One of the goals is to obtain a comprehensive set of fluid equations that incorporates any phenomena that leads to energy depletion. This means that in addition to viscous and thermal effects that transpire within their respective boundary layers, a quantification of the hysteresis effects due to the delay in thermal response should be obtained and incorporated. The most efficient tool to describe this process and for that matter any process is to apply statistical mechanic techniques. Once the Partition Function, Z, is obtained any macro level characteristic pertaining to the system can be derived. In this study Boltzmann’s
version of the combinatorial arrangement, C ˜ = N ! ∏ i − 1 m n i ! , is used. N represents
the number of states, and ni are the number of states at each energy level. Applying Sterling’s formula, this arrangement number becomes,
C = ln ( C ˜ ) = − ∑ i p i ln ( p i ) , where p i = n i N , is the probability of being in the ith
energy state that has energy E i . The statistical mechanic definition of Entropy, s, is the measure of information plus its distribution and is defined as finding the pis that will maximize the arrangement number, C, under the following constraints: ∑ i p i = 1 , and ∑ i p i E i = E (E = Average Energy). From this mathematical framework, the temperature, T ¯ , plays the role of a Lagrange Multiplier. The main results are that
p i = e − β E i Z (4)
where Z is the partition function defined as
Z = ∑ i e − β E i β = 1 T ¯ ¯ T ¯ = k T k = B o l t z m a n n C o n s t a n t (5)
and note that the units for T ¯ (“theorist” temperature) is in Joules. It is easily shown that the average energy and the Helmholtz Free Energy, A, are represented respectively as,
E = − ∂ ln ( Z ) ∂ β and A = E − T S = − T ¯ ln ( Z ) (6)
In quantum mechanics (see Section 2.2 below) the probability density is
defined as p ( r , t ) = ψ ψ ¯ = | ψ | 2 , where ψ is a state vector of the system. Applying Schrodinger’s equation, i ℏ ∂ Ψ ∂ t = H ^ Ψ ( r , t ) , and Hamilton’s operator,
defined in Equation (19a), the following differential equation is derived which expresses the propagation of probability
∂ p ∂ t + ∇ J = 0 , J = i ℏ 2 m [ ψ ∇ ψ ¯ − ψ ¯ ∇ ψ ]
J is the probability flux. For adiabatic processes the divergence of the probability flux is shown to be zero, ∇ J = 0 ; therefore the probability density is stationary with respect to time. This means that the pis are adiabatic invariant (which in turn means that entropy is constant) and thus the following expression for pressure holds
P → a d i a b a t i c ∂ E ∂ V | S = − ∂ E ∂ V | T + ∂ S ∂ V | T ∂ E ∂ S | V = − ∂ E ∂ V | T + ∂ S ∂ V | T T = − ∂ ( E − T S ) ∂ V | T = T ∂ ln ( Z ) ∂ V | T (7)
The average potential between the particles is defined as U 0 and under classical mechanic conditions the partition function is
Z = Z 0 ( β ) [ 1 − β N 2 U 0 2 V ] (8)
Z 0 ( β ) = ( e ρ ) N ( 2 m π β ) 3 N 2 is the partition function pertaining to the Ideas Gas.
