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.

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# 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.