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3 years ago

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At 500oC, the diffusion coefficient of Cu in Ni is 1.3 x 10-22 m2/s, and the activation energy for diffusion is 256,000 J/mol. F

rom this information, determine the diffusion coefficient of Cu in Ni at 900 oC . (R = 8.314 J/(mol K)) 0oC = 273 K

1 answer:

ipn [44]3 years ago

5 0

Answer:

the diffusion coefficient at 900°C is D₂=8.9*10⁻¹⁸ m²/s

Explanation:

The dependence of the diffusion coefficient is similar to the dependence of chemical reaction rate with respect to temperature , where

D=D₀*e^(-Q/RT)

where

D₀= diffusion energy at T=∞

Q= activation energy

T= absolute temperature

D= diffusion coefficient at temperature T

R=ideal gas constant

for temperatures T₁ and T₂

D₁=D₀*e^(-Q/RT₁)

D₂=D₀*e^(-Q/RT₂)

dividing both equations and rearranging terms we get

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)]

replacing values

D₂=D₁*e^[-Q/R(1/T₂-1/T₁)] = 1.3*10⁻²² m²/s*e^(-256,000 J/mol/8.314 J/(mol K)*(1/1073K-1/773K) = 8.9*10⁻¹⁸ m²/s

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Answer:

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$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

b) For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

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For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

c) A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

Explanation:

From the info given we know that a computer running a program that requires 250 s, with 70 s spent executing FP instructions, 85 s executed L/S instructions and 40 s spent executing branch instructions.

Part 1

For this case the new time to run the FP operation would be reduced 20% so that means 100-20% =80% from the original time

$(1-0.2)*70 s =56s$

The reduction on this case is $70-56 s=14s$

And since the new total time would be given by $250-14=236 s$

Part 2

For this case the total time is reduced 20% so that means that the new total time would be (1-0.2)=0.8 times the original total time $(1-0.2) *250s =200 s$

The original time for INT operations is calculated as:

$250 = 70+85+40 +t_{INT}$

$t_{INT}=55s$

For this part the only time that was changed is assumed the INT operations so then:

$200 = 70+85+40 \Delta t_{INT}$

And then: $\Delta t_{INT}= 200-70-85-40=5 s$

And we can quantify the decrease using the relative change:

$\% Change = \frac{5s}{55 s} *100 = 9.09\%$ of reduction

Part 3

A reduction of the total time implies that the total time would be 205 s from the results above. And the time for FP is 70, for L/S is 85 and for INT operations is 55 s, so then if we add 70+85+55=210s, we see that 210>205 so then we cannot reduce the total time 20% just reducing the branch intructions.

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3 years ago

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luda_lava [24]

Answer: ALL CAREFULLY ANSWERED CORRECTLY.

Explanation:

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A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 54 percent. Determine the rate of hea

Gennadij [26K]

Answer:

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Explanation:

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3 years ago

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Over [174]

Answer:

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Explanation:

Given data:

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discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

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$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

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3 years ago