Triphasic Model of Heat and Moisture Transport with Internal Mass Exchange in Paperboard

Abstract

Mixture theory is used to derive a triphasic model to describe processes in paperboard consisting of solid fiber, bound water and gas. The gas is viewed as a miscible mix of the two constituents dry air and water vapor. The governing equations are mass conservation laws for bound water, dry air, water vapor, and mixture energy balance. Constitutive relations are found by exploiting the macroscale dissipation inequality. Resulting constitutive equations include Fickian diffusion of water vapor and dry air, Darcian flow for gas and Fourier heat conduction for the mixture. Mass exchange between bound water, and water vapor due to adsorption/desorption is driven by the difference in chemical potential. The interaction function is based on equilibrium considerations for the bound water–water vapor system. From the description of the sorption isotherm, expressions for net isosteric heat and free energy related to water–fiber interaction are derived. The resulting thermodynamically consistent model is used to simulate moisture and heat dynamics for paperboard rolls. Simulation results are presented for a paperboard roll with anisotropic material properties subjected to a change in ambient relative humidity from 50 to 80 %.

Appendices

Appendix 1: Determining Pressure Relations

Assuming that the constitutive variables in curly brackets in (30) are not functions of its energy conjugated coefficient i.e., $\varvec{d}_g$, $\varvec{d}_l$, $\nabla \otimes \varvec{w}_{g_\mathrm {v}}$, $\nabla \otimes \varvec{w}_{g_\mathrm {a}}$, $\frac{D_s(\rho _{g_\mathrm {v}})}{Dt}$, $\frac{D_s(\rho _{g_\mathrm {a}})}{Dt}$ and $\frac{D_s(n_l)}{Dt}$, provide that the expressions inside the curly brackets are zero in order to satisfy the dissipation inequality. Therefore, it is possible to write

$$\begin{aligned}&-p_g + \rho _{g_\mathrm {v}} \lambda _{g_\mathrm {v}} + \rho _{g_\mathrm {a}} \lambda _{g_\mathrm {a}} = 0, \end{aligned}$$

(65a)

$$\begin{aligned}&\lambda _l - \frac{p_l}{\rho _l} = 0,\end{aligned}$$

(65b)

$$\begin{aligned}&\left( -\frac{p_{g_\mathrm {v}}}{\rho _{g_\mathrm {v}}} - A_{g_\mathrm {v}} + \lambda _{g_\mathrm {v}}\right) \varvec{I} + \varvec{\varLambda }_w = \varvec{0}, \end{aligned}$$

(65c)

$$\begin{aligned}&\left( -\frac{p_{g_\mathrm {a}}}{\rho _{g_\mathrm {a}}} - A_{g_\mathrm {a}} + \lambda _{g_\mathrm {a}}\right) \varvec{I} + \varvec{\varLambda }_w = \varvec{0}, \end{aligned}$$

(65d)

$$\begin{aligned}&\lambda _{g_\mathrm {v}} = \rho _g \frac{\partial A_g}{\partial \rho _{g_\mathrm {v}}}, \end{aligned}$$

(65e)

$$\begin{aligned}&\lambda _{g_\mathrm {a}} = \rho _g \frac{\partial A_g}{\partial \rho _{g_\mathrm {a}}}, \end{aligned}$$

(65f)

$$\begin{aligned}&\rho _l \lambda _l - \rho _{g_\mathrm {v}} \lambda _{g_\mathrm {v}} - \rho _{g_\mathrm {a}}\lambda _{g_\mathrm {a}} - n_l \rho _l \frac{\partial A_l}{\partial n_l} = 0, \end{aligned}$$

(65g)

where the deviatoric part of the fluid stress tensors as well as the second-order terms are neglected. Inserting (65b), (65c), (65d) into the definition of phase stress (15) subjected to the same assumptions yields

$$\begin{aligned} \varvec{\varLambda }_w = A_g \varvec{I}. \end{aligned}$$

(66)

