33-SFA-2022 Ramesh Natrajan_ver17_13 June 2022 - epub by Natarajan Ramesh

Author:Natarajan, Ramesh
Language: eng
Format: epub
Publisher: Wings Publication
Published: 2022-04-11T00:00:00+00:00

H

3500 VA

48 Volts

Step 3

Since the requirement based on the load in the above case comes to 2109 VA, we shall consider the 2500 VA inverter as suitable for our use.

Step 4

Now we can determine the current that shall be drawn by the load ideally. We have the option of using inverter models F and G.

Model F 2500 VA divided by 36 Volts = Draws 69.44 Amps DC

Model G 2500 VA divided by 48 Volts = Draws 52.08 Amps DC

Step 5

Find out the efficiency of the inverter. Usually, it is 80% or 85%.

If efficiency is 80%, it means the efficiency is 80 divided by 100 = 0.80.

If efficiency is 85%, it means the efficiency is 85 divided by 100 = 0.85.

Step 6

Now we can determine the current that shall be required to be considered, for our operational needs, based on the inverter model, for the two different efficiencies.

Model F (36 Volts) 69.44 / 0.80 = 86.3 Amps. for 80% inverter efficiency.

Model F (36 Volts) 69.44 / 0.85 = 81.6 Amps. for 85% inverter efficiency.

Model G (48 Volts) 52.08 / 0.80 = 85.1 Amps. for 80% inverter efficiency.

Model G (48 Volts) 52.08 / 0.85 = 81.2 Amps. for 85% inverter efficiency.

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