I have been asked several times what is the energy produced by a 1 kW panel in a day or how many units of energy are produced over 24 hours. Let me first state that the product of power and time gives you the energy produced. So for a 24-hour period, we have 24 kWhr of energy produced, if the solar panel was producing peak power throughout the day. But we know that that is not the case. So for the time component of the above equation, we can use 5 hours, also called peak sun hours. This gives a total of 5 kWhr (1kW x 5 hours) of energy produced over a 24-hour period i.e. 5 units of energy are produced. In a month you can possibly produce 150 units. If the per unit rate is Rs.10, you can produce Rs.1500 worth of electricity with a 1 kW panel in a month.

Continue reading Energy Produced by 1kWatt Solar Panel# Tag Archives: Solar

# Calculation of Solar Panel Spacing for India (New Dehli)

I have been asked many times that what is the area required for a solar array with a particular rating (such as 20 KW) and I have given a simple calculation method in one of my earlier post. But real life is never so simple. There are a bunch of inputs that we need to get an exact answer. In particular we want to know what is the required spacing between solar panels so that they are not in shade for a particular part of the day (typically 5 or 6 hours) and for this we need a number of inputs such as latitude, longitude, solar noon on the worst day of the year etc. Here we show the sequence of calculations for a typical city in Northern India. Similar calculations would hold true for many cities in Pakistan as well.

Let us first list down the data we have for New Dehli a typical city in Northern India (this data is available on ESRL website).

Latitude: 28.6139

Longitude: 77.2090

Time Zone: +5.5

Worst Solar Day: Dec 21

Start of Solar Charging: 09:19 am

Solar Noon: 12:19 pm

End of Solar Charging: 3:19 pm

Solar Azimuth: 135.5 degrees (Azimuth at 9:19 am, the time solar panel just comes out of the shade)

Solar Inclination: 22.33 degrees (Inclination at 9:19 am, the time solar panel just comes out of the shade)

Now lets assume that the length of the solar panel is 1.0 m and it is fixed at an angle of 30 degrees from the horizontal. Then the height can be calculated as:

h = 1.0 * sin(30) = 0.50

Now we are ready to calculate the distance between the panels.

D = h / tan(Inclination) = 0.50 / tan(22.33) = 1.22 m (this is the distance without Solar Azimuth Correction)

d = D * cos(180-Azimuth) = 1.22 * cos(44.5) = 0.87 (this is the distance with Solar Azimuth Correction)

So for a 1.0 m length panel we need to have a separation of about 0.87 m between the panels along the north-south line (panels facing south in the Northern hemisphere). To be on the safe side we can keep the distance a bit higher than this. We can estimate that the total area would be increased by a factor 2.0 (0.87 m is the separation of the panels and 0.87 m is the base of the triangle formed by the inclined solar panel) or we can simply multiply the area calculated by the simple calculations (for a panel lying flat on the ground) given in the earlier post by a factor of 1.74.

Note: These calculations are valid for solar panels placed on the ground, in straight lines, along the east-west line (facing south). For panels placed on inclined rooftops the calculations would be different (area required would be generally lower). Also note that solar panels are highly sensitive to shading and even if only 10% of the panel is in shade its performance without deteriorate appreciably. Lastly some installations have different panel tilts for winter and summer (latitude +/- 15 degrees) and this needs to be catered in the calculations.

# Calculate Solar Panel Tilt

1. Find the direction of magnetic North and consequently magnetic South.

2. Adjust for magnetic declination to find exact true South.

3. Point solar panels towards true South.

4. Find optimum tilt angle based on the latitude and the season.

Enter the value of latitude below to find the panel tilt in degrees.

**Winter**

### Latitude

**Spring and Fall**

### Latitude

**Summer**

### Latitude

Note:

1. The result above is the angle in degrees from the horizontal.

2. If you do not know the latitude of your city you can look it up here.

# How to Calculate the Surface Area Required by Solar Panels

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 :).

# Solar Analogy

All electromagnetic energy travels in the form of rays. The most obvious example is solar energy that is radiated by the sun in all directions. The further away a body is from the sun the lower the energy that it receives. Objects in the path of these rays cause shadows but not complete darkness as rays reflect from other objects and also diffract around the edges. These rays also have a phase and frequency that determines their behaviour when interacting with objects. The amount of rays that can be collected by a receiver depends upon its size and orientation. Solar energy can be harmful when a body is exposed to it for longer periods.

All these concepts are extendable to wireless communications. Wireless signals decay with distance, suffer from shadowing, reflect, refract, diffract, scatter, have phase and frequency, can be collected by appropriately designed antennas and can be harmful as well. The major difference being that modern transmitters are not isotropic radiators. Practical transmitters are like a sun that radiates solar energy to the earth in a narrow beam while ignoring the other planets.