Why inverter rating in KVA?
Answers were Sorted based on User's Feedback
Answer / seank
True rating of a power supply is the amount of Voltage and Current that is flowing through the device.
If inverter rating is in kVA then the manufacturer is publishing ratings without assuming a power factor so the inverter could be used by rating on loads with potentially widely varying power factors.
This kVA would be best described by the Apparent Power(S).
If Apparent Power (S) is under estimated then it is possible the power supply will be undersized.
kVA to kW calculation
The real power P in kilowatts (kW) is equal to the apparent power S in kilovolt-amps (kVA), times the power factor PF:
P(kW) = S(kVA) × PF
| Is This Answer Correct ? | 3 Yes | 0 No |
Answer / sreeknp
we cant predict the pf of load ,dependig upon the load kw
changes so we can say current supply and voltage only ,for
example a300va unit to which five tube can be connected
since its pf is .4 then ups will be over loaded if one ortwo
addtional tube will burn this ups i think ur doubt is cleared
| Is This Answer Correct ? | 11 Yes | 9 No |
The inverter rating is in KVA because of in any active
components (like transformers, D.G, inverters etc..), the
power is calculating based on the output of the devices. At
the output terminals, no losses are considered causing by
active/passive loads (R,L,C). This is not depend on the
input.
But in the passive components (like motors, engine etc..),
the output is taking by means of input.
The output is always lesser than the input due to losses.
So the motors are denoted in the unit of KW.
| Is This Answer Correct ? | 6 Yes | 8 No |
We have drive feeding an output contactor for a 5000 hp motor @ 5kV application. The motor FLA is 629amps x 1.25% service factor = 786.25 amps The cable that we current have in place between the drive and the output contactor is 350kcmil- MV-105 (parallel cables). Per the NEC table 310.60 (C) (69). This show the cable ampacity is at 615 amps per cable for a rating of 1230 amps. We have derated the cable for the cable tray and the multiple cables application and our calculations show that after the correction factor factor we are at 922amps. 1230 amps x .75% correction factor factor = 922.5 amps. The customer is driving us to then use another safety factor of .25% off of the already derated cable. Can you please provide a longhand calculation for your solution on this application?
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