$F_{\mathrm{t}}=u_{\mathrm{t}}S_{\mathrm{t}}$ must equal the sum of the flow rates through mask $F_{\mathrm{m}}$ and gap $F_{\mathrm{g}}$ (cf. Eq. (\[EQFtFgFm\]))

$$F_{\mathrm{m}} = F_{\mathrm{t}} - F_{\mathrm{g}} = F_{t} - u_{\mathrm{g}} H_{\mathrm{g}} B_{\mathrm{g}}\quad ,
\label{EQflowratesum}$$

and thus the flow velocity through the filter can be expressed as $$u_{\mathrm{m}} = \frac{F_{\mathrm{m}}}{S_{\mathrm{m}}} = \frac{ F_{t} - u_{\mathrm{g}} H_{\mathrm{g}} B_{\mathrm{g}}}{S_{\mathrm{m}}} \quad .
\label{EQvgfromflowratessum}$$ Inserting Eq. (\[EQvgfromflowratessum\]) into Eq. (\[EQdpsum1\]) gives

$$\begin{aligned}
\frac{u_{\mathrm{t}}}{|u_{\mathrm{t}}|}  \zeta  \frac{\rho }{2} u_{\mathrm{g}}^{2} + \left( \frac{12 \mu L_{\mathrm{g}}}{H_{\mathrm{g}}^{2}} + \frac{C_{\mathrm{m}}\rho H_{\mathrm{g}}B_{\mathrm{g}}}{S_{\mathrm{m}}} \right) u_{\mathrm{g}} \\ - \frac{C_{\mathrm{m}} \rho F_{\mathrm{t}}}{S_{\mathrm{m}}}    = 0 \quad .
\label{EQdpfull}
\end{aligned}$$

The solution for gap velocity $u_{\mathrm{g}}$ can be determined as follows: $$a  u_{\mathrm{g}}^{2} + b  u_{\mathrm{g}} + c = 0 \quad ,
\label{EQdpsumsol1}$$ $$a   = \frac{u_{\mathrm{t}}}{|u_{\mathrm{t}}|}  \frac{\zeta \rho}{2} \quad ,
\label{EQdpsumsol2}$$ $$b = \frac{12 \mu L_{\mathrm{g}}}{H_{\mathrm{g}}^{2}} + \frac{C_{\mathrm{m}}\rho H_{\mathrm{g}}B_{\mathrm{g}}}{S_{\mathrm{m}}} \quad ,
\label{EQdpsumsol3}$$

$$c = - \frac{C_{\mathrm{m}}\rho F_{\mathrm{t}}}{S_{\mathrm{m}}} \quad ,
\label{EQdpsumsol4}$$

$$u_{\mathrm{g}} = \frac{-b + \sqrt{b^{2}-4ac}}{2a}\quad .
\label{EQvgsol}$$

The average flow velocity through the mask $u_{\mathrm{m}}$, the pressure drop $\Delta p$ and the flow rates through the mask filter and through the gap can then be computed as $$F_{\mathrm{m}} = u_{\mathrm{m}} S_{\mathrm{m}} \quad ,
\label{EQFmsol}$$

$$\Delta p = C_{\mathrm{m}} \rho u_{\mathrm{m}} \quad ,
\label{EQdpsol}$$

$$F_{\mathrm{g}} = u_{\mathrm{g}} H_{\mathrm{g}} B_{\mathrm{g}} \quad .
