xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
what exactly are you showing us?
Power Triangle Challenge
- Thread starter xptpcrewx
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The two orthogonal sinewaves, the vector sum is the radius of a circle. The radius itself is relatively constant or stable, If it's a stable constant number the sinewave is hidden by or in the geometric representation.
Where is time in that representation. Every point on the same radius is the same instant of time. As soon as the radius rotates to a new radius, the magnitude of the radius may be stable or not changing but every value on the new radius is at the next instant of time. As the radius rotates, the sin and consine give all the points of value on the underlying sinewave. The speed of rotation of the radius is the frequency of the system.
The geometric representation was discovered hundreds of years before the findings that nature works this way.
Where is time in that representation. Every point on the same radius is the same instant of time. As soon as the radius rotates to a new radius, the magnitude of the radius may be stable or not changing but every value on the new radius is at the next instant of time. As the radius rotates, the sin and consine give all the points of value on the underlying sinewave. The speed of rotation of the radius is the frequency of the system.
The geometric representation was discovered hundreds of years before the findings that nature works this way.
kW, kVA, and orthogonal angles. Isn't what you were trying to do?
Carultch
Senior Member
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- Massachusetts
Start with the general equation of a sine wave:
y = A*sin(w*t + phi) + D
where:
A is amplitude
w is the radian frequency, traditionally represented with an omega, but I'll type w for simplicity.
phi is the phase offset
D is the offset of the midline of the sine wave from zero
For pure AC, the value of the DC offset (D) is by definition equal to zero:
y = A*sin(w*t + phi)
Using the trig identity, this sine term can be re-written as a combination of sine and cosine:
sin(alpha + beta) = sin(alpha)*cos(beta) + cos(alpha)*sin(beta)
Since beta in our previous equation is a constant (i.e. phi, the phase offset), we can write y as a sum of sine and cosine waves:
y = a*sin(w*t) + b*cos(w*t)
Lowercase a and b become the coefficients of sine and cosine of the w*t input, and are both functions of the overall amplitude (A), and the phase.
a = A*cos(phi)
b = A*sin(phi)
Now we will make V(t) and I(t) equal to variants of this equation. They will both be AC waveforms with two independent parameters, that ultimately determine the phase and amplitude. The difference is that these are instead written as coefficients of sine and cosine, instead of phase and amplitude.
V(t) = Va*sin(w*t) + Vb*cos(w*t)
I(t) = Ia*sin(w*t) + Ib*cos(w*t)
When V(t) and I(t) are represented as vectors, the values Va, Vb, Ia, and Ib become the components of these vectors, indicating the amplitudes of the sine and cosine components of their waveforms. <Va, Vb> is the vector of voltage, and <Ia, Ib> is the vector of current.
It is arbitrary where time is defined to zero. For simplicity, choose the time origin, such that the voltage waveform is the sine wave. This means Vb is by definition equal to zero.
V(t) = Va*sin(w*t)
I(t) = Ia*sin(w*t) + Ib*cos(w*t)
Multiply them together to get instantaneous power:
P(t) = V(t)*I(t)
P(t) = Va*sin(w*t)*(Ia*sin(w*t) + Ib*cos(w*t))
P(t) = Va*Ia*sin^2(w*t) + Va*Ib*sin(w*t)*cos(w*t)
Use trig identities to replace sin^2 and the product of sine and cosine:
sin^2(w*t) = 1/2*(1 - cos(2*w*t))
sin(w*t)*cos(w*t) = sin(2*w*t)/2
We are ultimately interested in the average value of the power of time function. If it adds up to zero over a cycle, it doesn't contribute any net power. It indicates storage and discharge of power in components such as capacitors and inductors, that just offset the phase. If power accumulates over the course of time with this function, then this means it contributes to real power. One cycle is called the period, and is equal to 2*pi/w.
integral of 1/2*(1 - cos(2*w*t)), from 0 to 2*pi/w =
1/2* (integral of 1, from 0 to 2*pi/w - integral of cos(2*w*t), from 0 to 2*pi/w)
1/2*(2*pi/w) - integral of cos(2*w*t), from 0 to 2*pi/w)
pi/w - integral of cos(2*w*t), from 0 to 2*pi/w)
pi/w - (sin(2*w*(2*pi/w)) - sin(0)))
pi/w - (sin(4*pi) - sin(0)))
pi/w
When taking an average value of the function, we have to divide by the total interval of integration.
