Compressible Effects
If density fluctuations are significant, compressible effects needs to be accounted for, but the
question is what significant means. Anderson suggests that fluctuations in the order of 5%
(∆ρ/ρ > 0.05) could be used as a threshold value. The question is still what that means.
Let’s try to figure out...
The compressibility for isothermal process is defined as
1
∂ρ
τT =
ρ
∂p T
τT =
(1)
∂ρ
∂
RT
p 1
= {p = ρRT, T = const} =
ρ
∂p T
p
∂p RT
T
and thus
τT =
1
p
(2)
We are trying to find an estimate of ∆ρ/ρ. Using the equation of state with constant temperature, ∆ρ/ρ can be expressed in terms of pressure
∆p RT
∆ρ
=
=
ρ
RT p
1
τT =
= τT ∆p
p
(3)
If we assume that compressible effects are not significant and use Bernoulli’s equation to get an
estimate of the pressure fluctuations generated by a flow
1
2
∆p ≈ ρ∞ U∞
2
(4)
With τT = 1/p∞ , we can rewrite Eqn. 4 as
∆ρ
1
≈
ρ∞
p∞
1
2
ρ∞ U∞
2
= {p∞ = ρ∞ RT∞ } =
The speed of sound in the freestream is obtained as a∞ =
2
2
2
ρ∞ U∞
U∞
=
2ρ∞ RT∞
2RT∞
√
γRT∞ , which gives
(5)
2
γU∞
∆ρ
γ 2
≈
= M∞
ρ∞
2a2∞
2
For air (γ = 1.4) and ∆ρ/ρ∞ < 0.05 we get M∞ < 0.27
3
(6)
Entropy
First law of thermodynamics:
de = δq − δw
(1)
For a reversible process: δw = pd(1/ρ) and δq = T ds
1
de = T ds − pd
ρ
(2)
Enthalpy is defined as: h = e + p/ρ and thus
1
1
dh = de + pd
+
dp
ρ
ρ
(3)
Eliminate de in Eqn. 2 using Eqn. 3
1
1
1
T ds = dh − pd
−
dp + pd
ρ
ρ
ρ
ds =
dh
dp
−
T
ρT
Using dh = Cp T and the equation of state p = ρRT , we get
ds = Cp
dp
dT
−R
T
p
(4)
Integrating Eqn. 4 gives
ˆ
s2 − s1 =
1
2
dT
Cp
− R ln
T
p2
p1
(5)
For a calorically perfect gas, Cp is constant (not a function of temperature) and can be moved
out from the integral and thus
2
s2 − s1 = Cp ln
T2
T1
− R ln
p2
p1
(6)
An alternative form of Eqn. 6 is obtained by using de = Cv dT Eqn. 2, which gives
ˆ
2
s2 − s1 =
Cv
1
dT
− R ln
T
ρ2
ρ1
(7)
Again, for a calorically perfect gas, we get
s2 − s1 = Cv ln
T2
T1
− R ln
ρ2
ρ1
(8)
Isentropic Relations
Adiabatic and reversible processes, i.e., isentropic processes implies ds = 0 and thus Eqn. 6
reduces to
Cp
ln
R
T2
T1
= ln
p2
p1
p2
p1
Cp
γ
=
R
γ−1
T2
T1
p2
=
p1
T2
T1
γ/(γ−1)
ρ2
=
ρ1
T2
T1
1/(γ−1)
γ
ln
γ−1
= ln
⇒
(9)
In the same way, Eqn. 8 gives
3
(10)
Eqn. 11 and Eqn. 10 constitutes the isentropic relations
p2
=
p1
ρ2
ρ1
γ
=
4
T2
T1
γ/(γ−1)
(11)
Specific Heat Relations
For thermally perfect and calorically perfect gases
dh
dT
de
Cv =
dT
Cp =
(1)
From the definition of enthalpy and the equation of state p = ρRT
h=e+
p
= e + RT
ρ
(2)
Differentiate Eqn. 2 with respect to temperature gives
de
d(RT )
dh
=
+
dT
dT
dT
(3)
Cp = Cv + R
(4)
Cp
R
=1+
Cv
Cv
(5)
Inserting the specific heats gives
