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A mechanistic derivation of the Monod bioreaction equation for a limiting nutrient

Abstract

We present a quasi-steady state mechanistic derivation of the Monod bioreaction equation based upon a conceptual model involving aqueous phase diffusive transport of substrate towards a spherical microbe; transport of the substrate across its surface membrane; and reaction depleting the substrate within the microbe. The resulting Monod coefficients \({K}_{S}\) and \({\mu }_{\mathrm{max}}\) are dependent upon substrate-species pairs and the mass transfer properties of the system. Two substrate transport scenarios are investigated: (1) a constant rate model that is a function of a constant flux across the surface of the microbe; and (2) a linear rate model that is the product of a constant transport velocity and the concentration of substrate in contact with the surface of the microbe. The model is verified and parameterized using benzene, toluene, and phenol depletion and biomass growth data obtained from Reardon et al. (Biotechnol Bioeng: 385–400, 2000). Calibration results indicate a normalized surface to bulk concentration ratio of nearly unity in all simulations for benzene, toluene, and phenol when paired with P. putida F1, implying that the process is not aqueous phase diffusion limited.

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Acknowledgements

Financial support for this research was provided by a NSERC CRDPJ 531663-18 Grant in participation with Chevron Technical Center (a Chevron U.S.A. Inc. division), and Chevron Canada Resources, Calgary, AB, Canada.

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Correspondence to Andre J. A. Unger.

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Appendices

Appendix A: Constant Rate Model

  1. (a)

    First, we solve for \({\left({k}_{L}\alpha \right)}_{dq}\).

    Let \({C}_{d}(r,t)\) be microscale concentration of donor \(d\) in the aqueous phase \(q\) for \(r>{r}_{b}\), and \(t>0\).

    $$\frac{{D}_{dq}^{*}}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial {C}_{d}}{\partial r}\right)=\frac{\partial {C}_{d}}{\partial t}$$
    (A.1a)

    and assuming steady state (no accumulation)

    $$\frac{{D}_{dq}^{*}}{{r}^{2}}\frac{d}{dr}\left({r}^{2}\frac{d{C}_{d}}{dr}\right)=0$$
    (A.1b)

    For \(r={r}_{b}\)

    $$\frac{{A}_{b}}{{V}_{b}}{\mathbb{J}}_{d}={R}_{db}$$
    (A.2a)
    $$\frac{3}{{r}_{b}}{\mathbb{J}}_{d}={R}_{db}$$
    (A.2b)

    where:\({A}_{b}=4\pi {r}_{b}^{2}\); \({V}_{b}=\frac{4}{3}\pi {r}_{b}^{3}\); \({R}_{db}\) is the microscale reaction rate within the volume of the microbe; and \({\mathbbm{n}}_{db}\) is the microscale flux of the substrate across the surface of the single spherical microbe shown on Fig. 1 at \(r={r}_{b}\) and with \({\mathbb{J}}_{d}={\mathbbm{n}}_{db}\) being constant, per Eq. (10a).

    We solve Eqs. (A.1b) and (A.2b) subject to the following conditions:

    $${C}_{dq,\mathrm{bulk}}\equiv {\left.{C}_{d}(r)\right|}_{\forall r\ge {r}_{c}}$$
    (A.3a)
    $${C}_{d}({r}_{b}^{+})={\kappa }_{db} {C}_{d}({r}_{b}^{-})$$
    (A.3b)
    $${D}_{dq}^{*}{\left.\frac{d{C}_{d}}{dr}\right|}_{r\to {r}_{b}^{+}}={\mathbb{J}}_{d}$$
    (A.3c)

    where \({C}_{d}(r)\) is the concentration of donor d in the biomass phase \(b\) for \(r<{r}_{b}\) and \(t>0\), and A.3b is our assumption of membrane transport processes.

