Maximum power transfer

[jumper]

This is a really cool thread Gar.:thumbsup:

 

[gar]

120110-2358 EST

K8MHZ:

If you read the radar references I provided you will see why the lower frequencies were first used for long range radar.

You need large tubes for high transmit power, that makes electron transit distances greater, thus transit time becomes a problem, and sets an upper frequency limit. Note: they were shooting for 100 kW peak pulse power in the 100 mHz range.

The receiving tubes could be small, like the 955, etc. Also crystal diodes were used for the mixer, and the superhet oscillator was below the signal frequency. Somewhere I have a surplus front end with an acorn oscillator tube. But that is a much larger tube than the 955. I am not sure what the inter-electrode distances were.

.

 

[ActionDave]

As a guy who pulls wire in pipe may I ask, What is Maximum Power Transfer and why would it not be efficient?

 

[Joethemechanic]

Think of maximum power transfer as a fleet of juiced up dump trucks hauling electrons. Fuel pumps turned up, drivers drinking red bull, speeding, trucks overloaded, no tarps and electrons blowing off the load like sand.

Most effecient transfer would be a different fleet of dump trucks hauling electrons. Lower fuel settings, sensible careful drivers, well tarped loads, no electrons blowing off the loads like sand.


Both fleets even though they are very different have their own uses. If put into use on the right job. One is not superior to the other.


Suppose you have something like a nuclear meltdown and you need to cover the thing with sand in a hurry in order to save human lives. You want to have something like the first one.

Now suppose you want to supply a concrete plant to make concrete for a steadily expanding infrastructure. You want an operation like the second one.

All engineering is compromise. you have to give up the less important qualities of something in order to get the qualities that are more important to you.


An electric powered freight locomotive would be designed for max power transfer during starting in order to overcome the inertia of the loaded train. Large losses of energy are accepted in order to get the torque from the traction motors for acceleration.

But once the train is in motion and up to speed we want to have the power transferred at highest efficiency to keep energy costs down.

That probably isn't the best example, but it's the best I could come up with at the moment. (Darn cat is walking on me)

 

[gar]

120111-0821 EST

ActionDave:

In my post #1 I suggested a 100 V source with a 5 ohm internal resistance.

#14 wire has a resistance of 2.525 ohms per 1000 ft. Round this to 2.5 ohms. A 1000 ft long extension cord will have a loop resistance of 5 ohms.

I want to heat some water at the end of the extension cord. I will use my power company meter to determine how much energy I am using to do this heating operation. The power company meter is what determines my cost.

Next assume the water tank is perfectly insulated so there is no heat loss from the tank. Suppose for this tank of water it will take 10 kWh of energy to raise the temperature of the water a desired amount.

First, I will try a 50 ohm resistor for this task. The current will be 100 / 55 = 1.82 A. The power dissipated in the load is 1.82 * 1.82 * 50 = 165 W and into the meter end of the extension cord the power is 1.82 * 100 = 182 W. This is an efficiency of 165/182 = 91%, and it will take 10,000 / 165 =60.6 hours to heat the water.

Second, make the load resistance 5 ohms. The current is 10 A, and the load power is 500 W. Total input power is 1000 W so the efficiency is 50%. The time to heat the water is 20 hours. The water is heated 3 times faster, but at a great cost for energy.

Third, make the load resistance 1 ohm. Now current is 100 / 6 = 16.7 A. Load power is 278 W, and efficiency is 278 / 1670 = 16.7%. Heating time 36 hours. Very much higher cost and it takes longer than at maximum power transfer.

You can check my calculations.

.

 

[gar]

120111-1300 EST

See the following site for the story of the first radar echo detected from the moon.
http://www.campevans.org/_CE/html/qst-1946-05.html

To determine when Google has indexed a new page that I put on my web site I use some search words that should be somewhat unique. Often times some of the results are surprising. As a result I encountered an old report created on a Chrysler project on an improved ignition system that I would not expect to be in public viewable files. This references a high voltage scope probe I created for the project.

