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3 years ago

5) Calculate the LMC wal thickness of a pipe and tubing with OD as 35 + .05 and ID as 25 + .05 A) 4.95 B) 5.05 C) 10 D) 15.025

Engineering

2 answers:

padilas [110]3 years ago

6 0

Answer:

B) 5.05

Explanation:

The wall thickness of a pipe is the difference between the diameter of outer wall and the diameter of inner wall divided by 2. It is given by:

Thickness of pipe = (Outer wall diameter - Inner wall diameter) / 2

Given that:

Inner diameter = ID = 25 ± 0.05, Outer diameter = OD = 35 ± 0.05

Maximum outer diameter = 35 + 0.05 = 35.05

Minimum inner diameter = 25 - 0.05 = 24.95

Thickness of pipe = (maximum outer wall diameter - minimum inner wall diameter) / 2 = (35.05 - 24.95) / 2 = 5.05

Thickness = (35 - 25) / 2 + 0.05 = 10/2 + 0.05 = 5 + 0.05 = 5.05

Therefore the LMC wall thickness is 5.05

weeeeeb [17]3 years ago

5 0

B I hope it helps for your question

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Water flows in a pipeline. At a point in the line where the diameter is 7 in., the velocity is 12 fps and the pressure is 50 psi

PolarNik [594]

Answer:

a) P₂ = 3219.11 lbf / ft² , b) P₂ = 721.91 lbf / ft² , c) P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

The area of a circle is

A = π r² = π/4 d²

v₂ = v₁ (d₁ / d₂)²

Let's calculate

v₂ = 12 (7/3)²

v₂ = 65 feet / s

Now let's use Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

y₁ = y₂

P₁ - P₂ = ½ ρ (v₂² –v₁²)

P₂ = P₁ - ½ ρ (v₂² –v₁²)

ρ = 62.43 lbf / ft³

P₁ = 50 psi (144 lbf/ ft² / psi) = 7200 lbf / ft²

P₂ = 7200 - ½ 62.43 / 32 (65² -12²)

P₂ = 7200 - 3980.89

P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

y₂ –y₁ = 40 ft

P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

P₂ = 7200 - 2497.2 - 3980.89

P₂ = 721.91 lbf / ft²

Case 3. Vertical water pipe flows down

y₂ –y₁ = -40

P₂ = 7200 + 2497.2 - 3980.89

P₂ = 5707.31 lbf / ft²

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3 years ago

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Nezavi [6.7K]

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3 years ago

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Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its the

Andre45 [30]

Answer:

Explanation:

The density of the unit cell of a material, Iron in this case, has to be approximately equal with its experimental value of 7.87 g/cm³.

The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.

For a BCC crystal structure the length of the side of the cube is given by:

a = 4r/√3

where a is the atomic radius of Iron

first we will convert this radius to cm since we want the density in g/cm³:

0.124 nm x 1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm

the volume of the cubic cell is:

v = a³ = ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³

The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:

BCC = 2 atoms / unit cell ( 1/8 from the 8 corners + 1 in the center)

m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms x 55.85 g/mol

= 1.85 x 10⁻²² g

Therefore,

d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³

An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.

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Sasha is using Python 3 as a calculator to divide 100 by 4 and types print 100 / 4. When run is typed, nothing happens. What nee

Nady [450]

Answer:

parentheses

Explanation:

print 100/4

SyntaxError: Missing parentheses in call to 'print'.

print (100/4)

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