Instantaneous 3-phase power:
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Ed MacLaren
Senior Member
Re: Instantaneous 3-phase power:
I think that is correct. At no time does the power equal zero, as is the case in a single phase system twice per cycle.
Ed
I think that is correct. At no time does the power equal zero, as is the case in a single phase system twice per cycle.
Ed
Re: Instantaneous 3-phase power:
I'm not sure if the power is constant, and I'm not willing to do the math to find out
I don't think it is completely constant, but it is more constant than single phase. In single phase, the inst. power varies from 0 to max.
If we look at a single phase service and the wiring from the POCO to the user, when the inst. power is zero, the wire is being under utilized. In effect, the wire must be sized for the peak power flow.
So actually, DC is the most efficient since the power is constant.
I'm not sure if the power is constant, and I'm not willing to do the math to find out
I don't think it is completely constant, but it is more constant than single phase. In single phase, the inst. power varies from 0 to max.If we look at a single phase service and the wiring from the POCO to the user, when the inst. power is zero, the wire is being under utilized. In effect, the wire must be sized for the peak power flow.
So actually, DC is the most efficient since the power is constant.
Re: Instantaneous 3-phase power:
Steve, I think the proof of the statement can be found in the trig identities, but I can't find my handbook right now.
I would argue that wire should be sized for the average power, that is for the heating effect. Another way to say this is that the effective AC current and effective AC voltage are equal numerically to the equivalent DC current and DC voltage.
Steve, I think the proof of the statement can be found in the trig identities, but I can't find my handbook right now.
I would argue that wire should be sized for the average power, that is for the heating effect. Another way to say this is that the effective AC current and effective AC voltage are equal numerically to the equivalent DC current and DC voltage.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
I have written out a proof, using the trigonometric identities, that power in a balanced three-phase system is constant. Its value is as follows:
I have a similar document that derives the value of ?Power as a function of time,? for a single phase system.
If anyone wants a copy of either file, then send me your email address in a Private Message, and I will send you the file(s).
I have written out a proof, using the trigonometric identities, that power in a balanced three-phase system is constant. Its value is as follows:
The proof is a contained in Microsoft Word file. It is two pages long, and takes up about 43 KB. I don?t think I could post it here, since some of the symbols (Greek letters) might not be readable by the Forum?s word processor.
I have a similar document that derives the value of ?Power as a function of time,? for a single phase system.
If anyone wants a copy of either file, then send me your email address in a Private Message, and I will send you the file(s).
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
Yes. I think it is the bigger of the two big advantages of 3-phase systems. The other is that, without having any moving parts, the stator of a 3-phase motor, with its three windings being 120 degrees apart in space, and with its three power lines having currents 120 degrees apart in time, creates a rotating magnetic field.
Re: Instantaneous 3-phase power:
This isn't a proof, but it is a start: Consider a 3 phase balanced source, with a 3 phase balanced resistive load. Total power is the sum of the power in each leg, which is V^2/R. If we assume R=1, things get simpler.
P=V(1)^2 + V(2)^2 + V(3)^2
where V(1) is the first phase and V(2) is the second phase. We also know V(1) =Vmax sin @. (where @ is theta or the phase angle). Again, if we make Vmax =1, things get simpler.
Finally, we know all three phases are 120 deg. apart. The equation for instantenous power in our simple circuit is:
P = (sin @)^2 + (sin (@+120))^2 + (sin (@+240)^2
If I plot this on my graphing calculator, I get a straight line, so the power is constant.
This isn't a proof, but it is a start: Consider a 3 phase balanced source, with a 3 phase balanced resistive load. Total power is the sum of the power in each leg, which is V^2/R. If we assume R=1, things get simpler.
P=V(1)^2 + V(2)^2 + V(3)^2
where V(1) is the first phase and V(2) is the second phase. We also know V(1) =Vmax sin @. (where @ is theta or the phase angle). Again, if we make Vmax =1, things get simpler.
Finally, we know all three phases are 120 deg. apart. The equation for instantenous power in our simple circuit is:
P = (sin @)^2 + (sin (@+120))^2 + (sin (@+240)^2
If I plot this on my graphing calculator, I get a straight line, so the power is constant.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
?Power,? as a function of time, is not given by V^2/R. That formula only gives the RMS value of power, a value that is itself a constant. But moment by moment, power can vary. In a single phase system, power varies up and down in a predictable pattern, and is never the same moment to moment.
