Appendix A: Supplements

Appendix A: Supplements A.1 Introduction to tensor notation In the derivation of balance equations the tensor notation is used because it allows the equations to be written in a clearer and simpler fashion. We have restricted ourselves in this to cartesian coordinates. In the following, the essential features of cartesian tensor notation are only illustrated to an extent required for the derivation of the balance equations; extensive publications are available for further reading. We will start with an example. The velocity w of a point of mass is known as a vector which can be set in a cartesian coordinate system using its components wx, wy, wz: w(wx,wy,wz) . If the unit vector in a cartesian coordinate system is indicated by ex, ey, ez,it holds that w = wx ex + wy ey + wz ez . In tensor notation, the indices x, y, z are replaced by the indices 1, 2, 3 and we write instead 3 w = w1 e1 + w2 e2 + w3 e3 = wi ei . i=1 The velocity vector w is characterised completely by its components wi, i = 1, 2, 3. In tensor notation, the velocity vector is indicated by the abbreviation wi with i =1, 2, 3. Correspondingly, the position vector x(x, y, z) is determined by its components x1 = x, x2 = y, x3 = z, and in tensor notation by the abbreviated xi with i =1, 2, 3. According to this, a vector is indicated by a single index. It is possible to differentiate between tensors of different levels. Zero level tensors are scalars. They do not change by transferring to another coordinate system. Examples of scalars are temperature ϑ, pressure p and density .No index is necessary for their characterisation. First level tensors are vectors. As explained above, they are indicated by one index. Second level tensors are characterised by two indices. The stress tensor is such a quantity. It has nine components τ11, τ12, τ13, τ21 ...τ33.The abbreviation normally written is τji,wherej and i each assume the values 1, 2, 3. In calculations with tensors, the following rule is used. If an index only appears Appendix A: Supplements 661 once in a term of an equation it is called a free index. It can be replaced by any other index. All terms in an equation must agree in their free indices. The relationship ai = c · bi , (A.1) whereby c is a constant (scalar), means that the vectors ai and bi only differ in their amount, they have the same direction. So a1 = cb1 ; a2 = cb2 ; a3 = cb3 . The internal product (scalar product) of two vectors a · b = a1 b1 + a2 b2 + a3 b3 looks like the following for the index notation 3 a · b = ai bi . i=1 As internal products appear frequently the following “summation convention”was settled: if an index appears twice in a term, then it should be summed over this index. This index is called a bound index. It can not be replaced by any other index. The summation symbol is left out. With that we have ai bi = a1 b1 + a2 b2 + a3 b3 . (A.2) Differentiation leads to a tensor that is one order higher. So the gradient of a scalar p ∂p ∂p ∂p gradp = ∇ p = e1 + e2 + e3 (A.3) ∂x1 ∂x2 ∂x3 is a vector with the three components ∂p/∂xi,i=1, 2, 3, which is abbreviated to just ∂p/∂xi in tensor notation. If we differentiate a vector wj,eachofthe three components w1, w2, w3 can be differentiated with respect to each position coordinate x1, x2, x3. This gives a second level tensor ∂wj (i =1, 2, 3;j =1, 2, 3) , (A.4) ∂xi that consists of 9 components. On the other hand, the divergence of a vector is a scalar, ∂wi ∂w1 ∂w2 ∂w3 divw = ∇·w = = + + . (A.5) ∂xi ∂x1 ∂x2 ∂x3 The formulation of a divergence produces a tensor one order lower than the original tensor. A sensible ‘operator’ is the Kronecker delta δij, defined by δij =1fori = j, δij =0fori = j. (A.6) 662 Appendix A: Supplements δij is also called the unit tensor. It further holds that δij bj = bi , (A.7) which can be confirmed by writing the whole equation out, because it is δij bj = δi1 b1 + δi2 b2 + δi3 b3 that is, for i =1: δ1j bj = δ11 b1 = b1,fori =2: δ2j bj = δ22 b2 = b2 and for i =3: δ3j bj = δ33 b3 = b3, and with that δij bj = bi. A.2 Relationship between mean and thermodynamic pressure By definition, the