PROOF for the air/electric hybrid!

[airhybrid]

Pardon me, I guess I forget to mention the 60 amps @ 14VDC. If that isn't enough for you, perhaps you would like to buy an 80 amp or 100 amp alternator. Or you could purchase a 200 amps more or less depending on type of alternator. How many solar panels would you need to equal that?

So you think that voltage and current are constant?

Go play with your alternator and see what kinds of current and voltages it has when it's not hooked up to anything and then when it's hooked up to various appliances that require 12 V. Or heck, even you system.

Let me try to help you and Eugene both at the same time. From the start I have said that you have to have fully charged batteries for this to work. Your car won't start with a dead battery either.

OK, the battery is like a storage tank that has ENERGY stored in a chemical form. On this system I have two batteries connected with an isolator to the alternator. The isolator allows me to pull that stored energy from both batteries to start and run my compressor. Although the compressor needs 15.5 amps @ 120 volts to start, this only takes a second and then only 7.4 amps is needed to keep my compressor running. This power is being drawn from both batteries via the isolator through the inverter.

Once the tank on my compressor has reached 115 pounds per square inch of stored ENERGY in the form of the compressed air in the tank, the compresor shut off due to the air pressure switch. Now I release the air in the tank through the tool . The tool starts spinning the belt and the alternator. As the alternator is spinning, the exciter post on the isolator that is connected to the field coil on the alternator, sends a small amount of energy from the batteries to the field coil and causes 60 amps @ 14 VDC to be supplied to both batteries.

Now if you had another inverter equal to the inverter connected to your compressor, you could connect 10-100 watt light bulbs to that inverter and keep them lit, because the stored energy in both batteries is being replaced by the spinning alternator through the isolator.

As the stored energy in the air tank drops from running the tool, the air pressure switch restarts the compressor and replaces it. So we have 3 different places to store the ENERGY being produced. 2 batteries and 1 air tank.

By the way Idahoev, if your alternator is not hooked up to a battery, your field coil in the alternator will not get it's charge and you won't get anything out of it.

[idahogev]

Let me try to help you and Eugene both at the same time. From the start I have said that you have to have fully charged batteries for this to work. Your car won't start with a dead battery either.

OK, the battery is like a storage tank that has ENERGY stored in a chemical form. On this system I have two batteries connected with an isolator to the alternator. The isolator allows me to pull that stored energy from both batteries to start and run my compressor. Although the compressor needs 15.5 amps @ 120 volts to start, this only takes a second and then only 7.4 amps is needed to keep my compressor running. This power is being drawn from both batteries via the isolator through the inverter.

Once the tank on my compressor has reached 115 pounds per square inch of stored ENERGY in the form of the compressed air in the tank, the compresor shut off due to the air pressure switch. Now I release the air in the tank through the tool . The tool starts spinning the belt and the alternator. As the alternator is spinning, the exciter post on the isolator that is connected to the field coil on the alternator, sends a small amount of energy from the batteries to the field coil and causes 60 amps @ 14 VDC to be supplied to both batteries.

Now if you had another inverter equal to the inverter connected to your compressor, you could connect 10-100 watt light bulbs to that inverter and keep them lit, because the stored energy in both batteries is being replaced by the spinning alternator through the isolator.

As the stored energy in the air tank drops from running the tool, the air pressure switch restarts the compressor and replaces it. So we have 3 different places to store the ENERGY being produced. 2 batteries and 1 air tank.

Good luck with that.

By the way Idahoev, if your alternator is not hooked up to a battery, your field coil in the alternator will not get it's charge and you won't get anything out of it.

Yes, that was my point. You'll have a voltage, but you will have no current. See the current depends on the load.

[airhybrid]

Let me try to help you and Eugene both at the same time. From the start I have said that you have to have fully charged batteries for this to work. Your car won't start with a dead battery either.

OK, the battery is like a storage tank that has ENERGY stored in a chemical form. On this system I have two batteries connected with an isolator to the alternator. The isolator allows me to pull that stored energy from both batteries to start and run my compressor. Although the compressor needs 15.5 amps @ 120 volts to start, this only takes a second and then only 7.4 amps is needed to keep my compressor running. This power is being drawn from both batteries via the isolator through the inverter.

