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Andrews [41]
3 years ago
13

BCC lithium has a lattice parameter of 3.5089 3 10–8 cm and contains one vacancy per 200 unit cells. Calculate (a) the number of

vacancies per cubic centimeter; and (b) the density of Li.
Engineering
1 answer:
Tanya [424]3 years ago
4 0

(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b) ρ = n X (AM) / v X Nₐ

<u>Explanation:</u>

<u />

Given-

Lattice parameter of Li  = 3.5089 X 10⁻⁸ cm

1 vacancy per 200 unit cells

Vacancy per cell = 1/200

(a)

Number of vacancies per cubic cm = ?

Vacancies/cm³ = vacancy per cell / (lattice parameter)³

Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³

Vacancies/cm³ = 1.157 X 10²⁰

Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰

(b)

Density is represented by ρ

ρ = n X (AM) / v X Nₐ

where,

Nₐ = Avogadro number

AM = atomic mass

n = number of atoms

v = volume of unit cell

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A cartridge electrical heater is shaped as a cylinder of length L=200mm and outer diameter D=20 mm. Under normal operating condi
lara31 [8.8K]

Answer:

T(water)=50.32℃

T(air)=3052.6℃

Explanation:

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To solve this problem we must use the equation that defines the transfer of heat by convection, which consists of the transport of heat through fluids in this case water and air.

The equation is as follows!

Q=ha(Ts-T\alpha )

Q = heat

h = heat transfer coefficient

Ts = surface temperature

T = fluid temperature

a = heat transfer area

The surface area of ​​a cylinder is calculated as follows

a=\pi D(\frac{D}{2} +L)

Where

D=diameter=20mm=0.02m

L=leght=200mm)0.2m

solving

a=\pi (0.02)(\frac{0.02}{2} +0.2)=0.01319m^2

For water

Q=2Kw=2000W

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a=0.01319m^2

Tα=20C

Q=ha(Ts-T\alpha )

solving for ts

Ts=T\alpha +\frac{Q}{ha}

Ts=20+\frac{2000}{(0.01319)(5000)} =50.32C

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Ts=20+\frac{2000}{(0.01319)(50)}=3052.6C

3 0
3 years ago
A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is
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Given

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Q=hA\left ( \Delta T\right )

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7 0
3 years ago
The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo
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Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)

consider the equation of bernoulli at the top of the pool

P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)

where P_1=P_0 atm pressure

At the top of the pool v_1=0m/s, substitute in V_1 in equation (2)

P_0+\rho gh_1 =constant ...(3)

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)

where P_2=P_0 atm pressure and h_2=0m

P_0+\rho v_1^2 =constant ...(6)

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

        P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)    

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e.) consider the equation of flow rate interval of v and t

flow(t)=\frac{dv}{dt}(m^3/s) hence this is the flow rate

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hence this serves as the cross sectional area.

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6 0
2 years ago
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Vinvika [58]

Answer:

Energy produce in one year =20.49 x 10¹⁶ J/year

Explanation:

Given that

Plant produce 6.50 × 10⁸ J/s of energy.

It produce  6.50 × 10⁸ J in 1 s.

We know that

1 year = 365 days

1 days = 24 hr

1 hr = 3600 s

1 year = 365 x 24 x 3600 s

1 year = 31536000 s

So energy produce in 1 year = 31536000 x  6.50 × 10⁸ J/year

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7 0
3 years ago
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