mean pressurep ¯ = −1/3 δji τkk only includes normal stresses. In order to create a link between mean and thermodynamic pressure we will consider a cubic fluid element at a temperature T and of specific volume v,Fig.A1.We willnowassumethatthecubeisatrestattimet = 0, so that the thermodynamic pressure p prevails inside the element. Now let us assume the mean pressurep ¯ is being exerted on the element from outside. Whenp>p ¯ the cube is compressed, shouldp<p ¯ then it expands. So, work −p¯ dV is carried out by the external pressurep ¯. This is equal to the work done during the volume change in the gas −p dV and the dissipated work. It therefore holds that dW = −p¯ dV = −p dV +dWdiss with the the dissipated work as dWdiss = −(¯p − p)dV. This is always positive according to the second law, because forp>p ¯ we have dV<0andfor¯p<p,dV>0. On the other hand, the increase in the volume dV is yielded from the transport theory (3.18)withZ = V and z = Z/M = V/M = 1/ to be dV ∂wi = dV. dt ∂xi V (t) The dissipated work can also be written as ⎛ ⎞ ⎜ ∂wi ⎟ dWdiss = −(¯p − p) ⎝ dV ⎠ dt. ∂xi V (t) It is clearly reasonable that the speed dV/dt of the volume change or ∂wi/∂xi is a monotonically decaying function ofp ¯− p, Fig. A2, as the larger the overpressure p¯ − p, the faster the volume of the cube reduces. It is therefore suggested, that Appendix A: Supplements 663 Fig. A.1: Interrelation between mean and thermo- Fig. A.2: Expansion as a function dynamic pressure of the over pressure where the speed of the volume change is not that fast, the curve in Fig. A2 may be replaced by a straight line: ∂wi p¯ − p = −ζ . ∂xi The factor defined by this, ζ>0, is the volume viscosity (SI units kg/s m). It has to be determined either experimentally or using methods of statistical thermodynamics, which is only possible for substances with simple molecules. It can be seen that the mean and thermodynamic pressures only strictly agree when ζ = 0 or the fluid is incompressible, ∂wi/∂xi =0. A.3 Navier-Stokes equations for an incompressible fluid of constant viscosity in cartesian coordinates The mass force is the acceleration due to gravity kj = gj. x1 = x-direction: ∂w1 ∂w1 ∂w1 ∂w1 + w1 + w2 + w3 = ∂t ∂x1 ∂x2 ∂x3 (A.8) 2 2 2 − ∂p ∂ w1 ∂ w1 ∂ w1 g1 + η 2 + 2 + 2 . ∂x1 ∂x1 ∂x2 ∂x3 664 Appendix A: Supplements x2 = y-direction: ∂w2 ∂w2 ∂w2 ∂w2 + w1 + w2 + w3 = ∂t ∂x1 ∂x2 ∂x3 (A.9) 2 2 2 − ∂p ∂ w2 ∂ w2 ∂ w3 g2 + η 2 + 2 + 2 . ∂x2 ∂x1 ∂x2 ∂x3 x3 = z-direction: ∂w3 ∂w3 ∂w3 ∂w3 + w1 + w2 + w3 = ∂t ∂x1 ∂x2 ∂x3 (A.10) 2 2 2 − ∂p ∂ w3 ∂ w3 ∂ w3 g3 + η 2 + 2 + 2 . ∂x3 ∂x1 ∂x2 ∂x3 A.4 Navier-Stokes equations for an incompressible fluid of constant viscosity in cylindrical coordinates The mass force is the acceleration due to gravity kj = gj. r-direction: 2 ∂wr ∂wr wθ ∂wr wθ ∂wr + wr + − + wz = ∂t ∂r r ∂θ r ∂z (A.11) 2 2 ∂p ∂ 1 ∂ 1 ∂ wr 2 ∂wθ ∂ wr gr − + η (rwr) + − + . ∂r ∂r r ∂r r2 ∂θ2 r2 ∂θ ∂z2 θ-direction: ∂wθ ∂wθ wθ ∂wθ wr wθ ∂wθ + wr + + + wz = ∂t ∂r r ∂θ r ∂z (A.12) 2 2 1 ∂p ∂ 1 ∂ 1 ∂ wθ 2 ∂wr ∂ wθ gθ − + η (rwθ) + + + . r ∂θ ∂r r ∂r r2 ∂θ2 r2 ∂θ ∂z2 z-direction: ∂wz ∂wz wθ ∂wz ∂wz + wr + + wz = ∂t ∂r r ∂θ ∂z (A.13) 2 2 ∂p 1 ∂ ∂wz 1 ∂ wz ∂ wz gz − + η r + + . ∂z r ∂r ∂r r2 ∂θ2 ∂z2 Appendix A: Supplements 665 A.5 Entropy balance for mixtures The Gibbs’ fundamental equation for mixtures, μK du = T ds − p dv + dξK ˜ K MK or du ds dv μK dξK = T − p + , (A.14) ˜ dt dt dt K MK dt taking into account dv − 1 d 1 ∂wi = 2 = dt dt ∂xi and (3.25) ∗ dξK ∂jK,i = − + Γ˙ K (A.15) dt ∂xi can be rearranged into ∗ du ds ∂wi μK ∂jK,i μK = T − p − + Γ˙ K . ˜ ˜ dt dt ∂xi K MK ∂xi K MK Due to μK /M˜ K = hK − TsK this delivers ∗ ∗ du ds ∂wi ∂jK,i ∂jK,i μK = T − p − hK + T sK + Γ˙ K . i i i ˜ dt dt ∂x K ∂x K ∂x K MK Putting this into the energy equation (3.81), taking into account the following, yields ∗ q˙ =˙qi + hK jK,i K ∗ ∂q˙i ∗ ds ∗ ∂hK ∂jK,i μK − +Φ+ j kK,i = T + j + T sK + Γ˙ K . K,i K,i ˜ ∂xi K dt K ∂xi K ∂xi K MK (A.16) We write the following for ∗ ∗ ∂jK,i ∂(jK,i sK ) ∗ ∂sK T sK = T − T jK,i K ∂xi K ∂xi K ∂xi and combine to give ∗ ∂hK ∂sK ∗ ∂μK 1 jK,i − T = jK,i .

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