Julian

[Dennis Alwon]

Okay, what am I missing? Why are we only looking at 49kw. Doesn't each phase have 2 loads of 49kw?

 

[gar]

190104-0837 EST

Dennis Alwon:

No.

For what you are thinking you have to more clearly define phase and its relation to the load.

When you have a delta load on a delta source the two currents at a node are not in phase with each other, and you need to use vector addition at the node to determine the third current at the node.

.

 

[electrofelon]

Okay, what am I missing? Why are we only looking at 49kw. Doesn't each phase have 2 loads of 49kw?

we have three single phase 49kw loads, so 147KW on the service or feeder. Current for each single phase load is 102 amps. Current on the service or feeder is 177A. The reason we dont have 2 times 102 amps on each phase the phase angle. Its similar to how when you have two L-N loads on opposite legs of a split phase system, the neutral currents cancel. In this case because of the phase angles, they dont completely cancel, you only go from 2 to 1.732. Another way to look at which may not be clear at first sight the way it is drawn in the OP, is that when you have three single loads connected A-B B-C C-A these three loads are essentially connected in a delta configuration. So you can think of it as a single delta load.

 

[Strathead]

Julian, others have alluded to this, but let me try to make it a little more straight forward. There are different codes for different situations. None of those are being spelled out. Fuses have to be sized a maximum of 125% of the transformer secondary output. Fuses have to be sized a % for a motor they feed. Fuses have to be sized for the wire size for resistive loads. Each of those could result in a different size, depending on the entire situation.

 

[gar]

190104-0859 EST

Dennis Alwon:

Power magnitudes are additive, current magnitudes are only additive if the currents are in phase.

Another viewpoint is logic. You have three source phase voltages, and you have three equal loads in a delta configuration. You can not say that each source voltage has two loads of the previous said value because this would mean the total load would be six loads.

.

 

[LMAO]

49000W/480=102A

That's the current through every conductor tapped off the transformer (vertical lines in your schematic).

 

[Dennis Alwon]

So if there were only 2 49 kw loads both going to phases A and B - loads are sp 480V--what is the load on A or B?

 

[gar]

190104-1031 EST

Dennis Alwon:

Your question is not clear.

What do you mean by phase A or B? What does load on A or B mean.

Each phase will has to be defined by two points. Your phases A and B are not defined.

A load has to be defined relative to the two points it connects to.

.

 

[Strathead]

So if there were only 2 49 kw loads both going to phases A and B - loads are sp 480V--what is the load on A or B?

Dennis I will try to explain. The load isn't on A or B. First keep in mind that a three phase transformer is a term we use but technically it is actually three individual transformers wired together.

The load is on a transformer. In the case of Delta, the A-B transformer In the case of Wye the load in your situation would be shared between the A-common and the B-common transformers. So in the Delta case we are discussing, two 49kw loads both connected to points A and B would put all of the load on that transformer and the load would be in phase and all additive. When one is on A-B and one is on A-C is when the whole cancellation phase angle brain wrenching, not as easy to grasp stuff happens and the square root of three comes in to play. In a three phase delta connection with three transformers and the above hook up I believe all three transformers actually see some of the load, because internally there is a connection between B and C so current flows internally. Remember that you can have an open delta connection. this is actually only two transformers 120º apart and that third coil doesn't exist, but you still have three phases because you have three points, A,B, C 120º apart from each other.

Does that make it as clear as mud?

 

[julian padilla]

Wow, that was enlightening. After all that wonderful discussion it appears the computer of augie47 was correct. I'm concluding, to size the conductor at 3/0 AWG per Table 310.15(B)(16), and OCP at 200 amps per Table 240.6(A) for Phase B. Thank you all for your help, I truly appreciate your input! Hopefully, I can find a few more interesting questions for y'all - were pretty busy around here! I'll try to study the "Basic Electrical Measurements" text recommended (since I prefer to do the math, I better study up).
Julian

 

[Dennis Alwon]

190104-1031 EST

Dennis Alwon:

Your question is not clear.

What do you mean by phase A or B? What does load on A or B mean.

Each phase will has to be defined by two points. Your phases A and B are not defined.

A load has to be defined relative to the two points it connects to.

.

Strathead explained it well, What I was saying was given the same example but with 2 49kw loads connected to Phase A and Phase B only. Phase C is not connected at all... What would the reading be on Phase A and Phase B--

 

[Besoeker]

I've been drawing out the circuits trying to see where the change in the load amps happens. I am very familiar with the 1.73 factor for three phase power. I'm seeing the B phase node at 102 + 102 = 204.

