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3 years ago

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The steel 4140 steel contains 0.4% C, however, it shows higher yield strength and ultimate strength than that of the 1045 (0.45%

C explain why)

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:

4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

Explanation:

we have given 4140 steel contains 0.4% C

we know here that 4140 steel is low steel alloy , and it have low amount of chromium , manganese etc alloying element

and these elements which are present in 4140 steel they increase yield strength and ultimate strength of steel

while in 1045 steel contains 0.45 % c is plain carbon steel

and it do not contain any alloying element

so that 4140 steel contains 0.4% C having higher yield strength and ultimate strength than the 1045 steel contains 0.45% C

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Compute the number of kilo- grams of hydrogen that pass per hour through a 6-mm-thick sheet of palladium having an area of 0.25

nydimaria [60]

Answer:

The number of kilo- grams of hydrogen that pass per hour through this sheet of palladium is $4.1 * 10^{-3} \frac{kg}{h}$

Explanation:

Given

x1 = 0 mm

x2 = 6 mm = 6 * $10^{-3}$ m

c1 = 2 kg/ $m^{3}$

c2 = 0.4 kg/ $m^{3}$

T = 600 °C

Area = 0.25 $m^{2}$

D = 1.7 * $10^{8} m^{2}/s$

First equation

J = - D $\frac{c1 - c2}{x1 - x2}$

Second equation

J = $\frac{M}{A*t}$

To find the J (flux) use the First equation

J = - 1.7 * $10^{8} m^{2}/s$ * $\frac{2 kg/m^{3} - 0.4 kg/m^{3}}{0 - 6 * 10^{-3} } = 4.53 * 10^{-6} \frac{kg}{m^{2}s }$

To find M use the Second equation

$4.53 * 10^{-6} \frac{kg}{m^{2}s}$ = $\frac{M}{0.25 m^{2} * 3600s/h}$

M = $4.1 * 10^{-3} \frac{kg}{h}$

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3 years ago

Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.

vekshin1

Solution :

<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>

3 1.62

2 2.17

$1\frac{1}{2}$ 3.62

$\frac{3}{4}$ 2.27

$\frac{3}{8}$ 1.38

PAN 0.21

Given :

Sieve weight % wt. retain % cumulative % finer

size retained wt. retain

No. 4 59.5 10.225% 10.225% 89.775%

No. 8 86.5 14.865% 25.090% 74.91%

No. 16 138 23.7154% 48.8054% 51.2%

No. 30 127.8 21.91% 70.7154% 29.2850%

No. 50 97 16.6695% 87.3849% 12.62%

No. 100 66.8 11.4796% 98.92% 1.08%

Pan <u> 6.3 </u> 1.08% 100% 0%

581.9 gram

Effective size = percentage finer 10% ( $$$D_{20}$ )

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x , 10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm) , 51.2%

N 8 (2.38 mm) , 74.91%

x, 60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

$Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4 and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

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