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- Thread starter Jai Sankar
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Nidum

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(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 I^{2}R 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

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 I

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:

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

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

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Mech_Engineer

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Your governing equations for heat loss per unit length would be:

- Convective Heat Loss: [itex]q'_c = \bar hπD(T_s - T_∞)[/itex]
- Radiation heat Loss: [itex]q'_r = επDσ(T_s^4 - T_∞^4)[/itex]

- [itex]q'_c[/itex] - Convective heat loss
- [itex]q'_r[/itex] - Radiative heat loss

- [itex]\bar h[/itex] - Convective coefficient

- [itex]D[/itex] - Wire diameter
- [itex]ε[/itex] - Wire emissivity

- [itex]σ[/itex] - Stefan-Boltzmann constant

- [itex]T_s[/itex] - Wire temperature
- [itex]T_∞[/itex] - Environment temperature

To get the convective coefficient follow these steps:

- Calculate Rayleigh number: [itex]Ra_{D}=\frac{g\beta\left(T_{s}-T_{inf}\right)D^{3}}{\nu\alpha}[/itex]
- Calculate Nusselt number: [itex]\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}[/itex]
- Calculate convective coeffcient: [itex]\overline{h}=\frac{k}{D}\overline{Nu}_{D}[/itex]

- [itex]\beta=1/T_{s}[/itex]
- Prandtl number: [itex]Pr = \frac{ν}{α}[/itex] (ratio of the kinematic viscosity and thermal diffusivity of air)

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