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nevsk [136]
2 years ago
9

3. (5%) you would like to physically separate different materials in a scrap recycling plant. describe at least one method that

might be used to separate polymers, aluminum alloys, and steels from one another.
Engineering
1 answer:
Likurg_2 [28]2 years ago
8 0

One of the methods that are used to separate polymers, aluminium alloys, and steels from one another is the Gravitation Separation method.

One straightforward technique is to run the mixture through a magnet, which will keep the steel particles on the magnet and separate them from the polymer.

What is the Gravitation Separation method?

When it is practicable to separate two components using gravity, i.e., when the combination's constituent parts have different specific weights, gravity separation is a technique used in industry. The components can be in suspension or in a dry granular mixture.

Polymers, Steel and Aluminium alloys can be readily split apart. The technique depends on how the two components are combined. The approach used is gravitational density. Due to the significant difference in relative specific mass values between steel and polymers (which range from 1.0 to 1.5), it is possible to separate them using flotation in a liquid that is safe and has the right density.

Therefore, the Gravitation Separation method is used to separate polymers, aluminium alloys and steels.

To learn more about the Polymer from the given link

brainly.com/question/2494725

#SPJ4

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If the channel-Length modulation effect is neglected, ID in the saturation region is considered to be independent of VDS
djverab [1.8K]
The answer is true because if the effect is neglected, the saturation id region is considered true
4 0
3 years ago
The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%
Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf)    ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

4 0
3 years ago
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
monitta

Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0
3 years ago
Which of the following is a valid reason why a scientist might reject a scientific theory
Flauer [41]
C I think hope it helps :)
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3 years ago
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Provide three examples of things engineers can implement when designing environmentally sustainable projects.
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Water supply.
Food production.
Housing and shelter.
Sanitation and waste management.
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