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Lena [83]
2 years ago
10

Which fields of engineering use fluid power? Explain how these fields make use of fluid power systems: water supply, agricultura

l equipment, construction, and earth-moving equipment.
Engineering
1 answer:
inysia [295]2 years ago
4 0

Answer:

Some fields of engineering that use fluid power are construction, heavy equipment, metallurgy, manufacturing, and oil and gas.

Water supply systems consist of these hydraulic components:

Pipes: Pipes transport water around the system. They are primary feeders (large-diameter pipes), secondary feeders (medium-diameter pipes), and distributors (small-diameter pipes).

Valves: They control the direction of the water flow. Indicating valves convey whether they are open or closed and non-indicating valves do not.

Hydrants: Hydrants appear at periodic intervals along the underground pipe network to provide water for firefighting.

Storage containers: Tanks located at a height provide water when demand is higher than supply. Gravity provides the required water pressure. Pumps can elevate water and refill the tank.

Agricultural equipment: Machines with hydraulic conveyors help transport harvested crops to wagon trailers.

In the construction industry, industrial forklifts lift and transport heavy weights. Hydraulic cylinders can tilt, lower, or elevate the required weight in a smooth motion.

Earth-moving equipment such as excavators have hydraulically powered buckets that can dig up ground, carry soil in the bucket, and drop the soil into trucks.

Explanation:

PLATO answer

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Serjik [45]

Answer:

The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.

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g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

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the radius = d/2 = 12 / 2 = 6 in

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Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

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\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

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