How many points do you need to define for the rectangle

The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical

Nookie1986 [14]

Solution:

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r = $\frac{a\sqrt{3} }{4}$

so, a = $\frac{4r}{\sqrt{3}}$

m = mass of each atom in a unit cell

mass of an atom = $\frac{M}{N_{A} }$ ,

where, $N_{A}$ is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ = $\frac{mass of unit cell}{volume of unit cell}$

density, ρ = $\frac{z\times M}{a^{3}\times N_{A}}$

ρ = $\frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}$

ρ = 10.215gm/ $cm^{3}$

5 0

4 years ago

11–17 A long, thin-walled double-pipe heat exchanger with tube and shell diameters of 1.0 cm and 2.5 cm, respectively, is used t

lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?

First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

$v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s$

It is necessary to get the Reynold's number:

$Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343$

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

$Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517$

The overall heat transfer coefficient:

$Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }$

Here

$h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C$

Substituting values:

$Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C$

5 0

4 years ago

True False. First angle projection type used in United states.

Sloan [31]

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

5 0

3 years ago

Can you identify the major theme illustrated by each of the following examples? If necessary, you may review the themes in Chapt

Bingel [31]

Where are the master plans stored and order given?

7 0

3 years ago

1. One of these is NOT a type of pneumatic tool. Which one?

Serggg [28]

Answer:Circular

Explanation:

It’s the only thing not list under pneumatic tools‍♂️

5 0

3 years ago