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# 10° T(20 + log t,)X°R-h) Using the Larson-Miller data for the S-590 alloy, predict the time to rupture for a component...

10° T(20 + log t,)X°R-h) Using the Larson-Miller data for the S-590 alloy, predict the time to rupture for a component that is subjected to a 100 stress of 140 MPa at 800C. 25 30 35 40 45 50 1000 100 10 10 12 16 20 24 28 10 T(20 + log t,(K-h) Stress (MPa) (Isd ot) ssaAs
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Given:

At 140 MPa value of the Larsson-miller parameter is 24 x 103

T (20+log tr ) = 24 x 103

1073 (20+log t) = 24 x 10

20+log t$\frac{24×{10}^{3}}{1073}=22.367$

log tr = 2.367

tr = (10)2.367

tr = 232.91 h or 9.7 days

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