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alekssr [168]
3 years ago
7

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

C. Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0.0 wt% C. At what position will the carbon concentration be 0.215 wt% after a 7 h treatment? The value of D at 1200°C is 7.5 × 10-11 m2/s.
z erf(z) z erf(z) z erf(z)
0.00 0.0000 0.55 0.5633 1.3 0.9340
0.025 0.0282 0.60 0.6039 1.4 0.9523
0.05 0.0564 0.65 0.6420 1.5 0.9661
0.10 0.1125 0.70 0.6778 1.6 0.9763
0.15 0.1680 0.75 0.7112 1.7 0.9838
0.20 0.2227 0.80 0.7421 1.8 0.9891
0.25 0.2763 0.85 0.7707 1.9 0.9928
0.30 0.3286 0.90 0.7970 2.0 0.9953
0.35 0.3794 0.95 0.8209 2.2 0.9981
0.40 0.4284 1.0 0.8427 2.4 0.9993
0.45 0.4755 1.1 0.8802 2.6 0.9998
0.50 0.5205 1.2 0.9103 2.8 0.9999
Engineering
1 answer:
Fantom [35]3 years ago
8 0

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})

erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

erf(\frac{x}{2\sqrt{DT}}) = 0.751

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

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The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8 What is the probability of observing
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Answer:

0.14% probability of observing more than 4 errors in the carpet

Explanation:

When we only have the mean, we use the Poisson distribution.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

The number of weaving errors in a twenty-foot by ten-foot roll of carpet has a mean of 0.8.

This means that \mu = 0.8

What is the probability of observing more than 4 errors in the carpet

Either we observe 4 or less errors, or we observe more than 4. The sum of the probabilities of these outcomes is 1. So

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P(X > 4) = 1 - P(X \leq 4)

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