Dividing Eqn. 4 by Cv gives
Introducing the ratio of specific heats defined as
Cp
Cv
(6)
R
γ−1
(7)
γ=
Now, inserting Eqn. 6 in Eqn. 5 gives
Cv =
2
In the same way, dividing Eqn. 4 with Cp gives
1=
R
1
Cv
R
+
= +
Cp Cp
γ Cp
(8)
and thus
Cp =
γR
γ−1
3
(9)
∂Ω
n
n
V
Ω
V·n
Figur 1: Generic control volume
Governing Equations
The governing equations stems from mass conservation, conservation of momentum and conservation of energy
The Continuity Equation
”Mass can be neither created nor destroyed, which implies that mass is conserved”
The net massflow into the control volume Ω in Fig. 1 is obtained by integrating mass flux over
the control volume surface ∂Ω
‹
−
ρv · ndS
∂Ω
Now, let’s consider a small infinitesimal volume dV inside Ω. The mass of dV is ρdV . Thus,
the mass enclosed within Ω can be calculated as
2
˚
ρdV
Ω
The rate of change of mass within Ω is obtained as
d
dt
˚
ρdV
Ω
Mass is conserved, which means that the rate of change of mass within Ω must equal the net
flux over the control volume surface.
d
dt
‹
˚
ρv · ndS
ρdV = −
∂Ω
Ω
or
d
dt
˚
‹
ρv · ndS = 0
ρdV +
Ω
∂Ω
which is the integral form of the continuity equation.
The Momentum Equation
”The time rate of change of momentum of a body equals the net force exerted on it”
d
(mv) = F
dt
What type of forces do we have?
• Body forces acting on the fluid inside Ω
– gravitation
– electromagnetic forces
– Coriolis forces
3
(1)
• Surface forces: pressure forces and shear forces
Body forces inside Ω:
˚
ρf dV
Ω
Surface force on ∂Ω:
‹
−
pndS
∂Ω
Since we are considering inviscid flow, there are no shear forces and thus we have the net force as
˚
‹
ρf dV −
F=
pndS
Ω
∂Ω
The fluid flowing through Ω will carry momentum and the net flow of momentum out from Ω
is calculated as
‹
‹
(ρv · ndS)v =
∂Ω
(ρv · n)vdS
∂Ω
Integrated momentum inside Ω
˚
ρvdV
Ω
Rate of change of momentum due to unsteady effects inside Ω
d
dt
˚
ρvdV
Ω
Combining the rate of change of momentum, the net momentum flux and the net forces we get
d
dt
˚
‹
(ρv · n)vdS =
ρvdV +
Ω
˚
∂Ω
ρf dV −
Ω
4
‹
pndS
∂Ω
combining the surface integrals, we get
d
dt
˚
‹
˚
[(ρv · n)v + pn] dS =
ρvdV +
Ω
ρf dV
Ω
∂Ω
which is the momentum equation on integral form.
5
(2)
The Energy Equation
”Energy can be neither created nor destroyed; it can only change in form”
E1 + E2 = E3
E1 : Rate of heat added to the fluid in Ω from the surroundings
– heat transfer
– radiation
E2 : Rate of work done on the fluid in Ω
E3 : Rate of change of energy of the fluid as it flows through Ω
˚
E1 =
q̇ρdV
Ω
where q̇ is the rate of heat added per unit mass
The rate of work done on the fluid in Ω due to pressure forces is obtained from the pressure
force term in the momentum equation.
