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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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Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

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large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ;  f_{Lw}=6.6 mi/h

5' on right   ;    f_{Lc} = 0.4 mi/hr

3' on left   ;      no adjustment

3 lanes in each direction    f n = 3 mi/h

v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}

f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}

f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}

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v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}

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FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}

      = (55 + 5) - 6.6 - 0.4 -3 -0

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D = \dfrac{V_P}{s}

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level of service is D using speed flow curves and LOS for basic free moving of vehicle

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