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Valentin [98]
3 years ago
6

An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein t

he surface carbon concentration is maintained at 1.0 wt%.
If after 50 h the concentration of carbon is 0.35 wt% at a position 3.5 mm below the surface, determine the temperature at which the treatment was carried out.
Engineering
1 answer:
weeeeeb [17]3 years ago
5 0

Answer:

the  temperature T at which the treatment is carried is out is  1274.24 K

Explanation:

Fick's second Law posits that the rate of change of concentration of diffusing species is directly proportional to the second derivative of the concentration.

Using the expression of the Fick's second Law:

\mathbf{\frac{C_x-C_o}{C_s-C_o} = 1- erf(\frac{x}{2\sqrt{Dt} })}

where;

C_o = initial concentration

C_x = the depth of the concentration

C_s = surface concentration

erf(\frac{x}{2\sqrt{Dt} })} = Gaussian error function.

Let variable z be used for the expression of the Gaussian error function.  erf(\frac{x}{2\sqrt{Dt} })}

Then, from the above equation:  replacing C_x with 0.35 ; C_o  with 0.2 and C_s  with 1.0; we have:

\mathbf{\frac{0.35-0.2}{1.0-0.2} = 1- erf( z)}

erf (z) = 0.8125

we obtain the error function value close to 0.8125 from the error function table and we did the  interpolation to obtain the exact value of variable  corresponding to 0.8125.

The table below shows the tabular form of the error function value close to 0.8125 .

Value for z                                                   Value for erf (z)

0.9                                                                0.797

z                                                                    0.8125

0.950                                                            0.8209  

From above; we can find  the value of variable  corresponding to the error function 0.8125 .

i.e

\frac{z-0.9}{0.95-0.9} =\frac{0.8125-0.797}{0.8209-0.797}

z = 0.932

However, the temperature dependence relation for the diffusion coefficient D can be expressed as:

z = \frac{x}{\sqrt{Dt} }

where;

z = 0.932

x = 3.5 mm = 0.0035 m

t = 50 h = 180000 sec

0.932 = \frac{0.35}{2\sqrt{D*180000} }

D = 1.958*10^{-11} m^2/s

Finally, the temperature T at which the treatment is carried is out is calculated as:

\mathbf{D=D_o \ exp \ (-\frac{Q_d}{RT}) }

From the table ‘Diffusion data’, we  obtain the values of temperature-independent pre exponential and activation energy for diffusion of carbon in FCC Fe.

D_o = 2.3*10^{-5} \ m^2/s

Q_d = 148, 000 \ J/mol

Replacing all values needed for the above equation; we have:

1.958*10^{-11}= (2.3*10^{-5})exp(\frac{-148,000}{(8.31)T})

8.51*10^{-7}=exp(\frac{-17,810}{T})

In(8.51*10^{-7})=(\frac{-17,810}{T})

-13.977 = -17,810/T

T = -17,810/ - 13.977

T = 1274.24 K

Hence, the  temperature T at which the treatment is carried is out is  1274.24 K

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Answer:

Local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.

Explanation:

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5 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
4 years ago
For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value
OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

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

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

0.50 = 1 - e^{-k^146^2.1}

0.50 = e^{-k306.6}

taking natural log on both side

ln(0.5) = -k(306.6)

k = 2.26\times 10^{-3}

for 86 % completion

0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}

e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14

-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)

t^{2.1} = 869.96

t = 25.10 sec

5 0
3 years ago
Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w
Dafna11 [192]

Answer:

\mathbf{h_2 =1021.9 \  mm}

Explanation:

Given that :

The initial volume of water V_1 = 1000.00 mL = 1000000 mm³

The initial temperature of the water  T_1 = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water T_2 = 70° C

The coefficient of thermal expansion for the glass is  ∝ = 3.8*10^{-6 } mm/mm  \ per ^oC

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature T_1 = 10 ^ 0C  the density of the water is \rho = 999.7 \ kg/m^3

At temperature T_2 = 70^0 C  the density of the water is \rho = 977.8 \ kg/m^3

The mass of the water is  \rho V = \rho _1 V_1 = \rho _2 V_2

Thus; we can say \rho _1 V_1 = \rho _2 V_2;

⇒ 999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2

V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }

V_2 = 1022.40 \ mL

v_2 = 1022400 \ mm^3

Thus, the volume of the water after heating to a required temperature of  70^0C is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

V_1 = A_1 *h_1

The area of the water before heating is:

A_1 = \dfrac{V_1}{h_1}

A_1 = \dfrac{1000000}{1000}

A_1 = 1000 \ mm^2

The area of the heated water is :

A_2 = A_1 (1  + \Delta t  \alpha )^2

A_2 = A_1 (1  + (T_2-T_1) \alpha )^2

A_2 = 1000 (1  + (70-10) 3.8*10^{-6} )^2

A_2 = 1000.5 \ mm^2

Finally, the depth of the heated hot water is:

h_2 = \dfrac{V_2}{A_2}

h_2 = \dfrac{1022400}{1000.5}

\mathbf{h_2 =1021.9 \  mm}

Hence the depth of the heated hot  water is \mathbf{h_2 =1021.9 \  mm}

4 0
3 years ago
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Dennis_Churaev [7]

Answer:

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Explanation:

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