‹
‹
E2pressure = −
(pndS) · v = −
pv · ndS
∂Ω
∂Ω
The rate of work done on the fluid in Ω due to body forces is
˚
E2body
f orces
˚
(ρf dV ) · v =
=
Ω
ρf · vdV
Ω
‹
E2 = E2pressure + E2body
f orces
=−
˚
pv · ndS +
∂Ω
ρf · vdV
Ω
The energy of the fluid per unit mass is the sum of internal energy e (molecular energy) and
the kinetic energy V 2 /2 and the net energy flux over the control volume surface is calculated
6
by the following integral
‹
V2
(ρv · ndS) e +
2
∂Ω
Analogous to mass and momentum, the total amount of energy of the fluid in Ω is calculated
as
˚
V2
ρ e+
dV
2
Ω
The time rate of change of the energy of the fluid in Ω is obtained as
d
dt
˚
V2
ρ e+
dV
2
Ω
Now, E3 is obtained as the sum of the time rate of change of energy of the fluid in Ω and the
net flux of energy carried by fluid passing the control volume surface.
d
E3 =
dt
˚
V2
ρ e+
2
Ω
‹
dV +
V2
(ρv · ndS) e +
2
∂Ω
With all elements of the energy equation defined, we are now ready to finally compile the full
equation
d
dt
˚
‹ ˚
˚
V2
V2
ρ e+
dV +
ρ e+
(v · n) + pv · n dS =
ρf · vdV +
q̇ρdV
2
2
Ω
∂Ω
Ω
Ω
(3)
The surface integral in the energy equation may be rewritten as
‹
‹
V2
p V2
ρ e+
(v · n) + pv · n dS =
(v · n)dS
ρ e+ +
2
ρ
2
∂Ω
∂Ω
and with the definition of enthalpy h = e + p/ρ, we get
‹
V2
ρ h+
(v · n)dS
2
∂Ω
7
Furthermore, introducing total internal energy eo and total enthalpy ho defined as
1
eo = e + V 2
2
and
1
ho = h + V 2
2
the energy equation is written as
d
dt
˚
‹
˚
ρho (v · n)dS =
ρeo dV +
Ω
∂Ω
˚
ρf · vdV +
Ω
q̇ρdV
(4)
Ω
Summary
The integral form of the governing equations for inviscid compressible flow has been derived
Continuity:
d
dt
˚
‹
ρv · ndS = 0
ρdV +
Ω
∂Ω
Momentum:
d
dt
˚
‹
[(ρv · n)v + pn] dS =
ρvdV +
Ω
˚
∂Ω
ρf dV
Ω
Energy:
d
dt
˚
‹
Ω
˚
ρho (v · n)dS =
ρeo dV +
∂Ω
Ω
8
˚
ρf · vdV +
q̇ρdV
Ω
Governing Equations on Integral Form
Eqns. 1 - 3 are the integral form of the continuity, momentum and energy equations, respectively. These equations may be rewritten with the corresponding equations on differential form
as a result.
d
dt
d
dt
d
dt
˚
‹
ρv · ndS = 0
ρdV +
Ω
˚
‹
˚
[(ρv · n)v + pn] dS =
ρvdV +
(2)
˚
˚
‹
ρf · vdV +
ρho (v · n)dS =
ρeo dV +
Ω
∂Ω
Ω
ρf dV
Ω
∂Ω
Ω
˚
(1)
∂Ω
q̇ρdV
(3)
Ω
Governing Equations on Differential Form
Conservation of Mass
Apply Gauss’s divergence theorem on the surface integral in Eqn. 1 gives
‹
˚
ρv · ndS =
∂Ω
∇ · (ρv)dV
Ω
Also, if Ω is a fixed control volume
d
dt
˚
˚
ρdV =
Ω
Ω
∂ρ
dV
∂t
The continuity equation can now be written as a single volume integral.
˚ Ω
∂ρ
+ ∇ · (ρv) dV = 0
∂t
Ω is an arbitrary control volume and thus
∂ρ
+ ∇ · (ρv) = 0
∂t
2
(4)
which is the continuity equation on partial differential form.
Conservation of Momentum
As for the continuity equation, the surface integral terms are rewritten as volume integrals using
Gauss’s divergence theorem.
˚
‹
∇ · (ρvv)dV
(ρv · n)vdS =
Ω
∂Ω
˚
‹
∇pdV
pndS =
Ω
∂Ω
Also, if Ω is a fixed control volume
d
dt
˚
˚
ρvdV =
Ω
Ω
∂
(ρv)dV
∂t
The momentum equation can now be written as one single volume integral
˚ Ω
∂
(ρv) + ∇ · (ρvv) + ∇p − ρf dV = 0
∂t
Ω is an arbitrary control volume and thus
∂
(ρv) + ∇ · (ρvv) + ∇p = ρf
∂t
(5)
which is the momentum equation on partial differential form
Conservation of Energy
Gauss’s divergence theorem applied to the surface integral term in the energy equation (Eqn.
