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Cake day: June 14th, 2025

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  • 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