2 Law of conservation of energy

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1 1 Newtonian viscous Fluid 1 Newtonian fluid For a Newtonian we already have shown that σ ij = pδ ij + λd k,k δ ij + 2µD ij where λ and µ are called viscosity coefficient. For a fluid under rigid body motion (i.e. zero rate of deformation) we have p = 1 3 σ ii and p = σ ii (nosum) i.e. pressure is the total compressive normal stress on any plane as well as the mean of the normal stresses. But in this case we have p + (λ + 2µ/3)D kk = 1 3 σ ii where k = λ + 2µ/3 is known as coefficient of bulk viscosity. It is clear that when D ij are non zero, p is neither the total compressive normal stress on any plane unless the viscous componenets happen to be zero as well as is not the mean of the normal stresses. Thus we can interpret pressure such that pδ ij is that part of σ ij which does not depend eplicitly on the rate of deformation. If we enforce the condition that the pressure is mean of the compressive normal stresses, we must have λ = 2µ/3 which is called Stoke s condition. Thus choosing µ as the only scaler constant we write σ ij = pδ ij + 2µ (D ij 13 ) D kkδ ij 2 Law of conservation of energy The law states that the material derivative of kinetic plus internal energies is equal to the sum of the rate of work of the surface and body forces, plus all other energies

2 that enter and leave the body per unit time. Other enrgies may include thermal, electrical, magnetical etc. Here we only cosidered mechanical and thermal energies. Let e be the specific internal energy or internal energy per unit mass. The conservation of energy is stated as D ρ (e + 12 ) Dt v iv i dv = ρb i v i dv + S t (n) i v i ds + ρ r dv q i n i ds S where r is specific rate at which heat is produced by internal sources and q is the heat flu vector. Now the term on the left hand side can be written as D ρ (e + 12 ) Dt v iv i dv ( = ρ De Dt + v iρ Dv ) i dv Dt ( = ρ De ) Dt + v i(ρb i + σ ij,j ) dv ( = ρ De ) Dt + ρv ib i + (σ ij v i ), j σ ij v i,j dv ( = ρ De ) Dt + ρv ib i σ ij D ij dv + t (n) i v i ds S 2 Hence we have Since the volume is arbitrary we get ρ De Dt dv = σ ij D ij dv + ρr dv q dv ρ De Dt = σ ijd ij + ρr q The term σ ij D ij is called stress power. For Newtonian iscous fluid we can eplicitly calculate the stress power. We have σ ij D ij = ( pδ ij + τ ij )D ij = p v + Φ where Φ is the dissipation function. The function Φ is positive definite since we have ( Φ = 2µ D ij 1 ) 3 ( v)δ ij D ij = 2µ (D ij D ij 13 ) ( v)2 which can be written as ( 2µ D ij 1 ) ( 3 ( v)δ ij D ij 1 ) 3 ( v)δ ij >

3 3 Alternative form of energy equation is also possible using definition of enthalpy h = e + p ρ We have Dh Dt = De Dt + 1 Dp ρ Dt p Dρ ρ 2 Dt Thus we have ρ Dh Dt = Dp Dt + Φ + q We may relate the heat flu to temperature by Fourier s law as q i = κt,i Then we get ρ Dh Dt = Dp Dt + Φ + (κt,i),i To close the system we use (i) a thermodynamic equation like e = e(t, p), the simplest being e = c v T or h = c p T and (ii) equation of state p = ρrt 3 Navier Stokes Equation These equations describe the conservation of mass, momentum and energy. These are where Dρ Dt + ρv k,k = ρ Dv i Dt = p,i + τ ij,j + ρb i ρ De Dt = pv k,k + Φ + (κt,i ),i τ ij = µ (v i,j + v j,i 13 ) v k,kδ ij Φ = τ ij D ij = τ ij v i,j These are supplemented with the thermodynamic relation and equation of state of a gas e = e(t, p) p = ρrt These gives seven equations in seven unknowns ρ, p, T, v i, e.

