How many water wheel can fit on a river until another one produces zero energy? Mostly what I'm after are the factors I need to know to do the problem.

Imagine we're in the days of Leonardo Da Vinci. And pizza is everyone's favorite food. And there's a river that goes on forever. Along the river, I build a water wheel that powers my pizza factory.
Then downstream I build another one.
Then another one.
etc.

What I managed to learn already from a similar question is that the water in the river had potential energy. Then as it moves along, it has kinetic energy, which is transferred to the wheel. After it passes the water wheel, the kinetic energy is less; it moves slower. If I just measure how much slower the river is moving after each water wheel, will that tell me how many more I can make?  
Is a water wheel's take of kinetic energy the only thing affecting the kinetic energy of the river afterward?
Or are there still other factors that allow me to build more/less water wheels? But that doesn't seem right because it seems like the water would pick up speed again if the river were at the same incline before and after the wheel. But that would mean the only thing limiting me is the water loss from evaporation, and the length of the river, and that doesn't seem right either.
 A: Consider the flow of a river:

$z$ is the height.
The total power equation of the river is:
$$\frac12\dot{m}v_1^2+\dot{m}gz_1=\frac12\dot{m}v_2^2+\dot{m}gz_2+P_f$$
Where $\dot{m}$ is the mass flow of the river ($\mathrm{kgs^{-1}}$) and $P_f$ a power loss due to friction, viscous losses, bends and other restrictions.
If, near $z_2$, we insert a number of energy plants that each extract $P_i$ of power, then the equation becomes:
$$\frac12\dot{m}v_1^2+\dot{m}gz_1=\frac12\dot{m}v_2^2+\dot{m}gz_2+P_f+\Sigma P_i$$
We can now write:
$$v_2^2=\frac{2}{\dot{m}}[\frac12\dot{m}v_1^2+\dot{m}gz_1-\dot{m}gz_2-P_f-\Sigma P_i]$$
For $v_2=0$ the power plants start acting like dam and no more energy can be extracted from the river.
From:
$$\frac12\dot{m}v_1^2+\dot{m}gz_1-\dot{m}gz_2-P_f-\Sigma P_i=0$$
the maximum $\Sigma P_i$ can be calculated, at least 'on paper'...
A: Having seen many old mill races, let me describe them: if the river is fowing rapidly, it is also flowing downhill, so you can put in mills one after the other, prhaps every 100 meters. However, rapidly flowing streams tend to have small sources, so there's not much total current --hence you find small mills in up in the hills.
As the number of tributaries increas, so does the current. You can build larger mills. But the current slows, and the river broadens, as the countryside flttens out. Then th mills tend to dam the stream, and use a mill race.  Thiscalso evens out the flow rate between wet and dry seasons, for the mill race supply is limited to the height of the dam or weir.
In the old days our neighbors downstream dammed the creek to power a sawmill.  The water level was raised less than a meter, but that was more than enough to power a saw to cut boards.
Just like dams today, the mills used falling water; if the current was fast, that can be used, but as it slows, the efficiency is too low. Then you see wind mills, or ox-powered mills.
As long as there is enough water, and enough potential energy in the water to be effective for your mill, you can install as many water wheels/turbines as will fit  ... leaving some space between them so that turbulence, etc., doesn't affect them.  An efficient extraction of power from a fast moving, but level stream implies that the water is backing up; in other words, your wheel is damming the flow. This just converts kinetic to potential energy by accident, and is usually undesirable. 
