Heating in a Vacuum

1. Apr 10, 2016

Jai Sankar

I have a flexinol wire, and it has a resistance of 0.11 ohms/inch and it requires 4000mA for a 1 second contraction. I was wondering how much energy was required for a 1 second contraction and if I were to heat the flexinol under a vacuum, how much less energy would be required. The flexinol is a cylinder with a radius of 0.020 in and is 3 inches long, a density of 0.235 lb/in^3, and a specific heat of 0.2 cal/g * °C.

2. Apr 10, 2016

Nidum

(1)

You have 3 inches of wire so total resistance is 0.33 ohms . Current is 4 Amps .

Power dissipated by a resistance is given by I2R Watts .

One Watt = One Joule / second - so how many Joules are needed for a 1 second ON time ?

(2)

Another calculation could tell you what voltage and Amp-Hour capacity a battery would need to be to supply required current and to give the required number of seconds ON time for your application .

(3)

I don't think that having wire in vacuum as opposed to air would make much difference to power consumption .

nb: Flexinol

Last edited: Apr 10, 2016
3. Apr 10, 2016

Jai Sankar

How much heat would be lost to air rather than a vacuum?

4. Apr 10, 2016

OldYat47

Paint it flat black and energy loss is greater, polish it like a mirror, energy loss would be less. Infrared radiates pretty close to equally well in air or in a vacuum. Air does add a convection component, but for 1 second isn't long enough to establish any significant convection current.

5. Apr 15, 2016

Physics_Kid

in a vacuum it will do it's thing slower because the heat can escape the boundary layer faster. the air-to-wire boundary layer is like "R" in thermal insulation equations, thus heat will escape slower so temp will rise faster.

6. Apr 18, 2016

Mech_Engineer

Natural convection is a tricky subject, but for a simple geometry like a wire in free air you can make some assumptions to make calculations a bit more manageable.

Your governing equations for heat loss per unit length would be:
1. Convective Heat Loss: $q'_c = \bar hπD(T_s - T_∞)$
2. Radiation heat Loss: $q'_r = επDσ(T_s^4 - T_∞^4)$
• $q'_c$ - Convective heat loss
• $q'_r$ - Radiative heat loss
• $\bar h$ - Convective coefficient
• $D$ - Wire diameter
• $ε$ - Wire emissivity
• $σ$ - Stefan-Boltzmann constant
• $T_s$ - Wire temperature
• $T_∞$ - Environment temperature
The dimensionless parameters used in the calculation of natural convection are where it gets a little tricky. You're looking for the convective coefficient but it depends on the temperature, so to start you'll guess a temperature and then calculate out the convective coefficient. You can follow an iterative process for estimating steady state temperature until you're able to balance heat loss with power input. So (guess temperature) → (calculate heat loss) → (compare to electrical power put in) → (repeat step 1).

To get the convective coefficient follow these steps:
1. Calculate Rayleigh number: $Ra_{D}=\frac{g\beta\left(T_{s}-T_{inf}\right)D^{3}}{\nu\alpha}$
2. Calculate Nusselt number: $\overline{Nu}_{D}=\left\{0.60+\frac{0.387Ra_{D}^{1/6}}{\left[1+\left(0.559/Pr\right)^{9/16}\right]^{8/27}}\right\}^{2}$
3. Calculate convective coeffcient: $\overline{h}=\frac{k}{D}\overline{Nu}_{D}$
where:
• $\beta=1/T_{s}$
• Prandtl number: $Pr = \frac{ν}{α}$ (ratio of the kinematic viscosity and thermal diffusivity of air)