And Now MVaR in words of one sylable or less 6

[JJayG]

Here's one for all of you clever electrical chappies out there.
Imagine trying to explain MVaRs to a new starter, ie. in the simplest way possible.
All analogies welcome (except maybe the horse and barge one).
Points may be awarded for creativity but deducted for over technicality.
This has always been one of those questions that,(along with what is entropy?)has been a source of consternation in power stations since Michael Faraday was a lad.
Go On, you know you want to............

JJ

 

[Yuma]

I've only heard it from a single person (a Canadian guy), and I thought maybe it's a common term in some parts of the world...

 

[waross]

Hi Gunnar;
To quote myself;

Should a German, an Italian, or a Russian challenge my use of the rules of the English grammar I would be offended. Should a Francophone challenge my or demand that I change my habits, Oh well, that's just the French.

I almost added, if a Swede challenges my use of the English language, tread cautiously, he may well be correct.
You are correct sir, as usual.
However, may I suggest an alternate point of view?

In a town lived a man and a Countess. The man despised the Countess and would only refer to her as a pig. (Please be patient, there is a point.)
Eventually the Countess had the man hauled into court. The judge explained to the man that he could not call the Countess a pig and he would be in serious trouble if he continued.
The man asked;
"If I am not allowed to call a Countess a pig, may I call a pig a Countess?"
"Yes you may." replied the judge.
The man then turned and said;
Good afternoon, Countess!"
That said,
I accept that I must call Volt Amps reactive vars.
But, may I call vars "Volt Amps reactive"?
If I am allowed to call vars "Volt Amps reactive" then by the accepted rules of English Grammar, the acronym VARs is acceptable.
I agree and accept that var is the proper name to use in a formula, and accept the responsibility for any loss or confusion that may result should I inadvertently capitalize any part of var.
However, may I use the grammatically correct acronym to replace the term Volt Amp reactive when writing conversationally?
Example:
You may speak about VARs all you wish, but if you are using the term in a mathematical formula the correct term is var.
I hope this suggestion is acceptable to most if not all of the parties to this discussion.

Yuma, I spent most of my life in Canada and have not heard the term.
Scotty, I hope your daughter has recovered from her cold.

On a lighter note;
Voltaire? What on earth has he to do with electricity, the Volt is named after Volta.
Voltaire had a knack for antagonizing those in authority, (possibly mostly those who were abusing their authority).
Despite his unpopularity banishments and exiles, he lived to a good old age and probably died of mostly natural causes. (Overexertion? An arduous journey of several days duration was the probable direct cause.)
Although Voltaire died when Ampere was a young boy and ten years before the French revolution, Voltaire's ideas and attacks on the French system of government of the time was/is seen by some as part of reason for the French revolution, despite the ten years or so that elapsed between his death and the start of the revolution.
Ampere's father died in that revolution. Was there any family animosity, probably not, but it is possible.
Forgive me for a poor attempt at humor based on such a tenuous connection. It's the probably the result of a brain damaged by too many years of exposure to too many electro static and electro magnetic fields.
var has been around since about 1965? Yikes, I missed that one. Probably because I didn't stop drinking until the late 60's.
Despite no longer drinking myself, I'll be glad to buy you one of your choice, Gunnar and have a coffee with you.
When oh when will I learn never to doubt the master?



Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[waross]

A question that I hope is related to this topic, by way of connection to the BIPM;
When I started to learn about electricity, from my father and uncles, to do with flashlights and bicycle generators and light switches, popular knowledge was that electricity flowed from positive to negative. Popular knowledge always lags behind scientific wisdom, witness the present discussion re: var, VARs.
As I grew older I learned about electrons and had to change my ideas. Electricity now travelled from negative to positive.
This was an advantage to dislectics everywhere. They could now use the right hand rule with their left hands.
Sometime later, word filtered down that electricity had been redefined from the movement of electrons to the movement of holes, or if you will, the movement of the absence of electrons, or the movement of nothing. Much ado about nothing?
I remember a question from one of my fellow instructors in the staff room back in the early 70's.
"Should we be teaching electron flow or conventional flow?"
There were a couple of non committal sentences and the subject was changed.
Question:
Was the BIPM the body responsible for disregarding electrons and redefining the flow of electricity as positive to negative?

