Friction loss in a vertical pipe flow I made this problem for more understanding of pressure and pressure loss in vertical flow.
Consider the following steady system, where a fluid enters a tank and exits through a vertical pipe of length $L$ and diameter $D=2R$. Height of fluid in the tank is constant and equals to $H$. Density and viscosity of liquid are $\rho$ and $\mu$, respectively. If flow is laminar find $Q$.

Now if I write Bernoulli's equation for the free surface of tank and exiting point of pipe, then I get$$\frac{P_{atm}}{\gamma}+\frac{v_0^2}{2g}+z_0=\frac{P_{atm}}{\gamma}+\frac{v^2}{2g}+z+h_l,$$where $h_L$ is the friction loss head of exiting pipe and $v=Q/(\pi R^2)$ and $\gamma= \rho g$. We know that $v_0 \approx 0$, thus$$H+L=\frac{v^2}{2g}+h_L$$
Now we need to find another relation between $v$ and $h_L$. Can we use Darcy–Weisbach equation? I think we can not due to vertical flow! I'm interested in writing momentum balance and derive the relation between friction loss and velocity (like Hagen–Poiseuille equation), but I don't know how to treat pressure terms! Is there a pressure distribution along the exiting pipe?
Edit1: Momentum balance for the laminar flow in the pipe gives the velocity as$$v_z(r)=\frac{R^2}{4 \mu}\left(-\frac{dp}{dz}+ \rho g \right) \left(1-\frac{r^2}{R^2} \right) $$And integrating over cross section of pipe for the flow rate gives$$Q=\pi r^2 v=\int_{0}^{R} 2 \pi r v_z(r) \ dr=\frac{\pi R^4}{8 \mu} \left(-\frac{dp}{dz}+ \rho g \right)$$And finally$$-\frac{dp}{dz}+ \rho g=\frac{32 \mu v}{D^2}$$Now which one of the following is right and why?
1) $h_L=L(-dp/dz)/ \gamma=\frac{32 \mu v L}{\gamma D^2}-L$
2) $h_L=L(-dp/dz+ \rho g)/ \gamma=\frac{32 \mu v L}{\gamma D^2}$
I don't have a sense of $p$ here! Can you give a physical sense of pressure within the exiting pipe?
Edit2: The answers and discussions in this question may solve the following similar questions:
Q1, Q2, Q3.
 A: This is not a complete answer, just an outline of how to go about to derive a formula for the velocity profile and hence the pressure loss across the cross section of a pipe. I'll model the pipe as infinitely long and horizontal to simplify the problem---the result should still be applicable, but needs to be used with a proper water pressure due to gravity in the case of a vertical pipe.
The viscous force is parametrized by a fluid's viscosity $\eta$. Expressed as pressure perpendicular to a velocity gradient in the $x$-direction, it is $\frac{\partial p}{\partial y} = \eta \frac{\partial v}{\partial x}$ where $v$ is the local velocity (essentially, $\dot{y}$). This expression needs to be transformed into cylindrical coordinates (for the cylindrical pipe). Together with the usual relations (and perhaps a continuity condition), that should allow to derive a flow profile $v(r)$, which is parabolic; the solution is giveb e.g. at Hyperphysics.
Having a flow profile $v(r)$ allows to calculate the viscous forces and to get the total pressure drop (per unit length of pipe) by integrating over the pipe's cross section. That would be the (partial) answer I set out to sketch.
A: You can use the Darcy-Weisbach equation, but you have to modify it a little for vertical flow.  In vertical flow, a differential force balance on the flow gives:
$$(P(z+\Delta z)-P(z))\frac{\pi D^2}{4}+\rho g \frac{\pi D^2}{4}\Delta z=\tau_w\Delta z \pi D$$where z is the elevation above the bottom of the tube and $\tau_w$ is the shear stress at the wall.  So, $$\frac{d(P+\rho gz)}{dz}=\frac{4}{D}\tau_w$$  For laminar flow,$$\tau_w=\frac{f}{4}\frac{\rho v^2}{2}$$where f is the Darcy-Weisbach friction factor.  So, combining the two equations, you get:
$$\frac{d(P+\rho gz)}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$
For a horizontal tube, you would have just:
$$\frac{dP}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$
So, for vertical flow, you simply replace the P in the horizontal flow equation by $P+\rho gz$.
A: Water is flowing in the pipe driven by a constant pressure gradient equal to $\rho g$. So you can apply D-W equation provided flow is laminar.
Response to question edit:
In writing $-\frac{dp}{dz}+\rho g$ you separated out contribution due to gravity to pressure gradient acting on the fluid. Therefore $-\frac{dp}{dz}$ is any pressure gradient applied by means other than due to gravity (for e.g. using pump), which in your case is zero.
