Derive laminar velocity in inclined pipe

Set-up. Pipe of radius $R$ inclined at angle $\alpha$ below the horizontal; the fluid flows down the pipe under gravity alone, so there is no externally imposed pressure gradient: $\dfrac{\partial p}{\partial x} = 0$ (the pressure along the pipe is atmospheric/constant). Flow is steady, laminar, fully developed, so $u = u(r)$ only, and $v_r = v_\theta = 0$.

        x (down the pipe)
     ___ \
    /     \  alpha
   / pipe  \
  /_________\ ____ horizontal
  gravity component along x: g sin(alpha)

Momentum equation along $x$ (Navier-Stokes in cylindrical coordinates, reduced for this case):

$$0 = \rho g\sin\alpha + \mu\,\frac{1}{r}\frac{d}{dr}\left(r\frac{du}{dr}\right)$$

(Equivalent force balance on a cylinder of radius $r$: weight component $\rho g \sin\alpha\,\pi r^2 L$ is balanced by the shear force $-\mu\dfrac{du}{dr}\,2\pi r L$.)

Integrate.

$$\frac{d}{dr}\left(r\frac{du}{dr}\right) = -\frac{\rho g\sin\alpha}{\mu}\,r$$ $$r\frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r^2 + C_1$$

$C_1 = 0$, since $du/dr$ must be finite at $r = 0$. Then

$$\frac{du}{dr} = -\frac{\rho g\sin\alpha}{2\mu}\,r \quad\Rightarrow\quad u = -\frac{\rho g\sin\alpha}{4\mu}\,r^2 + C_2$$

Boundary condition. No slip at the wall: $u = 0$ at $r = R$, so $C_2 = \dfrac{\rho g\sin\alpha}{4\mu}R^2$.

$$\boxed{u(r) = \frac{\rho g\sin\alpha}{4\mu}\left(R^2 - r^2\right)}$$

The profile is parabolic with $u_{max} = \dfrac{\rho g\sin\alpha}{4\mu}R^2$ at the axis. With $R = D/2$, $u = \dfrac{\rho g\sin\alpha}{16\mu}\left(D^2 - 4r^2\right)$. The mean velocity is $V = u_{max}/2 = \dfrac{\rho g D^2\sin\alpha}{32\mu}$.