Once the tank on my compressor has reached 115 pounds per square inch of stored ENERGY in the form of the compressed air in the tank, the compresor shut off due to the air pressure switch. Now I release the air in the tank through the tool . The tool starts spinning the belt and the alternator. As the alternator is spinning, the exciter post on the isolator that is connected to the field coil on the alternator, sends a small amount of energy from the batteries to the field coil and causes 60 amps @ 14 VDC to be supplied to both batteries.

Now if you had another inverter equal to the inverter connected to your compressor, you could connect 10-100 watt light bulbs to that inverter and keep them lit, because the stored energy in both batteries is being replaced by the spinning alternator through the isolator.

As the stored energy in the air tank drops from running the tool, the air pressure switch restarts the compressor and replaces it. So we have 3 different places to store the ENERGY being produced. 2 batteries and 1 air tank.

Good luck with that.

By the way Idahoev, if your alternator is not hooked up to a battery, your field coil in the alternator will not get it's charge and you won't get anything out of it.

Yes, that was my point. You'll have a voltage, but you will have no current. See the current depends on the load.

I guess the load of both inverters doesn't matter.

[airhybrid]

well,whats the ratings on the battery running the compressor and the rating on the compressor.
lets throw out a few numbers based on some typical items
Say your battery has a 100Ah capacity and the motor on the air compressor is 10A at 120V. That will take 100A at 12v minus a few more for waste in the inverter but we'll leave that out for the moment. Then say a 50% duty cycle on the compressor gives 2 hours from battery alone. Now add in the air tool and factor in its efficiency, the efficiency of the inverter, the efficiency of the alternator, etc. Since an inverter falls within power supply regulations if its above 80% efficient it gets the energy star label,and I haven't yet seen a normal store bought with the energy star so your looking at say 75% efficient, then loss everywhere else you'll probably get an overall 50% efficiency since your doing so many different conversions. So about 4 hours. Again this is with typical components and not yet adding in the second battery and loads on it.

According to the spring 2005 Real Goods resource guide and catalog, inverters lose at least 10% of the energy to give you that AC power. They also have an advertisement for a Xantrex SW series of inverters that claims a 95% efficiency.

I'll try this with an 80% efficency. That means that I will need to use a 1200 continuous watt/ 2500 watt surge capacity inverter to run my compressor. 80%= 960 watts cont./ 2000 watts surge.

My compressor requires 15.5 amps to start and 7.4 amps to run, or 1900 watts to start and 900 watts to run.
The output of this compressor is 5 cfm @ 90 psi. Compressed air is stored in this compressors tank @ 115 psi according to the attached gauge. Given the 80% efficency = an output of 4 cfm @ 90 psi.

My air cut-off tool requires 4 cfm @ 90 psi to spin a 4" disc @ 20,000 rpms. I have already reduced these rpms by half, by adding a bigger pulley and probably a little more by the pulley weighing more. A car alternator can provide full output at idle engine speeds of 1000 rpms, so any more losses here wouldn't matter.

My alternator is rated for 60 amps @14 VDC. Given the 80% efficency = 48 amps @ 14 VDC or 672 watts.

Isolators are sold according to the alternators they will be used with. Mine is rated for a 70 amp alternator so with the 80% =56 amps that it can handle.

Try using more accurate numbers.

[eugene]

At least your using a decent inverter, your average store bought one is much less efficient than an Xantrex. But Your missing a couple steps When you step voltage up or down the current goes in the inverse. So your 1900 watts to start and 900 watts to run works out to 135A at 14v,thats of so far, the battery will carry that starting load, but 64A at 14v to get your 900W, so your going to need a larger alternator to start with to just keep your system running and the battery topped off and account for other losses. Power in W = voltage x current so see how when one goes up the other goes down and power in = power out so if your drawing 900Wat 120v then your still pulling 900W at 12-14v on the in side (minus any losses).
now look up the equation to go from horsepower to watts (1 watt = 0.00134102209 horsepower not accounting for efficiency losses). so your 900W draw is 1.206919881HP. Now go look at your air tool and see what fractional hosrepower rating it has, they are usually like 1/3HP so first off your going to need a larger air motor to run things and therefore a larger compressor and therefore more current draw and therefore a larger alternator and so on.

