I'm trying to build a heater connected to 220 VAC, 50 Hz using a nichrome strip (2.3 Ω/m) and I'm going to use 1 meter or less (60 to 70 cm).
I know that this setup won't work at all with 220 VAC so my question is: can I add a resistor in series to reach the desired wattage which is 750 W to 1 kW? If yes, what is the resistor value? And are there any other solutions?
P.S. I also have nichrome wire with 1.75 Ω/m.
The total resistance \$R_N\$ of 60cm of 2.3Ω/m nichrome strip is:
$$ R_N = 0.6m \times 2.3\Omega = 1.4\Omega $$
When placed across 220V RMS, this will dissipate power:
$$ P = \frac{V_N^2}{R_N} = \frac{(220V)^2}{1.4\Omega} = 35kW $$
If for some reason the breakers didn't pop, that's likely to set something on fire, before melting. Your proposed solution is to reduce the power dissipated in the nichrome by using a series resistor. Let's see how well that would work.
Calculate what voltage across 60cm of nichrome would cause it to dissipate 1kW:
$$ \begin{aligned} P_N &= \frac{V^2}{R_N} \\ \\ V_N &= \sqrt{P_N \times R_N} \\ \\ &= \sqrt{1kW \times 1.4\Omega} \\ \\ V_N &= 37V \end{aligned} $$
Whatever resistor you place in series with the nichrome strip (as you suggested) would share the total 220V with the strip. In other words, a series resistor \$R_S\$ would have voltage \$V_S\$ across it:
$$ \begin{aligned} V_S + V_N &= 220V \\ \\ V_S &= 220V - V_N \\ \\ &= 220V - 37V \\ \\ V_S &= 183V \end{aligned} $$
Knowing that the nichrome strip and resistor will each drop a voltage in proportion to their resistance, we can write this simple relationship between the voltages across them and their resistances, and solve to find the extra resistance \$R_S\$ required:
$$ \begin{aligned} \frac{R_S}{R_N} &= \frac{183V}{37V} \\ \\ R_S &= R_N\frac{183V}{37V} \\ \\ &= 1.4\Omega \times \frac{183V}{37V} \\ \\ R_S &= 6.9\Omega \end{aligned} $$
Lastly, let's see what power that resistor \$R_S\$ would dissipate:
$$ \begin{aligned} P_S &= \frac{(V_S)^2}{R_S} \\ \\ &= \frac{(183V)^2}{6.9\Omega} \\ \\ P_S &= 4.9kW \end{aligned} $$
I think it's pretty clear that the resistor would probably explode as soon as you switch this contraption on, unless it were the size of large dog, so the short answer is no, you can't use a series resistor to limit power in the nichrome strip.
The take-away from all this is that if you wish to dissipate 1kW in something connected directly across the mains, it's the entire resistance that dissipates all the power, including any current-limiting device you place in the path, like \$R_S\$. Therefore you'd really need to find a length of nichrome with an appropriate resistance to do the whole job:
$$ \begin{aligned} P_N &= \frac{V^2}{R_N} \\ \\ R_N &= \frac{V^2}{P_N} \\ \\ &= \frac{(220V)^2}{1kW} \\ \\ R_N &= 48\Omega \end{aligned} $$
If you insist on using that particular nichrome strip, with that particular length, then the only solutions is to derive 37V RMS from the 220V RMS mains. The obvious way would be to use a transformer, but it would have to be able to handle 1kW, and have a secondary coil capable of carrying the necessary current without overheating. By Ohm's law:
$$ I_N = \frac{V_N}{R_N} = \frac{37V}{1.4\Omega} = 26A $$
That transformer would be the size of a small dog, so at least your contraption got smaller.
Alternatively, use a longer nichrome strip. The length of 2.3Ω/m nichrome you would need is:
$$ l_N = \frac{48\Omega}{2.3\Omega/m} = 21m $$
That seems excessive, so we're only left with the task of finding nichrome strip with higher resistivity. If you insist on using only 60cm, then the resistance-per-metre \$\sigma_N\$ of you nichrome must be:
$$ \sigma_N = \frac{R_N}{l_N} = \frac{48\Omega}{0.6m} = 80\Omega/m$$
I have no idea if that even exists to buy.
In summary, whatever heating element you design must be connected directly across the 220V mains, and will need to be 48Ω. It will probably be much longer than 60cm, and/or have a much greater resistivity than the nichrome strip you currently possess.
Maybe there's another way to employ something other than nichrome, but I have little experience with such things.
No. You must choose a different nichrome gauge to get the right resistance for the length you can use, or make many turns in that distance but spaced adequately apart on insulators, preferably Mica.
That is how toasters work. But you may want a lower temperature so please specify.
\$R_{hot}= V^2/P= 220^2/1000= 48.4 \$Ohms.
Cold (25'C) will be a much lower resistance if your power is accurate.
The rise in temperature controls the ratio in hot/cold resistance ratio which is determined also by your thermal resistance and voltage applied. So the goal or design spec, must specify both parameters before a wire and thermal design is chosen. Where are your design specs? Rth, Vrms, P and Temp rise.
Notice I did not specify the \$R_{cold}\$ which depends on both parameters above, delta T and Rth. for thermal resistance from junction to ambient.
It is true that when conductors with a large positive temperature coefficient for resistance are applied with constant rms voltage or DC, that the current tends towards a more constant power and current.
The temperature depends on the thermal resistance such that the centre of a heater will be hotter than the outside, so windings must be very loosely coupled to minimize the gradient temperatures of all windings.
The resistance of a 220V - 1 kW electric heater = 200² / 1000 = 48.4 Ω.
All that you require is a 48.4 Ω - 1 kW wire-wound resistor.
A standard 50 Ω - 1 kW wire-wound resistor would suffice.
For 750 W, a standard 64 Ω - 1 kW resistor may be used.
