Using 175.68 instead of 208 volts to determine the VA load for B Phase (Hi Leg)

[IWIRE52]

I am trying to understand why the 175.68 volts instead of 208 volts works on the Hi Leg to calculate the correct total load for a 120/240 Hi Leg Delta system?

Example: You have 100 Amps of 1 phase 120/240 load and 50 amps of 3 phase 240 volt load. You normally calculate the 100 amps of 120/240 load as 24,000VA, and the 50 Amps of 240, 3 phase load as 50x240x1.732=20,784VA, for a total of load of 44,784VA.

Alternately if you calculate the phases independently by their voltage, you get the following:
A phase 150Ax120V=18,000VA
B phase 50Ax208V=10,400VA (Using 120V here as well, is incorrect)
C phase 150Ax120V=18,000VA
Total 18,000VA+10,400VA+18,000VA=46,400VA which is incorrect.

But if you calculate the phases independently this way the result is correct.
A phase 150Ax120V=18,000VA
B phase 50Ax175.68V=8,784VA
C phase 150Ax120V=18,000VA
Total 18,000VA+8,784VA+18,000VA=44,784VA, which is correct.

I have tried it on other configurations and it seems to work using the 175.68 Volts for B Phase (Hi Leg) but I am not sure why this works mathematically?

 

[wwhitney]

175.69 = 240 * (sqrt(3) - 1)

If the 3 phase line current is X, and if you include X in your A and C currents, you are already including 240 * X. The full 3 phase power is 240 * sqrt(3) * X. So the discrepancy is 240 * (sqrt(3) - 1) * X. Which is the amount your method assigns to the B phase.

Cheers, Wayne

 

[IWIRE52]

I am trying to understand why the 175.68 volts instead of 208 volts works on the Hi Leg to calculate the correct total load for a 120/240 Hi Leg Delta system?

Example: You have 100 Amps of 1 phase 120/240 load and 50 amps of 3 phase 240 volt load. You normally calculate the 100 amps of 120/240 load as 24,000VA, and the 50 Amps of 240, 3 phase load as 50x240x1.732=20,784VA, for a total of load of 44,784VA.

Alternately if you calculate the phases independently by their voltage, you get the following:
A phase 150Ax120V=18,000VA
B phase 50Ax208V=10,400VA (Using 120V here as well, is incorrect)
C phase 150Ax120V=18,000VA
Total 18,000VA+10,400VA+18,000VA=46,400VA which is incorrect.

But if you calculate the phases independently this way the result is correct.
A phase 150Ax120V=18,000VA
B phase 50Ax175.68V=8,784VA
C phase 150Ax120V=18,000VA
Total 18,000VA+8,784VA+18,000VA=44,784VA, which is correct.

I have tried it on other configurations and it seems to work using the 175.68 Volts for B Phase (Hi Leg) but I am not sure why this works mathematically?

175.69 = 240 * (sqrt(3) - 1)

If the 3 phase line current is X, and if you include X in your A and C currents, you are already including 240 * X. The full 3 phase power is 240 * sqrt(3) * X. So the discrepancy is 240 * (sqrt(3) - 1) * X. Which is the amount your method assigns to the B phase.

Cheers, Wayne

Wayne, Thanks for the info. I am still not sure how you came up with that math. Could you link me to a website or explain the how we get “-1” in the equation in a little more detail.

 

[wwhitney]

Could you link me to a website or explain the how we get “-1” in the equation in a little more detail.

You need to add sqrt(3) * something [where something = 240V * 3 phase line current]. You've already added something [which is the same as 1 * something]. And sqrt(3) * something - 1 * something = (sqrt(3) - 1) * something [multiplication distributes over addition/subtraction]. So that latter amount is how much more you need to add.

Cheers, Wayne

 

[IWIRE52]

Wayne,

I found a helpful response in Gemini Pro AI as follows, but had to rephrase the question as AI routinely came back with no answer to the question I asked, or 208 volts (the wrong answer).

My Question to GeminiPro AI: I am trying to understand why 175.69 volts instead of 208 volts is the voltage multiplier times the amperage that has to be used for the the B Phase (Hi Leg) to calculate the correct load on a 120/240 Hi Leg Delta system, phase by phase, when you use 120 volts for A and C Phase?

