13.42 LECTURE 1:REVIEW OF BASIC PRINCIPLES OF FLUID MECHANICSSPRING 2003cA.H. TECHET & M.S. TRIANTAFYLLOU1. Basic Fluid MechanicsIn rigid body mechanics the motion of a body is described in terms of the body’s position in time.This body can be translating and possibly rotating, but not deforming. This description, followinga particle in time, is a Lagrangian description. In a fluid there are many particles and unlike rigidbodies, parcels of fluid and tend to deform continuously as they move. In order to fully describe theflow we must account for these deformations. Thus it is useful to use the Eulerian description, orcontrol volume approach, and describe the flow at every fixed point in space (x, y, z) as a functionof time, t.2. Governing Laws2.1. Conservation of Mass: Basic fluid mechanics laws dictate that mass is conserved within acontrol volume for constant density fluids. Thus the total mass entering the control volume mustequal the total mass exiting the control volume. Using figure 1 we can write a 2D mass balanceequation for the fluid entering and exiting the control volume δxδz. In the x-direction the massbalance equation is shown in equation 2.1,(2.1) {u +12∂u∂zδz} δz − {u +∂u∂xδx +12∂u∂zδz} δz = −∂u∂xδxδz.The first term on the LHS of eq. 2.1 represents the mass entering the control volume and the secondterm (on the LHS) the fluid exiting the control volume. A similar equation can be written for the13.42 Spring 2003.12 SPRING 2003cA.H. TECHET & M.S. TRIANTAFYLLOUδzuu+dudxδxu+dudxδxdudzδz+dudzδzu+δxc.v.Figure 1. Control Volume δxδz.mass balance in the z direction resulting in a net vertical fluid flux of(2.2) −∂w∂zδxδz.For an incompressible fluid, the sum of these two balances must be zero for mass to be conserved.(2.3) −(∂u∂x+∂w∂z)δxδ z = 0.This equation is valid for all δx and δz and can be simplified to arrive at the two-dimensionalequation for conservation of mass:(2.4)∂u∂x+∂w∂z= 0.Similarly in three-dimensions, the equation for mass conservation can be written as:(2.5)∂u∂x+∂v∂y+∂w∂z= 0.13.42 BASIC FLUID MECHANICS 3Recalling the gradient operator from vector calculus: ∇ = (∂∂x,∂∂y,∂∂z), we can abbreviate equa-tion 2.5 as(2.6) ∇ · V = 0.2.2. Newton’s Second Law: Newton’s second law is simply the law of conservation of momentum.It states that the time rate of change of momentum of a system of particles is equal to the sum ofexternal forces acting on that body.(2.7) ΣFi=ddt{MV}where M = ρδxδz is the mass of the fluid parcel (in two dimensions) and M V is the linearmomentum of the system (V is the velocity vector). Since the fluid density is constant, the time-rate of change of linear momentum can be written as(2.8)ddt{MV} = ρ δxδzdVdt.The rate of change of velocity of the fluid parcel can be found, for small δt, as(2.9)dVdt=limδt → 01δt{V(x + δxp, z + δzp, t + δtp) − V(x, z, t)}where we can substituteδxp= u δt and δzp= w δtinto equation 2.9 and cancel terms to arrive at a more familiar form of the momentum equation.The total derivative of the velocity is written as:(2.10)DVDt=∂V∂t+∂V∂xu +∂V∂zw.4 SPRING 2003cA.H. TECHET & M.S. TRIANTAFYLLOUwhich can be simplified, using the vector identity,(2.11) V · ∇ = u∂∂x+ v∂∂y+ w∂∂z.The total (material) derivative of the velocity is the sum of the conventional acceleration,∂V∂t, andthe advection term, (V · ∇)V:(2.12)DVDt=∂V∂t+ (V · ∇)V.Finally, the momentum equation from 2.7 can be rewritten as(2.13) ΣFi= ρDVDtδxδy3. ForcesThe LHS of equation 2.7 is the sum of the forces acting on the control volume. Contributionsfrom gravity and pressure both play a role in this term as well as any applied external forces.3.1. Gravity Forces: Force on a fluid volume due to gravity:(3.1) Fg= −(ρg δxδz)ˆk3.2. Pressure Forces: FP= P · A.Pressure force in x-direction:(3.2) FPx= {(p +12∂p∂zδz)δz − (p +12∂p∂zδz +∂p∂xδx)δ z = −∂p∂xδxδzPressure force in z-direction:(3.3) FPz= −∂p∂zδxδz13.42 BASIC FLUID MECHANICS 5Total pressure force in two dimensions:(3.4) FP= −(∂p∂x,∂p∂z)δxδ z = −∇p δxδz.4. Euler EquationSubstituting relations 3.1 and 3.4 for the gravity and pressure forces acting on the body, into themomentum equation 2.12 we arrive at(4.1) ρ{∂V∂t+ (V · ∇)V}δxδz = (−ρg δxδz)ˆk − ∇p δxδzfor any δx, δz. The final result is the Euler equation in vector form:(4.2) ρ{∂V∂t+ (V · ∇)V} = −ρgˆk − ∇p.We can further manipulate this equation with the vector identity: (V · ∇)V =12∇(V ·V), such thatthe Euler equation becomes(4.3) ρ{∂V∂t+12∇(V · V)} = −ρgˆk − ∇p.5. Irrotational flowFor irrotational flow the curl of the velocity must be zero.(5.1) ω = ∇ × V =ˆiˆjˆk∂∂x∂∂y∂∂zu v w= 0.(5.2) ω =ˆi(∂w∂y−∂v∂z) +ˆj(∂u∂z−∂w∂x) +ˆk(∂v∂x−∂u∂y) = 0.6 SPRING 2003cA.H. TECHET & M.S. TRIANTAFYLLOUFor 2D flow this reduces to∂u∂z=∂w∂x.6. Potential FlowDefine a potential, φ(x, z, t), that satisfies the basic laws of fluid mechanics: conservation of massand momentum, assuming incompressible and irrotational flow.u =∂φ∂xand w =∂φ∂z(6.1)which satisfy the irrotational condition since∂u∂z=∂∂z∂φ∂x∂w∂x=∂∂x∂φ∂zand φ is continuous.∂u∂z=∂w∂x7. Unsteady Bernoulli’s EquationThe p otential function can be substituted into equation 4.3 resulting in the unsteady BernoulliEquation.ρ{∂∂t∇φ +12∇V2} + ∇p + pg∇z = 0∇{ρ∂φ∂t+12ρV2+ p + ρgz} = 0UnsteadyBernoulli ⇒ ρ∂φ∂t+12ρV2+ p + ρgz = c(t)(7.1)8. Laplace EquationReturning to the conservation of mass equation, we can substitute in the relationship betweenpotential and velocity and arrive at a new form for the equation of mass conservation. This equation13.42 BASIC FLUID MECHANICS 7is the Laplace Equation which we will revisit in our discussion on linear waves.∂u∂x+∂w∂z= 0 ⇒∂2φ∂x2+∂2φ∂z2= 0(8.1)(8.2) LaplaceEquation ⇒ ∇2φ = 09. Useful ReferencesThe material covered in this section should be review. If you have not taken 13.021 or a similarclass dealing with basic fluid mechanics and water waves please contact the instructor. The referencesbelow are merely suggestions for further reading and reference.• J. N. Newman (1977) Marine Hydrodynamics MIT