Inserting (65e), (65f), (66) into (65c) and (65d) and utilizing that $\rho _g A_g(\theta ,\rho _{g_\mathrm {v}},\rho _{g_\mathrm {a}})= \rho _{g_\mathrm {v}} A_{g_\mathrm {v}}(\theta ,\rho _{g_\mathrm {v}}) + \rho _{g_\mathrm {a}} A_{g_\mathrm {a}}(\theta ,\rho _{g_\mathrm {a}})$ enable us to write the partial pressures as

$$\begin{aligned}&p_{g_\mathrm {v}} = (\rho _{g_\mathrm {v}})^2 \frac{\partial A_{g_\mathrm {v}}}{\partial \rho _{g_\mathrm {v}}}, \quad p_{g_\mathrm {a}} = (\rho _{g_\mathrm {a}})^2 \frac{\partial A_{g_\mathrm {a}}}{\partial \rho _{g_\mathrm {a}}}. \end{aligned}$$

(67)

Combination of (65a), (65b) and (65g) result in

$$\begin{aligned} p_l = p_g + n_l \rho _l \frac{\partial A_l}{\partial n_l}. \end{aligned}$$

(68)

Appendix 2: Derivations of Expressions for Seepage and Diffusion Velocities

Recall terms two to five in (32). Assuming that each term give nonnegative dissipation the following is obtained

$$\begin{aligned}&- \varvec{w}_{g_\mathrm {v}} \cdot \left\{ \nabla (n_g \rho _{g_\mathrm {v}} A_{g_\mathrm {v}}) + \hat{\varvec{\tau }}_{g_\mathrm {v}} - \lambda _{g_\mathrm {v}} \nabla (n_g \rho _{g_\mathrm {v}}) - \varvec{\varLambda }_w \cdot \nabla (n_g \rho _{g_\mathrm {v}}) \right\} \ge 0, \end{aligned}$$

(69a)

$$\begin{aligned}&- \varvec{w}_{g_\mathrm {a}} \cdot \left\{ \nabla (n_g \rho _{g_\mathrm {a}} A_{g_\mathrm {a}}) + \hat{\varvec{\tau }}_{g_\mathrm {a}} - \lambda _{g_\mathrm {a}} \nabla (n_g \rho _{g_\mathrm {a}}) - \varvec{\varLambda }_w \cdot \nabla (n_g \rho _{g_\mathrm {a}}) \right\} \ge 0, \end{aligned}$$

(69b)

$$\begin{aligned}&\varvec{v}^{l,s} \cdot \left\{ -n_l \rho _l \nabla (A_l) - n_l \rho _l \eta _l \nabla (\theta ) - \hat{\varvec{\tau }}_{l} + \lambda _l \rho _l \nabla (n_l) \right\} \ge 0, \end{aligned}$$

(69c)

$$\begin{aligned}&\varvec{v}^{g,s} \cdot \left\{ -n_g \rho _g \nabla (A_g) - n_g \rho _g \eta _g \nabla (\theta ) - \hat{\varvec{\tau }}_{g} + \lambda _{g_\mathrm {v}} \nabla (n_g \rho _{g_\mathrm {v}}) + \lambda _{g_\mathrm {a}} \nabla (n_g \rho _{g_\mathrm {a}}) \right\} \ge 0. \end{aligned}$$

(69d)

The linear momentum balances (10) and (17) written for the paperboard components with negligible inertial terms and non-deviatoric stresses are

$$\begin{aligned}&-\nabla (n_g p_{g_\mathrm {v}}) + \hat{\varvec{\tau }}_{g_\mathrm {v}} + n_g \rho _{g_\mathrm {v}} \varvec{b}_{g_\mathrm {v}} = 0, \end{aligned}$$

