Chapter 4 Differential Relations For Viscous Flow 4.1 Preliminary Remarks Two ways in analyzing fluid motion (1) Seeking an estimate of gross effects over a finite region or control volume Integral (2)Seeking the point-by-point details of a flow pattern by analyzing an infinitesimal region of the flow Differential
1 Chapter 4 Differential Relations For Viscous Flow 4.1 Preliminary Remarks * Two ways in analyzing fluid motion (1) Seeking an estimate of gross effects over a finite region or control volume. Integral (2) Seeking the point-by-point details of a flow pattern by analyzing an infinitesimal region of the flow. Differential
x Viscous flow Viscosity is inherent nature of real fluid straint(剪切) is very strong in internal flow. x Two forms of flow Turbulent(湍,紊)flow, laminar(层)flow Turbulent Flow Vs Laminar Flow Transition Reynolds tank Reynolds Re= number 惯性力/粘性力 Osbrone Reynolds
2 Turbulent Flow VS. Laminar Flow * Two forms of flow Turbulent（湍，紊） flow, laminar（层）flow * Viscous flow Viscosity is inherent nature of real fluid. Strain(剪切) is very strong in internal flow. Transition Re UL = Reynolds number Osbrone Reynolds Reynolds tank 惯性力/粘性力
4.2 The acceleration Field of a fluid V(r,t=iu(x, y,z, t)+jv(x,y, z, t)+hw(,y,z, t) C a dt +k dt at Ox ot oy at az at ou l一+1一+ Nonlinear terms Local acceleration Convective acceleration a ≠0 unsteady ≠0 nonuniform dt
3 4.2 The Acceleration Field of a Fluid V r t iu x y z t jv x y z t kw x y z t ( , ) ( , , , ) ( , , , ) ( , , , ) = + + dV du dv dw a i j k dt dt dt dt = = + + du u u x u y u z dt t x t y t z t = + + + u u u u u v w t x y z = + + + Local acceleration unsteady 0 t Convective acceleration nonuniform 0 i x Nonlinear terms
a +(.V at In the like manner Any property DΦad (.V)d Dt at =+(.V) Substantial(Material derivative Dt at 随体(物质、全)导数
4 V V t V a + ( ) = + = (V ) Dt t D In the like manner Any property Φ , ... T ( ) D V Dt t = + Substantial (Material) derivative 随体（物质、全）导数
Example Given v=3ti+xzj+ty2k Find the acceleration of a particle u=3t, v=xz, w=ty u ou l—+1+W dt a OX dy a 一+-+1-+1 3t2+tyx OX dw aw aw aw aw +u +1 2+2xz dt at ax a
5 Example Given . Find the acceleration of a particle. 2 V ti xzj ty k = + + 3 2 u t v xz w ty = = = 3 , , du u u u u u v w dt t x y z = + + + = 3 dv v v v v u v w dt t x y z = + + + 2 = + 3tz ty x dw w w w w u v w dt t x y z = + + + 2 = + y tyxz 2
4.3 Differential Equation of Mass conservation X inlet(mass flow) → mass flux per unit area 12+0(pu)b X outlet pudydz Ipu+cou dx]dya= z dz ax dx X flow out o(pu dxdydz Infinitesimal fixed cv In the like manner a(p Lxdvdz z Flow out off the Cv V. pl)dxdydz
6 X inlet (mass flow) udydz X outlet ( ) [ ] u u dx dydz x + dx y z x dz dy Infinitesimal fixed CV udydz ( ) [ ] u u dx dydz x + u mass flux per unit area → X flow out ( ) u dxdydz x 4.3 Differential Equation of Mass Conservation In the like manner ( ) v dxdydz y ( ) w dxdydz z Flow out off the CV ( ) V dxdydz
Loss of mass in the cv op ydvdz Ot a(pu dxdydz+o(pv) dxdydz (p) dxdyd az at at a(pu) dxdydx+ a(pv dxdydz a(pr OX Bp, d(pu) a(pv)*20 ),a( +V·(p)=0 at ax at p dp a Av.=0 dp+pv.V=0 dt
7 ( ) ( ) ( ) u v w dxdydz dxdydz dxdydz dxdydz x y z t + + = − ( ) ( ) ( ) 0 u v w t x y z + + + = ( ) 0 V t + = u v w V 0 t x y z + + + + = + V = 0 dt d Loss of mass in the CV dxdydz t − ( ) ( ) ( ) 0 u v w dxdydz dxdydz dxdydz dxdydz t x y z + + =
For steady flow V·()=0 For incompressible flow vv-0 Example 1 Under what conditions does the velocity field V=(a,x+b,y+Cz)i+(a,x+6,y+c22)j +(asx+bay+crzk represents an incompressible flow which conserves mass?( where ai,bi,c=const
8 For steady flow = ( ) 0 V For incompressible flow V = 0 Example 1 Under what conditions does the velocity field 1 1 1 2 2 2 3 3 3 ( ) ( ) ( ) V a x b y c z i a x b y c z j a x b y c z k = + + + + + + + + represents an incompressible flow which conserves mass? ( where ) , , a b c const iii =
Solution u=ax+,y+Cz v=a2x+b,y+C22 w=aix+bay+Cz Continuity for incompressible flow × +c=0→a1+b2+c3=0 Example 2 An incompressible velocity field u=a(x2-y,w=b, a, b are const, what v=? Solution ouX 02ax+ 0 v=-2axy+((x, t) An arbitrary function of x, z, t
9 Solution 0 u v w x y z + + = 1 1 1 u a x b y c z = + + 2 2 2 v a x b y c z = + + w a x b y c z = + + 3 3 3 1 2 3 a b c + + = 0 Continuity for incompressible flow Example 2 An incompressible velocity field: u=a(x2 -y 2 ),w=b, a,b are const,what v=? Solution 0 u v w x y z + + = 2 0 v ax y + = 2 v ax y = − v axy f x z t = − + 2 ( , , ) An arbitrary function of x,z,t
Assignment P264:P41(a)2P42,P4.4,P49(a)
10 Assignment: P264: P4.1(a), P4.2, P4.4 ,P4.9(a)
