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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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3 years ago
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A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig
Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

(C) the total time tAC needed for it to reach the ground at C from the instant it is released.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

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3 years ago
The change in specific internal energy depends on the path of a process. a)-True b)-False
Basile [38]

Answer:

(b) False

Explanation:

The specific internal energy of the system does not depend on the path of the process, it is a state function means its depend on only on the initial and the final position it does not depend on the path which it follow in the process.Internal energy is associated with the random motion of the molecules.

So it is false statement as internal energy is not a path function

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3 years ago
A rigid insulated tank is divided into 2 equal compartments by a thin rigid partition. One of the compartments contains air, ass
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3 years ago
Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
Liono4ka [1.6K]

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