You have estimated the size of the solar system that you need and are ready to get the equipment from the market to install it. But wait, are you sure you have enough space in your garden or your backyard or your rooftop to install the solar panels? How can you do a rough estimate of the area required by the solar panels? Here is a quick and easy way to go about it.

Lets assume that you want to install 10 solar panels rated at 100 Watts each and having a conversion efficiency of 18%. The total power output of the solar system can be calculated as:

**Total Power Output = Total Area x Solar Irradiance x Conversion Efficiency**

We know the required Total Output Power is 1000 Watts (10 panels x 100 Watts), the Solar Irradiance for a surface perpendicular to the sun’s rays at sea level on a clear day is about 1000 Watt/m^{2} and the Conversion Efficiency is 18%. Plugging these number in the above equation we get:

**1000 Watts = Total Area x 1000 Watts/m ^{2} x 0.18**

or

**Total Area = 1000/180 = 5.56 m ^{2}**

I you are going to install all the panels in one line you would need a space of approximately 1 m x 5.56 m (each panel having a size of 1 m x 0.556 m) on your rooftop. There you go. You have a rough estimate of the space required by the solar panels of your system.

Note:

1. Do remember that solar panels are usually installed at an angle to the earth’s surface and this may change the results somewhat. For an example of detailed calculation see the following post.

2. Appliances typically operate on AC voltage, whereas, solar panel produces DC voltage and battery also operates on DC. Therefore an inverter is needed to convert DC to AC and there can be substantial losses in conversion.

3. Imagine a solar panel has a conversion efficiency of 100% i.e. it converts all the solar energy into electrical energy then all you would need is a 1 m^{2} solar panel to produce 1000 Watts of electrical energy :).

Hi. Can you provide solar system design for 5kw system step by step…

Hi Faheem, let me try.

1. So if you are using 250Watt panels you would need 20 of those (20 x 250Watts = 5000Watts)

2. Assuming Peak Sun Hours of 5 hours, energy produced would be 25KWhr (5hr x 5KW = 25KWhr)

3. If you want to use half of that energy and store half of that energy you will need 5 batteries of 200Ah each, at least

4. You will also need a medium sized inverter (depends upon load) and a charge controller

5. This system should be sufficient for a typical home with lights, fans, TV and refrigerator (but maybe not sufficient for an AC)

Hope if helps!

With the bright light conditions and the efficiency as measured, calculate the size of solar panel required to power:

▪ A radio of average power demand approximately 0.1 Watt.

For the bright light the power was 59.09 watts and the efficiency was (59.09/1)/400 = 0.15. The solar cell active area was 1m2.

How do I use this to solve the question?

Peak Radiation x Area x Efficiency = Power Produced in Watts

400*A*0.15=0.1 Watts

60*A=0.1 Watts

A=0.001667 meter squared

A=16.67 centimeter squared

Approximately 4cm x 4cm panel would do the job.

Hi, I have a question that I am stuck on and I can’t seem to find any formula to calculate it.

To illustrate the amount of solar energy available to us, calculate how many electric power plants could be closed if an area the size of Cyprus was turned into Photo Voltaic panels.

Assume the following:

Solar power input = 220Wm²

Area of Cyprus = 9.25x10⁹ m²

Power plant output = 1000 MW

solar cell efficiency = 20%

First calculate the output of the Solar System as:

Peak Output = Solar Radiation x Exposure Area x Solar Efficiency

Peak Output = 220 W/m² x 9.25x10⁹ m² x 0.2 = 407 GW

Power Plants closed = 407000 MW / 1000 MW = 407

Hope this makes sense.

What will be the area of the solar panel given the following:

Annual average insolation = 350 W/m^2

Annual electricity usage = 13000 kWh

Conversion efficiency = 17%

Average cost = S 0.4 kWh

350*A*0.17*24*365=13000*1000

521*1000*A=13000*1000

A=13000/521

A=25m2

Hope this helps. I do not know why you gave the cost.

Do we consider 12hrs (on an average) or 24 hours for kwh calculation related to PV?

Let’s say you have peak production of 1000 Watts, then you multiply it with Peak Sun Hours, lets say 5 in this case and you get 5KWHr production over 24 hours. But if you want to see the average production over 24 hours you simply divide the KWHr calculated by 24. So you get 5000/24 = 200 Watts (approx). That’s how much continuous load you can support over a 24 hour period.

