How to calculate how much energy a pneumatic system would need?

[fulano]

TL;DR Summary

I found the culprit of my calculations: the calculator itself. I tested others and it gave way more reallistic results.

Well, I was trying to predict how much energy a pneumatic system would need in order to actuate in a specific force and speed, but it seems that I would need too little energy for too much work.

So, I was wondering how I would go on about making a pneumatic system with artificial muscles, they would essentially work like an horizontal pneumatic cylinder pushing against two ropes, simulating a contraction, but I don't know if I'm correct.

Source of image: "A Novel Soft Pneumatic Artificial Muscle with High-Contraction Ratio"

Using an online Hydraulic Cylinder Calculator:
600 lpm for a 50cm² area cylinder with 80 mm of diameter and 300mm of stroke.
at 0.1 bar = 5kg of force
3000 kg target / 5 kg = 600
600 LPM x 600 actuators = 360,000
360,000 x 82 actuators = 29,520,000 LPM

I found a data sheet of an 80cm wide industrial ducted fan that produces 400,000 liters per minute of airflow and 560 pascals of pressure at 3.3 kilowatts.

29,520,000 lpm / 400,000 lpm = 73.8
73.8 x 3.3 kilowatts = 243.54 kilowatts = 324.72 horsepower.
Since 0.1 bar = 10,000 pascals
10,000 / 560 = 17.8571428571 x 324.72 = 5,798.57 hp in total

If you take the fact that ropes suffer 5 times more tensil strength around 170º degrees of angle, then you would need 5 times more energy:
5,798.57 x 5 = 28,992.8


But if I use a torque and rpm calculator, where these fluidic muscles would be attached to, they would basically output around 10000 nm of torque with almost 300 rpm of speed.
And these would result in a 300 kilowatt motor.
300 kw per actuator x 82 actuators in total = 24,600 kilowatts = 32,800 horsepower.

Obviously, both values would be different anyway due to differences and inefficiencies, but the pneumatic one should at least be slightly higher than the second equation; An actuator that requires less horsepower to rotate a mechanical arm with 400 horsepower seems to break the laws of physics to me, but I don't know what I did wrong...
Maybe this is due to differences in the laws of physics being used? One is about fluid physics, for example...
But… Even if you don’t have the cable into consideration. It would still be just 5,798 horsepower…

I put these values to a kinetic energy calculator and it calculated it would output around 6000 joules of energy.
And since joule second = watt second...

 

[Baluncore]

So, I was wondering how I would go on about making a pneumatic system with artificial muscles, they would essentially work like an horizontal pneumatic cylinder pushing against two ropes, simulating a contraction, ...

We need to know the fundamental dimensions and forces required by your hypothetical actuator. Without a stable definition of the problem, all your numbers are meaningless.
What tension force is required, over what length of movement, in how much time ?

Forces are measured in newtons, N, and pressure in pascals, Pa. Energy is in joules, J, and watts, W, as joules per second.

Hydraulic systems usually operate with a constant fluid flow, with pressure rising only when flow is obstructed by a load.
Fluid pressure in Pa * flow in m³ = power in watts.

Pneumatic systems operate by compressing air, which gets hot and then cools in a pressurised reservoir. High pressure is used to move a volume of air quickly, the air applies pressure, but then expands, and is released to the atmosphere. That is all very inefficient.

When it comes to producing slow-speeds, with high-forces, pneumatics is inefficient compared to hydraulics. Use pneumatics only when you want high-speeds with low-forces.

 

[fulano]

We need to know the fundamental dimensions and forces required by your hypothetical actuator. Without a stable definition of the problem, all your numbers are meaningless.
What tension force is required, over what length of movement, in how much time ?

I was calculating to find out first how much force it would produce, and then work with it to either increase or decrease.
But I was thinking of 1/5 of a second 0.2 contraction with a constant tension force with at least 10 cm of linear movement.

Forces are measured in newtons, N, and pressure in pascals, Pa. Energy is in joules, J, and watts, W, as joules per second.

