺘﻘﺎﻝ ﻣﻤﻨﺘﻮﻡ ﻣﻮﻟﻜﻮﻟﻲ ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﻛـﺮﺩﻥ ﺍﺳـﺖ ﻧﺴـﺒﺖ ﺑـﻪ
ﺍﻧﺘﻘﺎﻝ ﻣﻤﻨﺘﻮﻡ ﻣﺘﻼﻃﻢ ). (eddy transport
′ )2
�������
𝑢(√
𝑦𝑡𝑖𝑠𝑛𝑒𝑡𝑛𝑖 𝑒𝑐𝑛𝑒𝑙𝑢𝑏𝑟𝑢𝑡 =
∞𝑢
����
,
� =0
𝑢′
���� ﻭ′ 𝑣′
�����
ﺑﻴﺸﺘﺮﻳﻦ ﻣﻘﺪﺍﺭ ﻣﺮﺑﻮﻁ ﺑﻪ ����2
𝑢 −ﺩﺭ ﻧﺰﺩﻳﻜﻲ ﺩﻳﻮﺍﺭ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳـﺪ ﻛـﻪ ﺩﺭ ﺁﻧﺠـﺎ ﮔﺮﺍﺩﻳـﺎﻥ
���� ﻭ 𝑤′2
𝑢′ﻭ 𝑣′2
ﻫﺎﻱ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﺗﻮﻟﻴﺪ ﺁﺷﻔﺘﮕﻲ ﺯﻳﺎﺩ ﺍﺳﺖ.
۱٦۹
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
Energy spectrum of Turbulence :
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﻳﻚ ﺭﻧﺞ ﮔﺴﺘﺮﺩﻩ ﺍﻱ ﺍﺯ ﻣﻘﻴﺎﺱ ﻃﻮﻝ 1ﺩﺍﺭﺩ ﻫﻤﺎﻧﻨﺪ ﺷﻜﻞ ﺯﻳﺮ:
F5
1
ﻛﻪ ) E(kﻃﻴﻒ ﺍﻧﺮژﻱ ﻭ kﻋﺪﺩ ﻣﻮﺝ ﺍﺳﺖ 𝑘 = .ﻭ 𝑘 ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﻌﻜﻮﺱ ﻃﻮﻝ ﻣﻮﺝ.
𝑙
ﺍﻧﺮﺯﻱ ﻧﻮﺳﺎﻧﺎﺕ ﺩﺭ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﺑﺰﺭگ )ﺑﺎ ﻋﺪﺩ ﻣﻮﺝ ﻛﻮﭼﻚ( ﺗﻮﻟﻴﺪ ﻣﻲ ﺷﻮﻧﺪ ﻭ ﻣﻜﺎﻧﻴﺰﻡ ﻛﺸـﺶ ﮔﺮﺩﺍﺑـﻪ 2ﺑﺎﻋـﺚ
ﺗﻮﻟﻴﺪ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﻛﻮﭼﻜﺘﺮ ﻣﻲ ﺷﻮﺩ ﻭ ﺍﻧﺮژﻱ ﺑﻪ ﺳﻤﺖ ﻧﺎﺣﻴﻪ ﺑﺎ ﻋﺪﺩ ﻣﻮﺝ ﺑﺎﻻ ﻛﺎﻫﺶ ﭘﻴﺪﺍ ﻣﻲ ﻛﻨﺪ ﻭ ﺍﻧـﺮژﻱ ﺩﺭ
ﻛﻮﭼﻜﺘﺮﻳﻦ ﮔﺮﺩﺍﺑﻪ ﻫﺎ ﺗﺒﺪﻳﻞ ﺑﻪ ﮔﺮﻣﺎ ﻣﻲ ﺷﻮﺩ.
F56
ﺣﻀﻮﺭ ﮔﺮﺍﺩﻳﺎﻥ ﻫﺎﻱ ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺩﺭ ﺟﺮﻳﺎﻥ ﺑﺮﺷﻲ؛ ﺳﺒﺐ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﺧﺰﺷﻲ ﻣﻲ ﺷﻮﻧﺪ .ﺩﺭ ﺍﻣﺘـﺪﺍﺩ ﺟﺮﻳـﺎﻥ
ﮔﺮﺩﺍﺑﻪ ﺑﻄﻮﺭ ﻣﻨﺎﺳﺒﻲ ﻛﺸﻴﺪﻩ ﻣﻲ ﺷﻮﻧﺪ ﭼﻮﻥ ﺍﺯ ﻳﻚ ﻃﺮﻑ ﻓﺸﺎﺭ ﺍﻋﻤﺎﻝ ﻣﻲ ﺷﻮﺩ ﺗﺎ ﺳﺮﻳﻌﺘﺮ ﺍﺯ ﺑﻘﻴﻪ ﺣﺮﻛﺖ ﻛﻨﺪ.
Dissipation rate (ℰ)≈ fluctuation energy producte rate
U= large – scale velocity fluctualence of turbulence
𝛬=corresponding length scale
𝑢۳
=ℰ
𝑦𝑔𝑟𝑒𝑛𝑒 𝑛𝑜𝑖𝑡𝑎𝑢𝑡𝑐𝑢𝑙𝑓 𝑓𝑜)𝑛𝑜𝑖𝑡𝑎𝑝𝑖𝑠𝑠𝑖𝑑 𝑟𝑜( 𝑛𝑜𝑖𝑡𝑎𝑢𝑡𝑐𝑢𝑑𝑜𝑟𝑝 𝑓𝑜 𝑒𝑡𝑎𝑟 =
𝛬
𝛬
ﻳﻌﻨﻲ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﺑﺰﺭگ ﺩﺭ ﭘﺮﻳﻮﺩ ﺯﻣﺎﻧﻲ ﻣﻘﺪﺍﺭﻱ ﺍﺯ ﺍﻧﺮژﻱ ﺧﻮﺩ ﺭﺍ ﺍﺯ ﺩﺳﺖ ﻣﻲ ﺩﻫﻨﺪ.
𝑢
𝛬𝑢
𝑣
= 𝑣𝑒𝑅
Kolmogorov Scales :
Length Scale
Vortex Stretch
۱۷۰
۱
۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺣﺮﻛﺖ ﻣﺘﻼﻃﻢ ﺩﺭ ﻣﻘﻴﺎﺱ ﺑﺰﺭگ 1ﻣﺴﺘﻘﻞ ﺍﺯ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺍﺳﺖ ﻭﻟﻲ ﺩﺭ ﻣﻘﻴﺎﺱ ﻛﻮﭼﻚ 2ﺑﻮﺳﻴﻠﻪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ
ﻛﻨﺘﺮﻝ ﻣﻲ ﺷﻮﺩ.
F58
F57
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻧـﺎﻟﻴﺰ ﺑﻌـﺪﻱ ﻛـﻪ ﺩﺭ ﻣﻘﻴـﺎﺱ ﻛﻮﭼـﻚ ﺗﻮﺭﺑﺎﻟﻨـﺖ ﻭﺍﺑﺴـﺘﻪ ﺑـﻪ υ,ℰﺍﺳـﺖ Kolmogorovﺑـﺮﺍﻱ
ﻛﻮﭼﻜﺘﺮﻳﻦ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﭘﻴﺪﺍ ﻛﺮﺩ :
𝑣3 1
𝑒𝑙𝑎𝑐𝑠 𝜂 = ( )4 = 𝑙𝑒𝑛𝑔𝑡ℎ −
𝜀
𝑣 1
˶ 𝜏 = ( )2 = 𝑡𝑖𝑚𝑒 −
𝜀
1
˶ 𝑣 = (𝑣𝜀)4 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 −
For a dissipation rate of ۱ (w/kg) of water ⇒ η=۳۰ μm
)Turbulent kinetic energy (TKE
ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺳﻪ ﻣﻮﻟﻔﻪ ﻧﻮﺳﺎﻧﻲ ﺳﺮﻋﺖ ﻣﻲ ﺑﺎﺷﺪ ﺑﻪ ﺍﺯﺍﻱ ﻭﺍﺣﺪ ﺟﺮﻡ.
���� 1
�����
���� ′ 2 +
)′2
𝑢( = 𝐾
𝑤 𝑣 ′2 +
2
ﺩﺭ ﺟﺮﻳﺎﻥ ﺗﻮﺭﺑﺎﻟﻨﺖ ﻳﻚ energy cascaleﺩﺍﺭﻳﻢ ﻛﻪ ﺍﻧﺮژﻱ ﺍﺯﻣﻘﻴﺎﺱ ﺑﺰﺭﮔﺘﺮ ﺑﻪ ﻣﻘﻴﺎﺱ ﻛﻮﭼﻜﺘﺮ ﺟﺮﻳﺎﻥ ﻣﻲ
ﻳﺎﺑﺪ ﻭ ﺳﺮﺍﻧﺠﺎﻡ ﺑﻪ ﻭﺳﻴﻠﻪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺗﺒﺪﻳﻞ ﺑﻪ ﮔﺮﻣﺎ ﻣﻲ ﺷﻮﺩ ﻭ ﺍﻧﺮژﻱ ﺑﺮﺍﻱ ﺣﺮﻛﺖ ﺩﺭ ﻣﻘﻴﺎﺱ ﺑﺰﺭگ ﺍﺯ ﺟﺮﻳﺎﻥ
ﺧﺎﺭﺟﻲ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ.
ﻓﺮﺁﻳﻨﺪ energy cascadeﺩﺭ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ ﺑﻮﺳﻴﻠﻪ ﺷﻌﺮRichard Sonﺧﻼﺻﻪ ﻣﻲ ﺷﻮﺩ :
"Big whirls have little whirls that feed on their velocity little whirls have lesser
"whirls, and so on to viscosity.
ﺩﺭ ﻫﺮ ﺟﺮﻳﺎﻥ ﺑﺮﺷﻲ ﻣﺘﻼﻃﻢ ﺍﻧﺮژﻱ ﻣﺘﻼﻃﻢ ﺑﻔﺮﻡ convectionﻭ diffusionﻭ dissipationﻭﺟﻮﺩ ﺩﺍﺭﺩ.
،Dissipationﺑﻪ ﺧﺎﻃﺮ ﻋﻤﻞ ﻳﺎ ﻛﺎﺭ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻻﻣﻴﻨﺎﺭ ﺭﻭﻱ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﻛﻮﭼﻚ ﺭﺥ ﻣﻲ ﺩﻫﺪ.
Turbulent energy balance in the boundary on a smooth flat plate :
Large Scale
Small Scale
۱۷۱
۱
۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺩﺭ ﻧﺘﻴﺠﻪ ﻣﻲ ﺗﻮﺍﻥ ﮔﻔﺖ ﻛﻪ ﺳﻬﻢ ﺍﺻﻠﻲ ﺍﻧﺮژﻱ ﺩﺭ ﺟﺮﻳﺎﻥ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺭﺍ ﺗﻮﻟﻴﺪ ﻭ ﺍﺗﻼﻑ ﺩﺍﺭﺍ ﻣﻲ ﺑﺎﺷﺪ ﻭ ﺩﺍﺭﻳﻢ:
𝑛𝑜𝑖𝑡𝑎𝑝𝑖𝑠𝑠𝑖𝑑 ≈ 𝑛𝑜𝑖𝑡𝑐𝑢𝑑𝑜𝑟𝑝
ﻧﻜﺘﻪ :ﺑﺮﺍﻱ ﻳﻚ ﻟﻮﻟﻪ ﺑﺰﺭﮔﺘﺮﻳﻦ ﮔﺮﺩﺍﺑﻪ ﻫﺎ ﺍﻧﺪﺍﺯﻩ ﺁﻧﻬﺎ ﺣﺪﻭﺩ ﻧﺼﻒ ﻗﻄﺮ ﻟﻮﻟﻪ ﻣﻲ ﺑﺎﺷﺪ.
ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺁﺏ ﺑﺎ ﺳﺮﻋﺖ 1.8m/sﺩﺭ ﻳﻚ ﻟﻮﻟﻪ ﺑﻪ ﻗﻄﺮ: 50mm
𝑅𝑒 ≈ ۱.٥
Size of largest eddy =۲٥mm
Size of most dissipation eddies =۰.۱۲٥ mm =۱۲٥μm
②
𝑖 = 1,2
����
𝑝𝜕�
���
𝚤𝑢𝜕
𝜕
′
���������
=−
+ µ∇2 𝑢�𝚤 +
𝑢𝜌�−
� 𝚥 𝑢𝚤 ′
𝑗𝑥𝜕
𝑖𝑥𝜕
𝑗𝑥𝜕
���
= 𝚤𝑢
����
𝚤𝑢𝜕
=0
𝑖𝑥𝜕
ﺗﻔﺎﻭﺕ ﻣﻌﺎﺩﻻﺕ②)ﻣﻠﻘﺐ ﺑﻪ ﻣﻌﺎﺩﻻﺕ ﺭﻳﻨﻮﻟﺪﺯ(ﺑﺎ ﻣﻌﺎﺩﻻﺕ ﻧﺎﻭﻳﺮ ﺍﺳﺘﻮﻛﺲ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺁﺭﺍﻡ ﺩﺭ ﻭﺟـﻮﺩ ﺟﻤـﻼﺕ
′
���������
𝑢𝜌−ﻣﻲ ﺑﺎﺷﺪ ﻛﻪ ﺑﻪ ﺩﻟﻴﻞ ﻣﻐﺸﻮﺵ ﺑﻮﺩﻥ ﺟﺮﻳﺎﻥ ﺑﻮﺟﻮﺩ ﺁﻣﺪﻩ ﺍﻧﺪ.ﺩﺭﺣﻘﻴﻘﺖ ﻫﻤﺎﻧﻄﻮﺭ ﻛﻪ ﺣﺮﻛﺖ ﺍﺗﻔﺎﻗﻲ
𝚥 𝑢𝚤 ′
ﻭ ﻧﺎﻣﻨﻈﻢ ﻣﻮﻟﻜﻮﻟﻬﺎ ﺳﺒﺐ ﻭﺟﻮﺩ ﻟﺰﺟﺖ ﻣﻮﻟﻜﻮﻟﻲ ﻭ ﺗﻨﺶ ﻣﻲ ﮔﺮﺩﺩ ﺣﺮﻛﺖ ﺍﺗﻔﺎﻗﻲ ﻭ ﻧﺎﻣﻨﻈﻢ ﺗﻮﺩﻩ ﻫﺎﻱ ﺳـﻴﺎﻝ ﺩﺭ
ﺟﺮﻳﺎﻧﻬﺎﻱ ﻣﺘﻼﻃﻢ ﻧﻴﺰ ﻣﻲ ﺗﻮﺍﻧﺪ ﻋﺎﻣﻞ ﭘﺪﻳﺪ ﺁﻭﺭﻧﺪﻩ ﻧﻮﻋﻲ ﺩﻳﮕﺮ ﺍﺯ ﺗﻨﺶ ﺑﺎﺷﺪ .ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﺗﺮﻡ ﻫﺎﻱ ﺍﺿـﺎﻓﻲ
ﻣﻮﺟﻮﺩ ﺩﺭ ﻣﻌﺎﺩﻻﺕ 2ﺑﻪ ﺗﻨﺶ ﻫﺎﻱ ﻇﺎﻫﺮﻱ ﻣﻌﺮﻭﻑ ﮔﺮﺩﻳﺪﻩ ﺍﻧﺪ)ﺗﻨﺶ ﻫﺎﻱ ﺭﻳﻨﻮﻟﺪﺯ (.ﺗـﻨﺶ ﻫـﺎﻱ ﻇـﺎﻫﺮﻱ ﺑـﻪ
ﺳﺒﺐ ﺩﺍﺭﺍ ﺑﻮﺩﻥ ﻣﻘﺪﺍﺭ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪ ﺍﻱ ﻧﺴﺒﺖ ﺑﻪ ﺗﻨﺶ ﻫﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ﻟﺰﺟﺖ ﻣﻮﻟﻜﻮﻟﻲ ﺩﺭ ﻗﺴﻤﺖ ﺍﻋﻈـﻢ ﺩﺍﻣﻨـﻪ
۱۷۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺟﺮﻳﺎﻥ ﺍﺯ ﺍﻫﻤﻴﺖ ﺑﺴﻴﺎﺭ ﺯﻳﺎﺩﻱ ﺑﺮﺧﻮﺭﺩﺍﺭﻧﺪ ﻭ ﺩﺭ ﻭﺍﻗﻊ ﻋﺎﻣﻞ ﺍﺻﻠﻲ ﺗﻔﺎﻭﺕ ﻣـﺎﺑﻴﻦ ﺭﻓﺘـﺎﺭ ﻣﺘﻮﺳـﻂ ﻛﻤﻴـﺖ ﻫـﺎ ﺩﺭ
ﺟﺮﻳﺎﻥ ﻣﻐﺸﻮﺵ ﺑﺎ ﺟﺮﻳﺎﻥ ﻻﻳﻪ ﺍﻱ ﻣﻲ ﺑﺎﺷﺪ.
ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﺍﻳﻦ ﺗﻨﺶ ﻫﺎ ﺍﻃﻼﻋﺎﺕ ﻣﺴﺘﻘﻴﻤﻲ ﺩﺭ ﺩﺳﺘﺮﺱ ﻧﻤﻲ ﺑﺎﺷﺪ.ﺑﻨﺎﺑﺮﺍﻳﻦ ﺣﻞ ﻣﻌـﺎﺩﻻﺕ ﺭﻳﻨﻮﻟـﺪﺯ ﺑـﺪﻟﻴﻞ
ﺍﻓﺰﺍﻳﺶ ﺗﻌﺪﺍﺩ ﻣﺠﻬﻮﻻﺕ ﺩﺭ ﺷﺮﺍﻳﻂ ﻓﻮﻕ ﺍﻣﻜﺎﻥ ﭘـﺬﻳﺮ ﻧﺨﻮﺍﻫـﺪ ﺑـﻮﺩ.ﺑﺮﺍﻱ ﺑﺮﺭﺳـﻲ ﺿـﺮﻭﺭﻱ ﺍﺳـﺖ ﺍﺯ ﺭﻭﺍﺑـﻂ ﻳـﺎ
ﻣﻌﺎﺩﻻﺗﻲ ﺍﺳﺘﻔﺎﺩﻩ ﺷﻮﺩ ﻛﻪ ﺑﺘﻮﺍﻧﻨﺪ ﺗﻨﺶ ﻫﺎﻱ ﻇﺎﻫﺮﻱ ﺭﺍ ﺑﻪ ﻛﻤﻴﺖ ﻫﺎﻱ ﻣﺘﻮﺳﻂ ﺟﺮﻳﺎﻥ ﻣﺮﺗﺒﻂ ﺳﺎﺯﺩ ﻳﺎ ﺑﻨﺤﻮﻱ
ﺁﻧﻬﺎ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻧﻤﺎﻳﻨﺪ.ﺍﻳﻦ ﺭﻭﺍﺑﻂ ﻭ ﻣﻌﺎﺩﻻﺕ ﻛﻪ ﺑﻪ ﻣﺪﻟﻬﺎﻱ ﺗﻮﺭﺑﻮﻻﻧﺲ ﻣﻌﺮﻭﻑ ﻣﻲ ﺑﺎﺷﺪ ﻋﻤﻮﻣﺎ ﺗﻘﺮﻳﺒﻲ ﺑـﻮﺩﻩ ﻭ
ﻣﻴﺰﺍﻥ ﺩﻗﺖ ﺁﻧﻬﺎ ﺑﺴﺘﮕﻲ ﺑﻪ ﺷﺮﺍﻳﻂ ﻣﺴﺌﻠﻪ ﺩﺍﺭﺩ.ﺩﺭ ﻗﺴﻤﺖ ﺑﻌﺪﻱ ﺗﻌﺪﺍﺩﻱ ﺍﺯ ﺍﻳﻦ ﻣﺪﻟﻬﺎ ﺑﺼﻮﺭﺕ ﺧﻼﺻـﻪ ﺑﺮﺭﺳـﻲ
ﻣﻴﮕﺮﺩﻧﺪ.