A PV cell that measures 156x156mm can produce a maximum power of 3.2W at a solar insolation of 800W/m 2 and at a temperature of 25degC. Calculate the cell efficiency (in percentage) under solar insolation of 800W/m 2 and at a temperature of 45degC. Consider temperature coefficient of maximum power is -0.47%/degC.

At 25degC

800*0.156*0.156*Efficiency=3.2

19.469*Efficiency=3.2

Efficiency=0.1644 (or 16.44%)

At 45degC we can downward adjust (do check it from some other source)

16.44-20*0.47=7.04%

hi I want to ask, how to calculate the solar area requirement if the energy used is around 300 kWh per month (average), average daily solar radiation is 18.92 MJ/m^2/day, and efficiency of solar conversion is 12 %

Lets first list the data we have:

Energy requirement=300kWh/30days=10kWh/day

Solar radiation=18.92MJ/m^2/day=5.25kWh/m^2/day

Efficiency=0.12

Energy generation=Radiated Energy*Area*Efficiency

10kWh/day=5.25kWh/m^2/day x Area x 0.12

Area=15.87m^2

Hope this helps!

hi!

if Voc=45V, Vmpp=39.2V, Isc=8A, Impp=7.5A, P=294W, t(-12,75)*C, and B=-0.1221 V/*C.

For 12KW, I’ll need 41 panels if I did right the calculations. My question is how can I know the temperature coefficient and find the total area if panels have this dimensions 630x520x205 mm?

thank you!

For area calculation:

12000W=1000*A*0.20

A=60m2 (approx 1.5m2 per panel, if your calculations are right)

Problem: The energy consumed by the average household per day is 60 KWh. The solar power per square meter at the Earth’s surface is (1,000 W/m^2). Assuming that this power is available for 8 hours each day and that energy can be stored to be used when needed, what is the total surface area of solar panels that will cover all the household’s needs? You can take the efficiency of the solar panels for capturing solar power as 15%.

This problem is confusing me as I don’t know if I can use the equation above with what’s given in this problem. Thanks

Yes the equation can be used as follows:

60,000=1000*A*0.15*8

60,000=1200A

A=60,000/1200=50 meter squared

Although peak energy is usually available for five hours and not eight hours. In that case area required would be 80 meter squared.

Alright, that makes sense, the 8 hours was throwing me off. Thank you for explaining it!

In my country Iraq the solar irradiance calculated is about 5000 Wh/m2 according to (ERNA) data, so if we have have a load of 1000 Wh, what will be the formula to calculate the area.

Area Required=A=1000Wh/(5000Wh*0.2)=1m2

Yes that is correct way to do the calculation.

The requirement of the house is 3 kW and the efficiency of the solar panel is 15%. Total solar panel installation area =?

Total Power Output = Total Area x Solar Irradiance x Conversion Efficiency

3000 = A x 1000 x 0.15

A = 3000 / 150

A = 20 square meters

But to be on the safe side you should have an area of 30 square meters available. Solar panels sometimes have to be put at an angle and shading can cause problems.

Suppose that there are solar panels with 20% conversion efficiency. The size of each panel is 1m x 1.5m the output is 3000 watts. When finding out how many panels are needed.

Will this formula work?

Total Power Output = Total Area x Solar Irradiance x Conversion Efficiency

Thank you so much for responding.

Yes this formula works. Plugging in the values:

Total Power Output = Total Area x Solar Irradiance x Conversion Efficiency

3000 = Total Area x 1000 x 0.20

Total Area = 3000 / 200 = 15 meter squared

Number of panels = 15 / 1.5 = 10 panels of 1.5 meter squared each

You must remember that this is the best case calculation. Actual power production would be less than 3000 Watts. It would only be at the peak of 3000 Watts around noon time when solar radiation is falling directly on the panels.

Hi kindly confirm what will be the calculations if I have 330 watt panels that is 1.92m^2 for 30 kw what I will put in conversion efficiency as all the above examples are taking as standard 100 watt panels with 17 to 20 percent thanks

Hi, let me see if I got your question…you have 1.92m^2 panel generating 330 watt…that means they have roughly 17% efficiency as shown below.

1000 Watts/m^2 * 0.17 * 1.92 m^2 = 326.4 Watts

The area required for 30 kW can be calculated as follows.

30,000 Watts / 170 Watts/m^2 = 176 m^2

Hope this helps!

