Determining Load in 3 Phase System

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I'm a system engineer responsible for specifying power quantities for
my systems. There are a lot of ways that are accepted in my industry
to do this that I know to be incorrect, so I'd like to hear from
someone with more specific expertise.

The systems are comprised of a combination of 120V resistive loads,
and 208V resistive and inductive loads, although the devices that have
inductive loads actually operate at various voltages, their power
supplies accept 208VAC. Most of the power available is 208/120VAC
60hz, usually in 400A increments.

What a lot of people do is add up the load P, divide by operating
voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at
208V, it would look something like

((5000/110)+(2500/203))/3 = 20A needed.

I can't imagine that's correct. What I have done based on advice and
reading is divide the load P by 1.73, a relevant power factor and the
supply voltage, so this:

(5000+2500)/(1.73*0.9*203) = 24A needed

But I can imagine that being wrong, also. What is really the way to do
this?

Thanks in advance.

 

[Paul E. Schoen]

I'm a system engineer responsible for specifying power quantities for
my systems. There are a lot of ways that are accepted in my industry
to do this that I know to be incorrect, so I'd like to hear from
someone with more specific expertise.

The systems are comprised of a combination of 120V resistive loads,
and 208V resistive and inductive loads, although the devices that have
inductive loads actually operate at various voltages, their power
supplies accept 208VAC. Most of the power available is 208/120VAC
60hz, usually in 400A increments.

What a lot of people do is add up the load P, divide by operating
voltage and divide by 3, so if we had 10 500w loads at 120V and 5 at
208V, it would look something like

((5000/110)+(2500/203))/3 = 20A needed.

I can't imagine that's correct. What I have done based on advice and
reading is divide the load P by 1.73, a relevant power factor and the
supply voltage, so this:

(5000+2500)/(1.73*0.9*203) = 24A needed

But I can imagine that being wrong, also. What is really the way to do
this?

Thanks in advance.

I think the correct way to do this is to figure the current for the single
phase and three phase loads separately. Also, your example will be an
unbalanced load with different current in each phase, because your loads
are not integral multiples of 3. But an approximation of average current
can be made.

The 10 500W 120 V resistive loads will draw an average of:
5000/(3*120) = 13.9 A per phase.

The 5 500W 208V reactive loads will draw an average of:
2500/(1.73*0.9*208) = 7.72 A per phase.

The total current will approximately 13.9 + 7.72 = 21.6 A.

This is an approximation. A proper calculation would involve the actual
impedances of the reactive loads, and a vector sum for the total current.
With 0.9 PF, the resistive current would be 6.94 A, and the reactive
current 3.38 A. The vector sum would be sqrt( (13.9+6.94)^2 + 3.38^2 ) =
21.11 A.

This still assumes a balanced load, which is not possible with your numbers
of resistive and reactive loads.

Paul

 

[Charles]

Some things to investigate:

The square root of 3 is oft used in 3-phase systems to determine line
currents and phase voltages.
Power factor is equal to apparent power/power
Power is V x I x Cos theta
A poor power factor can cause a surcharge on your energy bill and extra
current loads on your wires
Non-linear loads can cause high neutral currents (sometimes called triplens)
An involved subject area ... but doable

 

[Richard Seriani]

Some things to investigate:

The square root of 3 is oft used in 3-phase systems to determine line
currents and phase voltages.
Power factor is equal to apparent power/power
Power is V x I x Cos theta
A poor power factor can cause a surcharge on your energy bill and extra
current loads on your wires
Non-linear loads can cause high neutral currents (sometimes called
triplens)
An involved subject area ... but doable

Probably a slip of thr fingers, but you wrote, "Power factor is equal to
apparent power/power".
Actually, power factor is active power/apparent power, where active power id
VIcos(theta) and is measured in Watts and apparent power is VI and is
measured in Volt-Amps.

Richard

 

[Phil Allison]

"Richard Seriani"

"Charles"
Probably a slip of thr fingers, but you wrote, "Power factor is equal to
apparent power/power".
Actually, power factor is active power/apparent power, where active power
id VIcos(theta) and is measured in Watts and apparent power is VI and is
measured in Volt-Amps.

** You are both wrong:

Power Factor (PF) = Watts / VA


where -

Watts = the reading you get on a watt meter.

VA = rms volts x rms amps.

rms = reading on a "true rms" volt or amp meter of suitable type.


Cos theta is totally useless, when there is no theta.




....... Phil

 

[Charles]

Probably a slip of thr fingers, but you wrote, "Power factor is equal to
apparent power/power".
Actually, power factor is active power/apparent power, where active power
id VIcos(theta) and is measured in Watts and apparent power is VI and is
measured in Volt-Amps.

Thanks for the correction Richard. Inverted fractions are an on-going
problem here :>)

 

[Charles]

Thanks for the correction Richard. Inverted fractions are an on-going
problem here :>)

Oh, I forgot to mention that polite corrections are rare here ... I do
appreciate your demeanor!

 

[Richard Seriani]

Oh, I forgot to mention that polite corrections are rare here ... I do
appreciate your demeanor!

Charles,

I've had nearly 60 years experience making errors and I'm not done, yet. No
sense getting upset if someone else makes one.

Richard