OT, Hydraulic accumulators

[ndzied1]

I Need more time

Ok Peter,

For the 1 cubic inch of oil at 1000 psi I came up with 0.284 Joules of energy, which of course, is potential energy.

I don't have time to play with the air right now but see why it would end up with more stored energy.

Am I close? (I used 200,000 psi for the bulk modulus of oil).

 

[TConnolly]

.28 J is the right answer. A flashlight uses several times more energy per second.

Now least anyone get careless, put that much energy per cu. in. in a medium sized hydraulic cylinder and get careless and it'll still hurt you bad - or maybe you'll never feel a thing.

 

[rsdoran]

This part is too advanced for me, just an old bubba ya know.

The thing to me is I can see a "potential" for more energy to be stored with oil i.e. greater pressures may be obtained but do not see how 1 cubic inch of 1000 psi static pressure could have a greater power capability whatever the fluid medium may be.

Energy is a product of force x distance in time? The bulk modulus is the resistance of compressibility of a fluid so y'all are saying that since air has a lower resistance it would be slower (in time) to release pressure then oil...all other factors being equal?

I said resistance but bulk modulus is more of an elasticity, which determines how much pressure may be applied and retain the energy properties being transfered. I am not sure I am saying this properly.

 

[ndzied1]

Here are two scenarios:

1. You push on a 1" square steel rod with 1000 pounds thereby generating 1000 psi in the bar.
2. You push on a 1 in ^2 air-spring with 1000 pounds thereby generating 1000 psi in the air-spring.

Now, you are told to put your body infront of each device and immediately "let go". Which one do you want to be standing in front of?

 

[kamenges]

This is oversimplifying the a great degree, but I think one way to look at it is stick with the 'Force x Distance' idea, considering it from the perspective of the cylinder rod end. For the purposes of our discussion you can assume the force profiles (change in force per percentage of stroke) will be similar (even they they aren't exactly). However, the distance covered in the air example is much greater than in the oil example. The added distance given a similar force profile is what increases the energy storage.

What if you set a 1000 lb weight on a solid block of steel versus a coil spring from a pick-up truck. Which do you think would store more energy?

Keith

 

[ndzied1]

Keith,

Here's another thought you spurred in my mind, and probably another simplificaiton....

The reason you can control position better with oil than air is because oil does not store energy as well as air. Rather it transmits it. Air wants to store the energy so there is this extra store and release thing going on that mucks up the position control.

A steel rod is a great transmitter of force, so much so that in engineering school it was called a "rigid body." You could consider any force you push on at one end to be transmitted immediately to the other end.

 

[TConnolly]

Here are two scenarios:

1. You push on a 1" square steel rod with 1000 pounds thereby generating 1000 psi in the bar.
2. You push on a 1 in ^2 air-spring with 1000 pounds thereby generating 1000 psi in the air-spring.

Now, you are told to put your body infront of each device and immediately "let go". Which one do you want to be standing in front of?

Good example. The larger the bulk modulus the less likely it is to act like a spring at a given load. The bulk modulus of the steel rod (>1.6x10^11Pa) is more than a million and a half times higher that air (~1.01x10^5Pa STP). The bulk modulus of oil is ~1.8x10^9Pa, roughly 13,000 times that of air, but the steel b.m. is only about a hundred times higher than the oil.

Think about it this way: You need to lift 1,000 lbs. You have a hydraulic bottle jack and you have a similar sized air cylinder connected to a bicycle pump. Which one is going to cause you to sweat more, pumping it up with the bottle jack or the bicycle pump? If you sweat more then its a fair bet you put more energy into it.

Just thinking about the extra work that makes me want to reach for a different kind of "accumulator" containing a chilled fluid and pressurized gas.

 

[Peter Nachtwey]

I reduced the bulk modulus to 100000 psi

The volume of oil that must be compressed go to 1.01005 cubic inches. Since it pushes twice as far to get to the same pressure the energy required doubles as expected. Just think how many cubic inches of air need to be compressed to get 1 cubic inch at 1000 psi.

I notice that force x distance = energy or work. This is true.
This is how force x distance = pressure x volume

Force * distance = work
Pressure * area * distance = work
Pressure * volume = work

Doing this in your head. ( RadCon Math for Dan Bentler and other ex nukes )

delta p = B * delta v / v or
delta v = delta p * v / b or 1000 * 1 / 200000 = 0.005
This is a rough estimate of how much the volume was compresed.
work = avg pressure x delta v or (1000/2) * 0.005
Combine the two equations
work = 0.5*((delta p)^2)*v/B this is simpler on a calculator but I prefer the two equations if I were doing this in my head.

Notice I ignored the log function and assume the pressure increase is linear. When one starts compressing the pressure is 0. When done the pressure is 1000. I divided by 2 to get the average pressure. Again I assume it is linear. My quick estimate yields 2.5 in*lbf. Close enough for making estimates when talking to customers over the phone about their applications. One can also see that the work gets bigger as the bulk modulus gets smaller.

 

[leitmotif]

OK Peter so here goes with off head radcon math.
gage pressure 0 to 1000
absolute 14.7 to 1014.7
compression ratio is 69.027

SO if I take a vertical piston 69.027 inches long with area of one square inch and place a 1000 lb weight on it my final volume is one cubic inch, pressure inside cylinder is 1000 gage and there is one inch of travel left.
W = 68.027 in x 1000 lbs
= 68027.2 inlbs
or 5668 ft lbs
ASSUMING I did this in one second
AND I did this with no temp change (difficult in real world)
1HP = 550 lbs one foot one second.
I used 10.3 HP
Am I right sure seems high to me.
I don't do this engineering stuff often enough to be all that confident. One thing for sure is that if I did this correctly it surely justifies my high respect for compressed gases.

Dan Bentler

 

[Peter Nachtwey]

OK Peter so here goes with off head radcon math.
gage pressure 0 to 1000
absolute 14.7 to 1014.7
compression ratio is 69.027

SO if I take a vertical piston 69.027 inches long with area of one square inch and place a 1000 lb weight on it my final volume is one cubic inch, pressure inside cylinder is 1000 gage and there is one inch of travel left.
W = 68.027 in x 1000 lbs
= 68027.2 inlbs
or 5668 ft lbs
ASSUMING I did this in one second
AND I did this with no temp change (difficult in real world)
1HP = 550 lbs one foot one second.
I used 10.3 HP
Am I right sure seems high to me.
I don't do this engineering stuff often enough to be all that confident. One thing for sure is that if I did this correctly it surely justifies my high respect for compressed gases.

Dan Bentler

Yes, that is too high. 0.284 joules in 1 second is .284 watts. One can easily generate that much power. I have showed how to calculate the energy stored already. Your
W = 68.027 in x 1000 lbs is an error. Study my .pdf file.

 

[Peter Nachtwey]

Oops, I screwed up.

Here is a correction. The answer is still the same but the formula only works because I used ONE cubic inch.
ftp://ftp.deltamotion.com/public/PDF/Mathcad%20-%20EnergyInOil.pdf

Some day I will learn not to up load files to this site where I can't change or correct them.