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Slav-nsk [51]
3 years ago
7

A. Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R.

Engineering
1 answer:
Radda [10]3 years ago
3 0

Answer:

A) i) LD_110 = √3/(4R√2)

ii) LD_111 = 1/(2R)

B)i) LD_110 = 2.4 × 10^(9) m^(-1)

ii) LD_111 = 4 × 10^(9) m^(-1)

Explanation:

A) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;

√((4R)² - (4R/√3)²)

This reduces to; 4R√(1 - 1/3) = 4R√(2/3)

Now, the expression for the linear density of this direction is;

LD_110 =

Number of atoms centered on (110) direction/vector length of 110 direction

In this case, there is only one atom centred on the 110 direction. Thus;

LD_110 = 1/(4R√(2/3))

LD_110 = √3/(4R√2)

ii) The length of the vector for the direction 111 is equal to 4R, since

all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;

LD_111 = 2/(4R) = 1/(2R)

B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear

density for the [110] direction is;

LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))

LD_110 = 2.4 × 10^(9) m^(-1)

ii) for the 111 direction, we have;

LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))

LD_111 = 4 × 10^(9) m^(-1)

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Answer:

He wore his black suit, another color of shirt (not purple) and shoes

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Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

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3.8 LAB - Select lesson schedule with multiple joins
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The database has three tables for tracking horse-riding lessons: Horse with columns: ID - primary key; RegisteredName; Breed; Height; BirthDate.

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2 years ago
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
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Answer:

Explanation:

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C = 4.80 μF, R = 301 Ω resistor. V = 120volts

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= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

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Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra
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Answer:

a) T_{2}=837.2K

b) e=91.3 %

Explanation:

A) First, let's write the energy balance:

W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})  (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 \frac{kJ}{kgK} And its specific R constant is 0.287 \frac{kJ}{kgK}.

The only unknown from the energy balance is T_{2}, so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K

B) The isentropic efficiency (e) is defined as:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}

Where {h_{2s} is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because h_{2}-h_{1} can be obtained from the energy balance  \frac{W}{m}=h_{2}-h_{1}

h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}

An entropy change for an ideal gas with  constant Cp is given by:

s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})

Applying logarithm properties:

ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}

Then,

T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K

So, now it is possible to calculate h_{2s}-h_{1}:

h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}

Finally, the efficiency can be calculated:

e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3 %

4 0
3 years ago
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