[airhybrid]

At least your using a decent inverter, your average store bought one is much less efficient than an Xantrex. But Your missing a couple steps When you step voltage up or down the current goes in the inverse. So your 1900 watts to start and 900 watts to run works out to 135A at 14v,thats of so far, the battery will carry that starting load, but 64A at 14v to get your 900W, so your going to need a larger alternator to start with to just keep your system running and the battery topped off and account for other losses. Power in W = voltage x current so see how when one goes up the other goes down and power in = power out so if your drawing 900Wat 120v then your still pulling 900W at 12-14v on the in side (minus any losses).
now look up the equation to go from horsepower to watts (1 watt = 0.00134102209 horsepower not accounting for efficiency losses). so your 900W draw is 1.206919881HP. Now go look at your air tool and see what fractional hosrepower rating it has, they are usually like 1/3HP so first off your going to need a larger air motor to run things and therefore a larger compressor and therefore more current draw and therefore a larger alternator and so on.

What you are saying is that my air tool won't turn my alternator with enough torque to get the rated power out of it. I would agree except that the alternator is not directly connected to the air tool. These two are connected through a belt and pulleys. If the tool was mounted directly to the shaft of the alternator, you would be absolutly correct.
I explained a while back that I chose an 8" pulley for the tool because it reduced the speed by half and doubled the torque. If I find this insufficient, I still have 10,000 rpms to work with. Also getting a larger alternator is simple since they usually start around 60 amps. Thanks to those booming stereos in cars these days, they are being produced up to 200 amps and more.

[eugene]

You can calculate that too. If you have a 4" pulley on the air tool and 8" on the alternator then you half the amount of torque the alternator is requiring. But remember alternators also don't produce their full power unless at full rpm, the 60A rating is its max. A typical car engine will max out around 5000-6000 rpm and you have a large crankshaft pulley turning a smaller alternator pulley so your alternator may be running 10,000rpm at its max output so you may not reach max output is you reverse the pulley sizes, say your air tool runs at 10k and your pulleys are reducing the alternator speed to 5k then you may only get 75% of its max output. But remember you may need 2 HP at the alternator from a 1/4 HP air tool which means an 8:1 ratio in pulley sizes which now drops your alternator rpm down to near idle speed where it may only produce 25% of its rated output. Just because you can vary pulley sizes doesn't mean other factors don't vary as well, you still can't get something from nothing.

[airhybrid]

You can calculate that too. If you have a 4" pulley on the air tool and 8" on the alternator then you half the amount of torque the alternator is requiring. But remember alternators also don't produce their full power unless at full rpm, the 60A rating is its max. A typical car engine will max out around 5000-6000 rpm and you have a large crankshaft pulley turning a smaller alternator pulley so your alternator may be running 10,000rpm at its max output so you may not reach max output is you reverse the pulley sizes, say your air tool runs at 10k and your pulleys are reducing the alternator speed to 5k then you may only get 75% of its max output. But remember you may need 2 HP at the alternator from a 1/4 HP air tool which means an 8:1 ratio in pulley sizes which now drops your alternator rpm down to near idle speed where it may only produce 25% of its rated output. Just because you can vary pulley sizes doesn't mean other factors don't vary as well, you still can't get something from nothing.

The air tool is rated at 20,000 rpms to begin with. I said I would mount an 8" pulley in place of the 4" disc. This would reduce my speed by half, which is 10,000 rpms, but double my torque. Alternators can produce full output at 1000 rpms engine speed so that your car can idle at red lights and still keep the electrical system going.

Your right that you can't get something for nothing, so I wouldn't suggest trying this where there is no air.

[eugene]

Alternators can produce full output at 1000 rpms engine speed so that your car can idle at red lights and still keep the electrical system going.

This isn't so. Alternators produce full output at full rpm, at low rpm they produce less. Some cars with a lot of electrical load you can watch the voltage droop at rpm as the alternator can't keep up with the load and the battery has to provide. You'll see the car audio guys with capacitors and multiple batteries for those low rpm times.

