= Solution
Let $z$ increase down the pipe, $a=0.015\,\mathrm m$ be its radius and $r$ the distance from its axis. For steady, fully developed <laminar flow> of a <Newtonian fluid>, the axial <Navier-Stokes equations> reduce to
$$
0=-\frac{dp}{dz}+\eta\frac1r\frac d{dr}\left(r\frac{du}{dr}\right)+\rho g\sin20^\circ.
$$
Write $K=-dp/dz+\rho g\sin20^\circ$. Integrating twice, regularity at $r=0$ excludes a logarithmic term, and the <no-slip boundary condition> $u(a)=0$ gives the <Hagen-Poiseuille flow> profile
$$
\boxed{u(r)=\frac{K}{4\eta}(a^2-r^2).}
$$
The <volumetric flow rate> follows by integrating over the circular cross-section:
$$
Q=2\pi\int_0^a u(r)r\,dr
=\frac{\pi Ka^4}{8\eta},\qquad
\overline u=\frac{Q}{\pi a^2}=\frac{Ka^2}{8\eta}=\frac{u(0)}2.
$$
For the intended gravity-driven case with no imposed axial <pressure gradient>, $dp/dz=0$. Substitution gives
$$
\boxed{u(r)\simeq0.2095\left[1-\left(\frac{r}{0.015\,\mathrm m}\right)^2\right]\mathrm{m\,s^{-1}},}
$$
and
$$
\boxed{Q\simeq7.40\times10^{-5}\,\mathrm{m^3\,s^{-1}}
=74.0\,\mathrm{mL\,s^{-1}}\simeq4.44\,\mathrm{L\,min^{-1}}.}
$$
The corresponding mean speed is about $0.1047\,\mathrm{m\,s^{-1}}$ and the diameter-based <Reynolds number> is $\rho\overline u(2a)/\eta\simeq3.49$, consistent with <laminar flow>. The original PDF does not specify inlet and outlet pressures. If they impose an additional <pressure gradient>, the general result with $K$ above applies; gravity alone does not determine that extra contribution.
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