2 Continuum-Scale Fluid Flows
2.3 Conserved Quantities
Conserved Quantities 11
result is a bit more complicated than Equation 2.7:
∂ρv
∂t + div(ρv⊗v) =e+qmom. (2.11) For this equation, we also used a termecapturing forces and a termqmomfor the remaining sources and sinks of momentum in the system.
We can reformulate Equation 2.11 by taking advantage of the product rule for the time derivative
∂ρv
∂t =v∂ρ
∂t +ρ∂v
∂t and for the divergence term
div(ρv⊗v) =ρv·gradv+vdiv(ρv) , and get
v ∂ρ
∂t +ρ∂v
∂t +ρv·gradv+v div(ρv) =e+qmom. Reordering the left-hand side yields
ρ∂v
∂t +ρv·gradv+v ∂ρ
∂t + div(ρv)
=e+qmom. (2.12)
In Equation 2.12, we now notice that the third additive term corresponds to the left-hand side of the mass conservation Equation 2.7, but multiplied with the velocity. Assuming that the source term for massqmassis zero leads us to
ρ∂v
∂t +ρv·gradv =e+qmom. (2.13)
Otherwise –i.e., if we do not assume the source term of the mass balance equation to be zero – the third term of Equation 2.12 can be brought to the right-hand side and integrated into the momentum source termqmom. This means that the general form of the equation for the conservation of momentum is given by Equation 2.13.
Let us now have a closer look at the force termeof the right-hand side of Equation 2.13. It should be clear that we can spliteinto a termf capturing the forces which are exercised upon the surface of a considered domain, and a termh, representing forces which attack in the interior:
e=f +h
The former are calledsurface forcesand are exerted on the material by its environment, the latter are calledbody forcesand are caused by force fields like gravity or electromagnetism.
It can be shown [72] that the surface forcesf can be expressed as
f = divτ (2.14)
Conserved Quantities 13
Figure 2.4: The COUETTEthought experiment comprises two infinitely large parallel planar plates with the space between them occupied by a fluid; in the experiment, the upper plate moves with a velocity ofvxrelative to the lower one. At each plate, the observed force per areaFxis proportional to the relative velocity of the platesvx and anti-proportional to the distanceybetween the plates. The proportionality coefficientµis determined by the fluid between the two plates, and is called the dynamic viscosity.
whereτ ∈R3×3is calledstress tensor. Taking advantage of Equation 2.14 and neglecting all body forces except gravity, we get
ρ∂v
∂t +ρv·gradv= divτ+ρg+qmom (2.15) as the momentum balance equation wheregis the gravitational acceleration.
2.3.3 NEWTONIANFluids
For NEWTONIANfluids, we can substantiate the momentum conservation Equation 2.15 as follows: First, we assume that the stress tensorτ can be split into
τ =−pI+T (2.16)
where the termpI represents pressure, and the termT represents the shear stresses [9]. The reason why the termpI is negative is that the tensorτrepresents the stresses which act upon the material within the observed domain, and not the stresses which this material exercises upon its environment.
Inserting Equation 2.16 into the momentum balance equation (2.15), we get ρ∂v
∂t +ρv·gradv =−gradp+ divT +ρg+qmom. (2.17) We now consider the fact that the shear stress tensorT of NEWTONIAN fluids does not depend on the absolute deformation of the material relative to its initial position, but only on the rate at which the fluid gets displaced relative to its environment as illustrated in Figure 2.4. SinceT needs to be symmetric [9], we get a shear stress tensorT of the form
Tij =µ ∂vj
∂xi
+ ∂vi
∂xj
+δijλdivv , (2.18)
using the proportionality coefficientsµ∈ R+ andλ∈ R, withδij being the KRONECKER
delta. The factorλcaptures the stress due to expansion or contraction of the volume of the fluid. Since there are many difficulties for determiningλ, it is usually either assumed [9] to be zero as in the following, or−3µ/2.
If we take advantage of Equation 2.18 in the momentum balance equation (2.17), we get ρ∂v
∂t +ρv·gradv=−gradp+ div µ gradv+ (gradv)T
(2.19) +ρg+qmom,
the NAVIER-STOKESequations for compressible NEWTONIANfluids.
For incompressible fluids with constant dynamic viscosity we can simplify Equation 2.19 further: First, we use the relation
div (gradv)T =graddivv
and then consider the fact thatdivvis identical to zero if the densityρis constant. We thus get
ρ∂v
∂t +ρv·gradv =−gradp+µdivgradv+ρg+qmom (2.20) as the conservation equation for momentum.
