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

Answer:

a) Q = 251.758 kJ/mol

b) creep rate is    $= 1.751 \times 10^{-5} \% per hr$

Explanation:

we know Arrhenius expression is given as

$\dot \epsilon =Ce^{\frac{-Q}{RT}$

where

Q is activation energy

C is pre- exponential constant

At 700 degree C creep rate is$\dot \epsilon = 5.5\times 10^{-2}$% per hr

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

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

$\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}$

$\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}]$

solving for Q we get

Q = 251.758 kJ/mol

b) creep rate at 500 degree C

we know

$C = \epsilon e^{\frac{Q}{RT}}$

    $=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr$

$\epsilon_{500} = C e^{\frac{Q}{RT}}$

                         $= 1.804 \times 10^{12}  e{\frac{251758}{8.314(500+273}}$

                         $= 1.751 \times 10^{-5} \% per hr$