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

A study reports data on the effects of the drug tamoxifen on change in the level of cortisol-binding globulin (CBG) of patients during treatment. With age = x and ACBG = y, summary values are n = 26, Σx, = 1612, Σ(x, - x)² = 3756.96, Σy, = 281.9, Σ(y, - y)² = 465.34, and Ex,y,= 16,745. (a) Compute a 90% CI for the true correlation coefficient p. (Round your answers to four decimal places.) (b) Test Hop=-0.5 versus H: p< -0.5 at level 0.05. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. ◇ Reject Ho. There is no evidence that p < -0.5. ○ Fail to reject Ho. There is evidence that p < -0.5. Reject Ho. There is evidence that p < -0.5. Fail to reject Ho. There is no evidence that p < -0.5. (c) In a regression analysis of y on x, what proportion of variation in change of cortisol-binding globulin level could be explained by variation in…

For the frame and loading shown, determine the reactions at A and C. 24 Last 2 student ID+50 lbs 24 A 3 in. B A=Last 2 student ID+10 Inch B=Last 2 student ID+40 Inch A B D

The figures below shows the framing plan and section of a reinforced concrete floor system. Floor beams are shown as dotted lines. The weight of the ceiling and floor finishing is 6 psf, that of the mechanical and electrical systems is 7 psf, and the weight of the partitions is 180 psf. The floor live load is 105 psf. The 7 in. thick slab exterior bay (S-1) is reinforced with #5 rebars @ 10 in. o.c. as the main positive reinforcement at the mid span, and #4 @ 109 in. for the shrinkage and temperature reinforcement. The panel is simply supported on the exterior edge and monolithic with the beam at the interior edge. Check the adequacy of the slab. Use the ACI moment coefficients. fc’ = 6,000 psi and fy = 60,000 psi. The slab is in an interior location. Hint: • Estimate total dead load. Find factored maximum positive bending moment in the end span. • Find design positive moment capacity. • Compare and determine adequacy, including safety and economy. C D

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