Lecture 3: 1 Lecture 3.
Lecture 3: 2 Plan for today Summary of the key points of the last lecture. Review of vector and tensor products : the dot product (or inner product ) and the cross product (or vector product ). Part 1 : Description of fluid motion. (Chapter 2 of Arzel s notes.) Eulerian vs Lagrangian description Some terminology : streak lines, streamlines, stream-tubes Eulerian and material time derivative Conservation of mass Incompressibility Part 2 : Dynamics of fluid motion. (Chapter 3 of Arzel s notes.)
Lecture 3: 3 Navier-Stokes Equations
summary of lecture 2 4 Key points of last lecture (Lecture 1) In the planar Couette flow, where only one component of the flow, u x 0, and it varies linearly between the plates u x / y = U/L, there was a force F x on an area δa y of the plate, due to viscosity, given by F x δa = σ xy = µ u x y = µu L, (1) where the constant µ is the dynamic viscosity. We define the kinematic viscosity by ν µ ρ. (2) Mechanical pressure, p mec, is the pressure you would measure with a barometer. p mec includes the fluid static pressure plus that due to motion, of order kinetic energy density ρu 2 /2,
summary of lecture 2 5 where U is a typical velocity scale, and a small contribution from the viscose forces. It is calculated as the average normal stress in the fluid using the stress tensor σ ij : p mec = 1 3 σ ii = 1 3 (σ xx + σ yy + σ zz ). (3) Small Mach number flows M U/c s 1 are incompressible because ρ ρ = 1 2 M 2, (4) where ρ is the scale of (a typical size for) the density variations, U is a velocity scale, and c s is the speed of sound in the fluid. [For the atmosphere c s 340m/s and the fastest wind speed not related to tornadoes ever recorded was 113m/s, M = 1/3 (in a tornado, 135m/s, M = 0.4).] A static fluid has a balance between body forces and surface forces, the latter being only the normal stresses σ ij = δ ij p,
summary of lecture 2 6 where p is the fluid static pressure. The force balance on a (simply connected) volume V is expressed in integral form and one uses Gauss theorem to convert the surface integral over A to a volume integral, A F i n i da = V F i x i dv. Gauss theorem (5) In words, the outward flux of a tensor field F i n i through a closed surface A is equal to the volume integral of the divergence F i x i over the region V inside the surface. The most important force balance is the hydrostatic balance : p = ρf i (6) x i where p is the fluid static pressure, ρ is the density, F i is the component of the body force per unit mass along the Ox i -axis. This implies for the vertical component, when the
summary of lecture 2 7 body force is gravity F i = δ i3 g, p z = ρg (7) Archimedes principle was left as an exercise.
Math background 8 The dot product (or inner product ) and the cross product (or vector product ) In vector calculus we have the dot product (which I call sometimes by its more general term inner product ). For cartesian vectors we write u v = u x v x + u y v y + u z v z, (8) which in tensor notation becomes u i v i. Remember it is important that the indices are the same because that implies a summation 3 u i v i = u i v i = u 1 v 1 + u 2 v 2 + u 3 v 3 = u x v x + u y v y + u z v z, i=1 (9)
Math background 9 the final equality simply means we have labelled our three axes Ox, Oy, Oz with Ox 1, Ox 2, Ox 3 because this is often more convenient. Note that the dot product reduces the rank of the tensor by one. We started with two vectors (tensors of rank 1) and the inner product between them left us with a scalar (tensor of rank zero). In vector calculus we also have the cross product (sometimes referred to as vector product ). For cartesian vectors we write i j k u v = u x u y u z, v x v y v z = (u y v z u z v y ) i + (u z v x u x v z ) j + (u x v y u y v x ) k (10)
Math background 10 which in tensor notation becomes ɛ ijk u i v j or ɛ ijk u j v k where ɛ ijk is the alternating (or totally antisymmetry or Levi-Civita) tensor. Note here we start with two vectors and end up with a third vector, which is orthogonal to the first two vectors.
lecture 3 : kinematics 11 Lecture 3 : Fluid Kinematics : description of fluid motion Supplementary resources : Basic reference : Arzel notes, Chapter 2. Advanced reference : (Batchelor, 2000, Chapter 2).
