single vs. 3 phase

[mivey]

Mivey,
OK, can you answer #387? Or a least phrase it so it could be understood?

If you have balanced L-N loads, you will have 83.333 amps in each load for both cases. The difference is that in the 208Y/120V 3P4W case, you have zero neutral current, while in the 120/208V 1P3W case you will have 83.333 amps of neutral current.

FWIW: I don't have a problem using the term "leg" as I'm not really picky (I am trying to get better). If we want to know if there are any issues with using the term, I would ask jim dungar as he is more careful with his terminology.

[edit: "he" is, not "his" is :roll:]

 

[rattus]

Rattus,
Could you develop that a little further, and apply it to the post "test for apprentice" (#387 this thread) I am having problems getting the picture.
Thanks

Transformers are rated in apparent power (KVA) rather in real power (KW) because the transformer losses are functions of voltage and current, Power factor does not apply here unless one is computing real power.

Then, the apparent power, Pa, delivered by each transformer is,

Pat = 120V x Iphase

Multiply by two to get total apparent power delivered to a 208V load,

240V x Iphase

For 208V load between the two legs, the apparent power to the load is,

Pal = 208V x Iphase

The ratio of apparent power delivered to apparent power used is,

OFF = 240V/208V = 1.154

This question is asked in the Oregon supervisor's test.

This discrepancy occurs because the there is a +/- 30 degree phase angle between current and voltage in the two transformers.

Real power delivered and used is equal,

Pr = 240V x Iphase X cos(30) = 208V x Iphase

 

[rattus]

And this is the TRUMTH.

carl

Carl, TRUMTH?

 

[mivey]

Carl, TRUMTH?

A truth that trumps the others?

 

[coulter]

Bless you both for asking - And I'm laughing, cause this is susposed to be humorous for all of us.

"TRMTH" - Why that's CC more times true than "TRVTH"

Edit - Oh, I just noticed the extra "U" - maybe more humorous with out that.

carl :grin: :smile: :roll:

 

[dwcaveney]

Thanks Mivey

Thanks Mivey

If you have balanced L-N loads, you will have 83.333 amps in each load for both cases. The difference is that in the 208Y/120V 3P4W case, you have zero neutral current, while in the 120/208V 1P3W case you will have 83.333 amps of neutral current.

FWIW: I don't have a problem using the term "leg" as I'm not really picky (I am trying to get better). If we want to know if there are any issues with using the term, I would ask jim dungar as he is more careful with his terminology.

[edit: "he" is, not "his" is :roll:]

Mivey,

So is this right for part two of post #387?

10000/120=83.33

20000/208=96.15

Total =179.48 Amps OR: 30000/120=250 amps

Thanks

 

[mivey]

Mivey,

So is this right for part two of post #387?

10000/120=83.33

20000/208=96.15

Total =179.48 Amps

Thanks

You can't count the wattage twice. Let's see if I can get this right at this time of night:

For 120 volt L-N loads with 10kW per coil you would have 83.33 amps for the coil and load. This is true for one or three L-N loads.

For a single, 2-wire, 208 L-L load with 20kW you would have 96.23 amps for the coils and load (no connection to 3rd coil).

If you want to have three L-L loads, and still have 10 kW per coil, you are back to 83.33 amps per coil but 48.11 amps per load (two loads are tied to each coil).

[edit: used coil instead of leg]

 

[hardworkingstiff]

hummm

hummm

So,

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the load is connected at 208-volts you have 16.64 KVA.

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the loads are connected at 120-volts you have 19.2 KVA.

When A and B are only connected to 120-volt loads, will the neutral send any current back on C from the secondary side of the transformer to the primary side of the transformer? I guess it depends of whether its a Delta/Wye or Wye/Wye ?

 

[mivey]

So,

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the load is connected at 208-volts you have 16.64 KVA.

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the loads are connected at 120-volts you have 19.2 KVA.

Don't forget the original constraint we were given about 10 kW per coil. The single or three L-N loads would be 1.44 ohms, the three L-L loads would be 4.32 ohms, the single L-L load would be 2.16 ohms.