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ScottyUK]

Bill,

Thanks for the kind thoughts for our little 'un. I fear she is going to spend her first birthday quite ill, which is a shame. Fingers crossed that she perks up a little in the next couple of days.


----------------------------------

If we learn from our mistakes I'm getting a great education!

 

[JJayG]

There would appear to be quite a few trainee politicians in our midst. What was the original question again?
All I know is that it wasn't supposed to be an excuse for an argument on CaPitAl bLEediNg LetTers, how very very sad.
Why didn't you just tell me (as I suspected)that there isn't a simple answer and left it at that instead of going off on one?
Thanks to those who did take it in the spirit that it was written, and to those who merely used it as a vehicle to demonstrate their perceived intelectual superiority..........words fail me (But you don't quite spell it lIkE tHAt.
So, just to recap then:-

If I get two managers and tie them up in a vat of beer with a slinky on a railroad track...............

Never realised that my finger inadvertantly hitting/missing the shift key would cause such a fuss.

 

[waross]

Hi JJayG;
Sorry for that. No reflection on you whatsoever, we do this to each other.
How to explain power factor to a starter?
An analogy works by transferring knowledge from one field to another. Water pumping analogies for electrical effects only work if the learner is familiar with water pumping.
An analogy about analogies.
We were playing with a small motor bike once. We encouraged my sister-in-law to try it. I can't ride that she kept saying.
We kept assuring her that it was "Just like riding a bicycle."
Lynn finally tried it. Off she went, full throttle. Didn't make the first turn and plowed full throttle into a stand of saplings. She kept the throttle pinned and we could hear her screaming, the engine put-put-putting and the saplings breaking. Finally she stopped, in the center of the stand of saplings. When we finally got her back out in the clear, she was not happy with us.
"But, it's just the same as riding a bicycle."
"BUT I CAN"T RIDE A BICYCLE!!!"
Well, using an analogy from an area that the learner doesn't understand may be just as futile if less exciting.
Analogies work best one on one, or at least with a group that is familiar with the field referenced by the analogy.
An analogy based on the wheat fields of Kansas may be lost on a learner from the Everglades as will an analogy based on the Everglades may be lost on a learner who has never left Kansas.
Power factor and Volt amps reactive;
For the purists let's stick with displacement power factor.
When an alternating current is applied to a circuit, it will cause a current to flow. The current will be alternating. But, some circuit elements affect the flow of the current. Inductors inhibit the flow of current so that the current in an inductor, while still a sine wave, is phase shifted and reaches its peak value later than the voltage reaches its peak value. The peak current through a capacitor reaches it's peak before the voltage reaches its peak.
This effect may be described by a right triangle, The actual power consumed by the circuit is represented by the base, the Volts times Amps or VA is represented by the hypotenuse, and the altitude of the triangle is represented by an imaginary quantity called Volt-Amps-reactive. The angle will be the angle of the phase shift, or the degrees of phase shift described as one degree equals 1/360 of a complete cycle.
The right triangle and Pythagoras' theorem.
Your mission JJayG, should you choose to accept it, is to find out what fields your learner is familiar with where there is an effect that may be described by the right triangle and Pythagoras' theorem and make an analogy to transfer that knowledge to the electrical field.
One analogy that will certainly give your learner a lot of practice with the mathematics of power factor is the railway analogy.
An engine on one track is pulling a car on a parallel track. It takes a force of X units to move the car. The tracks are Y units apart. A rope Z units long is used to pull the car. What is the tension in the rope?
Solve for varying values of X, Y, and Z
Then have a beer and speculate on the ratio of beer to foam in relation to the volume of the mug.
Hope this helps.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[PHovnanian]

An analogy works by transferring knowledge from one field to another. Water pumping analogies for electrical effects only work if the learner is familiar with water pumping.

But we can safely assume that everyone knows about beer.

Beer is the Rosetta Stone of engineering.

