Pneumatic-electric motorcycle/bike.

Will'smotobikes19

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Front hub motor, battery and controller.
2 diving tanks rated 4500 psi and a 2hp piston type air motor.
Have one of those water cooled compressors on the rig with a retractable power cable. To charge the tanks stop at a gas station and plug it in. Tandem bike modified?
 
Front hub motor, battery and controller.
2 diving tanks rated 4500 psi and a 2hp piston type air motor.
Have one of those water cooled compressors on the rig with a retractable power cable. To charge the tanks stop at a gas station and plug it in. Tandem bike modified?
It takes a special expensive air compressor and equipment to fill up a scuba tank
A 120 psi gas station air pump ain't gonna fill those scuba tanks.
Also scuba tank should be put in water when ya fill it up or it can explode
 
willis PV = NRT...
The energy stored in a gas is proportional to its volume, pressure, and is really not much. A lot of energy but no power. Power is energy x time. A tank has a lot of energy, but no time.
I believe a scuba tank may store 5watt-hours of energy or around 17,000 watt seconds or 17 kilo joules. I could calculate the exact number using a different formula but i dont feel like it.

EDIT: Work = -P delta V
Assume a 20 liter tank at 2000psi for a scuba tank. I think those around what they are, IDK...
So 20 liter scuba tank x (136 atm - 1atm) (2000psi) x 101.3 Joules / Liter atm = 272 Kilo Joules = 0.075 kwh = 75 watt hours
I guess I was wrong. I thought scuba tank would be only 5 wh but its actually 75, my bad... But still 75wh is pretty low!!!!!!!!!!!!!!!!!
I guess 5wh would be closer to an aircompressor tank at 200psi, and I was equating an air compressor with a scuba tank, conceptualy, so my intuition was wrong....
Edit 2: Lol my brain was dead. Power is Energy / Time. Energy = Power x time. Got them mixed up. A scuba tank has a lot of power, but no energy... You can explode it in one second and release all 75 watt hours in one second, or 272,000 watts in one second which is going to blow up crap, but when that energy is extracted slowly over time, its really not much aka 75 watt hours.
 
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Could you run a water cooled 4500 psi compressor off batteries? for about 5 to 6 min? I think that would roughly compress 1000 psi. Also guys do you know where exhaust air exits one of these?

99914
I'm guessing one of the two threaded ports.

I have a folder with engineering ideas to increase torque and efficiency using a semi sealed impeller connected to the shaft and the exhaust port(s). The exhaust air would blow into the spinning fins it actually would look more like a water wheel than a blower impeller. I also thought 2 staggered rows of curved fins with flanges on both sides would produce more rotational force. A nozzle similar to a leaf blower would face rearward maybe making a tiny bit of thrust. The original design used a two stroke type design where exhaust air would blow out the port like a regular 2 stroke but with no transfer ports. I thought oiling and valves both pneumatic and mechanical to limit the amount of air in. A decompression valve on the upstroke could help with rpm and it would close right before TDC and air blast. In theory it would reduce amount of resistance since initial compression is not needed on an air motor. On a two stroke the gas and fuel would be compressed on the up stroke but I feel it would slow the rpm in my design hence the properly timed decompression cycle.

99915


110v 60HZ is the power required for the compressor I saw on amazon I don't know electrical too well probably need some components for an on board run.
 
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I am very sorry Willis. I used the wrong formula. The correct formula I should have used was
99917

I will recalculate it for you...
 
here is a good video for you willis

Assume 20 Liter sized tank at 2000psi.
136 atm = 2000psi,
When using other formulas, P1 * V1 = P2 * V2,
from 20 Liters x 136 atm = 2720 Liters x 1 atm

Now I can use the W = -P * Delta V. My mistake I made in the beggining was solving it like W =- delta P * V, which gave 45wh, my bad HAHA.
W = 1 atm x (2720 Liters - 20 Liters) x 101.3 Joules / Liter Atm = 272,000 Joules = 75 watt hours....

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
However some other formulas seem to get different results.

Assume a 20 liter scuba tank. = 0.02 meters cubed
Assume 3000psi pressure = 20.6 MPa


Work = 20.6 MPa * 0.02 cubic meters x ln(0.1 Mpa / 20.6 Mpa) + (20.6 Mpa - 0.1 MPa) x 0.02 cubic meters = -1.78 MJ = 0.494kwh = 494 watt hours.

Edit: Actually this formula assumes an Isothermal Process, and air compression is Adiabatic. I have no clue what that really means, but I think Isothermal only applies to constant temperature things like phase changes. For example, when boiling water into a gas, the temperature is constant at 212*, it does not increase; that extra heat energy is used in the phase change and expansion. I think the formula for abiadic is much more complex, but that would cut the 494 watt hours to around 213 watt hours. Which is way way better than the tiny 45 watt hours I told you.

Here is a good calculator for you Willis.
View attachment 99918

Keep everything as DEFAULT, dont mess with any of the values except Cylinder Volume V1, which is measured in liters, as 10^-3 m^3 = 1 Liter.
20MPA is 2900 psi and if you want to make it 4500psi, change it to 31 MPa.

Then copy and paste the KJ value into google and type in kilojoule to kwh calculator.... And 1000wh = 1kwh for your convienence.

The main thing Is is that these formulas seem to conflict with eachother.... I dont have the expertise to understand why I am getting very different answers with different formulas... But I will say Willis, if you want it to work, the sky is the limit man, go for it.
 
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here is a good video for you willis

Assume 20 Liter sized tank at 2000psi.
136 atm = 2000psi,
When using other formulas, P1 * V1 = P2 * V2,
from 20 Liters x 136 atm = 2720 Liters x 1 atm

Now I can use the W = -P * Delta V. My mistake I made in the beggining was solving it like W =- delta P * V, which gave 45wh, my bad HAHA.
W = 1 atm x (2720 Liters - 20 Liters) x 101.3 Joules / Liter Atm = 272,000 Joules = 75 watt hours....

//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
However some other formulas seem to get different results.

Assume a 20 liter scuba tank. = 0.02 meters cubed
Assume 3000psi pressure = 20.6 MPa


Work = 20.6 MPa * 0.02 cubic meters x ln(0.1 Mpa / 20.6 Mpa) + (20.6 Mpa - 0.1 MPa) x 0.02 cubic meters = -1.78 MJ = 0.494kwh = 494 watt hours.

Edit: Actually this formula assumes an Isothermal Process, and air compression is Adiabatic. I have no clue what that really means, but I think Isothermal only applies to constant temperature things like phase changes. For example, when boiling water into a gas, the temperature is constant at 212*, it does not increase; that extra heat energy is used in the phase change and expansion. I think the formula for abiadic is much more complex, but that would cut the 494 watt hours to around 213 watt hours. Which is way way better than the tiny 45 watt hours I told you.

Here is a good calculator for you Willis.
View attachment 99918

Keep everything as DEFAULT, dont mess with any of the values except Cylinder Volume V1, which is measured in liters, as 10^-3 m^3 = 1 Liter.
20MPA is 2900 psi and if you want to make it 4500psi, change it to 31 MPa.

Then copy and paste the KJ value into google and type in kilojoule to kwh calculator.... And 1000wh = 1kwh for your convienence.

The main thing Is is that these formulas seem to conflict with eachother.... I dont have the expertise to understand why I am getting very different answers with different formulas... But I will say Willis, if you want it to work, the sky is the limit man, go for it.

That video hits the nail on the head for the idea of air power.
 
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