Hydraulic systems usually operate with a constant fluid flow, with pressure rising only when flow is obstructed by a load.
Fluid pressure in Pa * flow in m³ = power in watts.

Pneumatic systems operate by compressing air, which gets hot and then cools in a pressurised reservoir. High pressure is used to move a volume of air quickly, the air applies pressure, but then expands, and is released to the atmosphere. That is all very inefficient.

When it comes to producing slow-speeds, with high-forces, pneumatics is inefficient compared to hydraulics. Use pneumatics only when you want high-speeds with low-forces.

I was calculating, let's say, that you had an X amount of air in a pressure vessel that opened up and reached equilibrium between the actuator and the reservoir tank.
The compressor, however, was supposed to be a diaphragm piston, since these are considered to be extremely efficient. On top of that, the heat generated through compression would pass through a heat exchanger that would flow air from the reservoir, keeping the system efficient.

Right now I'm not focused on how to control contraction and what not, I'm more concerned with the amount of energy required to supply it first.

 

[Baluncore]

Right now I'm not focused on how to control contraction and what not, I'm more concerned with the amount of energy required to supply it first.

You are going about it backwards. How movement is achieved, will decide efficiency. Any inefficiency will dominate the energy requirement.

But I was thinking of 1/5 of a second 0.2 contraction with a constant tension force with at least 10 cm of linear movement.

One way movement, in 1/5 of a second, I can understand.
"0.2 contraction", needs more explanation.
A "constant tension force" will need a number and a unit.
10 cm of movement, I understand.

 

[fulano]

You are going about it backwards. How movement is achieved, will decide efficiency. Any inefficiency will dominate the energy requirement.

Well, I don't know how to go correctly, I was going trial and error. But no matter what I tried, I couldn't find the correct values on anything.
I was trying to reach a bundle of filament artificial muscles with 3000 kilograms of output force with at least 5 to 10 centimeters of contraction length.

One way movement, in 1/5 of a second, I can understand.
"0.2 contraction", needs more explanation.
A "constant tension force" will need a number and a unit.
10 cm of movement, I understand.

I meant "contraction in 0.2 seconds", which is 1/5 of a second. Sometimes I think weird.

 

[Baluncore]

Sometimes I think weird.

That is why we ask follow-up questions.

I was trying to reach a bundle of filament artificial muscles with 3000 kilograms of output force ...

I don't think that could be implemented as a pneumatic system. As a single actuator, that would certainly need to be implemented as a hydraulic system. It would be difficult to implement 3000 actuators, each with a 1 kg force.
Are you describing a single actuator, or a bundle of independent actuators?

Structural vertebrate skeletons have fixed length tubular bones under compression, opposing tension from muscles and tendons, plus a few ligaments to restrain joint alignment. There is an inversion of the structure, where the bones effectively do the work. Such a structure, implemented with hydraulics, might replace the muscles and tendons with fixed length flat straps, and the bones with hydraulic cylinders, which are twice as forceful in extension as retraction, and can, at most, almost double their length.

 

[fulano]

That is why we ask follow-up questions.


I don't think that could be implemented as a pneumatic system. As a single actuator, that would certainly need to be implemented as a hydraulic system. It would be difficult to implement 3000 actuators, each with a 1 kg force.
Are you describing a single actuator, or a bundle of independent actuators?

Semi-independent actuators, Kinda like filament mckibben muscles.

Source: "Musculoskeletal lower-limb robot driven by multifilament muscles"

I tried to simplify the problem by calculating the muscles as horizontal cylinders and then making something like in the pic. making these big things into thin filaments.

Structural vertebrate skeletons have fixed length tubular bones under compression, opposing tension from muscles and tendons, plus a few ligaments to restrain joint alignment. There is an inversion of the structure, where the bones effectively do the work. Such a structure, implemented with hydraulics, might replace the muscles and tendons with fixed length flat straps, and the bones with hydraulic cylinders, which are twice as forceful in extension as retraction, and can, at most, almost double their length.