lecture 3 : kinematics 12 Specification of the flow field Thanks to the continuum hypothesis, we have a definite notion of the local velocity of the continuous fluid at a point in space P at an instant in time t, u i (P, t). How do we describe the whole flow field? There are two possible descriptions, 1) the Eulerian description and 2) the Lagrangian description. The Eulerian description uses continuous functions u i (x, y, z, t) that describe the fluid velocity at each location (x, y, z, t) in the fluid. In our succinct notation we can write this u i (x j, t). (I had to use a second index, j, because i is already taken for the component of the velocity. The expression u i (x i, t) is meaningless.) The Lagrangian description focuses on the trajectories of
lecture 3 : kinematics 13 fluid parcels. The flow quantities are then functions of time and a parameter x0 i that labels fluid parcels, v i (t, x0 j ). We shall henceforth focus on the Eulerian description.
lecture 3 : kinematics 14 Some terminology : streak lines, path lines, streamlines, stream-tubes At a fixed instant in time, t P, the movement of the flow is described completely by the vector function u( x, t P ) of position x. Imagine starting at a given point in the field, x start and then start moving with the velocity there, u( x start, t P ) but then instantaneously changing to the local velocity as the position evolves. The resulting trajectory is called a streamline. A stream-tube is the surface formed by all the streamlines passing through a closed curve in the fluid at a given instant. At a fixed instant in time, t P, imagine releasing a tiny test particle that flows with the fluid. (A test particle is a useful concept in physics corresponding to a particle so tiny that it
lecture 3 : kinematics 15 does not affect its environment ; the electric, magnetic, gravitational fields are all undisturbed by the test particle, as is the flow field of the fluid, its density, temperature, etc. are all unaffected by the purely imaginary test particle.) The resulting trajectory is called a path line. Finally we define a streak line as the curve on which lie all fluid elements that pass through a given point x start at some earlier time. Imagine slowly releasing a dye at a fixed point x start. The dye will form a streak line. Only in the special case of a steady flow would the 3 lines, streamline, path line and streak line, correspond.
lecture 3 : kinematics 16 Exercises Explain why only in the special case of a steady flow would the 3 lines, streamline, path line and streak line, correspond.
lecture 3 : kinematics 17 Fluid parcel acceleration For the equations of motion, we will need the acceleration of a fluid parcel. In the Eulerian framework that we use, we must take a time derivative at a fixed point in space, say P, at an instant in time, t P. But this calculation must take into account that the fluid parcel is moving relative to our coordinate system, with velocity u i (x j (t), t). Note we have written the spatial coordinates following the fluid parcel as functions of time x j (t). Then any property of the fluid parcel, including its velocity, becomes a function of time, u i (x j (t), t). We can then differentiate this function of time to find the local
lecture 3 : kinematics 18 acceleration of the fluid parcel d dt u i(x j (t), t) = t u i(x j (t), t) + u i (x j (t), t) dx j(t) x j dt = t u i(x j, t) + u j u i (x j, t) Du i x j Dt (11) The term involving the rate of strain tensor u i / x j are called the velocity advection terms : u j x j u i (x j, t) (12) Notice this involves the product of a rank 2 tensor, u i / x j, with the vector u j. But the indices tell us how to multiply the components together, see exercise below. The same consideration applies for temperature or any other point-like property θ of fluid parcels. To find the local time rate of change of the quantity at a fixed point, we take the
lecture 3 : kinematics 19 partial derivative, t θ(x i, t) Eulerian derivative (13) To find the time rate of change of the quantity following the fluid parcel at the instant it passes a fixed point, we must include the advective terms, d dt θ(x i(t), t) = t θ(x j, t) + u j x j θ(x j, t) material derivative (14) To distinguish these two derivatives we use the terms Eulerian and material or substative and we use capitals D/Dt to emphasize that it is special. For a scalar quantity θ (or a vector quantity with component u i = θ ) the material
lecture 3 : kinematics 20 derivative is D Dt θ(x j, t) d dt θ(x j(t), t) = t θ(x j, t) + u j x j θ(x j, t). (15) Table 1 Different time derivatives math physical interpretation standard name t θ(x i, t) D Dt θ(x j, t) time rate of change of θ at point x i time rate of change following fluid parcel Eulerian time derivative Material (or substantive or total) time derivative
lecture 3 : kinematics 21 Exercise : material derivative of a vector quantity Write out all three velocity advection terms u j u i / x j, each of which is the sum of three terms. Try to write this in vector notation. Hint : use parentheses to clarify how the products are to be interpreted. An anemometer measures the wind speed s on the top of a tower and a wind vane measures the angle θ from true north. Consider an inertial reference frame with origin at the tower, and Ox axis in the eastward direction, Oy axis in the northward direction. Write expressions for the u x and u y components of the horizontal velocity in terms of s and θ. (The vertical velocity is generally much much less.)
