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levacccp [35]
3 years ago
14

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)
Engineering
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}

4 0
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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