Yes, but with hydraulics you need to carry the working fluid with you. The valves, cylinders and pipes need to be extremely strong, which in turn makes them just as heavy.
In Either way, I'm trying to figure out what I can do with pneumatics, if it ends up being impractical I will go for the next option. Then repeat the whole process with another type of actuator and so on.

 

[Baluncore]

I tried to simplify the problem by calculating the muscles as horizontal cylinders and then making something like in the pic.

Considering the pneumatic cylinder, between two tensile filaments, as a structure that would need to be repeated many times in each fibre, at a minute scale, to get a magnified reduction in length. That assembly repetition needs to be simplified.

Now, if the ersatz muscle fibres, were soft silicone tubes, counter-wrap reinforced at between ±30° and ±45° with tensile filaments, then internal pressure would cause a shortening of the tube. Note that the hoop stress in a tube, due to internal pressure, is twice the extension stress between the ends, so hoop stress wins, and the filament restrained tube will become shorter, with a constant surface area.

 

[erobz]

I'm not thinking deeply about the application, but I think if you are looking to see how this might perform you could try analysis of something like this with linear springs:

 

[fulano]

I'm not thinking deeply about the application, but I think if you are looking to see how this might perform you could try analysis of something like this with linear springs:

View attachment 357396

My issue is not the structure per se, it is how much energy I need to input into the system and how much energy it actuates with.

 

[erobz]

My issue is not the structure per se, it is how much energy I need to input into the system and how much energy it actuates with.

Well, try with rigid bars to form the geometry first. You should be able to relate the desired output velocity of the bottom where $F_{out}$ is applied to the rate of extension of the pneumatic cylinder. All those forces and velocities determine the input power via the geometry of that mechanism

For instance, you could want to pull something vertically at constant velocity, the weight of that object times its velocity you desire is the desired output power (an underestimate if you account for accelerations to get to $v$). It could end up being a bit of analysis, but you should be able to work your way through the linkage and relate that velocity to how fast the cylinder would need to extend. The forces needed to do that are a function of that.

You have an arm and its geometry to consider also, So its probably a bit challenging to link it all up, but you could just focus on the scissor jack mechanism for starters to get a handle on it.

You work with Pyhagorean, implicit differentiate, then integration to find $x(t)$ depending on what you demand for $\dot h$ (the velocity of the weight you are lifting). If $\dot h$ is a constant, then treat it as a constant. If you want it to be a function of time $\dot h(t)$ then treat it as such in the ensuing math. Keep your eyes on the prize; ultimately you want to find the velocity $\dot x (t)$ as a function of time, given you want the weight you are lifting has some desired $\dot h(t)$. That is what leads you in the direction of the power $P(t)$ to complete such a task. What the task is needs to be defined before you can say how much power you need from the cylinder over the time of completing the task.

 

[jrmichler]

Here's how I solve similar pneumatic actuator problems. Each artificial muscle is a pneumatic actuator. A pneumatic actuator has a retracted internal volume and an extended internal volume. When retracted, the empty volume is at atmospheric pressure. When extended, the extended volume is at the pressure of the compressed air supply. The volume of compressed air that flowed into the extended actuator is equal to the extended volume.

Calculate the amount (ACF) of air to fully extend each actuator separately. Convert to standard cubic feet. Multiply by the number of cycles per minute. The result is the total air requirement for the system. Now that you know the system air pressure and air flow requirement, you can choose an air compressor from a catalog. That catalog will specify the power required.

Be careful. Air compressors are rated in SCFM, while the volume of air into an actuator is calculated in ACF. SCFM is Standard Cubic Feet Per Minute. A standard cubic foot is a cubic foot of air at STP - Standard Temperature and Pressure. ACF is Actual Cubic Feet at the system pressure. An actual cubic foot is the actual volume of the air. For example, a one cubic foot pressure vessel at 147 PSIA (1.01 MPa) will contain ten standard cubic feet of air.