3) gives
‹
˚
ρho (v · n)dS =
∂Ω
∇ · (ρho v)dV
Ω
3
Fixed control volume
d
dt
˚
˚
ρeo dV =
Ω
Ω
∂
(ρeo )dV
∂t
The energy equation can now be written as
˚ Ω
∂
(ρeo ) + ∇ · (ρho v) − ρf · v − q̇ρ dV = 0
∂t
Ω is an arbitrary control volume and thus
∂
(ρeo ) + ∇ · (ρho v) = ρf · v + q̇ρ
∂t
(6)
which is the energy equation on partial differential form
Summary
The governing equations for compressible inviscid flow on partial differential form:
∂ρ
+ ∇ · (ρv) = 0
∂t
∂
(ρv) + ∇ · (ρvv) + ∇p = ρf
∂t
∂
(ρeo ) + ∇ · (ρho v) = ρf · v + q̇ρ
∂t
The Differential Equations on Non-Conservation Form
The Substantial Derivative
The substantial derivative operator is defined as
D
∂
=
+v·∇
Dt
∂t
(7)
where the first term of the right hand side is the local derivative and the second term is the
convective derivative.
4
Conservation of Mass
If we apply the substantial derivative operator to density we get
Dρ
∂ρ
=
+ v · ∇ρ
Dt
∂t
From before we have the continuity equation on differential form as
∂ρ
+ ∇ · (ρv) = 0
∂t
which can be rewritten as
∂ρ
+ ρ(∇ · v) + v · ∇ρ = 0
∂t
and thus
Dρ
+ ρ(∇ · v) = 0
Dt
(8)
Eqn. 8 says that the mass of a fluid element with a fixed set of fluid particles is constant as the
element moves in space.
Conservation of Momentum
We start from the momentum equation on differential form derived above
∂
(ρv) + ∇ · (ρvv) + ∇p = ρf
∂t
Expanding the first and the second terms gives
ρ
∂v
∂ρ
+v
+ ρv · ∇v + v(∇ · ρv) + ∇p = ρf
∂t
∂t
Collecting terms, we can identify the substantial derivative operator applied to the velocity
vector and the continuity equation.
5
∂ρ
∂v
+ v · ∇v +v
+ ∇ · ρv +∇p = ρf
∂t
∂t
|
{z
}
|
{z
}
ρ
=0
= Dv
Dt
which gives us the non-conservation form of the momentum equation
Dv 1
+ ∇p = f
Dt
ρ
(9)
Conservation of Energy
The last equation on non-conservation differential form is the energy equation. We start by
rewriting the energy equation on differential form (Eqn. 6), repeated here for convenience
∂
(ρeo ) + ∇ · (ρho v) = ρf · v + q̇ρ
∂t
Total enthalpy, ho , is replaced with total energy, eo
ho = e o +
p
ρ
which gives
∂
(ρeo ) + ∇ · (ρeo v) + ∇ · (pv) = ρf · v + q̇ρ
∂t
Expanding the two first terms as
ρ
∂eo
∂ρ
+ eo
+ ρv · ∇eo + eo ∇ · (ρv) + ∇ · (pv) = ρf · v + q̇ρ
∂t
∂t
Collecting terms, we can identify the substantial derivative operator applied on total energy,
Deo /Dt and the continuity equation
∂ρ
∂eo
ρ
+ v · ∇eo +eo
+ ∇ · (ρv) +∇ · (pv) = ρf · v + q̇ρ
∂t
∂t
|
{z
}
|
{z
}
o
= De
Dt
=0
6
and thus we end up with the energy equation on non-conservation differential form
ρ
Deo
+ ∇ · (pv) = ρf · v + q̇ρ
Dt
7
(10)
The Governing Equations on Differential Non-Conservation Form
Continuity:
Dρ
+ ρ(∇ · v) = 0
Dt
(1)
Dv 1
+ ∇p = f
Dt
ρ
(2)
Momentum:
Energy:
ρ
Deo
+ ∇ · (pv) = ρf · v + q̇ρ
Dt
(3)
Internal Energy Formulation
Total internal energy is defined as
1
eo = e + v · v
2
Inserted in Eqn. 3, this gives
ρ
De
Dv
+ ρv ·
+ ∇ · (pv) = ρf · v + q̇ρ
Dt
Dt
Now, let’s replace the substantial derivative Dv/Dt using the momentum equation on nonconservation form (Eqn. 2).
