The figure below shows the new control volume that is between the regulated pressure from the tank supply ## P_r ## and the tire at pressure ##P##. I'm applying (1) the control volume
$$ \sum F = \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} +\int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A \tag{1} $$
##\underline{\sum F}##
Since we have from the earlier result that the Shear Force resisting the flow was porportional to the flow velocity within the variable ## \tau = \varphi v##, we now have to consider the total shear force acting the control volume as integral because the velocity is varying over the length of the tube.
$$ \sum F = \left( P_r + P_a \right) A - \left( P(t) + P_a \right) A - \mu \pi D \varphi \int v(x,t) \, dx $$
Momentum Accumulation Term:
This time the only quantity that can be taken outside the integral is the cross sectional area and we can drop the vector notation.
$$ \frac{d}{dt} \int_{cv} \boldsymbol v \rho d V\llap{-} = A \frac{d}{dt} \int v(x,t) \rho (x,t) dx $$
Net Efflux of Momentum Term:
I'm considering the properties ## \rho ## and ##v ## as uniformily distributed across the area of the pipe at section 1 and 2.
$$ \int_{cs} \boldsymbol v \rho \boldsymbol V \cdot d \boldsymbol A = A \left( \rho_2(t) {v_2(t)}^2 - \rho_r {v_1(t)}^2 \right) $$
$$ P_r A - P(t)A - \mu \pi D \varphi \int v(x,t) \, dx = A \frac{d}{dt} \int v(x,t) \rho (x,t) dx + A \left( \rho_2(t) {v_2(t)}^2 - \rho_r {v_1(t)}^2 \right) $$
As far as I can see this is step 1.