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Gabriel_2018

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Everything posted by Gabriel_2018

  1. You can use a resistor, 2200 Ohm 2 W, and quickly reconect.
  2. But, in my opinión and if It is possible, It is better to control density of each cell, when It is fully charged.
  3. Hi Lionheart! You can accept 10 mV per cell, so if your batteries are three cell, a good value is 30 mV.
  4. Ya, another brick in the wall... It is a cylindrical fuse. You can try a NH Fuse.
  5. I think It could be done in a Big scale. Thousands of panels close to a hydro-electric plant.
  6. You are right about using a gear or a servo valve. But there is a problem, you have to consider that the source is delayed respect the load. So, if I were you, I would install a small battery in order to stabilize the system, targetting battery Voltage using a PID control. Also, as Elbow mention, you need a Big mass of water. Finally, with a PSMM pump and a Kaplan turbine you could have +-60% efficiency.
  7. A spicy fuse then...
  8. What kind of fuse is it? Sometimes cylindrical fuses damages are caused by temperature at nominal, or lower, current. You could use a NH fuse if it happens again.
  9. Hi PanB, i´ll tell you my PlanA for this kind of installs. Here, in Spain, is a very repetitive case when someone who has a house and some trees, need a borehole to water. We are going to suppose your pump is a 1.5 kW 3X400 VAC motor. In that case, peak start current would be 6 or 8 times nominal current, so you would need a very powerful low frecuency, which use to be not cheap. SMA, Schneider... What I ussually do is to install an isolated transformer, which rises voltage from monophase inverter to 380-400 V, which feed a VFD, driver, which obtains 3x400 VAC from 1x400 Vac, that comes from the output of transformer. This scheme allows you to reduce start current to 1.5-2.0 nominal current. Also, it allows you to modify the frecuency of pump motor, in order to adjust solar power to pump power. As example, if your pump is taking 2 Kw at 50 Hz, and you modify frequency via VFD to 45 Hz, the power needed would reduce to (45/50)x(45/50)*(45/50)x2 Kw = 1.45 Kw. 80
  10. Gabriel_2018 replied to Gordon's topic in Inverters
    I thouht the same, but... Even using a boost, it wouldn´t be enough to maintain quiescent draw with only one panel!
  11. Gabriel_2018 replied to Gordon's topic in Inverters
    Only one? If It is a 36 cell panel, you will need, at least, four pannels.
  12. It is a good feature, if i remenber well, SMA can´t do it
  13. Then, if we have 3 Multiplus in a triphase system, can we use a monophase generator to charge batteries? @plonkster
  14. Well, since this kind of meters are cheap, a good choice can be to have a pair of them. One for PV and other for grid.
  15. If it is time for math ∫Pdt + ∫P2dt + ∫P3dt = ∫(P1+P2+P3)dt 😉
  16. I think the procedure is the same and the only difference is in the way they show those "Acumulated Data"... It can be seen, as example, in this protocol. KPM33+MODBUS+PROTOCOL (1).pdf
  17. I would say it is the same...
  18. I agree @plonkster 51.8 V, or 3.45 V per cell, is enough to have a 80% of LiFePo4 capabilities.
  19. To be hot or not, is that the question?
  20. Yes it is. You have to consider cloud edge effect, which can add 20-40% extra radiation.
  21. I would say one cell, or some, is short. What about warranty?
  22. Yes. It should be "How to connect solar pannels to inverter"
  23. 47 V is Vo panel. That´s the voltage of the panel in open circuit. 38 V is Vm panel. That´s the voltage in max power point.

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