Using ?w? to represent the frequency (generally we use 2 times Pi times 60 hertz for ?w?), and using your same ?@? to represent the phase angle between voltage and current, we get the following:
</font>
Adding those three gives the value of total power, as a function of time. It takes some fancy trig to reduce this mess to the simple expression that I gave in my first post.
Your attempt at a proof falls apart in this step. You left out the variable ?t,? for ?time.?
?Power,? as a function of time, is not given by V^2/R. That formula only gives the RMS value of power, a value that is itself a constant. But moment by moment, power can vary. In a single phase system, power varies up and down in a predictable pattern, and is never the same moment to moment.
Your attempt at a proof also falls apart in this step, for the same reason. Your use of @ as the argument of the sine function also leaves out the variable of time. The difficult part of the math is that the voltage curve (voltage as a function of time) and the current curve (current as a function of time) do not reach their peaks at the same time. They would, for a purely resistive system, like the one you are describing. But for the more common case of an inductive circuit (i.e., with motors or with fluorescent lights), the voltage curve reaches its peak value some time earlier than does the current curve.
Using ?w? to represent the frequency (generally we use 2 times Pi times 60 hertz for ?w?), and using your same ?@? to represent the phase angle between voltage and current, we get the following:
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">For Phase A, power equals Vmax sin (wt) times Imax sin (wt - @)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">For Phase B, power equals Vmax sin (wt + 120) times Imax sin (wt - @ + 120)</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">For Phase C, power equals Vmax sin (wt + 240) times Imax sin (wt - @ + 240)</font>
Adding those three gives the value of total power, as a function of time. It takes some fancy trig to reduce this mess to the simple expression that I gave in my first post.
Re: Instantaneous 3-phase power:
OK, you guys are making this too hard again. Consider this:
We can define the 3-phase voltages as:
1) v1 = Vp x sin(u);
2) v2 = Vp x sin(u +120);
3) v3 = Vp x sin(u -120)
where u is 2*pi*f*t
We can respresent the load as a parallel combination of resistance and reactance, then all we have to do is prove that the sum of the squares, (normalized voltage):
4) vn^2 = v1^2 + v2^2 +v3^2 = K
We are told that:
5) sin(u +/- v) = sin(u)cos(v) +/- cos(u)sin(v);
6) sin^2(u) + cos^2(u) = 1
Then, expanding eqn. 4 and using trig identities:
7) vn^2 = sin^2(u) + sin^2(u)cos^2(120) + cos^2(u)sin^2(120) + sin^2(u)cos^2(-120) + cos^2(u)sin^2(-120)
= sin^2(u) + sin^2(u)(0.25) + cos^2(u)(0.75) + sin^2(u)(0.25) + cos^2(u)(0.75)
= 1.5( sin^2(u) + cos^2(u)) = 1.5
Which proves that vn^2 is constant, nuf sed!
[ January 20, 2005, 08:44 AM: Message edited by: rattus ]
OK, you guys are making this too hard again. Consider this:
We can define the 3-phase voltages as:
1) v1 = Vp x sin(u);
2) v2 = Vp x sin(u +120);
3) v3 = Vp x sin(u -120)
where u is 2*pi*f*t
We can respresent the load as a parallel combination of resistance and reactance, then all we have to do is prove that the sum of the squares, (normalized voltage):
4) vn^2 = v1^2 + v2^2 +v3^2 = K
We are told that:
5) sin(u +/- v) = sin(u)cos(v) +/- cos(u)sin(v);
6) sin^2(u) + cos^2(u) = 1
Then, expanding eqn. 4 and using trig identities:
7) vn^2 = sin^2(u) + sin^2(u)cos^2(120) + cos^2(u)sin^2(120) + sin^2(u)cos^2(-120) + cos^2(u)sin^2(-120)
= sin^2(u) + sin^2(u)(0.25) + cos^2(u)(0.75) + sin^2(u)(0.25) + cos^2(u)(0.75)
= 1.5( sin^2(u) + cos^2(u)) = 1.5
Which proves that vn^2 is constant, nuf sed!
[ January 20, 2005, 08:44 AM: Message edited by: rattus ]
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
OK. If you insist. Here it is. Don't say I didn't warn you!