    First, we integrate Eq. A.1b w.r.t. \(r\) as:

    $${r}^{2}\frac{d{C}_{d}}{dr}={\beta }_{1}$$
    $$\frac{d{C}_{d}}{dr}=\frac{{\beta }_{1}}{{r}^{2}}$$
    $${C}_{d}=-\frac{{\beta }_{1}}{r}+{\beta }_{2}$$
    (A.4)

    Using Eq. (A.3a) under the assumption that \({r}_{c}\to \infty \) and then combining with Eq. (A.4) yields:

    $${C}_{dq,\mathrm{bulk}}={\beta }_{2}$$
    (A.5)

    Substituting Eq. (A.4) with Eq. (A.3b) at \(r={r}_{b}\) yields:

    $$-\frac{{\beta }_{1}}{{r}_{b}}+{C}_{dq,\mathrm{bulk}}={\kappa }_{db}{ C}_{d}({r}_{b}^{-})$$
    $${\beta }_{1}=\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right) {r}_{b}$$
    (A.6)

    We can now rewrite Eq. (A.4) as:

    $${C}_{d}=-\frac{{r}_{b}}{r}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)+{C}_{dq,\mathrm{bulk}}$$
    (A.7)

    Now, taking the derivative of Eq. (A.7) w.r.t. \(r\) and evaluating at position \(r\to {r}_{b}^{+}\) yields:

    $${\left.\frac{d{C}_{d}}{dr}\right|}_{r\to {r}_{b}^{+}}=\frac{1}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (A.8)

    Substituting Eq. (A.8) into (A.3c) yields:

    $${\mathbbm{n}}_{db}=\frac{{D}_{dq}^{*}}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (A.9)

    Next, substituting Eq. (A.2b) into Eq. (A.9) yields:

    $$\frac{{r}_{b}}{3}{R}_{db}=\frac{{D}_{dq}^{*}}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (A.10a)
    $$\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)\frac{{3D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (A.10b)

    From Eq. (A.3b), we have:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{{3D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (A.11a)
    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right){\left({k}_{L}\alpha \right)}_{dq}={R}_{db}$$
    (A.11b)

    where,

    $${k}_{L}=\frac{{D}_{dq}^{*}}{{r}_{b}}$$
    (A.12a)
    $$\alpha =\frac{3}{{r}_{b}}$$
    (A.12b)
  2. (b)

    Next, we solve for the surface concentration \({C}_{d}({r}_{b}^{+})\). We begin with Eq. (A.11a) as:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (A.13)

    but \(\frac{3}{{r}_{b}}{\mathbbm{n}}_{db}={R}_{db}\) from Eq. (A.2b). Therefore, we can rewrite Eq. (A.13) as:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}=\frac{3}{{r}_{b}}{\mathbbm{n}}_{db}$$
    (A.14a)
    $${C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})=\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}$$
    (A.14b)
    $${C}_{d}({r}_{b}^{+})={C}_{dq,\mathrm{bulk}}-\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}$$
    (A.14c)
    $${C}_{d}({r}_{b}^{-})=\frac{1}{{\kappa }_{db}}\left({C}_{dq,\mathrm{bulk}}-\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}\right)$$
    (A.14d)
  3. (c)

    Next, we solve the concentration \({C}_{d}\) as a function of radius \(r>{r}_{b}\). We begin with Eq. (A.7) as:

    $${C}_{d}=-\frac{{r}_{b}}{r}\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)+{C}_{dq,\mathrm{bulk}}$$
    (A.15)

    Now, substitute Eq. (A.14c) into (A.15) as:

    $$ \begin{aligned} C_{d} & = - \frac{{r_{b} }}{r}\left( {C_{{{\text{dq}},{\text{bulk}}}} - \left[ {C_{{{\text{dq}},{\text{bulk}}}} - \frac{{r_{b} }}{{D_{{{\text{dq}}}}^{*} }}\mathbbm{n}_{{{\text{db}}}} } \right]} \right) + C_{{{\text{dq}},{\text{bulk}}}} \\ & = C_{{{\text{dq}},{\text{bulk}}}} - \frac{{r_{b}^{2} }}{{r D_{{{\text{dq}}}}^{*} }}\mathbbm{n}_{{{\text{db}}}} \\ \end{aligned} $$
    (A.16)
  4. (d)

