Download 1 Conservation Equations

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1.1
Conservation Equations
Conservation of mass
Mass conservation states that the net rate of change of the mass of a control volume (CV) is equal to the
net rate of transport across the boundary of the CV. In vector form,
Z
Z
d
ρ dV +
n̂ · ρu dA = 0
(1)
dt V
A
in which n̂ is the outward pointing normal along the surface of the control volume. The quantity ρu is the
mass flux vector, and since the normal is defined as pointing out, the second term will be positive/negative
if a net amount of mass is leaving/entering the CV.
The divergence theorem of vector calculus transforms the surface integral into a volume integral, and the
order of differentiation and integration in the first term can be switched. This gives
Z ∂ρ
+ ∇ · (ρu) dV = 0
(2)
∂t
V
in which ∇· is the divergence operator; the specific form of this function will depend on the coordinate
system. The above relation must hold for any sub–volume of the CV, and thus the integrand must be zero:
∂ρ
+ ∇ · (ρu) = 0
∂t
(3)
This is the divergent form of the continuity equation.
The chain rule of calculus can be applied to the second term,
∇ · (ρu) = ρ ∇ · u + u · ∇ρ
(4)
Note that the density gradient, ∇ρ, is a vector, and u · ∇ρ gives a scalar. When replaced into Eq. (3) the
result is
Dρ
+ ρ∇ · u = 0
(5)
Dt
in which the material derivative is defined as
D
∂
=
+u·∇
Dt
∂t
(6)
Equation (5) is the conservative form of the continuity equation. The divergent and conservative forms
are completely equivalent; their distinction becomes more relevant when developing numerical methods for
computing flowfields.
1.2
Momentum conservation
Newton’s law states that the rate change of linear momentum of a system is equal to the net force acting
on a system. When applied to a CV, the rate of change of the CV momentum is given by the 1) the
instantaneous time derivative of the CV momentum, plus 2) the net rate of transport of momentum across
the CV boundaries. Note that these two contributions are analogous to the two sources of mass change in a
CV, i.e., the ”rate of storage” and the ”rate of transport” terms. Unlike mass, however, momentum can be
created or destroyed by application of forces to the CV. The forces arise from two contributions: 1) surface
forces which act on the surface of the control volume, and 2) body forces which act directly on the mass
within the CV. In vector form, the linear momentum balance on the CV appears as
Z
Z
Z
Z
d
ρ u dV +
n̂ · ρu u dA =
n̂ · σ dA +
ρ f dV
(7)
dt V
A
A
V
Note that this is a vector equation; it can be resolved into
the mass flux vector, and the dyadic tensor

ux ux
ρu u = ρui uj = ρ uy ux
uz ux
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three orthogonal components. As before, ρu is
ux uy
uy uy
uz uy

ux uz
uy uz 
uz uz
(8)
denotes the flux of j–directed momentum in the ith direction. The quantity σ is the stess tensor in the fluid.
This is a second–order tensor which has the stucture