ﻣﺪﻟﻬﺎﻱ ﺟﺮﻳﺎﻥ ﻣﻐﺸﻮﺵ
1
F59
ﺑﻄﻮﺭ ﻛﻠﻲ ﻣﺪﻟﻬﺎﻱ ﺍﺭﺍﺋﻪ ﺷﺪﻩ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻣﻐﺸﻮﺵ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺳﻪ ﺩﺳﺘﻪ ﺍﺻﻠﻲ ﺗﻘﺴﻴﻢ ﻧﻤﻮﺩ:
.1ﻣﺪﻟﻬﺎﻱ ﺟﺒﺮﻱ ﻟﺰﺟﺖ ﺗﻼﻃﻢ
۲
60F
..2ﻣﺪﻟﻬﺎﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﻟﺰﺟﺖ ﺗﻼﻃﻢ
ﺍﻟﻒ.ﻣﺪﻟﻬﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ
ﺏ.ﻣﺪﻟﻬﺎﻱ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺍﻱ
3
F61
4
F62
.3ﻣﺪﻟﻬﺎﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺑﺮﺍﻱ ﺗﻨﺶ ﻫﺎﻱ ﺑﺮﺷﻲ )ﻣﺪﻟﻬﺎﻱ ﭼﻨﺪ ﻣﻌﺎﺩﻟـﻪ ﺍﻱ (ﺩﺭ ﻣﻌﺎﺩﻟـﻪ ﻫـﺎﻱ ﺩﺳـﺘﻪ ﺍﻭﻝ ﻭ ﺩﻭﻡ
ﺗﻨﺶ ﻫﺎﻱ ﻇﺎﻫﺮﻱ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻧﻈﺮﻳﻪ ﺑﻮﺯﻳﻨﺴﻚ ﻣﺤﺎﺳـﺒﻪ ﻣﻴﮕﺮﺩﻧـﺪ .ﺩﺭ ﺣﺎﻟﻴﻜـﻪ ﺩﺭ ﻣـﺪﻟﻬﺎﻱ ﺩﺳـﺘﻪ ﺳـﻮﻡ ﺍﺯ
ﻧﻈﺮﻳﻪ ﺑﻮﺯﻳﻨﺴﻚ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻲ ﺷﻮﺩ ﺑﻠﻜﻪ ﺑﺮﺍﻱ ﺗﻌﻴﻴﻦ ﻫـﺮ ﻳـﻚ ﺍﺯ ﺗـﻨﺶ ﻫـﺎﻱ ﺗﻼﻃـﻢ ﻣﺴـﺘﻘﻴﻤﺎ ﻳـﻚ ﻣﻌﺎﺩﻟـﻪ
ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺍﻧﺘﻘﺎﻝ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ.
ﻧﻈﺮﻳﻪ ﺑﻮﺯﻳﻨﺴﻚ:
ﺑﻮﺯﻳﻨﺴﻚ ﺩﺭ ﺳﺎﻝ 1877ﺍﻇﻬﺎﺭ ﻧﻤﻮﺩ ﻛﻪ ﺗﻨﺶ ﺑﺮﺷﻲ ﻇﺎﻫﺮﻱ ﺗﻼﻃﻢ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺑﺼﻮﺭﺕ ﺣﺎﺻﻠﻀﺮﺏ ﮔﺮﺍﺩﻳﺎﻥ
ﺳﺮﻋﺖ ﻣﺘﻮﺳﻂ ﺩﺭ ﻛﻤﻴﺘﻲ ﺑﻨﺎﻡ ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﺑﻴﺎﻥ ﻛﺮﺩ.
����
𝑢𝜕
𝑦𝜕
𝜇 = ′𝑣 ′
�������
𝑢𝜌𝜏 = −
𝑡
۱
Turbulence Modeling
)Algebraic Models(Zero –Equation
۳
One-Equation Models
٤ Two-Equation Models
۲
۱۷۳
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻛﻪ 𝑡𝜇 ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﻣﻲ ﺑﺎﺷﺪ.ﺭﺍﺑﻄﻪ ﻓﻮﻕ ﺷﺒﺎﻫﺖ ﻛﺎﻣﻠﻲ ﺑﺎ ﻗﺎﻧﻮﻥ ﻟﺰﺟﺖ ﻣﻮﻟﻜﻮﻟﻲ ﻧﻴﻮﺗﻦ ﺩﺍﺭﺩ.ﻣﻨﺘﻬﺎ ﺑﺮﺧﻼﻑ
ﻟﺰﺟﺖ ﻣﻮﻟﻜﻮﻟﻲ ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﻳﻚ ﺧﺎﺻﻴﺖ ﺳﻴﺎﻝ ﻧﺒﻮﺩﻩ ﻭ ﻣﻘﺪﺍﺭ ﺁﻥ ﺍﺯ ﻧﻘﻄﻪ ﺍﻱ ﺑﻪ ﻧﻘﻄﻪ ﺩﻳﮕﺮ ﺑﺮﺣﺴﺐ ﺷﺮﺍﻳﻂ
ﺟﺮﻳﺎﻥ ﺗﻐﻴﻴﺮ ﻣﻲ ﻧﻤﺎﻳﺪ ﻭ ﻣﻘﺪﺍﺭ ﺁﻥ ﺑﻌﻨﻮﺍﻥ ﻳﻚ ﻣﺠﻬﻮﻝ ﺗﻠﻘﻲ ﻣﻴﮕﺮﺩﺩ .ﺑﺎ ﺍﻳﻦ ﺗﻘﺮﻳﺐ ﻧﻈﺮﻳﻪ ﺑﻮﺯﻳﻨﺴﻚ ﺑﻪ
ﺗﻨﻬﺎﻳﻲ ﻳﻚ ﻣﺪﻝ ﻛﺎﻣﻞ ﻧﻤﻲ ﺑﺎﺷﺪ.
ﺩﺭ ﻣﺪﻟﻬﺎﻱ ﺩﺳﺘﻪ ﺍﻭﻝ ﻭ ﺩﻭﻡ ﺍﺑﺘﺪﺍ 𝑡𝜇 ﺑﺮﺣﺴـﺐ ﭘﺎﺭﺍﻣﺘﺮﻫـﺎﻱ ﻗﺎﺑـﻞ ﻣﺤﺎﺳـﺒﻪ ﻳـﺎ ﻣﻌﻠـﻮﻡ ﺟﺮﻳـﺎﻥ ﻣﺤﺎﺳـﺒﻪ ﻣـﻲ
ﺷﻮﺩ.ﺳﭙﺲ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺭﺍﺑﻄﻪ ﺑﻮﺯﻳﻨﺴﻚ ﺗﻨﺸﻬﺎﻱ ﻇﺎﻫﺮﻱ ﺗﻌﻴﻴﻦ ﻣﻲ ﮔﺮﺩﻧﺪ.
ﺭﺍﺑﻄﻪ ﺑﻮﺯﻳﻨﺴﻚ ﺑﻪ ﻓﺮﻡ ﻛﺎﻣﻞ ﻭ ﺑﻪ ﺷﻜﻞ ﺗﺎﻧﺴﻮﺭﻱ ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻴﺎﻥ ﻣﻴﮕﺮﺩﺩ.
𝚥�𝑢𝜕 𝚤�𝑢𝜕
2
+
𝑗𝑖𝛿𝑘𝜌 � −
𝑖𝑥𝜕 𝑗𝑥𝜕
3
ﻣﻌﺎﺩﻟﻪ ﻫﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺟﺒﺮﻱ
� 𝜇 = ′ 𝑢′
�������
𝑢𝜌−
𝚥 𝚤
𝑡
ﭘﺎﻳﻪ ﺍﺳﺎﺳﻲ ﻣﺪﻟﻬﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺟﺒﺮﻱ ﺑﺮ ﻧﻈﺮﻳﻪ ﻃﻮﻝ ﺍﺧـﺘﻼﻁ ﭘﺮﺍﻧﺘـﻞ ) (1925ﺍﺳـﺘﻮﺍﺭ ﺍﺳـﺖ.ﺍﻭ ﺑـﺎ ﺑﻴـﺎﻥ
ﻧﻈﺮﻳﻪ ﺍﻱ )ﻣﺸﺎﺑﻪ ﺑﺎ ﻧﻈﺮﻳﻪ ﺟﻨﺒﺸﻲ ﮔﺎﺯﻫﺎ( ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﺭﺍ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﺩﺍﻧﺴﻴﺘﻪ ﺳﻴﺎﻝ ﻭ ﻳﻚ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﻭ
ﻳﻚ ﻃﻮﻝ ﻣﺸﺨﺼﻪ ﺩﺍﻧﺴﺖ.
1
∄�𝑢𝜌 ∝ 𝜇 →∄�𝑢𝜌 = 𝜇 𝑡𝑣𝜌𝑙 = 𝑡𝜇
3
ﺩﺭ ﺭﺍﺑﻄﻪ ﻓﻮﻕ 𝑙 ﻭ 𝑡𝑣 ﺑﺘﺮﺗﻴﺐ ﻧﻤﺎﻳﺸﮕﺮ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﻭ ﻃﻮﻝ ﻣﺸﺨﺼﻪ ﻣﻲ ﺑﺎﺷﻨﺪ.
ﺍﻛﺜﺮ ﻗﺮﻳﺐ ﺑﻪ ﺍﺗﻔﺎﻕ ﻣـﺪﻟﻬﺎﻱ ﺩﺳـﺘﻪ ﺍﻭﻝ ﻭ ﺩﻭﻡ ﺩﺭ ﻭﺍﻗـﻊ ﺍﺯ ﻫﻤـﻴﻦ ﻣﻌﺎﺩﻟـﻪ ﺑـﺮﺍﻱ ﻣﺤﺎﺳـﺒﻪ 𝑡𝜇 ﺍﺳـﺘﻔﺎﺩﻩ ﻣـﻲ
ﻧﻤﺎﻳﻨﺪ.ﻣﻨﺘﻬﻲ ﻫﺮ ﻳﻚ ﺍﺯ ﺁﻧﻬﺎ ﺭﻭﺍﺑﻂ ﻭ ﺭﻭﺷﻬﺎﻱ ﻣﺨﺘﻠﻔﻲ ﺭﺍ ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ 𝑙 ﻭ 𝑡𝑣 ﺍﺭﺍﺋﻪ ﻣﻲ ﻧﻤﺎﻳﻨﺪ.
ﭘﺮﺍﻧﺘﻞ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﺭﺍ ﺑﺎ ﺭﺍﺑﻄﻪ ﺯﻳﺮ ﺑﻴﺎﻥ ﻧﻤﻮﺩ.
����
𝑢𝜕
�
𝑦𝜕
� × 𝑚𝑙 = 𝑡𝑣
ﺩﺭ ﺭﺍﺑﻄﻪ ﻓﻮﻕ 𝑚𝑙 ﻧﻤﺎﻳﺎﻧﮕﺮ ﻃﻮﻝ ﺍﺧﺘﻼﻁ ﻣﻲ ﺑﺎﺷﺪ ﻛﻪ ﭘﺮﺍﻧﺘﻞ ﻣﻘﺪﺍﺭ ﺁﻥ ﺭﺍ ﻣﺘﻨﺎﺳﺐ ﺑﺎ ﻓﺎﺻـﻠﻪ ﺍﺯ ﺩﻳـﻮﺍﺭ ﺩﺭ ﻧﻄـﺮ
ﮔﺮﻓﺖ ﻭ ﺩﺭ ﻓﻮﺍﺻﻞ ﺩﻭﺭ ﺍﺯ ﺩﻳﻮﺍﺭ ﺁﻥ ﺭﺍ ﺛﺎﺑﺖ ﻓﺮﺽ ﻧﻤﻮﺩ.ﺑﻨﺎﺑﺮﺍﻳﻦ ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﺑﺮﺍﺳﺎﺱ ﻧﻈﺮﻳﻪ ﭘﺮﺍﻧﺘـﻞ ﺑـﺎ ﺭﺍﺑﻄـﻪ
ﺯﻳﺮ ﺑﻴﺎﻥ ﻣﻲ ﮔﺮﺩﺩ.
����
𝑢𝜕
� 𝑘 = 0.36
𝜕𝑦 1
� 𝜇𝑡 = 𝜌(𝑦𝑘1 )2
ﺍﺯ ﺍﻳﺮﺍﺩﺍﺕ ﻣﺪﻝ ﻓﻮﻕ ﻧﺤﻮﻩ ﺗﻮﺯﻳﻊ ﺍﺭﺍﺋﻪ ﺷﺪﻩ ﺑﺮﺍﻱ 𝑚𝑙 ﺩﺭ ﻧﺰﺩﻳﻜﻲ ﺩﻳﻮﺍﺭﻩ ﻣﻲ ﺑﺎﺷﺪ ﻛـﻪ ﻣﻨﺠـﺮ ﺑـﻪ ﺑـﺮﻭﺯﺍﺧﺘﻼﻑ
ﺯﻳﺎﺩ ﻣﺎ ﺑﻴﻦ ﻧﺘﺎﻳﺞ ﺗﺌﻮﺭﻱ ﻭ ﺁﺯﻣﺎﻳﺶ ﺩﺭ ﻧﺰﺩﻳﻜﻲ ﺩﻳﻮﺍﺭﻩ ﻣﻲ ﮔﺮﺩﺩ.
۱۷٤
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺍﺯ ﺍﻳﻦ ﺭﻭ ﻣﺤﻘﻘﻴﻦ ﺗﻮﺯﻳﻊ ﻫﺎﻱ ﺩﻳﮕﺮﻱ ﺑﺮﺍﻱ 𝑚𝑙 ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺍﻧﺪ ﻛـﻪ ﻣﻨﺠـﺮ ﺑـﻪ ﭘﻴـﺪﺍﻳﺶ ﻣـﺪﻟﻬﺎﻱ ﻣﺨﺘﻠﻔـﻲ
ﮔﺮﺩﻳﺪﻩ ﺍﺳﺖ.
ﺍﺯ ﻣﺰﺍﻳﺎﻱ ﻣﺪﻟﻬﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺟﺒﺮﻱ ﻣﻲ ﺗﻮﺍﻥ ﺑﻪ ﺳﺎﺩﮔﻲ ﻛﺎﺭﺑﺮﺩ ﺁﻧﻬﺎ )ﻛﻪ ﻧﻴﺎﺯ ﺑﻪ ﺣـﻞ ﻣﻌﺎﺩﻟـﻪ ﺩﻳﻔﺮﺍﻧﺴـﻴﻞ
ﺍﺿﺎﻓﻲ ﻧﺪﺍﺭﻧﺪ(ﺍﺷﺎﺭﻩ ﻧﻤﻮﺩ.ﻫﻤﭽﻨﻴﻦ ﺩﺭ ﺻﻮﺭﺗﻴﻜﻪ ﺗﻮﺯﻳﻊ ﻧﺴﺒﺘﺎ"ﺩﻗﻴﻘﻲ ﺍﺯ 𝑚𝑙 ﺩﺭ ﺩﺳﺘﺮﺱ ﺑﺎﺷﺪ ﺍﻳﻦ ﻣﺪﻟﻬﺎ ﻧﺘـﺎﻳﺞ
ﺧﻮﺑﻲ ﺑﺮﺍﻱ ﺗﻮﺯﻳﻊ ﺳﺮﻋﺖ ﻭ ﺗﻮﺯﻳﻊ ﺗﻨﺶ ﺑﺮﺷﻲ ﺍﺭﺍﺋﻪ ﻣﻲ ﺩﻫﻨﺪ.ﺍﻣﺎ ﺩﺭ ﺟﺮﻳﺎﻧﻬﺎﻱ ﭘﻴﭽﻴﺪﻩ )ﻣﺜﻼ" Recirulating
(flowﻛﻪ ﺍﺭﺗﺒﺎﻁ ﻣﺎﺑﻴﻦ ﻣﻮﻟﻔﻪ ﻫﺎﻱ ﺗﻨﺶ ﻭ ﻣﻴﺪﺍﻥ ﺳﺮﻋﺖ ﺑﺴﻴﺎﺭ ﭘﻴﭽﻴـﺪﻩ ﺍﺳـﺖ ﭘـﻴﺶ ﺑﻴﻨـﻲ ﻭ ﻣﺤﺎﺳـﺒﻪ ﺍﻳـﻦ
ﻣﺪﻟﻬﺎﺍﺯ ﻣﻴﺪﺍﻥ ﺟﺮﻳﺎﻥ ﺍﻏﻠﺐ ﺻﺤﻴﺢ ﻧﻤﻲ ﺑﺎﺷﺪ.ﺑﻪ ﻋﻼﻭﻩ ﺩﺭ ﺍﻳﻦ ﻣﺪﻟﻬﺎ ﺍﺯ ﻓﺮﺍﻳﻨﺪ ﺍﻧﺘﺸﺎﺭ ﻭ ﺟﺎﺑﺠﺎﻳﻲ ﺗﻼﻃﻢ ﺻـﺮﻑ
ﻧﻈﺮ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺍﻳﻦ ﺍﻣﺮ ﺑﺎﻋﺚ ﺍﻳﺠﺎﺩ ﺧﻄﺎﻱ ﺯﻳﺎﺩ ﺩﺭ ﺗﺤﻠﻴﻞ ﺟﺮﻳﺎﻧﻬﺎﻱ ﺍﺯ ﻧـﻮﻉ ﻻﻳـﻪ ﻣـﺮﺯﻱ )ﻧﻘـﺎﻁ ﺟﺮﻳـﺎﻥ ﺍﺯ
ﻧﻘﺎﻁ ﺑﺎﻻﺩﺳﺖ ﺍﺛﺮ ﻣﻲ ﭘﺬﻳﺮﻧﺪ(ﻭ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﺍﺯ ﻧﻮﻉ ﺟﺮﻳﺎﻥ ﺑﺮﮔﺸﺘﻲ )ﻧﻘﺎﻃﻲ ﺍﺯ ﺟﺮﻳﺎﻥ ﺍﺯ ﻧﻘﺎﻁ ﭘﺎﻳﻴﻦ ﺩﺳﺖ ﺍﺛﺮ
ﻣﻲ ﭘﺬﻳﺮﻧﺪ(ﻣﻲ ﮔﺮﺩﺩ.ﻋﻼﻭﻩ ﺑﺮ ﺍﻳﻨﻬﺎ ﺩﺭ ﻣﺪﻟﻬﺎﻱ ﻓﻮﻕ ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﺍﺭﺗﺒﺎﻁ ﻣﺴﺘﻘﻤﻲ ﺑﺎ ﮔﺮﺍﺩﻳـﺎﻥ ﺳـﺮﻋﺖ ﺩﺍﺭﺩ.ﺍﺯ
ﺍﻳﻨﺮﻭ ﺩﺭ ﻣﺪﻟﻬﺎ ﺑﺮﺍﻱ ﻣﻜﺎﻧﻬﺎﻳﻲ ﻛﻪ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﺑﺎﺷﺪ)ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺩﺭ ﺭﻭﻱ ﺧﻂ ﻣﺮﻛﺰﻱ ﻳﻚ ﻟﻮﻟﻪ ﺑـﺎ
ﺷﺮﺍﻳﻂ ﻣﺮﺯﻱ ﻣﺘﻘﺎﺭﻥ(ﻟﺰﺟﺖ ﺗﻼﻃﻢ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﻧﺘﻴﺠﻪ ﺍﻱ ﻛﻪ ﺑﺎ ﻭﺍﻗﻌﻴﺖ ﺗﻄﺎﺑﻖ ﻧـﺪﺍﺭﺩ ﻭ ﺑﺎﻋـﺚ ﻋـﺪﻡ
ﺍﻧﺘﻘﺎﻝ ﻣﻮﻟﻔﻪ ﻣﻐﺸﻮﺵ ﻣﻮﻣﻨﺘﻢ ﺩﺭ ﺍﻣﺘﺪﺍﺩ ﺍﻳﻨﭽﻨﻴﻦ ﺻﻔﺤﺎﺗﻲ ﻣﻲ ﮔﺮﺩﺩ.
