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If I know my single phase wattage total (adding up the wattage of all my loads (2500 watt lamps) for a power total) what do I do to get a 3-phase wattage total.

Do I divide that by 3 or do I divide that by 1.73 (square root of 3)?

I can find 3-phase to single phase power calculations (on the net) but not the other way (single phase totals to 3-phase total).

Thanks
KJ

2007-03-09 06:21:38 · 7 answers · asked by Ken J 1 in Science & Mathematics Engineering

I didn't phrase my question quite right, so if watts are watts (3-phase or single phase) so I divide the total watts by 3 to get the watts per leg (or phase) (This is on a purely resistive load). So as Helmut put it P= P1 + P2 + P3 so I can divide my total watts by 3 to get watts per phase and then get my current per phase by I = P/E? Thanks KJ

2007-03-09 08:05:29 · update #1

7 answers

Given resistive loads,
P = P1 + P2 + P3
If your phases are balanced
P = 3Pφ

P = V(line-line)*I(line)√3
V(line-neutral) = V(line-line)√3
Pφ = V(line-line)*I(line)/√3

2007-03-09 07:41:01 · answer #1 · answered by Helmut 7 · 0 0

Watts are Watts. there is no difference between single phase watts and 3 phase watts.

Your 3-phase amps will be less, but your watts will be the same, if you are concerned about energy usage.

2007-03-09 07:21:43 · answer #2 · answered by H_A_V_0_C 5 · 0 0

3 phase power =1.732 X voltage x current x power factor(1)
is the correct Answer for 3 phase power.power factor will be 1 in case of resisitive load, it will vary depends upon inductive load,if capacitance increase it wil go to unity

2007-03-09 07:08:48 · answer #3 · answered by ricky414 5 · 0 0

3 area device in many situations has 3 warm leads, a nuetral and a floor...the three stages could be related to the nuetral to form 3 seperate circuits one hundred twenty ranges out of area with one yet another... 2 area device is an identical element with 2 warm leads...a 240 volt connection on your place for a intense modern-day equipment could be in many situations hooked as much as a 2 area furnish...you need to use 2 of the three hots from a three area device to potential a 2 area gadget. in spite of if, the device is extremely not balanced simply by fact the load is on 2 of the three circuits. this suggests you will in many situations have a sprint modern-day on the nuetral connection...

2016-10-17 23:25:01 · answer #4 · answered by ? 4 · 0 0

Single phase KW= (V*A*Power Factor)/1000

3 Phase KW=(V*A*Power Factor*1.732)/1000

Power factor = Watts/(V*A)

2007-03-09 07:05:54 · answer #5 · answered by uisignorant 6 · 0 0

The watts should remain the same. It will be the voltage and the amps that vary.

2007-03-09 06:27:10 · answer #6 · answered by Anonymous · 0 0

for 3 ph delta
line amps=1.73*phase amps
line volts=phase volts
line watts=1.73*phase watts
for 3 ph wye
line amps=phase amps
line volts=1.73*phase volts
line watts=1.73*phase watts

2007-03-09 09:42:16 · answer #7 · answered by Anonymous · 0 0

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