Actuators provide a known force from a known air pressure. A properly designed system has each actuator sized to provide the desired force at the design air pressure. Most pneumatic actuators are designed for air pressure about 100 PSI (0.7 MPa). I don't know about your pneumatic artificial muscles, you will have to find that information. The 560 Pa blower mentioned in the OP is only 0.00056 MPa, so is not relevant to a system with pneumatic actuators.

 

[erobz]

@fulano When I get a "wow", I feel like I might have failed to explain myself well enough. If any of what I said about that didn't land, please ask. It's hard to tell how people are used to solving problems.

It sounds like you want the object being lifted to cover some distance under constant acceleration (you mentioned constant tension). If its like a bicep curl there probably isn't a 1:1 relationship between distance traveled by the object being lifted and the contraction of that linkage, that needs to be considered in an analysis. I'm just trying to be blunt about what you do to find a Power for a particular task. If, its too costly a time investment to do the calculus, I understand.

 

[Lnewqban]

View attachment 357387
If you take the fact that ropes suffer 5 times more tensile strength around 170º degrees of angle, then you would need 5 times more energy:

Not really.
Energy or generated work remains about the same during the stroke.
As the cylinder expands and muscle contracts, the mechanical advantage of each dramatically changes.

Think of how the input force and output velocity of movement change when you use an automobile scissor jack to change a flat tire.

 

[fulano]

@fulano When I get a "wow", I feel like I might have failed to explain myself well enough. If any of what I said about that didn't land, please ask. It's hard to tell how people are used to solving problems.

It sounds like you want the object being lifted to cover some distance under constant acceleration (you mentioned constant tension). If its like a bicep curl there probably isn't a 1:1 relationship between distance traveled by the object being lifted and the contraction of that linkage, that needs to be considered in an analysis. I'm just trying to be blunt about what you do to find a Power for a particular task. If, its too costly a time investment to do the calculus, I understand.

I tried again:
When asking around I still didn’t get the answer, but someone suggested taking the volume of the initial state and subtracting it by the volume of the compressed state.
On top of that, they also suggested doubling that value since in pneumatics you can only apply a constant force if the actuator is fully actuated. From the initial flow of air to the final state, the air works like a spring being released.
So, again, assuming 5mm of width, 5mm of length and 50mm of height when expanded, 5mm when contracted I would have:

• Actuated: 1250mm³
  Rectracted: 250mm³
  Volume Calculator
• 1250mm³ - 250mm³ = 1000mm³
  1000mm³ = 0.001 liters
• 0.001 liters x 1.1 bars (if I multiply by 0.1 bars it will decrease volume even though you are adding 0.1) = 0.0011 liters
• So, converting this to a pneumatic cylinder by taking the area of 25mm² and taking the diameter of what that would be: 5.642mm of diameter.
  Area Calculator
• Using the correct cylinder calculators: 0.25 newtons
  Piston Cylinder Pressure and Diameter to Force Calculator
  Pneumatic Cylinder Force Calculator
• So, assuming parallel actuators in a 1 meter long filament fiber:
  5mm/1000mm = 200 actuators
• 200 actuators x 0.25 newtons = 50 newtons = 5 kilograms
• 200 actuators x 0.0011 liters = 0.22 liters to fully actuate all 200 actuators.
• Like I said before, you need to take into equilibrium the pressure vessel and the actuators.
  IF you have 2 chambers, one with high pressure and another with lower pressure and open a valve connecting both, you will get an equilibrium.
  So, 0.22 liters x 2 = 0.44
• 3000 kilograms of actuation force / 5 kilograms of actuation force by the actuators = 600 parallel filaments
• 600 parallel filaments x 0.44 liters = 2640 liters
• Even though the actuators can lift all of this weight, they are in parallel, thus, you need to increase their length to compensate for the lack of actuation length/stroke.
  Assuming that the actuation is around half of the length of the pneumatic cylinder, around 25mm, I would need:
  100mm of travel / 25mm = 4
  200mm of travel / 25mm = 8
• Thus, 2640 x 8 = 21,120 liters per minute.
• fan with 400,000 liters per minute with 560 pascals and 3.3 kilowatts
  10,000 pascals / 560 pascals = 17.8571428571
  3.3 kw x 17.8571428571 = 58.928 kilowatts
• Since pneumatics needs to stay in equilibrium, I need to multiply it by 2
  58.928 x 2 = 117,856
• 400,000 / 21,120 liters per minute = 18.9393939394 times less energy
• 117,856 watts / 18.9393939394 times less energy = 6,222.7 watts

