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SCORPION-xisa
29 days ago
10

(a) For BCC iron, calculate the diameter of the minimum space available in an octahedral site at the center of the (010) plane,

and compare this to the diameter of a carbon atom. Assume that the iron atoms in the BCC structure are hard spheres that touch along the [111] direction.
(b) Compare the space available with the size of a carbon atom from Appendix C, and comment about the potential solubility of carbon in BCC iron in octahedral interstitial sites.
Engineering
1 answer:
choli [191]29 days ago
7 0

Response:

a) The diameter available is 0.0384 nm

b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.

Clarification:

For BCC iron

According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm

BCC iron encompasses 4 atomic radii, thus the body diagonal length = a(3)^\frac{1}{2}

which represents the atomic radius of BCC iron

4r = a(3)^\frac{1}{2}

Substituting the value of (a) from Appendix B, set as 0.2866 nm

4r = 0.2866 nm (3)^\frac{1}{2}

leading to  r =  0.4964 nm / 4 = 0.1241 nm

Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.

Thus, the radius ratio = 0.62

According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm

The diameter of the minimum space available

d_{a} = a - r_{a}

r_{a} = atomic radii = 0.2482 nm

With a = 0.2666 nm

therefore

d_{a} = 0.2866 nm - 0.2482 nm = 0.0384 nm

When comparing with the diameter of a carbon atom

This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.

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A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o
Daniel [215]

Answer:

The change in entropy of the steam is 2.673 kJ/K

Explanation:

The mass of the liquid-vapor mixture is 1.5 kg

The mass in the liquid phase is calculated as 3/4 × 1.5 kg = 1.125 kg

The mass in the vapor phase is calculated as 1.5 - 1.125 = 0.375 kg

According to the steam tables

At a pressure of 200 kPa (200/100 = 2 bar), the specific entropy of steam is found to be 7.127 kJ/kgK

The entropy of steam can be calculated as specific entropy multiplied by mass = 7.127 × 0.375 = 2.673 kJ/K

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1 day ago
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A technician has been dispatched to assist a sales person who cannot get his laptop to display through a projector. The technici
mote1985 [204]

Answer:Ensure the correct cable is connected between the laptop and the projector. Check for HDMI inputs or 15-pin video output interfaces.

Also, make sure the laptop is set to project to the correct display output.

Explanation:

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21 day ago
estimate the area for a landfill for 12000 p producing waste for 10 y. assume that the national average is
alex41 [274]

Answer:

1.015 ha.

Explanation:

To calculate the landfill area required for 12,000 people producing waste over 10 years, follow these steps:[STEP ONE: Calculate the average solid waste generated per person per year (kg p^-1 ^y(kg/py)).

According to the problem, the average solid waste produced is 2.78 kg per person daily (kg/pd), hence converting to kg/py involves:

2.78 × 365 days = 1014.7 kg/py.

STEP TWO: Determine yearly volume of refuse per person.

Thus, volume = 1014.7 kg/py ÷ 500 kg/m^3 = 2.03 m^3 per person per year.

STEP THREE: Calculate total solid waste volume over 10 years for 12,000 individuals.

Total waste volume over 10 years = 10 × 12,000 × 2.03 = 243,600 m^3.

STEP FOUR: Find the required area for the landfill.

Note: The total height for the landfill should be 20 + 4 = 24m.

Thus, the area for the landfill = 243,600 m^3 / 24m = 10,150 m^2.

If 10,000 m^2 equals 1 ha, then 10,150 m^2 ÷ 10,000 m^2 = 1.015 ha.

(f). Ensure to expand the landfill area for enhancements.

4 0
21 hour ago
The rate of flow of water in a pump installation is 60.6 kg/s. The intake static gage is 1.22 m below the pump centreline and re
mote1985 [204]

Answer:

The power of the pump is 23.09 kW.

Explanation:

Parameters

gravitational constant, g = 9.81 m/s^2

mass flow rate, \dot{m} = 60.6 kg/s

flow density, \rho = 1000 kg/m^3

efficiency of the pump, \eta = 0.74

output gauge pressure, p_o = 344.75 kPa

input gauge pressure, p_i = 68.95 kPa

cross-sectional area of output pipe, A_o = 0.069 m^2

cross-sectional area of input pipe, A_i = 0.093 m^2

height of discharge, z_o = 1.22 m - 0.61 m = 0.61 m (evaluated at pump’s maximum height of 1.22 m)

input height, z_i = 0 m

hydraulic power of the pump,P =? kW

Initially, the volumetric flow (Q) needs to be determined

Q = \frac{\dot{m}}{\rho}

Q = \frac{60.6 kg/s}{1000 kg/m^3}

Q = 0.0606 m^3/s

Next, compute the velocity (v) for both input and output

v_o = \frac{Q}{A_o}

v_o = \frac{0.0606 m^3/s}{0.069 m^2}

v_o = 0.88 m/s

v_i = \frac{Q}{A_i}

v_i = \frac{0.0606 m^3/s}{0.093 m^2}

v_i = 0.65 m/s

Subsequently, the total head (H) can be calculated

H = (z_o - z_i) + \frac{v_o^2 - v_i^2}{2 \, g} + \frac{p_o - p_i}{\rho \, g}

H = (0.61 m - 0 m) + \frac{{0.88 m/s}^2 - {0.65 m/s}^2}{2 \, 9.81 m/s^2} + \frac{(344.75 Pa-68.95 Pa)\times 10^3}{1000 kg/m^3 \, 9.81 m/s^2}

H = 28.74m

Finally, the computation of pump power is done as follows

P = \frac{Q \, \rho \, g \, H}{\eta}

P = \frac{0.0606 m^3/s \, 1000 kg/m^3 \, 9.81 m/s^2 \, 28.74m}{0.74}

P = 23.09 kW

6 0
25 days ago
A hydrogen-filled balloon to be used in high altitude atmosphere studies will eventually be 100 ft in diameter. At 150,000 ft, t
mote1985 [204]

Answer:

The calculated result is 11.7 ft

Explanation:

You can apply the combined gas law, which incorporates Boyle's law, Charles's law, and Gay-Lussac's Law, because hydrogen demonstrates ideal gas behavior under these specific conditions.

\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}

where the subscripts indicate "p" for pressure, "V" for volume, and "T" for temperature (in Kelvin) at varying moments. Let's denote t_1 as the balloon at 150,000 ft so

p_1 = 0.14 \ lb/in^2

V_1 = \frac{4}{3} \pi R_1^3 = 523598.77 \ ft^3

and T_1 = -67^\circ F = 218.15\ K.

Then t_2 represents the point at which the balloon is on the ground.

p_2 = 14.7 \ lb/in^2 and T_2 = 68^\circ F = 293.15\ K.

Based on the first equation

V_2 = \frac{p_1 V_1 T_2}{T_1 p_2}, we find

V_2 = 6701.07 ft^3 and consequently the radius turns out to be

R_2 = \sqrt[3]{\frac{3 V_2}{4 \pi}} = 11.7 \ ft.

5 0
5 days ago
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