HVAC · Heat-Transfer · Problem 1PDFSolution in PDF ↓
HVAC · Heat-Transfer · Problem 1
Problem & Solution
Video Synthesis
Problem: Hi, welcome to the Heat Transfer Practice Module.
Given: 10 pounds of pork shoulder are cooled from 68 degrees to 10 degrees for storage in a freezer; 0.86 BTU per pound degr...
Approach: So with that, let's dive right into the first problem.
Calc: 10 pounds of pork shoulder are cooled from 68 degrees to 10 degrees for storage in a freezer.
Calc: And one more thing worth mentioning that specific to pork, it's freezing point isn't necessarily 32 degrees because it isn't wa...
Result: So the key takeaway from this problem is that the specific heat capacity of a food or probably any substance above and below its freezing point are...
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Student questions asked in live office hours about this problem
OH 59: HVAC: Heat Transfer #1
Q: How do you handle a composite wall heat transfer problem with a large 2-foot air gap? The thermal resistance table in the handbook only goes up to 3.5 inches.
A: A sealed air gap that's large relative to the table range essentially reaches equilibrium with adjacent surfaces and contributes no thermal resistance—think of it as another outdoor surface where the air temperature equals the surrounding surface temperature and there's no temperature gradient to drive heat transfer. The total resistance of the wall is just the sum of the two solid material layers (A and B), ignoring the air gap entirely. Once you have total resistance, you can find interface temperatures using the proportional equation: T_x = T_i − (R_x/R_total)·(T_i − T_o).