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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 0
3 years ago
A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th
inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = \int\limits^a_b P \, dV  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 0
2 years ago
Three groups of students are given study outlines for 6 weeks. One group studies 2 hours a night, a second group studies 1 hour
katrin2010 [14]

Answer:

The constant here is the study outline

Explanation:

In scientific research, the constant variable is that part/variable of the experiment that does not change or is set not to change. Examples include temperature, environment or height.

Assuming the scenery described in this question is an experiment. All the groups presented are bound by a constant during the experiment. The constant here is the study outline. The study outline provided to the students is not going to change.

NOTE: There could be confusion as regards the answer being the final exam grade but that will be the dependent variable as that will be the outcome of the experiment while the time spent to study will be the independent variable.

8 0
2 years ago
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of
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Answer: Hello the question is incomplete below is the missing part

Question:  determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

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<em>first step : calculate the value of h2 using the equation below </em>

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From Table A-17

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2 years ago
How does a 2.5 MW wind turbine costing $ 4 million compare to a 5-kw wind turbine $3 /W? a) Same $/w b) Smaller $/w c) Larger $/
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MW means megawatt, and one megawatt is a million Watts.
The 2.5 MW turbine is 4/2.5=1.6 $/w
Answer B
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