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# Category Archives: Solar Energy

# Ibn al-Haytham – The Father of Optics

Whenever we read about the history of optics or in general electromagnetics we hear two names again and again, these are James Clerk Maxwell and Heinrich Rudolf Hertz. The discoveries of these two gentlemen although very important to development of Physics were very recent (about 150 years old). But the history of optics is much older and the basics of this field were etched in stone about a 1000 years back. One of the people responsible for the early development of this field was Arab mathematician, astronomer, and physicist ibn al-Haytham (c. 965 – c. 1040).

Ibn al-Haytham in his book referred to as the Book of Optics (Kitāb al-Manāẓir) postulated that light travels in form of rays in straight lines and these rays are reflected by objects which makes these objects visible when the rays enter the human eye. Ibn al-Haytham rejected an earlier theory proposed by Ptolemy and widely accepted by the scholars of that time that light rays emanate from the human eye which makes objects visible. Ibn al-Haytham studied the phenomenon of reflection and refraction of rays of light in his laboratory (he studied lenses, experimented with different mirrors: flat, spherical, parabolic, cylindrical, concave and convex). He also used the Camera Obscura in his experiments to show that light rays travel in straight lines. He studied the working of the human eye and was able to make some valuable contributions.

Other than optics ibn al-Haytham made significant contributions to number theory, geometry, astronomy and natural philosophy. Ibn al-Haytham wrote more than 200 works on a wide range of subjects but most of his works are now lost. Only about 96 are known and about 55 have survived to some extent. Those related to the subject of light include The Light of the Moon, The Light of the Stars, The Rainbow and the Halo, Spherical Burning Mirrors, Parabolic Burning Mirrors, The Burning Sphere, The Shape of the Eclipse, The Formation of Shadows, Discourse on Light, as well as his masterpiece, Book of Optics. It was through Latin and Hebrew translations that most of his important work survived.

UNESCO declared 2015 the International Year of Light and ibn al-Haytham was dubbed as the ‘the father of optics’. This was to celebrate ibn Al-Haytham’s achievements in optics, mathematics and astronomy. In honor of him, the Aga Khan University (Pakistan) named its Ophthalmology endowed chair as “The Ibn-e-Haytham Associate Professor and Chief of Ophthalmology”. The crater Alhazen (latin form of his name) on the Moon is named in his honour, as is the asteroid 59239 Alhazen.

PS: An interesting story about ibn al-Haytham is that he proposed to build a dam on the Nile to restrict the flooding it caused but realized soon that it was impossible to solve this problem with the available resources. Ibn al-Haytham feigned madness so as to not upset the ruler (al-Hakim, the Fatimid caliph in Egypt). Legend has it that he was put into confinement in the city of Cairo near Al Azhar University and it is here that he worked on his theories of optics. Al Azhar University still survives today as one of the greatest and oldest university in the world.

# 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.

# Sizing Up a Solar System for a Cellular Base Station

Many operators are thinking of moving from the main grid to alternative energy sources such as wind and solar. This is especially true in third world countries where electricity is not available 24/7 and is also very expensive. This has forced operators to switch their base stations to diesel generators (which is also a costly option).

In this article we do a rough estimation of the size a solar system required to run a cellular base station. We start with the assumption that 20 Watts of power are transmitted from a single antenna of base station. For a 3 sector site there are 3 antennas giving us total transmitted power of 60 Watts. Now if 50% of the power is lost in cables and connections we would have to boost up the transmitted power to 120 Watts.

We know that power amplifiers are highly in-efficient (depending upon the load) and a large amount of power is lost in this stage. So we assume an efficiency of 12 % giving us a total input power of 1000 Watts. Another 500 Watts are given to Air Conditioning (200 W), Signal Processing (150 W) and Rectifier (150 W). So the combined AC input to the base station is 1500 Watts. Now we turn our attention to sizing up the solar system.

If we assume that the BS is continuously consuming 1500 Watts over a 24 hour period we have a total energy consumption of 36 kWh. If the solar panels receive peak sun hours of 5 hours/day we would require solar panels rated at 7200 Watts. This could mean 72 solar panels of 100 Watts each or 36 solar panels of 200 Watts each or any other combination. It must be noted that we have not considered any margins for cloudy days when peak sun hours would be reduced. Also, we have not considered any reduction in power consumption when there is no load (or very less load) on the BS.

Next we calculate the amount of batteries required. We assume that the batteries are rated at 200 AH and 12 V. This gives us a total energy storage capacity per battery of 2.4 kWh. So the number of batteries required is calculated as 36 kWh/2.4 kWh = 15. It must be noted that some of the energy would be consumed in real-time and the actual number of batteries required would be lesser. Furthermore we would need an inverter of at least 1500 Watts and charge controller of 125 Amps.

# 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 :).