The "Power dissipation" of a MOSFET is the maximum theoretical allowed dissipation in the MOSFET if the case temperature was held at 25°C. It has no relation to the inverter's output power at all, so don't make this mistake.How does the MOSFETS work? It looks like they're rated upto like 150w and even if that's the case, there's 16 of them... that'd mean the inverter could only supply 2400w...am I missing something?
This number is used to quickly estimate thermal design. It's basically (Maximum junction temperature - 25°C) divided by the internal thermal resistance of the MOSFET from silicon chip to case (the metal back of the package). This power dissipation is never reached (not even close) in real use, because in real use, the case is never at 25°C but much hotter than that, and you would also derate it to keep it far from its maximum temperature. How much depends on the total cost. Heat sinks cost money, so if the gear is too cheap, you can expect the devices to operate closer to the thermal limit and have shorter lives.
The use of this number is just to compare between two MOSFETs. For example the IPP050N10 quotes 150W, and the IRFB4410 quotes 230W. I'll explain what it means.
Maximum junction temperature is the same at 175°C, the difference is Thermal Resistance Junction to Case is 1°C/W for IPP050N10 and 0.65°C/W for IRFB4410. So for each watt of heat burned in the MOSFET (this is your switching and conduction losses, not your inverter output power), the chip will get hotter than the case by 1°C for the former or 0.65°C for the latter. How much the case will get hotter than the heat sink is pretty much the same for both, because the case is the same.
What the number means is simply that the second MOSFET has a bigger chip, so it transfers heat better to the metal back. That's all there is to this spec. Of course, being bigger, it also has more capacitance, so it is slower (total gate charge is 51-78nC for the first and 83-120nC for the second) and the driver will have to pump more current into it to turn it on at every cycle. So IRFB4410 can get rid of more heat, but it will also have higher switching losses and higher driver losses. IRFB4410 will also have higher conduction losses due to the higher RdsON. But it is an older generation, so it is cheaper and the bigger chip means it will probably be harder to destroy. And... the diode in the IRFB4410 has much lower recovery losses, so if that's an important part of the total losses, then it may even have lower losses than the other two. But if it's not, then it won't. There's no clear winner between the three, it depends on the circuit and how it's used.