Dimensioning of PV installation
This chapter describes the dimensioning of the different components of the electrical installation, we make a choice based on the demand, availability and technical specifications.
We start off by determining the loads the system has to be able to support. When the loads are fixed, all other components follow in a natural way. This method will thus allow us to design a battery-based system that can carry all the loads. The danger of this method is that the costs can increase considerably. Sometimes it is more efficient to undersize the system forcing the user to consume the available energy in a more economical manner.
The main building with the complete electrical system, including the pump at the left. There are two grounding rods in green (AC and DC).
Determining the loads
This table allows us to calculate two important quantities. The maximal continuous power usage is Pmax = 1590W, and the total energy used on a daily basis is Eday = 6850 Wh.
However, when the weather is bad for some days, the local people should be aware of the fact that they should spend less energy.
The batteries should be able to store the complete energy consumption of one day. In other words, this means that there will be one day of autonomy. For safety reasons we limit the voltage of the battery bank to Ubank = 24 V. The DOD is the depth of discharge. We start by taking a DOD of 0,6. 𝛼 is the efficiency of the inverter, which is more or less equal to 0,9. 𝛽 is the efficiency of the battery bank, which we also take to be about 0,9. The general formula to calculate the capacity needed is the following:
𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 𝐸𝑑𝑎𝑦/ 𝑈𝑏𝑎𝑛𝑘 ∙ 𝛼 ∙ 𝛽 ∙ 𝐷𝑂𝐷
-> 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 590 𝐴ℎ
It is difficult to determine how many cycles will occur in one year, as the batteries are being charged and discharged at the same time. This is why the discharging rate will be low during the day. There will only be a significant discharge in the evening, the early morning or during cloudy periods. The capacity needed with a DOD of up to 40% is about 850 Ah. Four parallel strings of 200 Ah seem to be the best solution.
The battery study conducted by last year’s Humasol project students led to an important conclusion. It is not because batteries should theoretically have a longer lifetime, which this lifetime will be reached in reality. It all depends on how the batteries are treated. If the user doesn’t maintain the battery bank in a disciplined manner, the negative effect on these batteries will increase. Highly fluctuating charging and discharging will decrease the lifetime considerably. Because a lead-acid battery is somehow robust, it can deal with a chaotic environment much better than other types of batteries. Even though the school will have fixed charging and discharging patterns, we should account for possible random fluctuations and bad maintenance by the local people.
During the installation of the batteries, they should be shielded away from the sun as much as possible. Increased temperature will result in a reduction of the lifetime. This applies to every type of battery. The batteries should definitely not be placed on the ground, as moisture and other filthy particles could damage them.
Detail of the power electronics
The inverter should always be able to deliver the maximum continuous power. We use the Phoenix Inverter 24/3000 with a maximum continuous power of 2500 W at 25°C seems to be the best option.
A solar panel has a certain peak power, that shows the monthly peak sun hour for the Phnom Penh region. This is measured in kWh/m2/day onto a horizontal surface, facing directly to the south.
Peak sun hours Phnom Penh region
The general rule of thumb used to determine the optimal angle of the solar panel array is that it should be equal to the latitude of the location.
In Cambodia, this is about 15°.
In order to calculate the number of panels necessary, we use a minimal peak sun hours (SHmin = 4,5). There should also be a safety factor that accounts for the shading of the panels and the efficiency of the whole system, including the Joule losses in the electrical wiring. We take this factor µ to be between 0,75 and 0,80. The number of panels can then be determined by the following formula.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑛𝑒𝑙𝑠 = 𝐸𝑑𝑎𝑦 /𝑃𝑝𝑒𝑎𝑘 ∙ 𝑆𝐻𝑚𝑖𝑛 ∙ 𝜇
We use 240 Wp Yinlgi Solar panels that yield a voltage of about 35 Voc. There can only be two of these in series as the input voltage of the charge controller is limited to 100V. Using the formula above gives a number of solar panels equal to 8. This is split up into 4 parallel chains of each 2 panels in series.