    Next, we solve for \({\left({k}_{L}\alpha \right)}_{db}\). We begin by rewriting Eq. (4a) as:

    $${J}_{db}={J}_{dq}$$
    (A.17a)
    $${\left({k}_{L}\alpha \right)}_{db}{C}_{d}({r}_{b}^{-})={\left({k}_{L}\alpha \right)}_{dq}\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)$$
    (A.17b)
    $${\left({k}_{L}\alpha \right)}_{db} ={\left({k}_{L}\alpha \right)}_{dq}\frac{\kappa \left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)}{{C}_{d}({r}_{b}^{+})}$$
    (A.17c)
    $${\left({k}_{L}\alpha \right)}_{db} =\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}\frac{{\kappa }_{db}\left({C}_{dq,\mathrm{bulk}}-\left[{C}_{dq,\mathrm{bulk}}-\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}\right]\right)}{{C}_{d}({r}_{b}^{+})}$$
    (A.17d)
    $${\left({k}_{L}\alpha \right)}_{db} =\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}\frac{{\kappa }_{db}\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}}{{C}_{d}({r}_{b}^{+})}$$
    (A.17e)
    $${\left({k}_{L}\alpha \right)}_{db}=\frac{3}{{r}_{b}} \frac{{\kappa }_{db} {\mathbbm{n}}_{db}}{{C}_{d}({r}_{b}^{+})}$$
    (A.17f)
  5. (e)

    Next, we solve for the Monod \({K}_{S}\). We begin by rewriting Eq. (8b) as:

    $$ \begin{aligned} K_{S} & = \frac{1}{{\kappa_{{{\text{db}}}} }}\frac{{\left( {k_{L} \alpha } \right)_{{{\text{db}}}} }}{{\left( {k_{L} \alpha } \right)_{{{\text{dq}}}} }}C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{1}{{\kappa_{{{\text{db}}}} }}\frac{{\frac{{3 D_{{{\text{dq}}}}^{*} }}{{r_{b}^{2} }}}}{{\frac{{3 D_{{{\text{dq}}}}^{*} }}{{r_{b}^{2} }}}}\frac{{\kappa_{{{\text{db}}}} \frac{{r_{b} }}{{D_{{{\text{dq}}}}^{*} }}\mathbbm{n}_{{{\text{db}}}} }}{{C_{d} \left( {r_{b}^{ + } } \right)}}C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{{r_{b} }}{{D_{{{\text{dq}}}}^{*} }} \mathbbm{n}_{{{\text{db}}}} \frac{{C_{{{\text{dq}},\,{\text{bulk}}}} }}{{C_{d} \left( {r_{b}^{ + } } \right)}} \\ \end{aligned} $$
    (A.18)

    For the condition that \(\frac{{C}_{dq,\mathrm{bulk}}}{{C}_{d}({r}_{b}^{+})}\cong 1\), \({K}_{S}\) can be interpreted as a constant:

    $${K}_{S}\cong \frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{n}}_{db}$$
    (A.19)
  6. (f)

    Finally, we solve for the Monod \({\mu }_{\mathrm{max}}\). We begin by rewriting Eq. (8c) as:

    $$ \begin{aligned} \mu_{\max } & = \left( {k_{L} \alpha } \right)_{{{\text{db}}}} C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{{Y_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} \rho_{b} }} \frac{3}{{r_{b} }} \frac{{\kappa_{{{\text{db}}}} \mathbbm{n}_{{{\text{db}}}} }}{{C_{d} \left( {r_{b}^{ + } } \right)}}C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{{Y_{{{\text{db}}}} }}{{\rho_{b} }} \frac{3}{{r_{b} }}\mathbbm{n}_{{{\text{db}}}} \frac{{C_{{{\text{dq}},{\text{bulk}}}} }}{{C_{d} \left( {r_{b}^{ + } } \right)}} \\ \end{aligned} $$
    (A.20a)