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺿﻐﻒ ﻫﺎﻱ ﺑﺮﺷﻤﺮﺩﻩ ﺩﺭ ﻓﻮﻕ ﻣﺤﻘﻘﻴﻦ ﺑﺮﺍﻱ ﺑﺮﺭﺳﻲ ﺩﻗﻴﻘﺘﺮ ﺟﺮﻳﺎﻧﻬﺎﻱ ﭘﻴﭽﻴﺪﻩ ﺍﻧﻮﺍﻉ ﺩﻳﮕﺮﻱ ﺍﺯ ﻣﺪﻟﻬﺎ
ﺭﺍ ﺍﺑﺪﺍﻉ ﻧﻤﻮﺩﻩ ﺍﻧﺪ ﻛﻪ ﺩﺭ ﺍﺩﺍﻣﻪ ﺑﺮﺧﻲ ﺍﺯ ﺁﻧﻬﺎ ﻣﻌﺮﻓﻲ ﻣﻲ ﺷﻮﻧﺪ.
ﻣﺪﻟﻬﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ:
ﻫﻤﺎﻧﻄﻮﺭ ﻛﻪ ﺩﺭ ﻗﺴﻤﺖ ﻗﺒﻞ ﺑﻴﺎﻥ ﺷﺪ ﻣﺮﺗﺒﻂ ﻧﻤﻮﺩﻥ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﺑﺎ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﻣﺘﻮﺳـﻂ ﺑﺎﻋـﺚ ﺍﻳﺠـﺎﺩ
ﻣﺸﻜﻼﺗﻲ ﺩﺭ ﻣﻜﺎﻧﻬﺎﻳﻲ ﻛﻪ ﻣﻘﺪﺍﺭ ﮔﺮﺍﺩﻳﺎﻥ ﺳﺮﻋﺖ ﺑﺮﺍﺑﺮ ﺻﻔﺮ ﺍﺳﺖ ﻣﻲ ﮔﺮﺩﺩ.ﺍﺯ ﺍﻳﻦ ﺭﻭ ﻋﺪﻩ ﺍﻱ ﺍﺯ ﻣﺤﻘﻘﻴﻦ ﺑـﺮﺁﻥ
ﺷﺪﻧﺪ ﻛﻪ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﺭﺍ ﺑﻪ ﺟﺬﺭ ﻣﻴﺎﻧﮕﻴﻦ ﮔﻴﺮﻱ ﺷﺪﻩ ﺍﺯ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﻧﺴﺒﺖ ﺩﻫﻨﺪ.ﺑﻪ ﺑﻴﺎﻥ ﺭﻳﺎﺿﻲ
𝑘√ ∝ 𝑡𝑣
������� 1
���� 1
����
�����
′2
′2
′2
𝑘 = (𝑢′
) 𝑤 𝚤 𝑢′𝚤 = (𝑢 + 𝑣 +
2
2
ﺍﺯ ﻃﺮﻓﻲ ﺩﺭﺟﺮﻳﺎﻧﻬﺎﻱ ﭘﻴﭽﻴﺪﻩ ﺍﺛﺮﺍﺕ ﺍﻧﺘﺸﺎﺭ ﻭ ﺟﺎﺑﺠﺎﻳﻲ ﺗﻼﻃﻢ ﻗﺎﺑﻞ ﻣﻼﺣﻈﻪ ﺑـﻮﺩﻩ ﺑـﺮﺍﻱ ﺩﺭ ﻧﻈـﺮ ﮔـﺮﻓﺘﻦ ﺍﻳـﻦ
ﺍﺛﺮﺍﺕ ﺑﺮ ﺭﻭﻱ ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﺑﺎﻳﺪ ﺑﺠﺎﻱ ﻣﻌﺎﺩﻻﺕ ﺟﺒﺮﻱ ﺍﺯ ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻮﺩ.ﺑﺮ ﺍﻳـﻦ ﺍﺳـﺎﺱ
ﻣﺪﻟﻬﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﭘﺪﻳﺪ ﺁﻣﺪﻩ ﻛﻪ ﺩﺭ ﺁﻧﻬﺎ ﺍﻧﺮﺯﻱ ﺟﻨﺒﺸـﻲ ﺗﻼﻃـﻢ ﻛـﻪ ﺩﺭ ﻭﺍﻗـﻊ ﺑﻴـﺎﻥ ﻛﻨﻨـﺪﻩ
ﺳﺮﻋﺖ ﻣﺸﺨﺼﻪ ﻣﻲ ﺑﺎﺷﺪ ﺍﺯ ﺭﻭﻱ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﺤﺎﺳﺒﻪ ﻣﻲ ﮔﺮﺩﺩ ﻭﻟﻲ ﻃﻮﻝ ﻣﺸﺨﺼﻪ ﺑـﺎ ﺍﺳـﺘﻔﺎﺩﻩ ﺍﺯ
ﻳﻚ ﺭﺍﺑﻄﻪ ﺟﺒﺮﻱ ﺗﻌﻴﻴﻦ ﻣﻲ ﮔﺮﺩﺩ.
Shortcoming of mixing length Hipothesis:
= ۰ →µt=۰
۱۷٥
�
𝑢𝜕
𝑦𝜕
For
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺻﻔﺮ ﻧﻴﺴﺖ .ﺑﺮﺍﻱ ﻣﺜﺎﻝ µtﻣﻘﺪﺍﺭ = 0
ﺩﺭ ﻣﺮﻛﺰ ﻟﻮﻟﻪ ﻣﺸﺎﻫﺪﻩ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ:
�
𝑢𝜕
𝑦𝜕
ﺩﺍﺩﻩ ﻫﺎﻱ ﺗﺠﺮﺑﻲ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﺟﺎﻳﻲ ﻛﻪ
= ۰ →µt=۰µt|max
�
𝑢𝜕
𝑦𝜕
𝚥�𝑢𝜕 𝚤�𝑢𝜕
��������⎧
( 𝑡𝑣 = −𝑢𝚤′ 𝑢𝚥′
+
)
⎪
𝑖𝑥𝜕 𝑗𝑥𝜕
���
𝜕ɸ′
⎨
′ ′
��������
𝑢−
⎪
𝑡𝜇 = 𝚤 ɸ
𝑖𝑥𝜕
⎩
Turbulent diffusivity of heat or mass
𝑣
𝑟𝑒𝑏𝑚𝑢𝑛 𝑡𝑑𝑖𝑚𝜎 = 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑝𝑟𝑎𝑛𝑡𝑙 𝑆𝑐ℎ
→ 𝑡 = 𝑡𝜇,
𝜎
𝜎=1
|
�
𝑢𝜕
𝑦𝜕
| 𝛾𝑡≈ 𝑣𝑡 = lim
Heat or mass diffusivity
ﺍﮔﺮ = 0
�
𝑢𝜕
𝑦𝜕
ﻣﻌﺎﺩﻟﻪ ﺑﺎﻻ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ 𝛾𝑡 = 0ﻛﻪ ﺍﺯ ﻧﻈﺮ ﺗﺠﺮﺑﻲ ﻣﺸﻜﻞ ﺍﻳﺠﺎﺩ ﻣﻲ ﻛﻨﺪ .ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﺜـﺎﻝ
ﺩﺭ ﻣﺒﺪﻝ ﺣﺮﺍﺭﺗﻲ ﺑﺎﻻ ﺟﺎﻳﻲ ﻛﻪ 𝛾𝑡 = 0ﺍﺳﺖ ﺍﺯ ﻧﻈﺮ ﺗﺠﺮﺑﻲ ﻣﺸﺎﻫﺪﻩ ﻧﺸﺪﻩ ﺍﺳﺖ.
ﺍﮔﺮ ﺟﺮﻳﺎﻥ ﺑﺮﮔﺸﺘﻲ 1ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﻢ ﺑﺎﺯ ﺗﺌﻮﺭﻱ ﺑﺎﻻ ﺟﻮﺍﺏ ﻧﻤﻲ ﺩﻫﺪ.
F63
Recirculatingflow
۱۷٦
۱
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺩﺍﺩﻩ ﻫﺎﻱ ﺁﺯﻣﺎﻳﺸﮕﺎﻫﻲ ﻧﺸﺎﻥ ﻣﻴﺪﻫﺪ ﻛﻪ ﻣﺎﻛﺰﻳﻤﻢ ﻓﻼﻛﺲ ﺣﺮﺍﺭﺗﻲ ﺩﺭ ﻧﻘﻄﻪ R.Pﺭﺥ ﻣﻲ ﺩﻫﺪ ﺩﺭ ﺣﺎﻟﻴﻜﻪ ﺩﺭ
ﻣﺪﻝ ﺑﺎﻻ:
𝛾𝑡 = 𝑣𝑡 = 0ﭼﻮﻥ = 0
�
𝑢𝜕
𝑦𝜕
One Equation turbulence models:
ﺑﺴﻴﺎﺭﻱ ﺍﺯ ﻣﻌﺎﺩﻻﺕ ﻳﻚ ﻳﺎ ﭼﻨﺪ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺑﺮ ﭘﺎﻳﻪ ﺫﻳﻞ ﻫﺴﺘﻨﺪ:
۱
𝜇𝑡 = 𝜌𝑘 ۲ 𝑙 , k=Kinetic energy of turbulence , l=turbulent length scale
ﺍﻧﺘﻈﺎﺭ ﻣﻲ ﺭﻭﺩ ﻛﻪ 𝑘√ ﻳﻚ ﻣﻌﺮﻑ ﺑﻬﺘﺮﻱ ﺑﺮﺍﻱ vtﺩﺭ ﻣﻘﺎﻳﺴﻪ ﺑﺎ |
ﺭﺍﺑﻄﻪ ﻱ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ:
�
𝑢𝜕
𝑦𝜕
| limﺑﺎﺷﺪ.
ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ﻱ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻛﺲ ﺩﺭ ﺟﻬﺖ xﺭﺍ ﺑﻪ ﺍﺯﺍﻱ ﻧﻴﺮﻭﻫـﺎﻱ ﺟﺮﻣـﻲ ﺑﺮﺍﺑـﺮ ﺻـﻔﺮ،ﻗﺒﻞ ﺍﺯ ﻣﻴـﺎﻧﮕﻴﻦ
ﮔﻴﺮﻱ ﻧﺴﺒﺖ ﺑﻪ ﺯﻣﺎﻥ ﺩﺭ ) (𝑢� + 𝑢′ﺿﺮﺏ ﻛﻨﻴﻢ ﺩﺍﺭﻳﻢ:
𝜕
𝜕
𝜕
) 𝜌(𝑢� + 𝑢′ ) + (𝑢� + 𝑢′ )۲ 𝜌(𝑢� + 𝑢′ ) + (𝑢� + 𝑢′ )(𝑣̅ + 𝑣 ′ ) 𝜌(𝑢� + 𝑢′
𝑡𝜕
𝑦𝜕
𝑥𝜕
𝜕
𝑤() + (𝑢� + 𝑢′
) � + 𝑤 ′ ) 𝜌(𝑢� + 𝑢′
𝑧𝜕
𝜕
) (𝑝̅ + 𝑝′ ) + 𝜇(𝑢� + 𝑢′ )∇۲ (𝑢� + 𝑢′
) = −(𝑢� + 𝑢′
𝜕𝑥𝜕
) (𝑢� + 𝑢′
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺭﻭﺍﺑﻂ ﺫﻳﻞ:
𝜕
𝜌 𝜕
� 𝜌(𝑢� + 𝑢′ ) = � (𝑢� + 𝑢′ )2
𝑡𝜕
𝜕𝑡 2
𝜕
𝜌 𝜕
� 𝜌(𝑢� + 𝑢′ ) = (𝑢� + 𝑢′ ) � (𝑢� + 𝑢′ )2
𝑥𝜕
𝜕𝑥 2
۱۷۷
) (𝑢� + 𝑢′
(𝑢� + 𝑢′ )2
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝜕
𝜌 𝜕
� 𝜌(𝑢� + 𝑢′ ) = (𝑣̅ + 𝑣 ′ ) � (𝑢� + 𝑢′ )2
𝑦𝜕
𝜕𝑦 2
ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺑﺮﺍﻱ ﺟﻬﺖ xﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ:
) (𝑢� + 𝑢′ )(𝑣̅ + 𝑣 ′
𝜌𝜕
𝜌 𝜕
𝜌 𝜕
) (𝑢� + 𝑢′ )2 + (𝑢� + 𝑢′
) (𝑢� + 𝑢′ )2 + (𝑣̅ + 𝑣 ′
(𝑢� + 𝑢′ )2
𝜕𝑡 2
𝜕𝑥 2
𝜕𝑦 2
𝜌 𝜕
(𝑢� + 𝑢′ )2
𝑤( +
)� + 𝑤 ′
𝜕𝑧 2
𝜕
) = −(𝑢� + 𝑢′ ) (𝑝̅ + 𝑝′ ) + 𝜇(𝑢�∇2 𝑢� + 𝑢�∇2 𝑢′ + 𝑢′ ∇2 𝑢� + 𝑢′ ∇2 𝑢′
𝑥𝜕
ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺩﻳﮕﺮ ﺩﺭ ﺟﻬﺖ ﻫﺎﻱ yﻭ zﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ ﻛﻪ ﺷﻜﻞ ﻧﻬﺎﻳﻲ ﺁﻧﻬﺎ ﻋﺒﺎﺭﺕ ﺍﺳﺖ ﺍﺯ:
𝜌 𝜕
𝜌 𝜕
𝜌𝜕
) (𝑣̅ + 𝑣 ′ )2 + (𝑢� + 𝑢′
) (𝑣̅ + 𝑣 ′ )2 + (𝑣̅ + 𝑣 ′
(𝑣̅ + 𝑣 ′ )2
𝜕𝑡 2
𝜕𝑥 2
𝜕𝑦 2
𝜌 𝜕
(𝑣̅ + 𝑣 ′ )2
𝑤( +
)� + 𝑤 ′
𝜕𝑧 2
𝜕
)= −(𝑣̅ + 𝑣 ′ ) (𝑝̅ + 𝑝′ ) + 𝜇(𝑣̅ ∇2 𝑣̅ + 𝑣̅ ∇2 𝑣 ′ + 𝑣′∇2 𝑣̅ + 𝑣′∇2 𝑣′
𝑦𝜕
𝜌𝜕
𝜌 𝜕
𝜌 𝜕
𝑤(
𝑤(
𝑤(
) � + 𝑤 ′ )2 + (𝑢� + 𝑢′
) � + 𝑤 ′ )2 + (𝑣̅ + 𝑣 ′
� + 𝑤 ′ )2
𝜕𝑡 2
𝜕𝑥 2
𝜕𝑦 2
𝜌 𝜕
𝑤(
𝑤( +
)� + 𝑤 ′
� + 𝑤 ′ )2
𝜕𝑧 2
𝜕
𝑤(= −
𝑤(𝜇 � + 𝑤 ′ ) (𝑝̅ + 𝑝′ ) +
𝑤� +
𝑤 �∇2 𝑤 ′ + 𝑤′∇2
) � + 𝑤 ′ ∇2 𝑤 ′
𝑤 �∇2
𝑧𝜕
ﺳﻪ ﺭﺍﺑﻄﻪ ﻱ ﺑﺎﻻ ﺭﺍ ﺑﺎ ﻫﻢ ﺟﻤﻊ ﻣﻲ ﻛﻨﻴﻢ:
𝜌𝜕
𝜌𝜕
𝜌𝜕
𝜌 𝜕
(𝑢� + 𝑢′ )2 +
(𝑣̅ + 𝑣 ′ )2 +
𝑤(
(𝑢� + 𝑢′ )2
) � + 𝑤 ′ )2 + (𝑢� + 𝑢′
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑥 2
𝜌 𝜕
𝜌 𝜕
𝜕
) (𝑣̅ + 𝑣 ′ )2 + (𝑢� + 𝑢′
𝑤(
� + 𝑤 ′ )2 + (𝑣̅ + 𝑣 ′ ) (𝑢� + 𝑢′ )2
𝜕𝑥 2
𝜕𝑥 2
𝑦𝜕
) +(𝑢� + 𝑢′
𝜌 𝜕
𝜌 𝜕
𝜌 𝜕
) (𝑣̅ + 𝑣 ′ )2 + (𝑣̅ + 𝑣 ′
𝑤(
(𝑢� + 𝑢′ )2
𝑤( � + 𝑤 ′ )2 +
)� + 𝑤 ′
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑧 2
𝜌 𝜕
𝜌 𝜕
𝑤( (𝑣̅ + 𝑣 ′ )2 +
𝑤(
� + 𝑤 ′ )2
)� + 𝑤 ′
𝜕𝑧 2
𝜕𝑧 2
𝑤(+
)� + 𝑤 ′
𝜕
𝜕
𝜕
𝑤( (𝑝̅ + 𝑝′ ) − (𝑣̅ + 𝑣 ′ ) (𝑝̅ + 𝑝′ ) −
) � + 𝑤 ′ ) (𝑝̅ + 𝑝′
𝑥𝜕
𝑦𝜕
𝑧𝜕
) +(𝑣̅ + 𝑣 ′
) = −(𝑢� + 𝑢′
) +𝜇(𝑢�∇2 𝑢� + 𝑢�∇2 𝑢′ + 𝑢′ ∇2 𝑢� + 𝑢′ ∇2 𝑢′ ) + 𝜇(𝑣̅ ∇2 𝑣̅ + 𝑣̅ ∇2 𝑣 ′ + 𝑣′∇2 𝑣̅ + 𝑣 ′ ∇2 𝑣 ′
𝑤(𝜇 +
𝑤 �∇2
𝑤� +
𝑤 �∇2 𝑤 ′ + 𝑤′∇2
) � + 𝑤 ′ ∇2 𝑤 ′
۱۷۸
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﻓﺼﻞ ٦
ﺣﺎﻝ ﻣﻌﺎﺩﻟﻪ ﻗﺒﻞ ﺭﺍ ﻧﺴﺒﺖ ﺑﻪ ﺯﻣﺎﻥ ﻣﻴﺎﻧﮕﻴﺮﻱ ﻣﻲ ﻛﻨﻴﻢ:
����
̅𝑣′ 2 + 2
𝑣 ���2 +
�����
̅𝑣�
𝑣 ′� +
+
𝜌 𝜕
����
𝑢′ 2 + 2
���
�����
𝑢�
𝑢 �2 +
� 𝑢′ � +
𝜌 𝜕
��������������������������������
𝜕
𝜕
𝜕
=) ( ( )+ ( )+
𝜕𝑡 2
𝜕𝑡 2
𝜕 𝜌 2
𝜕 𝜌 2
𝜕 𝜌 2
𝜕 𝜌 ′2
𝜕 𝜌 ′2
𝑢� +
𝑣̅ +
𝑤
� +
𝑢� +
�𝑣
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑡 2
𝜕𝑡 2
2
2
𝜌 𝜕
𝑘𝜌 𝜕 ���′ 2 � = 𝜕 𝜌𝑘� +
𝑤
� 2) +
𝑤 �𝑢�′ + 𝑣�′ +
𝜕𝑡 2
𝑡𝜕
𝑡𝜕
𝑡𝜕
𝑡𝜕
�����
𝑤′ 2 + 2
����
������
𝑤�
𝑤 �2 +
= � �𝑤 ′
(𝑢�2 + 𝑣̅ 2 +
𝜌 𝜕
𝜕𝑡 2
= ���′ 2
𝑤
𝑡𝜕
𝜌 𝜕
𝜕𝑡 2
𝜌 𝜕
𝜕𝑡 2
�����������������������������������������������������������������������������������������������
𝜌 𝜕
𝜌 𝜕
𝜌 𝜕
) (𝑢� + 𝑢′
) (𝑢� + 𝑢′ )2 + (𝑢� + 𝑢′
) (𝑣̅ + 𝑣 ′ )2 + (𝑢� + 𝑢′
𝑤(
