Remote Area Power Supply Systems (MIET 2009)
Renewable Energy System for an Autonomous House
Teacher’s Name: Professor Aliakbar Akbarzadeh
Student Name: ZHANG Xiao
Student Number: S3175958
This report is about Remote Area Power Supply System. The word ‘remote’ here does not necessarily refer to long distances from the existing electricity grid net work. In terms of power supply, any location for which the grid extension is not as economical as alternative options can be termed ‘remote’.
There is no doubt that in most cases the electricity form of electric energy. It is least probable that the production of power and the generation of electricity in small scales such as powering a house, a farm, and even a small community could be done more cheaply than by using the grid. On the other hand, even a location which is not very far from the electricity grid might make the grid extension costs prohibitive. In these cases generation of power in remote areas becomes an attractive option.For example, for a house is 2km far from the grid and connection costs are estimated to be $25,000 per km, assuming a discount rate of 12% and inflation rate is 8%. If we consider the yearly electric energy consumption of this 4 people house to be 1277.5KWh, the real cost of electricity is the normal electricity cost (say 24 cents/kWh), therefore the grid extension translates into electric power at the cost of $1.18/kwh. It is with this high cost of electricity that the options of local power generation become attractive and economically viable.
In my design, I use PV panel and petrol generator set to simplest this forms and reduce the cost of electric power. In these days, petrol generator sets are commercially available for remote area applications for a wide range of power. These machines are relatively inexpensive to install 1000$/kWh. If they operate at their rated power, the fuel cost would be in order of 1.2$/Liter. They generally have a life time of about 10,000 hours. But in most applications, the load is intermittent and the generator set operates at a more reduced level than its rated power. This reduces the life time of the machine and at same time decreases the efficiency of conversion. This problem has been largely overcome by add an energy storage to the power supply system. The most common type common type of energy storage system for remote area
applications is batteries. In this case, everyday or every other day, the generator operates for a few hours at it is rated power and chargers the battery bank. Power is extracted from the battery system on demand.
The output power of engine generators is usually alternating current; therefore it needs to be rectified, regulated and controlled for battery charging. Also, most electric applications in my design operate with 240V AC, 50HZ. Therefore the DC power obtainable from the batteries needs to be conditioned. This is done by inverters which convert DC into AC power at the desired voltage and frequency.
It can be seen that the addition of a storage system to a petrol generator set requires extra components and extra costs. On the other hand, although initial capital investment increases, the overall cost of electric energy can be considerably decreased as a result of the efficient operation of the generator set. It should be noted that need for less frequent operation of the engine results in longer life time, less maintenance, less noise and in general, more convenience.
Generation of electricity by engine generators has a big disadvantage in that it relies on fuel as the main energy source. This total dependency of power generation on transported fuel, especially in remote areas, can be a source of inconvenience and extra costs. Locally available natural sources of energy can provide new options, so that the dependency on transported fuel can be decreased or even totally eliminated. Therefore, I add to PV panel into system, combines these two systems in my option.Solar radiation is another main source which can be effectively utilized for the production of power. Direct conversion of solar energy to electricity through solar cells was initially utilized for space applications, but now, there are infinite numbers of installations all around the world which utilize sunlight and produce power through photovoltaic systems. The photovoltaic module, which is the building block of photovoltaic energy systems is now, an off the shelf item. Each individual module can convert solar radiation into electricity with reasonable efficiency without and moving parts. The rated life time of PV modules is generally assumed to be 20 years. However the modules have exceeded this life period while operating in harsh environment. The voltage produced by PV modules is almost constant and does not significantly vary with the level of solar radiation.
It should be noted that, in locations where there are large seasonal variations in available solar radiation, a system totally based on photovoltaic would be oversized in substantial time periods of the year and therefore would be less economical. In this situation, it is best if a small petrol generator acts as an auxiliary power source for periods of low solar radiation.
Powering remote areas using PV system is becoming a standard method. Depending on the energy consumption today it is possible to set up a PV based energy system with an initial capital cost is 10240 depending on the load. The installed cost per KW is about 9615.43$ including battery storage (12RP670 ,C/20), inverter (SPI 1500-
24SS) and charger which is Suntron 24volt battery chargers with code number is 11203 can provide currents up 50 amps, etc. Then I decide number of PV (BP164) panel is 16, because it needs not the largest number of PV panel for 1277.5 kWh/year, and add petrol generator (UB2 in Appendix E) to offset the rest of energy. Due to my design the electric power cost is only 1.02$/kWh.
Finally, I think my design is also not very good enough, because the generator is less time to work, I should reduce the number of PV panel, so that I can further reduce the electric power cost. Maybe I can use 11 or 12 PV panels next time.
Summer and winter electric load for the remote homestead
Average hour/dayAverage wh/day
Six 40w lights240.0 34720960
Microwave oven800.0 0.10.18080
Vacuum cleaner450.0 0.80.8360360
total power2140.0 35003000
Schematic diagram for design
Design PV panel (at first)
I choose the BP164 of PV in Typical Photovoltaic Modules.
′Table 9.3 Available solar radiation in Melbourne (latitude 37?, 49S) for a horizontal surface and a north facing surface inclined at local angle latitude angle and the
corresponding peak sun hours.
