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