Instructors Solutions Manual (PIEV)230Problem 7.12 SolutionFor stabilized zirconia k= 2.0 J/(m-s-K) at T= 100° C k= 2.3 J/(m-s-K) at T= 1000° C Using linear extrapolation, estimate kat 1500° C k= 10010000.23.2)1001500(0.2−−−+= 2.433333 J / (m-s-K) Use the average value ofkover the temperature range in the wall. k= 2433333.20.2+= 2.216667 J / (m-s-K) Rearrange the equation for kto solve for tQ∆∆/.k= )/(/xTAtQ∆∆∆∆−tQ∆∆= xTkA∆∆−= cm100m1)cm2()K1400)(cm1(K-s-mJ216667.22××−= í15.5 W
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