Lecture 1 - Equations - Princeton University · permissionoftheowner,MMatalon. 1 Lecture 1...
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6/26/11 Copyright ©2011 by Moshe Matalon. This material is not to be sold,reproduced or distributed without the prior wri@en permission of the owner, M Matalon. 1 Lecture 1 Conservation equations for chemically reacting gas mixtures Conservation of mass Dρ Dt convective derivative ∂ρ ∂t + v ·∇ρ + ρ∇· v =0 ⇒ Dρ Dt + ρ∇· v =0 time rate of change ... following a fluid particle ∇· v = − 1 ρ Dρ Dt Dρ Dt =0 ⇒ ∇ · v =0 incompressible flow ∂ρ ∂t + ∇· ρv =0 Conservation equations for a pure substance
Transcript of Lecture 1 - Equations - Princeton University · permissionoftheowner,MMatalon. 1 Lecture 1...
6/26/11
Copyright ©2011 by Moshe Matalon. This material is not to be sold,reproduced or distributed without the prior wri@en permission of the owner, M Matalon. 1
Lecture 1 Conservation equations for chemically
reacting gas mixtures
Conservation of mass
Dρ
Dtconvectivederivative
∂ρ
∂t+ v ·∇ρ+ ρ∇ · v = 0
⇒ Dρ
Dt+ ρ∇ · v = 0
time rate of change . . .following a fluid particle
∇ · v = −1
ρ
Dρ
Dt
Dρ
Dt= 0 ⇒ ∇ ·v = 0
incompressible flow
∂ρ
∂t+∇ · ρv = 0
Conservation equations for a pure substance
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Copyright ©2011 by Moshe Matalon. This material is not to be sold,reproduced or distributed without the prior wri@en permission of the owner, M Matalon. 2
Conservation of momentum ρDv
Dt= ρg +∇ · σ
σ = −pI+Σ stress tensor
Σ = µ[(∇v) + (∇v)T] + (κ− 2
3µ)(∇ · v)I
pressure
shearviscosity
bulkviscosity
viscous stress tensor (Newtonian Fluid)
σ is a symmetric tensor, i.e σij = σji.
(∇ · σ)i =∂σij
∂xjsummation convention adopted(∇v)ij =
∂vi∂xj
ρDv
Dt= −∇p+∇ ·Σ+ ρg
The first term is a measure of the deformation of the fluid element and thesecond term is the rate of change of its volume
Conservation of energy
ρDeT
Dt= ρg · v +∇ · (σ · v)−∇ · q
rate of work done on the fluid by
gravitational & viscous forces
rate of heat transferred to
the fluid
eT = e+ 12v
2 = total energy
ρDv
Dt= ρg +∇ · σ / · v ⇒ ρ
D
Dt
�1
2v2
�= ρg · v + (∇ · σ) · v
ρDe
Dt= [∇ · (σ · v)− (∇ · σ) · v]−∇ · q
σ : ∇v = σij∂vi∂xj
q - heat flux vector(positive for outward heat transport)
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ρDe
Dt= −p∇ · v + Φ−∇ · qσ = −pI+Σ
Φ =1
2µ
�∂vi∂xj
+∂vj∂xi
�2
+ (κ− 2
3µ)(∇ · v)2
Rate of dissipation of internal energy due to
viscous forces Dissipation function
ρDh
Dt=
Dp
Dt+ Φ−∇ · q
⇒
Rate of internal energy increase by compression
An alternative form, in terms of the enthalpy h = e+ p/ρ
Dh
Dt=
De
Dt+
1
ρ
Dp
Dt− p
ρ2Dρ
Dt
= ρDh
Dt− Dp
Dt− p∇ · v
ρDh
Dt=
Dp
Dt+ Φ−∇ · q
ρDv
Dt= −∇p+∇ ·Σ+ ρg
∂ρ
∂t+∇ · ρv = 0
energy
momentum
mass
ConsCtuCve relaCons:
Σ = µ[(∇v) + (∇v)T] + (κ− 2
3µ)(∇ · v)I
Summary
∇ · q = −λ∇T
Newtonian fluid
Fourier Law of heat conducCon (neglecCng radiaCon, etc.)
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Thermodynamics
State of the gas is uniquely determined by two variables p = p(ρ, T ), e = e(p, T ), etc.
Ideal gas:
e = e(T ), h = h(T ) only
Reference temperature
Specific heats at constant pressure/volume cp , cv
dh
dT= cp ⇒ h = ho +
� T
T o
cpdT
de
dT= cv ⇒ e = eo +
� T
T o
cvdT
h = e+ p/ρdh
dT=
de
dT+
d
dT
�p
ρ
�cp = cv +R/W⇒
raCo of specific heats γ = cp/cv > 1
p =ρRT
W
universal gas constant
specific gas constant
R = 8.3145 cal/mol·KR̄ = R/W
TdS = dh− 1
ρdpentropy
ρDh
Dt=
Dp
Dt+ Φ−∇ · q ρ
DS
Dt=
1
T(Φ−∇ · q)
reversible process: irreversible process: In an isentropic flow (adiabaCc + reversible) the entropy remain constant
Φ = 0Φ > 0
dh = dp/ρ
dh = cpdT⇒ dp
p= γ
dρ
ρpρ−γ = const.
∆S ∼ heat acquired/T∆S larger than when Φ = 0
speed of sound a =
��∂p
∂ρ
�
S
=�γp/ρ =
�γRT
TDS
Dt=
Dh
Dt− 1
ρ
Dp
Dt
⇒