In order to obtain a general expression for the average energy of the gas, Equation (6) is applied
E = − ∂ ln ( Z ) ∂ β = N 3 2 T ¯ + N 2 2 V U 0 = [ 3 2 T ¯ + ρ ˜ 2 U 0 ] N (9)
Diatomic molecules like O2 and N2 can be modeled as stiff oscillators. As a simple representative example, a model of a simple oscillator with stiffness k will be studied. The total energy for this system is
E T = p 2 2 m + k x 2 2 (10)
Via the Boltzmann distribution, the Partition Function pertaining to the oscillator, Z o s , is
Z o s = ∬ e − β p 2 2 m e k x 2 2 d p d x = ∫ e − β p 2 2 m d p ∫ e k x 2 2 d x (11)
By applying Gaussian integrals the Partition Function for an undamped oscillator is
Z o s = 2 m π β 2 π β k = 2 π β m k = 2 π ω n 1 β (12)
As before, Equation (6) is used in order to obtain the average energy of the oscillating system
− ∂ ln ( Z o s ) ∂ β = − ∂ [ ln ( 2 π ω n ) − ln ( β ) ] ∂ β = 1 β = T ¯ (13)
The analysis above, especially Equation (11), assumes classical mechanic conditions. Equation (13) states that the average energy of the oscillating system is T ¯ which, based on prior analysis, is connected to kinetic energy. This result is somewhat suspect since it states that regardless of how stiff the system is, the average energy is always T ¯ . In classical mechanics, the path integrals measured with respect to the Max Planck constant, ℏ , are relatively large. An observable in classical mechanics is simply a real function in Euclidean space. In quantum mechanics every possible state of a given system corresponds to a separable Hilbert Space over the complex number field. Additionally, to every physical observable there corresponds, in the Hilbert space, a linear Hermitian operator that has a complete set of orthogonal eigenvectors. The following limit applies to systems oscillating at relatively high energy levels,
lim n → ∞ Δ E E n = lim n → ∞ ℏ ω / [ ( n + 1 2 ) ℏ ω ] → 0 thus, in classical mechanics, it can
be assumed that the energy forms a continuous spectrum; as compared to quantum mechanics where there is now a quantization (discretization) of the energy states [
E n = n ℏ ω (14)
Thus, the Partition Function is
Z q = ∑ n e − β n ℏ ω = ∑ n ( e − β ℏ ω ) n = 1 1 − e − β ℏ ω
Equation (6) is utilized once more in order to obtain the average energy of the stiff oscillator
E = − ∂ ln ( Z ) ∂ β = − 1 Z ∂ Z ∂ β = ℏ ω e − β ℏ ω 1 − e − β ℏ ω (15)
At high temperatures β ≪ 1 and therefore Equation (15) becomes
E = ℏ ω e − β ℏ ω 1 − e − β ℏ ω ~ ℏ ω β ℏ ω = 1 β = T ¯ (16)
Therefore at high temperature levels there is enough energy such that the quantization of energy is not required. The criteria to follow are the following
β ℏ ω > 1 → Q u a n t u m M e c h a n i c s β ℏ ω < 1 → C l a s s i c a l M e c h a n i c s (17)
The cross-over point is ℏ ω = T ¯ . Lastly, it is possible to extract the Energy’s standard deviation from the Partition Function. It is easy to show that
〈 E 2 〉 = 1 Z ∂ 2 Z ∂ β 2 . What follows is the following definition
〈 ( Δ E ) 2 〉 = 〈 E 2 〉 − 〈 E 〉 2 = 1 Z ∂ 2 Z ∂ β 2 − 1 Z 2 ( ∂ Z ∂ β ) 2 = ∂ 2 ln ( Z ) ∂ β 2 = − ∂ E ∂ β = − ∂ E ∂ T ∂ T ¯ ∂ β (18a)
Recognize that ∂ E ∂ T ¯ = C V , the specific heat constant at constant volume and ∂ T ¯ ∂ β = − T ¯ 2
〈 ( Δ E ) 2 〉 = T ¯ 2 C V or 〈 Δ E 〉 = T ¯ C V fluctuation-susceptibility Theorem (18b)
The specific heat constant at constant volume is proportional to standard deviation of the energy.
The vibration due to the collision of molecules extract energy from the passing wave but releases the energy after some delay, thus the relaxation process. This delay, akin to mechanical damping systems leads to hysteretic energy losses. The internal energy can be partitioned into translated (defined relative to the average flow velocity), rotational kinetic energy and energy due to molecular natural frequency,
u = u t r + u r o t + ∑ ν u ν
u ν quantum energy level/mass
The temperatures, Ttr and Trot, related to translational and rotational energies are larger than the temperatures related to molecular oscillation. They usually satisfy the second criteria in Equation (17) and therefore can be considered using Classical Theory. For the molecular vibration case, ℏ ω > k T ν , and, based on Equation (17), quantum mechanics should be applied.
The first postulate in quantum theory states that everything that can be known about the state of a system can be extracted from its state vector (wave function) and this is represented as a vector in Hilbert Space; these are denoted as ket-vectors | Ψ ( x ) 〉 . The bra-vectors are denoted as 〈 Ψ ( x ) | , the vector space of bras is the dual space(one-to-one correspondence with the space of functionals) to the kets vector space. The Riesz representation theorem allows the following notation 〈 Ψ | Ψ 〉 .
