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egge24 October 23, 2012 11:58

Volume flow rate through actuator disk (Tidal turbine)

Im modelling an array of 3 marine current turbines using a porous jump boundary condition. The domain is a rectangular channel with 650. length, 400m width and 100m depth. Hub height is 30m from the bottom. Porous jump parameters are the same for the 3 turbines and are set to represent a thrust coeff. = 0.8. Parameter values are:

- Face permeability, m: 1e+10
- Porous medium thickness, m: 0.5
- Pressure Jum Coeff. C2, 1/m: 25

Solution converges after 260 iterations. Follows a figure showing velocity magnitude contours at hub height horizontal plane.

When I generate the reports for Volumetric and Mass Flow Rates for each turbine, I get negative values for side turbines and a positive value for center turbine. Values are as follows:

Volumetric Flow Rate (m3/s)
-------------------------------- --------------------
(Up) Turbine-1 -37.982517
(Center) Turbine-2 90.828041
(Down) Turbine-3 -38.849861
---------------- ------------------------
Net 13.995665

Mass Flow Rate (kg/s)
-------------------------------- --------------------
(Up) Turbine-1 -38932.082
(Center) Turbine-2 93098.742
(Down) Turbine-3 -39821.109
---------------- -------------------------
Net 14345.555

But surface velocity integral are very similar for each disk.

Does anyone knows why this is happening and how can I obtain real values?

Thanks in adavance.

OttawaCFD August 11, 2013 15:49


I was wondering how you managed to figure out the required pressure jump c2 coefficient of 25? I am modelling a similar tidal turbine problem and not sure what the appropriate porous jump parameters would be to model a turbine with 35% efficiency. Thanks!


egge24 August 12, 2013 10:27

Hi Derek,

The values of C2 = 25/m and a porous medium thickness of 0.5m are incorrect. I took thos values from a scientific paper, but now I disagree.
The best way to calculate C2 is using actuator disk theory. In order to do the calculations you need to assum some initial values like, free stream velocity, power coefficient and medium thickness. The procedure will be:

1 - For a given power coefficient you can find out the related axial induction factor, a.
2 - With a you get the velocity at the wake, Uw. Uw = Uo(1-2a).
3 - Knowing the free stream velocity and wake velocity you calculate the pressure jump.
4 - Finally, with the pressure difference at both sides of the disk you can apply equation 7.3-69 ( and calculate C2.

Note that the velocity in equation 7.3-69 corresponds to the flow velocity at the disk, Ud, Ud = Uo(1-a).

Once you do the run with Fluent you can check your result. By applying this methodology I get a difference with teory of 4.4%.

I hope I've helped you.



OttawaCFD August 12, 2013 12:24

Thanks Eduardo this is great! Really appreciate the help

OttawaCFD August 12, 2013 15:18

Hi Eduardo,

After calculating using the steps you gave me I got a c2 value of 0.5 with assumed thickness of 1 m. Is this within a similar range to the value you found? Thanks


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