(70a)

$$\begin{aligned}&-\nabla (n_g p_{g_\mathrm {a}}) + \hat{\varvec{\tau }}_{g_\mathrm {a}} + n_g \rho _{g_\mathrm {a}} \varvec{b}_{g_\mathrm {a}} = 0, \end{aligned}$$

(70b)

$$\begin{aligned}&-\nabla (n_g p_g) + \hat{\varvec{\tau }}_g + n_g \rho _g \varvec{b}_g = 0, \end{aligned}$$

(70c)

$$\begin{aligned}&-\nabla (n_l p_l) + \hat{\varvec{\tau }}_l + n_l \rho _l \varvec{b}_l = 0, \end{aligned}$$

(70d)

where definitions (24a–d) were used. Inserting (70c), (70d) into (69c), (69d) and expanding the gradients, recalling the dependencies of Helmholtz free energy in (41) and the definitions of entropy in (29) as well as equations (65b), (65e) and (65f), provide after simplifications

$$\begin{aligned}&\varvec{v}^{l,s} \cdot \left\{ -n_l \rho _l \frac{\partial A_l}{\partial n_l} \nabla (n_l) -n_l \nabla ( p_l) + n_l \rho _l \varvec{b}_l \right\} \ge 0, \end{aligned}$$

(71a)

$$\begin{aligned}&\varvec{v}^{g,s} \cdot \left\{ \left( \rho _{g_\mathrm {v}} \rho _g \frac{\partial A_g}{\partial \rho _{g_\mathrm {v}}}+ \rho _{g_\mathrm {a}}\rho _{g} \frac{\partial A_g}{\partial \rho _{g_\mathrm {a}}} -p_g \right) \nabla (n_g) - n_g \nabla ( p_g) + n_g \rho _g \varvec{b}_g \right\} \ge 0. \end{aligned}$$

(71b)

From (65a), it is seen that the first term in (71b) vanishes. One possible way to satisfy the dissipation inequality is

$$\begin{aligned}&\varvec{v}^{l,s} = -\frac{\varvec{K}^p_{l,s}}{\bar{\mu _l}} \cdot [ n_l \nabla (p_l) + n_l \rho _l \frac{\partial A_l}{\partial n_l}\nabla (n_l) -n_l \rho _l \varvec{b}_l ],\end{aligned}$$

(72a)

$$\begin{aligned}&\varvec{v}^{g,s} = -\frac{\varvec{K}^p_{g,s}}{\bar{\mu _g}} \cdot [ n_g \nabla (p_g) - n_g \rho _g \varvec{b}_g ], \end{aligned}$$

(72b)

where $\frac{\varvec{K}^p_{l,s}}{\bar{\mu _l}}$ and $\frac{\varvec{K}^p_{g,s}}{\bar{\mu _g}}$ are positive definite tensors. Insertion of (70a) and (70b) into (69a) and (69b) leads to

$$\begin{aligned}&-\varvec{w}_{g_\mathrm {v}} \cdot \left\{ \nabla \left( n_g \rho _{g_\mathrm {v}} \left[ A_{g_\mathrm {v}} + \frac{p_{g_\mathrm {v}}}{\rho _{g_\mathrm {v}}}\right] \right) - n_g\rho _{g_\mathrm {v}}\varvec{b}_{g_\mathrm {v}} -(\lambda _{g_\mathrm {v}}\varvec{I}+\varvec{\varLambda }_w) \cdot \nabla (n_g \rho _{g_\mathrm {v}}) \right\} \ge 0, \end{aligned}$$

(73a)

$$\begin{aligned}&-\varvec{w}_{g_\mathrm {a}} \cdot \left\{ \nabla \left( n_g \rho _{g_\mathrm {a}} \left[ A_{g_\mathrm {a}}+ \frac{p_{g_\mathrm {a}}}{\rho _{g_\mathrm {a}}}\right] \right) - n_g\rho _{g_\mathrm {a}}\varvec{b}_{g_\mathrm {a}} -(\lambda _{g_\mathrm {a}}\varvec{I}+\varvec{\varLambda }_w) \cdot \nabla (n_g \rho _{g_\mathrm {a}}) \right\} \ge 0. \end{aligned}$$

