4. Dissipation rates

### 4.1 Energy consumption per time

In classical mechanics, work done due to an infinitesimal displacement of a particle dr under the influence of force field F is

$$
\mathbf{d}W = \mathbf{F} \cdot \mathbf{d}r \tag{76}
$$

The time differentiation of Eq. (76) provides an energy consumption rate (i.e., power represented by P) as a dot product of the particle velocity v and the applied force F:

Non-Equilibrium Particle Dynamics

$$
\dot{W} = \frac{\text{d}W}{\text{d}t} = \boldsymbol{\nu} \cdot \boldsymbol{F} \tag{77}
$$

which represents the changing rate of the specific entropy as a dot project of flux J and driving force X. The subscript k in Eq. (84) is for physical quantities on which the respective entropy depends. For mathematical simplicity, a new quantity is defined as Yk ¼ TXk, where T is the absolute temperature in Kelvin to have

JkYk (85)

RklJl (86)

δjlJl ¼ Jj (87)

JjRji h i <sup>¼</sup> <sup>2</sup>Xi (89)

� �dNi (91)

(90)

Tσ ¼ ∑ k

Xk ¼ ∑ l

where Rkl represents an inverse matrix of Ljk, i.e., RklLjk ¼ δlj, which can be

∑ k

where h jJ and j iJ exist as the row and column vectors of J, respectively, and R represents the generalized resistance matrix. A partial derivative of σ with respect

k

In this case, the specific entropy dissipation rate σ is presented using the flux and

which indicates that the specific entropy increases with respect to the flux and

The second thermodynamic law represents the infinitesimal entropy change in

<sup>T</sup> <sup>d</sup><sup>V</sup> � <sup>∑</sup> i

where E represents the internal energy, P represents the system pressure within

μi T

P

a volume V, and μ<sup>i</sup> and Ni are the chemical potential and the mole number of species i, respectively. Eq. (91) implies the entropy S ¼ S Eð Þ ;V; Ni as a function of the internal energy E, the volume V, and the number of species i Ni.This gives ξ<sup>1</sup> ¼ E, ξ<sup>2</sup> ¼ V, and ξ<sup>i</sup> ¼ Ni (i ¼ 3 for water and i ¼ 4 for solute). The driving forces

RikJk ½ �þ ∑

j

Substitution of Eq. (86) to Eq. (84) represents σ in terms of flux J

LjkRkl � �Jl <sup>¼</sup> <sup>∑</sup>

l

JjRjkJk ¼ h i JjRjJ (88)

inverse relationship of Eq. (81) is

DOI: http://dx.doi.org/10.5772/intechopen.86607

∂σ ∂Ji ¼ ∑ j, k

4.2 Effective driving forces

the microcanonical ensemble:

Xi are particularly calculated as

93

proven by substituting Eq. (86) with Eq. (81):

Jj ¼ ∑ k

LjkXk ¼ ∑ l

Fundamentals of Irreversible Thermodynamics for Coupled Transport

σ ¼ ∑ j, k

to an arbitrary flux Ji is equal to twice the generalized driving force:

<sup>δ</sup>ijRjkJk <sup>þ</sup> JjRjkδik h i <sup>¼</sup> <sup>∑</sup>

σ differential with the flux by substituting Eq. (89) with Eq. (84):

proves that the systems is away from a pure, static equilibrium state.

<sup>T</sup> <sup>d</sup><sup>E</sup> <sup>þ</sup>

<sup>d</sup><sup>S</sup> <sup>¼</sup> <sup>1</sup>

<sup>σ</sup> <sup>¼</sup> <sup>1</sup> 2 ∑ k Jk ∂σ ∂Jk

Note that Tσ has a physical dimension equal to that of the specific power. An

For an arbitrary physical quantity Q, variation rate of its density can be represented as

$$\frac{1}{V}\frac{dQ}{dt} = \frac{1}{V}\frac{d\mathbf{r}}{dt} \cdot \nabla Q = \mathbf{v} \cdot \nabla q \tag{78}$$