� + 𝑤 ′ )2
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥 2
𝜕 𝜌 2
𝜕 𝜌 2
𝜕 𝜕 𝜌 ′2
2
() �𝑢� + 𝑢�′ + 2𝑢�𝑢′ � = (𝑢� + 𝑢′
𝑢� +
𝑢 +
) 𝜌𝑢�𝑢′
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥 2
𝑥𝜕
) (𝑢� + 𝑢′
𝜕 𝜌 2
𝜕 𝜌 2
𝜕
𝜕 𝜌 2
𝜕 𝜌 ′2
𝜕
�𝑢 𝑢� +
�𝑢 𝑢′ +
𝜌𝑢�𝑢′ + 𝑢′
𝑢� + 𝑢′
𝑢 + 𝑢′ 𝜌𝑢�𝑢′
𝜕𝑥 2
𝑥𝜕
𝜕𝑥 2
𝜕𝑥 2
𝑥𝜕
𝜕𝑥 2
�𝑢 =
������������
𝜕 𝜌 2
𝜕 𝜌 2
�𝑢
�𝑢 = �𝑢
�𝑢
𝜕𝑥 2
𝜕𝑥 2
������������
𝜕 𝜌 2
𝜕 𝜌 ����2
�𝑢
�𝑢 = 𝑢′
𝑢′
𝜕𝑥 2
𝜕𝑥 2
������������
𝜕
𝜕
������
�𝑢
�𝑢 = 𝜌𝑢�𝑢′
𝜌𝑢�𝑢′ = 0
𝑥𝜕
𝑥𝜕
������������
𝜕 𝜌 2
𝑢′
𝑢� = 0
𝜕𝑥 2
�������������
𝜕 𝜌 ′2
𝜕 𝜌 ����2
𝑢′
𝑢 = 𝑢�′
𝑢′
𝜕𝑥 2
𝜕𝑥 2
�������������
𝜕
𝜕
𝜕𝑢′
�𝑢𝜕
�𝑢𝜕
�𝑢𝜌 𝑢′ 𝜌𝑢�𝑢′ = 𝑢′ �𝑢′ 𝜌𝑢� +
� = 𝑢′2
+ 𝜌𝑢�𝑢′
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
�������
�������
�������������
𝜕
�
𝜕𝑢′
�
𝜕𝑢′
𝑢𝜌𝜕 2
𝑢𝜕 2
′
′
′
�
�
+ 𝜌𝑢�𝑢′
𝑢𝜌 =
+ 𝜌𝑢�𝑢′
𝑢 = 𝑢�𝑢𝜌 → 𝑢′
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
�������
�������������
����������������������������� ɪ
𝜌 𝜕
𝜕 𝜌 2
𝜕 𝜌 ����2
� 2 𝜕𝑢� + 𝑢′ 𝜕 𝜌 𝑢′2 + 𝜌𝑢�𝑢′ 𝜕𝑢′
) ⇒ (𝑢� + 𝑢′
�𝑢 = (𝑢� + 𝑢′ )2
�𝑢 𝑢� +
𝑢′ + 𝜌𝑢′
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥 2
𝑥𝜕
𝜕𝑥 2
𝑥𝜕
����������������������������� ɪɪ
𝜌 𝜕
) ⇒ (𝑢� + 𝑢′
(𝑣̅ + 𝑣 ′ )2
𝜕𝑥 2
������������
������������
������������ 𝜕 𝜌 2
������������ 𝜕 𝜌 2
𝜕
������������� 𝜕 𝜌 2
������������� 𝜕 𝜌 2
𝜕
�𝑢 =
�𝑢 𝑣̅ +
�𝑢 𝑣′ +
𝜌𝑣̅ 𝑣′ + 𝑢′
𝑣̅ + 𝑢′
𝑣′ + 𝑢′ 𝜌𝑣̅ 𝑣′
𝜕𝑥 2
𝜕𝑥 2
𝑥𝜕
𝜕𝑥 2
𝜕𝑥 2
𝑥𝜕
۱۷۹
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
٦ ﻓﺼﻞ
�������������
�����������
�������
𝜕𝜌𝑣̅
𝜕𝜌𝑣 ������������
𝜕𝑣′ �����������
𝜕𝜌𝑣̅
𝜕𝑣
𝜕
𝜕𝑣 ′
𝑢′ 𝜌𝑣̅ 𝑣′ = 𝑢′ �𝑣 ′
+ 𝜌𝑣̅
� = 𝑢′ 𝑣 ′
+ 𝜌𝑣� 𝑢′
= 𝑢′ 𝑣 ′
+ 𝜌𝑣̅ 𝑢′
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
𝜕𝑥
�����������
𝜕𝑣̅
= 𝜌𝑢′ 𝑣 ′
+˶
𝜕𝑥
⇒ ɪɪ = 𝑢�
�������
𝜕 𝜌 2
𝜕 𝜌 �2
𝜕𝑣̅ �������������
𝜕 𝜌 2
𝜕𝑣′
′ 𝑣′
�����
𝑣̅ + 𝑢�
𝑣′ + 𝜌𝑢
+ 𝑢′
𝑣′ + 𝜌𝑣̅ 𝑢′
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥
𝜕𝑥 2
𝜕𝑥
�������������������������������������
𝜕 𝜌
(𝑤
ɪɪɪ = (𝑢� + 𝑢′ )
� + 𝑤 ′ )2
𝜕𝑥 2
𝜕 𝜌 2
𝜕 𝜌 ���2
𝜕𝑤
� ��������������
𝜕 𝜌 2 ������������
𝜕𝑤′
′ 𝑤′
������
= 𝑢�
𝑤
� + 𝑢�
𝑤′ + 𝜌𝑢
+ 𝑢′
𝑤′ + 𝜌𝑤
�𝑢′
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥
𝜕𝑥 2
𝜕𝑥
𝜕 𝜌 ′2
𝜕𝑣̅
𝜕 𝜌 2
2
′ 𝑣′
� 2 𝜕𝑢� + 𝜌𝑢
�����
���′ 2 � + 𝜌𝑢′
(𝑢� + 𝑣̅ 2 + 𝑤
� 2 ) + 𝑢�
�𝑢� + 𝑣�′ + 𝑤
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥
𝜕𝑥
��������
�������
�������������������������������������������
𝜕𝑤
�
𝜕
𝜌
𝜕
𝜌
𝜕
𝜌
𝜕𝑢′
𝜕𝑣′
′ 𝑤′
������
+ 𝜌𝑢
+ 𝑢′ (
𝑢′ 2 +
𝑣 ′2 +
𝑤 ′ 2 ) + 𝜌𝑢�𝑢′
+ 𝜌𝑣̅ 𝑢′
𝜕𝑥
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥
𝜕𝑥
��������
𝜕𝑤′
+ 𝜌𝑤
�𝑢′
𝜕𝑥
(ɪ + ɪɪ + ɪɪɪ)𝑥 = 𝑢�
𝜕 𝜌 ′2
𝜕
𝜕 𝜌
𝜕 𝜌 2
(𝑢� + 𝑢′ )2 = (𝑣̅ + 𝑣 ′ ) �
𝑢� +
𝑢 +
𝜌𝑢�𝑢′ �
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕 𝜌 2
𝜕 𝜌 2
𝜕
𝜕 𝜌 2
𝜕 𝜌 2
𝜕
= 𝑣̅
𝑢� + 𝑣̅
𝑢′ + 𝑣̅
𝜌𝑢�𝑢′ + 𝑣 ′
𝑢� + 𝑣 ′
𝑢′ + 𝑣 ′
𝜌𝑢�𝑢′
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
(ɪ)𝑦 = (𝑣̅ + 𝑣 ′ )
����𝑦 = 𝑣̅
⇒ (ɪ)
��������
𝜕 𝜌 2
𝜕 𝜌 � 2 �������������
𝜕 𝜌 2
𝜕𝑢�
𝜕𝑢′
′𝑣 ′
������
𝑢� + 𝑣̅
𝑢′ + 𝑣 ′
𝑢′ + 𝜌𝑢
+ 𝜌𝑢�𝑣 ′
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕𝑦
(ɪɪ)𝑦 = (𝑣̅ + 𝑣 ′ )
𝜕 𝜌
(𝑣̅ + 𝑣 ′ )2
𝜕𝑦 2
����𝑦 = 𝑣̅
⇒ (ɪɪ)
�������
�������������
𝜕 𝜌 2
𝜕 𝜌 �2
� 2 𝜕𝑣̅ + 𝑣 ′ 𝜕 𝜌 𝑣′2 + 𝜌𝑣̅ 𝑣 ′ 𝜕𝑣′
𝑣̅ + 𝑣̅
𝑣′ + 𝜌𝑣′
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕𝑦 2
𝜕𝑦
�������
⇒ (ɪɪɪ)
𝑦 = 𝑣̅
��������
𝜕 𝜌 2
𝜕 𝜌 �����
𝜕𝑤
� ��������������
𝜕 𝜌 ′2
𝜕𝑤 ′
′𝑤 ′
����
������
𝑤 + 𝑣̅
𝑤 ′ 2 + 𝜌𝑣
+ 𝑣′
𝑤 + 𝜌𝑤
�𝑣 ′
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕𝑦 2
𝜕𝑦
(ɪɪɪ)𝑦 = (𝑣̅ + 𝑣 ′ )
𝜕 𝜌
(𝑤
� + 𝑤′)2
𝜕𝑦 2
(ɪ + ɪɪ + ɪɪɪ)𝑦 = 𝑣̅
𝜕 𝜌 2
𝜕 𝜌 ′2
𝜕𝑢�
2
′𝑣 ′
������
(𝑢� + 𝑣̅ 2 + 𝑤
� 2 ) + 𝑣̅
�𝑢� + 𝑣�′ + 𝑤
� 2 � + 𝜌𝑢
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
��������
�������
𝜕𝑣̅
𝜕𝑤
� �������������������������������������������
𝜕 𝜌 ′2
𝜕 𝜌 ′2
𝜕 𝜌 ′2
𝜕𝑢′
𝜕𝑣′
2
′
′
′
����
������
+𝜌𝑣′
+ 𝜌𝑣 𝑤′
+𝑣 (
𝑢 +
𝑣 +
𝑤 ) + 𝜌𝑢�𝑣
+ 𝜌𝑣̅ 𝑣 ′
𝜕𝑦
𝜕𝑦
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦
𝜕𝑦
��������
𝜕𝑤′
+ 𝜌𝑤
�𝑣 ′
𝜕𝑦
۱۸۰
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
(ɪ)𝑧 = (𝑤
� + 𝑤 ′)
����𝑧 = 𝑤
⇒ (ɪ)
�
𝜕 𝜌
(𝑢� + 𝑢′ )2
𝜕𝑧 2
��������
𝜕 𝜌 2
𝜕 𝜌 ����2 ��������������
𝜕 𝜌 2
𝜕𝑢�
𝜕𝑢′
′ 𝑤′
������
𝑢� + 𝑤
�
𝑢′ + 𝑤 ′
𝑢′ + 𝜌𝑢
+ 𝜌𝑢�𝑤 ′
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
𝜕𝑧
(ɪɪ)𝑧 = (𝑤
� + 𝑤 ′)
����𝑧 = 𝑤
(ɪɪ)
�
𝜕 𝜌
(𝑣̅ + 𝑣′)2
𝜕𝑧 2
��������
𝜕 𝜌 2
𝜕 𝜌 � 2 �������������
𝜕 𝜌 2
𝜕𝑣̅
𝜕𝑣′
′ 𝑤′
������
𝑣̅ + 𝑤
�
𝑣′ + 𝑤 ′
𝑣′ + 𝜌𝑣
+ 𝜌𝑣̅ 𝑤 ′
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
𝜕𝑧
(ɪɪɪ)𝑧 = (𝑤
� + 𝑤 ′)
����𝑧 = 𝑤
(ɪɪɪ)
�
𝜕 𝜌
(𝑤
� + 𝑤′)2
𝜕𝑧 2
���������
𝜕 𝜌 2
𝜕 𝜌 ���2
� ��������������
𝜕 𝜌 2
𝜕𝑤′
����2 𝜕𝑤
𝑤
� +𝑤
�
𝑤′ + 𝜌𝑤′
+ 𝑤′
𝑤′ + 𝜌𝑤
�𝑤 ′
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
𝜕𝑧 2
𝜕𝑧
𝜕 𝜌 2
𝜕 𝜌 ����
𝜕𝑢�
�����
′ 2 � + 𝜌𝑢
′𝑤 ′
����������������
������
(ɪ + ɪɪ + ɪɪɪ)𝑧 = 𝑤
(𝑢� + 𝑣̅ 2 + 𝑤
�
� 2) + 𝑤
�
�𝑢′ 2 + ����
𝑣 ′2 + 𝑤
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
��������
��������
𝜕𝑣̅
� ������������������������������������������
𝜕 𝜌 2 𝜕 𝜌 2 𝜕 𝜌 2
𝜕𝑢′
𝜕𝑣′
′ 𝑤′
������
����2 𝜕𝑤
+𝜌𝑣
+ 𝜌𝑤′
+ 𝑤 ′(
𝑢′ +
𝑣′ +
𝑤′ ) + 𝜌𝑢�𝑤 ′
+ 𝜌𝑣̅ 𝑤 ′
𝜕𝑧
𝜕𝑧
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧
𝜕𝑧
��������
𝜕𝑤′
+ 𝜌𝑤
�𝑤′
𝜕𝑧
𝜕
𝜕
𝜕
𝜕
���������������
(ɪ + ɪɪ + ɪɪɪ)𝑥 + (ɪ��������������
+ ɪɪ + ɪɪɪ)𝑦 + (ɪ��������������
+ ɪɪ + ɪɪɪ)𝑧 + 𝜌𝑘� + 𝜌𝑘 = 𝜌𝑘� + 𝜌𝑘
𝜕𝑡
𝜕𝑡
𝜕𝑡
𝜕𝑡
+𝑢�
+𝑢�
+𝑤
�
𝜕 𝜌 2
𝜕 𝜌 2
𝜕 𝜌 2
(𝑢� + 𝑣̅ 2 + 𝑤
(𝑢� + 𝑣̅ 2 + 𝑤
(𝑢� + 𝑣̅ 2 + 𝑤
� 2 ) + 𝑣̅
� 2) + 𝑤
� 2)
�
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
𝜕 𝜌 ����
𝜕 𝜌 ����
����
�����
�����
′2 + 𝑤
′ 2 � + 𝑣̅
′2�
�𝑢′ 2 + 𝑣
�𝑢′ 2 + ����
𝑣 ′2 + 𝑤
𝜕𝑥 2
𝜕𝑦 2
��������������������������������������������
𝜕 𝜌 ����
𝜕 𝜌 ′2 𝜕 𝜌 ′2 𝜕 𝜌 ′2
�����
′ 2 � + 𝑢′ �
�𝑢′ 2 + ����
𝑣 ′2 + 𝑤
𝑢 +
𝑣 +
𝑤 �
𝜕𝑧 2
𝜕𝑥 2
𝜕𝑥 2
𝜕𝑥 2
��������������������������������������������
��������������������������������������������
𝜕 𝜌 ′2
𝜕 𝜌 ′2
𝜕 𝜌 ′2
𝜕 𝜌 ′2 𝜕 𝜌 ′2 𝜕 𝜌 ′2
+𝑣 ′ �
𝑢 +
𝑣 +
𝑤 � + 𝑤′ �
𝑢 +
𝑣 +
𝑤 �
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑦 2
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑧 2
𝜕𝑢�
𝜕𝑣̅
𝜕𝑤
�
𝜕𝑣̅ 𝜕𝑢�
𝜕𝑤
� 𝜕𝑢�
����
����
�����
′2
′2
′2
′𝑣 ′ �
′𝑤 ′ �
������
������
+𝜌𝑢
+ 𝜌𝑣
+ 𝜌𝑤
+ 𝜌𝑢
+ � + 𝜌𝑢
+ �
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥 𝜕𝑦
𝜕𝑥 𝜕𝑧
′𝑤 ′ �
