Question

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% per hour at 800°C and 0.055% per hour at 700°C.

(a) Calculate the activation energy for creep in this temperature range.
(b) Estimate the creep rate to be expected at the service temperature of 500°C.

Answer 1

a) Q = 251.758 kJ/mol

b) creep rate is
`= 1.751 * 10^(-5) \% per hr`

Step-by-step explanation:

we know Arrhenius expression is given as

`\dot \epsilon =Ce^{(-Q)/(RT)`

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is
`\dot \epsilon = 5.5* 10^(-2)`% per hr

At 800 degree C creep rate is
`\dot \epsilon = 1`% per hr

activation energy for creep is
`(\epsilon_(800))/(\epsilon_(700))` =
`= \frac{C* e^{(-Q)/(R(800+273))}}{C* e^{(-Q)/(R(700+273))}}`

`(1\%)/(5.5 * 10^(-2)\%) = e^{[(-Q)/(R(800+273))] -[(-Q)/(R(800+273))]}`

`(0.01)/(5.5* 10^(-4)) = ln [e^{(Q)/(8.314)[(1)/(1073) - (1)/(973)]}]`

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

`C = \epsilon e^{(Q)/(RT)}`

`=- 1\% e{(251758)/(8.314(500+273)} = 1.804 * 10^(12) \% per hr`

`\epsilon_(500) = C e^{(Q)/(RT)}`

`= 1.804 * 10^(12)  e{(251758)/(8.314(500+273)}`

`= 1.751 * 10^(-5) \% per hr`