Maximum power transfer

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rattus

Senior Member
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" ?

/QUOTE]

R is the source resistance--5ohms. r is the load resistance which we must compute unless we already know that the load impedance must match the source impedance.
 

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" ?

R is the source resistance--5ohms. r is the load resistance which we must compute unless we already know that the load impedance must match the source impedance.


So R would be the electron's resistance to movement in the generator armature, and r would be the resistance to the electron's movement in a motor's armature?


The above just being an example, any power transfer needing to have resistance at both ends
 
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Joethemechanic

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So I am seeing max power transfer taking place at a relativity high voltage with a synchronous ac generator at one end absorbing mechanical energy, and a synchronous ac generator at the other end producing mechanical energy.

Highest field excitation on the generator end (without going into over-saturation) and the synchronous motor end having it's field adjusted to a PF of unity.

All this done at the lowest practical speed/frequency.


All kinda nice and smooth like a nice laminar flow pipeline
 

rattus

Senior Member
So R would be the electron's resistance to movement in the generator armature, and r would be the resistance to the electron's movement in a motor's armature?

You could look at it that way, but the source impedance encompasses everything from the generator to the point of use--transformers, wiring, etc. The load resistance, r, could be from lighting, heating, induction motors, universal motors, whatever. Or it could be from a combination of these devices.
The above just being an example, any power transfer needing to have resistance at both ends

That is half true. Any load has an effective resistance, but the source would ideally exhibit zero resistance. Of course in practice, that is never attained.

BTW, you would never try for maximum power transfer in a power system. You would try for maximum efficiency.
 
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Joethemechanic

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but the source would ideally exhibit zero resistance.


Would not that break the First Law of Thermodynamics?

Or is what you are talking about just removing the I^2R losses from the armature windings?
 
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topgone

Senior Member
120109-0927 EST

topgone:

I specified a source of a sine wave voltage of an RMS amplitude, a frequency, and a specific internal impedance. The question was what is the load for maximum power transfer to the load. No restriction was placed on the load because the very broad word load was used. Now it is up to you to figure out what that load is to achieve the desired goal. Implicit in the problem is that the load has no internal power generation source, and that the load consists of linear components.

Having nothing to do with this question tell me what is the maximum phase shift of an RC network?

.
Thanks for that. I rather cancel out the inductance and make sure my load is equal to the internal resistance, That makes my load capacitive plus a 5-ohm total resistance. Period.
 

rattus

Senior Member
Would not that break the First Law of Thermodynamics?

Or is what you are talking about just removing the I^2R losses from the armature windings?

I don't see that it would. We often assume zero source impedance where R is small compared to r. As a practical matter though the source impedance is never zero.
 

Joethemechanic

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Ok so what I am visualizing as source impedence is incorrect. Because if you removed what I visualized, you would be breaing the first law of thermodynamics. Good thing you mentioned what you did, and good that I questioned it before my mind went too far down the wrong path.

What would be a good read on source impedance? Anything on the net?


Lets say we discovered a superconductor. It was made out of a metal with zero resistance. We build an AC generator with a rotating field. We wind the stator (*armature) with wire made out of this superconductor. No other changes are made to the alternator

Would source impedance now be zero?

* I often hesitate to use the term armature in the field, because many make the mistake of assuming the part that spins is always the armature
 
T

T.M.Haja Sahib

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120109-1115 EST
it is a very important concept relative to photo-voltaic energy sources.

gar
Thanks for your reply.Is it possible to supply more details on above.
 

LMAO

Senior Member
Location
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Yes, but can you explain how it is calculated?
I think that's what Gar was getting at.

I am too lazy to read the rest of comments on this interesting thread so forgive me if someone else has already answered this: P_load=I^2 * R_load = (V^2/(R_internal + R_load)^2)*R_load.
So for P_load = (V^2/(R_internal + R_load)^2)*R_load to be max, derivative of P_load with respect to R_load has to be zero.

Derivative of P_load with respect to R_load is

V^2[R+R_load)^2-2*(R_load^2+R*R_load)]/(R+R_load)^4 = V^2[R^2-R_load^2]/(R+R_load)^4

so for the derivative to be zero R^2 has to be equal to R_load^2 which means R=?R_load. Negative resistance is impossible so R=R_load.
 

K8MHZ

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Michigan. It's a beautiful peninsula, I've looked
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BTW, you would never try for maximum power transfer in a power system. You would try for maximum efficiency.

I can think of one exception, using a car battery to start a car. For a few seconds, max transfer is desired. Efficiency is not much of an issue.

Outside of power systems, max power transfer is needed in radio to keep the transmitter and antenna matched. That is where conjugate matching is essential. If a transmitter and an antenna aren't matched, resulting in a high standing wave ratio, all kinds of nasty things can happen, including damage to the transmitter.
 

gar

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120110-1310 EST

A few quick points.

Maximum power transfer is just a good theoretical question to challenge one's thought process.

K8MHZ's starter motor was a good example.

However, predate 1879 and the scientific community thought that a power distribution system should operate under maximum power transfer conditions. Also dynamos were very inefficient. Electric lighting was by arc lamps and these required current limiting.