[edit: nevermind, I think you were responding to #388 & #394]

[edit: Here is an example of what you were talking about using the 10 kW setup. Use a single L-L load.
Vcb=207.846<90, R_load=2.16, I=96.225<90,
W_load=207.846 * 96.225 * cos(90-90) = 20 kW (also 20 kVA)
c_coil kVA=120<120 * 96.225<90 = 11.547 kVA
c_coil kW=120 * 96.225 * cos(120-90) = 10 kW]

[edit: Don't mix kVA delivered with transformer kVA. In your example:
For single L-L:
Vcb=207.846<90, I=80<90
kW_load=207.846 *80 *cos(90-90)=16.63 kW
kVA_load = 207.846<90 * 80<90 = 16628<180 = 16.63 kVA
c_coil kW=120 * 80 * cos(120-90) = 8.313 kW (16.63 kW for both coils)
c_coil kVA=120<120 * 80<90 = 9600<-150 = 9.6 kVA
b_coil kVA=120<-120 * 80<-90 = 9600<150 = 9.6 kVA
Note: transformer size is 19.2 kVA total but only 16.63 kVA delivered to load)

Similarly, for L-N:
Vcn=120<120, Vbn=120<-120, Ic=80<120, Ib=80<-120
kW_load_cn=120 * 80 * cos(120-120)=9.6 kW
c_coil kVA=120<120 * 80<120 = 9600<-120 = 9.6 kVA = load c-n kVA
c_coil kW=120 * 80 * cos(120-120) = 9.6 kW
kW_load_bn=120 * 80 * cos(-120--120)=9.6 kW
b_coil kVA=120<-120 * 80<-120 = 9600<120 = 9.6 kVA = load b-n kVA
b_coil kW=120 * 80 * cos(-120--120) = 9.6 kW
Note: transformer size is 19.2 kVA total and 19.2 kVA delivered to load)
]

[edit: For three L-L:
V_coil=120, I_coil=80, watts_coil = 9600 = watts per load, I_load=46.188 amps
Vcb=207.846<90, Icb=46.188<90
kW_load_cb = 207.846 *46.188 *cos(90-90) = 9.6 kW
kVA_load_cb = 207.846<90 * 46.188<90 = 9600<180 = 9.6 kVA
Vca=207.846<150, Ica=46.188<150
kW_load_ca = 207.846 *46.188 *cos(150-150) = 9.6 kW
kVA_load_ca = 207.846<150 * 46.188<150 = 9600<-60 = 9.6 kVA
Ic = Icb + Ica = 46.188<90 + 46.188<150 = 80<120 and Vcn = 120<120
c_coil kVA = 80<120 * 120<120 = 9600<-120 = 9.6 kVA
c_coil kW = 80 * 120 *cos(120-120) = 9.6 kW
Note: total kVA = total kW = 28.8
]
These examples were for wye banks and resistive loads but the same method would apply for delta and non-resistive loads. We usually simplify this because a lot of times we have balanced loads.

 

[winnie]

Thanks Jon,

Could you answer post #387 this tread. The question may not be phrased correctly. If I got a number I could work backwards to "prove" it? I'm having problems with the theory and the math!

So I decided to wait and see what others came up with, to 'prime' my response

With the _constraint_ that each coil provides 10 KW, the amount of current that each coil supplies will change depending upon the load characteristics.

For example, say you had a perfectly resistive load connected line to neutral. You then have a power factor of 1.0, 120V, and therefore 83.3A.

Now imagine that you connect something with a power factor of 0.5 from line to neutral. For 10KW to be delivered to the load, you would need 166.7A to flow. The reason for this is that current would not be in phase with voltage, so that to deliver _real power_, more current is needed.

This result seems counterintuitive, but it is correct given the question as asked. The reality is that transformers are rated in KVA, not KW, and the rating is the maximum that the transformer can deliver on a continuous basis, not what the transformer is required to deliver.

Ask the question like this:
given a transformer bank with 120V, 10KVA secondaries, connected wye. The transformers are rated at 10KVA.
What is the rated output current?
How much power is delivered to a single phase resistive load connected line to neutral?
How much total power is delivered to three single phase resistive loads connected line to neutral in a wye configuration?
How much power is delivered to a single phase resistive load connected line to line?
How much total power is delivered to three single phase resistive loads each connected line to line in a delta configuration?
Etc.

You will find the answers closely related to the question that you asked, but without the counterintuitive aspects of a fixed power output constraint.

