line current v. phase current

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ToolHound

Senior Member
From equation in the 'Delta Connected' image attached below...

What is the practical value of the equation

phase amps X 1.732 = line amps

Where to use that ? In many transformer problems in my electrician workbooks, that equation is nowhere to be seen. Yet I come across it in another publication. It must be good for something ???? What is 'phase amps' anyhow ?

I guess between 120/208, the 208 would be the line voltage.
120 would be phase voltage.

But what is 'phase amps' ? I am trying to make a first step in
grasping use/meaning of the 1.732 multiplier in amps calcualations. I am use to using the 1.732 multiplier only 'next to a voltage number'.

What gives here?
 

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rcwilson

Senior Member
Location
Redmond, WA
Think of three 10 kVA single phase transformers wired in a delta as your drawing shows. That would be a 30 kVA three-phase bank with each transformer winding delivering 10 kVA. Assume it is 480V three phase system.

What is the line current when the bank is fully loaded at 30 kVA?
I = KVA x 1000/(1.732 xVolts line-line). I= 30 kVA x1000/(1.732 x 480V) = 36 amps.

Clamp a meter on the line feeding the transformer and you will read 36 amps.

But what is the current in each transformer winding?
Treat each transformer as if it was a single phase unit fed by 480V.

Iphase= 10 kVA x1000/ 480V = 20.83 amps. Clamp on the H1 or H2 lead of any of the three transformers and you will read 20.8 amps. 20.83 x 1.732 = 36 Amps.

Or think of 3 1000W heaters wired in a delta as a 208V 3 kW heater. Total power = 3 kW at 208V.
What is the circuit amps feeding it? 3 kVA x1000/(1.732 x 208V) = 8.33 Amps.

What current is flowing in each 1 kW heater? (Remember they are single phase heaters connected across 208V).
I=1000w/208V = 4.8 Amps. Note that 4.8 Amps x 1.732 = 8.33Amps.

That A phase line wire has to carry the A-B phase 4.8 amps and the C-A phase 4.8 amps which add up to 8.33 amps vectorially.
 
jumper

jumper

Senior Member
Location
3 Hr 2 Min from Winged Horses
ToolHound said:
From equation in the 'Delta Connected' image attached below...

What is the practical value of the equation

phase amps X 1.732 = line amps

Where to use that ? In many transformer problems in my electrician workbooks, that equation is nowhere to be seen. Yet I come across it in another publication. It must be good for something ???? What is 'phase amps' anyhow ?

I guess between 120/208, the 208 would be the line voltage.
120 would be phase voltage.

But what is 'phase amps' ? I am trying to make a first step in
grasping use/meaning of the 1.732 multiplier in amps calcualations. I am use to using the 1.732 multiplier only 'next to a voltage number'.

What gives here?
Click to expand...

What is 120 x 1.732 ?
What is 277 x 1.732 ?
 
T

ToolHound

Senior Member
rcwilson said:
Think of three 10 kVA single phase transformers....
Click to expand...

Whoa, somebody really does understand this stuff. Thanks rcwilson.

I'm getting out my pencil, my calculator, and I'm gonna work thru that explanation.

There's hope. Here I go. Back here later. --ToolHound
 
T

ToolHound

Senior Member
rcwilson said:
Think of three 10 kVA single phase transformers....
Click to expand...

rcwilson, this is great. I am still perusing and pondering this nice reply/example from all angles, as I work through it. As I work through this the question occurs to me...when literature sez..."delta connected" I infer that means...the 2ndary of the xformer is a delta configuration.

As you see it, is that piece of artwork/attachment in my original post referring to the primary being delta...or to the secondary being delta ? ( in other words, what does 'delta connected' mean ? Is the 'delta' referring to the primary or to the secondary ? ). Thanks. --ToolHound
 
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hardworkingstiff

Senior Member
Location
Wilmington, NC
Delta and Wye are 2 different ways to connect a 3-phase circuit (line or load side of transformer, motor winding, heater coils, ....).


th
 
Smart $

Smart $

Esteemed Member
Location
Ohio
ToolHound said:
rcwilson, this is great. I am still perusing and pondering this nice reply/example from all angles, as I work through it. As I work through this the question occurs to me...when literature sez..."delta connected" I infer that means...the 2ndary of the xformer is a delta configuration.

As you see it, is that piece of artwork/attachment in my original post referring to the primary being delta...or to the secondary being delta ? ( in other words, what does 'delta connected' mean ? Is the 'delta' referring to the primary or to the secondary ? ). Thanks. --ToolHound
Click to expand...
Delta-connected can refer to either the configuration of a source or a load... where either is comprised of three single elements connected in a triangle fashion, as in your example, to form part of a 3? circuit
 
T

ToolHound

Senior Member
Smart $ said:
Delta-connected can refer to either the configuration of a source or a load... where either is comprised of three single elements connected in a triangle fashion, as in your example, to form part of a 3? circuit
Click to expand...

Smart $, ok, thanks. Could be either source or load. Thanks. --ToolHound
 
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ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Consulting Electrical Engineer - Photovoltaic Systems
This was a hard one for me to grasp at first.

In a 240 split phase pair of conductors with a balanced load, the current in one conductor is the same as the current in both conductors (phase current = line current), which seems counterintuitive until you think of it as a loop with the instantaneous current going one direction in one conductor and the other direction in the other conductor, and it's all the same current.

With three phase delta current it's a little more complex, but it's the same situation. At a point in time where the current is maxxed in one direction in phase A, half of it is going the opposite direction in phase B and the other half is going the opposite direction in phase C, but just as in split phase, it's all the same current.
 
G

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
ggunn said:
This was a hard one for me to grasp at first.
Click to expand...

Still is for me. Especially that frustrating 1.732.
One thing that helps me somewhat is to remember that if you are at one end of a four wire circuit, and there is no current in the neutral, you have no idea whether the other end is wired as wye or delta. All you can measure are the line currents and voltages, so whether there is a delta load or a balanced wye load at the other end, the power will come out exactly the same as a function of line current and voltage.
The type of connection made at the other end will affect what resistances or impedances are required in the load elements to produce those line currents, but nothing else.
That gives me a way to cross check my vector based calculations, etc.
 
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