I was not looking for the moon thing it just popped up when I was sidetracked and decided to see how well QST might be indexed.

Also, Google is getting a lot more correlations on patent information.

.

 

[Joethemechanic]

Somewhere in the back of my mind I keep coming up with a correlation between standing wave ratios and power factor. But brought to the forefront of my thinking I seem to be having trouble building a conceptual model.

Am I just wasting my time running down a dead end road?

 

[gar]

120111-1552 EST

Joethemechanic:

Should be a correlation. When an antenna is not tuned to the radiated frequency, or there is an impedance mismatch, then some energy is reflected back from the load producing standing waves and less efficiency.

.

 

[Joethemechanic]

Yes forward power / reflected power thing. I think that is what is going on in the back of my mind. I need to go back and look at some old ARRL stuff and see what triggers a mental recall I guess. I was a bit of a radio geek back before I quit high school. My penmanship was lousy, and one day I just had enough of getting my grades bashed up over it.

 

[topgone]

120111-1552 EST

Joethemechanic:

Should be a correlation. When an antenna is not tuned to the radiated frequency, or there is an impedance mismatch, then some energy is reflected back from the load producing standing waves and less efficiency.

.

A little O/T, if I may be allowed:

Assuming a well-tuned antenna array and using an ELF or ULF, how strong will the transmission power be at 50 km mark on a beam-up direction? (HAARP) Is 3MW of transmission power maximum or what? (to break through about 500km of the earth's atmosphere)

Please disregard if you don't feel like it.

 

[gar]

120113-0946 EST

topgone:

If you have an isotropic radiator, the the surface area at a distance from the origin is that of a sphere, 4*Pi*r². Where r is the sphere's radius, or in this case the distance from the radiator (antenna). It does not matter whether it is sound, radio, light, or some other radiation the equation is the same.

Isotropic means radiating uniformly in all directions. The power density reduction is a result of the surface area of a sphere increasing as the square of the sphere's radius, or diameter.

At 50,000 meters the surface area is 12.57*25*10⁸ or 3.1416*10¹¹ square meters.
Thus, if 3*10⁶ watts is radiated, then the power density at 50,000 meters is
3/3.1416 * 10^-5 watts per square meter, or about 10 microwatts per square meter.

To the extent that you can beam the signal you can increase the power density.

At very low frequencies the atmosphere has little effect on the energy density. Virtually no absorbtion.

Check my math.

.

 

[K8MHZ]

A little O/T, if I may be allowed:

Assuming a well-tuned antenna array and using an ELF or ULF, how strong will the transmission power be at 50 km mark on a beam-up direction? (HAARP) Is 3MW of transmission power maximum or what? (to break through about 500km of the earth's atmosphere)

Please disregard if you don't feel like it.

HAARP is HF. I know a guy that used to work there. HIPAS is a similar research center.

The HAARP antenna array consists of 180 antennas on a total land area of about 35 acres. The array, along with its integrated transmitters, has a total radiated power capability of about 3,600 kilowatts.

http://www.haarp.alaska.edu/haarp/faq.html

HIPAS is bigger

The HIPAS (HIgh Power Auroral Stimulation) Observatory is an ionospheric heater, which can radiate 70 MW ERP at either 2.85 MHz or 4.53 MHz.

http://en.wikipedia.org/wiki/HIPAS_Observatory

Now, 70 million watts seems like a huge amount of power that would have a significant effect on the ionoshpere, but in reality it's like tossing a pebble into a river and watching the ripples from the pebble.

Cool stuff about the ionoshpere has been discovered at these to research centers, neither of which, despite prevalent lore, is run by the military or is used to control the weather or thoughts or anything else of the like.

 

[K8MHZ]

120113-0946 EST

topgone:

If you have an isotropic radiator, the the surface area at a distance from the origin is that of a sphere, 4*Pi*r². Where r is the sphere's radius, or in this case the distance from the radiator (antenna). It does not matter whether it is sound, radio, light, or some other radiation the equation is the same.

Isotropic means radiating uniformly in all directions. The power density reduction is a result of the surface area of a sphere increasing as the square of the sphere's radius, or diameter.

