How many panels do I need?

[Doubledee]

I want to power a 1kva load for 24 hours, how many 215watt panels will I need?

[Mike90250]

Lots of panels, and batteries.

I'll asume 1KVA is in reality, a 1,000 watt load, continous run 24/7. That's about what my small, 1/2 hp pond pump draws.

I'll assume, in winter you only get 3 hours of sun, and will run a genset daily to make up for the missing solar, or cloudy days. To do this entirely on solar will be very expensive.

Spring/fall, lets use 4 hours good sun.
1,000w = 24KW hours in 1 day.
if we allow the battery bank to discharge to 50%, we need a 48KWH battery bank
At 48VDC, that's 1,000AH capacity. BIG batteries
Need 4 of these : http://store.solar-electric.com/crinba.html
Crown Battery, 1090 AH @ 20 hr rate, 12 volt, 850 lbs and a forklift to move them in with. Or get LOTS of smaller ones, and all the messy wires to connect them together.

Recharge - you have 4 hours to harvest about 35KWH of power (there are losses in charging and inverting the power)
35KWH / 4 = 12KW of installed PV (again, more losses, need 12KW of PV to harvest 9KW of power. here's a link to a 10KW kit - leave out the GT inverter, about $32K !
http://lib.store.yahoo.net/lib/wind-...kw-gridtie.pdf but you will need battery chargers instead.

Or get an extenstion cord !!

[Sunking]

I want to power a 1kva load for 24 hours, how many 215watt panels will I need?

A truck load of batteries and panels.

Around 56 solar panels, and 10,000 pounds of batteries. About $70,000 worth plus shipping. Any more questions?

If it were me I would just pay the utility company $90 per month and smile all the way to the bank.

[Gopal]

1 KVA = 0.8 KW

hence it will be 800watts, considering for 24hrs service..
we get 800X24=19200 watt hrs.

for a 215w panel we need to design a system of 24volts..

hence, 19200(watt hrs) / 24 (volts) = 800 amps.

considering a 210 watt Mitsubishi panel,
has a current max of 7.92amps = 8amps (appox)

thus we need 800/80=100 panels.

100 panels of 210watts would be a sufficient.

[we need to design a charge controller, battery and an inverter of the system to be complete]

[Gagan]

1 KVA = 0.8 KW

hence it will be 800watts, considering for 24hrs service..
we get 800X24=19200 watt hrs.

for a 215w panel we need to design a system of 24volts..

hence, 19200(watt hrs) / 24 (volts) = 800 amps.

considering a 210 watt Mitsubishi panel,
has a current max of 7.92amps = 8amps (appox)

thus we need 800/80=100 panels.

100 panels of 210watts would be a sufficient.

[we need to design a charge controller, battery and an inverter of the system to be complete]

i agree with what gopal said .. 100 panels will be reqd..

[Sunking]

1 KVA = 0.8 KW

Just exactly how did you come up with .8 Kw.? No PF was given. nor did you take into account any efficiencies of a battery system.

[Gopal]

1kva=kw*power factor
KVA is the non power measure of the voltage multiplied by the amperes.

KVA is not a measure of true power it is a measure of the level of apparent power.

To convert from apparent power to true power, you must take the KVA and multiply it by the power factor. For example, 100 KVA of measured apparent power serving an inductive load with a power factor of .8 would result in a real power of 80 KW.

If Kilowatts (KW) are the measure of true or real power available for work then KVA is a measure of apparent power needed to get the true power to the work.For direct current systems, there is no difference.

However, because the voltage and amperage in an alternating current system is constantly changing, the wattage is lower than the volt-amps, because when the power goes near zero, no power is actually transmitted.

The ratio of watts to volt-amps is called power factor.

[Gagan]

1kva=kw*power factor
KVA is the non power measure of the voltage multiplied by the amperes.

KVA is not a measure of true power it is a measure of the level of apparent power.

To convert from apparent power to true power, you must take the KVA and multiply it by the power factor. For example, 100 KVA of measured apparent power serving an inductive load with a power factor of .8 would result in a real power of 80 KW.

If Kilowatts (KW) are the measure of true or real power available for work then KVA is a measure of apparent power needed to get the true power to the work.For direct current systems, there is no difference.

However, because the voltage and amperage in an alternating current system is constantly changing, the wattage is lower than the volt-amps, because when the power goes near zero, no power is actually transmitted.

The ratio of watts to volt-amps is called power factor.

gopal considering 1kva=.8 kwa
now 1 kva load for 24 hours means 24000kva i.e .8*24000 =19200kwa

now considering that the peak hours i.e when the panel draws maximum sunlight is 5 hours(it may vary) so accordingly 19200/5=3840 watts of solar panels are required

now "doubledee" asked how any 215 watt panels he requires .
which means 3840/215=17.86 or we can say 18 solar panels are required to cope up with the load..

[solar_dave]

gopal considering 1kva=.8 kwa
now 1 kva load for 24 hours means 24000kva i.e .8*24000 =19200kwa

now considering that the peak hours i.e when the panel draws maximum sunlight is 5 hours(it may vary) so accordingly 19200/5=3840 watts of solar panels are required

now "doubledee" asked how any 215 watt panels he requires .
which means 3840/215=17.86 or we can say 18 solar panels are required to cope up with the load..

That would be if you could actually get 215 watts out of a panel. More likely you should derate by 0.77, then if your using batteries the derate is more like 0.55. Better plan on about double the panels.

[Sunking]

The ratio of watts to volt-amps is called power factor.

I am a electrical engineer so I know what Power Factor is, I work with it every day. But you did not answer the question, the PF was not given.

Next in a battery system even though the reactive power is not real power, the batteries and panels must generate the VARS + the real power. The reactive power will be burned off as heat on the wring between the inverter and load.

Next is the efficiency of the whole system where at best for a battery system is 66%, so the panels must generate 18 Kwh / .66 = 27.2 Kwh. So if in winter his solar insolation is 3 hours he needs a 27.2 Kwh / 3 hours = 9.09 Kw or 9090 watts of solar panels. So if using 210 watt panels he needs 9090 / 210 = 43 panels.

[Mike90250]

I am a electrical engineer so I know what Power Factor is, I work with it every day. But you did not answer the question, the PF was not given.

Next in a battery system even though the reactive power is not real power, the batteries and panels must generate the VARS + the real power. The reactive power will be burned off as heat on the wring between the inverter and load.

Next is the efficiency of the whole system where at best for a battery system is 66%, so the panels must generate 18 Kwh / .66 = 27.2 Kwh. So if in winter his solar insolation is 3 hours he needs a 27.2 Kwh / 3 hours = 9.09 Kw or 9090 watts of solar panels. So if using 210 watt panels he needs 9090 / 210 = 43 panels.

I love it when a plan comes together ! And nobody was killed !

43 real life panels is about double of 18 lab spec panels.
Batteries not included.

[russ]

You have to admit his calculator works even if it doesn't know what it is calculating.