 

[BJC]

Beer is still the best analogy.
The part people are missing is that Beer Mugs are very expensive and the price is proportional to volume.
So if your beer comes out of the tap with a healthy head of foam you have a bigger more expensive mug thna the people drinking out the tap that puts no foam on the beer.
To get the same beer you get they can buy a cheaper mug.

 

[electricpete]

I will refer to an analogy between mechancial m / k / c systems and electrical R / L / C systems which I have previously posted:

You need only 6 simple rules in order to solve any sinusoidal steady state lumped linear m/k/c mechanical vibration problem using electrical circuit impedance (R/L/C) analysis techniques:

1 - mechanical force f plays the role of electrical current I

2 - mechanical velocity difference v plays the role of electr voltage difference V

3 - mechanical mobility v/f plays the role of electrical impedance Z = V/I
(use the symbol Z throughout. We use the term impedance to represent either electrical impedance or mechanical mobility .. avoid the term mechanical impedance which is the reciprocal of mobility).

4 - A spring (k) acts like an electrical inductor of inductance L=1/k
The impedance is Zk = j*w/*k

5 - A damper (c) acts like an electrical resistor of resistance R = 1/c
The impedance is Zc = 1/c

6 - A mass (m) acts like an electrical capacitor of capacitance C =m **
The impedance is Zm = 1/(j*w*m)
** one terminal of this mass/capacitance is connected to ground.

Derivation of the impedances and their electrical equivalencies is as follows:
Spring:
f = k x
f = k (v/jw)
Zk = v/f = j*w / k
By comparison to ZL = j*w*L, we find L = 1/k

Damping
f = c v
Zc = v/f = 1/C
By comparison to ZR = R, we find R = 1/c

Mass
f = m a (where a is referenced to ground)
f = m* (jw * v) (where v is referenced to ground)
Zm = v/f = 1/ (jw*m)
By comparison to ZC = 1/(j*w*C), we find C = m
(where C is connected between the location of the mass and ground).

The above derivations are based on phasor analysis. A note about phasors:
V, I, v, f above represent phasors (magnitude and phase at assumed frequency w)
Differentiation of the associated time signal corresponds to multiplying the phasor by j*w
Integration corresponds to multiplication by j*w

This analogy also preserves the definition of instantaneous power.
Pmech(t) = V(t)*F(t) vs Pelec(t) = v(t) i(t)

Also if you look carefully at the max stored energy W in the spring and mass elements, you will find they are exactly what is predicted by electrical calc of Winductor = 0.5*L*i^2 and Wcapacitor = 0.5*C*v^2.

So let's use this analogy to try to develop analogy for power system in the m / k / c world.

I will start with a very simple mechanical system called 1m

System 1m:
Ground === vs(t) ===rigid bar===c_load ==== Ground
vs(t) = velocity source. Note if we apply a source displacement d(t)=d0* cos(w*t), this is equivalent to a velocity source v(t) = d./dt (d(t)) = w*d0*sin(w*t) = v0 * sin(w*t) where v0=d0*w is peak velocity.

c_load is a visous damper which removes energy from the system. Similar to a dashpot with viscuos fluid. It obeys c = f / v. f and v are the same as f and v of the source. A

There is no energy storage, so any energy input by the source vs is instantly converted to heat in the load c_load. f is in phase with v.

The average rate of energy transfer from source (vs) to the load (c_load) is:
Pm1 = 0.5 * v0 * f0 = 0.5 * v0^2*c
note vs as above, and fs is source force (which happens to be same as load force in this simple system.


and analogous electrical system will be called system 1e
System 1e:
Ground === Vs(t) ===ideal wire ===R_load ==== Ground
where Vs(t) = V0*sin(w(t), Is=I0*sin(w(t) and the ideal wire had neither resistance nor inductance.

Again there is no energy stored in the system, so any energy input by the source Vs is instantly converted to heat in the load R_load. I is in phase with V.

The average rate of energy transfer from source (Vs) to the load (R_load) is:
P1e = 0.5 * V0 * I0 = 0.5 * V0^2 / R

Now let us examine a second system Pm2
System 2m:
Ground === vs(t) ===k===c_load ==== Ground
where we have added spring k instead of bar.

