13.021 { Marine Hydrodynamics, Fall 2004 Lecture 3 Copyright °c 2004 MIT - Department of Ocean Engineering, All rights reserved. 13.021 { Marine Hydrodynamics Lecture 3 1.2 - Stress Tensor Stress Tensor ¿ij:. The stress (force per unit area) at a point in a fluid needs nine components to be completely speci¯ed, since each component of the stress must be de¯ned not only by the direction in which it acts but also the orientation of the surface upon which it is acting. The ¯rst index speci¯es the direction in which the stress component acts, and the second identi¯es the orientation of the surface upon which it is acting. Therefore, the ith component of the force acting on a surface whose outward th normal points in the j direction is ¿ij. X2 22 12 32 21 23 11 13 31 33 X1 X3 Figure 1: Shear stresses on an in¯nitesimal cube whose surface are parallel to the coordinate system. 1 X2 2 P nˆ A1 1 A3 X1 Q 3 R areaA 0 A2 X3 ¡ Figure 2: Consider an in¯nitesimal body at rest with a surface PQR that is not perpendicular to any of the Cartesian axis. The unit normal vector to that surface isn ^ = n1x^1 + n2x^2 + n3x^3. The area of the surface = A0, and the area of each surface perpendicular to Xi is Ai = A0ni, for i = 1; 2; 3. P Newton's law: all4faces Fi = (volume force)i for i = 1, 2, 3 If ± is the typical dimension of the body : surface forces » ±2 : volume forces » ±3 An example of surface forces is the shear force and an example of volumetric forces is the gravity force. At equilibrium, the surface forces and volumetric forces are in balance. As the body gets smaller, the mass of the body goes to zero, which makesP the volumetric forces equal to zero and leaving the sum of the surface forces equal zero. So, as ± ! 0; all4faces Fi = 0 for i = 1; 2; 3 and ) ¿iA0 = ¿i1A1 +¿i2A2 +¿i3A3 = ¿ijAj. ButP the area of each surface ? to Xi is Ai = A0ni. Therefore ¿iA0 = ¿ijAj = ¿ij(A0nj), where ¿ijAj is the notation (represents the sum of all components). Thus ¿i = ¿ijnj for i = 1, 2, 3, where ¿i is the th component of stress in the i direction on a surface with a normal ~n . We call ¿ i the stress vector and we call ¿ij the stress matrix or tensor. Example: Pascal's Law for hydrostatics In a static fluid, the stress vector cannot be di®erent for di®erent directions of the surface normal since there is no preferred direction in the fluid. Therefore, at any point in the fluid, the stress vector must have the same direction as the normal vector ~n and the same magnitude for all directions of ~n . 2 no summation z }| { Pascal's Law: for hydrostatics ¿ij = ¡ (pi)(±ij) 2 3 ¡p1 0 0 ¿ = 4 0 ¡p2 0 5 » 0 0 ¡p3 th where pi is the pressure acting perpendicular to the i surface. If po is the pressure acting perpendicular to the surface PQR, then ¿i = ¡nip0 , but ¿i = ¿ijnj = ¡(pi)±ijnj = ¡(pi)(ni). Therefore po = pi , i = 1, 2, 3 and ~n is arbitrary. Symmetry of the Stress Tensor To prove the symmetry of the stress tensor we follow the steps: j ij ji ji o i ij Figure 3: Material element under tangential stress. P 1. The of surface forces = body forces + mass£ acceleration. Assume no symmetry. Balance of the forces in the ith direction gives: 2 (±)(¿ij)T OP ¡ (±)(¿ij)BOT T OM = O(± ); since surface forces are » ±2, where the O(±2) terms include the body forces per unit depth. Then, as ± ! 0; (¿ij)T OP = (¿ij)BOT T OM . P 2. The of surface torque = body moment + angular acceleration. Assume no symmetry. The balance moment with respect to o gives: 3 3 (¿ji±)± ¡ (¿ij±)± = O(± ); 3 since the body moment is proportional to ± . As ± ! 