Derivation of the Formula for Three Phase Power
Terms
VaM Maximum value of line-to-neutral voltage on Phase A.
VbM Maximum value of line-to-neutral voltage on Phase B.
VcM Maximum value of line-to-neutral voltage on Phase C.
VM Maximum value of line-to-neutral voltage. It is assumed that the loads are balanced, so that VM = VaM = VbM = VcM.
Va(t) Voltage from Phase A to neutral, as a function of time ?t.?
Vb(t) Voltage from Phase B to neutral, as a function of time ?t.?
Vc(t) Voltage from Phase C to neutral, as a function of time ?t.?
V [L-L] RMS The Root-Mean-Square value of Voltage, measured line-to-line, assumed to be the same for all three phases.
IaM Maximum value of phase current on Phase A.
IbM Maximum value of phase current on Phase B.
IcM Maximum value of phase current on Phase C.
IM Maximum value of phase current It is assumed that the loads are balanced, so that IM = IaM = IbM = IcM.
Ia(t) Current on Phase A, as a function of time ?t.?
Ib(t) Current on Phase B, as a function of time ?t.?
Ic(t) Current on Phase C, as a function of time ?t.?
P(t) Total power on Phases A, B, and C, as a function of time ?t.?
w Frequency in radians per second, assumed to be the same on Phases A, B, and C. Equal to 2pi times the frequency in cycles per second.
u Phase angle between voltage and current, assumed to be the same on Phases A, B, and C.
a and b Arbitrary angles used in trigonometric identities.
Let Va(t) = VM cos (wt)
Vb(t) = VM cos (wt - 120)
Vc(t) = VM cos (wt - 240)
Let Ia(t) = IM cos (wt - u)
Ib(t) = IM cos (wt - u - 120)
Ic(t) = IM cos (wt - u - 240)
Then P(t) = Va(t) Ia(t) + Vb(t) Ib(t) + Vc(t) Ic(t)
= VM IM cos (wt) cos (wt - u)
+
VM IM cos (wt - 120) cos (wt - u - 120)
+
VM IM cos (wt - 240) cos (wt - u - 240)
Use the Trig Identities:
cos(a) cos(b) = ? cos(a - b) + ? cos(a + b)
cos(a - b) = cos(a) cos (b) + sin(a) sin (b)
Then P(t) = ? (VM IM) cos (u) [1+ cos (2wt)] + ? (VM IM) [sin (u) sin (2wt)]
+
? (VM IM) cos (u) [1+ cos (2wt - 240)] + ? (VM IM) [sin (u) sin (2wt - 240)]
+
? (VM IM) cos (u) [1+ cos (2wt - 480)] + ? (VM IM) [sin (u) sin (2wt - 480)]
= (3) ? (VM IM) cos (u)
+
? (VM IM) cos (u) { cos (2wt) + cos (2wt - 240) + cos (2wt - 480)}
+
? (VM IM) sin (u) { sin (2wt) + sin (2wt - 240) + sin (2wt - 480)}
Note that the second and third term are at all times equal to zero (because the system is balanced).
Then P(t) = = (3) ? (VM IM) cos (u)
= (3) (VM / sqrt 2 ) (IM / sqrt 2) cos (u)
Now take note that (VM / sqrt 2 ) is the same as VRMS, and that (IM / sqrt 2) is the same as IRMS.
P(t) = (3) (VRMS) (IRMS) cos (u)
Finally, take note that
1. Current are usually expressed as phase currents, as shown above.
2. Voltages are usually expressed as line-to-line values, which is not the value used above.
3. The line-to-neutral voltage is equal to (1 / sqrt 3) times the line-to-line voltage.
P(t) = (sqrt 3) (V [L-L] RMS) (I [PHASE] RMS) cos (u)
Note that power, as a function of time, does not have the variable "T" on the right hand side of this equation. Therefore, the total power in a balanced three phase circuit is constant!
QED
OK. If you insist. Here it is. Don't say I didn't warn you!
Derivation of the Formula for Three Phase Power
Terms
VaM Maximum value of line-to-neutral voltage on Phase A.
VbM Maximum value of line-to-neutral voltage on Phase B.
VcM Maximum value of line-to-neutral voltage on Phase C.
VM Maximum value of line-to-neutral voltage. It is assumed that the loads are balanced, so that VM = VaM = VbM = VcM.