σxx σxy σxz
σ = σij = σyx σyy σyz 
(9)
σzx σzy σzz
The stress tensor σij appears as a tensor (i.e., a 3× 3 matrix) because two distinct directions are involved in
specifying the stress (which is force/unit area) acting on a surface: the directional component of the force
(the second subscript) and the surface normal direction (the first subscript). Consequently, the force vector
acting on a surface element dA that has a normal in the x–direction would be x̂ · σ dA.
The stress tensor is often resolved into an isotropic part (independent of direction) and a deviatoric part
via
σ = −P I + τ
(10)
in which P is the thermodynamic pressure, I is the identity matrix, and τ denotes the shear stress tensor.
Strictly speaking, the diagonal elements of the shear stress tensor (i.e., τxx , etc.) do not denote shear stresses
– rather, they describe normal stresses – yet for the situations we encounter the only significant normal stress
will be due to the pressure, and the diagonal components of τ will be zero. The shear stress tensor will be
related to the rate of fluid deformation (i.e., velocity gradients) via constitutive relations; Newton’s viscosity
law is a simple example. Such relations will be deferred for later.
The quantity f denotes the body force per unit mass and has units of acceleration. Indeed, the most
typical body force component is gravity, for which f would be the gravitational acceleration times the
downwards unit vector. Other mechanisms can result in body forces, such as electric currents, magnetic
fields, and coriolis effects.
Application of the divergence theorem to Eq. (7) yields the differential form of linear momentum conservation:
∂
(ρ u) + ∇ · (ρu u) = ∇ · σ + ρ f
(11)
∂t
and combination of this with the continuity equation in Eq. (3) leads to
ρ
Du
= ∇ · σ + ρf
Dt
(12)
This is the most general form of the momentum equation. Again, a constitutive law is needed to connect σ
to the velocity gradients and thereby ”close” the equation.
1.3
Energy conservation
Let e denote the internal energy of the fluid per unit mass. The total energy, per unit mass, of the fluid is
internal plus kinetic energy, or e + u2 /2 where u2 = u · u is the square magnitude of the velocity. From the
first law, the rate of storage plus rate of transport of total energy in the CV must be balanced by the rate
of work and rate of heat transfer to the CV. Work is force over a displacement, and rate of work will be
force times a rate of displacement, or force times velocity. As was the case with the momentum equation,
the force components include surface stresses and body forces. Putting it all together, the first law for the
CV has
Z
Z
u2
u2
d
ρ e+
dV +
n̂ · ρu e +
dA
dt V
2
2
A
Z
Z
Z
Z
00
=
n̂ · σ · u dA +
ρ f · u dV −
n̂ · q dA +
q 000 dV (13)
A
V
A
V
In the above, q00 denotes the heat flux, and the surface integral containing this term is negative if there is a
net transfer of heat from the CV; hence the negative sign. The quantity q 000 denotes a volumetric source of
heat (W/m3 ) within the fluid; this term would arise from chemical reactions, Joule heating from an electric
current – but not from fluid friction: this effect will be accounted by the stress tensor work term.
It is left as an exercise to show that the differential form of the energy equation becomes
ρ
Di
DP
=
+ τ : ∇u − ∇ · q00 + q 000
Dt
Dt
2
(14)
in which i = e + P/ρ is the enthalpy per unit mass of the fluid (yes, the symbol h is commonly used in
thermodynamics to denote enthalpy, but h has been reserved for the heat transfer coefficient). Note that
the body force term has disappeared – this results from combining the momentum equation with the energy
equation. What is left, with regard to work, is the τ : ∇u dissipation term. The operator : denotes a
contraction between the two tensors τ and ∇u and this produces a scalar. In tensor notation, this operation
is written
∂
τ : ∇u = σij
uj
(15)
∂xi
In the tensor notation used above, summation over repeated subscripts is implied. That is, the operation
sums over i = 1, 2, 3 and over j = 1, 2, 3.
Equation (14) is a general form of the energy equation for a moving fluid. To close the equation,
constitutive relations are needed to related τ to velocity gradients (Newton’s laws) and q00 to temperature
gradients (Fourier’s law). We also need an equation of state to connect enthalpy i to temperature. All such
details will be addressed when we examine specific flows.
Also note that no explicit reference is made to a coordinate system. The differential relations for mass,
momentum, and energy conservation are presented in a general, vector calculus form. Of course, the specific
forms of the gradient (i.e., ∇ρ), the vector gradient (∇u), and the divergence (∇ · q00 ) would be needed to
apply the formulas to a given problem with a given coordinate system. Such information can be found in
the text, or on Mathematica.
Exercises
1. Fill in the details in the derivation of Eq. (12) from Eq. (11). This will take only line or two of formulas.
2. Derive Eq. (14) beginning with Eq. (13). To do this, you will need to (among other things)
(a) Apply the divergence theorem per the previous derivations.
(b) Use the vector identity
∇ · (σ · u) = (∇ · σ) · u + σ : ∇u
(16)
(c) Replace e with i − P/ρ.
(d) Dot the momentum equation into u, use the fact that
Du2 /2
Du
·u=
Dt
Dt
(17)
(note that the resulting equation represents a balance of mechanical energy in the CV), and use
this equation to eliminate the kinetic energy terms in the energy equation.
(e) Use the stress tensor relation in Eq. (10) to split σ into the normal and shear parts. Equation
(15) will give you a hint at what results from I : ∇u – note that I, in tensor notation, is the
Kronecker delta function δij = 1 if i = j and 0 if i 6= j.
3. Consider a sphere, of radius R, that is immersed in a stationary fluid. The surface temperature of the
sphere is maintained at TS , and the temperature far from the sphere in the fluid is T∞ . Using the
steady 1–D conduction equation for the fluid temperature,
1 ∂
2
2 ∂T
∇ T = 2
r
=0
(18)
r ∂r
∂r
determine the formula for the Nusselt number of the sphere, which is defined as
NuD =
hD
2hR
=
kf
kf
(19)
You will need to solve the conduction equation for this problem and use the solution to determine the
heat flux at the surface of the sphere. Your result, in the end, will be a pretty simple formula.
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