YA

Hi, your equation for total power output using area is super useful, do you happen to know the source of where you got it from?

I’m doing a research project and would like to confirm the origins of this equation thanks!

This is a pretty standard equation and you can find it in most texts. You can also find it here:

https://photovoltaic-software.com/principle-ressources/how-calculate-solar-energy-power-pv-systems

i need to know..how many panels and what KW i will get for a roof top area of 190 m2 terrace. if i use a 350Wp solar panels

1 m2 horizontal surface receives peak radiation of 1000 Watts. A 1 m2 solar panel with an efficiency of 18% produces 180 Watts. 190 m2 of solar panels would ideally produce 190 x 180 = 34,200 Watts = 34.2 KW. But inclined solar panels also need some spacing between them so practically you would be generating about half the power or 17.1 KW. Total number of panels required would be 17,100 / 350 = 48.85 or roughly 50 panels.

sir clarify this question– In house hold consumes 200units of energy an average in a month.design a roof top setup for zero net cost .

With the given data you can calculate that daily consumption is 6.66 kwhr or 1333 watts for 5 hours. So you can put 14 panels of Watt each. You would also need batteries if you need to store energy and use it later.

A 1 KW solar plant produces about 130 Units (KWh) of energy per month. If your consumption is 200 Units, you can think of installing 1.5 KW plant. But the problem is you get inverters of 1 KW or 2 KW rating, not 1.5 KW. (I am not very sure on this aspect) So, if you have net metering system in place and install 2 KW plant, you will be pumping energy into the grid and if you install 1 KW plant, you will be drawing power from grid. You may please look into financial aspect and decide.

That’s a very good analysis. For the given situation I would recommend going for 1.5 KWatt solar system with a 2.0 KWatt inverter as it is always good to have some margin. If you have net metering in your area…great…if not then you will have to go for batteries and there would be some losses in storage as well.

Hi

1kw solar panels produce 5kwh per day then how many Ah battery required for storing that energy , what type of connection I given that( series or parell).

Which type of invertor is good for 1kw solar panels?

How many solar panels required?

Sir please provide full details with calculated part also.

You would need 4 batteries of 12V and 100AH each at least. Or you can put together 2 batteries of 24V and 100AH each. Batteries would be connected in parallel usually but they can also be connected in series depending upon output of solar panels. You would also need a charge controller and an inverter. Inverter can be avoided if you want to run DC appliances only.

Hope this helps.

Plz help me i want to install….. 100 kw rooftop solar power plnt….how much area i need. Plz show me the calculation very clearly…. So that in future i can estimate n calculate myself

OK litu I will do it very simply for you. The formula is as follows.

Area = 13.33 x Wattage in Kilowatt

Area = 13.33 x 100

Area = 1333 meter squared

how many solar panels would you be able to fit in 13236 meters if they are

65 inches x 39 inches? power supply doesnt matter too much just want a general idea as its for a report. thanks

First things first lets convert the panel dimensions into meters.

65 inches = 1.65 meters

39 inches = 1 meter

So the area of a single panel is 1.65 squared meters. Divide the total area by this number and you get the number of panels.

Number of panels = 13236 / 1.65 = 8022 panels

Usually solar panels are placed at an inclination so that they get maximum radiation from the sun. This inclination depends upon the latitude of the location. So there needs to be some spacing between the panels so they do not cause shading. A good ball park estimate for the number of panels is about half the number that is calculated for the case if the panels were flat.

8022/2 = 4011 or approximately 4000 panels

what is the area required for 25KW plant for south facing arrangement and also give details about battery’s requirement?

Suppose the area is A square meters then the equation becomes.

1000 x 0.20 x A = 25000

200 x A = 25000

A = 25000 / 200

A = 125 square meters

This is for panels lying flat on the ground. We would suggest that an area of at least 200 square meters must be reserved due to the following three reasons.

– Panels are not usually flat, they are placed at an angle and need more spacing so they do not shade each other, depends on your location

– You might need more panels for achieving peak generation of 25 KW, again depends on your location

– Panels might not achieve 20% efficiency that we have assumed above (the 0.20 factor that we used above was to cater for panel efficiency)

http://affordable-solar.com/site/images/learn/calculatingarrayarticleimage1.jpg

For calculation of batteries we need to know the load and the backup hours required. If you want a simplistic answer I would suggest that you put in 50 batteries of 200 AH each for a backup of about 5 hours.