������
+𝜌𝑣
��������
𝜕𝑤
� 𝜕𝑣̅
𝜕𝑢′ ��������
𝜕𝑢′ ��������
𝜕𝑢′
+ � + 𝜌𝑢� �𝑢′
+ 𝑣′
+ 𝑤′
�
𝜕𝑦 𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
��������
��������
𝜕𝑣 ′ ��������
𝜕𝑣 ′ ��������
𝜕𝑣 ′
𝜕𝑤′ ��������
𝜕𝑤′ ���������
𝜕𝑤′
+𝜌𝑣̅ �𝑢′
+ 𝑣′
+ 𝑤′
� + 𝜌𝑤
�(𝑢′
+ 𝑣′
+ 𝑤′
)
𝜕𝑥
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
۱۸۱
٦ ﻓﺼﻞ
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝜕𝑢′ 𝑢′
𝜕𝑢′
𝜕𝑢′
= 𝑢′
+ 𝑢′
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝜕𝑢′ 𝑣′
𝜕𝑣′
𝜕𝑢′
= 𝑢′
+ 𝑣′
𝑦𝜕
𝑦𝜕
𝑦𝜕
𝜕𝑢′ 𝑤′
𝜕𝑤′
𝜕𝑢′
= 𝑢′
+ 𝑤′
𝑧𝜕
𝑧𝜕
𝑧𝜕
𝜕𝑢′
𝜕𝑢′
𝜕𝑢′ 𝜕𝑢′ 𝑢′
𝜕𝑢′ 𝜕𝑢′ 𝑣 ′
𝜕𝑣 ′ 𝜕𝑢′ 𝑤 ′
+ 𝑣′
+ 𝑤′
=
− 𝑢′
+
− 𝑢′
+
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑦𝜕
𝑧𝜕
⇒ 𝑢′
𝜕𝑤 ′ 𝜕𝑢′ 𝑢′ 𝜕𝑢′ 𝑣 ′ 𝜕𝑢′ 𝑤 ′
𝜕𝑢′ 𝜕𝑣′ 𝜕𝑤′
=
+
+
( �𝑢 −
+
+
)
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑦𝜕 𝑥𝜕
𝑧𝜕
−𝑢′
′ 𝑢′
′ 𝑣′
′ 𝑤′
�����
�����
������
��������
�������� 𝜕𝑢′
�������� 𝜕𝑢′
𝜕𝑢′
𝑢𝜕
𝑢𝜕
𝑢𝜕
+ 𝑣′
+ 𝑤′
(�𝑢𝜌 = �
+
+
)
𝜌𝑢� �𝑢′
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
′ 𝑣′
′ 𝑣′
′ 𝑤′
�����
�����
������
��������
�������� 𝜕𝑣 ′
�������� 𝜕𝑣 ′
𝜕𝑣 ′
𝑢𝜕
𝑣𝜕
𝑣𝜕
′
′
′
𝑢� ̅𝑣𝜌
𝑣+
𝑤+
( ̅𝑣𝜌 = �
+
+
)
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
′ 𝑤′
′ 𝑤′
𝑣𝜕 ������
������
������
��������
�������� 𝜕𝑤 ′
��������� 𝜕𝑤 ′
𝜕𝑤 ′
𝜕𝑢′𝑤′
𝑤𝜕
𝑤𝜌
� �𝑢′
+ 𝑣′
+ 𝑤′
𝑤𝜌 = �
(�
+
+
)
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
��������������������������
��������������������������
𝜕
𝜕𝑝̅ 𝜕𝑝′
(𝑢� + 𝑢′ ) (𝑝̅ + 𝑝′ ) = (𝑢� + 𝑢′ ) � +
�
𝑥𝜕
𝑥𝜕 𝑥𝜕
������
������� 𝜕𝑝′
�������� ̅𝑝𝜕
𝜕𝑝′
�������� ̅𝑝𝜕
𝜕𝑝′
������� ̅𝑝𝜕
�𝑢 +
+ 𝑢′
+ 𝑢′
�𝑢 =
+ 𝑢′
�𝑢 =
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝑥𝜕
�������������������������
𝜕
�������� ̅𝑝𝜕
𝜕𝑝′
̅𝑣 = ) (𝑣̅ + 𝑣 ′ ) (𝑝̅ + 𝑝′
+ 𝑣′
𝑦𝜕
𝑦𝜕
𝑦𝜕
��������������������������
𝜕𝑝′
𝜕
�������� ̅𝑝𝜕
′
′
𝑤(
𝑤 = ) 𝑝 � + 𝑤 ) (𝑝̅ +
�
+ 𝑤′
𝑧𝜕
𝑧𝜕
𝑧𝜕
)′ ∇2 𝑢′
�����������������������������������������������
��������
𝑢(𝜇
𝑢 �∇2 𝑢� + 𝑢�∇2 𝑢′ + +𝑢′ ∇2 𝑢� + 𝑢′ ∇2 𝑢′) = 𝜇(𝑢�∇2 𝑢� +
��������������������������������������������
���������
̅𝑣(𝜇
∇2 𝑣̅ + 𝑣̅ ∇2 𝑣′ + 𝑣 ′ ∇2 𝑣̅ + 𝑣 ′ ∇2 𝑣′) = 𝜇(𝑣̅ ∇2 𝑣̅ + 𝑣′
)∇2 𝑣′
) ′ ∇2 𝑤 ′
�������������������������������������������������
����������
𝑤(𝜇
𝑤 �∇2
𝑤� +
𝑤 �∇2 𝑤′ + 𝑤 ′ ∇2
𝑤(𝜇 = )� + 𝑤′ ∇2 𝑤′
𝑤 �∇2
𝑤� +
ﺑﺎ ﺟﺎﻳﮕﺰﻳﻨﻲ ،ﻣﻌﺎﺩﻟﻪ ﺑﺪﺳﺖ ﺁﻣﺪﻩ ﺭﺍ ﺑﻪ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺗﻔﻜﻴﻚ ﻣﻲ ﻛﻨﻴﻢ:
(1ﻣﻌﺎﺩﻟﻪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﻣﻴﺎﻧﮕﻴﻦ ﺟﺮﻳﺎﻥ:
۱۸۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝜕
𝜕 𝜌 2
𝜕 𝜌 2
𝜕 𝜌 2
𝑤 (𝑢� + 𝑣̅ 2 +
𝑤 (𝑢� + 𝑣̅ 2 +
𝑤 (𝑢� + 𝑣̅ 2 +
�𝑢 𝜌𝑘� +
̅𝑣 � 2 ) +
𝑤 � 2) +
�
)� 2
𝑡𝜕
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
′ 𝑢′
′𝑣 ′
′𝑤 ′
′𝑣 ′
′𝑣 ′
′𝑤 ′
������
������
������
������
������
������
𝑢𝜕
𝑢𝜕
𝑢𝜕
𝑢𝜕
𝑣𝜕
𝑣𝜕
� �𝑢𝜌 +
+
+
� ̅𝑣𝜌 � +
+
+
�
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
′𝑤 ′
′𝑤 ′
′𝑤 ′
������
������
�������
𝑢𝜕
𝑣𝜕
𝑤𝜕
𝑤𝜌 +
��
+
+
�
𝑥𝜕
𝑦𝜕
𝑧𝜕
̅𝑝𝜕
̅𝑝𝜕
̅𝑝𝜕
�𝑢= −
̅𝑣 −
𝑤−
�
𝑤 + 𝜇(𝑢�∇2 𝑢� + 𝑣̅ ∇2 𝑣̅ +
𝑤 �∇2
)�
𝑦𝜕
𝑧𝜕
𝑥𝜕
(2ﻣﻌﺎﺩﻟﻪ ﺍﻧﺮژﻱ ﻧﻮﺳﺎﻧﺎﺕ ﺟﺮﻳﺎﻥ:
𝜕
���� 𝜌 𝜕
���� 𝜌 𝜕
����
�����
�����
𝑤 ′2 +
̅𝑣 ′ 2 � +
�′2
�𝑢 𝜌𝑘� +
𝑣 �𝑢′ 2 +
���� �𝑢′ 2 +
𝑤 𝑣 ′2 +
𝜕𝑥 2
𝜕𝑦 2
𝑡𝜕
�������������������������������
���� 𝜌 𝜕
𝜕 𝜌 ′2
�����
′ 2 � + 𝑢′
) (𝑢 + 𝑣 ′ 2 + 𝑤 ′ 2
���� �𝑢′ 2 +
𝑤 𝑣 ′2 +
𝜕𝑧 2
𝜕𝑥 2
𝑤+
�
�������������������������������
����������������������������
𝜕 𝜌 ′2
𝜕 𝜌 ′2
�𝑢𝜕
̅𝑣𝜕
����
����
′2
′2
(𝑢 + 𝑣 ′ 2 + 𝑤 ′ 2 ) + 𝑤 ′
𝑢𝜌 (𝑢 + 𝑣 ′ 2 + 𝑤 ′ 2 ) +
+𝑣 ′
𝑣𝜌 +
𝜕𝑦 2
𝜕𝑧 2
𝑥𝜕
𝑦𝜕
�𝑢𝜕 ̅𝑣𝜕
𝑤𝜕
�𝑢𝜕 �
𝑤𝜕
̅𝑣𝜕 �
𝑤𝜕
�
�����
′2
� ′𝑣 ′
� ′𝑤 ′
� ′𝑤 ′
������
������
������
𝑤𝜌+
𝑢𝜌 +
𝑢𝜌 + � +
𝑣𝜌 + � +
� +
𝑦𝜕 𝑥𝜕
𝑧𝜕 𝑥𝜕
𝑧𝜕 𝑦𝜕
𝑧𝜕
�������� 𝜕
�������� 𝜕
��������� 𝜕
𝑢(𝜇 ′ 𝑝 ′ ) +
𝑣 ′ ∇2 𝑢′ +
𝑤 ′ ∇2 𝑣′ +
)′ ∇2 𝑤′
��������
��������
���������
(𝑢′ 𝑝′ ) −
𝑤( (𝑣 ′ 𝑝′ ) −
𝑥𝜕
𝑦𝜕
𝑧𝜕
=−
𝜕𝑝′
𝜕𝑢′
𝜕 ′ ′
(𝑢 𝑝 ) = 𝑢′
+ 𝑝′
𝑥𝜕
𝑥𝜕
𝑥𝜕
𝜕 ′ ′
𝜕𝑝′
𝜕𝑢′
(𝑣 𝑝 ) = 𝑣 ′
+ 𝑝′
𝑦𝜕
𝑦𝜕
𝑦𝜕
𝜕
𝜕𝑝′
𝜕𝑤′
(𝑤 ′ 𝑝′ ) = 𝑤 ′
+ 𝑝′
𝑧𝜕
𝑧𝜕
𝑧𝜕
𝜕𝑝′
𝜕𝑝′
𝜕𝑝′
𝜕 ′ ′
𝜕𝑢′
𝜕 ′ ′
𝜕𝑣 ′
(𝑢 𝑝 ) − 𝑝′
(𝑣 𝑝 ) − 𝑝′
+ 𝑣′
+ 𝑤′
=
+
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕 𝑥𝜕
𝑦𝜕
⇒ 𝑢′
𝜕
𝜕𝑤 ′
𝜕 ′ ′
𝜕 ′ ′
𝜕
(𝑤 ′ 𝑝′ ) − 𝑝′
(𝑢 𝑝 ) +
) (𝑣 𝑝 ) + (𝑤 ′ 𝑝′
=
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑧𝜕
+
𝜕𝑢′ 𝜕𝑣 ′ 𝜕𝑤 ′
𝜕 ′ ′
𝜕 ′ ′
𝜕
(𝑢 𝑝 ) +
) (𝑣 𝑝 ) + (𝑤 ′ 𝑝′
� 𝑝−
+
+
=�
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
�������������
0
′
1 ����2 ����2 ����2
𝐾 = (𝑢′
) + 𝑣′ + 𝑤′
2
۱۸۳
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝜕
𝜕
𝜕
𝜕
𝐷
𝑤 𝜌𝑘 + 𝑢� 𝜌𝑘 + 𝑣̅ 𝜌𝑘 +
= 𝑘𝜌 �
)𝑘𝜌(
𝑡𝜕
𝑡𝜕
𝑡𝜕
𝑡𝜕
𝑡𝐷
𝜕
𝑢′2 + 𝑣′2 + 𝑤′2
𝜕
𝑢′2 + 𝑣′2 + 𝑤′2
𝑢′2 + 𝑣′2 + 𝑤′2 𝜕𝑢′
�𝜌𝑢′ .
𝜌� � = 𝑢′
𝜌�+
.
𝑥𝜕
𝑥𝜕
2
𝑥𝜕
2
2
�����������������������
������������������������������
���������������������������������
𝜕 𝜌 2
𝜕 𝜌 ′
𝑢′2 + 𝑣′2 + 𝑤′2 𝜕𝑢′
⇒ 𝑢′
= ) (𝑢′ + 𝑣′2 + 𝑤′2
𝜌 ( . 𝑢 . (𝑢′2 + 𝑣′2 + 𝑤′2 ) −
𝜕𝑥 2
𝜕𝑥 2
2
𝑥𝜕
�����������������������
������������������������������
���������������������������������
𝜕 𝜌 2
𝜕 𝜌 ′
𝑢′2 + 𝑣′2 + 𝑤′2 𝜕𝑣′
′
2
2
2
2
2
𝑣
= ) (𝑢′ + 𝑣′ + 𝑤′
𝜌 ( . 𝑣 . (𝑢′ + 𝑣′ + 𝑤′ ) −
𝜕𝑦 2
𝜕𝑦 2
2
𝑦𝜕
������������������������
�������������������������������
����������������������������������
𝜕 𝜌 2
𝜕 𝜌 ′
𝑢′2 + 𝑣′2 + 𝑤′2 𝜕𝑤′
= ) (𝑢′ + 𝑣′2 + 𝑤′2
𝜌 ( . 𝑤 . (𝑢′2 + 𝑣′2 + 𝑤′2 ) −
𝑤′
�������������
𝜕𝑧 2
𝜕𝑧 2
2
𝑧𝜕
∑….=0
2
𝜕2 1 ′2
𝜕2 1 ′2
𝑢′
𝜕2 1 ′2
� 𝑤 �∇ � � = 2� 𝑢 �+ 2� 𝑣 �+ 2
2
𝜕𝑥 2
𝜕𝑦 2
𝜕𝑧 2
2
𝜕𝑢′
𝜕 2 𝑢′
𝜕 1 ′2
𝜕2 1 ′2
𝜕𝑢′ 2
′
′
⇒
𝑢
=
𝑢
+
(
)
𝑢=� 𝑢 �
�
�
𝜕𝑥 2
𝑥𝜕
𝜕𝑥 2 2
𝜕𝑥 2
𝑥𝜕
2
𝑢′
𝜕 2 𝑢′
𝜕𝑢′
𝜕 2 𝑢′
𝜕𝑢′ 2
𝜕 2 𝑢′
𝜕𝑢′
′
+
(
)
+
𝑢
+ ( )2
∇ � � = 𝑢′ 2 + ( )2 + 𝑢′
2
2
2
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑦𝜕
𝑧𝜕
𝑧𝜕
2
𝜕𝑢′ 2
𝜕𝑢′
𝜕𝑢′
) + ( )2 + ( )2
𝑥𝜕
𝑦𝜕
𝑧𝜕
2
2
( = 𝑢′ ∇2 𝑢′ +
2
𝑢′2
𝜕𝑢′
𝜕𝑢′
𝜕𝑢′
� �𝑢 ∇ 𝑢 =∇ � �−� � −� � −
2
𝑥𝜕
𝑦𝜕
𝑧𝜕
2
2
2
2
2
′ 2 ′
⇒
𝑣′2
𝜕𝑣 ′
𝜕𝑣 ′
𝜕𝑣 ′
� �𝑣 ∇ 𝑣 =∇ � �−� � −� � −
2
𝑥𝜕
𝑦𝜕
𝑧𝜕
2
2
2
′ 2 ′
𝑤′2
𝜕𝑤 ′
𝜕𝑤 ′
𝜕𝑤 ′
�𝑤 ∇ 𝑤 =∇ � �−
�� −
�� −
�
2
𝑥𝜕
𝑦𝜕
𝑧𝜕
2
′
′ 2
ﺩﺭ ﻧﺘﻴﺠﻪ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺍﻏﺘﺸﺎﺷﺎﺕ ﺑﺮﺍﻱ ﻫﺮ ﻭﺍﺣﺪ ﺟﺮﻡ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻣﻲ ﺁﻳﺪ:
ɪ
ɪɪ
ɪɪ
�
����𝐷
����������������� 𝜌 𝜕
) ′ 𝑝′
������
������������������������� (𝜌𝑘) = − ( .
𝑢 𝑢′ (𝑢′ 2 + 𝑣 ′ 2 + 𝑤 ′ 2 ) +
𝑡𝐷
𝜕𝑥 2
ɪɪ
����������������� 𝜌 𝜕
������������������� 𝜌 𝜕
) ′ 𝑝′
������
������������������������� − ( .