avg of sin^2(w*t) = (pi/w) / (2*pi/w)
avg of sin^2(w*t) =1/2
integral of sin(2*w*t)/2 from 0 to 2*pi/w =
1/2*(-cos(2*w*(2*pi/w) - - cos(0))
1/2*(-cos(4*pi) - - cos(0))
1/2*(-1 + 1)
0
As you can see, only the sin^2(w*t) has a nonzero average value. Therefore, the average power is given by
Pavg = Va*Ia/2
What exactly do we do about the 1/2? That's where RMS voltage and current come in to play, and this is why it is the square root of 2 between RMS amplitudes and actual amplitudes. Va = sqrt(2) * Vrms. Ia = sqrt(2) * Ia_rms. Ib = sqrt(2) * Ib_rms. When multiplying together the square roots of 2, we take care of this 1/2 factor, such that Pavg = Vrms * Ia_rms. Since we represent I as a vector sum of Ia and Ib, this means that I = sqrt(Ia^2 + Ib^2), or in terms of rms currents, I = sqrt(2)*sqrt(Ia_rms^2 + Ib_rms^2). Make an overall RMS value for I, which is I = Irms*sqrt(2). This means Irms = sqrt(Ia_rms^2 + Ib_rms^2)
This shows that Ib, the component of the current waveform that is orthogonal to the voltage waveform, contributes NOTHING to the average real power delivered by the source to the load. The component current waveform that is orthogonal to the voltage waveform contributes to the mathematical contstruct we call reactive power.
Real Power:
P = Vrms * Ia_rms
Reactive power:
Q = Vrms * Ib_rms
Total apparent power:
S = Vrms * Irms
S = Vrms * sqrt(Ia_rms^2 + Ib_rms^2)
This shows how the cosine component of current, relative to voltage being exclusively a sine wave, allows us to represent real power (Watts) and reactive power (VA reactive) as perpendicular components of the power triangle. And the hypotenuse as the apparent power (VA).
xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
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xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
- Occupation
- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
Sorry No.
Carultch
Senior Member
- Location
- Massachusetts
I think what the OP is getting at, is why are kW and kVAR orthogonal to each other? And what does orthogonal mean in this context?
The short answer to what orthogonal means, is perpendicular. The long answer is that it is a more generalized mathematical term. Perpendicular is included in the definition of orthogonal, but not vice versa. Perpendicular is a geometry term, while orthogonal is a term that applies even to concepts you don't immediately think of as being perpendicular, but have properties in common with concepts you'd think of as perpendicular. In the context of vectors being orthogonal, it means the dot product is zero, which geometrically means they are perpendicular. In the context of functions being orthogonal, it means their inner product is zero. A fancy way of multiplying together functions that is analogous to dot products of vectors.
When I found the average value of sin(w*t)*cos(w*t), I was taking the inner product of the two. You see that over a complete cycle, this combination of functions averages to zero. Sine and cosine are two well-known examples of functions that are orthogonal to each other, and this is the reason we can represent the amplitude and phase parameters that define a sine wave, as a phasor vector. When the vectors that define two sine waves are perpendicular, it means that multiplying the two sine waves will get you an average value of zero. Multiplying phasor vectors in the dot product is equivalent to calculating the inner product of the functions they represent.
Then explain what you mean?
Right angle.
Carultch
Senior Member
- Location
- Massachusetts
I don't have a protractor that I can put between electrical waveforms to show that they are at "right angles" to each other. The term orthogonal in this context means more than just at right angles to each other. It has properties in common with right angles, but it doesn't just mean "right angles". We have a vector representation of the amplitudes and phases of each waveform that are at right angles to each other, but waveforms are functions of time, and it really is a time delay between equivalent parts of the cycle.
What orthogonal really means in this context, is that the inner product of waveforms is zero. This will happen if the waveforms are an odd multiple of a quarter cycle apart in phase, which means a phase-space representation of their parameters, is at a right angle between them.
xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
I thought I did a pretty good job in post # 1.