If your panel efficiency is 16%, will produce 160 Watt/m2. Your panel’s power capacity is 25 KWatt, so you will need 25000 Watt/160 Watt/m2 = 156.25 m2. If the panel is 250 Watt and size is 1.63 m2. number of panels you need 25000/250 = 100 panels and total size is approx 163 m2.

hi,

total area of roof top is 3000 metre squre .i need 30000 KW power consumption per month.almost 2000 kw per day consumption.could you please give me the desighn data for solar panel.

we need

1) maximum amount of kw produced for one metre squre panel and the cost of one metre squre panel

2) finally we need 2000 kw per day please give the sample calculation .

Suppose you have A meter squared of area. Then you can generate

A x 1000 x 0.2 Watts of power

Set this equal to the required power of 2000,000 Watts

A x 1000 x 0.2 = 2000,000

=> A = 10,000 meter squared

So the area you have 3000 square meter is not sufficient to produce 2000 kW of power.

One square meter can produce about 200 Watts and the cost of the solar system is about $1 to $2 per Watt depending upon how much backup you want. Solar panels can produce peak power for about 5 hours daily.

With the area you have you can produce 3000 x 200 = 600,000 Watts (600 kW) of peak electric power.

Lastly power is in Watts and monthly generation of energy is in KWHr, so please be careful with calculations.

Just to add, with 600 kWatt of peak power you can produce 90,000 kWHr in a month.

600 kWatt x 5 hours/day x 30 days/month = 90,000 kWHr/month

I want in install solar panels in 2000 square meters. I wish to know how much it will cost to install solar panels and the cost. How much electricity can be produced.

2000 sq meters means you can capture 2000 x 1000 x 0.15 = 300 kWatts of solar power. In one month you can produce 45,000 kWhr of solar energy. Cost of the system depends upon a number of factors and can range from about $1 to $2 per Watt.

Hi

1280 sq feet= 118 sq meters.

118X1000X20=23KW.

What is the capacity of plant in Andhra Pradesh, India.

I would like to supply total to the discom company. Is it require for storage plant.

What is the total cost?

hi

this is my requirement.

Sr. # Product Units Wattage Total Watts Daily Usage Total Watts -hrs

( Hrs)

1 DC fan 12 36 432 6.00 2,592

2 LED Lights 12 12 144 6.00 864

how many solar panels required, battery size and charge controlers size

hello,

you can go for 10 panels of 320 Wp ( canadian solar) . the dimension would be 1954 ˣ 982 ˣ 40.

hello plz help me to improve my knowledge for solar panel how can calculate power ,current and voltage and erea so plz help me am Eric from Rwandese student in Mechanical Engineering

Derek you are right. These calculations are not exact. But sometimes you need quick “back of the envelope calculations” to get a rough idea.

Your calculations assume the panels are always producing 100W, which they won’t since they don’t track the sun and the location of the panels on the earth isn’t given either which can greatly effect a panels output.

Hello!!! I have a doubt. If I need the solar that provide 500w. How many areas of solar panel that I use? And how to calculate it?

Triangle are=hight-39.5

Base-39

Width-55

250 watt panel use.string to string gap not required.

Roof size 2000 sq ft. Can you please mail me the max capacity of plant work on it, and total cost after subsidy for ongrid plant.

2000 sq ft = 185 sq meters

185 sq meters x 1000 Watt/sq meter x 0.20 efficiency = 37 kilowatt

That is the plant can produce a maximum of 37 kilowatt during peak radiation around noon time.

The cost can be approximated as 37000 Watt x $1/Watt = $37000 = 2.5 million Indian Rupee.

Hi

I’m UG student, doing my final year project, we have 30KW power, for this wattage how many solar panels are required and also how to calculate the area?

You will have to make a few assumptions. Lets assume that you have 100 Watt panels. You will need 300 of these to generate 30 kW. To calculate the area you need to know the efficiency of the solar panels. Lets assume the solar panels have 20% efficiency so the energy produced is 200 Watt/m^2. So the total area required is 30,000/200 = 150 m^2. This is assuming that the panels are placed on a flat surface along the ground. Keeping some margins you can say that a 15 m x 15 m area would be sufficient.

how to get 200watts/meter2 for 20percent efficiency

I do not get your question. But what I can tell you is that peak radiation of solar energy is assumed to be 1000 Watt/m2. For a panel that converts 20% solar energy to electrical energy the output of a 1m2 panel would be 200 Watts.