Battery Array Calculations

The first thing we need to determine is how much usable energy we need to store.  This is a simple calculation based on intended energy usage and the amount of time you want to run without sunlight.

`Eu=Ed*days`

We know all of this information:

`Eu=50 KWh * 1 day` 

`Eu=50 KWh`

We need to adjust this for the battery type we are using.  Because of the Depth of Discharge complications, we will design the battery array to us a less complicated parameter.  We will just set a Maximum discharge target. We need 50 KWh of USABLE energy. I am going to start with a DoD of 100%.  Then, we can easily adjust the battery array size simply by dividing by the DoD target.  If we decide to set a DoD target of 50%, we can size the array as follows:

`Es=(Eu)/("DoD Target")`

`Es=(50 KWh) / (.50)`

`Es=100 KWh`

We can do the same thing with the number of batteries required.  Suppose we need 10 batteries at 100% DoD.  Well, we would need 40 batteries at 25% DoD.

`Bt = (Bc)/"DoD"`

`Bt = 10/0.25`

`Bt = 40` 

This will allow us to move forward with the design without having to nail down a battery at this time.  It will also allow us to change our mind without redesigning the entire system. 

Once the array is designed at 100% DoD, we can decrease the DoD by adding more batteries. It is a linear scale.  

Likewise, we can also use this same concept to add more time to the time we can supply power without the Sun.  For instance, suppose I need 10 batteries to carry me over for one day, if I decide to increase that to 3 days, all I need to do is increase the number of batteries by 3.  So I would need 30 batteries.

This is important.  Since we have increased the battery array to 3 days, we are now only discharging the batteries by 33% during normal use.  Only rarely, possibly never, are we cycling them 100%. That would only happen during 3 days of no energy collection AND we are using the maximum amount of power I have ever used for all 3 days.

Calculations for my Array

I am going to pick a random battery and calculate from there.  

Vmaxtanks VMAXSLR125 AGM
12V
125Ah
75 Lbs.
$290
6.8 x 12.9 x 8.7 inches

I will need to add these in groups of 4.  In this group of 4 they will be interconnected in series, thereby producing  the required 48 volts needed to feed the inverter.

`"Bn" = "Vin"/"Vb"`

`"Bn = (48 V)/(12 V)`

`"Bn" = 4`

Each battery can deliver 1500 watt-hours (amp-hours x voltage) of energy.  To meet the 50 KWh design specification we need to adjust the array for the efficiency of the power inverter.  To deliver 50 KWh of power to the house per day, we must design the battery array to deliver 50 KWh plus the energy wasted in the Inverter.  We will use an inverter efficiency of 95%.

`Et = (Ed)/"Inverter Efficiency"`

`Et = (50 "KWh")/(0.95)`

`Et = 52.632 KWh`

So we need to design the battery array to supply 52.632 KWh of energy to the Inverter

The number of batteries we need is calculated as follow:

`Bn=(Et)/(Eb)`

`Bn=52632/1500`

`Bn=35.088`

Since we need to add batteries in groups of 4, we would need 36 batteries to meet design specifications

The battery array would weigh 2700 lbs and cost $10,440

Now that we have a battery count, we can easily scale it up and down.  Suppose I want it to supply energy for 3 days.  I simply take the calculated battery count and multiply it by 3.  Notice I use the calculated battery count of 35.088 and not the adjusted count of 36.  So we multiply 35.088 by 3 and get 105.264 batteries.  So we need 108 batteries.

To get a 3-day array it would cost $31,320 and weigh 8100 lbs.