where V is the constant system volume and q ¼ Q=V is a volumetric density of Q or named specific Q. Eq. (78) indicates that a density changing rate of Q is equal to q operated by v � ∇. If we replace Q by the internal energy of the system, then the specific energy consumption rate is expressed as

$$
\dot{W} = \frac{1}{V} \frac{\text{d}W}{\text{d}t} = \boldsymbol{\nu} \cdot \nabla w = \frac{\boldsymbol{\nu}}{A\_c} \cdot \nabla w' \tag{79}
$$

where w and w<sup>0</sup> are specific work done and work done per length, respectively, and Ac is the cross-sectional area normal to ∇w<sup>0</sup> . For a continuous media, ∇w<sup>0</sup> causes transport phenomena in a non-equilibrium state, and v=Ac is generated as proportional to a flux. A changing rate can be quantified as a product of a driving force and a flux, as implicated from Eq. (77).

Let us consider a closed system possessing ξ<sup>1</sup> and ξ2, as some thermodynamic quantities characterizing the system state. The values of ξ<sup>i</sup> at a state of equilibrium are denoted ξ<sup>0</sup> <sup>1</sup> and ξ<sup>0</sup> <sup>2</sup> and values outside equilibrium ξ<sup>0</sup> <sup>1</sup> and ξ<sup>0</sup> 2. Within a static equilibrium, the entropy represented by S of the system is maintained as the maximum. For a system away from the static equilibrium, the generalized driving force is defined as

$$X\_k = \nabla \left(\frac{\partial \mathbb{S}}{\partial \xi\_k}\right) \tag{80}$$

which is obviously zero for all k at the static equilibrium. A flux Jj of ξ<sup>j</sup> is defined as

$$J\_j = \frac{\mathbf{1}}{A\_c} \frac{\mathbf{d}\xi\_j}{\mathbf{d}t} = \frac{\dot{\xi}\_j}{A\_c} = \sum\_k L\_{jk} X\_k \tag{81}$$

which assumes that Jj represents a linear combination of all the existing driving forces Xk. We take Onsager's symmetry principle [12, 13], which indicates that the kinetic coefficient Ljk for all j and k are symmetrical such as

$$L\_{jk} = L\_{kj} \tag{82}$$

The entropy production rate per unit volume, or the specific entropy production rate, is defined as

$$
\sigma = \frac{\text{ds}}{\text{dt}}\tag{83}
$$

where s ¼ S=V. We expand the specific entropy s with respect to infinitesimal changes of ξ<sup>k</sup> as an independent variable:

$$\sigma = \sum\_{k} \frac{\mathrm{d}\xi\_{k}}{\mathrm{d}t} \frac{\partial \mathbf{s}}{\partial \xi\_{k}} = \sum\_{k} (A\_{c} I\_{k}) \cdot \left(\frac{\mathbf{1}}{A\_{c}} \frac{\partial \mathbf{S}}{\partial \xi\_{k}}\right) = \sum\_{k} J\_{k} X\_{k} \tag{84}$$

Fundamentals of Irreversible Thermodynamics for Coupled Transport DOI: http://dx.doi.org/10.5772/intechopen.86607

which represents the changing rate of the specific entropy as a dot project of flux J and driving force X. The subscript k in Eq. (84) is for physical quantities on which the respective entropy depends. For mathematical simplicity, a new quantity is defined as Yk ¼ TXk, where T is the absolute temperature in Kelvin to have

$$T\sigma = \sum\_{k} f\_{k} Y\_{k} \tag{85}$$

Note that Tσ has a physical dimension equal to that of the specific power. An inverse relationship of Eq. (81) is

$$X\_k = \sum\_l R\_{kl} I\_l \tag{86}$$

where Rkl represents an inverse matrix of Ljk, i.e., RklLjk ¼ δlj, which can be proven by substituting Eq. (86) with Eq. (81):

$$J\_j = \sum\_k L\_{jk} X\_k = \sum\_l \left(\sum\_k L\_{jk} R\_{kl}\right) l\_l = \sum\_l \delta\_{jl} l\_l = J\_j \tag{87}$$

