Question

Calculate the battery energy and solar panel power required for the AC loads of 500 Wh per day. Battery DoD is 50%, Battery losses are 15%, Controller losses are 5%, inverter losses are 15%, Solar radiation is 5.25 kWh/m2/day, Solar panel losses are 25%

Answer 1

Total energy required by the  load = 500Wh per day

INVERTER

energy= 500wh/day

losses= 15%

losses = 500*15/100  = 75

Input power of inverter  = 500+75

                                        = 575Wh/day

BATTERY

losses= 15%

losses= 575*15/100= 86.25

Input power=  575+86.25= 661.25wh/day

DOD = 50%

=661.25/0.5 = 1322.5 wh/day

SOLAR PANEL

losses are 25%

losses = 1322.5*25/100= 1830.6wh/day

As Sun Peak Hour is in kwh we have to convert our total  energy in kwk so,

=1830.6/1000= 1.8Kwh/day

sun peak hour (sph)= 5.25kwh/m2/day

1.8/5.25= 0.342kw i.e. 342 watts

so if i used a 45 watt panel, i have to use

342/45= 7.4 approx 8 panels