Va(t) Voltage from Phase A to neutral, as a function of time ?t.?
Vb(t) Voltage from Phase B to neutral, as a function of time ?t.?
Vc(t) Voltage from Phase C to neutral, as a function of time ?t.?
V [L-L] RMS The Root-Mean-Square value of Voltage, measured line-to-line, assumed to be the same for all three phases.
IaM Maximum value of phase current on Phase A.
IbM Maximum value of phase current on Phase B.
IcM Maximum value of phase current on Phase C.
IM Maximum value of phase current It is assumed that the loads are balanced, so that IM = IaM = IbM = IcM.
Ia(t) Current on Phase A, as a function of time ?t.?
Ib(t) Current on Phase B, as a function of time ?t.?
Ic(t) Current on Phase C, as a function of time ?t.?
P(t) Total power on Phases A, B, and C, as a function of time ?t.?
w Frequency in radians per second, assumed to be the same on Phases A, B, and C. Equal to 2pi times the frequency in cycles per second.
u Phase angle between voltage and current, assumed to be the same on Phases A, B, and C.
a and b Arbitrary angles used in trigonometric identities.
Let Va(t) = VM cos (wt)
Vb(t) = VM cos (wt - 120)
Vc(t) = VM cos (wt - 240)
Let Ia(t) = IM cos (wt - u)
Ib(t) = IM cos (wt - u - 120)
Ic(t) = IM cos (wt - u - 240)
Then P(t) = Va(t) Ia(t) + Vb(t) Ib(t) + Vc(t) Ic(t)
= VM IM cos (wt) cos (wt - u)
+
VM IM cos (wt - 120) cos (wt - u - 120)
+
VM IM cos (wt - 240) cos (wt - u - 240)
Use the Trig Identities:
cos(a) cos(b) = ? cos(a - b) + ? cos(a + b)
cos(a - b) = cos(a) cos (b) + sin(a) sin (b)
Then P(t) = ? (VM IM) cos (u) [1+ cos (2wt)] + ? (VM IM) [sin (u) sin (2wt)]
+
? (VM IM) cos (u) [1+ cos (2wt - 240)] + ? (VM IM) [sin (u) sin (2wt - 240)]
+
? (VM IM) cos (u) [1+ cos (2wt - 480)] + ? (VM IM) [sin (u) sin (2wt - 480)]
= (3) ? (VM IM) cos (u)
+
? (VM IM) cos (u) { cos (2wt) + cos (2wt - 240) + cos (2wt - 480)}
+
? (VM IM) sin (u) { sin (2wt) + sin (2wt - 240) + sin (2wt - 480)}
Note that the second and third term are at all times equal to zero (because the system is balanced).
Then P(t) = = (3) ? (VM IM) cos (u)
= (3) (VM / sqrt 2 ) (IM / sqrt 2) cos (u)
Now take note that (VM / sqrt 2 ) is the same as VRMS, and that (IM / sqrt 2) is the same as IRMS.
P(t) = (3) (VRMS) (IRMS) cos (u)
Finally, take note that
1. Current are usually expressed as phase currents, as shown above.
2. Voltages are usually expressed as line-to-line values, which is not the value used above.
3. The line-to-neutral voltage is equal to (1 / sqrt 3) times the line-to-line voltage.
P(t) = (sqrt 3) (V [L-L] RMS) (I [PHASE] RMS) cos (u)
Note that power, as a function of time, does not have the variable "T" on the right hand side of this equation. Therefore, the total power in a balanced three phase circuit is constant!
QED
Re: Instantaneous 3-phase power:
All:
A balanced load draws power equally from all three phases, but when one of the voltages is instantaneously zero, the phase relationship requires that the other two must each be at half amplitude.
At no instant does the instantaneous power drawn by the total load reach zero; in fact, the total instantaneous power is constant.
Consider phase A, v = 200(sq.rt.2) Cos((120 * pi * t) + 0)) and the current is i = 2(sq.rt2) Cos((120 * pi * t) - 60)).
Using phase A as the reference.
Pa(t) = v(an)i(an) = 800 Cos (120* pi * t)Cos((120*pi*t)- 60))
Pa(t) = 400[Cos(-60) + Cos((240*pi*t) - 60)]
= 200 + 400Cos((240*pi*t) - 60))
the other two phases are done in the same way and the total power absorbed by the load is
p(t) = pa + pb + pc = 600 watts.
