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An ideal gas is contained in a closed assembly with an initial pressure and temperature of

adell [148]

Answer:

Two identical containers each of volume V 0 are joined by a small pipe. The containers contain identical gases at temperature T 0 and pressure P 0 .One container is heated to temperature 2T 0 while maintaining the other at the same temperature. The common pressure of the gas is P and n is the number of moles of gas in container at temperature 2T 0

Explanation:

3 0

3 years ago

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

weeeeeb [17]

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

5 0

3 years ago

Please help me with this. Picture

qwelly [4]

Answer:

^547.'":8765456\⇆⊇⊇ys p)(ay^547.'":8765456\⇆⊇⊇ys p)(*

iuytrT&*7654567iol;';";lk↓ωω*&65∴∀Hgtre

6и876&*b n™Ο65656&^cxCv876и876&*b n™‰‰ay^547.'":87765^&* kIKUYtb

Explanation:

=yx^z3

ÜÜÜ

5 0

3 years ago

what is the expected life 1 inch diameter bar machined from AISI 1020 CD Steel is subjected to alternating bending stress betwee

Alexeev081 [22]

Answer:

1.287 *10⁷ cycles.

Explanation:

See attached pictures.

3 0

3 years ago

There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains

Luba_88 [7]

Answer:

16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

(4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

__

<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

y = 8 -0.25x . . . . . . height in feet after x minutes (tank B)

These will be equal when ...

y = y

12 -0.5x = 8 -0.25x

4 = 0.25x . . . . . . . . . . add 0.5x -8

16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0

2 years ago