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taurus [48]
3 years ago
14

Help me i need this bad :)

Engineering
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

you cant put tests here

Explanation:

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Plz help electrical technology
oksano4ka [1.4K]

Answer:

OPTION A,Larger

HOPE IT HELPS

8 0
3 years ago
Read 2 more answers
Why or why not the following materials will make good candidates for the construction of
zvonat [6]

Answer:

Answer explained below

Explanation:

3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.

The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.

To protect blades from these high dynamic stresses, friction dampers are used.

b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.

These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.

In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.

In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.

Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.

3 0
3 years ago
The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta
tatyana61 [14]

Answer:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular  momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

H_o =r x mv=rxL

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change  of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular  momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2".

If we analyze the staritning point we see that the initial velocity can be founded like this:

v_o =\omega r_{OIC}=\omega (0.15m)

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af

0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)

And if we integrate the left part and we simplify the right part we have

1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega

And if we solve for \omega we got:

\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}

8 0
2 years ago
Các đặc điểm chính của đường dây dài siêu cao áp .
rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0
2 years ago
Assume the impedance of a circuit element is Z = (3 + j4) Ω. Determine the circuit element’s conductance and susceptance.
djyliett [7]

Answer:

B. G = 333 mS, B = j250 mS

Explanation:

impedance of a circuit element is Z = (3 + j4) Ω

The general equation for impedance

Z = (R + jX) Ω

where

R = resistance in ohm

X = reactance

R = 3Ω  X = 4Ω

Conductance = 1/R while Susceptance = 1/X

Conductance = 1/3 = 0.333S

= 333 mS

Susceptance = 1/4 = 0.25S

= 250mS

The right option is B. G = 333 mS, B = j250 mS

8 0
2 years ago
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