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Lilit [14]
3 years ago
11

For some metal alloy it is known that the kinetics of recrystallization obey the Avrami equation, and that the value of k in the

exponential is 2.60 x 10-6, for time in seconds. If, at some temperature, the rate of recrystallization is 0.0013 s-1, what total time (in s) is required for the recrystallization reaction to go to 90% completion?
Engineering
1 answer:
JulsSmile [24]3 years ago
5 0

Answer:

t = 1456.8 sec

Explanation:

given data:

contant k = 2.60*10^{-6}

rate of crystallization is 0.0013 s-1

rate of transformation is given by

r = \frac{1}{t_0.5}

use specifies value to solve t_0.5

it is ime required for 50% tranformation

r = \frac{1}{.0013}=769.2 sec

Avrami equation is given by

y = 1 - e^{-kt^n}

0.5 = 1 - e^{-kt_0.5^n}

1-0.5 = e^{-kt_0.5^n}

ln (1 - 0.5) = -kt_0.5^n

ln \frac{ln (1 - 0.5)}{-k} = nln t_0.5

n = \frac{ ln \frac{ln (1 - 0.5)}{-k}}{ln t_0.5}

n = \frac{ ln \frac{ln (1 - 0.5)}{-2.60*10^{-6}}}{ln 769.2}

n = 1.88

second degree of recrystalization may be determine by rearranging original avrami equation

t = [\frac{-ln(1-y)}{k}]^{1/n}

for 90%completion

t = [\frac{-ln(1-0.9)}{2.60*10^{-6}}]^{1/1.88}

t = 1456.8 sec

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3 years ago
According to fire regulations in a town, the pressure drop in a commercial steel, horizontal pipe must not exceed 2.0 psi per 25
bonufazy [111]

Answer:

6.37 inch

Explanation:

Thinking process:

We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.

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Using the Bernoulli equation for mass conservation:

\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1}  = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}

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\frac{P1-P2}{\rho }  = f\frac{l}{D} \frac{v^{2} }{2g}

The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.

Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F

from the tables

Re= 2.01 × 10⁵

Hence, f = 0.018

Therefore, pressure drop, (P1-P2)/p = 2.70 ft

This occurs at ae presure change of 1.17 psi

Correlating with the chart, we find that the diameter will be D= 0.513

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3 years ago
Consider that a system has two entities, Students, Instructors and Course. The Student has the following properties: student nam
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Explanation:

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3 years ago
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Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms  for f so that f has eight primary implicants with two literals and no other prime implicant are m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14} and m_{15}

b) The essential prime implicant are c' d',a'b',ab and cd

c) The minimum sum-of-product expression for f are

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Explanation:

The explanation is shown on the second third and fourth image

8 0
3 years ago
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