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pychu [463]
3 years ago
8

Describe three parts of a fluid power system and the roles played by each to make the device work.

Engineering
1 answer:
artcher [175]3 years ago
5 0

Three parts of a fluid power system and the roles played by each to make the device work.

Explanation:

1.Fluid power is subdivided into hydraulics using a liquid such as mineral oil or water, and pneumatics using a gas such as air or other gases.

2.Fluid power systems perform work by a pressurized fluid bearing directly on a piston in a cylinder or in a fluid motor.

Hydraulics Pumps

Dynamic (non positive displacement) Pumps

  • This type is generally used for low-pressure, high volume flow applications. Since they are not capable of withstanding high pressures, there is little use in the fluid power field. Their maximum pressure is limited to 250-300 psi. This type of pump is primarily used for transporting fluids from one location to another. Centrifugal and axial flow propeller pumps are the two most common types of dynamic pumps.

Positive Displacement Pumps

  • This type is universally used for fluid power systems. With this pump, a fixed amount of fluid is ejected into the hydraulic system per revolution of pump shaft rotation. These pumps are capable of overcoming the pressure resulting from the mechanical loads on the system as well as the resistance to flow due to friction. These two features are highly desirable in fluid power pumps. These pumps also have the following advantages over non positive displacement pumps:
  • High-pressure capability (up to 12,000 psi)

Small compact size

high volumetric efficiency

small changes in efficiency throughout the design pressure range

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Answer:

A safety margin is the space left between your vehicle and the next to provide room, time and visibility at every instant

Explanation:

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2 years ago
The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

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