Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 2 pps
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 2 pps
... one may write
d
2
S =
1
2
∂
2
S
I
∂U
2
I
+
∂
2
S
II
∂U
2
II
eq
(dU)
2
+2
∂
2
S
I
∂U
I
∂V
I
+
∂
2
S
II
∂U
II
∂V
II
eq
dUdV
+
∂
2
S
I
∂V
2
I
+
∂
2
S
II
∂V
2
II
eq
(dV )
2
≤ 0. (1.5.1)
Recalling ... phenomenological relations:
j = l tr X, (2. 22a)
J = A
1
x , (2. 22b)
T = B
1
(tr X)I + B
2
X, (2. 22c)
where the phenomenological scalar coefficients l,...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 11 pps
... irreversible thermodynamics,
37, 44, 58, 63, 65, 72, 99, 179,
198, 20 3, 21 6, 22 0, 23 3, 24 1, 27 6,
29 2 29 4, 304
Clausius’ inequality, 12, 29 2
Clausius–Duhem’s inequality, 22 0, 23 4,
23 8, 23 9, 24 1, 24 3, 24 4, ... 70, 72, 80, 189, 191, 197,
20 3, 21 3, 21 9, 22 2, 22 5, 23 0, 23 4,
23 5, 23 8, 24 4, 24 8, 24 9, 25 6, 28 1,
29 0, 29 3, 29 8
Gibbs’ fr...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 3 pdf
... (L
11
− L
12
L
21
/L
22
) (heat conductivity),
π = T (L
12
/L
22
) (Peltier coefficient),
ε = L
21
/L
22
(Seebeck coefficient),
r =1/L
22
(electrical resistivity).
56 2 Classical Irreversible Thermodynamics
Step ... X,namely
I
X
=trX,II
X
=
1
2
[I
2
X
− tr(X ·X)], III
X
=det X. (2. 2.8)
By restricting the analysis to linear laws, (2. 2.5)– (2. 2.7) will take the simple
for...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 4 ppt
... independent and A
1
+ A
2
+ A
3
=0.
It will be shown that, when the flux–force relations are of the form
w
1
− w
3
= L
11
A
1
+ L
12
A
2
, (4 .25 a)
w
2
− w
3
= L
21
A
1
+ L
22
A
2
, (4 .25 b)
the Onsager’s ... relations:
w
1
− w
3
= L
11
A
1
+ L
12
A
2
, (4 .29 )
w
2
− w
3
= L
21
A
1
+ L
22
A
2
. (4.30)
Comparison with relations (4.1.9) and (4.1.10) yields
L
12
= L...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 6 potx
... differential equations
(D
2
− k
2
)(D
2
− k
2
− σ)W = Ra k
2
Θ, (6.1.13)
(D
2
− k
2
− σPr)Θ = −W, (6.1.14)
where D stands for d/dZ and k
2
= k
2
x
+ k
2
y
, Ra and Pr denote the dimen-
sionless Rayleigh ... (6.70)
where the matrix L is given by
L =
⎛
⎝
k
3
B − k
4
k
2
1
k
2
k
2
4
A
2
−k
3
B −
k
2
1
k
2
k
2
4
A
2
⎞
⎠
, (6.71)
whose eigenvalues σ
1
,σ
2
, de...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 7 ppt
... al. 20 01)
λ(ω, k)=
λ(T )
1+iωτ
1
+
k
2
l
2
1
1+iωτ
2
+
k
2
l
2
2
1+iωτ
3
+
k
2
l
2
3
1+iωτ
4
+···
, (7.36)
where l
n
are characteristic lengths of the order of the mean free path, and
τ
1
, τ
2
, ... has
S(U
1
,U
2
)=S
1
(U
1
)+S
2
(U
2
) and therefore
dS
dt
=
dS
1
dt
+
dS
2
dt
= T
−1
1
dU
1
dt
+ T
−1
2
dU
2
dt
. (7 .2. 1)
Since the global system is isolated, dU...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 8 pot
... refer to the
21 5
7 .2 One-Component Viscous Heat Conducting Fluids 199
By setting
α
1
= τ
1
(λT )
−1
,α
0
= τ
0
ζ
−1
,α
2
= τ
2
(2 )
−1
, (7.53)
µ
1
=(λT
2
)
−1
,µ
0
=(ζT)
−1
,µ
2
= (2 T)
−1
, (7.54)
(7.50)–(7. 52) ... second-order derivatives of s
EIT
with respect to its state variables are
negative; in particular,
∂
2
s
EIT
∂q · ∂q
= −
vτ
1
λT
2
< 0,
∂
2
s
EIT
∂
0
P
v
:...
Understanding Non-Equilibrium Thermodynamics - Springer 2008 Episode 9 pptx
... σ
s
.
8 .2. 3.3 Restrictions Placed by the Second Law
of Thermodynamics
The total kinetic energy per unit mass can be written as
1
2
cv
2
p
+
1
2
(1 − c)v
2
f
=
1
2
v
2
+
1
2
c(1 − c)ξ
2
. (8.66)
To ... Λ
0
(≡ θ
−1
), 2
∂s
∂q
2
q = Λ
1
(≡ γ(u, q
2
)q) (9. 62)
and
∂J
s
∂u
=
∂a
∂u
Λ
1
,
∂J
s
∂q
= Λ
0
I +2
1
q
∂a
∂q
2
, (9.63)
wherein we have identified Λ
0
with θ
−1
,...
Khám phá windowns server 2008 - p 2 pps
... Roles-Based Access and Delegation to Provision
Virtual Machines . 321
Administrator Role in VMM 20 08. 321
Delegated Administrator Within VMM 20 08 . 322
Self-Service User as a Role in VMM 20 08. 322
Managing ... at www.wowebook.com
ptg64 326 87
Who Needs VMM 20 08? . 26 9
VMM 20 08 for Delegated Administration Environments . 26 9
VMM 20 08 for Structure ITIL-Based Organ...
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