Customer Has Been Dropping a Phase

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swish41

Member
Location
Denver, CO
Hi all!

One of our customers has been dropping a phase on a three phase system and the other interns and I have been told to simply multiply our data for our other phases by 0.5. I am confused on where this is coming from and why this would work.

Thank you!
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190613-1311 EDT

swish41:

Your question as presented is un-understandable.

What does "dropping a phase" mean? Apparently it does not mean loss of a phase, like a fuse blown on one leg.

What are you measuring?

Provide a complete description of your problem.

Note: on a 3 wire three phase system you can measure total power consumption with 2 wattmeters appropriately connected.

.
 

swish41

Member
Location
Denver, CO
RE

RE

Hello!

We are measuring the KW on each phase.

So we receive data from meters that has 3 phases. Each phase has a KW value assigned to it. Our customer has been loosing the data for one of the phases and they just tell us to multiply a different phases by 1.5 to get back t the phase that was lost.
 

hbiss

EC, Westchester, New York NEC: 2014
Location
Hawthorne, New York NEC: 2014
Occupation
EC
In this trade the term "dropping a phase" means the loss of one phase of a 3 phase service or supply. It could also mean the loss of one leg of a single phase 120/240v service.

This is vastly different than estimating the missing data of one phase from the other two phases.

-Hal
 

kwired

Electron manager
Location
NE Nebraska
Hello!

We are measuring the KW on each phase.

So we receive data from meters that has 3 phases. Each phase has a KW value assigned to it. Our customer has been loosing the data for one of the phases and they just tell us to multiply a different phases by 1.5 to get back t the phase that was lost.
So are you having actual power loss on one phase or just losing recording information for some reason on that one phase? Sounds like the latter is what you likely are having, but just trying to clarify.
 

mivey

Senior Member
Hi all!

One of our customers has been dropping a phase on a three phase system and the other interns and I have been told to simply multiply our data for our other phases by 0.5. I am confused on where this is coming from and why this would work.

Thank you!
It would have to be empirical data as there is no sound mathematical support for doing that given normal 3-phase unbalanced load.

If the load is well balanced, then adding 1/2 of phase A and 1/2 of phase B to get phase C would be a reasonable assumption. A lot of assuming going on though about load and power factors and would be limited in use.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
It would have to be empirical data as there is no sound mathematical support for doing that given normal 3-phase unbalanced load.

If the load is well balanced, then adding 1/2 of phase A and 1/2 of phase B to get phase C would be a reasonable assumption. A lot of assuming going on though about load and power factors and would be limited in use.

If it's balanced, wouldn't the power on phase C be the same as either A or B?
 

mivey

Senior Member
If it's balanced, wouldn't the power on phase C be the same as either A or B?
Yes. They may have a slight imbalance and want a better guess. I'm not a big fan of guessing without knowing the use for the result. Precision may or may not be important (quoted from Captain Obvious).
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Yes. They may have a slight imbalance and want a better guess. I'm not a big fan of guessing without knowing the use for the result. Precision may or may not be important (quoted from Captain Obvious).
It just seemed to me that if there were enough difference between the average between the power on A and the power on B for it to be significantly different from either that on A or B alone, the power on C could be just about anything.
 

mivey

Senior Member
It just seemed to me that if there were enough difference between the average between the power on A and the power on B for it to be significantly different from either that on A or B alone, the power on C could be just about anything.
Yup.
 
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