For a similar 'look at the counterintuitive side of a common problem', consider a simple DC resistive circuit, but with a _constant current_ power supply rather than the far more common 'voltage source'. Assume something simple like '1A will always flow through the circuit'. Now consider how power dissipation changes with load resistance, and what you need to do in order to make power dissipation _zero_, in other words, what you need to do to turn the power off.

-Jon

 

[coulter]

... consider a simple DC resistive circuit, but with a _constant current_ power supply rather than the far more common 'voltage source'. Assume something simple like '1A will always flow through the circuit'. Now consider how power dissipation changes with load resistance, and what you need to do in order to make power dissipation _zero_, in other words, what you need to do to turn the power off. ...

Just a guess - short circuit the power supply?

carl

 

[mivey]

Just a guess - short circuit the power supply?

carl

Sounds right. Zero heat dissipation would probably not be popular where you are at. You might like a good Gray-Black-Red.

BTW, TRMTH is funnier, I just forgot to respond earlier.:smile:

 

[winnie]

Just a guess - short circuit the power supply?

carl

Exactly.

-Jon

 

[hardworkingstiff]

So,

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the load is connected at 208-volts you have 16.64 KVA.

'A' coil draws 80-amps and 'B' coil draws 80-amps. If the loads are connected at 120-volts you have 19.2 KVA.

When A and B are only connected to 120-volt loads, will the neutral send any current back on C from the secondary side of the transformer to the primary side of the transformer? I guess it depends of whether its a Delta/Wye or Wye/Wye ?

Can someone who knows answer yes or no to whether 'C' coil will have current if 'A' and 'B' have 120-volt circuits connected as stated above?

Sorry Mivey, while your response was impressive, it was not dumbed down enough for me to comprehend.

 

[crossman]

why is 2 phase - 240 volt - called single phase???

Because, with this system connected to loads which have the same XL/XC/R ratios, it is impossible to create seperate currents which are not in phase or 180 degrees out of phase.*

*"In phase" and "180 out of phase" describe the exact same sine wave phase condition, only measured from different references.

 

[mivey]

Can someone who knows answer yes or no to whether 'C' coil will have current if 'A' and 'B' have 120-volt circuits connected as stated above?

Sorry Mivey, while your response was impressive, it was not dumbed down enough for me to comprehend.

For a delta-wye, yes.

 

[rattus]

No, in either case:

No, in either case:

Can someone who knows answer yes or no to whether 'C' coil will have current if 'A' and 'B' have 120-volt circuits connected as stated above?

Wye-wye:

C is open, therefore will be zero secondary or primary current in C.

Delta-wye:

EDIT: Changed my mind. Primary winding C sees an infinite reflected impedance, therefore it will draw magnetizing current only.

 

[mivey]

Wye-wye:

C is open, therefore will be zero secondary or primary current in C.

Delta-wye:

EDIT: Changed my mind. Primary winding C sees an infinite reflected impedance, therefore it will draw magnetizing current only.

Picky, picky, picky. Why can't we just ignore physics for once?:smile:

So for the primary side of the wye-wye you can have load current in two line conductors and in the neutral conductor, but no load current in one of the primary windings (because no load current in the secondary winding).

And for the primary of the delta-wye you can have load current in all three line conductors (sqrt(3) higher in one conductor) but no load current in one of the primary windings (because no load current in the secondary winding).

Now am I back on the straight and narrow?:smile:

 

[rattus]

Yup:

Yup:

Picky, picky, picky. Why can't we just ignore physics for once?:smile:

So for the primary side of the wye-wye you can have load current in two line conductors and in the neutral conductor, but no load current in one of the primary windings (because no load current in the secondary winding).

And for the primary of the delta-wye you can have load current in all three line conductors (sqrt(3) higher in one conductor) but no load current in one of the primary windings (because no load current in the secondary winding).

Now am I back on the straight and narrow?:smile:

Alright. Now what are the primary currents for a line to line load?

 

[hardworkingstiff]

Alright. Now what are the primary currents for a line to line load?

If I understand this question, you are talking about line to line load (no neutral connection) of a 120/208 'single-phase' circuit.

Delta/Wye would have current on all 3 primary conductors and 2 secondary conductors, but current would only be flowing through 2 primary coils and 2 secondary coils.

Wye/Wye would have current on 2 primary conductors (and the neutral of the primary, but not the 3rd phase conductor) and 2 secondary conductors (with no current on the neutral of the secondary), current on 2 primary coils and 2 secondary coils.

Right/Wrong?