• If I apply this muscle with 3000kg of actuation strength to a lever with the load arm 3 times longer than the effort arm, it will lift something with a force of 1000kg.
  Given that a mechanical arm moving at 300 RPM and 10,000 newton meters, it would require 300 kilowatts.
• But the expansion of the actuator would make that super fast, faster than 300 rpm.
• Wouldn’t that mean that I’m breaking the laws of physics?

 

[jrmichler]

Wouldn’t that mean that I’m breaking the laws of physics?

No. It means that there is at least one error in your assumptions or your calculations. It's a standard error checking technique. It is also known as a sanity check.

 

[Baluncore]

Wouldn’t that mean that I’m breaking the laws of physics?

You cannot break the laws of physics.

The system you are considering, is sufficiently unreal that its analysis is meaningless. It is also so complex, that you cannot understand and analyse it correctly. The mechanism used to convert compression to tension introduces an unnecessary reciprocal function into the mechanism.

Yes, but with hydraulics you need to carry the working fluid with you. The valves, cylinders and pipes need to be extremely strong, which in turn makes them just as heavy.

You have insufficient experience with hydraulics to understand why it is so much more efficient than pneumatics for this application. You are not recognising the impracticality of assembling so many components in parallel. The required strength of the hydraulic components is low, since pressures only need to be 200 psi, not 2000 psi. Hydraulic systems carry their working fluid with them, stored in their components. The fluid can be water based.

 

[fulano]

No. It means that there is at least one error in your assumptions or your calculations. It's a standard error checking technique. It is also known as a sanity check.

Probably because I considered the continuous flow or air and not the instantaneous flow of pressurized gas being released.
It would work like a spring, so just a few thousand joules would be used to compress it and it would release all of these joules at same time, while the wattage of rpm and torque assumes continuous operation of a motor/engine.

 

[fulano]

You cannot break the laws of physics.

The system you are considering, is sufficiently unreal that its analysis is meaningless. It is also so complex, that you cannot understand and analyse it correctly. The mechanism used to convert compression to tension introduces an unnecessary reciprocal function into the mechanism.

The calculation that I've made showed such little power to move things probably because I considered the continuous flow or air and not the instantaneous flow of pressurized gas being released.
It would work like a spring, so just a few thousand joules would be used to compress it and it would release all of these joules at same time, while the wattage of rpm and torque assumes continuous operation of a motor/engine.

You have insufficient experience with hydraulics to understand why it is so much more efficient than pneumatics for this application. You are not recognising the impracticality of assembling so many components in parallel. The required strength of the hydraulic components is low, since pressures only need to be 200 psi, not 2000 psi. Hydraulic systems carry their working fluid with them, stored in their components. The fluid can be water based.

I see...

 

[erobz]

I tried again:
When asking around I still didn’t get the answer, but someone suggested taking the volume of the initial state and subtracting it by the volume of the compressed state.
On top of that, they also suggested doubling that value since in pneumatics you can only apply a constant force if the actuator is fully actuated. From the initial flow of air to the final state, the air works like a spring being released.
So, again, assuming 5mm of width, 5mm of length and 50mm of height when expanded, 5mm when contracted I would have:

• Actuated: 1250mm³
  Rectracted: 250mm³
  Volume Calculator
• 1250mm³ - 250mm³ = 1000mm³
  1000mm³ = 0.001 liters
• 0.001 liters x 1.1 bars (if I multiply by 0.1 bars it will decrease volume even though you are adding 0.1) = 0.0011 liters
• So, converting this to a pneumatic cylinder by taking the area of 25mm² and taking the diameter of what that would be: 5.642mm of diameter.
  Area Calculator
• Using the correct cylinder calculators: 0.25 newtons
  Piston Cylinder Pressure and Diameter to Force Calculator
  Pneumatic Cylinder Force Calculator
• So, assuming parallel actuators in a 1 meter long filament fiber:
  5mm/1000mm = 200 actuators
• 200 actuators x 0.25 newtons = 50 newtons = 5 kilograms
• 200 actuators x 0.0011 liters = 0.22 liters to fully actuate all 200 actuators.
• Like I said before, you need to take into equilibrium the pressure vessel and the actuators.
  IF you have 2 chambers, one with high pressure and another with lower pressure and open a valve connecting both, you will get an equilibrium.
  So, 0.22 liters x 2 = 0.44
• 3000 kilograms of actuation force / 5 kilograms of actuation force by the actuators = 600 parallel filaments
• 600 parallel filaments x 0.44 liters = 2640 liters
• Even though the actuators can lift all of this weight, they are in parallel, thus, you need to increase their length to compensate for the lack of actuation length/stroke.
  Assuming that the actuation is around half of the length of the pneumatic cylinder, around 25mm, I would need:
  100mm of travel / 25mm = 4
  200mm of travel / 25mm = 8
• Thus, 2640 x 8 = 21,120 liters per minute.
• fan with 400,000 liters per minute with 560 pascals and 3.3 kilowatts
  10,000 pascals / 560 pascals = 17.8571428571
  3.3 kw x 17.8571428571 = 58.928 kilowatts
• Since pneumatics needs to stay in equilibrium, I need to multiply it by 2
  58.928 x 2 = 117,856
• 400,000 / 21,120 liters per minute = 18.9393939394 times less energy
• 117,856 watts / 18.9393939394 times less energy = 6,222.7 watts

• If I apply this muscle with 3000kg of actuation strength to a lever with the load arm 3 times longer than the effort arm, it will lift something with a force of 1000kg.
  Given that a mechanical arm moving at 300 RPM and 10,000 newton meters, it would require 300 kilowatts.
• But the expansion of the actuator would make that super fast, faster than 300 rpm.
• Wouldn’t that mean that I’m breaking the laws of physics?

Are you making any progress on analyzing the mechanism and the load of the task you are trying to accomplish ( like I talk about in post 11), or is that out of your comfort zone?

 

[fulano]

Are you making any progress on analyzing the mechanism and the load of the task you are trying to accomplish ( like I talk about in post 11), or is that out of your comfort zone?

I think I solved it, since it ended up as 6000 watts and the kinetic energy of 3000kg at a few meters per second.
But yes, everything is outo of my comfort zone given I literally don't know anything about physics.

 

[erobz]

I think I solved it, since it ended up as 6000 watts and the kinetic energy of 3000kg at a few meters per second.

Kinetic energy of a full-sized pickup truck moving at 7 mph... Seems a touch high?

But yes, everything is outo of my comfort zone given I literally don't know anything about physics.

Ok, well. I wasn't sure initially. Is this a "high school project" or something equivalent where you haven't necessarily had the requisite physics/engineering coursework?

 

[fulano]

Kinetic energy of a full-sized pickup truck moving at 7 mph... Seems a touch high?

Ok, well. I wasn't sure initially. Is this a "high school project" or something equivalent where you haven't necessarily had the requisite physics/engineering coursework?

Nope, a personal project to make a mech in real life.
Silly, I know, but the important part is that I have fun calculating it, even if I will never build it in real life.

 

[erobz]

Nope, a personal project to make a mech in real life.
Silly, I know, but the important part is that I have fun calculating it, even if I will never build it in real life.

I can't decide how to take this...have fun calculating - ok...But you have to understand some of what you are trying to calculate if you are going to learn anything about that aspect of it. Blindly calculating things doesn't seem particularly useful to your mission of building a robot.