# Measuring 3 phase power consumption

#### rups22

Joined Oct 26, 2022
16
Newbie question:
I am trying to calculate my (restaurant) total power consumption at the supply fuses using a clamp meter. The supply is three phase and an example reading would be:
P1: 3.1A
P2: 5.9A
P: 5.1A

Do simply sum these current readings and multiply by 230v (Spain) or is there a more complex calculation required

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#### Dodgydave

Joined Jun 22, 2012
11,304
It's the sum X 1.73 X power factor (pf)X Voltage.
If you don't have any inductive loads like motors, then the PF will be 1, .

Ideally you need to balance the phases.

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#### Ian0

Joined Aug 7, 2020
9,842
Assuming that you are mostly running 230V equipment and assuming unity power factor* then it is 230V * the sum of the currents.
Effectively, you have three single phase supplies, the first one supplies 3.1A : that is 713W
The second supplies 5.9A : that is 1357W
The third supplies 5.1A: that is 1173W
Total : 3243W

* Guessing at ovens which will definitely be unity power factor and a few small motors (fans, mixers etc) which will be less than unity but small enough loads not to make much difference.

#### rups22

Joined Oct 26, 2022
16
Assuming that you are mostly running 230V equipment and assuming unity power factor* then it is 230V * the sum of the currents.
Effectively, you have three single phase supplies, the first one supplies 3.1A : that is 713W
The second supplies 5.9A : that is 1357W
The third supplies 5.1A: that is 1173W
Total : 3243W

* Guessing at ovens which will definitely be unity power factor and a few small motors (fans, mixers etc) which will be less than unity but small enough loads not to make much difference.
Exactly what I needed to know thank you. Exactly it's ovens, air con unit and extractor fan motor etc. Thank you again

#### rups22

Joined Oct 26, 2022
16
It's the sum X 1.73 X power factor (pf)X Voltage.
If you don't have any inductive loads like motors, then the PF will be 1, .

Ideally you need to balance the phases.
Insightful, thank you mate

#### michael8

Joined Jan 11, 2015
415
I'm really confused by this. Let's ignore power factor and assume 3 phase 230 volt to neutral which is about 400 volt
phase to phase. OK, say I have the following:
p1: 1A
P2: 1A
P3: 0A

I've got a 400 ohm resistor between P1 and P2 and nothing connected to P3. So my resistor has 400 volts across it and one ampere flowing through it. The power is easily VI = 400 * 1 -> 400 W. The currents in P1 and P2 are the same (but opposite sign) as it has nowhere else to go...

Your saying 230 * sqrt(3) * (1+1) -> 230 * 1.73 * (1 + 1) -> 400 * 2 -> 800 watts? Huh?

#### Ian0

Joined Aug 7, 2020
9,842
y
I'm really confused by this. Let's ignore power factor and assume 3 phase 230 volt to neutral which is about 400 volt
phase to phase. OK, say I have the following:
p1: 1A
P2: 1A
P3: 0A

I've got a 400 ohm resistor between P1 and P2 and nothing connected to P3. So my resistor has 400 volts across it and one ampere flowing through it. The power is easily VI = 400 * 1 -> 400 W. The currents in P1 and P2 are the same (but opposite sign) as it has nowhere else to go...

Your saying 230 * sqrt(3) * (1+1) -> 230 * 1.73 * (1 + 1) -> 400 * 2 -> 800 watts? Huh?
Your phase currents are no longer in phase with your live-to-neutral voltages because you have connected the load between two lives.
That's why I checked to see if the TS had only 230V live-to-neutral loads.
Having a single 400V load would be very unusual.

#### Ian0

Joined Aug 7, 2020
9,842
Ideally you need to balance the phases.
That's not so easy when you have many different single phase loads which can be switched on randomly!