When two metallic pieces are in contact, because of the intrinsic high conductivity of metals, the resistance to heat flow through them is dominated by the contact resistance at their interface. Use aluminum surface pairs in contact with three different surface roughnesses (i) \left\langle \delta ^{2}\right\rangle^{1/2} = 1.2 to 1.7 \mu m, and (ii) \left\langle \delta ^{2}\right\rangle^{1/2} = 0.20 to 0.46 \mu m, for a harder aluminum (2024-T3), and (iii) \left\langle \delta ^{2}\right\rangle^{1/2} = 0.25 \mu m for a softer aluminum (75S-T6). The contact conductances for these are shown in Figure. Assume that the effect of the temperature of the gap fluid is negligible (this is justifiable when the temperature is far from the softening temperature and for low conductivity gap fluid, i.e., other than liquid metals).
(a) Then using a contact pressure p_{c} = 10^{5} Pa, compare the contact resistance A_{k}R_{k,c} among these three contact interfaces.
(b) For a heat flux of q_{k} = 10^{4} W/m^{2}, determine the temperature jump across the contact interface \Delta T_{c}, for the three surface roughnesses.
Question 3.11: When two metallic pieces are in contact, because of the intr...
Since we have assumed a negligible temperature and gap-fluid dependence, we can use the results of Figure 3.25 which are for various temperatures.
(a) From Figure, at p_{c} = 10^{5} Pa, we read the contact conductance per unit area, 1/A_{k}R_{k,c}, for Al-Al for the three different RMS roughness as
(i) \frac{1}{A_{k}R_{k,c}}\equiv \left(\frac{k}{L} \right)_{c}=3\times 10^{2}(W/m^{2})/^{\circ }C for \left\langle \delta ^{2}\right\rangle^{1/2} = 1.2 to 1.7 \mu m harder Al
(ii) \frac{1}{A_{k}R_{k,c}}\equiv \left(\frac{k}{L} \right)_{c}=4 \times 10^{2}(W/m^{2})/^{\circ }C for \left\langle \delta ^{2}\right\rangle^{1/2} = 0.20 to 0.46 \mu m harder Al
(iii) \frac{1}{A_{k}R_{k,c}}\equiv \left(\frac{k}{L} \right)_{c}=8 \times 10^{3}(W/m^{2})/^{\circ }C for \left\langle \delta ^{2}\right\rangle^{1/2} = 0.25 \mu m softer Al
Note that the softer aluminum with the smallest roughness has the smallest contact resistance (i.e., highest contact conductance).
(b) From \Delta T_{c}\equiv Q_{k}R_{k,c}=q_{k}A_{k}R_{k,c}=q_{k}\left(\frac{L}{k} \right)_{c} we have
\Delta T_{c}\equiv q_{k}\left(\frac{L}{k}\right)_{c} =q_{k}\frac{1}{(k/L)_{c}}Using this, for the three contact resistances we have the temperature jumps as
(i) \Delta T_{c}=10^{4}(W/m^{2})\frac{1}{3\times 10^{2}(W/m^{2}-^{\circ }C)}=33.3^{\circ }C or K for \left\langle \delta ^{2}\right\rangle^{1/2} = 1.2 to 1.7 \mu m harder Al
(ii) \Delta T_{c}=10^{4}(W/m^{2})\frac{1}{4\times 10^{2}(W/m^{2}-^{\circ }C)}=25.0^{\circ }C or K for \left\langle \delta ^{2}\right\rangle^{1/2} = 0.20 to 0.46 \mu m harder Al
(iii) \Delta T_{c}=10^{4}(W/m^{2})\frac{1}{8\times 10^{3}(W/m^{2}-^{\circ }C)}=1.25^{\circ }C or K for \left\langle \delta ^{2}\right\rangle^{1/2} = 0.25 \mu m softer Al