Question

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the blade and the turbine housing. Given that the blade cannot hit the housing, determine the time to failure of the blade considering the following information.1)The stress on the blade is 100 MPa.2)The yield strength of the blade is 175 MPa3)The Young’s modulus for the blade is 50 GPa4)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

Answer 1

Step-by-step 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,
`\sigma` /Strain, ∈

initial Strain,
`\epsilon_i = (\sigma)/(E)`

`\epsilon_i = (100* 10^(6) Pa)/(50* 10^(9) Pa)`

`\epsilon_i = 0.002`

creep rate in the steady state

`(\delta \epsilon)/(\delta t) = (1 * {10}^(-5))\sigma^4 exp^((-2eV)/(kT) )`

`(\epsilon_(initial) - \epsilon _(primary))/(t_(initial)-t_(final)) = 1 * 10^(-5)(100)^(4)exp((-2eV)/(8.62*10^(-5)((eV)/(K) )(800+273)K) )`

but Tinitial = 0

`\epsilon_(initial) - \epsilon _(primary)} = 0.002 - 0.003 = -0.001`

`(-0.001)/(-t_(final)) = 1 * 10^(-5)(100)^(4)* 10^{((-2eV)/(8.62*10^(-5)((eV)/(K) )1073K) )}`

solving the above equation,

we get

Tfinal = 2459.82 hr