Student, have a few questions about breakers/ power in general

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Hello fellow electrical enthusiasts,

I am currently a student studying electrical engineering, currently pursuing an engineering internship and very interested in the field of power systems, especially in low voltage/ building applications.

I have a few questions about some concepts within power engineering.

Let's say we have a 3 phase circuit breaker that is supposed to be delivering power to a building. The power is calculated for each phase (a,b,c) and then the total power is adjusted (using NEC/ other code) based on demand factor, and that allows us to calculate whether or not we're overloading the breaker (for example, some breakers have 225 amp rating).

Let's say I calculate 20kVA per phase (say we're doing something simple like receptacles/ lights). Does this mean the total power being used by the breaker is 60kVA, or is it 20kVA * sqrt(3), which is 34kVA. Is the 225 amp rating per phase, or is it the total current being absorbed at once by the system?

Now let's also assume the breaker is connected to a 3 phase motor. This motor draws 14,411 VA per phase. The strange thing is, that the motor is rated at 20 Horsepower, which is equal to 14,411 Watts. But if you add up all the phases (14,411*3), this gives you 58 horsepower.

Another thing that is confusing me. If we have a panel that isn't connected to anything, let's say it's rated at 225 Amps at 120 volts. Does that mean this panel is using power, even though it's connected to nothing?

Thank you for any help
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Another thing that is confusing me. If we have a panel that isn't connected to anything, let's say it's rated at 225 Amps at 120 volts. Does that mean this panel is using power, even though it's connected to nothing?

If no current is flowing then no power is being used. P = VI, and if I = 0, then P = 0. How far along are you in your EE studies? That's first semester circuits stuff.
 

__dan

Senior Member
Generally if you're calculating breaker and wire size, the unit or dimension you want is Amps. You're talking about knowing if you're overloading the equipment, it's an Amp rating.

Calculating for the load, it's probably mostly all kW at that point. Determine or calculate the total. Next you may have a choice of system voltage. System voltage might be relatively fixed once the choice has been made, then you can convert to Amps to get equipment sizes.

I've never used kW per phase. It would be load total kW, then the available or choice of system voltage, 480, 240, 208, then the choice or availability of single or three phase supply connection. Once you know load total kW and system voltage, single or three phase, then you can convert to Amps per phase which your equipment is rated for.
 
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Iron_Ben

Senior Member
Location
Lancaster, PA
Hello fellow electrical enthusiasts,

I am currently a student studying electrical engineering, currently pursuing an engineering internship and very interested in the field of power systems, especially in low voltage/ building applications.

I have a few questions about some concepts within power engineering.

Your thirst for knowledge (along with your enthusiasm) is laudable. This tells me that the electrical field *may* be a good fit for you. Here’s a tip for you, from a 62 year old retired electrical engineer who worked in heavy industry and power his whole career:

With only a very few exceptions, the best engineers have hands-on experience as grunts, helpers, apprentices, etc. Get yourself in with an electrical contractor or find work in an industrial setting, or wiring houses. Such experience is invaluable. Good luck to you.
 

mbrooke

Batteries Included
Location
United States
Occupation
Technician
Hello fellow electrical enthusiasts,

I am currently a student studying electrical engineering, currently pursuing an engineering internship and very interested in the field of power systems, especially in low voltage/ building applications.

I have a few questions about some concepts within power engineering.

Let's say we have a 3 phase circuit breaker that is supposed to be delivering power to a building. The power is calculated for each phase (a,b,c) and then the total power is adjusted (using NEC/ other code) based on demand factor, and that allows us to calculate whether or not we're overloading the breaker (for example, some breakers have 225 amp rating).

Let's say I calculate 20kVA per phase (say we're doing something simple like receptacles/ lights). Does this mean the total power being used by the breaker is 60kVA, or is it 20kVA * sqrt(3), which is 34kVA. Is the 225 amp rating per phase, or is it the total current being absorbed at once by the system?

Now let's also assume the breaker is connected to a 3 phase motor. This motor draws 14,411 VA per phase. The strange thing is, that the motor is rated at 20 Horsepower, which is equal to 14,411 Watts. But if you add up all the phases (14,411*3), this gives you 58 horsepower.

Another thing that is confusing me. If we have a panel that isn't connected to anything, let's say it's rated at 225 Amps at 120 volts. Does that mean this panel is using power, even though it's connected to nothing?

Thank you for any help


Per phase. So a 225 amp breaker is 225 amps per phase.


225 amps x 208 volts x 1.73= 81kva.



If no current is flowing no power or work is being done.


Technically you will have capacitance from the busbars to ground (panel tub), and because that reactive current is flowing through conductors with resistance, a very minuet amount of power is used/dissipated. But we are talking micro watts and its negligible.
 

mbrooke

Batteries Included
Location
United States
Occupation
Technician
Your thirst for knowledge (along with your enthusiasm) is laudable. This tells me that the electrical field *may* be a good fit for you. Here’s a tip for you, from a 62 year old retired electrical engineer who worked in heavy industry and power his whole career:

With only a very few exceptions, the best engineers have hands-on experience as grunts, helpers, apprentices, etc. Get yourself in with an electrical contractor or find work in an industrial setting, or wiring houses. Such experience is invaluable. Good luck to you.

Could have done without the first few sentences.
 

jim dungar

Moderator
Staff member
Location
Wisconsin
Occupation
PE (Retired) - Power Systems
Now let's also assume the breaker is connected to a 3 phase motor. This motor draws 14,411 VA per phase. The strange thing is, that the motor is rated at 20 Horsepower, which is equal to 14,411 Watts. But if you add up all the phases (14,411*3), this gives you 58 horsepower.