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I did a pretty good of my calculations. They are accurate and actual.
Would you like some more examples?
xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
You provided a table with numbers with no context. Accurate and actual as they may be, this is not a derivation and doesn't provide any insight to any of the questions posed in the OP.
The context is exactly what they are. I m sorry you can't seem to grasp that.
xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
Look at it this way, if you were asked this same question on an exam in college, your table of numbers wouldn't be considered proof and you would get an F.
But they ARE proof, sunshine.
Carultch
Senior Member
- Location
- Massachusetts
You don't prove by example alone. You can disprove by counterexample, but you can't prove a positive statement by example alone.
You can demonstrate a mathematical relationship by example, and you can get an idea of what a solution might be by example, or by series of examples called tables and graphs. But this isn't proof. Proof can only come when solving the problem in the generalized case, to show that it is valid for all possible conditions for which you claim it is valid.
Consider the equation y = (x^2 - 5^2)/(x - 5). If you graph this, you will seemingly get a line that equivalent to y = x + 5. Graphing will evaluate the equations with a large number of points, and connect the dots between them to fill in the relationship between x and y. And yes, in general, most values of x will demonstrate that (x^2 - 5^2)/(x - 5) = x + 5.
However, at the specific point of x=5, something interesting and frustrating happens. We end up with a zero divided by zero in the first equation, and the first equation is undefined at this point. If we only tried even numbers from 0 thru 10 to plot this graph, the issue at x=5 is a blindspot to our attempt to prove by example. If we claimed the domain of the equation (x^2 - 5^2)/(x - 5) = x + 5, is all real numbers, we would be incorrect. The point is, mathematics requires proofs without blindspots, to show that our proof is complete.
This property where equations have a pole and a zero that coincide at the same input, is called a removable zero, or removable singularity. It ends up meaning that the terms can cancel and the function surrounding this point will remain the same as if the removable singularity never existed. It just means the removable singularity is a problematic point, where the function is undefined. For all values of x other than 5, it is true that (x^2 - 5^2)/(x - 5) = x + 5
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It is proof. It isn't an example. It is actually a drive that worked and is working in real life. The calculations are for that specific system.
wwhitney
Senior Member
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- Berkeley, CA
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- Retired
I haven't looked at the numbers, but it may be proof for that one drive. That doesn't mean it works for all drives. Proving that would require a general argument along the lines of what Carultch or I gave.
Cheers, Wayne
Cheers, Wayne
xptpcrewx
Power System Engineer
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- Las Vegas, Nevada, USA
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- Licensed Electrical Engineer, Licensed Electrical Contractor, Certified Master Electrician
So far wwhitney and Carultch have demonstrated the concept, so thank you for participating. Attached is my concise version of the derivation. I don't go into the details between steps, since I will assume you all know basic trigonometry and algebra (and can work through and look up the identities for yourselves). The key thing here is the set-up (shift analysis and voltage to standard position θv = 0) and the final result: Vrms·Irms [cos(θv-θi) + cos(2ωt)·cos(θv-θi) + sin(2ωt)·sin(θv-θi)]
Since the terms Vrms·Irms [cos(2ωt)·cos(θv-θi)] and Vrms·Irms [sin(2ωt)·sin(θv-θi] are orthogonal, they can be expressed in the form of a pythagorean identity. Moreover, by omitting the time-varying sinusoid factors (cos(2ωt) and sin(2ωt)) and focusing on the component amplitudes (Vrms·Irms [cos(θv-θi)] and Vrms·Irms [sin(θv-θi]), one could represent these geometrically as the adjacent and opposite sides of a right triangle, where the component amplitudes Vrms·Irms [cos(θv-θi)] and Vrms·Irms [sin(θv-θi] are equal to the quantities P and Q respectively.
Analysis:
The result Vrms·Irms [cos(θv-θi) + cos(2ωt)·cos(θv-θi) + sin(2ωt)·sin(θv-θi)] is the instantaneous power and the only actual concept of electrical power for sinusoids as defined by physics. This however could be resolved into arbitrary components as follows:
P = Vrms·Irms [cos(θv-θi)] can be thought of as the maximum or peak magnitude of the instantaneous power component occurring in phase with the voltage, or it can be thought of as the average/net power transferred in one direction. Either way, they are equivalent in value.