lecture 3 : kinematics 22 The plots of u x and u y are shown in Fig.1. The slope of the u x plot at instant t P has value u x / t. This corresponds to which time derivative, the Eulerian or material?
lecture 3 : kinematics 23 Wind velocity u x Dt Du x u y t t p Figure 1 Time series of two components of wind velocity. At instant t P the plot of u x is decreasing with slope u x / t, as indicated by the tangent line drawn at that point.
lecture 3 : kinematics 24 The net force on an infinitesimal fluid parcel of mass δm at point P at instant t P is δf i. By Newton s 2nd law, the acceleration of fluid parcel δa i = δf i /δm. Furthermore, the acceleration is the time derivative of the velocity u i. Which time derivative (Eulerian or material) applies here?
Conservation of mass the continuity equation 25 Conservation of mass Consider a closed surface A, fixed relative to our inertial reference system, that encloses a volume V entirely inside the fluid. ρ(x i, t) is the fluid density at instant t at the point with coordinates x i. We temporarily suppress the arguments (x i, t) since they play no role here. Consider further the case where we have no sources or sinks of mass (as is most often the case in typical fluid mechanics problems). The mass of the fluid enclosed by A is ρdv (16) The velocity of a fluid has units of a volume flux per unit V
Conservation of mass the continuity equation 26 area : m s = m3 m 2 t. Indeed we can think of velocity this way. Consider a straight tube of cross-sectional area δa filled with homogenous fluid moving uniformly at velocity u along the axis of the tube, see Fig.2. The blue-shaded fluid, and only this fluid, has passed the dashed line at P within the time period δt. The volume of this fluid is δv = δaδl = δa(uδt), (17) which implies the volume flow rate per unit area is u = δv δtδa.
Conservation of mass the continuity equation 27 da P n dv = da dl = daudt u dl = udt Figure 2 Tube of crossectional area δa filled with homogenous fluid moving uniformly at velocity u along the axis of the tube. The blue-shaded fluid, and only this fluid, has passed the dashed line at P within the time period δt.
Conservation of mass the continuity equation 28 The volume flow rate per unit area across a surface that is not normal to the fluid velocity, i.e. n is not parallel to u but rather there is an angle θ between them, is calculated by See Fig.3. u i n i = u n = u cos(θ) = u δa normal δa oblique. (18)
Conservation of mass the continuity equation 29 da oblique da normal q n u dv = da normal dl = da normal udt =da oblique cosq udt dl = udt Figure 3 The volume flow rate is identical to that in Fig2 the volume of blue-shaded fluid has not change because the two triangular regions lost are compensated exactally by the two gained. But the flow rate per unit area δa oblique is less by a factor cos θ = δa normal δa oblique.
Conservation of mass the continuity equation 30 The flux of mass through the surface per unit area is ρ(u i n i ), evaluated at the boundary A where the normal to the boundary is the unit vector n i so that the total flux through the boundary is ρu i n i da. (19) This flux represents the only way for the mass in V to A
Conservation of mass the continuity equation 31 change (the flux is positive for mass leaving) : 0 = ρdv + ρu i n i da, conservation of mass t V A ρ = V t dv + ρu i n i da, bndy fixed, ρ continuous A ρ = V t dv + (ρu i )dv, Gauss theorem V x i ρ = t + (ρu i )dv, simplify x i V = ρ t + x i (ρu i ) = 0. (20) The result Eq(20) is of fundamental importance in fluid mechanics and is usually called conservation of mass or the continuity equation. The equation can be derived in several different ways, that are equally informative, see references. It can be written in another form ; by expanding the second
Conservation of mass the continuity equation 32 term we find Dρ Dt = ρ u i x i (21) The physical interpretation is that, the time derivative of the density of the fluid measured while following the fluid motion, Dρ/Dt, is equal to minus the rate of increase of the volume per unit volume u i x i times the mass per unit volume ρ.
