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MrRa [10]
3 years ago
7

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% pe

r 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.
Engineering
1 answer:
LenaWriter [7]3 years ago
4 0

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

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