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There is a type of McKibben air-muscle called "untethered artificial muscles", this type in specific has a liquid inside of it that is heated through an specific method, generating steam that in turn generates pressure inside the muscle, finally making an actuation.

Here is a video of an untethered artificial muscle that boils the water through magnetic induction.


I was trying to calculate how much energy I would need to input in one order to achieve full actuation. However, I used the unit of Calorie, which is a unit of energy required to heat 1 gram of water by 1ºC.

And it gave me an incredibly low value of 0.06 watts to fully actuate the muscle, which seems unusually low, so I would like to double check.


So, assuming that a muscle with 1 inch of inner diameter and 30cm of length is full of water and a heating element, I would have more or less 0.152 liters of water.

Since 1 liter of water generates 1600 liters of steam at room pressure, in order to make 0.152 liters 5 times to achieve 5 bar/atm of pressure, I would need 0.76 liters of steam.

(1 liter of water)/ (1600 liters of steam) = (x liters of water) / (0.76 liters of steam)

x = 0.000475 liters of water

So, since to heat 1 gram of water in 1ºC I would need 1 calorie, this means that in order to heat 0.475 milliliters (or 0.475 grams) of water, I would need

0.475x124 = 58.9 calories = 0,06845489 watt-hour of energy.

The "124" comes from the heating of water from 27ºC to 150ºC, so an increase in 124 degrees. The 150 ºC is due to the fact that the boiling temperature of water increases to 150ºC at 5 atm/bar of pressure.


Does the math checks out or I'm really making some awful math mistake? I don't really know how else to check the math.

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    $\begingroup$ I did not check your calculation in detail, but to transform liquid water into steam you need to take into account the latent heat of vaporization. $\endgroup$
    – JRTG
    Sep 9, 2023 at 15:41
  • $\begingroup$ How do I calculate that? $\endgroup$
    – Fulano
    Sep 9, 2023 at 16:00
  • $\begingroup$ Google on 'latent heat of vaporization'. $\endgroup$
    – JRTG
    Sep 9, 2023 at 18:52

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Besides the heat of vaporisation noted by JRTG, realise you need to heat all the water to boiling point(or at least a significant portion of it), not just the amount of water that must be boiled. So at this point the energy requirements are roughly 1000x your initial.

Also consider how quickly that assembly will lose heat to the environment. This will require more energy so you're well above 1000x initial.

You can get the theoretical lowest value by using E = mGh. mass x gravity x distance moved vertically. That is the amount of energy added to the thing being lifted. That value assumes 100% efficiency and a massless actuator. I'm guessing that method as you describe it is less than 5% efficient.

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