GeminiPro A Answer: This is an excellent and highly specific question. The short answer is that 175.69 volts is not a real, measurable voltage on your system; it is a mathematical "trick" (a weighting factor) used in panel schedules to make the math balance out perfectly when calculating 3-phase
loads. If you were to use the actual measured voltage of the High Leg (208V), your total calculated
load would be artificially high. Here is the step-by-step breakdown of exactly
why 175.69 is the magic number.
1. The Goal of the Calculation When filling out a panel schedule, you are trying to find the total Apparent Power (Volt-Amps, or VA) of the system by adding up the individual
loads on Phase A, Phase B, and Phase C. For a balanced 3-phase load (like a 3-phase motor) running on a 240V Delta system, the
true total power formula is:
Total VA = Line Voltage × Line Current × /3 Let's assume we have a balanced 3-phase load
drawing 1 Amp per phase.
• Total VA = 240V × 1A × 1.73205
• Total VA = 415.69 VA Therefore, for every 1 Amp of 3-phase load,
your panel schedule needs to add up to exactly
415.69 VA.
2. What happens on Phase A and Phase C? In your panel schedule, you are treating Phase A and Phase C as if they are standard line-to- neutral loads, multiplying their amperage by
120V. Remember, our load is drawing 1 Amp per
phase:
• Phase A calculated load: 120V × 1A = 120 VA
• Phase C calculated load: 120V × 1A = 120 VA If we add Phase A and Phase C together, we
have accounted for 240 VA of our total power. 3. Finding the "Missing" Power for Phase B We know the true total power of this 1-Amp load is 415.69 VA. We have already accounted
for 240 VA on phases A and C. To find out what multiplier we must use for Phase B to make the total correct, we simply
subtract:
• Total True Power: 415.69 VA
• Minus (Phase A + C): - 240.00 VA
• Remaining Power: 175.69 VA Because the current on Phase B is 1 Amp, the
voltage multiplier must be 175.69 Volts to yield
175.69 VA.
Why not use 208 Volts? It is true that if you take a multimeter and measure from the B-Phase (High Leg) to Neutral, you will read 208 Volts. However, if you use 208V as your multiplier in a panel schedule for a 3-phase load, look at what happens to the
math:
• Phase A: 120V × 1A = 120 VA
• Phase B: 208V × 1A = 208 VA
• Phase C: 120V × 1A = 120 VA
• Total Calculated VA = 448 VA But we already proved using the standard 3-
phase formula that the true power is only
415.69 VA. If you use 208V as your multiplier, you will overestimate your 3-phase loads by about 8%. While overestimating is generally safer than underestimating in electrical engineering, it can lead to unnecessarily upsizing breakers,
wires, and transformers, which costs money.
Summary • 120V is the physical voltage of A and C to
neutral.
• 208V is the physical voltage of B to neutral. • 175.69V is a purely mathematical multiplier used only on paper/software. It forces the sum of (A + B + C) to perfectly match the true 3- phase power equation ($240 \times | \times
(sqrt(3}$).

 

[wwhitney]

I found a helpful response in Gemini Pro AI as follows

Glad you found an answer that works for you. That answer agrees with mine, but with more detail and working out the 1A load case.

Curious why you want to find the total VA?

For the purposes of conductor or transformer sizing, you'd want to calculate the 3 line currents A, B, and C separately, and then take the largest. A method that goes through a total VA will only do that in the fully balanced case, i.e. if there are any 120V loads, they are balanced on A-N and C-N so you can consider them a 240V load. And then the 240V single phase loads are balanced on the 3 phases, AB, BC, and CA (which has the balanced 120V loads).

For any unbalanced loading, calculating the total VA and dividing by 240V * sqrt(3) will give you a lower bound for the current on the most heavily loaded line conductor. I.e. it would be non-conservative for sizing purposes.

Cheers, Wayne

 

[tortuga]

I'd add a caviat to this that it has to be a three transformer full delta for this:

And then the 240V single phase loads are balanced on the 3 phases, AB, BC, and CA (which has the balanced 120V loads).

On the common open hi-leg delta where BC has no transformer, 240V single phase load on BC (the open set) would add to both transformers.