New partial off grid system help needed

[axis11]

Good quality MPPT CC are expensive. It depends on your purchasing capability and your future plans for upgrade.

[daz]

Ok great two panels then is it worth getting a 40a cc then or uprate that in a few years as well

Just remember that the amount of panels you require will be dependant on whether you are going to use a MPPT or PWM charge controller. PWM is only about 66% efficient whereas MPPT is about 95% efficient. What this basically means is that if you use a PWM controller, you will need more solar panels compared to if you used a MPPT controller (for the same amount of power).

[Tobym]

Ok on the subject of future proofing how do i calculate the size of the cable( inc voltage drop as its about 20-25 meters from panel to battery) for all 4 panels when I eventually put the others on. Is it the amps ie 4.85x4 so 20amp cable the. Add say another 10 amp for drop. If so 30 amp cable then is it ac or dc cable if ac then some domestic 240v 4mm twin core flex will be heaps wont it ?

[Tobym]

Ill defo use an mppt cc

[axis11]

There is a program to calculate the sizes of cables and losses with respect to loads and length. Try to search for it. IIRC " Voltage drop calculator". For future estimates, your panels will put out 4.86 amps @140v. That's about 50 amps @ 12v output from your MPPT.

[Tobym]

Ok how do you get the 50 amp and is that the amp/h that will be going into the batteries

[axis11]

50 amps, thats current output from the CC. That's 50AH in 1 hour, 100AH in two hours and so on. Do some reading on " How MPPT works".

[daz]

Ok on the subject of future proofing how do i calculate the size of the cable( inc voltage drop as its about 20-25 meters from panel to battery) for all 4 panels when I eventually put the others on. Is it the amps ie 4.85x4 so 20amp cable the. Add say another 10 amp for drop. If so 30 amp cable then is it ac or dc cable if ac then some domestic 240v 4mm twin core flex will be heaps wont it ?

You calculate voltage drop by using Ohms Law, which states that Voltage is directly proportional to the current mulitplied by the resistance.

V=IR
where:
V=voltage (in Volts)
I=current (in Amps)
R=resistance (in Ohms)


As an example, lets say you have a 140V 4.8A solar array, with 20metres of cable.
The cable's resistance can be found on its datasheet. In this case let's assume it is 1.0 Ohms per 100 metres. So for 20 metres the resistance is (1.0/100)20=0.2 Ohms.

Now.......
V=IR
V=(4.8A)(0.2R)
V=0.96V

So in our example, over a 20metre cable carrying 4.8A, you would have a Volt drop of 0.96V. So our solar array comes into the CC at 140V-0.96V=139.04V


This is why is is more efficient to use a higher voltage over a lower voltage, as a higher voltage equals a lower current, which means less Volt drop!

[inetdog]

Ok how do you get the 50 amp and is that the amp/h that will be going into the batteries

Please do not write amp/h. That literally means amps per hour, and since an amp is already a rate (proportional to electrons going by per second) the rate divided by time is meaningless.

If you keep a current of one amp going for one hour, you have transferred one amp-hour (AH) worth of electrons. It is AH that measure a batteries capacity. And AH times Volts is watt-hours (Wh) which is unit of energy.

[Tobym]

Ok thanks again so am I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours (watts per hour yes) into the cc so when 680.4 is divided 12v the batteries are having 56.7 amps put in to them an hour. Please tell me I've got it. So if I run all 4 panels as series pairs then linked in parallel the amps would be 9.72 and the volts would 70. So 70x9.72=680.4/12=56.7 again but there potentially more loss as the voltage is lower so cables would have to increase in size . Sorry for being a bit thick its been a very long time since I used my brain in this way

[daz]

Ok thanks again so am I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours (watts per hour yes) into the cc so when 680.4 is divided 12v the batteries are having 56.7 amps put in to them an hour. Please tell me I've got it. So if I run all 4 panels as series pairs then linked in parallel the amps would be 9.72 and the volts would 70. So 70x9.72=680.4/12=56.7 again but there potentially more loss as the voltage is lower so cables would have to increase in size . Sorry for being a bit thick its been a very long time since I used my brain in this way

I am not quite sure where you got the 50A idea from?....

am I correct then in saying the way you got 50a is 35v x4 panels =140v but the amps stay at 4.86 as they are in series so 140x4.86 = 680.4 watt hours

You are mixing up your units here.....

A 35W panel will put out about 1.94A (@18V). When you put panels in series, you add up the voltage, but the current stays the same. When you place panels in parallel, you add the current, but the voltage stays the same.

So 4x35W panels in series will give you 140W (1.94A @ 72V). Using a MPPT CC, you will be looking at about 140W/12V=11.66A going into the battery.

Sorry for being a bit thick its been a very long time since I used my brain in this way

Thats why we come here....to learn!

[Tobym]

Im a bit confused my panels are 170w and35v is that correct the 50a

[inetdog]

Im a bit confused my panels are 170w and35v is that correct the 50a

One panel at 175 watts could be 35 volts and 5 amps, not 50.

[Sunking]

Im a bit confused my panels are 170w and35v is that correct the 50a

No a 175 watt panel with a Vmp of 35 volts will have a Imp of 5 amps.

[Tobym]

Yep I get that it's 4.86a so with 4 of them it means 50a into battery. Ps say I have 2 12v batteries I assume that's 25a amp in each not 50 into each