������ 𝑣 ′ (𝑢′ 2 + 𝑣 ′ 2 + 𝑤 ′ 2 ) +
������������������������� 𝑣 ′ 𝑝′ ) − ( .
𝑤 𝑤 ′ (𝑢′ 2 + 𝑣 ′ 2 + 𝑤 ′ 2 ) +
𝜕𝑧 2
𝜕𝑦 2
۱۸٤
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ɪɪɪ
ɪɪɪ
ɪɪɪ
��
�
��
𝑢𝜕
̅𝑣𝜕
𝑤𝜕
′ 𝑢′
′𝑣 ′
′𝑤 ′
������
������
�������
𝑢𝜌−
𝑣𝜌 −
𝑤𝜌 −
𝑥𝜕
𝑦𝜕
𝑧𝜕
ɪɪɪ
𝐼𝑉
�����������
�����������
�����������
�𝑢𝜕 ̅𝑣𝜕
𝑤𝜕
�𝑢𝜕 �
𝑤𝜕
̅𝑣𝜕 �
′
′
′
′
′
′
𝑘2
�
������
������
������
𝑢𝜌 −
𝑢𝜌 𝑣 � + � −
� 𝑤
𝑣𝜌 + � −
� 𝑤
∇𝜇 + � +
𝑦𝜕 𝑥𝜕
𝑧𝜕 𝑥𝜕
𝑧𝜕 𝑦𝜕
��������
��������
��������
��������
��������
��������
��������
𝜕𝑢′
𝜕𝑢′
𝜕𝑣′
𝜕𝑣′
𝜕𝑣′
�������� 𝜕𝑤′ 2
�������� 𝜕𝑤′ 2
𝜕𝑤′ 2
𝜕𝑢′
() +
() +
])
( −𝜇[( )2 + ( )2 + ( )2 + ( )2 + ( )2 + ( )2 +
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑉 = ] [𝜇
𝐷
𝑛𝑜𝑖𝑡𝑐𝑒𝑣𝑛𝑜𝑐 = )𝑘𝜌(
𝑡𝐷
=𝐼
ﺳﺮﻋﺖ ﺗﻐﻴﻴﺮ ﺟﺮﻳﺎﻥ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﺩﺭ ﺭﺍﺳﺘﺎﻱ ﺧﻂ ﺟﺮﻳﺎﻥ ﺩﺭ ﺍﺛﺮ ﻛﻮﻧﻮﻛﺴﻴﻮﻥ ﻭﺩﺭ ﺍﺛﺮ ﻣﻴﺎﻧﮕﻴﻦ ﺟﺮﻳﺎﻥ
ﻣﻲ ﺑﺎﺷﺪ.
ﺳﻪ ﺗﺮﻡ :
�������������
𝚥𝜌𝑢′ 𝚤 𝑢′ 𝚤 𝑢′
𝜕
′ 𝑝′ +
������
𝑢�
𝑛𝑜𝑖𝑠𝑢𝑓𝑓𝑒𝐷 𝑒𝑐𝑛𝑒𝑙𝑢𝑏𝑟𝑢𝑇 = �
𝚥
𝑗𝑥𝜕
2
ﻧﺮﺥ ﺟﺮﻳﺎﻥ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﻧﻮﺳﺎﻧﺎﺕ ﺑﻮﺳﻴﻠﻪ ﺍﻧﺘﺸﺎﺭ ﻳﺎ ﻧﻔﻮﺫ
= 𝐼𝐼
���������
𝑜𝑡 𝑒𝑢𝑑 𝑒𝑚𝑢𝑙𝑜𝑣 𝑙𝑜𝑟𝑡𝑛𝑜𝑐 𝑙𝑎𝑖𝑡𝑛𝑒𝑟𝑒𝑓𝑓𝑖𝑑 𝑎 𝑛𝑜 𝑒𝑛𝑜𝑑 𝑘𝑟𝑜𝑤 𝑤𝑜𝑙𝑓 ≡ 𝚥−𝑝′ 𝑢′
𝑠𝑛𝑜𝑖𝑡𝑎𝑢𝑡𝑐𝑢𝑙𝑓 𝑒𝑟𝑢𝑠𝑠𝑒𝑟𝑝 𝑒𝑡ℎ
−𝜌𝑢′ 𝑖 𝑢′ 𝑖 𝑢′𝑗 = 𝑟𝑎𝑡𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑠 𝑡𝑟𝑎𝑛𝑠𝑝𝑜𝑟𝑡𝑒𝑑 𝑡ℎ𝑟𝑜𝑢𝑔ℎ
ﺷﺶ ﺗﺮﻡ:
𝑠𝑛𝑜𝑖𝑡𝑎𝑢𝑡𝑐𝑢𝑙𝑓 𝑡𝑛𝑒𝑙𝑢𝑏𝑟𝑢𝑡 𝑎𝑖𝑣 𝑤𝑜𝑙𝑓 𝑒𝑡ℎ
𝚥�𝑢𝜕 𝚤�𝑢𝜕
+
𝑛𝑜𝑖𝑡𝑐𝑢𝑑𝑜𝑟𝑝 = �
𝑖𝑥𝜕 𝑗𝑥𝜕
ﻧﺮﺥ ﺗﻮﻟﻴﺪ ﺍﻧﺮژﻱ ﺗﻼﻃﻢ ﺑﻮﺳﻴﻠﻪ ﺗﻨﺶ ﻫﺎﻱ ﺭﻳﻨﻮﻟﺪﺯ
� ′ 𝑢′
�������
𝑢𝜌𝐼𝐼𝐼 = −
𝚥 𝚤
≡represents the specific kinetic energy per unit volume that an eddy will gain
per unit time due to the mean (flow) strain rate.
۱۸٥
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺍﮔﺮ ﺑﻪ ﻣﻌﺎﺩﻟﻪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﻣﺘﻮﺳﻂ ﺟﺮﻳﺎﻥ ﻧﮕﺎﻩ ﻛﻨﻴﻢ ﻣﻲ ﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﺍﻳﻦ ﺗﺮﻡ ﻫﻤﺎﻧﻨـﺪ ﻳـﻚ ﭼـﺎﻩ 1ﺩﺭ
ﺁﻥ ﻣﻌﺎﺩﻟﻪ ﻇﺎﻫﺮ ﻣﻲ ﺷﻮﺩ ﻭ ﺍﻳﻦ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺑﺮﺍﺳﺘﻲ ﻳـﻚ ﻧﺘﻴﺠـﻪ ﺍﻱ ﺍﺯ ﺩﺳـﺖ
ﺩﺍﺩﻥ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻮﺳﻂ mean flowﺍﺳﺖ.
F64
𝑘𝜕2
𝜇 = 𝑘 ∇𝜇 = 𝐼𝑉
𝑛𝑜𝑖𝑠𝑢𝑓𝑓𝑖𝑑 𝑠𝑢𝑜𝑐𝑠𝑖𝑣 =
𝑖𝑥𝜕 𝑗𝑥𝜕
2
ﻧﺮﺥ ﺗﻮﻟﻴﺪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﺑﻮﺳﻴﻠﻪ ﺗﻨﺶ ﻫﺎﻱ ﻧﺎﺷﻲ ﺍﺯ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﻣﻮﻟﻜﻮﻟﻲ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲ ﺩﻫﺪ ﻛﻪ ﺍﻳﻦ ﺗﺮﻡ
ﺍﻏﻠﺐ ﺩﺭ ﺍﻋﺪﺍﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺯﻳﺎﺩ ﺻﻔﺮ ﺍﺳﺖ.
This term represents the diffusion of turbulent energy by the molecular motion
that is equally responsibly for diffusing the mean flow momentum.
ﺳﻪ ﺗﺮﻡ:
ﺗﻠﻔﺎﺕ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ
𝑧 𝑖 = 𝑥 → 𝑗 = 𝑥, 𝑦,
�����������
𝚤 𝜕𝑢′ 𝚤 𝜕𝑢′
𝑧 = 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑜𝑛 = ℰ𝑖 = 𝑥, 𝑦,
𝚥𝑥𝜕 𝚥𝑥𝜕
𝑧 𝑖 = 𝑦 → 𝑗 = 𝑥, 𝑦,
𝜇𝑉 = −
𝑧 𝑗 = 𝑥, 𝑦,
𝑧 𝑖 = 𝑧 → 𝑗 = 𝑥, 𝑦,
���������
��������� 𝜕𝑢′ 2
��������� 𝜕𝑢′ 2
��������� 𝜕𝑢′ 2
��������� 𝜕𝑣 ′ 2
��������� 𝜕𝑣 ′ 2
𝜕𝑣 ′ 2
] ⋯ = −𝜇[� � + � � + � � + � � + � � + � � +
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
This term represents the mean rate at which the kinetic energy of the smallest
turbulent eddies is transferred to thermal energy at the molecular level.
ﭼﻮﻥ ﺗﺮﻡ ﻫﺎ ﺑﻪ ﺗﻮﺍﻥ 2ﻣﻲ ﺭﺳﻨﺪ ﻫﻤﻴﺸﻪ ﻣﺜﺒﺖ ﺍﺳﺖ ﻭﺩﺭ µﻛﻪ ﺿﺮﺏ ﻣﻲ ﺷﻮﻧﺪ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﻧﻘﺶ ﻳﻚ
ﭼﺎﻩ)(sinkﺭﺍ ﺑﺎﺯﻱ ﻣﻲ ﻛﻨﻨﺪ.
ﺍﮔﺮ ﻋﺪﺩ ﺭﻳﻨﻮﻟﺪﺯ ﺑﺎﻻ ﺑﺎﺷﺪ fluctuating strain rateﺧﻴﻠﯽ ﺑﺰﺭﮔﺘﺮ ﺍﺯ mean rate of strainﺍﺳﺖ.
Modeled Turbulent kinetic Energy Equation
ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺑﺮﺍﻱ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﺗﺮﻡ ﻫﺎﻳﻲ ﺷﺎﻣﻞ ﻧﻮﺳﺎﻧﺎﺕ ﻣﺸﺨﺼـﻪ ﻫـﺎﻱ ﺟﺮﻳـﺎﻥ ﻭﺟـﻮﺩ
ﺩﺍﺭﺩ ﻭﺩﺭ ﺁﻧﺠﺎﻳﻲ ﻛﻪ ﺭﻭﺵ ﻣﺴﺘﻘﻴﻤﻲ ﺑﺮﺍﻱ ﻣﺤﺎﺳﺒﻪ ﺍﻳﻦ ﺗﺮﻡ ﻫﺎ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ ،ﺑﺎﻳﺪ ﺁﻧﻬﺎ ﺭﺍ ﺑـﺮ ﺣﺴـﺐ ﺗـﺮﻡ ﻫـﺎﻱ
ﻗﺎﺑﻞ ﻣﺤﺎﺳﺒﻪ ﺗﻘﺮﻳﺐ ﺯﺩ.
Sink
۱۸٦
۱
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﺯﺩﻥ ﺗﺮﻡ ﻫﺎﻱ ﺍﻧﺘﺸﺎﺭ ﻣﻲ ﺗﻮﺍﻥ ﻓﺮﻡ ﮔﺮﺍﺩﻳﺎﻧﻲ ﻛﻪ ﺗﻮﺳـﻂ Prandtlﻭ Kolmogorovﺍﺭﺍﺋـﻪ ﺷـﺪﻩ
ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ!
ﻓﺮﺽ :ﺍﻧﺘﺸﺎﺭ ﺗﻼﻃﻢ ﻣﺸﺎﺑﻪ ﺍﻧﺘﺸﺎﺭ ﻣﻮﻟﻜﻮﻟﻲ
𝑤𝑎𝑙 𝑡𝑛𝑒𝑖𝑑𝑎𝑟𝑔
𝑘𝜕 𝑡𝜇
1 ′ ′ ′
′ ′
�����������
𝑢𝜌 � −
= � 𝑗 𝑢 𝑝 𝚤𝑢 𝚤𝑢 𝚥 +
2
𝑗𝑥𝜕 𝑘𝜎
𝑘 𝑟𝑜𝑓 𝑟𝑒𝑏𝑚𝑢𝑛 𝑡𝑑𝑖𝑚𝜎𝑘 = 𝑃𝑟𝑎𝑛𝑑𝑡𝑙 − 𝑆𝑐ℎ
ﺩﺭ ﺗﺮﻡ ﻫﺎﻱ ﺗﻮﻟﻴﺪ 1ﺣﺎﺻﻠﻀﺮﺏ ﺩﻭﺗﺎﻳﻲ ﻧﻮﺳﺎﻧﺎﺕ ﺳﺮﻋﺖ ،ﻣﺠﻬﻮﻻﺕ ﻏﻴﺮ ﻗﺎﺑﻞ ﻣﺤﺎﺳﺒﻪ ﻫﺴﺘﻨﺪ ﻛﻪ ﺍﺯ ﺁﻧﺠﺎﻳﻲ ﻛﻪ
ﺍﻳﻦ ﺗﺮﻡ ﻫﺎ ﺩﺭ ﻭﺍﻗﻊ ﻫﻤﺎﻥ ﺗﺮﻡ ﻫﺎﻱ ﺗﻨﺶ ﻫـﺎﻱ ﺗﻼﻃـﻢ ﻫﺴـﺘﻨﺪ.ﺑﺮﺍﻱ ﺗﻘﺮﻳـﺐ ﺯﺩﻥ ﺁﻧﻬـﺎ ﻣـﻲ ﺗـﻮﺍﻥ ﺍﺯ ﻧﻈﺮﻳـﻪ
Boussinesqﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ.
F65
𝚥�𝑢𝜕 𝚤�𝑢𝜕
2
+
𝑗𝑖𝛿𝑘𝜌 � −
𝑖𝑥𝜕 𝑗𝑥𝜕
3
� 𝜇 = ′ 𝑢′
�������
𝑢𝜌−
𝚥 𝚤
𝑡
𝑒𝑙𝑎𝑐𝑠 𝜇𝑡 = 𝜌√𝑘 𝑙 , 𝑙 = 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑐𝑒 𝑙𝑒𝑛𝑔𝑡ℎ
Isotropic relation ≡ assumes that momentum transport in all directions is
the same at a given point in space.
ﺑﻌﻨﻮﺍﻥ ﺍﻭﻟﻴﻦ ﺗﻘﺮﻳﺐ ﺑﺮﺍﻱ ﭘﻴﺪﺍ ﻛﺮﺩﻥ ﺭﺍﺑﻄﻪ ﺍﻱ ﺑﺮﺍﻱ ℰﺑﺮﺍﻱ thin shear layerﺩﺭ ﺣﺎﻟﺖ ﭘﺎﻳـﺪﺍﺭ ﻭ ﻫﻤـﻮژﻥ
ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﻣﻲ ﺷﻮﺩ ﻛﻪ production=dissipationﻳﻌﻨﻲ ﺟﺮﻳﺎﻥ ﺑﺼﻮﺭﺕ ﻣﺤﻠﻲ ﺩﺭ ﺗﻌـﺎﺩﻝ ﺍﺳـﺖ .ﺑـﺎ ﺍﻳـﻦ
ﻓﺮﺽ ﺍﺯ ﺭﺍﺑﻄﻪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺩﺍﺭﻳﻢ:
�����������
𝚤�𝑢𝜕
𝚤 𝜕𝑢′ 𝚤 𝜕𝑢′
𝑣=
=ℰ
𝑗𝑥𝜕
𝑘𝑥𝜕 𝑘𝑥𝜕
𝑗−𝑢′ 𝑖 𝑢′
ﺍﺯ ﺁﻧﺠﺎﻳﻲ ﻛﻪ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺍﻳﺰﻭﺗﺮﻭپ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺑﺎ ﻳﻚ (l) lengthscaleﻭﻳﻚ (u)velocity scaleﺗﻌﺮﻳـﻒ
ﻣﻲ ﺷﻮﺩ؛ ﺍﺯ ﻣﻌﺎﺩﻟﻪ ﻗﺒﻠﻲ ﻧﺘﻴﺠﻪ ﻣﻲ ﺷﻮﺩ ﻛﻪ:
𝑢3
≈ℰ
𝑙
ﻧﺮﺥ ﺍﺗﻼﻑ ﺍﻧﺮژﻱ ﺑﻪ ﻭﺳﻴﻠﻪ ﺑﺰﺭﮔﺘﺮﻳﻦ ﮔﺮﺩﺍﺑﻪ ﻫﺎ ﻛﻨﺘﺮﻭﻝ ﻣﻲ ﺷﻮﺩ .ﺩﺭ energy cascadeﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﺑـﺰﺭگ
ﺑﻪ ﮔﺮﺩﺍﺑﻪ ﻫﺎﻱ ﻛﻮﭼﻚ ﺍﻧﺮژﻱ ﻣﻲ ﺩﻫﻨﺪ ﻛﻪ ﻧﻬﺎﻳﺘﺎ ﺍﺗﻼﻑ ﻣﻲ ﺷﻮﻧﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ dissipationﺑﻪ ﻭﺳﻴﻠﻪ ﺑﺰﺭﮔﺘﺮﻳﻦ
ﮔﺮﺩﺍﺑﻪ ﻫﺎ scaleﻣﻲ ﺷﻮﻧﺪ ﻭ ﺍﮔﺮﺁﻧﺎﻟﻴﺰ ﺑﻌﺪﻱ ﺭﺍ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺗﻮﺭﺑﺎﻟﻨﺖ ﺑﺎ ﺗﻚ scaleﻫﺎﻱ uﻭ lﺑﻜﺎﺭ ﺑﺒﺮﻳﻢ ﺑـﺎﺯ
ﻫﻢ ﺭﺍﺑﻄﻪ
𝑢3
𝑙
≈ ℰﺣﺎﺻﻞ ﻣﻲ ﺷﻮﺩ.
Production
۱۸۷
۱
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺑﺤﺚ): white (۱۹۹۱
ﺍﮔﺮ ﻳﻚ ﮔﺮﺩﺍﺑﻪ ﺑﻪ ﺍﻧﺪﺍﺯﻩ lﺑﺎ ﺳﺮﻋﺖ uﺣﺮﻛﺖ ﻛﻨﺪ ﺍﻧﺮژﻱ ﺁﻥ ﺍﺗﻼﻑ ﻣﻲ ﺷﻮﺩ ﺑﻮﺳﻴﻠﻪ ﺭﺍﺑﻄﻪ ﺗﻘﺮﻳﺒﻲ ﺫﻳﻞ:
(𝑑𝑟𝑎𝑔)(𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦) (𝜌𝑢2 𝑙2 )(𝑢) 𝑢3
=
≈
𝑠𝑠𝑎𝑚
𝜌𝑙3
𝑙
1
≈ℰ
𝑢 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑠𝑐𝑎𝑙𝑒 = √𝑘 = 𝑘 2
3
2
𝜕𝑢�𝚤 2
) = 𝜌. 𝑐𝐷 . ℰ
𝑗𝑥𝜕
𝑘
𝑙
≈⇒ℰ
(� � 𝜇 ⇒
ﻛﻪ CDﻳﻚ ﺛﺎﺑﺖ ﺗﺠﺮﺑﻲ ﺍﺳﺖ.