ρ
De
− v · ∇p + ρf
·
v + ∇ · (pv) = ρf
·
v + q̇ρ
Dt
Now, expand the term ∇ · (pv) gives
ρ
De
+ p(∇ · v) = q̇ρ ⇒ ρ De + p(∇ · v) = q̇ρ
·
−v
∇p
+
v
· ∇p
Dt
Dt
2
Divide by ρ
De p
+ (∇ · v) = q̇
Dt ρ
(4)
Conservation of mass gives
Dρ
1 Dρ
+ ρ(∇ · v) = 0 ⇒ ∇ · v = −
Dt
ρ Dt
Insert in Eqn. 5
De
p Dρ
De
D
−
= q̇ ⇒
+p
Dt ρ2 Dt
Dt
Dt
1
= q̇
ρ
De
Dν
+p
= q̇
Dt
Dt
(5)
Compare with the first law of thermodynamics: de = δq − δw
Enthalpy Formulation
p
Dh
De 1 Dp
D
h=e+ ⇒
=
+
+p
ρ
Dt
Dt ρ Dt
Dt
1
ρ
with De/Dt from Eqn. 5
D 1
1 Dp
D 1
Dh
= q̇ − p +
+ p Dt
ρ Dt Dt
ρ
Dt ρ
Dh
1 Dp
= q̇ +
Dt
ρ Dt
3
(6)
Total Enthalpy Formulation
1
Dho
Dh
Dv
ho = h + vv ⇒
=
+v·
2
Dt
Dt
Dt
From the momentum equation (Eqn. 2)
Dv
1
= f − ∇p
Dt
ρ
which gives
Dho
Dh
1
=
+ v · f − v · ∇p
Dt
Dt
ρ
Inserting Dh/Dt from Eqn. 6 gives
Dho
1 Dp
1
1 Dp
= q̇ +
+ v · f − v · ∇p =
− v · ∇p + q̇ + v · f
Dt
ρ Dt
ρ
ρ Dt
The substantial derivative operator applied to pressure
∂p
Dp
=
+ v · ∇p
Dt
∂t
and thus
Dp
∂p
− v · ∇p =
Dt
∂t
which gives
Dho
1 ∂p
=
+ q̇ + v · f
Dt
ρ ∂t
If we assume adiabatic flow without body forces
4
Dho
1 ∂p
=
Dt
ρ ∂t
If we further assume the flow to be steady state we get
Dho
=0
Dt
This means that in a steady-state adiabatic flow without body forces, total enthalpy is constant
along a streamline.
5
The momentum equation without body forces
ρ
Dv
= −∇p
Dt
Expanding the substantial derivative
ρ
∂v
+ ρv · ∇v = −∇p
∂t
The first and second law of thermodynamics gives
T ∇s = ∇h −
∇p
ρ
Insert ∇p from the momentum equation
T ∇s = ∇h +
∂v
+ v · ∇v
∂t
Definition of total enthalpy (ho )
1
ho = h + v · v ⇒ ∇h = ∇ho − ∇
2
1
v·v
2
The last term can be rewritten as
∇
1
v·v
2
= v × (∇ × v) + v · ∇v
which gives
∇h = ∇ho − v × (∇ × v) − v · ∇v
Insert ∇h in the entropy equation gives
2
T ∇s = ∇ho − v × (∇ × v) − v
· ∇v
+
T ∇s = ∇ho − v × (∇ × v) +
3
∂v
+
v
· ∇v
∂t
∂v
∂t