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lisabon 2012 [21]
3 years ago
9

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

Engineering
1 answer:
AysviL [449]3 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.

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g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What
earnstyle [38]

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity (v), measured in meters per second, is determined by the following expression:

v = \frac{4\cdot \dot V}{\pi \cdot D^{2}} (1)

Where:

\dot V - Flow rate, measured in cubic meters per second.

D - Diameter, measured in meters.

If we know that \dot V = 0.01\,\frac{m^{3}}{s} and D = 0.05\,m, then the flow velocity is:

v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}

v \approx 5.093\,\frac{m}{s}

The density and dinamic viscosity of the glycerin at 20 ºC are \rho = 1260\,\frac{kg}{m^{3}} and \mu = 1.5\,\frac{kg}{m\cdot s}, then the Reynolds number (Re), dimensionless, which is used to define the flow regime of the fluid, is used:

Re = \frac{\rho\cdot v \cdot D}{\mu} (2)

If we know that \rho = 1260\,\frac{kg}{m^{3}}, \mu = 1.519\,\frac{kg}{m\cdot s}, v \approx 5.093\,\frac{m}{s} and D = 0.05\,m, then the Reynolds number is:

Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }

Re = 211.230

A pipeline is in turbulent flow when Re > 4000, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (f), dimensionless, is determined by the following expression:

f = \frac{64}{Re}

If we get that  Re = 211.230, then the friction factor is:

f = \frac{64}{211.230}

f = 0.303

The friction factor is 0.303.

4 0
2 years ago
Water is contained in a rigid vessel of 5 m3 at a quality of 0.8 and a pressure of 1 MPa. If the pressure is reduced to 270.3 kP
professor190 [17]

Answer:

russ ruwabsd ljabnlndszdnjndfsmsdf,bsd

Explanation:

4 0
3 years ago
Compute the volume percent of graphite, VGr, in a 3.2 wt% C cast iron, assuming that all the carbon exists as the graphite phase
Yanka [14]

Answer:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%

Where:

V_{gr} - Volume occupied by the graphite phase, measured in cubic centimeters.

V_{fe} - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%

Where:

m_{fe}, m_{gr} - Masses of the ferrite and graphite phases, measured in grams.

\rho_{fe}, \rho_{gr} - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%

\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%

If \rho_{gr} = 2.3\,\frac{g}{cm^{3}}, \rho_{fe} = 7.9\,\frac{g}{cm^{3}}, m_{gr} = 3.2\,g and m_{fe} = 96.8\,g, the volume percentage of graphite is:

\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%

\%V_{gr} = 10.197\,\%V

The volume percentage of graphite is 10.197 per cent.

5 0
3 years ago
I) A sag vertical curve is to be designed to join a 4% grade to a 2% grade. If the design
Burka [1]

Answer:

=4/5 because I'm not going to go back in a year meaning that they are you are

4 0
2 years ago
A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is
Dafna1 [17]

Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

v^2-u^2=2as

where,

s = distance covered by the object = 6.93 m

u = initial velocity  = 2.99 m/s

v = final velocity = ?

a = acceleration = 9.8m/s^2

Now put all the given values in the above equation, we get the final velocity of the ball.

v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)

v=12.03m/s

Thus, the final velocity of the ball is, 12.03 m/s

7 0
3 years ago
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