Substitution of Eq. (86) to Eq. (84) represents σ in terms of flux J

$$
\sigma = \sum\_{j,k} J\_j R\_{jk} J\_k = \langle J|\mathbf{R}|J\rangle \tag{88}
$$

where h jJ and j iJ exist as the row and column vectors of J, respectively, and R represents the generalized resistance matrix. A partial derivative of σ with respect to an arbitrary flux Ji is equal to twice the generalized driving force:

$$\frac{\partial \sigma}{\partial I\_i} = \sum\_{j,k} \left[ \delta\_{\vec{\eta}} R\_{jk} I\_k + J\_j R\_{jk} \delta\_{ik} \right] = \sum\_k \left[ R\_{ik} I\_k \right] + \sum\_j \left[ J\_j R\_{\vec{\eta}} \right] = 2 \mathbf{X}\_i \tag{89}$$

In this case, the specific entropy dissipation rate σ is presented using the flux and σ differential with the flux by substituting Eq. (89) with Eq. (84):

$$
\sigma = \frac{1}{2} \sum\_{k} f\_k \frac{\partial \sigma}{\partial f\_k} \tag{90}
$$

which indicates that the specific entropy increases with respect to the flux and proves that the systems is away from a pure, static equilibrium state.

### 4.2 Effective driving forces

The second thermodynamic law represents the infinitesimal entropy change in the microcanonical ensemble:

$$\mathbf{dS} = \frac{1}{T}\mathbf{d}E + \frac{P}{T}\mathbf{d}V - \sum\_{i} \left(\frac{\mu\_{i}}{T}\right)\mathbf{dN}\_{i} \tag{91}$$

where E represents the internal energy, P represents the system pressure within a volume V, and μ<sup>i</sup> and Ni are the chemical potential and the mole number of species i, respectively. Eq. (91) implies the entropy S ¼ S Eð Þ ;V; Ni as a function of the internal energy E, the volume V, and the number of species i Ni.This gives ξ<sup>1</sup> ¼ E, ξ<sup>2</sup> ¼ V, and ξ<sup>i</sup> ¼ Ni (i ¼ 3 for water and i ¼ 4 for solute). The driving forces Xi are particularly calculated as

<sup>W</sup>\_ <sup>¼</sup> <sup>d</sup><sup>W</sup>

For an arbitrary physical quantity Q, variation rate of its density can be

where V is the constant system volume and q ¼ Q=V is a volumetric density of Q or named specific Q. Eq. (78) indicates that a density changing rate of Q is equal to q operated by v � ∇. If we replace Q by the internal energy of the system, then the

<sup>d</sup><sup>t</sup> <sup>¼</sup> <sup>v</sup> � <sup>∇</sup><sup>w</sup> <sup>¼</sup> <sup>v</sup>

where w and w<sup>0</sup> are specific work done and work done per length, respectively,

transport phenomena in a non-equilibrium state, and v=Ac is generated as proportional to a flux. A changing rate can be quantified as a product of a driving force and

Let us consider a closed system possessing ξ<sup>1</sup> and ξ2, as some thermodynamic quantities characterizing the system state. The values of ξ<sup>i</sup> at a state of equilibrium are

rium, the entropy represented by S of the system is maintained as the maximum. For a system away from the static equilibrium, the generalized driving force is defined as

> ∂S ∂ξk

which is obviously zero for all k at the static equilibrium. A flux Jj of ξ<sup>j</sup> is defined as

\_ ξj Ac ¼ ∑ k

which assumes that Jj represents a linear combination of all the existing driving forces Xk. We take Onsager's symmetry principle [12, 13], which indicates that the

