Sizing a 10 KW off-grid solar power plant
The first step is to estimate the energy consumption of a typical domestic consumer.
Working time (Hour)
Energy reqd. (W-hour)
Total daily requirement
The load estimated is 7.48 KW and it is always recommended to install higher capacity than the actual consumption as a thumb rule. The solar panels are installed for 10 KW to meet the load requirements.
The second step in sizing the plant is determining the inverter rating. In an off-grid system, the load requirement is met by the power supplied from Battery via Inverter. The power supplied from panels is sent to inverter for conversion. There are few losses to be considered such as switching losses, conversion losses, and heat loss. The over-all efficiency of an inverter is considered to be 70%.
Energy required = 7480/0.7 = 10685 Watt-hour
The third step is to size the battery bank for energy storage and discharge when there are any power outages. One of the characteristic to be taken into account while designing a battery bank is Depth of Discharge (DOD) which is the maximum amount of energy extracted from a battery before discharging. The DOD of a Li-ion battery is considered to 80%.
There is need for balancing of cost as well as the system losses are also taken into account while designing. Assuming 12 V, 100 Ah capacity battery with 80% depth of discharge and system voltage is 96 V.
Charge capacity of battery bank = Energy required / System voltage considered
= 10685/96 = 111.3 Ah = 111 Ah
Number of batteries connected in series = System considered voltage / Battery voltage
= 96/12 = 8 batteries
The final step is to determine the energy output from the solar modules and the number of modules to be installed. The power from is transferred to the battery via charge controller. Considering, 95% for Li-ion battery the energy required to supply to the battery is 7480 Watt-hour.
Energy required = 7480/0.95 = 7874 Watt-hour
The charge controller is required if the energy storage device is included in the system to monitor the conditioning of the battery. The efficiency of a charge controller is considered to be 90%.
Energy required = 7874/0.90 = 8749 Watt-hour
The energy required from the solar panels every day is 8749 Watt-hour. The losses in solar panels are assumed to be 30% due to dust deposition, damage of cells in a module and while operating at higher temperatures i.e., heat loss.
Energy required = 8749/0.7 = 12499 Watt-hour
Now, it is the time to select the panel for installation. There are lot of panel manufacturers available in the market and choose wisely based on the ratings and the space availability. Here, considering 330 Wp Waree solar panel with maximum voltage of 36.55 V and 9.03 A maximum current. The required voltage output from the module array is 96 V and the selected module has 36.55 V. Total number of modules in the system are 30 modules. In such cases, the charge controller would convert such extra voltage into current, hence we do not require any additional modules in series. The charge capacity required from the module array can be calculated as follow:
Charge capacity = 12499/36.55 = 342 Ah
Assuming, 5 sunny hours on each day. The required current output in the array is
Current output = Charge capacity / No. of sunny hours
= 342/5 = 68.4 A
The no. of modules required to produce the current output (in parallel)
Current output / Maximum current
= 68.4/9.03 = 7.57 ~ 8 modules
The pictorial representation of 10 kW off-grid system is set up as follows: