Answer:
Explanation:
Given conditions
1)The stress on the blade is 100 MPa
2)The yield strength of the blade is 175 MPa
3)The Young’s modulus for the blade is 50 GPa
4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.
5)The temperature of the blade is 800°C.
6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)
where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K
Young Modulus, E = Stress, [tex]\sigma[/tex] /Strain, ∈
initial Strain, [tex]\epsilon_i = \frac{\sigma}{E}[/tex]
[tex]\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}[/tex]
[tex]\epsilon_i = 0.002[/tex]
creep rate in the steady state
[tex]\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )[/tex]
[tex]\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )[/tex]
but Tinitial = 0
[tex]\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001[/tex]
[tex]\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}[/tex]
solving the above equation,
we get
Tfinal = 2459.82 hr