Incompressibility solenoidal flow 33 Incompressibility This course is about incompressible fluid flow. This means, at least to a reasonable approximation, that the volume of fluid parcels does not change. Mathematically this translates to the flow divergence being negligible : u i x i = 0. (22) But when is a fluid flow incompressible? How can we anticipate this important property? Rewrite the continuity equation Eq(21) by isolating the divergence term 1 ρ Dρ Dt + u i x i = 0. (23) Analyzing the order of magnitude of each term, we note that
Incompressibility solenoidal flow 34 Dρ Dt = O( ρ/t ). That is, we anticipate the material derivative of density to have an order of magnitude given by the scale of the variations in density ρ divided by a time scale, T. For the latter, we use the advective timescale, T = L/U, where L is a typical length scale and U is a typical velocity scale. So the order of magnitude of the first term is O ( 1 ρ Dρ Dt ) = ρ ρ The same advective time scale gives the order of magnitude of the individual terms in the divergence O ( u x u i = u x i x + v y + w z, ) ( ) v = O = O y U L ( w z Recall it was the mean normal stress, the so-called ) (24) = L U. (25)
Incompressibility solenoidal flow 35 mechanical fluid pressure p mec that determined the amount of compression, ρ/ρ = O(M 2 /2) 1 for flows much slower than the sound speed, U c s. Substituting these scalings in Eq(23) and multiplying by L/U we have ρ/ρ = O(M 2 /2) = L U ( u x + v y + w ) z (26) The RHS of Eq(26) contains the sum of three terms that, when multiplied by L/U are of O(1). If these three terms we unrelated to each other, then the RHS would be order O(1) as well, yet the LHS is O(M 2 /2) 1, which seems to be a contradiction. The way out of this contradiction is the possibility, in fact the inevitability for some flows, that ( ) ( ) ( ) ( ) ui u v w O O = O = O = U (27) x i x y z L
Incompressibility solenoidal flow 36 because the individual terms in the divergence are organized to cancel : ( u v x y + w ). (28) z Fluid flows for which the flow divergence is negligible, are said to be incompressible or solenoidal. u i x i = 0, (29)
Dynamics 37 Part 2 : Dynamics of fluid motion
Dynamics 38 Navier-Stokes equation We now derive the fundamental governing equation of fluid motion, i.e. the Navier-Stokes equation, by applying Newton s second law and the continuity equation Eq(21) to a volume of fluid. Finally we will simplify the viscose term by assuming an incompressible, Eq(29), and Newtonian fluid. Consider a fixed volume V of fluid bounded by a surface A that is stationary with respect to our inertial reference frame. Applying Newton s law, the total time rate of change of the i th component of momentum in the volume must balance the net i th component of force on the fluid within the volume : V ρ Du i Dt dv = V ρf i dv + A σ ij n j da, (30)
Dynamics 39 where F i is the i th component of force per unit mass, and, as usual, σ ij is the fluid stress tensor, and n j is the outward facing unit vector normal to the surface A. Consider the first term : ρ Du i Dt dv = ρ V V ( ui t + u j ) u i dv x j used Eq(11) (31) Consider the last term : σ ij σ ij n j da = dv Gauss theorem (32) x j A Combining these two results we have [ ( ) ui ρ t + u u i j ρf i σ ij x j x j V V ] dv = 0. (33)
Dynamics 40 Because the volume is arbitrary, this implies ( ) ui 0 = ρ t + u u i j ρf i σ ij, x j x j u i t + u j u i x j = F i + 1 ρ σ ij x j, (34) = F i + 1 ( pδ ij + 2µe ij 2 ρ x j 3 µe kkδ ij ), ( ) = gδ iz 1 p 2 u i + ν ρ x i x 2 + 1 u j j 3 x i x j (35) In going from the second to third line we assumed a Newtonian fluid and used Eq(14) from Lecture 1. In going from the third to final line we assumed the body force was gravity in the z direction and used Eqs(15,16) from Lecture 1.
Dynamics 41 Exercise Verify in detail the last result, Eq(35).
Dynamics 42 Références Batchelor, G. K. (2000), An Introduction to Fluid Dynamics, 615 pp., Cambridge University Press, Cambridge, UK, 615 + xviii pp + 24 plates.