# All Kinds of Turbines

Dave asked an interesting question, commenting on the heat-from-the-tunnel project:

Has anyone considered the fact that the water can be used to first drive turbines and then distributed to supply the input source for the heat pumps?

I am a water turbine fan, and every time I spot a small hydro power plant on a hiking map, I have to find it. Pelton turbine. The small regional utility has several of them, the flow rate is typically a few 100 liters per second. The NSA should find an image of myself in the reflections.

This does not mean I have developed intuition for the numbers, so I have to do some cross-checks.

You can harvest either kinetic or potential energy from a flowing river in a hydro power plant. Harvesting kinetic energy could be done by something like the ‘under-water version’ of a wind turbine:

The tunnel produces a flow of 300 liters per second but this information is not yet sufficient for estimating mechanical power.

The kinetic energy of a mass $m$ moving at velocity $v$ is: $\frac{mv^{2}}{2}$. From the mean velocity in a flow of water we could calculate the energy carried by flow by replacing $m$ in this expression by mass flow.

If 300 liters per second flow through a pipe with an area of 1 m2, the flow velocity is equal to  0,3 m3/s divided by this area, thus 0,3 m/s. This translates to a kinetic energy of: $\frac{300 ^{.} 0,3^{2}}{2}$ W = 13,5 W

… only, just enough for a small light bulb.

If the cross-section of the pipe would be ten times smaller, the power would be 100 times larger – 1,35 kW.

(Edit: This is just speculating about typical sizes of the natural pipe determined by rocks or whatever. You cannot create energy out of nothing as increasing velocity by a sort of funnel would decrease pressure. I was rather thinking of a river bed open to ambient air – and ambient pressure – than a closed pipe.)

On the other hand, if that water would be allowed to ‘fall’, we could harvest potential energy: Also this mill wheel is utilizing potential energy from the height difference of a few meters. (Critically inspected by The Chief Engineer, photo by elkement)

This is how commercial hydro power plants work, including those located at rivers in seemingly flat lowlands.

The potential energy of a point mass at height $h$ is $mgh$, $g$ being the acceleration due to gravity (~ 10m/s2). Assuming a usable height of 10m, 300kg/s would result in about

300 . 10 . 10 W = 30kW – quite a difference!

Of course there are huge error bars here but the modest output of kinetic energy is typical for the topography of planet earth.

Mass flow has to be conserved, and it enters both expressions as a factor. If I am interested in comparing potential and kinetic energies relative to each other, it is sufficient to compare $\frac{v^{2}}{2}$ to $gh$.

Cross-checking this for a flow of water we know more about:

The Danube flows at about 3-10 m/s, so $\frac{v_{Danube}^{2}}{2}$ = 4,5 – 50m2/s2

But we cannot extract all that energy: The flow of water would come to a halt at the turbine – where should the water go then? For the same reasons there is a theoretical maximum percentage of wind power that turbines can harvest, even if perfectly frictionless.

In addition, such a turbine would need to be much smaller than the cross-section of the river. Mass flow needs to be conserved: when part of the water slows down, it gets spread over a larger cross-section.

So the realistic $\frac{v_{Danube}^{2}}{2}$ will be smaller.

I have stumbled upon an Austrian startup offering floating turbines, designed for operations in larger rivers and delivering about 70kW at 3,3m/s flow velocity (Images on the German site). This is small compared to the overall kinetic energy of the Danube of about several MW, calculated from 2.000m3/s (mass flow near Vienna) and about 3m/s. The first hydro power plant at the Danube in Austria, built in 1959 – an icon of post World War II reconstruction (Wikimedia). The plant is currently modernised, the rated power will be increased by 5% to 250MW. Utilized difference in height: 10m.

So the whole kinetic energy – that cannot be extracted anyway – is still small compared to the rated power of typical power plants which are several 100MW!

If the water of the Danube ‘falls’ about 10m then $gh_{Danube}$ ~ 100

… which is much larger than realistic values of $\frac{v_{Danube}^{2}}{2}$! Typical usable kinetic energies are lower than typical potential energies.

So if tunnel drain water should drive a turbine, the usable height is crucial. But expected powers are rather low compared to the heat power to be gained (several MW) so this is probably not economically feasible.

I was curious about the largest power plants on earth: Currently the Chinese Three Gorges Dam delivers 22GW. I have heard about plans in Sweden to build a plant that could deliver 50GW – a pumped hydro storage plant utilizing a 50km tunnel between two large lakes, with a difference in altitude of 44m (See the mentions here or here.)