

I would say trying to cook 3x faster isnt a surface temp 3x higher, but the rate of heat transfer is 3x faster. Because then your laat step is backwards.
Q = h a dT
If you just calculated H goes up 1.69x, then dT doesnt need to go up MORE, it needs to go up LESS. A higher HTC increases the rate of heat transfer. Which… now you need to recalc your free conv HTC because temp changed. But if it didnt…
3Q = 1.69h a 1.78DT
Old DT was 260, now its 463, which is an ovev temp of 503F
Edit: this is a transient problem we’re trying to reduce to a steady state calculation with pretty poorly defined requirements. Sounds like a typical day at work for me lol


Youre assuming these ass bags abide by due process, and youll ever see a lawyer