The entropy production rate per unit volume, or the specific entropy production

<sup>σ</sup> <sup>¼</sup> <sup>d</sup><sup>s</sup>

where s ¼ S=V. We expand the specific entropy s with respect to infinitesimal

AcJk ð Þ� <sup>1</sup>

Ac

∂S ∂ξk 

¼ ∑ k

Xk ¼ ∇

<sup>2</sup> and values outside equilibrium ξ<sup>0</sup>

Jj <sup>¼</sup> <sup>1</sup> Ac dξ<sup>j</sup> dt ¼

kinetic coefficient Ljk for all j and k are symmetrical such as

Ac

<sup>1</sup> and ξ<sup>0</sup>

1 V dQ <sup>d</sup><sup>t</sup> <sup>¼</sup> <sup>1</sup> V dr dt

<sup>W</sup>\_ <sup>¼</sup> <sup>1</sup> V dW

specific energy consumption rate is expressed as

and Ac is the cross-sectional area normal to ∇w<sup>0</sup>

a flux, as implicated from Eq. (77).

<sup>1</sup> and ξ<sup>0</sup>

represented as

Non-Equilibrium Particle Dynamics

denoted ξ<sup>0</sup>

rate, is defined as

92

changes of ξ<sup>k</sup> as an independent variable:

σ ¼ ∑ k

dξ<sup>k</sup> dt

∂s ∂ξk ¼ ∑ k

<sup>d</sup><sup>t</sup> <sup>¼</sup> <sup>v</sup> � <sup>F</sup> (77)

� ∇Q ¼ v � ∇q (78)

� ∇w<sup>0</sup> (79)

. For a continuous media, ∇w<sup>0</sup> causes

LjkXk (81)

Ljk ¼ Lkj (82)

<sup>d</sup><sup>t</sup> (83)

JkXk (84)

2. Within a static equilib-

(80)

$$X\_1 = X\_q = \nabla \left(\frac{\partial \mathbf{S}}{\partial \mathbf{E}}\right)\_{V, N\_l} = \nabla \left(\frac{1}{T}\right) \tag{92}$$

4.3 Applications

of Lqs and Lvs:

where D [m<sup>2</sup>

cient Lqq is represented as

proportionality between Lss and DSE is

95

4.3.1 Solute diffusion

which is equivalent to Fick's law of

DOI: http://dx.doi.org/10.5772/intechopen.86607

By Eqs. (103) and (104), one can find

of concentration gradient, we have

The primary driving force for the solute transport is Xs ¼ �∇ð Þ μs=T , if temperature and pressure gradients are not significant in solute transport. We consider the diffusive flux of solute only in an isothermal and isobaric process and neglect terms

<sup>T</sup> ¼ � Lss

∂μs ∂c

/s] is a solute diffusion coefficient. If Eq. (102) is expressed in terms

<sup>T</sup> <sup>∇</sup>μ<sup>s</sup> (102)

Js ¼ �D∇c (103)

<sup>∇</sup><sup>c</sup> (104)

¼ D (105)

<sup>∇</sup>μ<sup>s</sup> ð Þ<sup>2</sup> (106)

(107)

(108)

Js ¼ �Lss<sup>∇</sup> <sup>μ</sup><sup>s</sup>

Fundamentals of Irreversible Thermodynamics for Coupled Transport

Jss ¼ � Lss T

> Lss T

<sup>σ</sup><sup>s</sup> <sup>¼</sup> JsXs <sup>¼</sup> Lss

Next, we consider the Stokes-Einstein diffusivity:

Lss <sup>¼</sup> DSET ð Þ <sup>∂</sup>μs=∂<sup>c</sup> <sup>T</sup>

For weakly interacting solutes, the solute chemical potential is

∂μs ∂c 

T

Then, the entropy-changing rate based on the solute transport is calculated as

<sup>T</sup><sup>2</sup> <sup>∇</sup>μ<sup>s</sup> ð Þ<sup>2</sup> <sup>¼</sup> <sup>D</sup>=<sup>T</sup>

DSE <sup>¼</sup> kBT 3πηdp

where kB is Boltzmann constant, η is the solvent viscosity, and dp is the diameter of a particle diffusing within the solvent medium. The phenomenological coeffi-

