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slamgirl [31]
3 years ago
11

Should the ship breaking business continue why or why not?

Engineering
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

Ship-breaking or ship demolition is a type of ship disposal involving the breaking up of ships for either a source of parts, which can be sold for re-use, or for the extraction of raw materials, chiefly scrap. It may also be known as ship dismantling, ship cracking, or ship recycling. Modern ships have a lifespan of 25 to 30 years before corrosion, metal fatigue and a lack of parts render them uneconomical to operate.[1] Ship-breaking allows the materials from the ship, especially steel, to be recycled and made into new products. This lowers the demand for mined iron ore and reduces energy use in the steel making process. Fixtures and other equipment on board the vessels can also be reused. While ship-breaking is sustainable, there are concerns about the use of poorer countries without stringent environmental legislation. It is also labor-intensive, and considered one of the world's most dangerous industries.[2]

In 2012, roughly 1,250 ocean ships were broken down, and their average age was 26 years.[3][4] In 2013, the world total of demolished ships amounted to 29,052,000 tonnes, 92% of which were demolished in Asia. As of January 2020, India has the largest global share at 30%;[5] followed by Bangladesh, China and Pakistan.[6] Alang, India currently has the world's largest ship graveyard,[5] followed by Chittagong Ship Breaking Yard in Bangladesh and Gadani in Pakistan.[6]

The largest sources of ships are states of China, Greece and Germany respectively, although there is a greater variation in the source of carriers versus their disposal.[7] The ship-breaking yards of India, Bangladesh, China and Pakistan employ 225,000 workers as well as providing many indirect jobs. In Bangladesh, the recycled steel covers 20% of the country's needs and in India it is almost 10%.[8]

As an alternative to ship-breaking, ships may be sunk to create artificial reefs after legally-mandated removal of hazardous materials, or sunk in deep ocean waters. Storage is a viable temporary option, whether on land or afloat, though all ships will be eventually scrapped, sunk, or preserved for museums.

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A gaseous mixture contain 32.9 mol % He, 40.7 mol % N_{2} and 26.4 mol % Ar. Determine the composition of this mixture on a mass
Salsk061 [2.6K]

The composition of the mixture on a mass basis are as follows:

  • He = 131.70 g
  • N2 = 1140.41 g
  • Ar = 1054.68 g

number of moles = mass of substnce / molar mass

molar mass and molecular weight has equal numerical values therefore

Number of moles = mass of substnce / molecular weight

mass of substance = number of moles * molecular weight

the moles are expressed in percentage hence the total moles should be 100

check: 32.9 + 40.7 + 26.4 = 100

He

mass of substance = 32.9 * 4.003 = 131.6987 g

N_{2}

mass of substance = 40.7 * 28.02 = 1140.414 g

Ar

mass of substance = 26.4 * 39.95 = 1054.68 g

6 0
2 years ago
Given an orifice meter with the dimensions in lab, a flow rate of 60 lbm/min and a differential pressure of 200 inches of water,
monitta

Answer:

Theoretically impossible

Explanation:

Given that:

The mass flow rate = 60 lbm/min

The differential pressure = 200 inches

The flow rate Q is can be expressed by the formula:

Q = K.A\sqrt{2g \Delta h}

where;

K = Discharge coefficient

A = area

Δh = Head drop

g = gravity

From the given parameter, the area is unknown.

Therefore, the flow coefficient of the meter will be theoretically impossible to be determined.

4 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10
Fiesta28 [93]

Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m

ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

Substituting these values into the formula, we can calculate the max stress as

 ====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)

σₐ  = 24.4MPa

6 0
3 years ago
A cylindrical specimen of Aluminium having a diameter of 12.8 mm and gauge length of 50.8 is pulled in tension. Use the data giv
Katyanochek1 [597]

Answer:

Hello the needed data given is not properly arranged attached below is the properly arranged data

Answer:

b) 62.5 * 10^3 MPa

c) ≈ 285 MPa

d)  370Mpa

e)  16%

Explanation:

Given Data:

cylindrical aluminum diameter = 12.8 mm

Gauge length = 50.8 mm

A) plot of engineering stress vs engineering strain

attached below

B ) calculate Modulus of elasticity

Modulus of elasticity = Δб / Δ ε

                                   = ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa

C) Determine the yield strength

at strain offset = 0.002

hence yield strength ≈ 285 MPa

D) Determine tensile strength of the alloy

The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress  vs strain ( complete plot )

E) Determine approximate ductility in percent elongation

ductility in percent elongation = plastic strain at fracture * 100

total strain = 0.165 , plastic strain = 0.16

therefore Ductility in percent elongation = 0.16 * 100 = 16%

5 0
3 years ago
Refrigerant-134a is compressed from 2 bar, saturated vapor, to 10 bar, 90o C in a compressor operating at steady state. The mass
Ronch [10]

Answer:

Find attach the solution

Explanation:

Entropy production is almost the same in (b) and (c) as T_{avg} ~ T_{surroundings}

3 0
3 years ago
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