The 1-D Hamilton operator is given as
H ^ = p ^ 2 2 m + V ( x ^ ) = − h 2 2 m ∂ 2 ∂ x 2 + V ( x ^ ) (19)
where p ^ is the Linear Momentum operator, p ^ = − i ℏ ∂ ∂ x , and x ^ is the position
operator. p ^ and x ^ are both observables and thus are hermitian operators on the Hilbert Space, therefore H ^ has a purely discrete spectrum of eigenvalues, E n , that has a complete set of eigenstates, | E n ( x ) 〉 , i.e. ( H ^ | E n ( x ) 〉 = E n | E n ( x ) 〉 ). Hence, any state or ket-vector, | Ψ ( x ) 〉 , can be expanded by the set of eigenstates, | Ψ ( x ) 〉 = ∑ n a n | E n ( x ) 〉 . Due to Schrodinger’s equation the eigenstates pertaining to the energy operator are stationary with respect to time. For a diatomic molecule the oscillating like characteristics between the 2 atoms can be simply represented by a spring constant K. Given a particular natural frequency, fe, K can be represented as K = 4 π 2 f e 2 m . The potential energy pertaining to the simple oscillator is represented as
Ф ( x ) = − ∫ 0 ∞ F d x = ∫ 0 ∞ K x d x = ∫ 0 ∞ 4 π 2 f e 2 m x d x = 2 π 2 f e 2 m x 2
Therefore the potential operator becomes V ^ ( x ^ ) = 1 2 m ω 2 x ^ 2 . Hamilton’s operator can be pseudo separated as follows
a ^ = m 2 ω x ^ + i 2 m p ^ = ℏ ω 2 [ ( ℏ / m ω ) 1 / 2 ∂ ∂ x + ( m ω / ℏ ) 1 / 2 x ^ ] = ℏ ω 2 ( ∂ ∂ η + η ^ ) (20a)
a ^ * = m 2 ω x ^ − i 2 m p ^ = ℏ ω 2 [ − ( ℏ / m ω ) 1 / 2 ∂ ∂ x + ( m ω / ℏ ) 1 / 2 x ^ ] = ℏ ω 2 ( − ∂ ∂ η + η ^ ) (20b)
where η ^ = ( m ω / ℏ ) 1 / 2 x ^ . It can be shown that the commutator [ a ^ , a ^ * ] = ℏ ω and therefore
H ^ = a ^ * a ^ + 1 2 ℏ ω = a ^ a ^ * − 1 2 ℏ ω (20c)
Based on Equation (20c) the eigenvalue equation can be written as
a ^ * a ^ | E n ( x ) 〉 = ( E n − 1 2 ℏ ω ) | E n ( x ) 〉 , a ^ a ^ * | E n ( x ) 〉 = ( E n + 1 2 ℏ ω ) | E n ( x ) 〉 (21)
Built on the ideas in Equation (21) the concept of annihilators, a ^ , and creators, a ^ * , are established; the following relationships exists
a ^ | E n ( x ) 〉 = | E n − 1 ( x ) 〉 corresponding eigenvalue E n − 1 = E n − ℏ ω } annihilator
a ^ * | E n ( x ) 〉 = | E n + 1 ( x ) 〉 corresponding eigenvalue E n + 1 = E n + ℏ ω } creator
Another possible solution, tied to the annihilator equation, is that a ^ | E 0 ( x ) 〉 = 0 . Here, | E 0 ( x ) 〉 is known as the zero-point wave-function and E 0 is the zero-point energy level. Equation (20a) pertaining to the zero-point state results in
ℏ ω 2 ( ∂ ∂ η + η ^ ) ψ 0 = 0 , ∴ ∂ ψ 0 ∂ η + η ^ ψ 0 = 0 normalized solution ψ 0 = ( m ω π ℏ ) 1 / 4 e − η 2 / 2
The eigenstates pertaining to the higher energy levels can be generated by applying the Riesz representation notation.
