Maximum power transfer

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K8MHZ

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I would solve this problem as follows:

I = Vs/(r + R)

Where Vs is the source voltage, r is the load resistance, and R is the source resistance., then find the power, P.

P = r x Vs^2/(r + R)^2

then find the derivative,

dP/dr and set it to zero.

Now solve for r.

A little thought tells us that if the source is reactive, then the load must also be reactive, the complex conjugate of the source impedance so the reactances offset each other.

Why would we have to solve for 'r' if we know 'R'? (Jacobi's Law says they will be equal)

Where was it mentioned that the source was reactive?
 

Joethemechanic

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A little thought tells us that if the source is reactive, then the load must also be reactive, the complex conjugate of the source impedance so the reactances offset each other.

Now I'm thinking about how a asynchronous generator, such as an induction motor in a braking (over-driven) situation can not supply reactive power. It must absorb reactive power from the line. (or a bank of caps) It can only supply active power
 

Joethemechanic

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I'm a little lost as to just exactly what you aree looking for?

Max power transfer, meaning without loss?

Max power transfer , meaning over the shortest period of time

Or has this all got to do with dc power transmission which has large entropic problems at it's ends if we want to utilize it along with our present power distribution system.


I mean this whole thing about the photovoltaic installations. We have no choice but to generate it at that end as DC, so we might as well transmit it on high voltage DC transmission lines. High voltage dc can be underground also. Even the above ground structures do not have to be as large.
 

gar

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Joethemechanic:

The simple circuit to consider is the DC circuit. Then let x = rload / Rinternal

This is to make the equation to solve for maximum power moderately simple.

pload = rload* i2load ........ p, r, and i are all variables as distinguished from constants.
Vsource and Rinternal are constants.

pload = (V2source / Rinternal) * x / (1 + x)2


Thus, all we need to do is find the value of x that produces a maximum for x / (1+x)2 .

This can be rewritten as x / (x2 + 2*x + 1) .

By trial and error let x = 0, .1, .5, .75, .9, 1.0, 1.1, 2, 10, and infinity.
The results are 0, 0.08, 0.22, 0.244, 0.249307, 0.2500000, 0.249433, and you can go further if you want. With this knowledge it would be logical to try 0.999 and 1.001 and see if it still looks like 0.25 is the maximum.

Those familiar with calculus can differentiate the function relative to x and set equal to 0 and solve for x. The denominator is complex but always positive and non zero so it is only necessary to look at the numerator which is -x2 + 1 and for x non-negative the solution is x = 1. Thus, the load resistance for maximum power transfer is when the load resistance equals the internal resistance.

.
 

Joethemechanic

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Maximum power transfer over a given set of conductors takes place at the lowest frequency.

From what I understood, 60Hz was a compromise. It was the lowest frequency that they felt did not cause a noticeable flicker in incandescent lights
 

Joethemechanic

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Now class, let's not always see the same hands......


Yeah, what's up with that?

Calculus class was a long time ago for me. I'm still somewhat up on trig and geometry.

On a day to day basis, I work with guys that can't sum up all the weight of a lift if everything has the weight marked on it with a red paint stick.
 

gar

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wirenut1980:

I have to do some other work now. But I will try to remember your question.

The whole point of the thread was just to initiate thought by the readers and it seems to be doing that. i certainly don't expect the average reader to have any experience with the concepts of calculus. Thus, they need to use some other means to try to answer the questions. At lot of real world problems have to be solved by a trial and error or an iterative procedure.

I think that there is some justification that readers understand the concept of maximum power transfer, and I think that Joethemechanic is asking for some help on the concept. Maybe someone can assist him while I am off trying to do some work.

.

.
 

K8MHZ

Senior Member
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Michigan. It's a beautiful peninsula, I've looked
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Electrician
Maximum power transfer over a given set of conductors takes place at the lowest frequency.

From what I understood, 60Hz was a compromise. It was the lowest frequency that they felt did not cause a noticeable flicker in incandescent lights

From Wiki:

After observing flicker of lamps operated by the 40 Hz power transmitted by the Lauffen-Frankfurt link in 1891, AEG raised their standard frequency to 50 Hz in 1891.[5]

Westinghouse Electric decided to standardize on a lower frequency to permit operation of both electric lighting and induction motors on the same generating system. Although 50 Hz was suitable for both, in 1890 Westinghouse considered that existing arc-lighting equipment operated slightly better on 60 Hz, and so that frequency was chosen.[6] The operation of Tesla's induction motor required a lower frequency than the 133 Hz common for lighting systems in 1890. In 1893 General Electric Corporation, which was affiliated with AEG in Germany, built a generating project at Mill Creek, California using 50 Hz, but changed to 60 Hz a year later to maintain market share with the Westinghouse standard.
 

rattus

Senior Member
gar didn't give us the value of r, I think he wanted us to solve for it.

Later posts did specify a reactive component source impedance.
 

Joethemechanic

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That is why I always thought the railroad was on the right track (no pun intended) by going with 25Hz traction. Like they do from somewhere below Philly up to where the Hellgate 60Hz line starts in NYC.

When I was a kid riding with my dad in the car, I always wondered why the 60 cycle hum we encountered near a substation on the AM radio wasn't the same hum that interfered with the radio near one of the railroad's substations.
 
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iwire

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That is why I always thought the railroad was on the right track (no pun intended) by going with 25Hz traction. Like they do from somewhere below Philly up to where the Hellgate 60Hz line starts in NYC.

When I was a kid riding with my dad in the car, I always wondered why the 60 cycle hum we encountered near a substation on the AM radio wasn't the same hum that interfered with the radio near one of the railroad's substations.

Low frequency has less voltage drop problems but results in much larger heaver transformers, motors and generators. Fine for a train.

High frequency has a lot of voltage drop but has much smaller lighter transformers, motors and generators, great for planes.

For general purposes it seems 50 to 60 hz is the best balance. At least that is what I have read.
 

Joethemechanic

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Electro-Mechanical Technician. Industrial machinery
The reasons I always understood for 25Hz were

Higher power factor

Easier conversion to DC

Satisfactory operation of motors designed for DC on 25Hz AC

I believe it was a PRR thing. PRR at one time had the most extensive product testing and development laboratories in the world
 

K8MHZ

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Michigan. It's a beautiful peninsula, I've looked
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gar didn't give us the value of r, I think he wanted us to solve for it.

Later posts did specify a reactive component source impedance.

He did give us 'R', which has to be equal to 'r'.

Suppose I change the source to a 100 V RMS 60 Hz sine wave with an internal impedance of 5 ohms resistance in series with a 10 H inductor, then what load results in maximum power transfer to the load?

I took that as to mean the source was 100 V with an internal impedance of 5 ohms connected to a 10 H inductor, not the 10 H inductor being part of the source.
 

Joethemechanic

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You know my for lack of a better word "language" or vernacular is somewhat rusty from probably 35 years of inactivity.


How does "R" differ from "r" ?


Just too many years of calculating on the back of envelopes on the front seat of my truck I guess
 

rattus

Senior Member
I took that as to mean the source was 100 V with an internal impedance of 5 ohms connected to a 10 H inductor, not the 10 H inductor being part of the source.

Doesn't matter, the inductive reactance must be offset either way for max power.
 
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