The solar panels, grounding is green and goes to the DC grounding rod, three panels in series
The charge controller should be able to handle two things. The voltage from the solar panels applied at the input and the charging current is drawn by the batteries at the output. Standard charge controllers have a maximal output current of 50 A. This implies a power of 1200 W.
This is however not enough to handle all the power coming from the solar panels when they are working at their fullest potential. This implies the need for a second charge controller. The BlueSolar MPPT 100/50 (12/24V-50A) is available at Kamworks and has a maximum input voltage of 100V. This allows us to place solar panels in series as long as the Voc stays under the threshold of 100V. With the chosen 240 Wp panels, we are able to make strings of two panels resulting in a Voc of about 70V.
This section describes the cables from the charge controller to the batteries, the connections in- between the batteries and from the batteries to the inverter. For the DC cables from the panels to the charge controller, we need special ultraviolet resistant cables, which are more expensive.
The biggest DC current will flow from the battery bank to the inverter. We chose the inverter to be one capable of handling 3000VA at 25°C. This corresponds to a maximum current of about 125A. The cable area can easily be calculated by the following formula.
𝐴 = 𝜌 ∙ 𝑙 ∙ 𝐼𝑚𝑎𝑥 / 𝜀 ∙ 𝑈
𝜌 is the resistivity of copper, l is the length of cable between the batteries and the inverter, Imax is the maximum current, U is the voltage of the battery bank (24 V) and 𝜀 is the voltage drop due to losses in the wire, which is chosen to be about 0,01 to 0,02. The numerical result should be rounded up to standard wire size. It seems sensible to use the same area for all DC wiring. By doing so we do not need to buy different areas giving us more flexibility.
Once the maximum current stays fixed, the remaining factors are the voltage drop and the length of the cable. Let’s say that in the worst-case scenario we have a length of about three meters. The voltage drop can be chosen freely, as long as it remains under 3%. Because the length of cable for the DC wiring is small compared to the AC wiring lengths, we can easily neglect the surplus price of DC wiring coming from the difference between 16 and 35 mm2 wires. The price of armoured cable and submergible cable also outweighs the price of the DC cable. Keeping this in mind, we choose an area of 35 mm2. This corresponds to a voltage drop of 0,8 % in the worst case. With a cable area of 16 mm2 there would be a voltage drop of 1,7 % in the worst case.
The next table collects the results. The metric used to determine the price is $0,12/mm2/m. We estimate the total length of the DC cable to be 20 meters. This includes connecting the two charge controllers to the battery bank, connecting the batteries with each other and connecting the batteries with the inverter.
||Voltage drop [%]
As mentioned before, we take the same area for the DC cables throughout the whole DC system, apart from the interconnection between the solar panels and the charge controllers. Problems could arise if the cable is too thick to match the fittings of the charge controller or the inverter. Therefore, it is very handy to take specific tools with us to Cambodia in order to make small adjustments to cables. Stripping tools and connected devices such as WAGO connectors, wiring nuts and heat shrinks are essential and can spare a lot of our time.
The area of the AC cable is 2,5 mm2. This is a standard AC cable size. There are two big buildings with a total of 15 classrooms that need to be wired. As we dimensioned the system for 45 lamps, we are going to divide the 45 lamps uniformly over the 15 rooms, which gives three lamps in every room. The eight laptops are mobile. In this dimensioning, we choose to confine the outlets to the teachers’ room. We drew four outlets in the electrical scheme. Because the outlets have to be grounded, we use 3×2,5 mm2 cables for this. We don’t know where the teachers’ room is located, but we estimate a length of 30 meters is necessary. During the project, we will off by installing lamps in the main building. If we have enough time, and the budget for the armoured cable, we will also install lights in the second building.
Detail of the teachers’ room, only two lamps drawn for simplicity
The lamps do not need grounding. Cables of 2×2,5 mm2 are therefore in order. The AREI states that there may be 8 lamps in one circuit in parallel. When we place three lamps in every room, this means there will be four circuits in the main building and two in the other building. A combined length of 220 meters is estimated. The transition of the main building to the other building has to be made under the ground with armoured cable. This needs to be 4×2,5 mm2 armoured cable to minimize costs.
Both the DC and AC cable connections should be taped off with (black) electrical tape.
The cable from the solar panels to the inside of the building should have stronger insulation than normal cables. Cables that are exposed to the sun should be insulated this way making them UV and weather resistant. Because these PV cables are expensive, making smart routing can reduce the price significantly. The maximal current through these cables is equal to twice the short circuit current given in the datasheets of the solar panels, as there are two parallel chains for every charge controller: Imax = 17,5 A. Taking a maximal length of 10 meters, the previous formula from the section of DC cables yields an area of 6 mm2.
Every AC circuit requires a fuse with a nominal current of 20A. In an AC circuit, it doesn’t matter on which side the fuse is installed as the current constantly changes direction. DC cables with a surface area of 16 mm2 require fuses with a nominal current of 80A and 35 mm2 DC cables need 100A fuses. The previous information can be found in the literature study. Although both the charge controllers and the inverter have built-in security, there still has to be the possibility to decouple these components. This is why there should be a fuse between the solar panels and every charge controller and a fuse between the batteries and the charge controllers. In the DC circuits, it is obvious that there should always be a fuse on both the + and the – side of the circuit.
It is important to keep the AC and DC grounding separate, by using two different grounding rods. It seems practical to divide the AC grounding into two parts. One grounding device is used for the main building and another one for the smaller building. The price of three grounding rods with a length of 3m is about $85. The rods need to be installed as explained in the literature study.
As we saw in the literature study, a good grounding has to have a resistance of no more than 50Ω, if possible less than 30Ω. The following formula calculates the area of the grounding wire, and this is applicable for both DC and AC. In this formula, l is the length of the cable, which is unknown at this moment. R is the resistance and 𝜌 is the resistivity of copper as used in the previous formula.
𝐴 = 𝜌 ∙ 𝑙 / 𝑅
If you fill in some values you can see that it is possible to reach a very long distance, up to one kilometre, with the standard AC area of 2,5 mm2 while remaining under 20 Ω. However, making the length of the grounding cable longer than necessary, will increase the cost and the resistivity.
Hence, it is wise to minimize the amount of grounding cable in order to protect the system from surges as much as possible.
Every grounding cable should be isolated in a green/yellow colour. It is also smart to tape off the electrical connections with the ground by use of a green/yellow electrical tape.
It is crucial to ground the frame of the solar panels carefully by interconnecting the entire frame electrically and bringing the grounding wire to the grounding rod. Because this wire is running on top of the roof, it has to be PV cable.
The charge controllers don’t need to be grounded, neither should the batteries. When the batteries are stored on a conducting surface, this surface has to be grounded. Therefore, it is better to use a non-conducting surface. This surface should also be dry and thermally insulate the batteries.
The inverter has to be grounded on the AC side. The lamps do not have a grounding pin. Outlets, on the other hand, do need to be grounded.
Wiring on point
The lightning can destroy the entire system in a split second. Ideally, one would install a rod on the top of the roof, and connect this rod with the ground. But this has to be done decently and we have not examined this thoroughly. When it is not installed properly, the roof can be set on fire or heavily damaged. As we are no experts in this field, a rod on the roof is not a good idea unless we can find an expert in Cambodia. The fuses incorporated throughout the entire system will offer the necessary protection from surges due to lightning.
Dimensioning pump installation
It has been decided not to use batteries within the solar pump system. The reason for this is to avoid the use of a component that is expensive, not eco-sustainable and that requires a lot of maintenance.
In order to have water during the night or when the radiation of the sun is insufficient, the water is stored inside a tank. In this manner, the tank is equivalent to the batteries in the electrical system. The tank needs to be elevated up to a certain height in order to establish a reasonable flow rate at the taps. In the drawing below a simple scheme of a possible hydraulic system is shown. The hydraulic system can be divided into two sub-systems that are dimensioned separately. The subsystem A is the pump system that provides the necessary amount of water to the storage tanks. The subsystem B is the tank system that provides the water to all the outlets of the school by making use of gravity.