    For the condition that \(\frac{{C}_{dq,\mathrm{bulk}}}{{C}_{d}({r}_{b}^{+})}\cong 1\), \({\mu }_{\mathrm{max}}\) can be interpreted as a constant:

    $${\mu }_{\mathrm{max}}\cong \frac{{Y}_{db}}{{\rho }_{b}} \frac{3}{{r}_{b}}{\mathbbm{n}}_{db}$$
    (A.21)

Appendix B: Linear rate model

  1. (a)

    First, we solve for \({\left({k}_{L}\alpha \right)}_{dq}\).

    Let \({C}_{d}(r,t)\) be microscale concentration of donor \(d\) in the aqueous phase \(q\) for \(r>{r}_{b}\), and \(t>0\).

    $$\frac{{D}_{dq}^{*}}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial {C}_{d}}{\partial r}\right)=\frac{\partial {C}_{d}}{\partial t}$$
    (B.1a)

    and assuming steady state (no accumulation)

    $$\frac{{D}_{dq}^{*}}{{r}^{2}}\frac{d}{dr}\left({r}^{2}\frac{d{C}_{d}}{dr}\right)=0$$
    (B.1b)

    For \(r={r}_{b}\)

    $$\frac{{A}_{b}}{{V}_{b}}{\mathbb{J}}_{d}={R}_{db}$$
    (B.2a)
    $$\frac{3}{{r}_{b}}{\mathbb{J}}_{d}={R}_{db}$$
    (B.2b)

    where:\({A}_{b}=4\pi {r}_{b}^{2}\); \({V}_{b}=\frac{4}{3}\pi {r}_{b}^{3}\); \({R}_{db}\) is the microscale reaction rate within the volume of the microbe; and, \({\mathbbm{v}}_{db}\) is the microscale transport velocity across the surface of the single spherical microbe shown on Fig. 1 with \({\mathbb{J}}_{d}={\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{-})\) being variable and a function of the surface concentration \({C}_{d}({r}_{b}^{-})\).

    We solve Eqs. (B.1b) and (B.2b) subject to the conditions stated in Eq. (A.3).

    First, we integrate Eq. (B.1b) w.r.t. \(r\) as:

    $${r}^{2}\frac{d{C}_{d}}{dr}={\beta }_{1}$$
    $$\frac{d{C}_{d}}{dr}=\frac{{\beta }_{1}}{{r}^{2}}$$
    $${C}_{d}=-\frac{{\beta }_{1}}{r}+{\beta }_{2}$$
    (B.3)

    Using Eq. (A.3a) under the assumption that \({r}_{c}\to \infty \) and then combining with Eq. (B.3) yields:

    $${C}_{dq,\mathrm{bulk}}={\beta }_{2}$$
    (B.4)

    Substituting Eq. (B.3) with Eq. (A.3b) at \(r={r}_{b}\) yields:

    $$-\frac{{\beta }_{1}}{{r}_{b}}+{C}_{dq,\mathrm{bulk}}={\kappa }_{db} {C}_{d}({r}_{b}^{-})$$
    $${\beta }_{1}=\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right) {r}_{b}$$
    (B.5)

    We can now rewrite Eq. (B.3) as:

    $${C}_{d}=-\frac{{r}_{b}}{r}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)+{C}_{dq,\mathrm{bulk}}$$
    (B.6)

    Now, taking the derivative of Eq. (B.6) w.r.t. \(r\) and evaluating at position \(r={r}_{b}\) yields:

    $${\left.\frac{d{C}_{d}}{dr}\right|}_{r\to {r}_{b}^{+}}=\frac{1}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (B.7)

    Substituting Eq. (B.7) into (A.3c) yields:

    $${\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{-})=\frac{{D}_{dq}^{*}}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (B.8)

