Heat Transfer Mechanisms — Question 2

Section 18-11 · Physics pages 3014–3015

Question 2 A small thermal bridge can dominate a parallel wall A plane wall has total area 10.0 m2 and thickness .100 m. Insulation with ki = .0400 W/(m K) occupies 9.90 m2, while metal bridges with km = 50.0 W/(m K) occupy .100 m2. Both paths share a 20.0 K temperature difference. Neglect lateral coupling and surface resistance. insulation thin metal bridge Parallel paths share a temperature difference, not a heat-transfer rate. (a) Find each heat rate and the fraction carried by the metal. (b) Find the maximum metal area fraction f that would keep total conduction at or below 200 W, with the same total area and thickness.
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Question 2 - Solution A small thermal bridge can dominate a parallel wall insulation thin metal bridge (a) Method and calculation Pi = .0400(9.90)(20) .100 = 79.2 W, Pm = 50(.100)(20) .100 = 1000 W. Ptotal = 1079.2 W, Pm Ptotal = .926612 ≈92.66%. (b) Interpretation and further calculation P = A∆T L [ki(1 −f) + kmf] = 2000[.0400 + 49.96f]. P ≤200 ⇒f ≤.100 −.0400 49.96 = .00120096. Only about .1201% of the area may be metal under these idealized conditions. Physical check Area averaging conductivity is valid for this parallel geometry; averaging thermal resistances would describe a different arrangement.

Question 2

A small thermal bridge can dominate a parallel wall

A plane wall has total area 10.0 m2^2 and thickness .100 m. Insulation with ki=.0400k_i=.0400 W/(m K) occupies 9.90 m2^2, while metal bridges with km=50.0k_m=50.0 W/(m K) occupy .100 m2^2. Both paths share a 20.0 K temperature difference. Neglect lateral coupling and surface resistance.

Diagram shown in the original SVG worksheet.

  1. Find each heat rate and the fraction carried by the metal.

  2. Find the maximum metal area fraction ff that would keep total conduction at or below 200 W, with the same total area and thickness.

Question 2 - Solution

A small thermal bridge can dominate a parallel wall

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Pi=.0400(9.90)(20).100=79.2W,Pm=50(.100)(20).100=1000W.P_i=\frac{.0400(9.90)(20)}{.100}=79.2\,\mathrm W,\quad P_m=\frac{50(.100)(20)}{.100}=1000\,\mathrm W.Ptotal=1079.2W,PmPtotal=.926612≈92.66%.P_{\mathrm{total}}=1079.2\,\mathrm W,\quad\frac{P_m}{P_{\mathrm{total}}}=.926612\approx92.66\%.

(b) Interpretation and further calculation

P=AΔTL[ki(1−f)+kmf]=2000[.0400+49.96f].P=\frac{A\Delta T}{L}[k_i(1-f)+k_mf]=2000[.0400+49.96f].P≤200⇒f≤.100−.040049.96=.00120096.P\le200\ \Rightarrow\ f\le\frac{.100-.0400}{49.96}=.00120096.Only about .1201%.1201\% of the area may be metal under these idealized conditions.

Physical check

Area averaging conductivity is valid for this parallel geometry; averaging thermal resistances would describe a different arrangement.