This is a coil of nichrome wire.

How long should it be? MATH!

Candidate model aircraft “LiPo” battery

14.8 V

– voltage drop under load?

- estimate 14V

1.5 Ampere-Hour

"45C" rating which advises max continuous current of

45 x 1.5 = 67.5

Let’s say 60 to allow some contingency and be realistic.

Power

P = VI

Power (in Watts) = Voltage X Current

P = 14 x 60

P = 840 W

For this to happen, coil resistance is:

R = V divided by I

R = 14 divided by 60

R = 0.23 Ohm

Nichome wire diameter 0.5mm information supplied states 4.5 Ohm per metre.

Therefore length needed is:

L = 1 x 0.23 divided by 4.5

L = 0.051 m

L = 5cm

14.8 V

– voltage drop under load?

- estimate 14V

1.5 Ampere-Hour

"45C" rating which advises max continuous current of

45 x 1.5 = 67.5

Let’s say 60 to allow some contingency and be realistic.

Power

P = VI

Power (in Watts) = Voltage X Current

P = 14 x 60

P = 840 W

For this to happen, coil resistance is:

R = V divided by I

R = 14 divided by 60

R = 0.23 Ohm

Nichome wire diameter 0.5mm information supplied states 4.5 Ohm per metre.

Therefore length needed is:

L = 1 x 0.23 divided by 4.5

L = 0.051 m

L = 5cm

Let us say 6cm for connection at ends - also for the first prototype good to err on the higher resistance side.

I am thinking about the need to transfer energy from the wire to the water. I am guessing based on life experience with similar setups that this needs more wire surface area to be in contact with the water. We can get this by the idea of "2 resistors in parallel". eg 2 x higher resistances in parallel give a lower resistor. Each higher resistor is 12cm long, the 60A current is split to a more moderate 30A each and the total wire length is now 24cm.

Coil beginning. A centre "tap" of wire twisted together runs from halfway along the coil to the base. Also running down the centre - careful that they don't touch! is the top end of the coil going back to join the bottom end.

## No comments:

## Post a Comment