𝑘𝜕2𝑘 𝜕2𝑘 𝜕2
𝜕
𝑘𝜕
𝑘𝜕
𝜕
𝑘𝜕
𝜕
+
+
�
=
𝜇(
)
+
+
𝜇�
�
𝜇�
�
𝑥𝜕 𝑥𝜕
𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2
𝑦𝜕 𝑦𝜕
𝑧𝜕 𝑧𝜕
𝑗
𝑖
� 𝜇 = 𝑘 𝜇∇2
ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻨﻜﻪ ﺑﺮﺍﻱ ﺍﻛﺜﺮ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻣﺘﻼﻃﻢ 𝛔k=1ﻣﻲ ﺑﺎﺷﺪ.ﻣﻲ ﺗﻮﺍﻥ ﺍﻳﻦ ﺗﺮﻡ ﻫﺎ ﺭﺍ ﺑﺎ ﺗـﺮﻡ
ﻫﺎﻱ ﺍﻧﺘﺸﺎﺭ ﺟﻤﻊ ﻛﺮﺩ ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﺎ ﻛﻤﻚ ﮔﺮﻓﺘﻦ ﺍﺯ ﺗﻌﺮﻳﻒ. µeff=µ+µt
ﺩﺍﺭﻳﻢ:
𝜕
𝑘𝜕
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝑘𝜕 𝑡𝜇 𝜕
(
.
)
� .
�+
𝜇�
=�
𝑖𝑥𝜕 𝑖𝑥𝜕
𝑖𝑥𝜕 𝑘𝜎 𝑖𝑥𝜕
𝑖𝑥𝜕 𝑘𝜎 𝑖𝑥𝜕
ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺑﺎ ﺑﻬﺮﻩ ﮔﻴﺮﻱ ﺍﺯ ﺗﻘﺮﻳﺐ ﻫﺎﻱ ﺑﻪ ﻛﺎﺭ ﺭﻓﺘﻪ ﻣﻌﺎﺩﻟﻪ ﻧﻬﺎﻳﻲ ﺑﺮﺍﻱ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﺑﻔﺮﻡ ﺯﻳـﺮ ﺑـﻪ
ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ:
𝐷
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
= )𝑘𝜌(
. �+
� . �+
� .
�
�
𝑡𝐷
𝑥𝜕 𝑘𝜎 𝑥𝜕
𝑦𝜕 𝑘𝜎 𝑦𝜕
𝑧𝜕 𝑘𝜎 𝑧𝜕
𝜕𝑢� 2
𝜕𝑣̅ 2
𝑤𝜕
� 2
𝜕𝑣̅ 𝜕𝑢� 2
𝑤𝜕
� 𝜕𝑢� 2
� + 𝜇𝑒𝑓𝑓 �2 � � + 2 � � + 2 � � + � + � +
� +
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑦𝜕 𝑥𝜕
𝑧𝜕 𝑥𝜕
𝑤𝜕
� 𝜕𝑣̅ 2
�+
𝜀 𝐷𝑐𝜌 + � � −
𝑧𝜕 𝑦𝜕
ﺑﺎ ﺗﻮﺟﻪ ﺩﺍﺷﺖ ﻛﻪ ﺩﺭ ﺗﺮﻡ ﻫﺎﻱ ﻣﺮﺑﻮﻁ ﺑﻪ ﺗﻮﻟﻴﺪ µtﺑﻪ µeffﺗﺒﺪﻳﻞ ﺷﺪﻩ ،ﺯﻳﺮﺍ ﺩﺭ ﻧﻮﺍﺣﻲ ﻧﺰﺩﻳﻚ ﺑﻪ ﻣـﺮﺯµeff=µ
ﻣﻲ ﺑﺎﺷﺪ ﻭﺩﺭ ﻧﻮﺍﺣﻲ ﺩﻭﺭ ﺍﺯ ﻣﺮﺯ µeff≈µtﺍﺳﺖ.
۱۸۸
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﻄﺎﻟﺐ ﻓﻮﻕ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﺑﺮﺍﻱ ﻳﻚ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﺑﺼﻮﺭﺕ ﺯﻳـﺮ ﺳـﺎﺩﻩ
ﻣﻲ ﺷﻮﺩ:
𝐷
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝜕𝑢� 2
𝜕𝑣̅ 2
𝜕𝑣̅ 𝜕𝑢� 2
= )𝑘𝜌( 𝜌
. �+
� � . � + 𝜇𝑒𝑓𝑓 �2 � � + 2 � � + � +
�
�
𝑡𝐷
𝑥𝜕 𝑘𝜎 𝑥𝜕
𝑦𝜕 𝑘𝜎 𝑦𝜕
𝑥𝜕
𝑦𝜕
𝑦𝜕 𝑥𝜕
𝜀 − 𝜌. 𝑐𝐷 .
3
1
𝑙 𝜇𝑡 = 𝑐𝜇 𝜌𝑘 2
ﻛﻪ CDﻭ Cµﺛﺎﺑﺖ ﻫﺴﺘﻨﺪ.
,
𝑘2
𝐷𝑐 = 𝜀
𝑙
ﻛﻪ ﺑﺮﺍﻱ ﻣﺪﻝ ﻫﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ Kﺍﺯ ﻃﺮﻳﻖ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻭ 𝑙 ﺑﺎ ﺗﻮﺟـﻪ ﭘﺎﺭﺍﻣﺘﺮﻫـﺎﻱ ﺟﺮﻳـﺎﻥ ﻣﺘﻮﺳـﻂ
ﺑﺼﻮﺭﺕ ﺟﺒﺮﻱ ﻣﺤﺎﺳﺒﻪ ﻣﻲ ﺷﻮﺩ ﻭ ﺭﻭﺍﺑﻂ ﻣﺨﺘﻠﻔﻲ ﺍﺭﺍﺋﻪ ﺷﺪﻩ ﺍﺳﺖ.
ﺑﺮﺍﻱ ﻳﻚ ) thin shear layerﻻﻳﻪ ﻣﺮﺯﻱ(ﻣﻌﺎﺩﻟﻪ ﻗﺒﻞ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﺩﺭ ﻣﻲ ﺁﻳﺪ:
3
𝑘𝜕 𝑡𝑣 𝜕
�𝑢𝜕
𝑘𝐷
𝑘2
=
𝐷𝐶 � . � + 𝑣𝑡 ( )2 −
𝑦𝜕
𝑙
𝑦𝜕 𝑘𝜎 𝑦𝜕 𝑡𝐷
Near wall Distribution:
ﻧﺰﺩﻳﻚ ﻳﻚ ﺩﻳﻮﺍﺭ ،ﺗﺮﻡ ﻫﺎﻱ ﻛﻨﻮﻛﺴﻴﻮﻥ ﻭ ﻧﻔﻮﺫ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻮﺭﺑﺎﻟﻨﺖ ﻗﺎﺑﻞ ﺻﺮﻑ ﻧﻈﺮ ﺷﺪﻥ ﻫﺴﺘﻨﺪ.
𝑘𝜕 𝑡𝑣 𝜕
� . �=0
𝑦𝜕 𝑘𝜎 𝑦𝜕
,
𝑘𝐷
≈0
𝑡𝐷
3
⇒ Production = dissipatipn
′ 𝑣 ′ )2
������
𝑢(
𝑘2
𝐷𝐶 =
𝑡𝑣
𝑙
ﺍﺯ ﺩﺍﺩﻩ ﻫﺎﻱ ﺗﺠﺮﺑﻲ ﭘﻴﺪﺍ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ:
𝑦𝑙𝑙𝑎𝑢𝑠𝑢
= 0.25 ≈ 0.3
𝜎𝑘 = 1
,
①
������
𝑢′ 𝑣 ′
𝑘
−
3
𝑟𝑜
�𝑢𝜕
𝑘2
𝐷𝐶 = � � 𝑡𝑣 ⇒
𝑙
𝑦𝜕
′ 𝑣 ′ )2
������
𝑢(
= 𝐷𝐶 ⇒
𝑘2
𝑙 𝑘√ = 𝑡𝑣
⇒ 𝐶𝐷 = 0.07~0.08
Distribution of Length scale:
ﻧﺰﺩﻳﻚ ﺩﻳﻮﺍﺭ:
1
۱۸۹
𝑘 ′ 𝑣′� = 𝐶 2
�����
𝑢� ⇒ ①
𝐷
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
�𝑢𝜕
�
𝑦𝜕
−1
� √𝑘 = 𝐶𝐷2 . 𝑙.
⇒
𝑙 𝑘√ = 𝑡𝑣
� ′ ′
�𝑢𝜕
������
𝑢−
𝑡𝑣 = 𝑣
𝑦𝜕
𝜕𝑢� 2
)
𝑦𝜕
1
�𝑢𝜕
𝑘 � = 𝐶𝐷2
𝑦𝜕
−1
� 𝑙𝑘√
( ′ 𝑣 ′ = 𝐶 2 . 𝑙2
������
𝑢⇒ −
𝐷
For inertial sublayer:
∗𝑢 �𝑢𝜕
=
𝑦𝑘 𝑦𝜕
2
������ , −
𝑘√ = 𝑒𝑙𝑎𝑐𝑠 𝑦𝑡𝑖𝑐𝑜𝑙𝑒𝑣 = ∗𝑢 ∗𝑢 = 𝑢′ 𝑣 ′
�𝑢𝜕
∗𝑢
2
= √𝑘 𝑙.
∗𝑢 = ∗𝑢 = √𝑘.
𝑦𝜕
𝑦𝑘
𝜕𝑢� 2
)
𝑦𝜕
𝑣 = ′𝑣 ′
������
𝑢−
𝑡
−1
2
∗2
( ⇒ 𝑢 = 𝐶𝐷 . 𝑙2
①
1
4
−1
𝜕𝑢� 2
�𝑢𝜕
) = 𝐶𝐷2 . 𝑙2 ( )2
𝑦𝜕
𝑦𝜕
1
4
2
( ⇒ (𝑘𝑦)2
1
2
2
𝑚𝑙 ⇒ 𝑙 = 𝐶𝐷 . (𝑘𝑦) ⇒ 𝑙 = 𝐶𝐷 . 𝑘𝑦 = 𝐶𝐷 .
𝑦𝑓𝑜𝑟 𝐶𝐷 = 0.08 , 𝑘 = 0.4 ⇒ 𝑙 = 0.2
ﺁﻥ ﭘﻴﺸﻨﻬﺎﺩ ﺷﺪﻩ ﻛﻪ ﻣﻌﺎﺩﻟﻪ ① ﺭﺍ ﺑﺮﺍﻱ ﻛﻞ ﺩﺍﻣﻨﻪ ﺟﺮﻳﺎﻥ ﺑﺎ ℓmﺩﺍﺩﻩ ﺷﺪﻩ ﺑﻪ ﻭﺳﻴﻠﻪ ﻳﻚ ﺑﻴﺎﻥ ﺗﺠﺮﺑﻲ ﺑﻜﺎﺭ
ﺭﻭﺩ.
ﺑﺮﺁﻭﺭﺩ ﺍﻳﻦ ﻣﺪﻝ ﺍﺯ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻻﻳﻪ ﻣﺮﺯﻱ ﺧﻮﺏ ﺑﻮﺩﻩ ﺍﺳـﺖ.ﺑﺎﻳﺪ ﺗﻮﺟـﻪ ﻛـﺮﺩ ﻛـﻪ ﺩﺭ ﺑﻌﻀـﻲ ﺍﺯ ﺟﺮﻳـﺎﻥ ﻫـﺎﻱ
ﭘﻴﭽﻴﺪﻩ ﺗﺮ ،ﺍﺛﺮﺍﺕ ﺟﺎﺑﻪ ﺟﺎﻳﻲ ﻭ ﺍﻧﺘﺸﺎﺭ ﺑﺮ ﺭﻭﻱ ﻣﻘﻴﺎﺱ ﻃﻮﻝ ﻧﻴﺰ ﺩﺍﺭﺍﻱ ﺍﻫﻤﻴﺖ ﺯﻳـﺎﺩﻱ ﺍﺳـﺖ ﻭ ﺍﺭﺍﺋـﻪ ﻣﻌـﺎﺩﻻﺕ
ﺟﺒﺮﻱ ﻛﻪ ﺑﺘﻮﺍﻧﺪ ﺗﻮﺯﻳﻊ ℓﺭﺍ ﺩﺭ ﻫﺮ ﻧﻮﻉ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻤﻲ ﺑﻪ ﺩﻗﺖ ﻣﺸﺨﺺ ﻛﻨﺪ؛ ﺍﻣﻜﺎﻥ ﭘـﺬﻳﺮ ﻧﻴﺴـﺖ .ﺍﺯ ﺍﻳـﻦ ﺭﻭ
ﻛﺎﺭﺑﺮﺩ ﻣﺪﻝ ﻫﺎﻱ ﺗﻚ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺗﻘﺮﻳﺒﺎ ﺑﻪ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻻﻳﻪ ﻣﺮﺯﻱ ﻣﺤﺪﻭﺩ ﺍﺳﺖ.
ﺑﻄﻮﺭ ﻛﻠﻲ ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﭘﻴﭽﻴﺪﻩ ﺍﺯ ﻧﻮﻉ ﭼﺮﺧﺸﻲ ﻣﺪﻝ ﻫﺎﻱ ﺗـﻚ ﻣﻌﺎﺩﻟـﻪ ﺍﻱ ﺍﺯ ﺗﻮﺍﻧـﺎﻳﻲ ﺧـﻮﺑﻲ ﺑﺮﺧـﻮﺭﺩﺍﺭ
ﻧﻴﺴﺘﻨﺪ؛ ﻟﺬﺍ ﺑﺎﻳﺪ ﺳﺮﺍﻍ ﻣﺪﻝ ﻫﺎﻳﻲ ﺭﻓﺖ ﻛﻪ ﺍﺛﺮ ﺍﻧﺘﻘﺎﻝ ﻭ ﺍﻧﺘﺸﺎﺭ ﺭﺍ ﺭﻭﻱ ﻣﻘﻴﺎﺱ ﻃﻮﻝ ﻣﺪﻧﻈﺮ ﻗﺮﺍﺭ ﻣﻲ ﺩﻫﻨﺪ.
Short comings:
-Transport of the turbulent length scaleis not accounded for.
-The model coffers little advantage over the mixing length model.
۱۹۰
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
Two models Equations
ﺩﺭ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﭘﻴﭽﻴﺪﻩ ،ﺑﺎﻳﺪ ﺍﺛﺮﺍﺕ ﺍﻧﺘﻘﺎﻝ ﻭ ﺍﻧﺘﺸﺎﺭ ﻃﻮﻝ ﻣﺸﺨﺼﻪ ﺭﺍ ﻧﻴﺰ ﺑﺎ ﻓﺮﻡ ﺩﻗﻴﻖ ﺗﺮﻱ ﻣﺪ ﻧﻈـﺮ ﻗـﺮﺍﺭﺩﺍﺩ .ﺍﺯ
ﺍﻳﻦ ﺭﻭ ﻣﺪﻝ ﻫﺎﻱ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﺍﺯ ﺟﻬﺖ ﺗﻼﻃـﻢ ﻣﻄـﺮﺡ ﮔﺮﺩﻳـﺪﻩ ﺍﻧـﺪ .ﺗﻤـﺎﻣﻲ ﺍﻳـﻦ ﻣـﺪﻝ ﻫـﺎ ،ﺩﺭ
ﻣﺤﺎﺳﺒﻪ ﻭﻳﺴﻜﻮﺯﻳﺘﻪ ﺗﻼﻃﻢ ﺍﺯ ﺭﺍﺑﻄﻪ µt=ρℓk1/2ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲ ﻧﻤﺎﻳﻨﺪ .ﺑﻌﻼﻭﻩ ﺩﺭ ﺍﻳﻦ ﻧﻮﻉ ﻣـﺪﻝ ﻫـﺎ ،ﻫـﻢ ﺑـﺮﺍﻱ
ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﻭ ﻫﻢ ﺑﺮﺍﻱ ﻣﻘﻴﺎﺱ ﻃﻮﻝ ،ﻣﻌﺎﺩﻻﺕ ﺍﻧﺘﻘﺎﻝ ﺍﺳﺘﺨﺮﺍﺝ ﻣﻲ ﮔﺮﺩﺩ.
ﺭﻭﻳﻪ ﻛﻠﻲ ﺗﺸﻜﻴﻞ ﻣﻌﺎﺩﻟﻪ ﺩﻭﻡ ﺩﺭ ﺍﻳﻦ ﻣﺪﻝ ﻫﺎ ﺑﻘﺮﺍﺭ ﺫﻳﻞ ﺍﺳﺖ:
-1ﻣﺘﻐﻴﺮ Zﺑﺼﻮﺭﺕ ﻣﻘﺎﺑﻞ ﺗﻌﺮﻳﻒ ﻣﻲ ﮔﺮﺩﺩ Z=kmℓnﻛﻪ mﻭ nﺍﻋﺪﺍﺩ ﺛﺎﺑﺘﻲ ﻫﺴﺘﻨﺪ.
-2ﻣﻌﺎﺩﻟﻪ ﺩﻗﻴﻘﻲ ﺑﺮﺍﻱ Zﺍﺯ ﻣﻌﺎﺩﻻﺕ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻛﺲ ﺍﺳﺘﺨﺮﺍﺝ ﻣﻲ ﮔﺮﺩﺩ.
-3ﺭﻭﺍﺑﻄﻲ ﺑﺮﺍﻱ ﺗﻘﺮﻳﺐ ﻧﻤﻮﺩﻥ ﻣﻘﺎﺩﻳﺮ ﻣﺠﻬﻮﻝ ﺗﻼﻃﻢ ﺑﺮ ﺣﺴﺐ ﻣﻘﺎﺩﻳﺮ ﻗﺎﺑﻞ ﻣﺤﺎﺳﺒﻪ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ.
-4ﭘﺲ ﺍﺯ ﺗﻌﻴﻴﻦ kﻭ Zﺍﺯ ﻣﻌﺎﺩﻻﺕ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﻣﺮﺑﻮﻃﻪ ℓ ،ﺍﺯ ﻣﻘﺎﺩﻳﺮ kﻭ Zﻣﻤﺤﺎﺳﺒﻪ ﻣﻲ ﮔﺮﺩﺩ.
