0
\$\begingroup\$

I have a 220v/50Hz/420w motor, I'm planning to use the simple design
Solar Panel --> Charge Controller --> Inverter --> Motor

and i calculate battery requirements as follows:
P = I * V,
420 = I * 220,
I = 1.9A ~ 2A

does this mean for a 40AH Battery, my motor will run for 20 Hours (10 Hours to not go below 50% of the Battery) ?

I think my calculations are wrong thanks.

\$\endgroup\$
  • \$\begingroup\$ NBNBNB!!!! Look at www.gaisma.com . Find page for your location. Googling eg: gaisma houston for Houston normally takes you there directly. Figures in chart of kWh/m^2/day is effectively "hours of full sun per average day" by month. These are the hours of effective sunlight available. Us these when calculating panel energy available. Note these are AVERAGE for the month concerned. || GooBling: gaisma houston gives gaisma.com/en/location/houston-texas.html as 1st hit. etc \$\endgroup\$ – Russell McMahon May 12 '14 at 2:31
3
\$\begingroup\$

Your answer would be approximately correct only if your battery voltage was ~= 220V.
As that is probably not what you had in mind, your answer is probably wrong.

IF you are using a lower voltage battery then you need to use energy rather than current capacity to calculate equivalent amounts at different voltages.

A 420 Watt motor requires 420 Wh (Watt Hours) of energy per hour. That's the easy part :-).

The Wh capacity of a battery is V_battery x Ah_capacity_battery.
Battery energy capacity in Wh = V x Ah
If voltage is changed energy will be lost in the process.
Call the efficiency factor Zbm = Z_battery_motor, where Zbm < 1.
eg if Zbm = 0.85 then 15% of the energy is lost during conversion.
figure of Zbm = 0.8 is an OK starting point from say 12V or 24V to 220V.

Motor energy requirement per hour = Wm = Vbat x Ibat / Zbm Ibat = Wmotor / Vbat / Zbm For a 24 V battery Ibat = Wm/Vb/Zbm = 420/24/.8 ~= 22A.
To run the system for one hour you'd need a nominally 22Ah battery. BUT batteries are usually rated at the 10h rate or some other period of some hours, so at the 1h rate they will have much lower capacity. You would need to look at specifications for the battery you had in mind but a factor of say 1.5 is probably not overly pessimistic. So Ah required from 1h operation ~= 1.5 x 22 = 33Ah.
If you want to not go below 50% capacity (which is very wise for lead acid batteries) you need double that so say 66Ah.

So your originally stated 40 Ah battery would run a 420W motor at 220V for about 60 minutes x 40/66 ~+ 36 minutes IF it was a 24V battery. A 12V battery with the same assumptions would run the motor for about 18 minutes.

A lot depends on assumptions made.
Above I have used:

  • Up converter efficiency low voltage to 220V = 80%

  • Battery 1h Ah rate = Specification sheet rate / 1.5

  • Discharge depth = 50%

  • 12V or 24V system.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ thanks, and for charging, the solar panel produces 25W, 17v, so sunshine for 7 hours (25*7*0.8) = 140 WH, so for 12v/62AH battery, it would require (12 * 62 = 744 WH required to charge the battery) so (744 / 140 = 5.3 Hours) until it's completley charged? \$\endgroup\$ – user41468 May 11 '14 at 11:19
  • \$\begingroup\$ Sorry, no. You missed a factor. 140 WH is per day, not per hour. So your simple calculation says 5.3 days. And assuming an equivalent of 7 hours per day full sunlight is wildly optimistic. You might come close with a tracker which keeps the panel pointed toward the sun, but not otherwise. Depending on your latitude and the time of year, 4 hours per day may be more like it. In which case, it's more like 9 days. \$\endgroup\$ – WhatRoughBeast May 11 '14 at 11:46
  • \$\begingroup\$ oh, thank you, so 9 days consisting of 4 hours of charging each, or 5.3 days consisting of 7 hours of charging each. thanks a lot, i understood it perfectly now :), oh sorry, one last thing, those 9 days, divided by 2? since battery is only discharged to 50%? so only 50% needs refilling? \$\endgroup\$ – user41468 May 11 '14 at 12:04
  • \$\begingroup\$ Yes, sort of. This assumes 100% charging efficiency - and you won't get that. The 7 hours assumes some sort of tracking unit for the solar panels - just letting them sit there won't work. And the factor of 2 is well-spotted. \$\endgroup\$ – WhatRoughBeast May 11 '14 at 13:23
  • \$\begingroup\$ so if i got a second panel? in parallel with this one? this will reduce the charging time to half? same with 4 panels? \$\endgroup\$ – user41468 May 11 '14 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.