(73b)

Insertion of (65c), (65d) and (66) into (73a) and (73b) results in

$$\begin{aligned}&-\varvec{w}_{g_\mathrm {v}} \cdot \left\{ n_g \rho _{g_\mathrm {v}} \nabla \left( A_{g_\mathrm {v}} + \frac{p_{g_\mathrm {v}}}{\rho _{g_\mathrm {v}}}\right) - n_g\rho _{g_\mathrm {v}}\varvec{b}_{g_\mathrm {v}} \right\} \ge 0, \end{aligned}$$

(74a)

$$\begin{aligned}&-\varvec{w}_{g_\mathrm {a}} \cdot \left\{ n_g \rho _{g_\mathrm {a}} \nabla \left( A_{g_\mathrm {a}}+ \frac{p_{g_\mathrm {a}}}{\rho _{g_\mathrm {a}}}\right) - n_g\rho _{g_\mathrm {a}}\varvec{b}_{g_\mathrm {a}} \right\} \ge 0. \end{aligned}$$

(74b)

One possible way to satisfy the dissipation inequality is the choice

$$\begin{aligned}&\varvec{w}_{g_\mathrm {v}} = - \varvec{D}_{g_\mathrm {v}} \cdot [ \nabla (\mu _{g_\mathrm {v}}) -\varvec{b}_{g_\mathrm {v}}], \end{aligned}$$

(75a)

$$\begin{aligned}&\varvec{w}_{g_\mathrm {a}} = -\varvec{D}_{g_\mathrm {a}} \cdot [\nabla (\mu _{g_\mathrm {a}}) - \varvec{b}_{g_\mathrm {a}}], \end{aligned}$$

(75b)

where $\mu _{g_\mathrm {v}}= A_{g_\mathrm {v}} + \frac{p_{g_\mathrm {v}}}{\rho _{g_\mathrm {v}}}$, $\mu _{g_\mathrm {a}}= A_{g_\mathrm {a}} + \frac{p_{g_\mathrm {a}}}{\rho _{g_\mathrm {a}}}$ and $\varvec{D}_{g_\mathrm {v}}$, $\varvec{D}_{g_\mathrm {a}}$ are positive definite tensors.

Appendix 3: Derivation of the Energy Balance

Combining the energy balances (11) for all component of the system ($s,l,g_\mathrm {v},g_\mathrm {a}$) and eliminating the terms $n_\alpha \rho _{\alpha _j} \hat{E}_{\alpha _j}$ and $n_\alpha \rho _{\alpha _j} \hat{Q}^\beta _{\alpha _j}$ with summation constraints, that arise if the interfaces are assumed to not have any thermodynamic properties (e.g., no mass) (cf. Bennethum and Cushman 1996a), provide

$$\begin{aligned}&n_s \rho _s \frac{D_s(u_s)}{Dt} + n_l \rho _l \frac{D_s(u_l)}{Dt} + n_g \rho _{g_\mathrm {v}} \frac{D_s(u_{g_\mathrm {a}})}{Dt} + n_g \rho _{g_\mathrm {a}} \frac{D_s(u_{g_\mathrm {a}})}{Dt} = \nonumber \\&\quad -n_g \rho _{g_\mathrm {v}} \nabla (u_{g_\mathrm {v}}) \cdot \varvec{v}^{g,s} - n_g \rho _{g_\mathrm {a}} \nabla (u_{g_\mathrm {v}}) \cdot \varvec{v}^{g,s} + u_{g_\mathrm {v}} \nabla \cdot (n_g \rho _{g_\mathrm {v}} \varvec{w}_{g_\mathrm {v}}) +u_{g_\mathrm {a}} \nabla \cdot (n_g \rho _{g_\mathrm {a}} \varvec{w}_{g_\mathrm {a}}) \nonumber \\&\quad - \nabla \cdot (n_g p_g \varvec{v}^{g,s}) - \nabla \cdot (\tilde{\varvec{q}}) - \hat{m}_v (u_{g_\mathrm {v}}-u_l). \end{aligned}$$