2.3.4 Creeping Flows
For creeping incompressible flows, we can also neglect the inertia term of the NAVIER-STOKES
Equation 2.20, and get the STOKESequation
−gradp+µdivgradv+ρg+qmom= 0. (2.21) In order for this assumption to be applicable, we need to be able to define what “creeping”
means. Usually, this property is defined using the REYNOLDSnumber Re := vcLc
νc
where vc is thecharacteristic velocity of the considered physical system (for example, the absolute value of the maximum velocity of the fluid),Lcrepresents thecharacteristic length of the system (for example, the diameter of the pipe for pipe-flow problems), andνc=µc/ρc
is thecharacteristic kinematic viscosityof the fluid. We now define a flow as “creeping” if it exhibits a REYNOLDSnumber smaller than1.
Since the characteristic length and the characteristic velocity can be chosen arbitrarily, we cannot assign too much meaning to the absolute value of the REYNOLDSnumber. Having said that, for many important classes of flow problems there are standard conventions of how to determineLcandvc. This means that we can compare the absolute value of the REYNOLDS
number onlywithin a given class of flow problems. Such classes include, for example, pipe-flows or flows around airfoils.
Conserved Quantities 15
2.3.5 Conservation of Energy
The last conservation quantity which we will consider in this work is energy. We may think of energy as a scalar quantity that has kinetic, potential, and thermal contributions. Assuming that the gravitational acceleration is constant and that gravity is the only body force, we can express conservation of energy [9] as
∂
∂tρ
u+ 1
2kvk2+z·g
+ div(hρv+τ v−λgradT) =qenergy−ρv·g (2.22) whereuis the specific internal energy of the substance as described in Section 2.1,z=x−xref is the distance of a spatial position relative to an arbitrary but fixed reference point in space,λ is the heat conduction coefficient,h=u+p/ρis thespecific enthalpyof the substance, andqenergy is the source or sink term for energy.
If we neglect the kinetic energy and friction, we get
∂
∂tρ(u+z·g) + div(hρv−λgradT) =qenergy−ρv·g
as the equation for the conservation of energy. Amongst others, these assumptions are valid for creeping fluid flows. If we also assume that the considered system only exhibits small variations of its heightz, we get
∂ ρu
∂t + div(hρv−λgradT) =qenergy . (2.23) The fact that we need to consider the specific internal energyuin the accumulation term but the specific enthalpyhin the flux term is due to the fact that transported material needs to displace other material before it can occupy a given volume. As illustrated in Figure 2.5, the energy required to displace the other material is equivalent to the volume occupied by the transported material times the force with which the displaced material pushes back. In fluids this “push-back force” is the pressure that the displaced material exercises upon the surface of the transported material.
Dissipation
The third law of thermodynamics states that, in a closed system, all spontaneously occuring processes increase the entropy of the system as a whole2. In the context of the conservation of energy, this means that some energy is always converted into heat,i.e., internal energy. In Equation 2.23, we account for this effect by transporting enthalpy, while accumulating only the internal energy. To illustrate the point, let us consider Figure 2.5: There, some material gets transported from the left to the right of a cylinder. The energy which is on the right side at the end is the internal energy which the material originally possessed when it was on
2There might be parts of the system where entropy is reduced, but this is always compensated by additional entropy elsewhere.
Figure 2.5: To move the substance with the specific internal energyufrom the left to the right, one has to displace the material on the right of the cylinder. This can be imagined as a four-step process: First, the right piston creates a vacuum, then the vessel with the substance moves to the right. The work required by the piston to create the vacuum is called the volume changing workWv = Rs0+∆s
s0 F ds.
Assuming a constant cross-sectionAof the cylinder and constant pressureprightof the environment of the right side, this is equivalent toWv,right = ∆sA pright. After the substance has been transferred to the right, the left piston can occupy the void space and “recovers” the volume changing workWv,left= ∆sA pleft.
the left side. But in addition, a piston had to displace the material originally occupying the space on the right side, which requires a physical work of∆s A prightto be done. Assuming the pressure to be constant at a fixed spatial location,i.e., the moved substance reduces its pressure fromplefttopright, requires work of(pleft−pright)A∆s. Since energy is conserved, this work gets converted into internal energy if the transport of material does not happen in a closed vessel that cannot expand.
Porous Media 17
Figure 2.6: An example of a porous medium where two fluid phases are co-located with a solid.