    Next, substituting Eq. (B.2b) into Eq. (B.8) yields:

    $$\frac{{r}_{b}}{3}{R}_{db}=\frac{{D}_{dq}^{*}}{{r}_{b}}\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)$$
    (B.9a)
    $$\left({C}_{dq,\mathrm{bulk}}-{\kappa }_{db} {C}_{d}({r}_{b}^{-})\right)\frac{{3D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (B.9b)

    From Eq. (A.3b), we have:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{{3D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (B.10a)
    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right){\left({k}_{L}\alpha \right)}_{dq}={R}_{db}$$
    (B.10b)

    where,

    $${k}_{L}=\frac{{D}_{dq}^{*}}{{r}_{b}}$$
    (B.11a)
    $$\alpha =\frac{3}{{r}_{b}}$$
    (B.11b)
  2. (b)

    Next, we solve for the surface concentration \({C}_{d}({r}_{b}^{+})\). We begin with Eq. (B.10a) as:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}={R}_{db}$$
    (B.12)

    but \(\frac{3}{{r}_{b}}{\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{-})={R}_{db}\) from Eq. (B.2b). Therefore, we can rewrite Eq. (B.12) as:

    $$\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)\frac{3 {D}_{dq}^{*}}{{r}_{b}^{2}}=\frac{3}{{r}_{b}}{\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{-})$$
    (B.13a)
    $${C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})=\frac{{r}_{b}}{{D}_{dq}^{*}}{\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{-})$$
    (B.13b)
    $${C}_{dq,\mathrm{bulk}}=\frac{{r}_{b}}{{\kappa D}_{dq}^{*}}{\mathbbm{v}}_{db}{C}_{d}({r}_{b}^{+})+{C}_{d}({r}_{b}^{+})$$
    (B.13c)
    $${C}_{dq,\mathrm{bulk}}={C}_{d}({r}_{b}^{+})\left(\frac{{r}_{b}}{{{\kappa }_{db} D}_{dq}^{*}}{\mathbbm{v}}_{db}+1\right)$$
    (B.13d)
    $${C}_{d}({r}_{b}^{+})=\frac{{C}_{dq,\mathrm{bulk}}}{1+\frac{{r}_{b} {\mathbbm{v}}_{db}}{{\kappa }_{db} {D}_{dq}^{*}}}$$
    (B.13e)
  3. (c)

    Next, we solve the concentration \({C}_{d}\) as a function of radius \(r>{r}_{b}\). We begin with Eq. (B.6) as:

    $${C}_{d}=-\frac{{r}_{b}}{r}\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)+{C}_{dq,\mathrm{bulk}}$$
    (B.14)

    Now, substitute Eq. (B.13e) into (B.14) as:

    $$ \begin{aligned} C_{d} & = - \frac{{r_{b} }}{r}\left( {C_{{{\text{dq}},{\text{bulk}}}} - \frac{{C_{{{\text{dq}},{\text{bulk}}}} }}{{1 + \frac{{r_{b} \mathbbm{v}_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}}}} \right) + C_{{{\text{dq}},{\text{bulk}}}} \\ & = C_{{{\text{dq}},{\text{bulk}}}} - \frac{{r_{b} }}{r}\left( {C_{{{\text{dq}},{\text{bulk}}}} - \frac{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} C_{{{\text{dq}},{\text{bulk}}}} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} + r_{b} \mathbbm{v}_{{{\text{db}}}} }}} \right) \\ & = C_{{{\text{dq}},{\text{bulk}}}} \left( {1 - \frac{{r_{b}^{2} }}{r} \cdot \frac{{v_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} + r_{b} \mathbbm{v}_{{{\text{db}}}} }}} \right) \\ \end{aligned} $$
    (B.15)
  4. (d)

    Next, we solve for \({\left({k}_{L}\alpha \right)}_{db}\). We begin by rewriting Eq. (4a) as Eqs. (6a) and (6b):

    $${J}_{db}={J}_{dq}$$
    $${\left({k}_{L}\alpha \right)}_{db} {C}_{d}({r}_{b}^{-})={\left({k}_{L}\alpha \right)}_{dq}\left({C}_{dq,\mathrm{bulk}}-{C}_{d}({r}_{b}^{+})\right)$$

