Frequently Asked Questions
General questions about the operation of the Power Sets
Which consumers can I operate?
How long can I operate consumers?
How long does it take to recharge the batteries?
Which consumers can I operate?
First, determine whether DC and / or AC units should be operated.
DC units:
The power consumption of direct current devices is usually measured in amperes. Together with the necessary operating voltage, the power consumption can be calculated.
A water pump with 12V 5A power supply requires 12V x 5A = 60W.When selecting DC consumers make sure that the load does not exceed the max. power output (Imax/Pmax) of the charge controller. The output voltage generally corresponds to the battery voltage. Larger systems may have a battery voltage up to 48V. In order to operate consumers at a lower voltage level, e.g. 24 or 12V, a DC-DC converter is required.
You should avoid connecting DC consumers directly to the battery bank, as this may lead to a deep discharge of the battery.
AC units:
The power of AC devices is usually directly given in watts (W), the real power, or less frequently in volt-amperes (VA), the apparent power. With the power factor cosphi you can calculate from the apparent power the active power or vice versa.
The real power of a CRT monitor is 70W, the power factor according to the manufacturer 0.7. The apparent power is therefore 70W / 0.7 = 100 VA.For resistive loads, such as for example Light bulbs, heaters, hot plates, etc. the real power corresponds to the apparent power (power factor = 1). Inductive loads, i.e. devices that are powered by an electric motor, such as Drills, compressors, water pumps, etc. have in general a power factor <1. This type of consumers may have a high surge power demand, depending on the device type, 3 - 6 times of the nominal power. The same applies to capacitive loads. These consumers like for example flashes are rather rare.
In order to use the energy from the battery a 230VAC inverter is required. The output power (Pnom) of the inverter should comply with the required power demand of the consumers. For consumers with a high surge power demand it is important to ensure that the inverter has an appropriate overload capability.
Inverters with a modified sine wave output are not be compatible to all consumers. Especially electronically controlled tools or household appliances may not work correctly.
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How long can I operate consumers?
The max. operation time of a consumer depends largely on the capacity of the battery. The capacity is expressed in amp hours (Ah) and depends on the discharge current. For this reason, several capacities for various applications are given by the manufacturer. In isolated operation the battery is discharged over a long period, so that the manufacturer's instructions for 10h (C10) and 20h (C20) have to be chosen, to determine the applicable battery capacity.
The battery life is significantly influenced by the depth of discharge (DoD). In particular, sealed gel or AGM batteries (VRLA) should not be discharged more than 50% in cyclic applications. A higher depth of discharge (> 50%) shortens the lifetime of the battery enormously.
Is the battery fully charged, you can calculate the available amount of energy based on the rated capacity (C20) and battery voltage.
The usable amount of energy of a 100Ah 24V battery bank is 100 Ah (C20) x 50% DoD x = 24VDC 1200Wh.The calculation of the possible operation time of a DC consumer is now very easy by using the above calculated amount of energy divided by the power of the consumer.
A water pump with 60W 12VDC power can operated according to the above calculated capacity as long as follows: 1200Wh / 60W = 20h (hours).For AC loads, the calculation of the operating time can be done analogous with the real power, but still needs to take the efficiency loss of the inverter into account with about n = 0.9 (10%).
A TV with 80W 230VAC can be operated according to the above calculated capacity as long as follows: 1200Wh / (80W / 0,9) = 13.5 h (hours).[ to top ]
How long does it take to recharge the batteries?
The charging of the batteries by the photovoltaic generators is influenced by many factors. The most important here are the output of the solar generator and the solar radiation (sun hours simplified). Due to seasonal variations you get individual energy for every months. The average radiation power in Germany varies between 0.8 kWh / m² / day (sunshine) in December and 5.5 kWh / m² / day (sunshine) in July / August. You can calculate the output of a solar generator with the product of watt peak indication and daily global radiation (solar hours).
In December, a 100Wp solar module generates approx. 0.8 hours x 100 Wp = 80Wh, in July / August about 5.5 hours x 100 Wp = 550Wh.The mean daily irradiance (global radiation) of each months for certain locations in Europe can be taken from the radiance data bank of the EU (Photovoltaic Geographical Information System). Click on the right side on "Monthly Radiation and select your location. After you pressing "Calculate", a table appears in a new window with the global radiation values (Gopt) of each month at the optimal inclination angle of the modules. Note that the values are given in watt-hours. For the above calculation, you must divide the values by 1000 to get kWh reps. hours of sunshine.
Using the above mentioned calculation method of the photovoltaic yield, you can now see how long it may take to recharge the battery. Please note, that not the entire energy output of the solar modules can used, because of losses due to cabling, mismatch and storage. The overall efficiency is about n = 0.7 (30%). The charging time is calculated by dividing the battery capacity through the yield of the solar panels less 30% losses.
For an isolated system with 100 Wp solar generator and a 24V / 100Ah battery bank the recharging time in December is about 1200Wh / (80Wh/dx 0,7) = 21.4 d (days) and in the summer about 1200Wh / (550Wh/dx 0,7) = 3, 1 days.[ to top ]
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