2.28*64=145.92Wh/day (horizontal surface)
3.06*64=195.84Wh/day (facing north)
So we put PV panel facing north and tilted at local latitude.We assume the largest power we need, which is 3500W.In winter,
So we can make a table about the modules we need at different months
peak sun hoursnumber of PV panel
Then I decide number of PV panel is 16, because it needs not the largest number of PV panel.
Here, we have to decide on the arrangement of the modules. Let us assume that we have been able to implement energy conservation and we need a maximum number of 16 modules. If the battery bank is 24V, we should arrange the modules such that every string has 2 modules in series. Therefore we have 8 strings in a parallel arrangement. Looking at the specifications of BP164, The maximum power point the voltage and current of selected module is 17.1V and 3.74 amps respectively.
2*17.1=34.2V and 8*3.74=29.92 amps,
So the total installed power is 34.2*29.92=1023.264W.
Battery storage for PV system
In most cases, the battery system should be capable of providing the required energy for 3-4 days without any input from the PV modules. For most batteries, it is recommended that they should not be discharged below 30% level.
Therefore, in calculation above, I consider a 3KW/day demand for the month of July, 3.5KW/day demand for the month of Jan and 3 day storage, the enquired battery capacity would be: 3*3/0.7*1.3=16.71KW
The factor 1.3 takes into account the inefficiencies and losses in different components. It is also assumed that maximum permissible depth of discharge is 0.7.Assume a 24 V battery bank,
So 19.5*1000/24=812.5 amp-hour/day
So we look at the specification of batteries given in Appendix G, we see that 24RP900 Exide battery rated at C/100 has a capacity of 900amp hours at 24V (temporary). I choose this battery as my storage system.
Design petrol generator
In order to decrease the number of PV modules required, we can add an auxiliary energy source such as petrol or diesel generator to a PV system reduce the cost of the installed system. In PV design modules, we limit the total number of PV modules to 16, and then the system would be able to produce enough energy for most of the year without needing the engine generator. During the rest of the year, where the load exceeds the produced energy by the 16 PV modules, the engine generator should run a few hours to charge the batteries.
The largest power the generator system need to be charge in winter is 3000-16*64*0.7*3.06=3000-2193=807W
The size of inverter is directly determined by peak power. The total power in the table of application has a value of 3500W at peak power. However, not all lights and appliances are likely to be in used at the same time. Therefore, the peak power in demand will be 0.5*3500=1750W. The inverter for the application would then need to be at least 2KW.so we can use the SPI 1500-24SS inverter. By Appendix H
We assume the efficiency for the inverter to be 85%. Therefore, the battery bank should be able to provide daily the following amount of electrical energy:807/0.85=949.4W/day
The system Voltage is 24V,
So the battery bank is 949.4/24=40 amp-hour/day and the daily depth of discharge of 0.5: 40/0.5=80 amp-hour/day.
80 amp <812.5 amp, so we use PV battery storage which is 24RP900. However,
the C100 is too long to charging, so we choose the 12RP670 (C/20). And use 4 of
them. 2 in series in each parallel. So the Voltage is 24V and the current is 423*2=846 amp-hour/day.
A 20 hour charging period result in a charging rate of;
The battery charger should be selected on the basis of the battery bank voltage and the charging current. The battery voltage is 24V and a current of 42.3 amps is the maximum charging rate. Looking at Appendix I, we see that Suntron 24volt battery
chargers with code number is 11203 can provide currents up 50 amps. The
maximum of output power by charger will be 24*50=1200 W. the efficiencies of the selected battery charger is specified to be 0.6 for AC to DC conversion. Therefore, 1200/0.6=2000W
The maximum AC input to battery charger is 2000W. For this purpose we select 2.5KVA. Dunlite petrol generator (specified as UB2 in Appendix E). At 50HZ and 240V, and the AC output is 2000W. The generator should be turn off when the batteries are fully charged. The average current output would be: (50%+12%)/2=31% of the maximum rated current. The maximum DC for selected charger is 50 amps. The average is 50*31%=15.5 amp. And the battery bank is 40 amp-hours for winter.So the average running time is
At 40/15.5 = 2.6 hours/day for winter time.
However, in summer, there need not to use generator to charging, because in summer time, the 16*64*0.7*7.09>3500W.
Cost the system using information available from product suppliersDepends on APPENDIX L
Array frames: we use 4 modules in a frame: 245*4=980$
Petrol generators: 2.5KVA*400=1000$
Batteries :( 12V*423/1000) =5.076KW.
Battery chargers: 24*50/1000*1000=1200$
The house is 2km far from the grid and connection costs are estimated to be 25,000/km, there is a demand the most of 3500*365=1277.5kw electricity per annum. There are two options (the inflation rate and discount rate can be assumed to be 8% and 12% respectively, assume the system use 20 years of life time).
Option 1 (grid connection): cost of 25,000$/km, and paying a normal tariff of 24c/kwh (depends on electricity bill from company).
Initial cost: 2*25000=50,000$
The present value of the extension would be:
As stated above, the electricity consumptions 1277.5KW/year, in the first year, the electricity cost is: 1277.5*0.24=306.6$. This cost should be inflated at rate of 8% for subsequent years, discounted back to the present and added up. In other words, the present worth of electricity costs over a 20 year period would be a 450*PWF (8%, 12%, 20).
PWF (8%, 12%, 20) =1100820101220101211008-+.+.+.-+.= 13.95
Therefore, the present money over 20 years