Which sizing method is correct?(solar power system design)~ URGENT !!!

[Sunny55]

I have try to size a standalone solar power system here and i wish to know which method is correct and gives the highest accuracy as well as working properly as designed.

Method 1


DC loads= 70Wh/day
AC loads= 288Wh/day
inverter efficiency @ 0.9 = 288wh/0.9 = 320wh/day

total power used = 320 + 70 = 390wh/day

solar panel input = (390/4hours)*1.4
= 136.5W

Solar regulator = 5.7A*1.25 = 7.125A
Add 25% extra capacity gives around 9A.

Inverter = 48W*2 = 96W (to take 150W inverter)

Battery required = (390*1day/12V) / 0.7(DoD)*1.1 = 51.07Ah.

Method 2

DC loads= 70Wh/day
AC loads= 288Wh/day
Total power = 288+70 = 358wh/day

add 20% for system losses = 71.6+358 = 429.6Wh/day

solar panel size = (429.6wh/day / 4hours) = 107.4W

solar regulator and inverter size same as method 1.

Thanks.

 

[BradtheRad]

I believe you should track AC losses separately from DC losses.

The losses tend to vary more when an AC inverter is involved...
Depending on:
* what waveform it puts out
* whether the load is resistive, or is a motor
* how much surge current your appliances draw, and how often they turn on per day.

Your inverter needs the capacity to put out sufficient current as appliances start up.
Likewise your battery needs to be sized to put out sufficient current to power the inverter at such times.