〈 ψ n + 1 | ψ n + 1 〉 = 〈 a ^ * ψ n | a ^ * ψ n 〉 = 〈 ψ n | a ^ a ^ * ψ n 〉 = 〈 ψ n | ( H ^ + ℏ ω 2 ) ψ n 〉 = ℏ ω ( n + 1 ) 〈 ψ n | ψ n 〉
Thus, if ψ n is normalized so is
ψ n + 1 = ( ℏ ω ( n + 1 ) ) − 1 / 2 a ^ * ψ n = ( n + 1 ) − 1 / 2 1 2 ( − ∂ ∂ η + η ^ ) ψ n
As a consequence, the recurring formula for the eigenfunctions/states becomes
ψ n = ( 2 n ! ) − 1 / 2 ( − d d η + η ) n e − η 2 2 = ( 2 n ! ) − 1 2 ( m ω π ℏ ) 1 4 H n ( η ) e − η 2 2 , n = 0 , 1 , 2 , … (22)
where H n ( η ) are the Hermite polynomials. Additionally due to Equation
(19a) the pertaining eigenvalues of H ^ are E n = ℏ ω ( n + 1 2 )
An alternative approach is to apply Schrodinger’s Equation i ℏ ∂ Ψ ∂ t = H ^ Ψ ( r , t ) .
Plugging in the definition for the Hamilton Operator defined in Equation (19a) results in the following simple oscillator equation
d 2 Ψ v d x 2 + 2 m ℏ 2 ( E j − Φ ( x ) ) Ψ v = d 2 Ψ v d x 2 + 2 m ℏ 2 ( E j − 1 2 m ( ω x ) 2 ) Ψ v = 0 (23)
Setting
z = m ω ℏ x , α = 2 m E j ℏ 2 , and Ψ v = e − z 2 2 w (24)
Schrodinger’s Equation simplifies to
d 2 w d z 2 − 2 z d w d z + ( α ( d z d x ) 2 − 1 ) w = d 2 w d z 2 − 2 z d w d z + ( 2 E j ℏ ω − 1 ) w = 0 (25)
When 2 E j ℏ ω − 1 = 2 j = 0 , 1 , 2 , 3 , … the solutions to this equation are the Hermite
Polynomials. Note that the solution is equivalent to Equation (22). Applying this definition and solving for E j , the energy levels are defined as
E j = ℏ ω ( j + 1 2 ) = j = 0 , 1 , 2 , … (26)
Due to the assumption of independence of the different energies the partition function pertaining to internal energies can be represented as
Z i n t = Z r o t Z v Z c h e m (27)
The partition function for the vibrational portion pertaining to a particular type of molecule v (e.g. O2, N2) is represented as
Z v = ∑ v j e − [ ( ϵ v j − ϵ v 0 ) k T v ] β = 1 T ¯ ¯ = 1 k T v (28)
ϵ v 0 is the ground vibrational state energy (obtained by setting j = 0 in Equation (26)). Using Equation (26) in (28), the vibrational partition function can be rewritten as
Z v = ∑ j = 0 ∞ e − [ ℏ ω k T v ] j = 1 1 − e − [ ℏ ω k T v ] = 1 1 − e − θ e * T v (29)
Such that θ e * = ℏ ω k .
Applying Equation (6), an expression for the average energy is obtained
u i n t = R T 2 ( ∂ ln Z i n t ∂ T ) V (30)
Therefore the vibrational internal energy becomes
( u v − u v 0 ) = − R T v 2 ( e − θ e * T v 1 − e − θ e * T v ( − θ e * T v 2 ) ) = R θ e * e θ e * T v − 1 (31)
Differentiating this equation with respect to temperature one obtains [
d u v d T v = C v v = θ e * 2 T v 2 R e θ e * T v ( e θ e * T v − 1 ) 2 (32)
Applying the Boltzmann distribution definition for entropy
s = − ∑ i p i ln ( p i ) = β E + ln ( Z ( β ) )
results in
s v = R ( − ln ( 1 − e − θ e * T v ) R + θ e * k T v e θ e * T v − 1 ) (33)
Recall that T v ≪ T t r , T r o t and the temperatures pertaining to Translational and rotational energy levels are high enough to be considered by classical
mechanics. The combination of kinetic energies ( p 2 2 m + L 2 2 I ) and positions are
considered as possible states, the application of the Boltzmann distribution results in the following equation for average energies.
