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Alisiya [41]
3 years ago
12

Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De

termine the flow rate of the kerosene
Engineering
1 answer:
Anettt [7]3 years ago
7 0

Answer:

Q=4.98\times 10^{-3}\ m^3/s

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

For Copper tube is 3/4 standard type K drawn tube

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

Pressure difference given as

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where

L is length of tube

μ is dynamic viscosity

Q is volume flow rate

d is inner diameter of tube

ΔP is pressure drop

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50\times Q}{\pi\times 0.0189^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s

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Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g

Silt also contains more of spherical particles with weights around 0.0029g.

Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.

BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.

Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.

Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.

Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS

Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.

Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.

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3 years ago
All of the following have the same units except: Group of answer choices resistance capacitive reactance. inductance. impedance.
Brilliant_brown [7]

Answer:

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Explanation:

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2 years ago
which of the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick
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<h3>What is fiber reinforced plastic?</h3>

Fiber reinforced plastic is defined as a polymer-based composite material that is strengthened by fiber support. Fiber-reinforced plastics (FRP) are composite materials that use glass or carbon fibers as reinforcement and polymer resins as a matrix.

Because laser cutting is a far less aggressive and abrasive process than water jet cutting, it is much more accurate. A laser can carefully and safely cut materials as thin as 0.006 inches, whereas water jet cutters can't handle cutting through surfaces smaller than 0.02 inches.

Thus, laser beam machining and water jet cutting are the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick sheet of fiber-reinforced plastic. Hence option d and f is correct.

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3 0
1 year ago
Discuss the organizational system that you believe would be the most effective for the safety officer in a medium-sized (100-200
marin [14]

Answer:

A safety manager is a person who designs and maintains the safety elements at workplace. A balance should be required for production and the job in providing work environment. As a safety officer in a medium sized manufacturing facility the following organizational system can be designed and maintained:

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Write torsion equation and explain the importance of each components.<br>​
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4) The bar is stressed within its elastic limit.

Nomenclature

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l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
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