流体力学

τ=µ dα dt
if µ =const, the fluid is called the Newtonian fluid. e.g. all gases and water. µ ≠ const, the fluid is the Non-Newtonian fluid. (¦p°ª¤À¤l )
1-2 Forces in the fluid
τ ∝α τ = Gα
The proportional constant is G, called shear modulus. For solid: α ≠ α (t )
by rhyen
chap1_v200_Introduction.doc
1/8
if G = constant , hook solid or elastic solid. if G ≠ const
5/8
l
Sign convention : A stress component is considered positive when the direction of the stress component and the plane on which it acts are both positive or both negative. A stress component is considered negative when the direction of stress component and the plane on which it acts are of opposite sign.(¥k¤â©w«h )
α
t=0
t>0
Deformation vs. shear force: ¹ï shear force ¤Þ°_ªºÅܧÎ,¥Î angular deformation( α )©w¸q¤§¡CDeformation ªº ¤j¤p»P shear force ªº¤j¤p¦³Ãö . ±q¹êÅçÆ[¹î±o¨ì¤U-ÓÃö«Y¦¡ : Mathematical Model:
F j i : surface direction Ai → 0 A j j : force direction

σ xx τ xy τ xz Notation τˆ ≡ τ yx σ yy τ yz (τ ˆ: tensor) τ zx τ zy σ zz
²Ä
1
³¹
introduction
1-1 Solid and Fluid
l l ©TÅé¤O¾Ç (§÷®Æ¤O¾Ç )¡G ±´°Q ©TÅé¨ü¤O»PÅܧΪºÃö«Y¥H¤Î²£¥Íªº¹B°Ê¡C Fluid mechanics: (¬yÅé¤O¾Ç)¡G±´°Q ¬yÅé¨ü¤O»PÅܧΪºÃö«Y¥H¤Î²£¥Íªº¹B°Ê¡C n Question: ¨âªÌªº®t²§¦ó¦b¡H n There are three possible phases in which all matter can exist: solid , liquid and gas. Liquid and gas are called fluid. Substance deforms if a force applied on it. e.g. shear deformation of solid and fluid. u solid can resist a shear stress u fluid will continuously deform under the action (³o-Ó§Ú-̦b¥»¬y ¤O½Ò±ý¤ÀªRªº¬yÅé )of shear stress no matter how small the shear stress may be. u ¥t¤@ºØ¬yÅé , ¦p Clay suspension and tooth paste require a minimum shear to cause motion. Question: What is the shear stress? Ans: τ = F A : the direction of force is perpendicular to the direction of the surface. shear force ºû«ù¤£ÅÜ,«h α (angular deformation)¥ç¤£ÀH®É¶¡ÅÜ .
l l Body force: distributed over the volume of fluid . n e.g. gravitational, magnetic, electric forces. Surface force: acting on the boundary through direct contact. r ˆδA , n e.g. friction(shear), pressure(normal) , δA = n ˆ :direction of the surface n n
3. 4.
If µ = 0 ⇒ inviscid fluid. Therefore, the shear stress is zero, even if the velocity gradient exists.(³o¬O ²z·Qªº±¡ªp,¹ê»Ú¤W¨¥ , µ ≠ 0 always). In general, a surface direction can be decomposed into 3 directions. Each surface direction has 3 stress components. Therefore, it forms 9 stress components for an infinitesimal volume element. r F r stress ≡ lim r A→ 0 A r ˆ A = iˆAx + ˆ jAy + k Az r ˆF ˆF + ˆ F =i j Fy + k x z In general :τ ij = lim
σ xx = lim τ xy = lim
δFx δ Ax →0 δA x δF y
δAx δF z τ xz = lim δA x →0 δA x
δ Ax →0
Figure: Stress components in rectangular coordinates: (a) two-dimensional stress components; (b) three-dimensional stress components.
5.
For the infinitesimal element shown below, there are six planes (two x planes, two y planes, two z planes)on which stresses may act.
by rhyen
chap1_v200_Introduction.doc
(2)Fluid
α
α
t = 0
α = α (t ) for athematical model: Rate of deformation is proportional to the shear stress.
τ ∝ dα dt
where the rate of deformation is defined as the rate of angular deformation. The proportional constant is called the dynamic viscosity.
by rhyen
chap1_v200_Introduction.doc
3/8
1-3 Relationship between stress and velocity gradient
dl B C dα dy B C
→ u + du
A dx t = 0
D
→ u
A t = ∆t
D
At t=0, no force applied on the surface element ABCD. At the time interval ∆t , shear force is applied on the surface BC normal to the y direction. Assume the velocity at A is u A , and the velocity at B is u B . Then the difference of the velocity at B from A can be obtained by the Taylor’s series expansion ∂u du = u B − u A = ( ) A dy ∂y The displacement of particle B is dl = (u + du − u ) × dt = dudt . The shear strain(or angular deformation): ∂u dl dudt du dα = tan −1 = tan −1 ≈ dt = ∂y dt dy dy dy dα du ∂u The rate of shear strain: α &= = = . dt dy ∂y i.e. The rate of shear strain =velocity gradient. ∂u ⇒τ = µ ← Newton' s Viscosity Law ∂y Express the shear stress more specifically,
1-4 Unit of viscosity
τ yx = µ du dy
F 2 τ FT M [µ ] = yx = L = = du L T L2 LT L dy [µ ] = N ⋅ S m 2 or kg m ⋅ s