At 50,000 meters the surface area is 12.57*25*10⁸ or 3.1416*10¹¹ square meters.
Thus, if 3*10⁶ watts is radiated, then the power density at 50,000 meters is
3/3.1416 * 10^-5 watts per square meter, or about 10 microwatts per square meter.

To the extent that you can beam the signal you can increase the power density.

At very low frequencies the atmosphere has little effect on the energy density. Virtually no absorbtion.

Check my math.

.

An isotropic radiator does not exist in real life (except for the sun). The antennas at HIPAS and HAARP were designed to be directional and put as much signal as possible straight up. 180 antennas in an array covering 35 acres presents a significant amount of gain.

Your comment on absorption is in error. The ionosphere's layers all work differently and go back and forth between absorbing and refracting radio waves in the 3 - 30 mHz range and is greatly affected by the sun. For example, right now because of solar activity, 10 meters (28 mHz) can be used during the day to talk around the world with 100 watts or less. At night, the ionosphere absorbs 10 meters and it's not good for more than a few miles, nearly line of sight.

This web site has a chart that shows how each part of the HF bands are currently being affected by the ionosphere.

http://www.qrz.com/index.html

Notice it changes from band to band and from day to night as the ionosphere goes back and forth from absorption to refraction.

Gar,

Sorry, I just noticed your calcs were for 50 kilometers. That's a bit shy of the ionosphere. D layer starts at around 60 km and won't be an issue with your calcs.

There is path loss with any radio wave, but it's worse as we go higher in frequency. It does exist on HF, but we tend to dismiss it and focus on the ionosphere (pun intended) instead.

 

[topgone]

Y'all:
Thank you for the responses, gar and K8MHz.

gar,
I found an error in the calcs you presented using 4.pi.r² = 4 x pi x 50,000² = 3.1416 X 10¹⁰ m². The density at 50 kms using you spherical formula would then be = 3,600,000W/3.1416X10¹⁰ = 0.1146 mW/m².

If i go with K8MHz, that the double-dipole antenna array at HAARP is capable of beaming straight up (directional), with 35 acres = 141,638 m² and a transmitting power of 3,600,000 watts, my density at source will be 3,600,000 W/141,638 m² = 25.417 W per m²

Using the inverse-square law, I calculated that the power density @ 50km will just be about = 25.417 W/m² x (1/50,000)² = 1.44mW/m²!

This is higher than gar's calcs! (At 60 kms mark = 7.1 uW/m²)

Did I miss a lot here?

 

[K8MHZ]

Y'all:
Thank you for the responses, gar and K8MHz.

gar,
I found an error in the calcs you presented using 4.pi.r² = 4 x pi x 50,000² = 3.1416 X 10¹⁰ m². The density at 50 kms using you spherical formula would then be = 3,600,000W/3.1416X10¹⁰ = 0.1146 mW/m².

If i go with K8MHz, that the double-dipole antenna array at HAARP is capable of beaming straight up (directional), with 35 acres = 141,638 m² and a transmitting power of 3,600,000 watts, my density at source will be 3,600,000 W/141,638 m² = 25.417 W per m²

Using the inverse-square law, I calculated that the power density @ 50km will just be about = 25.417 W/m² x (1/50,000)² = 1.44mW/m²!

This is higher than gar's calcs! (At 60 kms mark = 7.1 uW/m²)

Did I miss a lot here?

From HAARP's web site:

The IRI would transmit radio waves over the frequency range 2.8 to 10 MHz. The transmitted radio wave beam would occupy a conical volume roughly 30 miles in diameter at an altitude of 300 miles. The transmitted radio waves would have up to 3.3 MW of power, only slightly higher than waves transmitted by radio and television stations.

