Transistor
JF-Expert Member
- Aug 8, 2013
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5KV SOLAR SYSTEM COST AND POWER DELIVERENCE ESTIMATION
1. Size of battery required (A.H)
Data
Let battery efficiency be 60% =0.6
Maximum safe road for 5kv inverter is 4.5kv
Battery voltage is 12v
Solution
Actual wattage for 4.5kv load with efficiency of 0.6 is
Actual wattage =4500w/0.6=7500 watt(7.5kv)
Actual current for efficiency wattage of 7.5kv
Actual current=7500w/12v= 625 Ampere
Battery size(A.H)=Ampere x Time
Battery size(A.H)=625A x 12 hours= 7500 A.H approximate 8000A.H
Therefore we need about 8000A.H so as we can get approximately continuous power of 5kv for 12 hours with inverter of 5kv with efficiency of 60%.
2.Size of panel required
Since we need to replace 625 Ampere after every one hour which means we need to replace 7500Ampere for 12 hours.
Since solar panels they can provide full power only during full sun where in tropical region we have approximately of 8hours of full sun therefore the size of the panel will be
The panel should be able to deliver 950 Ampere per hour which is equal to
7600 Ampere for 8 Hours
Now the power of panel require is
Power=Current x Voltage
Power=950 x 17.8=16910 approximately 17000 Wattage
Therefore the solar panel required is
17000 Wattage
17.8 Voltage
950 Ampere
3.Control Charger required
12 Voltage
950 Ampere
REQUIREMENTS
- Nataka kuwa na solar system ambayo itaweza kunipa independent power supply (AC) ya kati ya saa 8 hadi 12, natakiwa kuwa na:
- Solar Panels ngapi?
- Battery Bank yenye LOAD kiasi gani?
- Ningependa kuwa na system ya 5,000 Watts, je, itaweza kutoa umeme wa kutosheleza kwa nyumba nzima ya kawaida, yenye 2 Fridges, 2 Deep Freezers, TV set, two computers (1 desktop, 1 laptop), taa kwenye vyumba tofauti 8, in two wings
Power required | 5000w=5kv |
Max time Consumption | 12 Hours |
Rate of consumption estimated | 4.5kv continuous power for 12 hours |
CALCULATION FOR ESTIMATION
1. Size of battery required (A.H)
Data
Let battery efficiency be 60% =0.6
Maximum safe road for 5kv inverter is 4.5kv
Battery voltage is 12v
Solution
Actual wattage for 4.5kv load with efficiency of 0.6 is
Actual wattage =4500w/0.6=7500 watt(7.5kv)
Actual current for efficiency wattage of 7.5kv
Actual current=7500w/12v= 625 Ampere
Battery size(A.H)=Ampere x Time
Battery size(A.H)=625A x 12 hours= 7500 A.H approximate 8000A.H
Therefore we need about 8000A.H so as we can get approximately continuous power of 5kv for 12 hours with inverter of 5kv with efficiency of 60%.
2.Size of panel required
Since we need to replace 625 Ampere after every one hour which means we need to replace 7500Ampere for 12 hours.
Since solar panels they can provide full power only during full sun where in tropical region we have approximately of 8hours of full sun therefore the size of the panel will be
The panel should be able to deliver 950 Ampere per hour which is equal to
7600 Ampere for 8 Hours
Now the power of panel require is
Power=Current x Voltage
Power=950 x 17.8=16910 approximately 17000 Wattage
Therefore the solar panel required is
17000 Wattage
17.8 Voltage
950 Ampere
3.Control Charger required
12 Voltage
950 Ampere
REQUIREMENTS
EQUIPMENTS | POWER/SIZE | COST |
Inverter | 5KV 12V | |
Solar Panels | 17000 Wattage 17.8 Voltage 950 Ampere | |
Charger Control | 12 Voltage 950 Ampere | |
Parallel connected Batteries | 8000 A.H (You can divide with any number of battery Ah) For example if you have 1000 A.H battery 8000 A.H/1000A.H=8 batteries of 1000 A.H are needed and so on. Which will be connected parallel | |