> <sup>¼</sup> kBT 3πηdp

where μ<sup>0</sup> is generally a function of T and P, which are constant in this equation, R is the gas constant, and a is the solute activity. For a dilute solution, the activity represented by a is often approximated as the concentration c (i.e., a≃c). The

ð Þ <sup>∂</sup>μs=∂<sup>c</sup> <sup>T</sup>

� <sup>T</sup> <sup>∂</sup>μ<sup>s</sup> ∂c �<sup>1</sup>

T

μ ¼ μ<sup>0</sup> þ RT ln a (109)

$$X\_2 = X\_v = \nabla \left(\frac{\partial \mathcal{S}}{\partial V}\right)\_{E, N\_r} = \nabla \left(\frac{P}{T}\right) \tag{93}$$

$$X\_3 = X\_t = \nabla \left(\frac{\partial \mathcal{S}}{\partial \mathcal{N}\_s}\right)\_{E,V} = \nabla \left(-\frac{\mu\_s}{T}\right) \tag{94}$$

where subscripts q, v, and s of X indicates heat, volume of solvent, and solute, respectively. In Eq. (92), entropy S is differentiated by energy E, keeping V, and Ns invariant, which are applied to Eqs. (93) and (94). Eq. (94) indicates that the driving force is a negative gradient of the chemical potential divided by the ambient temperature. Within the isothermal-isobaric ensemble, Gibbs free energy is defined as

$$\mathbf{G} = \mathbf{H} - \mathbf{T} \mathbf{S} \tag{95}$$

where Hð Þ ¼ E þ PV is enthalpy. If the solute concentration is diluted (i.e., Nw ≫ Ns), it is referred to as a weak solution. As such, the overall chemical potential can be approximated as

$$
\mu = \frac{\partial G}{\partial N} = \frac{G}{N\_w + N\_s} \simeq \frac{G}{N\_w} = \mu\_w = \overline{H} - \overline{\mathcal{S}}T \tag{96}
$$

where H and S represent molar enthalpy and entropy, respectively. An infinitesimal change of Gibbs free energy is, in particular, written as

$$\mathbf{d}G = -\mathbf{Sd}T + \mathbf{Vd}P + \mu\_s \mathbf{dN}\_s \tag{97}$$

which is equivalent to

$$\mathrm{d}\left(\frac{G}{N\_w}\right) \simeq \mathrm{d}\mu\_w = -\overline{\mathrm{S}}\mathrm{d}T + \overline{\mathrm{V}}\mathrm{d}P + \mu\_\mathrm{s}\mathrm{d}c \tag{98}$$

where V is a molar volume of the system, μ<sup>s</sup> is the solute chemical potential, and c ¼ Ns=Nw is the molar fraction of solute molecules. The gradient of the solvent chemical potential was rewritten as a linear combination of gradients of temperature, pressure, and molar solute fraction:

$$
\nabla \mu\_w = -\overline{\mathbf{S}} \nabla T + \overline{\mathbf{V}} \nabla P + \mu\_\sharp \nabla c \tag{99}
$$

where the following mathematical identity was used

$$
\nabla \frac{\mu\_k}{T} = \mu\_k \nabla \left(\frac{1}{T}\right) + \frac{1}{T} (\nabla \mu\_k) \tag{100}
$$

In general, fluxes of heat, solvent volume, and solute molecules are intrinsically coupled to their driving forces, such as

$$
\begin{bmatrix} J\_q \\ J\_v \\ J\_s \end{bmatrix} = \begin{bmatrix} L\_{qq} & L\_{qv} & L\_{qs} \\ L\_{qv} & L\_{vv} & L\_{vs} \\ L\_{qs} & L\_{vs} & L\_{qs} \end{bmatrix} \begin{bmatrix} X\_q \\ X\_v \\ X\_s \end{bmatrix} \tag{101}
$$

where Onsager's reciprocal relationship, Lij ¼ Lji, is employed.

Fundamentals of Irreversible Thermodynamics for Coupled Transport DOI: http://dx.doi.org/10.5772/intechopen.86607