− σ n = p − μ B ∇ ⋅ v such that μ B = p β r o t N c u r o t 2 u 2 = b u l k v i s c o s i t y (34)
The instantaneous change of entropy can be represented in the following form
T d s = d u + p d ρ − 1 + ∑ υ A υ d T υ (35)
The ∑ υ A υ d T υ terms represent the small deviation from equilibrium. The affinities A υ are to be determined. The portion devoid of vibration energy is denoted from now on with subscript fr (frozen), and the portion pertaining to vibrational energy is indicated with subscript v [
s = s f r ( u t r + u r o t , ρ − 1 ) + ∑ υ s υ ( T υ ) (36)
T d s f r = d ( u t r + u r o t ) + p d ρ − 1 (37)
Set
u t r + u r o t = C V , f r T assuming ideal gas P = ρ R T (38)
C V , f r = R γ − 1 , C V , f r = coefficient of specific heat at constant volume. (39)
Substituting Equation (38) into (37) and solve for s f r the following expression is obtained
s f r = C V , f r ln ( u − ∑ υ u υ ) + R ln ρ − 1 + c o n s t r e c a l l t h a t u − ∑ υ u υ = u t r + u r o t (40)
Utilizing Equation (37) - (40) and applying them into Equation (35) yields
A υ = T ( − C V , f r u t r + u r o t d u υ d T υ + 1 T υ d u υ d T υ ) = ( T T υ − 1 ) d u υ d T υ = ( T T υ − 1 ) C V , v (41)
The expression in Equation (32) can be substituted into Equation (41). Since T υ ≪ θ e * , C V , v can deduce to
C V , v = d u υ d T υ = n υ n R ( θ e * T υ ) 2 e − θ e * T υ (42)
C V , v act as the specific heat pertaining to each molecular type v.
The inclusion of these energy terms into the standard energy equations results in
D u D t − σ n D ρ − 1 D t = T D s v D t − ∑ υ s υ D T υ D t − μ B ∇ ⋅ v D ρ − 1 D t (43)
Applying the conservation of mass and the heat flux the entropy balance equations become
ρ D s D t + ∇ ⋅ q T = 1 T μ B ( ∇ ⋅ v ) 2 + 1 2 T μ D ∇ v ↔ : D ∇ v ↔ + κ T 2 ( ∇ T ) 2 + ρ T ∑ υ A υ D T υ D t (44)
The Navier Stokes Equation with the inclusion of the bulk viscosity becomes
ρ D v D t = − ∇ p + ∇ ( μ B ∇ ⋅ v ) + μ ∇ ⋅ D ∇ v ↔ (45)
Applying conservation of energy results in the following expression for the vibration energy
D u υ D t = C V , v D T υ D t = n υ N c υ Δ ϵ υ (46)
N c υ represents the number of collisions a molecule of type υ has per unit time, n υ is the number of molecules of type υ , and Δ ϵ υ is the average energy gained per collision. It is known that most of the high temperature resides in the translational and rotational portions, therefore Δ ϵ υ can best be expressed as
Δ ϵ υ = β υ k ( T − T υ ) (47)
and yields
D T υ D t = ( T − T υ ) 1 τ υ such that τ υ = C V , v n υ N c υ k β υ (48)
More elaborate modeling for Δ ϵ υ can be achieved by applying Discrete Boltzmann Equations but these are outside the scope of this project [
T υ = T − ( T − T υ ( 0 ) ) e − t / τ υ (49)
If T is changed by an amount Δ T it will take a time of τ υ for the incremental change in T υ is ( 1 − e − 1 ) Δ t . Therefore τ υ is called the relaxation time. And it is this relaxation time that introduces the hysteretic energy losses [
The energy conservation dissipation equation is in the form of
∂ w ∂ t + ∇ ⋅ I = − D
where
w = 1 2 ρ o v 2 + 1 2 p 2 ρ o c 2 + 1 2 ( ρ T C p ) o s f r 2 + ∑ υ 1 2 ( ρ C V , v T ) o T υ 2 (50a)
I = p v − μ B v ( ∇ ⋅ v ) − μ v ⋅ ( D ∇ v ↔ ) − κ T 0 − 1 T ∇ T (50b)
D = μ B ( ∇ ⋅ v ) 2 + 1 2 μ ( D ∇ v ↔ ) : ( D ∇ v ↔ ) + κ T 0 − 1 ( ∇ T ) 2 + ∑ υ ( ρ o C V , v T τ υ ) o ( T − T υ ) 2 (50c)