The conversion between HP and kW will always result in a lower value than what you calculate for the input power.
Remember motors are rated in output power, you must account for the losses (e.g.. Power Factor and Efficiency) in the conversion of power between electrical and mechanical.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190614-0815 EDT

sengineer101:

I have questions about your education.

At what school are you attending for your EE program?
At what level are you in this program?
Have you previously been an electrician, or studied in some trade area?
What is an engineering internship? Doing what, and where?

There are some strange things in your first post. An EE student is likely to have had high school physics, and certainly is required to have college physics before taking EE courses. This should mean that you know power is additive, and in most cases does not involve vector or phasor addition.

Three phase concepts are not likely taught in physics, or first year EE courses. Early courses are going to cover basics. Where would you have been exposed to an equation with 1.732 in it without a derivation of where that constant came from?

I think electricians view and think about how circuits work differently than do electrical engineers. Electrical engineers are taught about basic concepts and how to apply these basics to more complex problems. Electricians are taught more along the line of --- here is a rule, follow it.

How are multipole breakers rated? Each pole has exactly the same rating of each other pole, and that is the rating of the breaker. If two poles carry no current and one pole exceeds its trip time curve does the breaker trip?

Suppose you have a 3 phase wye source with 120 V line to neutral, and you load line to line with a resistance of value R. What is the current in each line including neutral? What value of R will cause the breaker to trip? You have to make assumptions and fill in the missing information to provide an answer.

For this same source you have 3 equal resistors connected in delta dissipating 1500 W each. What are the various line currents including neutral?

Let's say I calculate 20kVA per phase (say we're doing something simple like receptacles/ lights). Does this mean the total power being used by the breaker is 60kVA, or is it 20kVA * sqrt(3), which is 34kVA. Is the 225 amp rating per phase, or is it the total current being absorbed at once by the system?
So that you can easily visualize the problem consider the source is wye and the loads are line to neutral.

First, kVA is not necessarily equal to power. But for a resistive load power does equal kVA. So assume resistive. Since power is additive the total power is 60 kW. Where does sqrt(3) come from? Out of the blue sky?

Now let's also assume the breaker is connected to a 3 phase motor. This motor draws 14,411 VA per phase. The strange thing is, that the motor is rated at 20 Horsepower, which is equal to 14,411 Watts. But if you add up all the phases (14,411*3), this gives you 58 horsepower.
How many watts per horse power? See https://www.google.com/search?q=hp+...rome..69i57.6634j0j8&sourceid=chrome&ie=UTF-8
745.7 * 20 = 14,914 or
746 * 20 = 14,920
Where did 14,441 come from? 722.05 is the result of 14.441/20.
But as pointed out by jim dungar that is not the electrical input.

Breakers trip on current not power.

Another thing that is confusing me. If we have a panel that isn't connected to anything, let's say it's rated at 225 Amps at 120 volts. Does that mean this panel is using power, even though it's connected to nothing.
A physics class background would answer this question.

.




.
 

Besoeker3

Senior Member
Location
UK
Occupation
Retired Electrical Engineer
The conversion between HP and kW will always result in a lower value than what you calculate for the input power.
Remember motors are rated in output power, you must account for the losses (e.g.. Power Factor and Efficiency) in the conversion of power between electrical and mechanical.
Why not just stick with kW?
Power is power.
For me, and most of the rest of the world, HP is a redundant unit.
 
I probably should have paraphrased my question. I know that when no load is connected, we don't draw power. But we still need to account for the load operating at maximum capacity when we do load calculations. For example, a 180 VA plug may be connected to nothing (thus drawing no power), but we still need to add it into our panel schedules when calculating the rating. That's what was confusing me. But I get it now.

Also, the sqrt(3) is derived from the relationship between VL-L and VL-N in a 3 phase system. No need to insult my education. I've never worked with power systems outside a class setting (analysis, etc) so this is very confusing to me. I've been mostly exposed to single phase circuits so the whole 3 phase systems, delta to wye, induction motors etc is still a learn in progress.
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190614-1334 EDT

sengineer101:

The problem I saw in your first post was that you spoke like an electrician rather than like an engineering student. If I don't really know what your background is, then I am not in a good position to communicate with you.

Also, the sqrt(3) is derived from the relationship between VL-L and VL-N in a 3 phase system.
That really tells me nothing, and it does not tell me if you can derive that relationship, nor what assumptions are necessary for it to be true. It does not tell me if you understand the relationship. If the system is unbalanced is it true?

.
 
understandable, the square root of 3 comes from the phase relationships when you sum up phasors. For example in a wye configuration the line to neutral voltages could have the form 1<0 and 1<-120. For VAN and VBN. Now if we wanna find VAB we subtract, and doing simple phasor addition, we realize that this is equal to 1- (-.5-.87j) which equals to 1.5+ .87j. Magnitude of this is 1.73 which is square root 3.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
190615-2007 EDT

sengineer101:

Consider the following question, and if you desire to analyze it, then provide all details on your analysis.

A Simpson 260 or 270 meter is a Weston D'Arsonval type movement with Simpson meters after about 1960 being taut-band.

1. Use a DC regulated power supply to supply voltage to the Simpson.

2. Use the 50 V range.

3. Select AC and adjust for an exact reading of 50 V.

4. Switch to DC and record the reading.

5. Do a theoretical analysis of what the ratio should be.

6. How does an actual measurement compare, and why might it be different?

.
 
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