[WiredForStereo]

airhybrid

The real difficulty here is trying to show you that your air tool WILL NOT turn your alternator and produce the rated output of the alternator. You cannot use a 1/4 hp air tool and make 2 hp (alternator output) with it no matter the gearing. The best you will get is about 1/8 hp output (in electricity.) If 1/4 horsepower goes in, less than 1/4 horsepower will come out.

It's like turning a shaft at 10 rpms on one end and expecting the other end to be turning at 80 rpms. I'm not talking about gearing here, I'm talking about energy. Gearing does not matter.

If you replace your 20k disk with something twice as big, you'll get the twice the rpms (on the receiving end,) just half the torque. Half the torque at twice the rpms is EQUAL power, but twice as useless, because there's no torque. If you try hard enough to cut something, you'll be able to stop the air tool. Your alternator will be able to provide 8 times as much stopping power as it takes you to stop the wheel. The air tool doesn't have a chance.

[airhybrid]

airhybrid

The real difficulty here is trying to show you that your air tool WILL NOT turn your alternator and produce the rated output of the alternator. You cannot use a 1/4 hp air tool and make 2 hp (alternator output) with it no matter the gearing. The best you will get is about 1/8 hp output (in electricity.) If 1/4 horsepower goes in, less than 1/4 horsepower will come out.

It's like turning a shaft at 10 rpms on one end and expecting the other end to be turning at 80 rpms. I'm not talking about gearing here, I'm talking about energy. Gearing does not matter.

If you replace your 20k disk with something twice as big, you'll get the twice the rpms (on the receiving end,) just half the torque. Half the torque at twice the rpms is EQUAL power, but twice as useless, because there's no torque. If you try hard enough to cut something, you'll be able to stop the air tool. Your alternator will be able to provide 8 times as much stopping power as it takes you to stop the wheel. The air tool doesn't have a chance.

The air tool has a 1/2 HP air motor. Since 745.7 watts = 1 HP, I should be able to get 372.85 watts from the alternator. Placing a larger pulley on the end of the air tool and connecting it with a belt to the alternator will provide more torque to the alternator pulley. Are all the gears on a 10 speed bicycle the same size?
Using a larger diameter pulley, is like placing a longer piece of pipe over a breaker bar. More torque.
I'm not connecting the tool directly to the front of the alternator.

[airhybrid]

Alternators can produce full output at 1000 rpms engine speed so that your car can idle at red lights and still keep the electrical system going.

This isn't so. Alternators produce full output at full rpm, at low rpm they produce less. Some cars with a lot of electrical load you can watch the voltage droop at rpm as the alternator can't keep up with the load and the battery has to provide. You'll see the car audio guys with capacitors and multiple batteries for those low rpm times.

Your right, full output was a misprint. Other vehicles with large electrical loads, use hi-output alternators.
Mean Green Alternators produce alts with a maximum of 2800 watts to keep systems going without extra batteries etc...

[eugene]

Remember though those high output alternators don't produce full power at low speeds, in fact some of the higher output ones put out the same or less at idle speed and don't get to producing power until they are spinning pretty fast. And those take more power to turn, so say you buy the huge alternator and it takes 4 hp at 4000rpm and your taking your 1/4hp at 20,000 rpm your going to need to gear it way down to get the power needed but ruin into the issue again that it now spinning the alternator too slow.

[WiredForStereo]

No, you've got it backwards. If you put a larger pulley as the driver, the driven shaft receives less torque, not more. I think you may not understand the concept of exactly what torque is.

When you switch gears on a bicycle to go faster, say downhill, you are providing less torque to the back wheel, but more speed. When you shift gears to go up a really steep hill, you are providing more torque at less speed. What counts is what force is being applied to the receiving shaft. If you have a small sprocket or pulley on the receiving shaft, you are applying less torque at a greater speed. If you have a large sprocket, you apply a large amount of torque to the receiving shaft but at low speed.

However, at no point do you change the amount of power (horsepower, watts) being transferred through your system. Which leads me to my next point which is that your understanding of efficiency is also flawed.

