How to calculate hydropower in kW without water head

I am trying to do some calculations to figure out how much power output would be produced from a water powered generator that uses horizontal current flow to move the generators coils rather than vertical water drop.

The formula to calculate power is: $$\text{Power} = \rho gh\,Q\eta$$

where:

$$\rho = \text{density (kg/m3) (~ 1000 kg/m3 for water)}$$

$$Q = \text{water flow (m3/s)}$$

$$g = \text{acceleration of gravity}$$

$$h = \text{falling height, head (m)}$$

$$\eta = \text{efficiency}$$

However, I am unsure how to calculate when gravity and falling head are out of the equation, as setting them to 1 or 0 will give incorrect answers.

the water flow is $3.1m^3/second$

the efficiency is $0.78$

If anyone could explain how to calculate the power output I would be extremely grateful.

Below is a link to a similar device to what I am describing: Waterwheel Documentary- Harnessing Electricity on the Zambezi River

• The answer is zero unless you have a pressure head. Apparently you have some upstream compressor/pump which is pushing the water horizontally, so just replace the pressure/force due to gravity with the pressure head. – Carl Witthoft Mar 2 '17 at 14:23

If you have horizontal water flow (no head) you are extracting its kinetic energy (from its velocity) not its potential energy (from its head).

So you need to know the velocity of that flow as well as the flow rate to calculate the energy available. (You can calculate it from the flow rate and channel dimensions)

You cannot extract all the available energy - that would imply the water stopping downstream of the turbine! For wind power (compressible fluid), the maximum energy you can extract is the Betz limit or 59% of the original kinetic energy. Something similar will apply to incompressible fluids - apparently Betz itself, as it applies to Newtonian fluids, which do include incompressible fluids - and it implies slower flow in a larger (wider,deeper) channel after the turbine.

In practice if you extract half the Betz limit (between 25 and 30% of the available energy) you're doing pretty well, but given you have a figure for efficiency, you are probably meant to calculate power as efficiency * Betz * available kinetic energy.

• Thanks this has helped however I have been reading that the efficiency of a common design is 0.78, so would the efficiency be a separate variable to the betz limit ? e.g $waterflow * density * velocity * efficiency * betz limit$ or $waterflow * density * velocity * betz limit$ – Aztec warrior Mar 2 '17 at 15:28
• Efficiency and Betz are separate variables, i.e. 100% efficiency would extract the betz limit, see last paragraph. – user_1818839 Mar 2 '17 at 15:32
• And by the way that's not the correct equation ... you do know how to calculate kinetic energy, right? – user_1818839 Mar 2 '17 at 15:42
• oh ok thanks iv incorporated both into my equation, I thought i did but if your telling me its incorrect then perhaps not, can you show me where the error is ? – Aztec warrior Mar 2 '17 at 15:51
• Use E = 0.5 * mv^2 ... so available power = dE/dt and mass flow rate is dm/dt ... – user_1818839 Mar 2 '17 at 16:45

Just measure what comes out of the generator and take an average ,peak and low.

• This is a design concept idea so i cant physically measure it. – Aztec warrior Mar 2 '17 at 14:58
• He wants to say that manufacture it. Measure the quantity, manufacture it again and repeat this process over and over again. – Fennekin Mar 7 '17 at 12:41