From the coupled system of acoustic equation, asymptotic techniques result in different dispersion relationships for 3 different regimes; for the viscous and entropy boundary layers, and the equations pertaining to the acoustic mode. These translate to different set of partial differential equations pertaining to the 3 different regimes. The vibrational relaxation terms will mostly affect the acoustic mode regime. In order to obtain the new dispersion relationship a solution in the form of e i k x e − i ω t is plugged into the equation of motion (see Appendix A.1 - A.5), the ensuing is obtained
k 2 p ^ = ( ω 2 + i ω k 2 μ B + 4 3 μ ρ 0 ) ρ ^ (51a)
i ω ρ 0 s ^ f r = [ ( κ T ) o k 2 − ( i ω ρ T ) 0 ∑ v C V , v 1 − i ω τ v ] T ^ (51b)
ρ ^ = p ^ c 2 − ( β T / C p ) 0 2 ( s ^ f r / T ^ ) p ^ 1 − ( T / C P ) 0 ( s ^ f r / T ^ ) (51c)
Plugging this last equation into Equation (51a) the dispersion relation is obtained [
k 2 ω 2 + i ω k 2 μ B + 4 3 μ ρ 0 = 1 c 2 − ( γ − 1 ) T 0 ( s ^ f r / T ) / C p c 2 1 − ( T / C p ) 0 ( s ^ f r / T ^ ) (51d)
For the acoustic mode and the frequency range of interest the following
approximations exist: k 2 = ( ω c ) 2 , C V , v C p ≪ 1 , ω κ ρ 0 C p c 2 ≪ 1 , ω μ ρ 0 c 2 ≪ 1 the last 2
expressions can be found in Kirchhoff’s dispersion relation and were considered small in [
k = ω c + i ω 2 μ 2 ρ 0 c 3 [ 4 3 + μ B μ + ( γ − 1 ) κ C p μ ] + 1 2 ( γ − 1 ) ω c ∑ v C V , v / C p 1 − i ω τ v (51e)
The absorption coefficients are the imaginary portion of Equation (51e)
ω 2 μ 2 ρ 0 c 3 [ 4 3 + μ B μ + ( γ − 1 ) κ C p μ ] + 1 2 ( γ − 1 ) ω 2 c ∑ v τ v C V , v / C p 1 + ( ω τ v ) 2 (51f)
where
α v = π 2 λ ( γ − 1 ) ω 2 c τ v C V , v / C p 1 + ( ω τ v ) 2 (51g)
Recall that λ = 2 π c ω . At relatively low frequencies, ω ≪ 1 τ υ Equation (51g) indicates that
α v → τ υ c C V , v C p ( γ − 1 ) ω 2 quadratic increase w.r.t. ω (51h)
and at relatively high frequencies, ω ≫ 1 τ υ , results in the following
α v → τ υ c C V , v C p ( γ − 1 ) approaches a constant value (51i)
In [
∇ y 2 T 0 + i ω ρ 0 P r υ T 0 = i ω P 0 κ in D f (52)
Before deriving the thermal exchange for a general porous medium (with pores of arbitrary geometry), the thermal exchange for a typical cylindrical pore with cross-sectional radius R will be analyzed. The form this solution takes will be used as a pattern to emulate when considering solutions for the general case. Equation (52) in cylindrical form is
1 r d d r ( r d T 0 d r ) + i ω ρ 0 P r υ T 0 = i ω P 0 κ in D f (53)
The Bessel function generator is defined as
e i x sin ( θ ) = J 0 ( x ) + i 2 J 1 ( x ) sin θ + 2 J 2 ( x ) cos 2 x + … (54a)
The J 0 Bessel function is efficiently represented via Hankel’s integral
J 0 ( x ) = 1 π ∫ − 1 1 e i x t ( 1 − t 2 ) d t = 2 π ∫ 0 1 cos ( x t ) ( 1 − t 2 ) d t (54b)
It can be deduced from Equation (54a) that
2 n x J n ( x ) = J n − 1 ( x ) + J n + 1 ( x ) and 2 ∂ J n ( x ) ∂ x = J n − 1 ( x ) − J n + 1 ( x ) (55a)
The ensuing 2 equations are obtained by combining the equations in (55a)
J n − 1 ( x ) = n x J n ( x ) + ∂ J n ( x ) ∂ x and J n + 1 ( x ) = n x J n ( x ) − ∂ J n ( x ) ∂ x (55b)
Applying Equation (55b) results in
d d x { x J 1 ( x ) } = x J 0 ( x ) (55c)
Equation (55c) is integrated from 0 to R, this generates an equation that will be utilized shortly
∫ 0 R x J 0 ( γ x ) d x = R γ J 1 ( γ ) (55d)
Teagle [