4 4 Incompressible Navier Stokes Equation 4 For incompressible flow we shall always treat flow with density constant (tecnically not most general!). For an incompressible fluid D kk =. Thus we have σ ij = pδ ij + 2µD ij and we now have p = σ ii /3 which has the meaning of mean normal compressive stress. Substituting into the equations of motion ( ) vi ρ t + v v i j = ρb i + σ ij j j which upon simplification gives ( ) vi ρ t + v v i j j = ρb i p i + µ 2 v i j j In componentwise we write the above as ( ) v1 ρ t + v v 1 v 1 v v 2 + v 3 = ρb 1 p ( ) 2 + µ v ( ) v2 ρ t + v v 2 v 2 v v 2 + v 3 = ρb 2 p ( ) 2 + µ v ( ) v3 ρ t + v v 3 v 3 v v 2 + v 3 = ρb 3 p ( ) 2 + µ v There are four unknowns v 1, v 2, v 3 and p in the three equations. The fourth equation is supplied by the condition on incompressibility i.e. mass conservation div v = i.e. v 1 + v 2 + v 3 = These four equations are called Navier Stokes equations of motion for a incompressible fluid. Since the energy equations decouple from the mass conservation and momentum equations and energy equation need to be solved only when we need temperarure. For energy equation we have ρ De Dt = pv k,k + Φ + (κt,i ),i Almost always safe to assume that Φ 1. And since the fluid is incompressible we have (using e = c v T ) DT ρc v Dt = (κt,i),i The Navier-Stokes equation in invariant form can be written as [ ] v ρ + (grad v)v = ρb grad p + µ div (grad v) T t div v =

5 5 Eact Solutions Couette Flow For steady flow in the absence of body forces the Navier-Stokes equation reduce to ρ( v)v = µ (( v) T ) p v = Consider the plane velocity field Now the mass conservation implies that v() = v 1 ( 1, 2 )e 1 v 1 1 = so that v 1 = v 1 ( 2 ). Further the matri of v is v 1 2 [ v] = Thus v = and ( v)v =. Thus the equation of motion reduces to µ 2 v This implies that p = p( 1 ). Also = p 1, p = p = ( ) p = µ v 1 1 =. This implies that p/ 1 is a constant. We now consider two speicfic cases consistent with the above flow. Problem 1 (Plane Couette Flow). Consider the flow between two infinite flat plates, one at 2 = and one at 2 = h. The bottom plate is stationary and the top plate is moving in the 1 direction with velocity v. Thus the boundary conditions are v 1 () =, v 1 (h) = v Let us assume that p/ 1 = i.e. p = constant. Thus we have v 1 = α + β 2

6 6 Using the boundary condition we get v 1 = v 2 /h The stress components are σ 11 = σ 22 = σ 33 = p, σ 12 = σ 21 = µv/h Thus the force per unit area eerted by the fluid on the top plate has a normal component p and a tangential component µv/h. This fact furnishes a method of determining the viscosity of a fluid. Problem 1 (Plane Poiseuille Flow). Now the two plates are fied in space. Then the relevent boundary conditions are v 1 () = v 1 (h) = If the presuure is constant then velocity field would be identically zero. We therfore allow pressure drop. In particular let p/ 1 = δ with δ (the pressure drop per unit length) constant. Now the solution is given by which using the boundary condition gives v 1 = δ2 2 2µ + α + β 2 v 1 = δ 2 2µ (h 2) and the velocity distribution is parabolic. Also the stress components are given by σ 11 = σ 22 = σ 33 = p, σ 12 = σ 21 = δ(h/2 2 ) Thus the shearing stresses are maimum at the wall and vanishes at the centre. The volume discharge Q through a cross section is Q = h v 1 d 2 = h 3 δ/12µ Since the dischage can be easily measured as well as the pressure drop, this formula yields a convenient method of determing the viscosity.

7 Problem 1 (General Couette Flow). In this case either of the two surfaces is moving at constant speed and there is also eternal pressure gradient. The solution can be obtained by superimposing the two solutions obtined earlier since the governing equations are linear. Thus we get v 1 = v 2 /h + δ 2 2µ (h 2) If δ > then the pressure gradient will assist the motion induced by the upper surface. On the other hand it will resist the induced motion if δ <. In this case we may have reverse flow near the lower surface Poiseuille Flow The steady flow of viscous fluid in a pipe of arbitrary but constant cross section is referred to as Poiseuille flow. Let the cross section is in the y plane and the steady flow is in the z direction. a z y b z y c z y Since the flow is unidirectional, we have v = (,, w). Now equation of mass conservation gives w z = = w = w(, y)

