Mathematical modeling to study influence of porosity on apple and potato during dehydration

Abstract

Several structural and physical changes in foodstuffs are the consequence of water removal during the drying process. Porosity (volume fraction of pores) is one of the key parameter that affects the quality and other properties of foods (such as apple and potato). To understand the effect of dehydration in apple and potato, in the present study an arbitrary small cubic volume element is considered which contains pores (intracellular spaces) distributed in it. Further, it is assumed that each pore in the cubic volume element is spherical. A mathematical relation is developed between porosity (volume fraction of pores) and pressure generated (due to contraction of cells during water removal) in outward direction on the surface of spherical elements containing pore. The developed relation is satisfactory in respect of experimental observations given in the literature. For the given pressure range, acquired porosity range is 0.1 to 0.92 for apple and 0.03 to 0.89 for potato which is matched with the existing experimental values. The results showed that the porosity is increasing with the increasing values of pressure, as expected, during moisture removal. Further, it is observed that the current porosity is depended on the initial porosity for both apple and potato.

Appendices

Appendix A

In the terms of components, the deformation gradient tensor can be written as

$$ \mathbf{F}=\frac{\partial \mathbf{y}}{\partial \mathbf{x}}=\left[\begin{array}{ccc}\hfill \frac{\partial {y}_1}{\partial {x}_1}\hfill & \hfill \frac{\partial {y}_1}{\partial {x}_2}\hfill & \hfill \frac{\partial {y}_1}{\partial {x}_3}\hfill \\ {}\hfill \frac{\partial {y}_2}{\partial {x}_1}\hfill & \hfill \frac{\partial {y}_2}{\partial {x}_2}\hfill & \hfill \frac{\partial {y}_2}{\partial {x}_3}\hfill \\ {}\hfill \frac{\partial {y}_3}{\partial {x}_1}\hfill & \hfill \frac{\partial {y}_3}{\partial {x}_2}\hfill & \hfill \frac{\partial {y}_3}{\partial {x}_3}\hfill \end{array}\right] $$

(A1)

where y = (y ₁, y ₂, y ₃), x = (x ₁, x ₂, x ₃), $R=\left|\mathbf{x}\right|=\sqrt{\left({x}_1^2+{x}_2^2+{x}_3^2\right)}$ and

$$ \frac{\partial {y}_i}{\partial {x}_j}=\frac{x_i{x}_j}{R^2}\left({F}^{\prime }(R)-\frac{F(R)}{R}\right)+\frac{F(R)}{R}{\delta}_{ij}\kern0.24em ,\kern0.48em i,\;j=1,\;2,\;3 $$

(A2)

on substituting Eq. (A2) in Eq. (A1) and after some simplifications we can find

$$ Det\left(\mathbf{F}\right)={F}^{\prime }(R){\left(\frac{F(R)}{R}\right)}^2 $$

(A3)

By using incompressibility condition Det(F) = 1 (from Eq. 5), Eq. (A3) can be solved to get

$$ {F}^{\prime }(R)={\left(\frac{R}{F(R)}\right)}^2 $$

(A4)

Appendix B

From Eq. (5) we have B ³ − B ³₀  = A ³ − A ³₀ , substituting the value of A ³ − A ³₀ in Eq. (13), we can get

$$ {\lambda}_{\theta }={\left(1-\frac{B^3-{B}_0^3}{r^3}\right)}^{-1/3} $$

(B1)

Now, for r = B, Eq. (B1) reduced to

$$ {\lambda}_{\theta }={\left(\frac{B_0^3}{B^3}\right)}^{-1/3}={\left(\frac{B_0}{B}\right)}^{-1} $$

(B2)

Again, substituting the value of B from Eq. (5) into Eq. (B2)

$$ {\lambda}_{\theta }={\left(\frac{B_0^3}{B_0^3+{A}^3-{A}_0^3}\right)}^{-1/3}={\left(\frac{B_0^3+{A}^3-{A}_0^3}{B_0^3}\right)}^{1/3} $$

(B3)