## 3 Problem Description
### Batch-Processing Time
#### 3.2.2 Development  
**Part 1: Condition on [PERSON]'s flow rate**  
During the initial phase, the flow rate is given by:  
\[Q_{\text{Darcy}}=\frac{K\cdot A_{0}\cdot\Delta P}{\mu} \tag{1}\]  
There are two possible cases:  
**Case 1: \(Q_{\text{Darcy}}\geq Q_{\text{op}}\).** In this case, the flow rate is forced to \(Q_{\text{op}}\) and the volume filtered during this phase is:  
\[V_{1}=Q_{\text{op}}\cdot t_{1} \tag{2}\]  
**Case 2: \(Q_{\text{min}}<Q_{\text{Darcy}}<Q_{\text{op}}\).** In this case, the flow rate is determined by [PERSON]'s law, i.e., \(Q(t)=Q_{\text{Darcy}}\), and the filtered volume is:  
\[V_{1}=Q_{\text{Darcy}}\cdot t_{1}=\frac{K\cdot A_{0}\cdot\Delta P}{\mu}\cdot t _{1} \tag{3}\]  
If \(Q_{\text{[PERSON]}}\leq Q_{\text{min}}\), filtration does not occur.  
**Part 2: Decreasing flow rate due to clogging**  
From \(t_{1}\) onwards, the flow rate starts to decrease due to clogging. The effective filtration area decreases exponentially:  
\[A(t)=A_{0}\cdot e^{-r_{s}(t-t_{1})} \tag{4}\]The flow rate during this phase follows [PERSON]'s law:  
\[Q(t)=\frac{K\cdot A_{0}\cdot e^{-r_{s}\cdot(t-t_{1})}\cdot\Delta P}{\mu} \tag{5}\]  
The volume filtered during the second phase \(V_{2}\) is:  
\[V_{2}=\int_{t_{1}}^{t_{\text{total}}}Q(t)\,dt=\int_{t_{1}}^{t_{\text{total}}} \frac{K\cdot A_{0}\cdot e^{-r_{s}\cdot(t-t_{1})}\cdot\Delta P}{\mu}\,dt \tag{6}\]  
Solving the integral:  
\[V_{2}=\frac{K\cdot A_{0}\cdot\Delta P}{\mu\cdot r_{s}}\Big{(}1-e^{-r_{s}\cdot (t_{\text{total}}-t_{1})}\Big{)} \tag{7}\]  
**Total filtered volume**  
The total filtered volume is the sum of the volumes filtered in both phases:  
\[V_{\text{total}}=V_{1}+V_{2} \tag{8}\]  
Substituting the expressions for \(V_{1}\) and \(V_{2}\), we have two cases:  
**Case 1: \(Q_{\text{Darcy}}\geq Q_{\text{op}}\)**:  
\[V_{\text{total}}=Q_{\text{op}}\cdot t_{1}+\frac{K\cdot A_{0}\cdot\Delta P}{ \mu\cdot r_{s}}\Big{(}1-e^{-r_{s}\cdot(t_{\text{total}}-t_{1})}\Big{)} \tag{9}\]  
**Case 2: \(Q_{\text{min}}<Q_{\text{Darcy}}<Q_{\text{op}}\)**:  
\[V_{\text{total}}=\frac{K\cdot A_{0}\cdot\Delta P}{\mu}\cdot t_{1}+\frac{K\cdot A _{0}\cdot\Delta P}{\mu\cdot r_{s}}\Big{(}1-e^{-r_{s}\cdot(t_{\text{total}}-t_ {1})}\Big{)} \tag{10}\]  
**Solving for total filtration time \(t_{\text{total}}\)**  
To solve for \(t_{\text{total}}\), we begin by the right-hand side of the equation:  
\[V_{\text{total}}-V_{1}=\frac{K\cdot A_{0}\cdot\Delta P}{\mu\cdot r_{s}}\Big{(} 1-e^{-r_{s}\cdot(t_{\text{total}}-t_{1})}\Big{)} \tag{11}\]  
Multiplying both sides by \(\frac{\mu\cdot r_{s}}{K\cdot A_{0}\cdot\Delta P}\):  
\[r_{s}\cdot\frac{V_{\text{total}}-V_{1}}{K\cdot A_{0}\cdot\Delta P/\mu}=1-e^{- r_{s}\cdot(t_{\text{total}}-t_{1})} \tag{12}\]  
Solving for \(e^{-r_{s}\cdot(t_{\text{total}}-t_{1})}\):  
\[e^{-r_{s}\cdot(t_{\text{total}}-t_{1})}=1-r_{s}\cdot\frac{V_{\text{total}}-V_{ 1}}{K\cdot A_{0}\cdot\Delta P/\mu} \tag{13}\]  
Finally, taking the natural logarithm for both sides:  
\[t_{\text{total}}=t_{1}-\frac{1}{r_{s}}\ln\!\left(1-r_{s}\cdot\frac{V_{\text{ total}}-V_{1}}{K\cdot A_{0}\cdot\Delta P/\mu}\right) \tag{14}\]  
**Final equation**  
\[t_{\text{total}}=t_{1}-\frac{1}{r_{s}}\ln\!\left(1-r_{s}\cdot\frac{V_{\text{ total}}-V_{1}}{K\cdot A_{0}\cdot\Delta P/\mu}\right) \tag{15}\]where \(V_{1}\) will depend on whether \(Q_{\rm{Darcy}}\geq Q_{\rm{op}}\) or \(Q_{\rm{min}}<Q_{\rm{Darcy}}<Q_{\rm{op}}\).