# Three Phase System

Discussion in 'General Electronics Chat' started by jcpark0911, Aug 6, 2007.

1. ### jcpark0911 Thread Starter New Member

Aug 6, 2007
3
0
I am designing a three-phase system and trying to find the total amperes per phase after hooking up all the equipment. There is one thing that I don't agree with my client, that is to calculate the total amperes per phase. Let me give you an example.

I have a 480V three-phase system that has 42 slots. There is an A/C unit connected to one of the slots available. The specifications of the A/C unit are the following.

Voltage: 460[v]/3[phase]/60[hz]
HP(Horse Power): 2.4
MCA(Minimum Circuit Amps): 30 [amps]
MOCP(Maximum Overcurrent Protection): 50 [amps]
Heat: 25 [kw]
As far as I understand, my boss has been using MCA as the operating current for the A/C unit, and he has been calculating the total current per phase using MCA. In order to find amperes on a three-phase system, I have been using the following fomula.

amperes = (kilowatts x 1000) / (volts x power-factor x 1.73) based on Ugly's Electrical References

Using the equation above, I have ameres = (25 x 100)/(480 x 1 x 1.73) = 30.106 [amps].

Question: Would it be safe to say that the amperes per phase is 30.106 [amps]? If so, would it also be safe to say that phase-B & phase-C have the exact amperes with different phase (radian or degree)?

Let's say that there are three more circuits with the same specifications of the A/C unit above on circuit number 1, 2, 3, and the one above is on 4. Would the total amperes per phase be 120 [amperes] per phase?

I have a client, Senior Electrical Engineer, who wants to divide the current by three. I am thinking that she considers a three-phase system as a parallel circuit with the same resistance. Do you think she is right, or we (my boss & I) are right?

This will cause a big problem because if our client is right, we may have oversized our wire size and breaker size (they are already installed on the rig). But, if we are right, then we need to tell our client that she is wrong. If we install wire and breaker based on her calculation, it will burn out the panel (worst-case-senario).

If I can prove whether she is right or wrong with proof, that would be a relief.

Thank you,

Jason

2. ### NAO1243 New Member

Aug 6, 2007
1
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I concur with your assessment. You are correct that 30.106 Amps is the current per phase on a three phase system. With three more loads of the same kind, the total would be 120 Amps per phase. The total power would therefore be 120*0.48*1.732*Power Factor of 1.0 or total P=100 KW. Unfortunately, your client is wrong and should not size the wires and breakers based upon her understanding, because they will be undersized and she will have big problems.

3. ### spar59 Active Member

Aug 4, 2007
51
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Voltage: 460[v]/3[phase]/60[hz]
HP(Horse Power): 2.4
MCA(Minimum Circuit Amps): 30 [amps]
MOCP(Maximum Overcurrent Protection): 50 [amps]
Heat: 25 [kw]

I have a rather awkward question - does the heat 25kW refer to built in resistive heating for winter use or is it the heat transfer capabilty of the heat pump system when driven by the 2.4 HP motor?

If it is the latter then this would be consistent with the 2.6A per phase full load current. In this case the requirement for a 30A supply circuit is to cope with motor starting current which can easily be 10x full load current and must not inadvertently trip the supply circuit breakers. You would still need to install decent size cables since you need to consider the voltage drop in them under motor start conditions.

However when considering the incoming feed to a distribution board feeding 3 or 4 of these units allocating 30A per unit is may not really be necessary since unless they are connected to a common controller they are unlikely to start simultaneously.

However when current is specified in a three phase system it is the current drawn on each phase not total current (unless specifically listed on a phase by phase basis) - you are definitely right in that respect!

Steve.

4. ### jcpark0911 Thread Starter New Member

Aug 6, 2007
3
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That is a relief. What I have to face from now on is how to tell my client that her way of doing it is not right. In other words, how can I prove it?

To answer spar59's question, I believe it is a built-in heater.

5. ### spar59 Active Member

Aug 4, 2007
51
0
As a matter of interest if you convert your 2.5 HP to kW it is not possible for this to be only 0.8A per phase (unless the motor is oversized and only runs at part load).

Since the full load current is given as 2.6A (neglecting any possible heater elements) and this is obviously quite different to the 0.8A your client expects how about just powering one up and sticking an ammeter in circuit or using a clip-on (jaw type) ammeter to prove the load is 2.6A per phase with a smug look on your face.

This does of course assume a unit is available for testing.

Test it in your clients absence first to avoid embarassment but I don't think there will be any.

Steve.

6. ### jcpark0911 Thread Starter New Member

Aug 6, 2007
3
0
That's a good one. I have to ask one of the field electricians to do it for me.
I gotta see the look on our client's face.

Thanks

7. ### recca02 Senior Member

Apr 2, 2007
1,211
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the three appliances wud be in parallel and hence the resistance eq decreases,
hence current increases to 120 as assessed by u .
simply put each circuit takes its own requirement independent of other till they can draw the power. so more circuit in parallel more current to be added