The spring is an energy storage device and we now can have energy storage in the sytem within the spring. The energy that comes out of the source does not instantly appear at the load. The load can also pull energy out of the spring at times when the source isn't adding energy. The force is no longer in phase with the voltage.

The average rate of energy transfer from source (vs) to the load (c_load) can be shown to be:
Pm2 = 0.5 * v0 * f0 * cos(theta) where theta is the angle by which fs lags vs.
* I'm not going to dwell on the proof of this, but will come back later to disuss why it's important. *

We can also see that due to the presence of the new cos(theta) term resulting from energy storage within the system, we now need a higher magnitude f0 to transfer the same average power from the same source.

The electrical analogy to System 2m is System 2e
System 2e:
Ground === Vs(t) ===L ===R_load ==== Ground
Where we have replaced the "ideal" wire with one that has some inductance L.

We now have energy storage in the system (within the magnetic fields of L). The power leaving the source no longer instantaneously enters the load. The source current is no longer in phase with the source voltage.

The average rate of energy transfer from source (Vs) to the load (R_load) is:
P1e = 0.5 * V0 * I0 * cos(theta) where cos(theta) is power factor.

*** Most EE's are probably saying by now – this is worthless.... you haven't made the system easier you've made it harder. And I agree 100%. But turn it around and assume you were given the above mechanical system 2m and asked to analyse it! It would take a little bit of work I imagine and results would not be intuitive. But transfer it to electrical system 2e and it's a piece of cake.

Presumably, the situation is reversed for an ME. He looks at system P2e and sees a tricky system, but he looks at system P2m and sees something recognizeable. I think that is the same point waross was making.

By the way, turning this sytem from a mechanical system to a fluid system is fairly easy. Velocity is replaced with volume flow rate. Force remains force. Ideal wires connecting components are replaced with rigid pipe filled with incompressible fluid. Spring is replaced with elastic pipe or vessel containing incompressible fluid. Damper can still be a dashpot since that is sort of a fluid device.


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[electricpete]

Correction in bold:
"P1e = 0.5 * V0 * I0 * cos(theta) where cos(theta) is power factor."
should have been
"P2e = 0.5 * V0 * I0 * cos(theta) where cos(theta) is power factor."

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[jghrist]

electricpete,

If I understand your analogy correctly, beer is damp and foam is springy. I think I'll need another one to fully comprehend the situation.

 

[Bronzeado]

Hi folks,

Just to keep the "fire" on!

Gunnar is correct. The unit of reactive power is "var", not "VAR" nor "VAr". by SI convention. This was decided a long time ago. So, if we want to use correctly the SI units, we may use “var” for reactive power.

Also, we should write "ampère" not "Ampère" for current unit (the symbol is A). The same for the voltage: "volt" not "Volt" (the symbol is "V"). Also, "kV" (not "KV") and "kvar" (not "Kvar" nor "KVAR"). This is all by convention (according to SI).

I think that all this confusion is because we, as engineers, have the tendency to simplify the model of everything and. sometimes, we forget to go back to the basics.

Also, we always keep trying to visualize a mathematical entity as being physical entity. Most of the time, this is not possible.

In reality, there exist fundamentally the electric and the magnetic filed, which are associated to voltage (V) and current (A), respectively. The electrical engineering "quantities" S, P and Q are derived mathematically from these two physical entities (electric and the magnetic field, so V and A) and they have different units, VA, W and var, respectively. So we can not just add P and Q to get S. Mathematically, S is the square root of the sum of P squared plus Q squared.
All these quantities (S, P and Q) were defined based on linear system and sinusoidal waves. If the voltage and/or current waveform are distorted, another electrical quantity D (distorted power) has been defined (I do not remember its unit). So, S will be now (?) the square root of the sum of P squared plus Q squared plus D squared. This quantities is only maths, not physical.

To finalize, I would like to stress and remind you all that the apparent (VA), real (watt), reactive (var) and (now) distorted (?) power are only mathematical models created to simplify (?) power system analysis. So, S, P, Q and D are defined as being components of the instantaneous power wave, These components do not exist as separate entities but they might be conveniently considered for purpose of engineering analysis.

