Wall friction force for developing pipe flow

Control volume. Take the fluid in the pipe between the inlet section 1 (uniform velocity $V$, pressure $p_1$) and the section 2 (velocity $u = 2V(1 - r^2/R^2)$, pressure $p_2$). The pipe is horizontal. Let $F$ be the friction force exerted by the wall on the fluid (equal and opposite to the friction force of the fluid on the wall).

Continuity check. Mean velocity at section 2:

$$\bar u = \frac{1}{\pi R^2}\int_0^R 2V\left(1 - \frac{r^2}{R^2}\right)2\pi r\,dr = \frac{4V}{R^2}\left[\frac{R^2}{2} - \frac{R^2}{4}\right] = V$$

So the discharge is the same at both sections.

Momentum flux at section 2. The momentum flux is $\rho\int u^2\,dA$, since $u$ is not uniform:

$$\int_0^R u^2\,2\pi r\,dr = 4V^2\cdot 2\pi\int_0^R\left(1 - \frac{r^2}{R^2}\right)^2 r\,dr = 8\pi V^2\left[\frac{R^2}{2} - \frac{R^2}{2} + \frac{R^2}{6}\right] = \frac{4}{3}\pi R^2V^2$$

(using $\int_0^R (r - 2r^3/R^2 + r^5/R^4)\,dr = R^2/2 - R^2/2 + R^2/6 = R^2/6$).

Momentum flux at section 1 (uniform): $\rho V^2\,\pi R^2$.

Momentum equation (flow direction). Net force = rate of change of momentum flux:

$$p_1\pi R^2 - p_2\pi R^2 - F = \rho\left(\frac{4}{3}\pi R^2V^2 - \pi R^2V^2\right) = \frac{\rho \pi R^2V^2}{3}$$ $$\boxed{F = \pi R^2\left(p_1 - p_2 - \frac{\rho V^2}{3}\right)}$$

This is the friction force of the wall on the fluid, and by Newton's third law it is also the friction drag on the pipe walls from the inlet to the section. The term $\rho V^2/3$ is the pressure drop used to build up the parabolic profile (increase in momentum flux), not lost by friction.