the particular solution for Equation (53) is υ κ P r P 0 = ρ 0 C p P 0 . The homogeneous solution is A 1 J 0 ( i ω ρ 0 P r μ r ) , where A 1 is a derived so that boundary conditions
at the wall of the cylinder pores are met, T 0 ( R ) = 0 . Combining the particular and homogenous solutions and satisfying boundary conditions, the following expression for temperature is obtained
T 0 ( r ) = ρ 0 C p P 0 ( 1 − J 0 ( χ r ) J 0 ( χ R ) ) ∋ χ = i ω ρ 0 P r μ (56)
For statistical energy analysis calculations, average temperature of the pore cross-section suffices, therefore
〈 T 0 〉 = ∫ 0 R 2 π T 0 ( r ) r d r π R 2 = ρ 0 C p P 0 ( 1 − 2 J 1 ( χ R ) R χ J 0 ( χ R ) ) = ρ 0 C p P 0 ( 1 − ( 1 − i ) 2 J 1 ( χ R ) R J 0 ( χ R ) μ ω ρ 0 P r ) (57)
For the more general case, the focus is on analyzing the energy flux escaping the acoustic mode into the entropy boundary layer. Johnson et al. [
Trilling [
Vorticity Mode field
∇ 2 v v o r = ρ μ ∂ v v o r ∂ t , ∇ ⋅ v v o r = 0 , s v o r = p v o r = T v o r = 0 (58)
It is important to point out that Equation (58) is a classical diffusion equation
The entropy Mode field
∇ 2 s e n t = ρ c P κ ∂ s e n t ∂ t (59a)
p e n t = 0 , v e n t = ( β T κ ρ c P ) o ∇ s e n t , ∇ × v e n t = 0 , T ′ e n t = ( T c P ) s e n t , ρ ′ e n t = ( β T ρ c P ) o s e n t (59b)
Acoustic Mode field
ρ ∂ v a c ∂ t = − ∇ p a c , s a c = 0 , ∇ × v a c = 0 , T ′ a c = ( β T ρ c P ) o p a c , ρ ′ a c = p a c c 2 (60)
Due to the superposition principle the microscopic velocity flow field can be written as
v = v v o r + v e n t + v a c (61)
The dispersion for the vorticity and entropy modes are
k v o r 2 = i ω ρ μ and k e n t 2 = i ω ρ c P κ (62)
Therefore the vorticity and entropy mode field extend to 1 | k v o r | and 1 | k e n t | so
δ v o r = ( 2 μ ω ρ ) 1 / 2 and δ e n t = ( 2 κ ω ρ c P ) 1 / 2 = δ v o r ( P r ) 1 / 2 (63)
Near the solid surface it can be assumed that the solution has the form
Ψ ( x , y , z ) = Ψ ( x , y , 0 ) e − ( 1 − i ) z / δ e n t . (64)
where this relation satisfies the following equation
∂ 2 Ψ ( x , y , z ) ∂ z = − i 2 δ 2 Ψ ( x , y , z ) . Additionally, as in the aforementioned
cylindrical case, the temperature fluctuation at the wall is zero. The temperature conditions in Equations (58) (29b) (60) produces the following
T ′ a c + T e n t = 0 (65)
Applying Equation (64) into Equations (58), (59a), (59b) and splitting the “del” operator into tangential and normal components (where the tangential portion will have a subscript T) yields the following
∇ T ⋅ v v o r , T ( x , y ) − ( 1 − i ) v v o r ( x , y ) ⋅ n δ v o r = 0 (66)
v e n t , T ( x , y ) = β κ ρ c P ∇ T T e n t = 0 (67)
v e n t ⋅ n = − β κ ρ c P ( 1 − i ) T e n t δ e n t (68)
At z = 0 plane Equation (61) will give the following boundary condition
v w a l l = v v o r + v e n t + v a c (69)
and applying the ∇ T ⋅ (horizontal divergence) along with Equation (67), Equation (69) can be rewritten the following way
∇ T ⋅ v a c , T ( x , y ) + ( 1 − i ) v v o r ( x , y ) ⋅ n δ v o r = 0 (70)
Remember that
γ − 1 = T β ˜ 2 γ K T ρ C p = T β ˜ 2 K S ρ C p = T β ˜ 2 c 2 C p K T / K S
= isothermal/adiabatic bulk modul (71)
Additionally, using Equation (68) and (70) the subsequent conditions are obtained in the direction normal to the surface
v w a l l ⋅ n = − ( 1 + i ) δ v o r 2 ∇ T ⋅ v a c , T ( x , y ) + ( 1 − i ) ( γ − 1 ) ω c 2 δ e n t 2 p a c ρ + v a c ⋅ n (72)