(76)

Here the material time derivative is taken to follow the solid phase. The bound water velocity ($\varvec{v}^{l,s}$) and second-order velocity terms are neglected. No external sources ($r_{\alpha _j}, \, r_{\alpha }$) are present and gravity is omitted. The terms involving diffusive velocities are eliminated with the help of the balances of mass for vapor (37) and dry air (38), and the rate of evaporation is eliminated using the bound water balance of mass (39)

$$\begin{aligned}&u_{g_\mathrm {v}} \nabla (n_g \rho _{g_\mathrm {v}} \varvec{w}_{g_\mathrm {v}}) = -u_{g_\mathrm {v}} \frac{D_s(n_g \rho _{g_\mathrm {v}})}{Dt} + u_{g_\mathrm {v}} \hat{m}_v - u_{g_\mathrm {v}} \nabla \cdot (n_g \rho _{g_\mathrm {v}} \varvec{v}^{g,s}), \end{aligned}$$

(77a)

$$\begin{aligned}&u_{g_\mathrm {a}} \nabla (n_g \rho _{g_\mathrm {a}} \varvec{w}_{g_\mathrm {a}}) = -u_{g_\mathrm {a}} \frac{D_s(n_g \rho _{g_\mathrm {a}})}{Dt} - u_{g_\mathrm {a}} \nabla \cdot (n_g \rho _{g_\mathrm {a}} \varvec{v}^{g,s}), \end{aligned}$$

(77b)

$$\begin{aligned}&u_l \hat{m}_v = - \rho _l u_l \frac{D_s(n_l)}{Dt}. \end{aligned}$$

(77c)

Insertion of (77a), (77b) and (77c) into (76) results in

$$\begin{aligned}&n_s \rho _s \frac{D_s(u_s)}{Dt} + n_l \rho _l \frac{D_s(u_l)}{Dt} + n_g \rho _{g_\mathrm {v}} \frac{D_s(u_{g_\mathrm {a}})}{Dt} + n_g \rho _{g_\mathrm {a}} \frac{D_s(u_{g_\mathrm {a}})}{Dt} + n_g u_{g_\mathrm {v}} \frac{D_s(\rho _{g_\mathrm {v}})}{Dt} \nonumber \\&\quad + n_g u_{g_\mathrm {a}} \frac{D_s(\rho _{g_\mathrm {a}})}{Dt} + \nabla \cdot (\tilde{\varvec{q}} + (n_g \rho _{g_\mathrm {v}} u_{g_\mathrm {v}} + n_g \rho _{g_\mathrm {a}} u_{g_\mathrm {a}} + n_g p_g) \varvec{v}^{g,s} ) \nonumber \\&\quad +\rho _l u_l \frac{D_s(n_l)}{Dt} + \rho _{g_\mathrm {v}} u_{g_\mathrm {v}} \frac{D_s(n_g)}{Dt} + \rho _{g_\mathrm {a}} u_{g_\mathrm {a}} \frac{D_s(n_g)}{Dt} = 0 , \end{aligned}$$

(78)

For a rigid incompressible solid $n_s = n_{s0}$ hence the time derivative of (2) becomes

$$\begin{aligned} \frac{D_s(n_g)}{Dt} = - \frac{D_s(n_l)}{Dt}. \end{aligned}$$

(79)

Further, with Helmholtz free energies given by (41) the dependencies of the internal energies are

$$\begin{aligned} u_{g_\mathrm {v}} = u_{g_\mathrm {v}}(\theta ), \quad u_{g_\mathrm {a}} = u_{g_\mathrm {a}}(\theta ), \quad u_s = u_s(\theta ), \quad u_l = u_l(\theta ,n_l). \end{aligned}$$

