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Thepotemich [5.8K]
3 years ago
6

Fluid power is a. The technology that deals with the generation, control, and transmission of power-using pressurized fluids b.

muscle that moves industry. c. used to push, pull, regulate, or drive virtually all the machine of modern industry d. probably as old as civilization itself e. all of the above
Engineering
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:  a) The technology that deals with the generation, control and transmission of power using pressurized fluids

Explanation: Fluid power is defined as the fluids which are under pressure and then are used for generation,control and transmit the power. Fluid power systems produces high forces as well as power in small amount . These systems usually tend to have better life if maintained properly. The force that are applied on this system can be monitored by gauges as well as meter.

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Water is being heated in a closed pan on top of a range whilebeing stirred by a paddle wheel. During the process 30kJ of heat is
My name is Ann [436]

Answer:

38 kJ

Explanation:

The solution is obtained using the energy balance:  

ΔE=E_in-E_out

U_2-U_1=Q_in+W_in-Q_out

U_2=U_1+Q_in+W_in-Q_out

      =38 kJ

4 0
3 years ago
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5 0
3 years ago
Which of the following are not related to a materials structure? (Mark all that apply) a)- Atomic bonding b)- Crystal structure
belka [17]

The answer is c) atomic number

3 0
3 years ago
What are some aircraft aging problems? How can you as an Aviation Maintenance Manager monitor problems that relate to aircraft f
julia-pushkina [17]

Answer:

Answered

Explanation:

The two key processes that lead to aircraft ageing are fatigue and corrosion. These processes generally affect the aircraft structure, but can also affect wiring, flight controls, power plants, and other components. Fatigue and corrosion can work independently from one another, or they can interact. The interaction between fatigue and corrosion can increase the rate of ageing to a greater extent than that due to either process alone.

Fatigue predominately takes place in metal components, but it can also affect non-metallic materials. Fatigue occurs through cyclic loading patterns, where a component is repeatedly loaded. Bending a metal paper clip backwards and forwards is an example of fatigue; the paper clip will not break if only bent once, however, if it is repeatedly loaded, it will eventually break. Fatigue failures will often take place at loads much lower than the materials ultimate strength.

Generally, the initiation point for fatigue will be a microscopic crack that forms at a location of high stress, such as a hole, notch, or material imperfection. The crack will then grow as loads are repeatedly applied. If not detected and treated, the crack will eventually grow to a critical size and failure will occur at loads well below the original strength of the material.

The relationship between repetitive loading and fatigue crack growth, creates a link between fatigue related ageing, the number of flight cycles, and the number of flight hours that an aircraft has accumulated.

Aircraft components that are susceptible to fatigue include most structural components such as the wings, the fuselage, and the engine.

During initial manufacturing the research department is responsible for designing the aircraft to withstand fatigue. During operations the fatigue risk management of aviation maintenance will try to rectify the problems due to fatigue.

4 0
3 years ago
A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as
marysya [2.9K]

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

4 0
3 years ago
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