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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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Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe
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Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

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Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

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∂1 = 150x10³x0.075/2x0,0001

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∂2 = 1/2[84.37-28.33]

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This is a 2 d diagram donut is analyzed in 2 direction.

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A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the
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Answer:

Explanation:

Given data in question

mean stress  = 50 MPa

amplitude stress  = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first  maximum stress  and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

\sigma _{m} =  \sigma _{maximum} + \sigma _{minimum}  / 2

50 =  \sigma _{maximum} + \sigma _{minimum}  / 2    ...........1

and

225 =  \sigma _{maximum} + \sigma _{minimum}  / 2      ...........2

from eqution 1 and 2 we get maximum and minimum stress

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and \sigma _{minimum} = -175 MPa     ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

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ratio = -175 / 227

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now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = \sigma _{maximum} - \sigma _{minimum}  

magnitude = 275 - (-175) = 450 MPa

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