The activation enthalpy between 973 K and 1053 K .

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

During a creep test (constant stress equal to 3.5 MPa) a material has an initial strain of 10 and a strain of 2 x 10* after 30 days. The same material was also submitted to a relaxation test (constant strain equal to 1.5 x 104). During the relaxation test, determine the stress after 30 days. Assume that the material follows the Maxwell model. 4 X

Which of the following statements is incorrect? Both the grain boundary and bulk diffusion creep mechanism lead to elongated grains. For linear creep, the creep exponent in the steady-state creep rate equation is around 1. S-N curves are a plot of the stress versus the number of cycles to failure. The relaxation modulus is a time-dependent elastic modulus used for viscoelastic polymers. Crystalline polymers show a large drop in the relaxation modulus around the glass transition temperature.

Triangular wedges stress condition.

Answer 2

Question

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Answer 6

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Answer 7

Chapter 9, Problem 9.4P

To determine

The activation enthalpy between 973 K and 1053 K .

Answer 8

Expert Solution & Answer

Answer to Problem 9.4P

The activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The logarithm of stress is 7.10 .

Concept used:

Write the expression for tensile stage II creep rate.

dεIIdt=Aσnexp−(ΔHpkT) …… (1)

Here, dεIIdt is the tensile stage II creep rate, A is the constant and σ is the stress, ΔHp is the activation enthalpy, k is the Boltzmann constant and T is the temperature.

Calculation:

Refer to Figure 9.5 “The log of the stage II creep strain rate of zirconium as function of the log of stress, at three temperatures” to obtain the value of strain rate corresponding to logarithm of stress as 7.10 at 973 K and 1053 K as 1.0×10−5 s−1 and 1.55×10−4 s−1 respectively.

Substitute 1.0×10−5 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 973 K for T in equation (1).

1.0×10−5 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 973 K)) …… (2)

Substitute 1.55×10−4 s−1 for dεIIdt , 8.62×10−5 eV/atom⋅K for k and 1053 K for T in equation (1).

1.55×10−4 s−1=A(σ)nexp−(ΔHp( 8.62× 10 −5 eV/atom⋅K)( 1053 K)) …… (3)

Divide equation (2) by equation (3).

1.0× 10 −5 s −11.55× 10 −4 s −1=A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 973 K ) )A ( σ )nexp−( Δ H p ( 8.62× 10 −5 eV/atom⋅K )( 1053 K ) )0.0645=exp( −Δ H p ( 8.62× 10 −5 eV/atom⋅K )[ 1 973 K− 1 1073 K])ΔHp=−ln( 0.0645)( 8.62× 10 −5 eV/atom⋅K)0.8× 10 −4 K −1ΔHp=2.953 eV/atom

Conclusion:

Thus, the activation enthalpy between 973 K and 1053 K for zirconium is 2.953 eV/atom .

Answer 12

Ch. 9 - Prob. 1CQ Ch. 9 - Prob. 2CQ Ch. 9 - Prob. 3CQ Ch. 9 - Prob. 4CQ Ch. 9 - Prob. 5CQ Ch. 9 - Prob. 6CQ Ch. 9 - Prob. 7CQ Ch. 9 - Prob. 8CQ Ch. 9 - Prob. 9CQ Ch. 9 - Prob. 10CQ

The activation enthalpy between 973 K and 1053 K .

Videos

The activation enthalpy between 973 K and 1053 K .

Answer to Problem 9.4P

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions