Posted by pete on March 1, 2009, 4:45 pm
Hi,
Can you all let me know if my understanding is correct:
If I have a device that consumes 120W of power and its needs to run
for 2hrs at night, then I will need the following:
- a battery that runs at 12v with a minimum capacity of 200Ah (120W/
12v * 2hrs) or a 6v battery with a capacity of 400Ah
- a 120W PV module in an area with 2hrs of solar insulation...or a 60W
PV module placed in an area with 4hrs of solar insulation?
I realize, I will need all of the auxiliary stuff like a charge
controller and what not, but I'm just wondering if my math is on the
right track.
Thanks,
Pete
Posted by Ron Rosenfeld on March 1, 2009, 6:11 pm
On Sun, 1 Mar 2009 13:45:26 -0800 (PST), pete@p373.net wrote:
>Hi,
>Can you all let me know if my understanding is correct:
>If I have a device that consumes 120W of power and its needs to run
>for 2hrs at night, then I will need the following:
>- a battery that runs at 12v with a minimum capacity of 200Ah (120W/
>12v * 2hrs) or a 6v battery with a capacity of 400Ah
>- a 120W PV module in an area with 2hrs of solar insulation...or a 60W
>PV module placed in an area with 4hrs of solar insulation?
>I realize, I will need all of the auxiliary stuff like a charge
>controller and what not, but I'm just wondering if my math is on the
>right track.
>Thanks,
>Pete
Sort of. But 120/12 * 2 = 20 and not 200.
You need to account for losses in your system; and you also need to account
for the fact that you don't want to discharge a battery 100%.
Let us assume that your 120 Watt load is DC, and your nominal system
voltage is 12V
120 watts * 2 hrs/day = 240 watt-hrs/day
12V System nominal voltage -- 20Ah
While you might expect, with 4 hrs insolation, a 60W panel to be
sufficient, the fact of the matter is that, for most panel ratings, the 60W
is quite optimistic (due to the standard used to determine the output of
the panel). So you might want to derate this 20% or so to account for
other than STC conditions, aging panels, dirt, losses in battery charging,
etc. Panel array size might be 60/0.8 or 75 watts.
And your adjusted total daily consumption would now by 24Ah
With regard to the batteries, the type you might use in a small, 12V system
might be able to be discharged to 50%. (Higher quality batteries might be
OK down 80%, but you get the point).
We've determined that your daily load is about 24Ah, so, to discharge the
batteries no more than 50%, you would need a 48Ah bank.
However, you also have to account for numbers of days without any sunlight.
That 4 hr figure is an average over the course of a month; it is not a
guaranteed daily figure.
For most places, it is relatively safe to figure on a 5 day battery bank.
So 48 * 5 = 240Ah bank.
If you are using AC, you will also need to account for inverter related
losses.
--ron
Posted by Martin Riddle on March 1, 2009, 8:19 pm
> On Sun, 1 Mar 2009 13:45:26 -0800 (PST), pete@p373.net wrote:
>>Hi,
>>
>>Can you all let me know if my understanding is correct:
>>
>>If I have a device that consumes 120W of power and its needs to run
>>for 2hrs at night, then I will need the following:
>>
>>- a battery that runs at 12v with a minimum capacity of 200Ah (120W/
>>12v * 2hrs) or a 6v battery with a capacity of 400Ah
>>- a 120W PV module in an area with 2hrs of solar insulation...or a 60W
>>PV module placed in an area with 4hrs of solar insulation?
>>
>>I realize, I will need all of the auxiliary stuff like a charge
>>controller and what not, but I'm just wondering if my math is on the
>>right track.
>>
>>Thanks,
>>Pete
> Sort of. But 120/12 * 2 = 20 and not 200.
> You need to account for losses in your system; and you also need to
> account
> for the fact that you don't want to discharge a battery 100%.
> Let us assume that your 120 Watt load is DC, and your nominal system
> voltage is 12V
> 120 watts * 2 hrs/day = 240 watt-hrs/day
> 12V System nominal voltage -- 20Ah
> While you might expect, with 4 hrs insolation, a 60W panel to be
> sufficient, the fact of the matter is that, for most panel ratings,
> the 60W
> is quite optimistic (due to the standard used to determine the output
> of
> the panel). So you might want to derate this 20% or so to account for
> other than STC conditions, aging panels, dirt, losses in battery
> charging,
> etc. Panel array size might be 60/0.8 or 75 watts.
> And your adjusted total daily consumption would now by 24Ah
> With regard to the batteries, the type you might use in a small, 12V
> system
> might be able to be discharged to 50%. (Higher quality batteries
> might be
> OK down 80%, but you get the point).
> We've determined that your daily load is about 24Ah, so, to discharge
> the
> batteries no more than 50%, you would need a 48Ah bank.
> However, you also have to account for numbers of days without any
> sunlight.
> That 4 hr figure is an average over the course of a month; it is not a
> guaranteed daily figure.
> For most places, it is relatively safe to figure on a 5 day battery
> bank.
> So 48 * 5 = 240Ah bank.
> If you are using AC, you will also need to account for inverter
> related
> losses.
> --ron
I agree with Ron, but I would derate the overall system %30, not %20. So
instead of 60W use 42W etc.
Cheers
Posted by Synth on March 1, 2009, 8:46 pm
> > On Sun, 1 Mar 2009 13:45:26 -0800 (PST), p...@p373.net wrote:
> >>Hi,
> >>Can you all let me know if my understanding is correct:
> >>If I have a device that consumes 120W of power and its needs to run
> >>for 2hrs at night, then I will need the following:
> >>- a battery that runs at 12v with a minimum capacity of 200Ah (120W/
> >>12v * 2hrs) or a 6v battery with a capacity of 400Ah
> >>- a 120W PV module in an area with 2hrs of solar insulation...or a 60W
> >>PV module placed in an area with 4hrs of solar insulation?
> >>I realize, I will need all of the auxiliary stuff like a charge
> >>controller and what not, but I'm just wondering if my math is on the
> >>right track.
> >>Thanks,
> >>Pete
> > Sort of. But 120/12 * 2 = 20 and not 200.
> > You need to account for losses in your system; and you also need to
> > account
> > for the fact that you don't want to discharge a battery 100%.
> > Let us assume that your 120 Watt load is DC, and your nominal system
> > voltage is 12V
> > 120 watts * 2 hrs/day = 240 watt-hrs/day
> > 12V System nominal voltage -- 20Ah
> > While you might expect, with 4 hrs insolation, a 60W panel to be
> > sufficient, the fact of the matter is that, for most panel ratings,
> > the 60W
> > is quite optimistic (due to the standard used to determine the output
> > of
> > the panel). So you might want to derate this 20% or so to account for
> > other than STC conditions, aging panels, dirt, losses in battery
> > charging,
> > etc. Panel array size might be 60/0.8 or 75 watts.
> > And your adjusted total daily consumption would now by 24Ah
> > With regard to the batteries, the type you might use in a small, 12V
> > system
> > might be able to be discharged to 50%. (Higher quality batteries
> > might be
> > OK down 80%, but you get the point).
> > We've determined that your daily load is about 24Ah, so, to discharge
> > the
> > batteries no more than 50%, you would need a 48Ah bank.
> > However, you also have to account for numbers of days without any
> > sunlight.
> > That 4 hr figure is an average over the course of a month; it is not a
> > guaranteed daily figure.
> > For most places, it is relatively safe to figure on a 5 day battery
> > bank.
> > So 48 * 5 = 240Ah bank.
> > If you are using AC, you will also need to account for inverter
> > related
> > losses.
> > --ron
> I agree with Ron, but I would derate the overall system %30, not %20. So
> instead of 60W use 42W etc.
> Cheers
cool, thanks to both of you for your input
>Can you all let me know if my understanding is correct:
>If I have a device that consumes 120W of power and its needs to run
>for 2hrs at night, then I will need the following:
>- a battery that runs at 12v with a minimum capacity of 200Ah (120W/
>12v * 2hrs) or a 6v battery with a capacity of 400Ah
>- a 120W PV module in an area with 2hrs of solar insulation...or a 60W
>PV module placed in an area with 4hrs of solar insulation?
>I realize, I will need all of the auxiliary stuff like a charge
>controller and what not, but I'm just wondering if my math is on the
>right track.
>Thanks,
>Pete