Te currents are not in phase so you can't just add them arithmetically.

 

[Besoeker]

Been too many years since college & computers have ruined me..
My computer, if it can be trusted, shows the load to be 178 amps, next standard size fuse would be 200 amps.
I'll let one of the young 'uns show you the math(.and check me)

I agree. Assuming the loads are all unity power factor.

 

[Strathead]

Strathead explained it well, What I was saying was given the same example but with 2 49kw loads connected to Phase A and Phase B only. Phase C is not connected at all... What would the reading be on Phase A and Phase B--

Thank you Dennis. That means a lot to me, coming from you. I still remember seeing Jim Morrison as my greeter when i first joined the forum. And as a new second year apprenticeship instructor, I am trying every day to find fresh ways to express the complexities of electricity without causing one's head to explode. I am of the opinion that most of the guys who go through my class are better served understanding a few applicable but not in depth things that get them 98% of the way than trying to filling that last 2% and alienating 90% of them.

By this I mean to imply that there is 10%, or possibly less, that hang out here because we ain't scared and then the 1% explain to us that last 2% .... over and over until we get it.

 

[gar]

190104-1643 EST

Dennis Alwon:

Relative to your post #31

Strathead explained it well, What I was saying was given the same example but with 2 49kw loads connected to Phase A and Phase B only. Phase C is not connected at all... What would the reading be on Phase A and Phase B--

I still really don't know what your question is.

But if the question is:

If two 2.49 kW resistive loads are connected in parallel, total load power is 4.98 kW, and this parallel combination is connected between two points labeled A and B which differ in potential by 480 V, then what is the current in the wire to either A or B?

Obviously this is a single phase unity power factor problem, and the current is simply 4980 / 480 = 10.4 A. Thus, 5.2 A flows thru each 2.49 kw load. All currents are in phase, and in phase with the A to B voltage.

.

 

[Dennis Alwon]

So if I took an amp meter and put it around phase A or Phase B conductors,then what would I read-- would I read 5.2 amps on each wire or would I read 10.4 amps? We are assuming a delta trany..

What I understood was we would only read 5.2 amps

 

[gar]

190104-1848 EST

Dennis Alwon:

A three phase transformer has nothing to do with your question. Two output terminals of a three phase transformer constitutes a single phase source. It is the same as two terminals of a battery in a DC circuit, or two terminals on a doorbell transformer.

Clamp your meter around the lead going to terminal A and the reading is 10.4 A . Same for terminal B.

Clamp your meter around one lead to one 2.49 kW load and the reading is 5.2 A.

Do the experiment in some fashion on your bench.

.

 

[Dennis Alwon]

190104-1848 EST

Dennis Alwon:

A three phase transformer has nothing to do with your question. Two output terminals of a three phase transformer constitutes a single phase source. It is the same as two terminals of a battery in a DC circuit, or two terminals on a doorbell transformer.

Clamp your meter around the lead going to terminal A and the reading is 10.4 A . Same for terminal B.

Clamp your meter around one lead to one 2.49 kW load and the reading is 5.2 A.

Do the experiment in some fashion on your bench.

.

Well that is what I originally thought and with all the chatter I got confused. In the original post there is a 3 phase delta and for some reason you don't add the loads on one feeder conductor. Do you see what I am saying?

 

[electrofelon]

Well that is what I originally thought and with all the chatter I got confused. In the original post there is a 3 phase delta and for some reason you don't add the loads on one feeder conductor. Do you see what I am saying?

Are you questioning the case of two single phase L-L loads connected A-B and B-C and the resulting currents on the three phase system?

 

[GoldDigger]

Strathead explained it well, What I was saying was given the same example but with 2 49kw loads connected to Phase A and Phase B only. Phase C is not connected at all... What would the reading be on Phase A and Phase B--

What convention are you using to relate the Phase letter to the wire designator?
I will call the wires L1, L2 and L3 to avoid any other association. And I will call the voltage from L1 to L2 Phase A, the voltage from L2 to L3 Phase B and the voltage from L3 to L1 Phase C.

So when you say a load is connected to A and B only, that means that one load is connected between L1 and L2, the other is connected from L2 to L3. So the current in L1 is 49kW divided by the Phase A voltage. The current in L3 is also 49kW/V. The current in L2, however, is the vector sum of the two currents from the two loads. That one will have a sqrt(3) factor in it, and will not be in phase with either load voltage.