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Since the speed of the water may vary across the river, let us focus on the speed at the river mouth.

When a water wheel is placed in a river, part of the kinetic energy of the water is stored. Therefore, the water should flow slower than without the water wheel. This is the case for every point between the water wheel and the river mouth.

On the other hand, the total amount of water that enters the river annually (through glaciers, rain, etc) remains the same. Let us call this amount A. Now the total amount of water that exits the river anually is also A. Let $\sigma$ be the area of the cross-section of the river at the mouth. Now the speed of the water at the cross section is proportional to $A/\sigma$. This is independent on the water wheel. Therefore, the river does not flow slower with the water wheel.

How can these two be reconciled?

UPDATE I would like to emphasize that my question is NOT about the flow rate of the water, but about the velocity. Here flow rate is in liters per second, while velocity is in meters per second.

Also, the question is about the water downstream of the water wheel, not upstream.

Finally, I would like to know what the difference is between the situation without the water wheel, and the situation with the water wheel when equilibrium is attained.

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    $\begingroup$ Why must the area remain the same? $\endgroup$
    – Jon Custer
    Jan 6, 2023 at 14:53
  • $\begingroup$ Good point, maybe the area becomes larger when you add a water wheel $\endgroup$
    – Riemann
    Jan 6, 2023 at 14:59
  • $\begingroup$ I think the river gets a little bit deeper. In the extreme case is the Hoover Dam, where the "water wheels" sit at the base. $\endgroup$
    – JEB
    Jan 6, 2023 at 15:43
  • $\begingroup$ This simulation model tells that velocity of water increases researchgate.net/publication/… i don;t know how correct is that. but, the phenomenon is conservation of flow rate (pressure), kinetics, potential energy when system is at steady state where wheel acts as an obstacle building some potential energy ini system and leeching the kinetic energy out as well. $\endgroup$
    – Haris
    Jan 15, 2023 at 2:53
  • $\begingroup$ However, according to my analysis, the kinetic energy of the water is increased when the velocity increases; but it seems contradictory that a water wheel would increase the kinetic energy of the water. $\endgroup$
    – Riemann
    Jan 15, 2023 at 9:06

6 Answers 6

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The water wheel cannot change the flow rate of the water. If it reduced the flow rate then water upstream of the wheel would flow towards the wheel faster than the water below the wheel was flowing away. That means the depth of the water upstream would increase steadily with time and eventually overflow the river banks.

What the wheel does is increase the depth of the water immediately upstream from it. That is, it behaves like a small dam. Then the energy to rotate the wheel and grind your corn (or whatever) comes from the change in gravitational potential energy of the water as is flows down the height difference between the upstream and downstream sides of the wheel.

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  • $\begingroup$ If it reduced the flow rate the water upstream of the wheel would build up and eventually overflow the river banks. What the wheel does is increase the depth of the water immediately upstream from it. Aren't those two assertions the same but using different words? $\endgroup$
    – Gert
    Jan 6, 2023 at 20:01
  • $\begingroup$ @Gert Better now? $\endgroup$ Jan 6, 2023 at 20:05
  • $\begingroup$ That's definitely less ambiguous. But I'm not sure your answer is correct. Might formulate one myself later on. $\endgroup$
    – Gert
    Jan 6, 2023 at 20:12
  • $\begingroup$ This might be improved by addressing the difference between the equilibrium flow rate and the instantaneous flow rate upon first lowering the water-wheel into the water. $\endgroup$
    – g s
    Jan 6, 2023 at 20:14
  • $\begingroup$ @g s Why, how would that improve the answer? Looking at steady state is always easier. $\endgroup$
    – Gert
    Jan 6, 2023 at 20:17
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Restricting to two dimensions (a canal with high vertical walls) and neglecting the angle of the river bed - the sum of the downstream horizontal flow rate and the upstream vertical flow rate from the water wheel is constant and equal to the upstream horizontal flow rate.

By upstream vertical flow rate I mean the time derivative of the integral of the space integral of the darkest region. (Height differences greatly exaggerated.)

enter image description here

The downstream horizontal flow rate increases with the upstream water level, which means that the rate of change of the upstream water level decreases as the magnitude of the upstream water level increases. This tends the system towards the steady state described in John Rennie's answer, in which the upstream water level has stopped increasing and the upstream and downstream flow rate have equalized. (Height differences greatly exaggerated.)

enter image description here


In the intermediate state, the energy balance is as follows:

The system costs energy to...

  • Spin the wheel against its load (e.g. a generator, friction, a water-mill...)
  • Pay for the increasing potential energy of the darkest blue region.

Energy is retrieved from the system to...

  • Reduce the downstream flow rate
  • Drop water from the darkest blue region into the pale blue region

In the steady state, the energy balance is as follows:

The system costs energy to...

  • Spin the wheel against its load

Energy is retrieved from the system to...

  • Drop water from the darkest blue region into the pale blue region

Note velocity and flow rate aren't quite the same thing. Since the upstream region is deeper and has the same flow rate, its velocity is smaller. The continuous renewal of the energy available to extract with the water wheel comes from that decrease in velocity.


Note that if we insert the water wheel at time $t_1$ and the system attains steady state at time $t_2$, then after $\Delta t$ time any point between $v(t_1+\Delta t)$ and $v(t_2+\Delta t)$ downstream, the flow rate has indeed been temporarily reduced.

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In a steadily flowing river, every section of a river reaches an equilibrium. Let's consider one section.

If the amount of water flowing into the section is larger than flowing out of it, the water level will rise at the entrance to the section. The higher potential energy difference between entrance and exit causes the water to flow faster, causing more water to flow out of the section at the low end. After a while, the section reaches an equilibrium where the flow rate at the entrance is the same as at the exit. In that equilibrium, any potential energy is only converted to heat because of friction.

