Tag Archives: PV

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)

Solar Panel Spacing Calculation

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

*+ = Degrees

Spring and Fall

Latitude

* = Degrees

Summer

Latitude

* = Degrees

 


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.