Herivelto Bronzeado

 

[Skogsgurra]

We are now very far from the OP's "One syllable explanation". But we have had a useful and sometimes refreshing discussion on topics that are usually considered less important.

Now, Herivelto, that D for distortion "power" is also something that needs discussion. Especially the way it is added perpendicular to P and Q. The D^2 in your expression for S.

That has been done for decades and I think that it is something one shouldn't do. It may very well be one of the "engineering simplifications" that you mention in your post.

The reason why I think it is wrong to do so is that "D" does not lend itself to a geometrical construction with fixed angles (because the frequencies involved are different). But, on the other hand, signal components with different frequencies have always been added as root mean square - or RMS. So, there may be a good reason for doing so. But, on still another hand, the RMS addition is used for components in noise where there is no relation between the different components and their frequencies.

Can anyone show that the geometrical addition of P, Q and D is correct? I am, of course, thinking of you, Pete.

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[edison123]

How are milliwatts, millivolts, milliamps, milliohms etc. represented in SI ?

Muthu

http://www.edison.co.in

 

[Skogsgurra]

mW, mV, mA etcetera. Capital M is mega (million) and "m" is milli (1/1000). Also, there is no "K" - only "k" for 1000.

OK, there is one "K" - used for 1024 in KB as opposed to "k" as in kB (1000 bytes).

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[edison123]

So, mega gets a M and kilo gets only a puny k ?

Muthu

http://www.edison.co.in

 

[Skogsgurra]

Yes. Right!

Gunnar Englund

http://www.gke.org

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100 % recycled posting: Electrons, ideas, finger-tips have been used over and over again...

 

[electricpete]

Gunnar, you succeeded in getting me to take an hour away from my family on a Sunday (I'm sure that was not your intent).

At any rate, I took the bait and here is my analysis of the relationships S, P, Q, D

First of all, D is sometimes defined as D is defined as sqrt(Strue^2 – P^2 – Q^2).


In that case, the conclusion that Strue = SRSS (P, Q, D) is trivial.
(SRSS = square root of sum of squares)

So I will instead use an alternate definition of D (more below)

Start with definition of apparent power:
Strue = Vrms * Irms [equation 1 – definition]

With harmonics present (and the waveforms periodic in w0)
I(t) = Sum Im(t) for m = 1 to m_max
where Im(t) = sqrt(2)*Im_rms_ * sin(m*w0*t + theta_m)

V(t) = Sum In(t) for n = 1to n_max
where Vn(t) = sqrt(2)*Vn_rms * sin(m*w0*t + theta_n)

Strue = Irms * Vrms = SRSS(Im_rms) * SRSS (Vn_rms) [equation 2]

If we multiply the terms of equation 1, we get a square root of products of the form Im_rms^2 * Vn_rms^2. But this represents SRSS combinations of the terms Im_rms and V_n_rms.

So we can write:
Strue = SRSS (In_rms*Vm*rms) where all n and m combinations are included [eq 3]

Now we define the "SRSS combination property"
SRSS Combination Property: SRSS(A,SRSS(B,C)) = SRSS(A, B, C).
This can be verified by simple algebra and is an important property which will be used again and again.

Now let us look at the pile of terms Im*Vn within equation 3 and sort them into buckets D, P, Q:

If m < > N, then the term goes into the D bucket and combined using SRSS:
D = SRSS (Imrms Vnrms) for m < > n
(By the way, directly above is promised alternate definition of D)

If m = n, then we need to break down Im further into two parts: Imrms_p=Im_rms*cos(theta_mm) and Im_rms_q = Im_rms*sin(theta_mm) where theta_mm angle between voltage and current for this harmonic. Then we use these two parts to fill the buckets P or Q accordingly
P = SRSS (Im_rms_p^2*Vnrms^2) where we include all m = n
Q = SRSS (Im_rms_q^2*Vnrms^2) where again we include all m = n

Define S1 =Sapparent power of fundamental components
S1 = SRSS(P,Q) should be obvious since each we have broken Im_rms into two pieces (Im_rms_p and Im_rms_q) whose SRSS combination is Im_rms based on sin^1+cos^2=1

Now at the end of all this, every term has ended up in one and only one bucket.
The terms where m=n have ended up in the P and Q bucket where S1 = SRSS(P,Q).
The terms were m<>n have ended up in the D bucket.

