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3 years ago

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Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al

loy that initially contains 0.10 wt% C. The surface concentration is to be maintained at 0.90 wt% C, and the treatment is to be conducted at 1100°C. Use the diffusion data for γ-Fe.

1 answer:

Free_Kalibri [48]3 years ago

7 0

Answer:

idk

Explanation:

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Water flows through a horizontal plastic pipe with a diameter of 0.15 m at a velocity of 15 cm/s. Determine the pressure drop pe

Sonja [21]

Answer:0.1898 Pa/m

Explanation:

Given data

Diameter of Pipe $\left ( D\right )=0.15m$

Velocity of water in pipe $\left ( V\right )=15cm/s$

We know viscosity of water is $\left (\mu\right )=8.90\times10^{-4}pa-s$

Pressure drop is given by hagen poiseuille equation

$\Delta P=\frac{128\mu \L Q}{\pi D^4}$

We have asked pressure Drop per unit length i.e.

$\frac{\Delta P}{L} =\frac{128\mu \ Q}{\pi D^4}$

Substituting Values

$\frac{\Delta P}{L}=\frac{128\times8.90\times10^{-4}\times\pi \times\left ( 0.15^{3}\right )}{\pi\times 4 \times\left ( 0.15^{2}\right )}$

$\frac{\Delta P}{L}$ =0.1898 Pa/m

4 0

3 years ago

Pedro holds a heavy science book over his head for 10 minutes. Petro is doing work during that time. True or False

algol [13]

Answer:

True because he is working his arms to lift and hold the weight

Explanation:

4 0

3 years ago

For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

3 0

3 years ago

Read 2 more answers

A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b

Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition Cβ = 87

we use here lever rule that is

Wα = Wβ ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

$\frac{C_\beta - Co}{C_\beta - Ca}$ = 0.5

$\frac{87 - 57}{87 - Ca}$ = 0.5

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0

4 years ago

A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr

Hoochie [10]

Answer:

$\Delta m = 102.25\,kg$

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

$m_{o} = \rho_{air}\cdot V_{tank}$

$m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{o} = 31.25\,kg$

The final mass inside the rigid tank is:

$m_{f} = \rho \cdot V_{tank}$

$m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})$

$m_{f}= 133.5\,kg$

The supplied air mass is:

$\Delta m = m_{f}-m_{o}$

$\Delta m = 133.5\,kg-31.25\,kg$

$\Delta m = 102.25\,kg$

4 0

3 years ago