Again expressing the product of two cosine functions as one-half the sum of the cosine of the difference angle and the cosine of the sum angle,
then
P(t) = 1/2VmIm cos (theta- phi) ...
+ 1/2VmIm cos( 240 pi t + theta + phi)
The first term is constant, and independent of t. the remaining term is a cosine function; p(t) is periodic, and its period is 1/2T.
By inspection the average value of the second term is zero over a period T (or 1/2 T) and the average of the first term, a constant must be that constant itself.
then the P(t) =1/2VmIm Cos(V theta - I phi)
The average power is one-half the product of the crest amplitude of the voltage, the crest amplitude of the current and the cosine of the phase-angle difference between the current and the voltage.
Gary
All:
A balanced load draws power equally from all three phases, but when one of the voltages is instantaneously zero, the phase relationship requires that the other two must each be at half amplitude.
At no instant does the instantaneous power drawn by the total load reach zero; in fact, the total instantaneous power is constant.
Consider phase A, v = 200(sq.rt.2) Cos((120 * pi * t) + 0)) and the current is i = 2(sq.rt2) Cos((120 * pi * t) - 60)).
Using phase A as the reference.
Pa(t) = v(an)i(an) = 800 Cos (120* pi * t)Cos((120*pi*t)- 60))
Pa(t) = 400[Cos(-60) + Cos((240*pi*t) - 60)]
= 200 + 400Cos((240*pi*t) - 60))
the other two phases are done in the same way and the total power absorbed by the load is
p(t) = pa + pb + pc = 600 watts.
Again expressing the product of two cosine functions as one-half the sum of the cosine of the difference angle and the cosine of the sum angle,
then
P(t) = 1/2VmIm cos (theta- phi) ...
+ 1/2VmIm cos( 240 pi t + theta + phi)
The first term is constant, and independent of t. the remaining term is a cosine function; p(t) is periodic, and its period is 1/2T.
By inspection the average value of the second term is zero over a period T (or 1/2 T) and the average of the first term, a constant must be that constant itself.
then the P(t) =1/2VmIm Cos(V theta - I phi)
The average power is one-half the product of the crest amplitude of the voltage, the crest amplitude of the current and the cosine of the phase-angle difference between the current and the voltage.
Gary
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
Gary: Your discussion is similar to the rigorous proof that I presented above. However, it differs in one significant way.
</font>
</font>
Gary: Your discussion is similar to the rigorous proof that I presented above. However, it differs in one significant way.
I agree with this specific example, and with the result of 600 watts. The difference between this specific example and the more general derivation that follows is that this example adds all three phases together. The general derivation talks only about one phase.
True, but not relevant to the issue at hand. The average value is not what is under discussion here. We are looking at the instantaneous power (i.e., total of all three phases), as a function of time. Look what happens when you add in the p(t) associated with the other two phases.
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">You get the same first term, no change at all, for each of the two other phases. The total contribution is therefore three times the first term.</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">More importantly, you get three cosine functions that look like your second term. Each of the three, as you correctly pointed out, has an average value of zero. However, the three have the same magnitude and are separated from each other by exactly 120 degrees of arc. In other words, you have { cos (a) plus cos (a-120) plus cos (a-240) }. This set of three terms will add up to zero at every instant in time!</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">As a result, the total p(t) equals three times the first term.</font>
Again, true, but not relevant to the issue at hand. The total power of all three phases, as a function of time, is a constant that is equal to three times that value. A simple arithmetic manipulation will bring you to the expression that I derived above:
</font>
- <font size="2" face="Verdana, Helvetica, sans-serif">P(t) = (sqrt 3) (V [L-L] RMS) (I [PHASE] RMS) cos (u)</font>
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
True, but not helpful. The mathematical manipulations you can use to prove that a function is constant at one specific angle cannot be used to prove that it is constant at all angles. That was the flaw in your logic. One would need to inset the variable ?time,? and perform a ?rigorous proof,? in order to show that this specific function (total power of a balanced three phase system) is indeed constant. My post of 8:43 last night is just such a rigorous proof.