8 8 Thus ( v)v =. The equation of motion reduces to p + µ 2 v = Writing components wise we get p = p y =. Thus p = p(z). Thus equation for w becomes ( ) 2 w + 2 w 2 y 2 = 1 dp µ dz Now taking the right hand side can atmost be function of z. Taking derivative with repsect to z results in dp/dz equals to a constant, G say. Thus we have 2 w + 2 w 2 y = G 2 µ For very special geometries, closed for solution can be found. For other, no closed form solution eist. In those cases we might sought solution using series form. Problem 1 (Poiseuille Flow: Circular cross section). We now use polar coordinate in the y-plane. This effectively means we are using cylindrical coordinate for the original problem. Let a be the radious of the cylinder. Then transformaing to = r cos θ and y = r sin θ we get Integrating this we get 1 r d dr w = G µ ( r dw ) = G dr µ r A log r + B The boundary condition is that w() is finite and w(a) =. Thus we have w = G 4µ (a2 r 2 ) The maimum velocity occurs at r = and w ma = G 4µ a2 The same solution could be obtained if we had guessed the solution in the form w(, y) = α( 2 + y 2 a 2 ) Note that this satisfies the boundary condition and the constant α is obtained by substituting this in the momentum equation for w.

9 9 The flu through the pipe is given by The mean flow velcity is Q = w av = 2π a w r dr dθ = πga4 8µ F lu Area of cross section = Ga2 8µ Problem 2 (Poiseuille Flow: Elliptic cross section). velocity of the form ( ) 2 w(, y) = α a + y2 2 b 1 2 Let us assume the aial It satisfies the boundary condition. Direct substitution into the governing equations for w gives α = G a 2 b 2 2µ a 2 + b 2 Thus the velocity profile is given by w(, y) = G ( [ ]) a 2 b µ a 2 + b 2 a + y2 2 b 2 Problem 3 (Poiseuille Flow: Rectangular cross section). Let the flow is confined within the channel given by a a and b y b. Thus the equation to be solved is subhject to 2 w + 2 w 2 y = G 2 µ w(±a) = w(±b) = We look for solution in the separable form i.e. w(, y) = X()Y (y) with X(±a) = Y (±b) =. Hence we choose w as w(, y) = m= n= [ A mn cos (2m + 1) π 2a ] [ cos (2n + 1) πy 2b Where the constants A mn are to be found. These can be found by substituting into the governing equations for w. ] 5.3 Flow down an inclined plane Consider two dimensional flow of a fluid of height h down an inclined plane. Fi the coordinate aes as shown in the figure. The boundary condition is that p = p at = h and no-slip at =.

10 1 h α p z We seek a steady solution of the w = w() which satisfy the equation of continuity identically. The momentum equations are From the first equation we get = p ρg sin α = p z + ρg cos α + w µd2 d 2 p = ρg sin α + f(z) Now using p( = h, z) = p, we get f(z) = p a + ρg sin αh and hence p = p + ρg sin α(h ) Substituting into the equation for w we get Integrating twice we get w = d 2 w d 2 cos α = ρg µ ρg cos α 2 + A + B 2µ No-slip at = implies B =. If we neglect surface tension, then the stress vector at = h for the fluid and air will be same. Thus we have For i = 1 we get σ air ij n j = σ ij n j, n = (1,, ), i = 1, 2, 3 σ air 11 = σ 11 or p + 2µ air D air 11 = p + 2µD 11 which is satisfied identically. For i = 2 we get σ air 21 = σ 21 or 2µ air D air 21 = 2µD 21

11 11 This is also satisfied identically. For i = 3 we have Since µ air µ we have at = h σ air 31 = σ 31 or 2µ air D air 31 = 2µD 31 D 31 = or dw d = This we used to get A and the solution w becomes w = The above solution can be written as From this we derive w = ρg cos α (2h ) 2µ at = h ρg cos α (h 2 (h ) 2 ) 2µ w ma = ρg cos αh2 2µ and Q = ρg cos αh3 3µ 6 Unsteady unidirectional flow For unsteady unidirectional flow we have to solve v = we 3, As in the steady cases, we have w = w(y, z, t), p = p y 6.1 Impulsively started plate v t = p ρ + ν 2 v = = p = p(z, t) Consider a semi-infinite fluid resting above a rigid plane. For t < the fluid is at rest. At t = the plane wall instanteneouly acuires a speed U in the z direction. Considering the figure, it is clear that the flow variables must be independent of and hence w = w(y, t). Now the momentum equation along z direction gives ρ w t = p z + w µ 2 y 2 Now the above equation implying p/ z must be independent of z. Thus we have p(y, z, t) = p (y, t) + p 1 (y, t)z