Edison realized that to build a power distribution system the source had to be efficient, have a low internal resistance, and the distribution wiring had to be of low resistance relative to the load. Further this also meant the distribution voltage had to be high, and thus light bulbs had to have a high resistance. My guess is that 110 V or thereabouts was a reasonable compromise of the various factors. The loads had to be switchable parallel loads. Had to work like gas lights.

Edison developed a relatively efficient dynamo using a laminated iron core in the armature, and a compound field for voltage regulation. They also discovered that ferromagnetic materials had the characteristic of saturation.

Back to maximum power transfer.

If you have a dynamic microphone you are limited in available input power, the person speaking. Thus, to get the maximum output signal you want to operate at maximum power transfer.

Suppose you have an engine powering a generator. This is a power limited system. You want to heat a tank of water as fast as possible, then you also want maximum power transfer to the heating resistance.

In a radio transmitter the FCC limits your maximum input power. You want from this system the maximum radiated power from the antenna.

A quick comment on photovoltaic. The source is a non-linear device and you want to get maximum power to your load. Thus, the inverter is designed to test for and adjust to operate at the maximum power point.

.
 

Besoeker

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Location
UK
I am too lazy to read the rest of comments on this interesting thread so forgive me if someone else has already answered this: P_load=I^2 * R_load = (V^2/(R_internal + R_load)^2)*R_load.
So for P_load = (V^2/(R_internal + R_load)^2)*R_load to be max, derivative of P_load with respect to R_load has to be zero.

Derivative of P_load with respect to R_load is

V^2[R+R_load)^2-2*(R_load^2+R*R_load)]/(R+R_load)^4 = V^2[R^2-R_load^2]/(R+R_load)^4

so for the derivative to be zero R^2 has to be equal to R_load^2 which means R=?R_load. Negative resistance is impossible so R=R_load.
Yes.
Writing equations within the constraints of boards like this can be something of a challenge.
 

Joethemechanic

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Electro-Mechanical Technician. Industrial machinery
Hey, were my wavelength calcs right for 60Hz being like 3000 and change long?

I kept having that standing wave ratio thing from my radio days in my head BTW


BTW, most automotive starters are series field devices, just like many traction motors, particularly in electric forklift trucks
 

gar

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1230110-1726 EST

Joethemechanic:

The velocity of light is 299,792,458 meters / second in a vacuum.

Thus, the wavelength in meters is approximately = 300,000,000 / frequency in CPS (Hz) = 300 / MC (mHz).

30 MC is a wavelength of 10 M.
60 CPS is a wavelength of 5,000,000 meters = 3104 miles.

20 M was a good band for communication with Australia.

First radar reflection from the moon was in the mid 1940s using about 150 MC (2 M).

Detection of the Japanese attack planes on Sunday 7 Dec 1941 was with radar in the 2 M to 2.5 M range. Ignored.

Reflected radar energy varies by the -4th power of the target range for an isotropic transmitter and reflector.
http://www.radartutorial.eu/01.basics/rb13.en.html

.
 
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Joethemechanic

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OOPs

I forgot units

I think the wavelength I calculated for 60Hz was 5,000 meters, or 3,000 plus miles.


OOPs again that should read 5,000 kilometers
 
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K8MHZ

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Michigan. It's a beautiful peninsula, I've looked
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Electrician
1230110-1726 EST

Joethemechanic:

The velocity of light is 299,792,458 meters / second in a vacuum.

Thus, the wavelength in meters is approximately = 300,000,000 / frequency in CPS (Hz) = 300 / MC (mHz).

30 MC is a wavelength of 10 M.
60 CPS is a wavelength of 5,000,000 meters = 3104 miles.

20 M was a good band for communication with Australia.

First radar reflection from the moon was in the mid 1940s using about 150 MC (2 M).

Detection of the Japanese attack planes on Sunday 7 Dec 1941 was with radar in the 2 M to 2.5 M range. Ignored.

.

MC?

Is someone showing their age?? :D

During the solar minimum, 20 meters is the best long distance band. I made a contact from Michigan to Japan on single sideband with 100 watts into a wire antenna during the last one. Right now, there is more solar activity so 10 meters, during the day, is the best band for distance. Japan could be had with 10 watts under the right circumstances and using an antenna half the physical size. A couple of years ago, 10 meters was pretty useless for long distance, but was fine for short comms.

During WWII, the Germans had mastered what was then 'UHF', or 10 meters for their short comms between tanks in Africa. All was well for them until the sun started acting up and their tank to tank comms could be heard by ham radio operators in the US.

Ham radio operators use 2 meter CW to bounce signals off the moon. It is called EME. Some of the stations are incredible, with 100,000 dollar antennas.

I didn't know that the Pearl radar was 2 meters. I guess I always associated radar with higher frequency bands, but thinking about it, OTH, or over the horizon radar uses HF bands, like 40 and 80 meters.
 

Joethemechanic

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Electro-Mechanical Technician. Industrial machinery
That would be quite an undertaking to build a folded dipole for.


I hate to say it, but I'm old enough to, at one time have held a CB radio (11 meter for those in other countries) license. KGQ-4370 I believe was the call sign.


And I got to say, I always loved the 40, and 80 meter stuff
 
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