Re: Instantaneous 3-phase power:
Charlie B:
I won't deny your post is the rigorous proof. Good job. But my graphing calculator proved the function I posted is constant for about 1000 points, not just one point (resistive circuit only). Not proof, but for me that eliminates any reasonable doubt. And with only a couple lines of math - the "engineering efficent way".
It takes a real mathmatician to argue a function that is constant for 1000 points might be different at one point inbetween. I think you missed your calling
Steve
Charlie B:
I won't deny your post is the rigorous proof. Good job. But my graphing calculator proved the function I posted is constant for about 1000 points, not just one point (resistive circuit only). Not proof, but for me that eliminates any reasonable doubt. And with only a couple lines of math - the "engineering efficent way".
It takes a real mathmatician to argue a function that is constant for 1000 points might be different at one point inbetween. I think you missed your calling
Steve
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
Now I see your point. I missed it earlier. Mea culpa.
You did not disregard the variable ?time,? as I had suspected. Nor did you attempt a derivation that omitted that variable. Rather, you are having your calculator select values for the variable ?time,? and calculating the power at each point. Certainly, it?s not rigorous, but it is a perfectly acceptable method of demonstrating the desired result. Bravo.
Now I see your point. I missed it earlier. Mea culpa.
You did not disregard the variable ?time,? as I had suspected. Nor did you attempt a derivation that omitted that variable. Rather, you are having your calculator select values for the variable ?time,? and calculating the power at each point. Certainly, it?s not rigorous, but it is a perfectly acceptable method of demonstrating the desired result. Bravo.
Re: Instantaneous 3-phase power:
OK Gents, looks like the point has been proven, but I would argue that the proofs are more complex than necessary.
See my edited post of 1/19 for a simpler, more direct approach to the problem. Criticisms accepted.
There was a textbook entitled, "DC Amplifiers and How to Avoid Building Them", by Argimbeau. He said, "Real advances are made by recognizing the difficulties and avoiding them". Same argument here, no need to get involved with peak values, power factor, calculus, sqrt(3), etc. Just trigonometry.
OK Gents, looks like the point has been proven, but I would argue that the proofs are more complex than necessary.
See my edited post of 1/19 for a simpler, more direct approach to the problem. Criticisms accepted.
There was a textbook entitled, "DC Amplifiers and How to Avoid Building Them", by Argimbeau. He said, "Real advances are made by recognizing the difficulties and avoiding them". Same argument here, no need to get involved with peak values, power factor, calculus, sqrt(3), etc. Just trigonometry.
- Location
- Lockport, IL
- Occupation
- Retired Electrical Engineer
Re: Instantaneous 3-phase power:
Criticism follows: See my post of Jan 19, 1:53 pm. Your equation 4 contains the same error in logic that I describe in the first section of that post. You have used the simplification that power can be represented as V^2/R. That is valid only for average power, not for the instantaneous value of power (i.e., power as a function of time). Your original question had to do with the suggestion that power was constant at every instant in time. You cannot prove that with trigonometric manipulations of the equation for average power.
Not true. If you want a valid proof, you have to go through the math. If you can be convinced by anything less, then I have some vacation property in Costa Rica that I would like to sell ya.
Re: Instantaneous 3-phase power:
Charlie B.,
Thank you for your response, but I beg to differ. The whole proof is written in terms of instantaneous power and voltage as indicated by the lower case letters. You cannot deny that:
p(t) = (v1^2 + v2^2 + v3^2)/R
and as you say:
Pavg = Vrms^2/R
If we can show that:
p(t) = K
that is proof enough.
If you can convince me otherwise, I will turn in my log-log-duplex-decitrig!
Old Dr. Hansen told us we would need those trig identities some day. Well 50 years later, I needed them.
[ January 20, 2005, 09:00 PM: Message edited by: rattus ]
Charlie B.,
Thank you for your response, but I beg to differ. The whole proof is written in terms of instantaneous power and voltage as indicated by the lower case letters. You cannot deny that:
p(t) = (v1^2 + v2^2 + v3^2)/R
and as you say:
Pavg = Vrms^2/R
If we can show that:
p(t) = K
that is proof enough.
If you can convince me otherwise, I will turn in my log-log-duplex-decitrig!
Old Dr. Hansen told us we would need those trig identities some day. Well 50 years later, I needed them.
[ January 20, 2005, 09:00 PM: Message edited by: rattus ]
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