Current Calc: 2 Phase 3 Wire

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dasarmin

Member
Location
Texas
This is a 3 phase 480Vac heater circuit with only 2 heaters wired in a delta.
Consider a purely resistive heater circuit; 1 - 600W between phases 1 & 2 and 1 ? 900W between phases 2 & 3; for a total of 1500W.
Calculate the current in the common leg (2).
My approach was to use the following:
I = 1500 / 480 * 2 = 1.56Amps
Ic (common leg current) = I * 1.41 = 1.56 * 1.41 = 2.20Amps

Is this correct?
 
J

jghrist

Senior Member
I get 2.724 by vector addition.

Assume V12=480V @ 0?, V23=480V @ -120?, V31=480V @ 120?

I12 = 600/480 = 1.25 @ 0? = 1.25
I23 = 900/480 = 1.875 @ -120? = -0.938 + 1.624j

I2 = I23 - I12 = -0.938-1.25 + 1.624j = -2.188 + 1.624j

sqrt(2.188? + 1.624?) = 2.724
 
J

jghrist

Senior Member
ibew501ed said:
? I=E/R? is that the difference
Click to expand...

I=P/V (with resistive loads)
Actually, I made a mistake because the voltage is in the denominator and I didn't adjust the current angles.

I12 = 600/480 = 1.25 @ 0? = 1.25

I23 = 900/480 = 1.875 @ +120? = -0.938 - 1.624j

I2 = I23 - I12 = -0.938-1.25 + 1.624j = -2.188 - 1.624j

sqrt(2.188? + 1.624?) = 2.724

Same answer
 
J

jghrist

Senior Member
I was thinking 900/(480 @ -120?) = 1.875 @ 120? not 1.875 @ -120?, an insignificant difference if you make the mistake consistently, but I thought I'd make the correction in case anyone wanted to follow the calculations. I originally made your assumption of current angle equal to voltage angle, but when someone brought up the E/R, P/V bit, it made me look at the equation. At this point I wish I hadn't said anything.
 
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Smart $

Smart $

Esteemed Member
Location
Ohio
jghrist said:
I was thinking 900/(480 @ -120?) = 1.875 @ 120? not 1.875 @ -120?, an insignificant difference if you make the mistake consistently, but I thought I'd make the correction in case anyone wanted to follow the calculations. I originally made your assumption of current angle equal to voltage angle, but when someone brought up the E/R, P/V bit, it made me look at the equation. At this point I wish I hadn't said anything.
Click to expand...
I'm simply not picking up what you're laying down. ;)

Your angles are correct in post #3, though you made the jump from vectorial addition to subtraction without explanation...???

The direction of current simply cannot be 120? (or 240?) out of sync with the voltage where there is unity power factor between voltage and current.
 
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jghrist

Senior Member
Smart $ said:
I'm simply not picking up what you're laying down. ;)

Your angles are correct in post #3, though you made the jump from vectorial addition to subtraction without explanation...???

The direction of current simply cannot be 120? (or 240?) out of sync with the voltage where there is unity power factor between voltage and current.
Click to expand...

It all depends on how you define the voltages and currents. If you define I12 as going from ?1 to ?2 and I23 as going from ?2 to ?3, then you have to subtract to get the line current.

The currents aren't 120? out of synch with the voltage, but one voltage is 120? out of synch with the other.
 
Smart $

Smart $

Esteemed Member
Location
Ohio
jghrist said:
It all depends on how you define the voltages and currents. If you define I12 as going from ?1 to ?2 and I23 as going from ?2 to ?3, then you have to subtract to get the line current.
Click to expand...
I understand that, but others may not.

Your voltage and current have to be measured to the common current node for vector addition. Yours, as stated, are not. As such, amidst your calculations, you negated the I23 current values.

jghrist said:
The currents aren't 120? out of synch with the voltage, but one voltage is 120? out of synch with the other.
Click to expand...
Correct... but that's not what you indicated in post #8 where you wrote, "I23 = 900/480 = 1.875 @ +120?". You posted that as a correction... yet it is not. I'm just trying to get the mud settled here ;)
 
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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
100112-1324 EST

Going back to post 3. It should be (1.25-.938)^2 = 0.97344. Add this to 1.62375^2 and take the sq-root and the result is 1.6535 A.

An intuitive check. If the one phase had been 180 instead of 120, then the result obviously would be 1.875-1.25 = 0.625. A value of 2.724 is way too big for the 120 angle. At 90 deg it would be 2.2535.

.
 
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dkarst

Senior Member
Location
Minnesota
74one said:
Here is my calculations:

Leg#1=600w/480v=1.25A
Leg#2=1500w/480v*1.73=1.8A
Leg#3=900w/480v=1.875A
Click to expand...

Sorry but the value you arrived at above for the combined leg is incorrect, you cannot multiply the phase current by 1.73 to get line current in this unbalanced delta case. The correct value is 2.72 A
 
J

jghrist

Senior Member
Please ignore my Post #8. It confuses the issue.

Maybe the attached sketch will help. With the currents as defined, load currents are subtracted to get line currents. The real part is -0.938 - 1.25 = -2.188
 
7

74one

Member
dkarst said:
Sorry but the value you arrived at above for the combined leg is incorrect, you cannot multiply the phase current by 1.73 to get line current in this unbalanced delta case. The correct value is 2.72 A
Click to expand...
I would really appreciate if you could elaborate on this.
 
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