nichrome wire formula

Dear Flx Cpy:

Same you would expect. However, the resistivity of nicrome is a function of its operating temperature.

Not unless you know its resistance at your operating tmperature.

I have never seen such a thing. The second "H" is supplying double current. How about a simple delta: H---H \ / H

Better measure the current. Better measure the voltage. That will give you the operating points necessary to duplicate it on a single voltage supply.

No. This might: H | ====== | N | ====== | | ====== | Six strands with total length as you note. Each strand drops just a little less power than you what I am guessing you are supplying above (I assumed 460vac, which is 277 vac phase to neutral).

If you supplied three phase 230 or 208 vac, then only four strands are necessary: H | ====== | N | ====== |

You have to get the same current flow in each segment of wire. If you reduce the applied voltage, then you need to create parallel paths.

David A. Smith

This is probably a better diagram of what I have now:

H------------------ \ /\ / \ / \ / \/ \/ N------------------

In the diagram below, if a, b, c, and d are equal length, would wire A be the hottest and D be the coldest?

120VAC... H---------------------------- | | | | a b c d | | | | N----------------------------

I guess my main question was whether or not there was a formula to calculate 1) the load across H and N given the sum of the lengths of a,b,c,d, and 2) the output generated by the entire array based on the properties of the Nichrome wire ( in this case 1.671 Ohms/Foot at

68=B0F). I think I might be better prepared to ask a question like this if I did some more reading about basic electronics...sorry. Thank you for your response, David.

Dear Flx Cpy:

What is the voltage between H and N here? Still 120vac?

No, neglecting losses in the "wires" and H and N, a - d will be the same temperature if they are the same length.

The formulae are: Watts law: Power = Voltage^2 / R_equiv

Resistance of parallel elements:

1 / R_equiv = 1 /R_a + 1/R_b + 1/R_c + 1/R_d

When the wire is at 68 deg F, for wire "a" R_a = 1.671 * length_a (in feet)

But, you *need* to know the total current drawn and voltage applied in the circuit that you are happy with. This will tell you what the resistance is of the nichrome running at your "satisfactory" temperature.

R_equiv = applied voltage / total current

David A. Smith

For reasonable assumptions about the resistance of the feed, all would dissipate much the same power, if they all show the same cold resistance.

The ultra-basic formulas: power (watts) = volts times amps Using R for the cold resistance of one length of wire say hot resistance = 2 x cold resistance, and amps = volts/R so the power in each wire is V x V / Rcold x 2 and the total power is 2 x V x V / Rcold for all four wires.

Alright: for a 1KW heater, what is the cold R of each wire?

1000 = 2 X 120 X 120 /Rcold

Rcold = 2 X 120 X 120 / 1000 = 29 ohms (roughly!) amps (per wire) = 120 / R x 2 = 120/58 = 2 amps For 4 wires, that's 8 amps, and 8 amps times 120 volts = 960 watts.

Hope the example helped. Remember that resistance multiplier that I used ( x2) depends strongly on the temperature of the hot wire. It varies.... For a start, you could try 4 wires, each 17.4 ft long for a 1 kW heater. MORE wire for LESS heat (!)

Brian Whatcott Altus OK