You have a half horsepower motor, therefore lets just say 375 watts. A car alternator is about 62% max efficiency (according to Wikipedia, see "alternator"), so we'll use that figure. That means using your 1/2 horsepower air tool hooked to an alternator, you'll produce maximum of about 233 watts. Your inverter may produce AC power at about 95% efficiency which leaves you making 221 watts to your compressor. You said it takes about 880 watts to run your compressor continuously, which your air tool hooked to an alternator charging a battery powering an inverter cannot provide. I neglected the losses through the battery because I assumed there wouldn't be any charging going on since the whole system is being drained to feed the air compressor.

My previous point still stands. Your alternator would be trying to produce power to replace the drain on the battery and would likely bring your air tool almost to a complete stop. 60 amps at 14 vdc equals 840 watts which is not even enough to run the compressor in the first place. Because of the efficiency of an alternator, the power you'd have to input into the alternator to get 840 watts out is about 1354 watts which is 1.8 horsepower, which again, your air tool cannot produce.

You see, it's all about power. It's about energy. The throughput of the system has little to do with rpms, torque, or cfms. If you were to increase the power of the air tool, you'd have to increase the power of the compressor, and the power feeding the compressor and it just goes in circles. The whole system is a complex battery drainer. That's all it will do for you.

I'm really sorry, but the physics just won't work for this one.

[airhybrid]

No, you've got it backwards. If you put a larger pulley as the driver, the driven shaft receives less torque, not more. I think you may not understand the concept of exactly what torque is.

When you switch gears on a bicycle to go faster, say downhill, you are providing less torque to the back wheel, but more speed. When you shift gears to go up a really steep hill, you are providing more torque at less speed. What counts is what force is being applied to the receiving shaft. If you have a small sprocket or pulley on the receiving shaft, you are applying less torque at a greater speed. If you have a large sprocket, you apply a large amount of torque to the receiving shaft but at low speed.

However, at no point do you change the amount of power (horsepower, watts) being transferred through your system. Which leads me to my next point which is that your understanding of efficiency is also flawed.

You have a half horsepower motor, therefore lets just say 375 watts. A car alternator is about 62% max efficiency (according to Wikipedia, see "alternator"), so we'll use that figure. That means using your 1/2 horsepower air tool hooked to an alternator, you'll produce maximum of about 233 watts. Your inverter may produce AC power at about 95% efficiency which leaves you making 221 watts to your compressor. You said it takes about 880 watts to run your compressor continuously, which your air tool hooked to an alternator charging a battery powering an inverter cannot provide. I neglected the losses through the battery because I assumed there wouldn't be any charging going on since the whole system is being drained to feed the air compressor.

My previous point still stands. Your alternator would be trying to produce power to replace the drain on the battery and would likely bring your air tool almost to a complete stop. 60 amps at 14 vdc equals 840 watts which is not even enough to run the compressor in the first place. Because of the efficiency of an alternator, the power you'd have to input into the alternator to get 840 watts out is about 1354 watts which is 1.8 horsepower, which again, your air tool cannot produce.

You see, it's all about power. It's about energy. The throughput of the system has little to do with rpms, torque, or cfms. If you were to increase the power of the air tool, you'd have to increase the power of the compressor, and the power feeding the compressor and it just goes in circles. The whole system is a complex battery drainer. That's all it will do for you.

I'm really sorry, but the physics just won't work for this one.

OK, your right about the pulleys. The larger one would have to be on the alternator. It's like removing the knob from an old car stereo. The post was very hard to turn without it.

I believe you are confusing some other things though. First I wouldn't consider wikipedia an authority on anything. Do you really think car manufacturers would settle for 62% efficient.

The 7.4 amps being used by the compressor is supplied from the battery @ 14 VDC, but the inverter raises this to 120 VAC. So the 888 watts AC required, is coming through the inverter from the 103.6 watts DC being drawn from the battery. The alternator raises the batteries voltage from 12 VDC to 14 VDC, which also raises the 12 VDC required for the inverter to 14 VDC. Now I have to be able to replace this 103.6 watts or 7.4 amps @ 14 VDC. Since my 1/2 hp air tool is capable of producing 372.85 watts, I think that will cover the efficency problem. Even at 62% thats still 233 watts to cover the 103.6 watts.

Also remember that the compressor will shut off once the pressure in the tank reaches 115 psi. The air tool will continue to spin the alternator, but the compressor won't restart untill the pressure in the tank falls to usually about 80 psi. So with the compressor not running the battery is still being charged.