All categories

2 years ago

What are the important factors needed to be considered while selecting a brake or clutch?

Engineering

1 answer:

AysviL [449]2 years ago

4 0

Answer:

The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:

-Engagement

-Friction

-Electromagnetic

-Mechanical

-Actuation

-Electric

-Fluid power

-Self-actuation

-Key concepts

-Application notes

-Selection criteria

Explanation:

Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.

You might be interested in

5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Ba

Elza [17]

As a multiple choice the only correct answer is D

3 0

2 years ago

Read 2 more answers

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.

Explanation:

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.

5 0

2 years ago

Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace

Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0

2 years ago

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur

Lina20 [59]

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.

8 0

2 years ago

The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su

VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

- We will calculate the total heat flux for the entire system q:

q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

q = ( 333 - 69 ) / 0.2900649351

q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

q = ( T_i - T_2 ) / R_i2

Where,

R_i2 = 1 / h_i + L_1 / k_1

R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

( T_i - T_2 ) = q*R_i2

T_2 = T_i - q*R_i2

Plug the values in:

T_2 = 333 - 910.141*0.14357

T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0

2 years ago