ﻣﺤﻘﻘﻴﻦ ﻣﺨﺘﻠﻒ Zﻫﺎﻱ ﻣﺨﺘﻠﻔﻲ ﺭﺍ ﺑﻜﺎﺭ ﺑﺮﺩﻩ ﺍﻧﺪ)ﺟﺪﻭﻝ ﺫﻳﻞ(:
ﻣﺤﻘﻘﻴﻦ ﺭﺍﺑﻄﻪ ﺩﻭﻡ ﺍﻧﺘﺨﺎﺑﻲ ﺧﻮﺩ ﺭﺍ ﺍﺯ ﺭﻭﺷﻲ ﺑﺴﻴﺎﺭ ﻣﺘﻔﺎﻭﺕ ﺑﺎ ﺩﻳﮕﺮﺍﻥ ﺑﺪﺳﺖ ﻣﻲ ﺁﻭﺭﻧﺪ .ﺍﻣﺎ ﺍﮔـﺮ ﻧﺘـﺎﻳﺞ ﻧﻬـﺎﻳﻲ
ﺑﺎﻳﻜﺪﻳﮕﺮ ﻣﻘﺎﻳﺴﻪ ﮔﺮﺩﺩ،ﺷﺒﺎﻫﺖ ﺯﻳﺎﺩﻱ ﻣﺎﺑﻴﻦ ﺁﻧﻬﺎ ﻣﺸﺎﻫﺪﻩ ﻣﻲ ﺷﻮﺩ؛ ﺑﻪ ﻓﺮﻣﻲ ﻛﻪ ﻫﻤﻪ ﺁﻧﻬﺎ ﺭﺍ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ ﻳـﻚ
ﻗﺎﻟﺐ ﺧﺎﺹ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺍﺭﺍﺋﻪ ﻧﻤﻮﺩ.
۱۹۱
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝚥�𝑢𝜕 𝚤�𝑢𝜕 𝚤�𝑢𝜕 𝑡𝜇
𝑘 𝜌2
𝑧𝐷
𝑧𝜕 𝑡𝜇 𝜕
𝜌
=
�
� � + 𝑧 �𝐶1
�
+
�� − 𝐶2
𝛷𝑆 � +
𝑗𝑥𝜕 𝑧𝜎 𝑗𝑥𝜕 𝑡𝐷
𝑖𝑥𝜕 𝑗𝑥𝜕 𝑗𝑥𝜕 𝑘
𝑡𝜇
ﺩﺭ ﺑﻴﺸﺘﺮ ﺟﺮﻳﺎﻥ ﻫﺎﻱ ﻛﻤﻴﺖ ﻫﺎﻱ C2، C۱ﻭ 𝛔zﻣﻘﺎﺩﻳﺮ ﺛﺎﺑﺖ ﻣﻲ ﺑﺎﺷﻨﺪ ﻭ Szﺗﺮﻡ ﻫﺎﻱ ﺛﺎﻧﻮﻳﻪ ﻣﻨﺒﻊ ﺭﺍ ﺷـﺎﻣﻞ
ﻣﻲ ﮔﺮﺩﺩ.
ﺩﺭ ﺣﻘﻴﻘﺖ ﺭﺍﺑﻄﻪ ﺑﺎﻻ ﻧﺸﺎﻥ ﺩﻫﻨﺪﻩ ﺍﻳﻦ ﺍﻣﺮ ﺍﺳﺖ ﻛﻪ ﺗﻐﻴﻴﺮﺍﺕ Zﺩﺭ ﺭﺍﺳﺘﺎﻱ ﻳﻚ ﺧﻂ ﺟﺮﻳﺎﻥ ﻧﺎﺷﻲ ﺍﺯ ﺍﻧﺘﻘﺎﻝ ﺩﺭ
ﺍﺛﺮ ﺍﻧﺘﺸﺎﺭ ﺍﺛﺮﺍﺕ ﻣﺘﻘﺎﺑﻞ ﺑﻴﻦ ﺗﻼﻃﻢ ﻭ ﻛﻤﻴﺖ ﻫﺎﻱ ﻣﺘﻮﺳﻂ ﺟﺮﻳﺎﻥ ﻭ ﻫﻤﭽﻨـﻴﻦ ﺍﺛـﺮﺍﺕ ﻣﺘﻘﺎﺑـﻞ ﻋﻮﺍﻣـﻞ ﻣﺨﺘﻠـﻒ
ﺗﻼﻃﻢ ﺑﺮ ﺭﻭﻱ ﻳﻜﺪﻳﮕﺮ ﻣﻲ ﺑﺎﺷﺪ.
ﺍﺯ ﻣﻴﺎﻥ ﻣﺪﻝ ﻫﺎﻱ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺍﻱ ﺩﻳﻔﺮﺍﻧﺴﻴﻠﻲ ﻟﺰﺟﺖ ﺗﻼﻃﻢ ،ﻣﺪﻝ K-ℰﻋﻤﻮﻣﻴﺖ ﺑﻴﺸﺘﺮﻱ ﺩﺍﺷﺘﻪ ﺍﺳﺖ.
ﻣﺪﻝ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ :K-ℰ3
𝑘2
ﺩﺭ ﺍﻳﻦ ﻣﺪﻝ Zﺑﺼﻮﺭﺕ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﻣﻲ ﺷﻮﺩ ﻛﻪ ﺁﻥ ﺭﺍ ﺑﺎ ﻧﻤﺎﺩ ℰﻧﺸﺎﻥ ﻣـﻲ ﺩﻫﻨـﺪ ﻭ ﻧﻤﺎﻳـﺎﻧﮕﺮﻧﺮﺥ ﺗﻠﻔـﺎﺕ
𝑙
ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺗﻼﻃﻢ ﻣﻲ ﺑﺎﺷﺪ ﻭ ﺑﺼﻮﺭﺕ ﺯﻳﺮ ﺗﻌﺮﻳﻒ ﻣﻲ ﺷﻮﺩ.
�����������
𝜕𝑢′
𝚤𝚤 𝜕𝑢′
𝚥𝑥𝜕 𝚥𝑥𝜕
ﻣﻌﺎﺩﻟﻪ ﺗﻠﻔﺎﺕ ﺍﻧﺮژﻱ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ):(ℰ
𝑣=ℰ
ﻣﻌﺎﺩﻟﻪ ﺩﻭﻣﻲ ﻛﻪ ﺗﻮﺳﻂ ﻣﺪﻝ K-ℰﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲ ﮔﻴﺮﺩ ،ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺑﺮﺍﻱ ﺗﺮﻡ ﺗﻠﻔﺎﺕ ﺍﻧﺮژﻱ
) (ℰﻣﻲ ﺑﺎﺷﺪ ﻛﻪ ﺑﻌﻨﻮﺍﻥ ﻣﺠﻬﻮﻝ ﺩﺭ ﻣﻌﺎﺩﻟﻪ ﺩﻳﻔﺮﺍﻧﺴﻴﻞ ﺍﻧﺮژﻱ ﺟﻨﺒﺸـﻲ ﻇـﺎﻫﺮ ﺷـﺪﻩ ﺍﺳـﺖ؛ ﻳﻜـﻲ ﺍﺯ ﺭﺍﻩ ﻫـﺎﻱ
ﺭﺳﻴﺪﻥ ﺑﻪ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻗﺎﻧﻮﻥ ﺑﻘﺎء ﻣﻲ ﺑﺎﺷﺪ ﻛﻪ ﺑﺮﺍﻱ ﻳﻚ ﻛﻤﻴﺖ ﺍﺳـﻜﺎﻟﺮ) (ɸﺩﺭ ﻳـﻚ ﺣﺠـﻢ ﻛﻨﺘـﺮﻝ
ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ.
۱۹۲
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
Accumulation = Input – output + Generation – consumption
𝑦𝑑𝑧𝑑 𝑥𝑑𝑥+
𝑧𝑑𝑦𝑑𝑥𝑑 + 𝑆ɸ .
⌊− 𝜌𝑢ɸ
𝑦𝑑𝑧𝑑 𝑥𝑑𝑥+
ﺑﺎ ﺗﻘﺴﻴﻢ ﻃﺮﻓﻴﻦ ﺑﺮ dxdydzﻭ ﺍﻋﻤﺎﻝ ﺟﻬﺖ ﻫﺎﻱ yﻭ: Z
𝑦𝑑𝑧𝑑 𝑥
𝜕ɸ
�)
𝑥𝜕
𝜕
⌊(𝜌ɸ)𝑑𝑥 𝑑𝑦 𝑑𝑧 = 𝜌𝑢ɸ
𝑡𝜕
𝑓𝑓𝑒𝜇𝐶( −
𝑦𝑑𝑧𝑑 𝑥
𝜕ɸ
⌊
𝑥𝜕
⇒
𝑓𝑓𝑒𝜇𝐶 +
𝜕
𝜕
𝜕
𝜕
(𝜌ɸ) = − (𝜌𝑢ɸ) −
)(𝜌𝑣ɸ) − (𝜌𝑤ɸ
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝜕
𝜕
𝜕
𝜕ɸ
𝜕ɸ
𝜕ɸ
𝑓𝑓𝑒𝜇 �𝐶𝑥 .
�+
𝑓𝑓𝑒𝜇 �𝐶𝑦 .
𝑓𝑓𝑒𝜇 � + �𝐶𝑧 .
� + 𝑆ɸ
𝑥𝜕
𝑦𝜕
𝑧𝜕
𝑥𝜕
𝑦𝜕
𝑧𝜕
ﺑﺮﺍﻱ ﻭﺿﻌﻴﺘﻲ ﻛﻪ ﺧﺼﻮﺻﻴﺖ ﺩﻳﻔﻴﻮژﻥ ﻣﻘﺪﺍﺭ ﺍﺳﻜﺎﻟﺮ ɸﺩﺭ ﺟﻬﺖ ﻫﺎﻱ ﻣﺨﺘﻠﻒ ﻳﻜﺴﺎﻥ ﺑﺎﺷﺪ ﻣﻲ ﺗﻮﺍﻥ ﺩﺭ ﻧﻈﺮ
+
ﮔﺮﻓﺖ ﻛﻪ:
ﻛﻪ 𝛔ɸﻋﺪﺩ ﭘﺮﺍﻧﻮﻝ ﺑﺮﺍﻱ ﺍﻧﺘﺸﺎﺭ )ﺩﻳﻔﻴﻮژﻥ( ɸﺍﺳﺖ .ﺑﺮﺍﻱ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﺩﺍﺭﻳﻢ:
1
𝜎ɸ
= 𝑧𝐶 = 𝑦𝐶 = 𝑥𝐶
𝐷
𝜕 𝜇𝑒𝑓𝑓 𝜕ɸ
𝜕 𝜇𝑒𝑓𝑓 𝜕ɸ
= )(𝜌ɸ
�
. �+
�
. � + 𝑆ɸ
𝑡𝐷
𝑥𝜕 𝜕𝑥 𝜎ɸ
𝑦𝜕 𝜕𝑦 𝜎ɸ
ﺩﺭ ﺭﺍﺑﻄﻪ ﺑﺎﻻ :ﺳﺮﻋﺖ ﺍﺗﻼﻑ – ﺳﺮﻋﺖ ﺗﻮﻟﻴﺪ = Sɸﺍﺳﺖ.
ﺍﮔﺮ ɸﺭﺍ ﺑﻌﻨﻮﺍﻥ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ ﺟﺮﻳﺎﻥ ﻣﺘﻼﻃﻢ) (Kﺩﺭ ﻧﻈﺮﺑﮕﻴـﺮﻳﻢ ﺑـﺎ ﻣﻘﺎﻳﺴـﻪ ﻣﻌﺎﺩﻟـﻪ ﺑـﺎﻻ ﺑـﺎ ﻣﻌﺎﺩﻟـﻪ ﺍﻧـﺮژﻱ
ﺟﻨﺒﺸﻲ ﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ:
۱۹۳
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
𝜕𝑣̅ 2
𝜕𝑣̅ 𝜕𝑢� 2
𝜕𝑢� 2
� 𝑆𝑘 = 𝜇𝑒𝑓𝑓 �2 � � + 2 � � + � + � � −
𝐶�𝜌.
𝜀��𝐷�.
𝑦𝜕
𝑦𝜕 𝑥𝜕
𝑥𝜕
���������������������������
ﺍﺗﻼﻑ
ﺗﻮﻟﻴﺪ
ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ɸ=ℰﺑﺎﺷﺪ ﻣﻌﺎﺩﻟﻪ ﺑﻘﺎء ℰﺑﺪﺳﺖ ﻣﻲ ﺁﻳﺪ:
𝜀𝜕 𝑓𝑓𝑒𝜇 𝜕
𝐷
𝜀𝜕 𝑓𝑓𝑒𝜇 𝜕
= )𝜀𝜌(
. �+
𝜀𝑆 . � +
�
�
𝑥𝜕 𝜀𝜎 𝑥𝜕
𝑡𝐷
𝑦𝜕 𝜀𝜎 𝑦𝜕
ﺑﺎ ﻗﻴﺎﺱ ﻛﺮﺩﻥ ﺑﺎ ﺗﻮﻟﻴﺪ ﻭ ﺗﻠﻔﺎﺕ Kﻭ Sℰﺑﺼﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺘﻪ ﻣﻲ ﺷﻮﺩ:
𝜕𝑢� 2
𝜕𝑣̅ 2
𝛿 𝜕𝑣̅ 𝜕𝑢� 2
𝜀2
𝜌 𝑆𝜀 = 𝐶1 . 𝜇𝑒𝑓𝑓 �2 � � + 2 � � + � + � � − 𝐶2
𝑘
𝑥𝜕
𝑦𝜕
𝑘 𝑦𝜕 𝑥𝜕
ﺑﺪﻳﻦ ﺗﺮﺗﻴﺐ ﻣﻌﺎﺩﻟﻪ ℰﻛﻪ ﺩﺭ ﻣﺪﻝ K-ℰﺑﺮﺍﻱ ﻳﻚ ﺟﺮﻳﺎﻥ ﺩﻭ ﺑﻌﺪﻱ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲ ﮔﻴﺮﺩ؛ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ:
-1ﻣﻌﺎﺩﻟﻪ ﺑﻘﺎء ﺟﺮﻡ:
-2ﻣﻌﺎﺩﻟﻪ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻛﺲ ﺩﺭ ﺟﻬﺖ : x
̅𝑣𝜕 �𝑢𝜕
+
=0
𝑦𝜕 𝑥𝜕
-3ﻣﻌﺎﺩﻟﻪ ﻧﺎﻭﻳﻪ ﺍﺳﺘﻮﻛﺲ ﺩﺭ ﺟﻬﺖ : y
𝐷
̅𝑝𝜕
𝜕
�𝑢𝜕
𝜕
̅𝑣𝜕 �𝑢𝜕
(𝜌𝑢�) = −
+ 2 �𝜇𝑒𝑓𝑓 � +
�� �𝜇𝑒𝑓𝑓 � +
𝑥𝜕
𝑡𝐷
𝑥𝜕
𝑥𝜕
𝑦𝜕
𝑥𝜕 𝑦𝜕
-4ﻣﻌﺎﺩﻟﻪ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ:
𝐷
𝜕 ̅𝑝𝜕
�𝑢𝜕 ̅𝑣𝜕
𝜕
̅𝑣𝜕
(𝜌𝑣̅ ) = −
+
� 𝑓𝑓𝑒𝜇� �𝜇𝑒𝑓𝑓 � + �� + 2
𝑡𝐷
𝑥𝜕 𝑦𝜕
𝑦𝜕
𝑦𝜕 𝑥𝜕
𝑦𝜕
𝐷
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
𝑘𝜕 𝑓𝑓𝑒𝜇 𝜕
= )𝑘𝜌(
. �+
𝜀 . � + 𝜇𝑒𝑓𝑓 𝐺 − 𝜌. 𝐶𝐷 .
�
�
𝑡𝐷
𝑦𝜕 𝑘𝜎 𝑦𝜕
𝑥𝜕 𝑘𝜎 𝑥𝜕
𝜕𝑢� 2
𝜕𝑣̅ 2
𝜕𝑢� 𝜕𝑣̅ 2
� 𝐺 = 2� � + 2� � + � +
𝑥𝜕
𝑦𝜕
𝑥𝜕 𝑦𝜕
-5ﻣﻌﺎﺩﻟﻪ ﺗﻠﻔﺎﺕ ﺍﻧﺮژﻱ ﺟﻨﺒﺸﻲ:
𝜀 𝑓𝑓𝑒𝜇
𝐷
𝜀𝜕 𝑓𝑓𝑒𝜇 𝜕
𝜀𝜕 𝑓𝑓𝑒𝜇 𝜕
𝜀2
= )𝜀𝜌(
. �+
. � + 𝐶1
− 𝐶2 𝜌.
�
�
𝑡𝐷
𝑥𝜕 𝜀𝜎 𝑥𝜕
𝑦𝜕 𝜀𝜎 𝑦𝜕
𝑘
𝑘
۱۹٤
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
-6
ﺩﺭ ﺭﻭﺍﺑﻂ ﻓﻮﻕ 𝜇𝐶ﻭ 𝑘𝜎 ﻭ 𝜀𝜎 ﻭ 𝐷𝐶 ﻭ 𝐶2ﻭ 𝐶1ﺛﺎﺑﺖ ﻫﺎﻱ ﺗﺠﺮﺑﻲ ﻫﺴﺘﻨﺪ.
𝑘2
𝜌 𝜇𝐶 = 𝜇 +
𝜀
𝑓𝑓𝑒𝜇
𝐶𝜇 = 0.09 , 𝐶1 = 1.45 , 𝐶2 = 1.9 ,𝜎𝜀 = 1.3 , 𝜎𝑘 = 1
ﻣﻌﺎﺩﻻﺕ ﺑﺎﻻ ﺍﺯ ﻧﻈﺮ ﺣﻞ ﻋﺪﺩﻱ ﺷﺒﻴﻪ ﺑﻪ ﻫﻢ ﻫﺴﺘﻨﺪ.
۱۹٥
ﻓﺼﻞ ٦
ﺟﺮﻳﺎﻥ ﻣﺘﻼﻁﻢ
۱۹٦
ﭘﻴﻮﺳﺖ
۱۹۷
ﭘﻴﻮﺳﺖ
۱۹۸
ﭘﻴﻮﺳﺖ
۱۹۹
ﭘﻴﻮﺳﺖ
۲۰۰
ﭘﻴﻮﺳﺖ
۲۰۱