Discussion
Started 1 March 2022

Kinetic Turbine (Hydro-Turbine)

Kinetic energy turbines, also known as free-flow turbines, produce electricity by using the kinetic energy of moving water rather than the potential energy of the head. Rivers, man-made channels, tidal streams, and ocean currents are all possible places for the systems to work. Because kinetic systems employ the natural flow of a water stream, they do not require water to be diverted through man-made channels, riverbeds, or pipelines, but they may have uses in these conduits. Because they may utilize existing infrastructure like bridges, tailraces, and channels, kinetic systems do not necessitate massive civil works.

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Bachir Achour
Université de Biskra
Most of the hydraulic turbines that are presently used for hydropower generation have been developed for installation in water dams across streams. However, dams damage the environment and interfere with fish migration. They also cannot be used for power systems extracting energy from such huge potential sources as ocean currents or lowgrade rivers. Thus, new hydraulic turbines are needed that can operate efficiently in free flow without dams. The principal difference between exploiting high-head and free flow turbines (Kinetic turbines) is that the latter need large flow openings to capture as many water masses as possible with low velocities and pressure. It was demonstrated that the efficiency of a free flow turbine is maximal when the resistance is of a free flow turbine rather small and a large part of the flow (~61 per cent) goes through.
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Power output of a micro Kaplan turbine that reduces linearly during 3 days from 16 kW to 13.65 kW with constant hydraulic and blades configurations?
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2 answers
  • Emanuele QuarantaEmanuele Quaranta
There is a micro Kaplan hydro plant:
Flow rate: 0.58 mc/s Head: 4.85 m RPM: 625 Max electrical power: 16.5 kW Generator+gearbox efficiency: 0.89 Maximum Turbine efficiency: 0.67 Draft tube: vertical/conical, 1.5 m high
The problem is the decrease in electrical power (measured during operation) with the time up to an asymptotic value, although all the external conditions (head, flow rate, rpm, blades configurations) remain constant.
I summarize here better, by using a picture (in attach). The picture depicts the power in kW versus the time in minutes.
Let's suppose that the turbine is switched off. Then it is switched on and the ELECTRICAL power reaches in few seconds 16.5 kW. But, as you can see from the picture, then it starts to decrease approaching, over three days, 13.65 kW, and remains at 13.65 kW. Then, suppose that it is again switched off and switched on (process that last few minutes, not again plotted in the figure),  again the ELECTRICAL POWER reaches 16.5 kW as at the beginning, but then it decreases approaching 13.65 kW (after 3 days) and remains 13.65 kW. If we repeat again the switch on/off (process done in few minutes), again it reaches 16.5 kW, and then decreases up to 13.65 kW. If the turbine is not switched on/off, the power remains 13.65 kW “forever” ….. until a new switched off/on is made. Therefore, the switched off/on process last few minutes, while the decreasing trend until the lowest power value of 13.65 kW lasts few days.
Have you got general ideas?
1) Formation of a big vortex in the draft tube. If this would occur, the decreasing trend should last few minutes (the time needed for the vortex development), and the power would immediately approach 13 kW. This phenomenun should last few minutes, not days. Furthermore, by CFD simulations, the vortex in the draft tube is not powerful, and at the outlet of the draft tube flow velocities are vertical.
2) Possible inclusion of air from the conical (vertical) draft tube towards the turbine. The draft tube is immersed of slightly more than 1.1 D into the downstream water level (D is maximum turbine diameter), as recommended by engineering practice. From CFD simulations flow velocities are downward, so this point should be ok.
3) Possible electro-mechanics/electrical load problem.
4) Cavitation: during operation there are not noise, neither appreciable vibrations, but there is hub erosion due to cavitation. But cavitation should reduce the maximum power output, not generate a decreasing trend during the time.
Thanks
How do I change the measurment unit in a Simscape custom component ?
Question
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  • Nils AubagueNils Aubague
I'm trying to create a custom block in Simscape that convert energy from pressurized water into a torque. Here is my code :
omponent pelton_turbine % Ce composant calcule le couple généré par l'eau sur la turbine. % 🔹 Déclaration des ports nodes H = foundation.hydraulic.hydraulic; % Port hydraulique R = foundation.mechanical.rotational.rotational; % Port mécanique rotatif end % 🔹 Déclaration des paramètres parameters eta = {0.85, '1'}; % Rendement de la turbine rho = {1000, 'kg/m^3'}; % Densité de l'eau r = {0.5, 'm'}; % Rayon moyen de la roue g = {9.81, 'm/s^2'}; % Gravité end % 🔹 Déclaration des variables internes variables Q = {0, 'm^3/s'}; T = {0, 'N*m'}; % Couple généré H_head = {0, 'm'}; % Hauteur d'eau équivalente end branches % Débit hydraulique pris directement depuis le port H Q : H.q -> *; end
equations % Calcul de la hauteur d'eau (pression convertie en mètre de colonne d'eau) H_head == H.p/ (rho * g); % Calcul du couple généré par l'eau T == {eta * rho * Q * r * sqrt(H_head * 2 * g), 'N*m'}; % Transmission du couple à l’axe mécanique R.t == T; end end
My problem is that I have this error when I try to build my component :
Invalid use of a value with unit cm^3*kg/(m*s^2) when attempting to bind a unit. The value to which the unit is bound must not have an associated unit. • In pelton_turbine.pelton_turbine (line 36) eta = 0.8500 rho = {1000, 'kg/m^3'} Q = {[1x1 double], 'cm^3/s'} r = {0.5000, 'm'} H_head = {[1x1 double], 'm'} g = {9.8100, 'm/s^2'}
I don't get why the flow rate (Q) is in cm^3/s instead of m^3/s and I don't know how to change it. Do you have an idea ?
I tried using Chatgpt but all his advices seemed to be useless. Since Q is a throught varaible Ican't define his unit. I also tried changing the units in the Configuration Parameters in my .slx file (I changed cm^3/s into m^3/s for the flow rate) but it didn't have any effect on my .ssc file and I keep getting the error.
Thanks,
Nils

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