Design of hybrid power supply system

I have a laptop of 19.5V 3.34A
and a receiver of 14V 4.5A..
i need to design a hybrid power supply system using solar panel and a windmill...what could be the specifications i should go for a solar panel and a battery required? also the windmill?

Suppose you want to use your equipment for 5 hours. It will consume (19.5 x 3.34) + (14 x 4.5)... Multiply the sum x 5...

Equals 641 Watt-hrs. (You will not necessarily use the maximum every moment but this is only a rough calculation.)

A 24-volt system will serve your voltage needs. It must be able to provide 7.84 A for 5 hours. The total A-Hrs is 39.2.

Suppose you get a charge session of 10 hours from your PV panel (windmill) per day. You need to replenish 641 W-Hrs. Therefore at 24V your source must provide 2.67 A for 10 hrs.

Again, just a rough calculation. Notice there's an important difference between W-hrs, and A-hrs.

Feel free to check my figures, they could be wrong.

I would only add to Brad's excellent analysis to take into account the convers1on and storage efficiencies.

I would -roughly- assume an overall efficiency of 0.7 to 0.8, meaning that Brad's calculations require to be increased by a factor of 25 to 40%

how did you calculate that i will need a 24Volt system...and how 7.84Amp is needed? could u elaborate on the formula and calculation part?

vinit13 said:

how did you calculate that i will need a 24Volt system

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A 12V system may be suitable, if you wish. It has a greater share of the market of appliances, inverters, power sources, tutorial support, etc. (24V has a small market.)

But then you must step up 12V to your specified 19.5 and 14V. There are converters (boost converter for example), but such converters lose some energy due to inefficiency.

It is best if you compare the options. Example, one option is to get three 6V batteries in series. Their real voltage might add up to 19.2V total, which should be okay to run your laptop directly. However 19V is unlikely to be compatible with your PV panels and windmill.

Therefore it is an option worth considering, to jump up to 24V.

and how 7.84Amp is needed? could u elaborate on the formula and calculation part?

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3.34 + 4.5 = 7.84. Approximate figures are okay in the planning stage. Later you can decide whether you wish to have separate supplies for your equipment.

Do you find these values on the labels? Are they maximum figures? You must test your equipment with meters if you want true measurements.

To get 19.5V at 7.84 A:

12V power source will need to provide 13A.

24V power source will need to provide 7A.

These are rough calculations. You must add some allowance for losses (as Schmitt trigger points out in post #3).

yes these figures are mentioned on the adapters of the devices...

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Laptop Specs: 19.5V x 3.34A = 65.13W
Receiver Specs: 15V x v4A=63W
Total Watts: 128Watts
Every solar panel has some losses even if they are receiving sunlight at the peak hours of the day. Generally 20% are to be considered. So we cannot use a solar panel of 128 Watt.
Losses=128 x 20%=25.6W
Therefore we need solar panel of higher power rating which could provide 128+25.6=153W
Solar Panels come in different specifications depending upon the power required. As mentioned below in specification, we can consider 2 panels of 100W in parallel.
Considering we need to use these for 6Hours Max: 100x6=768W Per day
Value in kW: 768/1000=0.76kW per day
So we need 0.76kW for 6 hours
So we need to choose a battery which can provide at least 100Ah.

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are my calculations appropriate?

vinit13 said:

Laptop Specs: 19.5V x 3.34A = 65.13W
Receiver Specs: 15V x v4A=63W
Total Watts: 128Watts
Every solar panel has some losses even if they are receiving sunlight at the peak hours of the day. Generally 20% are to be considered. So we cannot use a solar panel of 128 Watt.
Losses=128 x 20%=25.6W
Therefore we need solar panel of higher power rating which could provide 128+25.6=153W

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This is fine. You calculated units in Watts, which is the correct method. You allowed a reasonable percentage for losses.

Considering we need to use these for 6Hours Max: 100x6=768W Per day

Click to expand...

When you start figuring in terms of hours, your units need to have 'Hrs' in it.

Since you arrived at 153W above, then it makes sense if we continue with that.

Therefore 153W for 6 hrs, results in 918 W-Hrs (per day).

Value in kW: 768/1000=0.76kW per day
So we need 0.76kW for 6 hours
So we need to choose a battery which can provide at least 100Ah.

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are my calculations appropriate?

Click to expand...

Sorry, can't see where you got the 768 figure.

Anyway using my 918 W-Hr figure, you divide by 12 to get A-hrs from a 12V battery.

i need for 6 hours thats why i multiplied 153W by 6. so i need it for 6 hours only in a day. thats y 768. and then i converted it in kW by 1000.

?... As a check, divide 768 by 6, and you get 128 (not 153).

Anyway you're getting the idea.

Since you are considering having a wind generator, then you ought to consult a weather bureau, or almanacs, regarding wind speed and dependability. You'll want almost constant winds, in order to make a wind generator worthwhile. The higher you can mount the turbine, the better, in terms of wind speed and constancy.

Do you have a stream nearby? Constant water flow over a dam, 2 or 3 feet high? A hydro generator is reliable and constant.

no..no dam...but the wind is quite wild....at times i feel it can generate enough of energy to power up a battery