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September 23, 2020
If you mean energy (which is what you sell), read How much energy could I generate from a hydro turbine?.
If you mean power, read on.
Power is the rate of producing energy. Power is measured in Watts (W) or kiloWatts (kW). Energy is what is used to do work and is measured in kilowatt-hours (kWh) or megawatt-hours (MWh).
In simple terms, the maximum hydropower power output is entirely dependent on how much head and flow is available at the site, so a tiny micro-hydro system might produce just 2 kW, whereas a large utility-scale hydro system could easily produce hundreds of Megawatts (MW). To put this in context, a 2 kW hydropower system could satisfy the annual electrical energy needs of two average UK homes, whereas a utility-scale 200 MW system could supply 200,000 average UK homes.
If you don’t mind equations the easiest way to explain how much power you could generate is to look at the equation for calculating hydropower:
P = m x g x Hnet x η
P：power, measured in Watts (W).
m：mass flow rate in kg/s (numerically the same as the flow rate in litres/second because 1 litre of water weighs 1 kg)
g：the gravitational constant, which is 9.81m/s2
Hnet：the net head. This is the gross head physically measured at the site, less any head losses. To keep things simple head losses can be assumed to be 10%, so Hnet=Hgross x 0.9
η：the product of all of the component efficiencies, which are normally the turbine, drive system and generator
For a typical small hydro system the turbine efficiency would be 85%, drive efficiency 95% and generator efficiency 93%, so the overall system efficiency would be:
0.85 x 0.95 x 0.93 = 0.751 i.e. 75.1%
Therefore, if you had a relatively low gross head of 2.5 metres, and a turbine that could take a maximum flow rate of 3 m3/s, the maximum power output of the system would be:
First convert the gross head into the net head by multiplying it by 0.9, so:
Hnet = Hgross x 0.9 = 2.5 x 0.9 = 2.25 m
Then convert the flow rate in m3/s into litres/second by multiplying it by 1000, so:
3 m3/s = 3,000 litres per second
Remember that 1 litre of water weighs 1 kg, so m is the same numerically as the flow rate in litres/second, in this case 3,000 kg/s.
Now you are ready to calculate the hydropower power:
Power (W) = m x g x Hnet x η = 3,000 x 9.81 x 2.25 x 0.751 = 49,729 W = 49.7 kW
Now, do the same for a high-head hydropower site where the gross head is 50 metres and maximum flow rate through the turbine is 150 litres / second.
In this case Hnet = 50 x 0.9 = 45 m and the flow rate in litres/second is 150, hence:
Power (W) = m x g x Hnet x η = 150 x 9.81 x 45 x 0.751 = 49,729 W = 49.7 kW
What is interesting here is that for two entirely different sites, one with a net head of 2.25 metres and the other 45 metres, can generate exactly the same amount of power because the low-head site has much more flow (3,000 litres / second) compared to the high-head site with just 150 litres/second.
This clearly shows how the two main variables when calculating hydropower power output from a hydropower system are the head and the flow, and the power output is proportional to the head multiplied by the flow.
Of course the two systems in the example above would be physically very different. The low head site would need a physically large Archimedean Screw or Kaplan turbine inside a turbine house the size of a large garage because it would have to be physically large to discharge such a large volume of water with a relatively low pressure (head) across it. The high-head site would only need a small Pelton or Turgo turbine the size of a fridge because it only has to discharge 5% of the flow rate of the low-head system and under a much higher pressure.
It is interesting that in the real world the heads and flows in the example above aren’t too far from reality, because high-head sites tend to be at the heads of rivers in upland areas, so the ground slopes steeply enabling high heads to be created, but the rainfall catchment of the watercourse is relatively small, so the flow rate is small. That same upland stream 20 km downstream would have merged with countless small tributaries and formed into a much larger river with a higher flow rate, but the surrounding area would now be lowland agricultural land with only a modest gradient. It would only be possible to have a low head across a weir to avoid risking flooding the surrounding land, but the flow rate in the lowland river would be much larger to compensate.
The UK has a range of all types of high, medium and low head hydropower sites. England has more low-head sites, Scotland more high-head, and Wales a mixture of everything but still with significant medium and high-head opportunities.
Power and energy generation can be maximised by keeping the inlet screen clear of debris which maintains a maximum system head. This can be automatically acheived using our innovative GoFlo Travelling screen manufactured in the UK by our sister company . Discover the benefits of installing a GoFlo travelling screen on your hydropower system in this case study: Maximising the benefits of hydropower technology using innovative GoFlo travelling screen technology.