The storage tank acts as a buffer between sub-system A and subsystem B, which guarantees the availability of water during cloudy periods or even at night. Since we will try to place the tank as close as possible to the borehole the hydraulic losses due to friction in the pipes of subsystem A will be minimalized, reducing the necessary power of the pomp.
Example of a hydraulic system, split into two subsystems
Determining the need
First of all, we need to determine the daily water consumption of the school.
There are two sanitary buildings. Each building has six faucets and three toilets. There has to be enough pressure on the water to transfer the water from the tank to these loads and enough pressure has to be left to flush the toilets. The worst-case scenario is that all toilets are flushed at the same time, and all the taps are open. Obviously this won’t occur in a real-life situation.
We assume usage of 10 litres for every flush and 50 cl for each time a child washes his hands. We take into account, the 600 children in the school and a margin, when local people also use these sanitary buildings. A reasonable assumption is that every one of these people flushes once and washes his/her hands three times a day. This yields a water consumption of 7050 litres every day. When the amount of people consuming water raises only a little bit or when a tap remains open for a while due to undisciplined children for water consumption can quickly rise up to 8000 litres a day.
Furthermore, the climate of Cambodia is such that in the months of June, July and August the pump has less output because there is less irradiation in these months, but the pump will have more output in other months. So it is necessary to have enough volume in the tank to be able to store this extra water.
For this reason, we have eventually chosen for a tank of 10 000 litres.
Making the well for the pump
We start off by dimensioning the pumping system by hand. By doing so we will be able to understand the formulas and principles used in the Lorentz dimensioning tool. When choosing a pump, the two most important parameters that need to be specified are the lift and the flow rate. The flow rate is directly related to the size of the water tank you are planning to fill throughout the course of one day.
To calculate the flow rate it has been supposed that the pump will work at its fullest potential for 8 hours and that the volume of the tank is 10 m3. In this manner, the output of the pump has to be around 1,2 m3/h. This is an approximation since the irradiation varies throughout the course of the day.
The total lift can be split up into three different parts:
- Lift given by the distance from the pump up to the earth’s surface. There is no data available yet as the new borehole still needs to be drilled. We have assumed this lift to be 30
- Lift given by the elevation of the tank. This lift is around 2 or 3 m. The height of the tank will be determined later on, during the dimensioning of the subsystem
- Lift given by the hydraulic losses. The flow inside the pipes and through the valves causes friction, which converts kinetic energy into thermal energy. The pump will have to be able to compensate for these
The hydraulic losses can be divided into two types:
- Concentrate loss due to valves and changing of direction in the piping system. (Minor losses)
- Distributed losses due to the friction between the water and the pipelines. (Major losses)
Choosing the pump with Lorentz tool
With the Lorentz dimensioning tool it is easy to dimension the system for a certain need. The most important parameter is how deep the pump is placed relative to the height of the tank. Because the depth of the pump is currently unknown we dimension for a length of 45 meters. The distance from the pump controller to the pump is a less important factor. This distance includes the depth of the hole and is assumed to be about 60 meters. For these setting, two solutions appear. The first solution is the best as the pumps output is more uniform throughout the year.
The pump system
The next figures show the pump characteristics. The power needed for the pump is 600 Wp. This means that we need three Yingli Solar panels of 240 Wp need in series. A PS600 controller can handle a Voc of 150 V, so three panels in series aren’t a problem. Lorentz also sells their own solar panels to deliver energy to the pumps, but generally, they are rather overpriced.
PS600 HR-07 working line
The upper graph in the following picture shows the output of the pump throughout the year. During four months, the output is greater than the capacity of the tank. This is why a float switch will have to be implemented in the system. The graph below shows the output from the pump during an average day in August.
Monthly and daily output PS600 HR-07