That equilibrium depends only on the flow rate of the water, and the shape of the section. If the water flow rate stays the same and the shape of the river stays the same, there will not be any change to water levels or velocity. Whether a water wheel is deployed upstream has no influence on that.

So the velocity of the water will not change. Only directly after the water wheel will there be some influence (where the shape of the river has been changed to accommodate the water wheel).

The energy that is transferred to the water wheel, would otherwise have been lost by friction upstream from the water wheel. The water wheel holds up the water, which raises the water level, and lowers velocity upstream. The friction of the river water with the river bank is proportional to the square of the velocity. This restores the equilibrium speed of the river, some distance downstream of the water wheel.

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  • $\begingroup$ If the water flow rate and shape of the river and velocity stay the same, how do you account for the energy that is being extracted from the water by the water wheel? $\endgroup$
    – Riemann
    Jan 15, 2023 at 13:14
  • $\begingroup$ @Riemann It is energy that would otherwise be converted to heat due to friction. In a steadily flowing river, ALL potential energy is converted to heat. $\endgroup$
    – fishinear
    Jan 15, 2023 at 13:21
  • $\begingroup$ In the section, the water flowing into the section has some velocity / kinetic energy. This velocity influences the eventual equilibrium in the section. So it would seem that when water with a higher velocity enters the section (but the same flow), then in the equilibrium the water level would be lower $\endgroup$
    – Riemann
    Jan 15, 2023 at 15:31
  • $\begingroup$ @Riemann Higher velocity at the same flow rate means a lower level of the water coming into the section. Therefore a lower potential difference between entry and exit. That lower potential difference causes the section to fill up, raising the water level at its entry. That will eventually lead to the old equilibrium (or to a higher flow rate if there is more water available upstream). $\endgroup$
    – fishinear
    Jan 15, 2023 at 15:49
  • $\begingroup$ @Riemann At constant flow rate, the velocity of the water is really determined by the shape and height difference of the river downstream. You can force a higher/lower velocity, but only by increasing/decreasing the flow rate. $\endgroup$
    – fishinear
    Jan 15, 2023 at 15:54
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Ater being inserted, the water wheel slows down the entire river, both up and down stream, until it reaches a new steady state where flow rate is roughly equal everywhere, and slower than it was before.

It's easy to see that water molecules impacting the wheel will transfer energy to it and slow downstream.

This creates a brief state where molecules are arriving faster than they are leaving, causing a buildup. This buildup acts as another obstacle for the incoming molecules, extending the effect all the way upstream, slowing the flow rate until it becomes equal to the one downstream and the buildup stops growing.

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  • $\begingroup$ Thank you for your answer! However, my question is not about the flow rate but about the velocity. $\endgroup$
    – Riemann
    Jan 15, 2023 at 9:10
  • $\begingroup$ Also, how can the flow rate be slower than before? I would think that the amount of water coming from the source of the river remains the same. $\endgroup$
    – Riemann
    Jan 15, 2023 at 12:37
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The water wheel acts a dam that is partially open. When the wheel is inserted the level of water upstream rises, the mass flow rate upstream decreases, and the velocity upstream decreases. With the addition of the wheel and the raising of the upstream level, the water upstream will overflow the riverbed, and land is dredged out around the riverbank to create a reservoir (lake).

The source of the water into the upstream part of the river is variable, depending on rain, snowmelt, etc. The level in the reservoir depends on the difference between source inflow rate and outflow rate through the water wheel, and the water level in the reservoir varies as these are not in balance. During times of great inflow to the reservoir, to keep the reservoir level from becoming too high, more water is allowed to exit by opening a diversion spillway if necessary. During times of sparse source flow the water level in the reservoir drops unless the flow through the wheel is reduced by preventing free rotation of the wheel (closing off flow through the dam).

In practice the water wheel in is placed at the base of the dam; the kinetic energy of the water that has fallen from the reservoir is converted into electrical energy.

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  • $\begingroup$ Thank you for your answer! I would also like to know what happens downstream of the water wheel. $\endgroup$
    – Riemann
    Jan 15, 2023 at 12:39
  • $\begingroup$ [Slight correction] The more power extracted from the kinetic energy of the falling water, the less the velocity of the water out of the wheel so the level of the water past the wheel increases to match the flow rate coming into the wheel. However, without extracting power the falling water creates more turbulence in the sink at the bottom, thereby also lowering the velocity so the difference is less than you might expect. $\endgroup$
    – John Darby
    Jan 15, 2023 at 14:20
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All power-extraction machines that work with moving water are driven by the head or pressure difference between the inlet and outlet sides of the machine. the driving head is the sum of two components: the static pressure head caused by the height difference between the inlet and the outlet, and the dynamic velocity head developed by the speed of the water through the device.

An undershot water wheel (like a sternwheeler steamboat) creates thrust by speeding up the water going through the wheel (i.e., by adding to its velocity head), or it can extract work from the flow (by subtracting from the velocity head flowing through it)- both of which it does at essentially zero pressure head as the flow regime in this case is open-channel flow.

This means that an undershot water wheel dipped into the flow of an unobstructed open channel actually does slow down the water passing through it, and that slow wake then mixes with the unobstructed flow nearby and eventually picks up speed again as gravity acts on it. The buildup or increase in water level immediately upstream of the wheel is very small and the decrease in water level immediately downstream of the wheel is also very small.

BTW in the opposite case of a flow consisting of hundreds of PSI of pressure head between the inlet and the outlet, relatively small mass flow rates, and closed-channel flow, the power pulled from the stream is almost entirely from the pressure head and not the velocity head. An example is the pelton wheel, which can have a thousand feet or more of pressure head on the inlet and atmospheric at the outlet.

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