(80)

Utilizing the results from (79) and (80) the energy balance (78) may be written as

$$\begin{aligned}&\left( n_s \rho _s \frac{\partial u_s}{\partial \theta } + n_l \rho _l \frac{\partial u_l}{\partial \theta } + n_g \rho _{g_\mathrm {v}} \frac{\partial u_{g_\mathrm {v}}}{\partial \theta } + n_g \rho _{g_\mathrm {a}} \frac{\partial u_{g_\mathrm {a}}}{\partial \theta } \right) \frac{D_s (\theta )}{D t} + n_g u_{g_\mathrm {v}} \frac{D_s (\rho _{g_\mathrm {v}})}{D t} \nonumber \\&+ n_g u_{g_\mathrm {a}} \frac{D_s (\rho _{g_\mathrm {a}})}{D t} - \left( \rho _{g_\mathrm {v}} u_{g_\mathrm {v}} + \rho _{g_\mathrm {a}} u_{g_\mathrm {a}} - \rho _l u_l - n_l \rho _l \frac{\partial u_l}{\partial n_l} \right) \frac{D_s (n_l)}{D t} \nonumber \\&+ \nabla \cdot \left( \tilde{\varvec{q}} + n_g (\rho _{g_\mathrm {v}} u_{g_\mathrm {v}} + p_{g_\mathrm {v}} + \rho _{g_\mathrm {a}} u_{g_\mathrm {a}} + p_{g_\mathrm {a}}) \varvec{v}^{g,s} \right) = 0. \end{aligned}$$

(81)

Appendix 4: Requirement of Exothermic Adsorption

With insertion of the mass balances (37), (38), (39) the energy equation (40) may be written for a homogeneous state (the gradient terms disappear) as

$$\begin{aligned} \frac{\partial \theta }{\partial t} = -\frac{\tilde{u}}{\tilde{c}} \hat{m}_v, \end{aligned}$$

(82)

where

$$\begin{aligned} \tilde{u}= & {} (u_{g_\mathrm {v}} - u^o_l) + (u^o_l-u_l) + W \frac{\partial (u^o_l-u_l)}{\partial W} , \end{aligned}$$

(83)

$$\begin{aligned} \tilde{c}= & {} n_s \rho _s c^v_s + n_l \rho _l c^v_l + n_g \rho _{g_\mathrm {v}} c^v_{g_\mathrm {v}} + n_g \rho _{g_\mathrm {a}} c^v_{g_\mathrm {a}}. \end{aligned}$$

(84)

The first parenthesis is the internal energy change for evaporation of free water

$$\begin{aligned} u_{g_\mathrm {v}} - u^o_l= (c^v_{g_\mathrm {v}}-c^v_{l})(\theta -\theta ^*)+(u^*_{g_\mathrm {v}} - u^*_l), \end{aligned}$$

(85)

which is positive in the expected temperature range. The second term is related to the function f via Legendre transformations. For an adsorption process $\hat{m}_v < 0$ provide that the temperature will increase due to the reaction if $\tilde{u} > 0$.

The specific format of the internal energy of sorption is related to the choice of isotherm description. For a format of the sorption isotherm according to

$$\begin{aligned} a_\mathrm{w} = \exp \left( -\frac{A}{R \theta } \gamma (W) \right) , \end{aligned}$$

(86)

the internal energy change associated with the solid–liquid interactions become

$$\begin{aligned} (u^o_l-u_l) + W \frac{\partial (u^o_l-u_l)}{\partial W} = \frac{1}{W} \frac{A}{R \theta } \gamma (W). \end{aligned}$$

(87)

Especifically for the Chung–Pfost-type isotherm $\gamma (W) = \exp (-BW)$, a sufficient condition for an exothermic adsorption is $A>0$.