Instructors Solutions Manual PIEV 230 Problem 712 Solution For stabilized

Instructors solutions manual piev 230 problem 712

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Instructors Solutions Manual (PIEV) 230 Problem 7.12 Solution For 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 k at 1500° C k = 100 1000 0 . 2 3 . 2 ) 100 1500 ( 0 . 2 + = 2.433333 J / (m-s-K) Use the average value of k over the temperature range in the wall. k = 2 433333 . 2 0 . 2 + = 2.216667 J / (m-s-K) Rearrange the equation for k to solve for t Q / . k = ) / ( / x T A t Q t Q = x T kA = cm 100 m 1 ) cm 2 ( ) K 1400 )( cm 1 ( K - s - m J 216667 . 2 2 × × = í 15.5 W
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Instructors Solutions Manual (PIEV) 231 Section 7.4 – Thermal Shock
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Instructors Solutions Manual (PIEV) 232 Problem 7.14 Solution For magnesia brick α = 13.5×10 í 6 /°C and E = 210 GPa (sintered) For constrained thermal expansion thermal ε = int constra ε or T α = E σ . Solve for σ σ = TE α = (13.5×10 í 6 /°C)(1200°C í 25°C)(210 GPa) = 3331 MPa Problem 7.15 Solution For silica glass α = 0.5×10 í 6 /°C and E = 72.4 GPa. For constrained thermal expansion thermal ε = int constra ε or T α = E σ . Solve for σ σ = TE α = (0.5×10 í 6 /°C)(1200°C í 25°C)(72.4 GPa) = 42.5 MPa Problem 7.13 Solution For mullite α = 5.3×10 í 6 /°C and E = 69 GPa.
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