12 12 y U z Since pressure is boundend at z = ± we must have p 1 (y, t) = and hence p/ z = Hence the equation reduces to w t = ν 2 w y 2 subject to the initial and boundary conditions w(, t) = U w(y, t) as y w(y, ) = We get rid of the constant U by defining W = Uw thus reducing the problem to W t = ν 2 W y 2 subject to the initial and boundary conditions W (, t) = 1 W (y, t) as y W (y, ) = Clearly we must have W = φ(ν, y, t). Solution by Laplace transform: Taking Laplace transform on both sides we get s W = ν d2 W y 2

13 13 subject to W (, s) = 1 s W (y, s) as y The solution is W = Ae y s/ν + Be y s/ν Using the boundary conditions we get The inverse transform gives where erfc(η) is defined by Thus the solution is given by W = e y s/ν W (y, t) = erfc(η), where η = y 2 νt erfc(η) = 1 2 π η s e t2 dt = 1 erf(η) u = U erfc(η), η = y 2 νt Solution by Similarity: Since W is nondimensional and W = φ(y, ν, t), the function φ must also be nondimensional. Since y, ν, t involve two independent dimension [L] and [T ] only one independent nondimensional variable is possible. Let us take the nondimensional variable as Thus let us choose η = y 2 νt W = f(η) Putting into the governing equations we have The boundary conditions are f = 2ηf f( ) = f() = 1

14 14 We solve the above as f = Ae η2 Integrating once again we get η f = B + A e η2 dη Putting into boundary conditions we get B = 1 and A = 2/ π. Thus we have the same solution. If we plot the u/u versus η then we observe that u/u =.5 at η 1.4. Thus the width of the boundary layer δ 2.8 νt. Thus the boundary layer (see later in the course) increases as t. The vorticity is given by ω = v = U π νt e Using Gamma function we can show that ω dy y 2 4νt e1 is a constant vector for t >. Thus there is no new vorticity generated in the flow. The vorticity at t = due to impulsive start is simply diffuse away with time. And the voritcity at time t is confined within the layer [, 2 νt]. 6.2 Oscillating flows Consider the flow similar to the above but the plate is now oscillating with frequency ω. The governing equations are subject to the boundary conditions w t = ν 2 w y 2 w(, t) = U cos(ωt) w(y, t) as y We can remove the dependence on U by defining w = UW. problem to W t = ν 2 W y 2 Thus reduces the

15 15 subject to the boundary conditions W (, t) = cos(ωt) W (y, t) as y Note that here W = φ(ν, y, t, ω) and φ must be function of two independent nondimensional variable. Thus similarity solution is not possible. Since the frquency of the wall is ω and the problem is linear, we assume that the frequency of the fluid flow w is also ω. To solve let us assume that W = Re(f(y)e iωt ) Substituting into the governing equations we get subject to the boundary conditions Thus we have f = iω ν f f() = 1 and f(y) as y f = Ae λy + Be λy, λ = iω ν Now i = (1 + i)/ 2 we must have A =. Using the other conditions gives B = 1. Thus we have f = e ω/2ν(1+i)y Hence w = Ue ( ω/2νy cos ωt ) ω/2νy The vorticity is given by ω = v = w y e 1 = U ω/νe ( ω/2νy cos ωt ) ω/2νy + π/4 e 1 Thus most of the vorticity is confined withing the a region of thickness ν/ω. The higher the frquence the thinner the layer. The wall shear stress is proportinal to w y cos(ωt) sin(ωt) = 2 cos(ωt + π/4) y= Thus there is a phase difference of π/4 between the wall velocity and wall shear.

Candidates must show on each answer book the type of calculator used. Log Tables, Statistical Tables and Graph Paper are available on request.

UNIVERSITY OF EAST ANGLIA School of Mathematics Spring Semester Examination 2004 FLUID DYNAMICS Time allowed: 3 hours Attempt Question 1 and FOUR other questions. Candidates must show on each answer book