By the combination property of SRSS
S = SRSS(everything in the bucket) [equation 3]
S = SRSS(D, S1)
again by the combination property we can split S1 into P and Q
S = SRSS(D, P, Q) [equation 4]

Sounds very complicated, but simple with the combination property. Each term gets added directly into S in eqation 3. We SRSS combine individual clumps of terms in equation 4. By the SRSS combination property we expect the same results either way.


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[electricpete]

A few corrections:

First of all, D is sometimes defined as D is defined as sqrt(Strue^2 – P^2 – Q^2).

In that case, the conclusion that Strue = SRSS (P, Q, D) is trivial.
(SRSS = square root of sum of squares)

So I will instead use an alternate definition of D (more below)

Start with definition of apparent power:
Strue = Vrms * Irms [equation 1 – definition]

With harmonics present (and the waveforms periodic in w0)
I(t) = Sum Im(t) for m = 1 to m_max
where Im(t) = sqrt(2)*Im_rms_ * sin(m*w0*t + theta_m)

V(t) = Sum In(t) for n = 1to n_max
where Vn(t) = sqrt(2)*Vn_rms * sin(m*w0*t + theta_n)

Strue = Irms * Vrms = SRSS(Im_rms) * SRSS (Vn_rms) [equation 2]

If we multiply the terms of equation 1, we get a square root of products of the form Im_rms^2 * Vn_rms^2. But this represents SRSS combinations of the terms Im_rms and V_n_rms.

So we can write:
Strue = SRSS (In_rms*Vm*rms) where all n and m combinations are included [eq 3]

Now we define the "SRSS combination property"
SRSS Combination Property: SRSS(A,SRSS(B,C)) = SRSS(A, B, C).
This can be verified by simple algebra and is an important property which will be used again and again.

Now let us look at the pile of terms Im*Vn within equation 3 and sort them into buckets D, P, Q:

If m < > N, then the term goes into the D bucket and combined using SRSS:
D = SRSS (Imrms Vnrms) for m < > n
(By the way, directly above is promised alternate definition of D)

If m = n, then we need to break down Im further into two parts: Imrms_p=Im_rms*cos(theta_mm) and Im_rms_q = Im_rms*sin(theta_mm) where theta_mm angle between voltage and current for this harmonic. Then we use these two parts to fill the buckets P or Q accordingly
P = SRSS (Im_rms_p*Vnrms) where we include all m = n
Q = SRSS (Im_rms_q*Vnrms) where again we include all m = n

Define S1 =Sapparent power of fundamental components
S1 = SRSS(P,Q) should be obvious since each we have broken Im_rms into two pieces (Im_rms_p and Im_rms_q) whose SRSS combination is Im_rms based on sin^1+cos^2=1

Now at the end of all this, every term has ended up in one and only one bucket.
The terms where m=n have ended up in the P and Q bucket where S1 = SRSS(P,Q).
The terms were m<>n have ended up in the D bucket.

By the combination property of SRSS
Strue = SRSS(everything in the bucket) [equation 3]

by the combination property (inverted), we can break that into terms where m<>n (show up in D and terms where m=1 (show up in S1):

Strue = SRSS(D, S1)
again by the inverse combination property we can split S1 into P and Q
Strue = SRSS(D, P, Q) [equation 4]

Sounds very complicated, but simple with the combination property. Each term gets added directly into S in eqation 3. We SRSS combine individual clumps of terms in equation 4. By the SRSS combination property we expect the same results either way.


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[electricpete]

Q = SRSS (Im_rms_q*Vnrms) where again we include all m = n
was not correct.

Should have been:
Q = SRSS (Im_rms_q*Vnrms) where m=n=1

The remaining terms (Im_rms_q*Vnrms where m=n <>1) should be thrown into the D bucket. We still have everything accounted for once and only once. Btw voltage THD typically low so these terms generally not very big anyway.

JJayG - sorry for the tangent.

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