# The Significance of the Sorption Isotherm on the Simulated Performance of Grain Driers
## 3 Mathematical Analysis  
The above description of the heat and mass transfer phenomena that occur in beds of drying grains is semi-qualitative. An explicit expression for the speed, \(\mathcal{V}\), of the heat and mass transfer fronts through ventilated beds of hygroscopic media, such as food grains, can be obtained if we consider the solid phase and the intergranular air to be in thermodynamic equilibrium. Previous analyses, such as that of [PERSON] [12], have assumed that the density and mean velocity of the intergranular air are constant throughout beds of drying grains. This is manifestly not the case because each region of a bed of ventilated material has a different temperature, hence the density of the intergranular air varies in time and space. Therefore, we consider a small region of the bed that has a length \(\Delta x\) and in which the intergranular air initially has a density, humidity, and velocity of \(\rho_{a}+\Delta\rho_{a}\), \(w+\Delta w\), and \(v+\Delta v\), respectively, as indicated in Figure 3. The initial moisture content of the grain is \(W+\Delta W\). After a time, \(\Delta t\), air is displaced so that its properties within the element become \(\rho_{a}\) and \(w\), and the interstitial velocity of the air is \(v\). The grain assumes a moisture content, \(W\), which is in thermodynamic equilibrium with the air entering the element. The conservation of mass applied to dry air (in which moisture is conveyed) demands that  
\[\varepsilon\rho_{a}v=\varepsilon(\rho_{a}+\Delta\rho_{a})(v+\Delta v)=G \tag{2}\]  
in which \(G\) is the mass flow rate of the component of dry air per unit cross-sectional area of the bed. A mass balance on the moisture in the element is expressed as  
\[ \begin{array}{ll}\text{\em Amount of moisture in the}\\ \text{\em air that enters the element}\end{array} \quad=\quad \begin{array}{ll}\text{\em Amount of moisture in the}\\ \text{\em air that leaves the element}\end{array} \quad+\quad \begin{array}{ll}\text{\em Amount of moisture in the}\\ \text{\em element that accumulates in}\\ \text{\em the air and solid}\end{array}  \tag{3}\]  
Algebraically, this is expressed as  
\[ \begin{array}{ll}\rho_{a}\varepsilon vw\Delta t=\varepsilon(\rho_{a}+ \Delta\rho_{a})(v+\Delta v)(w+\Delta w)\Delta t-\rho_{s}(1-\varepsilon)\Delta W \Delta x\\ \hskip 14.226378 pt-\varepsilon\rho_{a}\Delta w\Delta x-\varepsilon\Delta\rho _{a}\Delta w\Delta x-\varepsilon\rho_{a}\Delta w\Delta x\end{array}  \tag{4}\]  
Since the change in moisture that accumulates in the air and solid in the element of length \(\Delta x\) occurs in a time \(\Delta t\), the velocity, \(\mathcal{V}\), with which a drying or heating front traverses the element is  
\[\mathcal{V}=\frac{\Delta x}{\Delta t} \tag{5}\]  
which when used in Equation (4) results in  
\[\mathcal{V}_{w}=\frac{G}{\varepsilon\rho_{a}+\varepsilon\Delta\rho_{a}+ \varepsilon\rho_{a}\frac{w}{\Delta w}+\rho_{s}(1-\varepsilon)\frac{\Delta W} {\Delta w}} \tag{6}\]  
\(\mathcal{V}_{w}\) is the velocity of the transfer front based on a mass balance of moisture across the element portrayed in Figure 3. If we had performed an enthalpy balance on the element instead of a moisture balance and designated the change in the enthalpy of the solids and air as \(\Delta H\) and \(\Delta h\), respectively, we would obtain the same velocity, \(\mathcal{V}_{h}\). \[\mathcal{V}_{h}=\frac{G}{\varepsilon\rho_{a}+\varepsilon\Delta\rho_{a}+ \varepsilon\rho_{a}\frac{h}{\Delta h}+\rho_{s}(1-\varepsilon)\frac{\Delta H} {\Delta h}} \tag{7}\]  
noting that \(\mathcal{V}_{w}=\mathcal{V}_{h}\). The enthalpy, \(h\), of moist air is expressed as  
\[h=c_{a}(T-T^{o})+w(\left.h_{v}\right|_{T}+c_{W}(T-T^{o})) \tag{8}\]  
Figure 3: Air that enters an element of length \(\Delta x\), has a mean intergranular velocity \(v\), density \(\rho_{a}\), and humidity \(w\). It leaves with a velocity \(v+\Delta v\), density \(\rho_{a}+\Delta\rho_{a}\), and humidity \(w+\Delta w\). Initially, the moisture content of canola is in equilibrium with the air in the latter state, i.e., \(W+\Delta W\).