You are designing a cylindrical heater that you want to use to heat a room. The
ID: 2327273 • Letter: Y
Question
Explanation / Answer
From convectve transfer at surface: 319 = 30DT
DT= 10.63
Surface temp =35.63
Apply cyclindrical formula with composite conductivity
Using 290 as the heat rate
290 = (100-35.63)/ [ 1.225254 +ln rg (.651435) ]
rg =..214 m = 21,4 cm
Thermal resistance = r1/r4 *1/h + r1/kst ln(r2/r1) + r1/kg*ln(r3/r2) + r1/Kal*ln(r4/r3)]
Thickness of Al = 50 cm - 21.4.cm = 28.6 cm apprx
Thickness matters decided by glass thickness
Max temp 100 c at the heater surface (given)
(Formula Q(1-r/R0 )not needed since the heat transfer rate obtained thru given value.
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