It sounds like you need a heatsink.
Another thing to check is your gate voltage when you have the FET switched on. If your Gate-Drain voltage is too high, the MOSFET may not be properly biased on, a situation which would generate a lot of heat.
Also, how often is this switching? Is it a steady-state thing, or is this part of a switching supply? If it's a switching supply, you will also need to look at the rate at which the system switches.
Anyways, assuming you have everything biased properly (probably a safe assumption, but measure it anyways):
RDS(on) = 130mΩ @ G-D voltage of 5V
So, with 130 mΩ in series with 3A:
$$V = 0.130 * 3$$
$$V = 0.39$$
$$Power = V * A$$
$$Power = 0.39 * 3$$
$$Power = 1.17W$$
So you're going to be dissipating 1.17W of power in the MOSFET, in the best-case situation.
That will get very toasty without a heatsink. If you're just running this as a bare TO-220 device, it getting extremely hot isn't too suprising.
So, assuming we have a TO-220 in free-air:
TO-220 junction-to-air thermal to ambient equals 62.5 degree per watt.
(From here)
Therefore:
$$Δ°C = 62.5 * 1.17$$
$$Δ°C = 73.125$$
$$Device Temperature °C = 73.125 + Ambient$$
$$Device Temperature °C = 98.125$$
So assuming ideal thermal dissipation on a bare TO-220, it's still going to easily reach ~100°C.
Any environmental factors that further reduce the device's cooling will make it worse.
I'm not sure using an N-channel MOSFET for Q1 makes a lot of sense in this application. For one thing, you're going to have to be careful that you don't exceed the maximum gate-source voltage — pulling the gate to +120V while the source is held at less than +22V is almost certainly too much. Any pulling the gate all the way to ground with +22V on the source may also be too much.
You'd probably be better off to use a P-channel MOSFET in this application, connecting its source to the generator, putting a 15V zener diode between the source and gate, and then pulling the gate down through a resistor when you want to enable (rather than disable) the charging.
Best Answer
Your heater at 2.4 ohms with 12 V will draw a current of I = V/R = 12/2.4 = 5 A.
While that is within the capability of the MOSFET's 13 A rating, you are failing to account for the power dissipated in the MOSFET itself.
At 5 A and 0.280 ohms RdsON you will get P = I^2*R = 5^2 * 0.280 = 25 * 0.280 = 7 watts. The datasheet shows a Rthj-amb of 62.5 C/W.
So 7 watts will give you 7 * 62.5 = 437 C! That's WAY too hot. The max operating temperature is 150 C.
You need a heatsink capable of keeping the device below 150 C or you need a different MOSFET that will not dissipate so much heat.