Q = Vrms·Irms [sin(θv-θi] can be thought of as the maximum or peak magnitude of the instantaneous power component occurring in quadrature with the voltage.
Conclusion:
So there you have it. This is the connection of the power triangle to physics and arguably how the power triangle came to be. The P and Q components are in quadrature because the instantaneous power expression can be resolved into and written in the form of orthogonal components. Note: This is only a mathematical and somewhat arbitrary engineering construct because in nature, instantaneous power is the only thing that is really occurring (the superposition of components only exists in our analysis).
Opinion:
While inability to do this challenge is not grounds to disqualify anyone from calling themself an engineer, it does reveal something about that engineers level of understanding on basic concepts. Not being able to reason through this challenge would be justified if it required advanced education and analytical techniques, but not something as fundamental as this. Keep in mind, there was never an expectation to perform this challenge at the drop of a hat, but as engineers we should be able to do it and explain it. I also believe as an engineer, the attitude shouldn't be to dismiss such a challenge because its not useful to your job. At the end of the day, knowledge is only useful if you apply it, and the concepts behind this challenge do not fall short of useful applications.
Since the terms Vrms·Irms [cos(2ωt)·cos(θv-θi)] and Vrms·Irms [sin(2ωt)·sin(θv-θi] are orthogonal, they can be expressed in the form of a pythagorean identity. Moreover, by omitting the time-varying sinusoid factors (cos(2ωt) and sin(2ωt)) and focusing on the component amplitudes (Vrms·Irms [cos(θv-θi)] and Vrms·Irms [sin(θv-θi]), one could represent these geometrically as the adjacent and opposite sides of a right triangle, where the component amplitudes Vrms·Irms [cos(θv-θi)] and Vrms·Irms [sin(θv-θi] are equal to the quantities P and Q respectively.
Analysis:
The result Vrms·Irms [cos(θv-θi) + cos(2ωt)·cos(θv-θi) + sin(2ωt)·sin(θv-θi)] is the instantaneous power and the only actual concept of electrical power for sinusoids as defined by physics. This however could be resolved into arbitrary components as follows:
- Vrms·Irms is an expression for the apparent power,
- Vrms·Irms {cos(θv-θi) + cos(2ωt)·cos(θv-θi)} is an expression for the instantaneous real/active power,
- Vrms·Irms {cos(θv-θi) } is a constant term which represents the average/net power transferred in one direction,
- Vrms·Irms {cos(2ωt)·cos(θv-θi)} is an expression for the time-varying/oscillating power component occurring in phase with the voltage (net power per cycle = 0),
- Vrms·Irms [sin(2ωt)·sin(θv-θi)] is an expression for the instantaneous imaginary/reactive power. It is the time-varying/oscillating power component occurring in quadrature with the voltage (net power per cycle = 0),
P = Vrms·Irms [cos(θv-θi)] can be thought of as the maximum or peak magnitude of the instantaneous power component occurring in phase with the voltage, or it can be thought of as the average/net power transferred in one direction. Either way, they are equivalent in value.
Q = Vrms·Irms [sin(θv-θi] can be thought of as the maximum or peak magnitude of the instantaneous power component occurring in quadrature with the voltage.
Conclusion:
So there you have it. This is the connection of the power triangle to physics and arguably how the power triangle came to be. The P and Q components are in quadrature because the instantaneous power expression can be resolved into and written in the form of orthogonal components. Note: This is only a mathematical and somewhat arbitrary engineering construct because in nature, instantaneous power is the only thing that is really occurring (the superposition of components only exists in our analysis).
Opinion:
While inability to do this challenge is not grounds to disqualify anyone from calling themself an engineer, it does reveal something about that engineers level of understanding on basic concepts. Not being able to reason through this challenge would be justified if it required advanced education and analytical techniques, but not something as fundamental as this. Keep in mind, there was never an expectation to perform this challenge at the drop of a hat, but as engineers we should be able to do it and explain it. I also believe as an engineer, the attitude shouldn't be to dismiss such a challenge because its not useful to your job. At the end of the day, knowledge is only useful if you apply it, and the concepts behind this challenge do not fall short of useful applications.
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