Even if all the transmitted power from the IRI was absorbed by the ionosphere it would take more than 33,000 HAARP-scale IRIs, transmitting simultaneously to account for just 1 percent of the auroral ionosphere's energy budget. Another way of showing the vast difference between the amount of energy that would be dissipated in the atmosphere by the HAARP transmissions and natural processes is through a comparison of the local dissipation power in terms of power densities. The maximum power density of the IRI transmitted waves would be about 30 milliwatts per square meter (mW/m2) at 50 miles altitude decreasing to 1 mW/m2 at 186 miles altitude in the F region. In comparison, the densities of power dissipated by an aurora could exceed 2 W/m2, or roughly 2000 times greater then the expected maximum dissipation due to the absorption of the HAARP high frequency transmissions in the F region. Even the daily absorption of solar radiation easily exceeds the most intense, low altitude HAARP-ionduced energy deposition rate by a factor of ten.

So, using their antennas with 3.3 million watts, power density at 50 miles is 30 milliwatts per square meter.

You do know that HAARP is shut down and HIPAS is now doing the research, don't you? (They are bigger and badder, anyway!!)

Also, did you know that HAARP did a successful moon bounce in 2008 using their set up on 40 meters? If you heard them and sent a report, they would send you a QSL card. I listened and couldn't hear them.

http://k9zw.wordpress.com/2008/01/18/haarp-lwa-eme-bounce/

YouTube has videos of ham radio operators picking up the transmissions, which were in CW mode using Morse Code. Some of them could hear the transmitter in AK followed by a few second delay and then the weaker signal that bounced off the moon. It was a pretty big deal and got lots of press in the ham radio crowd.

 

[Joethemechanic]

I think you guys are starting to cross the threshold from what I have forgotten, and into what I never knew. Whilst I understand the concepts, I think I'm getting a little lost in the math.

Who would ever figure my mind would get challenged on an internet forum. :lol:

 

[K8MHZ]

Here are some pics of the DUGA-3 OTH radar antennas that tormented ham radio operators for years as Russia scanned the HF bands for missiles and the like from the US.

View attachment 6237

Think about the ERP of that monstrosity! And people lived in the shadow of it. The pics surfaced when Chernobyl blew up as the transmitter site is near the reactor.

 

[K8MHZ]

I think you guys are starting to cross the threshold from what I have forgotten, and into what I never knew. Whilst I understand the concepts, I think I'm getting a little lost in the math.

Who would ever figure my mind would get challenged on an internet forum. :lol:

Well, Gar should get the thanks and all the credit for the math. I just do this stuff as a hobby. I prefer measured data over math predictions, but my background is a tech and a troubleshooter, not an engineer.

By trade I'm just a licensed journeyman electrician.

I find this forum and QRZ to be VERY informative and challenging. I have been playing around with radio and electricity since I was a kid and have made it both my profession and hobby and both sites have something new to learn on a very regular basis.

 

[K8MHZ]

I gotta go for a bit, but think about ERP backwards for a second.

I used a 175 watt transmitter on SSB, which is pretty much a 40-80 percent signal (compared to an unmodulated AM carrier) due to the way it modulates the signal, connected to an antenna made entirely out of old coax I got off the top of our county building and made a contact to Japan with it. I had no amp, just the power from the back of the radio into the 'double bazooka' antenna strung between some trees in my back yard. The station was about 9,600 miles away. Think of how miniscule my signal must have been. To me, that is just as awesome as the big iron.

If you want to do the math, figure 155 watts into a 2dBi (2dB better than an isotropic radiator) gain antenna with a path of 9,600 miles. I would be interested in what that would work out to.

 

[gar]

120113-1633 EST

K8MHZ:

One needs a basic reference. Ultimately an isotropic radiator is a basic reference. Such a practical device does not exist for most radio measurements. A dipole can be defined relative to an isotropic radiator, and thus many measurements are referenced to a dipole.

See http://en.wikipedia.org/wiki/Antenna_gain

As you narrow the width of a radiated beam of energy for a given input power you will increase the power density, but unless you can perfectly collimate a beam, the energy density of the narrowed beam will diminish as the square of the radial distance.

When I was referring to the atmosphere having no effect I specified very low frequency radiation.

It is interesting that the results of the values given in the question produced a power density of 10 microwatts per sq-meter.

.