Heat Transfer Mechanisms — Question 1

Question 1

Series layers and contact resistance share one heat-transfer rate

A plane wall of area $A=2.00\,\mathrm{m^2}$ has layer 1 with $L_1=.100$ m, $k_1=.50$ W/(m K), followed by layer 2 with $L_2=.200$ m, $k_2=1.00$ W/(m K). Their contact adds resistance $R_c=.0500$ K/W. The exterior faces are maintained at 350 K and 290 K. Use $R=L/(kA)$; neglect lateral losses and internal generation.

1. Find the steady heat rate and both temperatures immediately adjacent to the contact.

2. If contact resistance is removed, find the new rate. Explain why an interface temperature jump is present in the original idealized model.

Question 1 - Solution

Series layers and contact resistance share one heat-transfer rate

(a) Method and calculation

$$R_1=.100\,\mathrm{K/W},\quad R_2=.100\,\mathrm{K/W},\quad R_{\mathrm{total}}=.250\,\mathrm{K/W}.$$$$P=\frac{350-290}{.250}=240\,\mathrm W.$$$$T_{1c}=350-240(.100)=326\,\mathrm K,\quad T_{2c}=326-240(.0500)=314\,\mathrm K.$$

(b) Interpretation and further calculation

$$P_{R_c=0}=60/.200=300\,\mathrm W.$$$$T_{1c}-T_{2c}=PR_c=12.0\,\mathrm K.$$The lumped contact resistance represents an unresolved thin resistive region. The ideal jump disappears when $R_c=0$; finite steady transfer remains.

Physical check

The drops $24+12+24=60$ K add to the specified face-temperature difference.