some comments:
> You need to be more careful with units. Your calculated answer is 474 kW/m^2, not 474 kW. That means that the answer must be multiplied by the appropriate area. If you have a larger or smaller area than 1 m^2, the temperature rise will be different.
> Your answer must be caveated by the fact that the contact thermal resistance to the aluminum might be quite high, relative to the aluminum itself. This would detract from the calculated heat flux capability.
> It's generally difficult to establish a fixed cold-side temperature. Unless you're running chilled water at a pretty drastic flow on the other side, the temperature of the cold side will not stay put. If there's air on the cold side, unless you've got a gigantic flow, you'll be hard pressed to maintain the heat flow. Even forced air convection would limit you to well less than 50 W/m^2/K, so you'd need a delta-T of 10,000 K to match your calculated heat flow
> Your comments suggest a slightly misaligned view of the equation. The equation basically says that IF you FLOW 474 kW/m^2 through that chunk of aluminum, you'l GET a 100-K temperature rise from the cold side. Note that this is a STEADY-STATE solution, meaning that the temperature rises exists, ONLY IF you maintain that heat flow. If the heat flow drops to 47 kW, you'll only get a 10-K rise. If your application cannot sustain that kind of heat output, then that temperature rise cannot exist. Generally speaking, getting no more than about 5 K or 20 K across an aluminum structure is more the norm.
TTFN
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