0 , ¿ij = ¿ji. 1.3 Mass and Momentum Conservation Consider a material volume #m and recall that a material volume is a ¯xed mass of material. A material volume always encloses the same fluid particles despite a change in size, position, volume or surface area over time. 1.3.1 Mass Conservation The mass inside the material volume is: ZZZ M(#m) = ½d# #m(t) Sm(t) ϑ m )t( Figure 4: Material volume #m(t) with surface Sm(t). Therefore the time rate of increase of mass inside the material volume is: RRR d d dt M(#m) = dt ½d# = 0; #m (t) which implies conservation of mass for the material volume #m. 4 1.3.2 Momentum Conservation th The velocity of fluid inside the material volume in the i direction is denoted as ui. Linear momentum of the material volume in the ith direction is ZZZ ½uid# #m(t) Newton's law of motion: The time rate of change of momentum of the fluid in the material control volume must equal the sum of all the forces acting on the fluid in that volume. Thus: d (momentum)i =(body force)i + (surface force)i dt ZZZ ZZZ ZZ d ½u d# = F d# + ¿ n dS dt i i |ij{z}j #m(t) #m(t) Sm(t) ¿i RRR RR Divergence Theorems: For vectors: |r{z ¢ ~v}d# = ½⊃ |{z}~v:n^ dS # S @vj vj nj RRR @xj RR @¿ij For tensors: d# = ½⊃ ¿ijnjdS @xj # S Thus using divergence theorems: RRR RRR ³ ´ d @¿ij ½uid# = Fi + d#, dt @xj #m(t) #m(t) which gives the conservation of the momentum for the material volume #m. 1.4 Kinematic Transport Theorems Consider a flow through some moving control volume #(t) during a small time interval ¢t. Let f (~x;t) be any (Eulerian) fluid property per unit volume of fluid (e.g. mass, momentum, etc.). Consider the integral 5 ZZZ I(t) = f (~x;t) d# #(t) According with the de¯nition of the derivative, we can write d I(t + ¢t) ¡ I(t) I(t) = lim dt ¢t!0 8 ¢t 9 >ZZZ ZZZ > 1 < = = lim f(~x;t + ¢t)d# ¡ f(~x;t)d# ¢t!0 ¢t :> ;> #(t+¢t) #(t) S(t+ ∆t) ϑ t( + ∆ )t ϑ ( t) S(t) Figure 5: Control volume # and its bounding surface S at instants t and t + ¢t. Next, we consider the steps 1. Taylor series expansion of f about t. @f f(~x;t + ¢t) = f(~x;t) + ¢t (~x;t) + O((¢t)2) @t RRR RRR RRR RRR RR 2. d# = d# + d# where d# = [Un(~x;t)¢t] dS and Un(~x;t) is the normal velocity #(t+¢t) #(t) ¢# ¢# S(t) of S(t). 6 S(t+ ∆t) S(t) G ∆ + ∆ 2 U n )t,x( t O( )t dS Figure 6: Element of the surface S at instants t and t + ¢t. So we have 8 9 <>ZZZ ZZZ ZZ ZZZ => d 1 @f 2 I(t) = lim d#f + ¢t d# + ¢t dSUnf ¡ d#f + O(¢t) dt ¢t!0 ¢t :> @t ;> #(t) #(t) S(t) #(t) Kinematic Transport Theorem (KTT) » Leibnitz rule in 3D RRR RRR RR d @f(~x;t) dt f(~x;t)d# = @t d# + f(~x;t)Un(~x;t)dS #(t) #(t) S(t) If the control volume is a material volume: #(t) = #m(t) and Un = ~v ¢ n^, where ~v is the fluid particle velocity. Then the Kinematic Transport theorem (KTT) assume the form ZZZ ZZZ ZZ d @f(~x;t) f(~x;t)d# = d# + f(~x;t)(~v ¢ n^) dS dt @t | {z } #m(t) #m(t) Sm(t) fvini(Einstein Notation) Using the divergence theorem: ZZZ ZZ |r{z ¢ ~®}d# = ½⊃ |~®{z¢ n^}dS (1) @ ® n # ®i S i i @xi 1st Kinematic Transport Theorem (KTT) 7 2 3 RRR RRR 6 7 d f (~x;t) d# = 6 @f(~x;t) + r ¢ (f~v)7 d#; dt 4 @t | {z }5 #m(t) #m(t) @ (fvi) @xi where f is a fluid property per unit volume. 1.5 Continuity Equation Let the fluid property per unit volume be mass per unit volume ( f = ½) ZZZ ZZZ · ¸ d @½ 0 = ½d# = + r ¢ (½~v) d# " dt " @t conservation # (t) 1stKTT # (t) of mass m m since #m is arbitrary, so the integrand ´ 0 everywhere. Therefore, the di®erential form of conservation of mass i.e. Continuity equation follows: @½ + r ¢ (½~v) = 0 @t @½ + [~v ¢ r½ + ½r ¢ ~v] = 0 |@t {z } D½ Dt Therefore, D½ Dt + ½r ¢ ~v = 0 In general, ½ = ½(p; T; : : :). We consider the special case of incompressible flow (Note, the density of the entire flow is not constant when we have more than one fluid, like water and oil, as illustrated in the picture above). 8 Constant ρ fluidparticle ρ2 fluidparticle oil water ρ1 Figure 7: Interface of two fluids (oil-water) Therefore, for an incompressible flow: D½ Dt = 0 @v Then r ¢ ~v or i = 0, which is the Continuity equation for incompressible fluid. @x | {z i } rate of volume dilatation 1.6 Euler's Equation (di®erential form of conservation of momentum) 2nd Kinematic Transport Theorem ( = 1st KTT + continuity equation). If G = fluid property per unit mass, then ½G = fluid property per unit volume ZZZ ZZZ · ¸ d @ ½Gd# = (½G) + r ¢ (½G~v) d# dt @t # (t) # (t) m m 2 3 ZZZ 6 µ ¶ µ ¶7 6 7 6 @½ @G 7 = 6G + r ¢ ½~v + ½ + ~v ¢ rG 7 d#; 4 @t @t 5 #m(t) | {z } | {z } DG 0 from mass conservation Dt and the 2nd Kinematic Transport Theorem (KTT) follows: RRR RRR d DG dt ½Gd# = ½ Dt d# #m #m 9 Application: th We consider G as the i momentum per unit mass ( vi).