The activation enthalpy between 973 K and 1053 K .

Question

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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 $2.953 eV/atom$ .

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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 $2.953 eV/atom$ .

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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 $2.953 eV/atom$ .

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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 $2.953 eV/atom$ .

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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).

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