Recall that the main goal is to obtain an expression for the energy that is being dissipated and this is dictated by the net power flowing out of the acoustic (ac) mode close to the boundary. Multiplying Equation (72) by the complex conjugate of the acoustic mode pressure, p a c * , and solving for − p a c * v a c ⋅ n , which is the intensity leaving the acoustic mode, provides the following equation
− ( I a c ⋅ n ) a v = 1 2 R e ( p a c * v w a l l ⋅ n ) − δ v o r 4 ∇ ⋅ { R e [ ( 1 + i ) p a c * v a c , T ( x , y ) ] } + δ v o r 4 R e [ ( 1 + i ) ∇ T p a c * ⋅ v a c , T ( x , y ) ] + R e [ 1 2 ( 1 − i ) ( γ − 1 ) ω c 2 δ e n t 2 | p a c | 2 ρ ] (73)
Recall that the relative velocity at the wall is equal to zero. From the first relation in Equation (60) it can be deduced that
∇ T p a c * = i ω ρ v a c , T * ( x , y ) (74)
The second term in the right hand side of Equation (73) will average out to zero given enough surface area. Equation (74) will be applied in the third term of Equation (73) to finally obtain the following relation
( d 2 E d A d t ) d i s s = ( ω ρ μ 2 ) 1 / 2 ( v a c , T 2 ( x , y ) ) a v + ( γ − 1 ) ( ω ρ κ 2 c P ) 1 / 2 | p a c | a v 2 ρ c 2 (75)
For the mechanical dissipation the first term in the right hand side is used by [
The thermal counterpart of dynamic permeability and tortuosity are
ϕ T o = − k ′ ( ω ) κ ∂ P o ∂ t (76)
α ′ ( ω ) c P ρ f ∂ T o ∂ t = ∂ P o ∂ t (77)
where T o and P o satisfy the following
The real constant
A volume of porous sample enclosed by a thin membrane is considered. In this particular case, it is assumed that the thin membrane acts like an adiabatic wall. The interest is to know where the singularities to
To guarantee that no thermal power is entering the element, the following has to hold
Since the membrane is adiabatic and since conditions listed in Equation (78) are being satisfied the integral in Equation (80) is consequentially fulfilled. Pre multiplying the Fourier equation of heat diffusion equation, Equation (79), by the complex conjugate
Making use of the usual identification
So
The first term in the right hand side of Equation (84) goes to zero because of boundary condition found in Equation (80). Inserting this result in Equation (81) and solving for
Due to the shifting rule, this indicates that the singularity for
In [
Following the same steps performed in [
sents the thermal portion of the energy flux exiting the main acoustic mode. The third term in the RHS of Equation (50a) is the thermal energy. Multiplying this term by the speed of sound, c, the energy flux entering the pore is attained. The ensuing equation represents the ratio of thermal energy that is leaving the acoustic mode to that of thermal energy that is entering the thermal element.
Recall temperature condition
applying Equation (71), Equation (87) simplifies to
In Equation (88),
defined as
for average temperature is defined as (assuming ideal gas conditions)
In Equation (89), as increases the thinner the entropy boundary layer,
plugged into Equation (3). The following expression for thermal permeability
is obtained
The second term in Equation (90) is obtained by applying the definition of
where,
Finally, this is the term that is used in Equation (2) to obtain the compressibility factor of the fluid of the porous medium.