    By isolating \({\left({k}_{L}\alpha \right)}_{db}\), and performing appropriate substitutions for \({C}_{d}\left({r}_{b}^{-}\right)\) from (A.3b), \({\left({k}_{L}\alpha \right)}_{dq}\) from (B.11), and \({C}_{d}({r}_{b}^{+})\) from (B.13e):

    $$ \begin{aligned} \left( {k_{L} \alpha } \right)_{{{\text{db}}}} & = \left( {k_{L} \alpha } \right)_{{{\text{dq}}}} \frac{{\kappa_{{{\text{db}}}} \left( {C_{{{\text{dq}},{\text{bulk}}}} - C_{d} \left( {r_{b}^{ + } } \right)} \right)}}{{C_{d} \left( {r_{b}^{ + } } \right)}} \\ & = \frac{{3\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}{{r_{b}^{2} }}\left( {C_{{{\text{dq}},{\text{bulk}}}} /C_{d} \left( {r_{b}^{ + } } \right) - 1} \right) \\ & = \frac{{3\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}{{r_{b}^{2} }}\left( {\frac{{C_{{{\text{dq}},{\text{bulk}}}} }}{{C_{{{\text{dq}},{\text{bulk}}}} }}\left( {1 + \frac{{r_{b} \mathbbm{v}_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}} \right) - 1} \right) \\ & = \frac{{3\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}{{r_{b}^{2} }}\left( {\frac{{r_{b} \mathbbm{v}_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}} \right) \\ & = \frac{3}{{r_{b} }}\mathbbm{v}_{{{\text{db}}}} \\ \end{aligned} $$
    (B.16)
  5. (e)

    Next, we solve for the Monod \({K}_{S}\). We begin by rewriting Eq. (8b) as:

    $$ \begin{aligned} K_{S} & = \frac{1}{{\kappa_{{{\text{db}}}} }}\frac{{\left( {k_{L} \alpha } \right)_{{{\text{db}}}} }}{{\left( {k_{L} \alpha } \right)_{{{\text{dq}}}} }}C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{1}{{\kappa_{{{\text{db}}}} }}\frac{{3\mathbbm{v}_{{{\text{db}}}} /r_{b} }}{{3D_{{{\text{dq}}}}^{*} /r_{b}^{2} }}C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{{r_{b} }}{{\kappa_{{{\text{db}}}} D_{{{\text{dq}}}}^{*} }}\mathbbm{v}_{{{\text{db}}}} C_{{{\text{dq}},{\text{bulk}}}} \\ \end{aligned} $$
    (B.17)
  6. (f)

    Finally, we solve for the Monod \({\mu }_{\mathrm{max}}\). We begin by rewriting Eq. (8c) as:

    $$ \begin{aligned} \mu_{\max } & = \frac{{Y_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} \rho_{b} }}\left( {k_{L} \alpha } \right)_{{{\text{db}}}} C_{{{\text{dq}},{\text{bulk}}}} \\ & = \frac{{Y_{{{\text{db}}}} }}{{\kappa_{{{\text{db}}}} \rho_{b} }} \frac{3}{{r_{b} }}\mathbbm{v}_{{{\text{db}}}} C_{{{\text{dq}},{\text{bulk}}}} \\ \end{aligned} $$
    (B.18)

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Enouy, R.W., Walton, K.M., Malton, I.I. et al. A mechanistic derivation of the Monod bioreaction equation for a limiting nutrient. J. Math. Biol. 84, 62 (2022). https://doi.org/10.1007/s00285-022-01760-0

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  • DOI: https://doi.org/10.1007/s00285-022-01760-0

Keywords

  • Monod
  • Mass transport
  • Biodegradation
  • Limiting nutrient

Mathematics Subject Classification

  • 92-10 Mathematical modeling or simulation for problems pertaining to biology