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2 years ago

What is the objective of phasing out an INDUCTION MOTOR before putting the machine into commission?

Engineering

1 answer:

enyata [817]2 years ago

8 0

The main objective of phasing out an INDUCTION MOTOR is to identify the ends of the stator coils.

<h3>What is an induction motor?</h3>

An induction motor is a device based on alternate electricity (AC) which is composed of three different stator coils.

An induction motor is a device also known as an asynchronous motor due to its irregular velocity.

In conclusion, the objective of phasing out an INDUCTION MOTOR is to identify the ends of the stator coils.

Learn more on induction motors here:

brainly.com/question/15721280

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3 years ago

A single square-thread screw has an input power of 3 kW at a speed of 1 rev/s. The screw has a diameter of 40 mm and a pitch of

Sonbull [250]

Answer:

Axial Resisting Load, F = 31.24kN

Efficiency = 16.67%

Explanation:

Given

Input Power = P,in = 3kW = 3,000W

Speed, S = 1rev/s

Pitch, p = 8mm

Thread frictional coefficient = μt = 0.18

Collar frictional coefficient = μc = 0.09

Friction radius of collar, Rc = 50mm

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Then we calculate the torque due to friction from the collar

T = Fμc * Rc

T = F * 0.09 * 50

T = 4.5F. Nmm

T = 4.5 * 10^-3F Nm

Then, we calculate the axial resisting load 'F' by using the the following power input relation.

P,in = Tw

P,in = (T1 + T2) * 2πN

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3,000 = (3.44 + 4.5) * 10^-3 * F * 10^-3 * 2 * π * 2

F = 3000/((3.44 + 4.5) * 10^-3 * 10^-3 * 2 * π * 2

F = 31,247.69N

F = 31.24kN

Hence, the axial resisting load is

F = 31.24kN

Calculating Efficiency

Efficiency = Fp/2πP

Efficiency = 2Fp/P,in

Substitute each value

Efficiency = 2 * 31,247.69 * 8 * 10^-3/3000

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8 0

3 years ago

Read 2 more answers

An inductor is connected to a voltage source and it is found that it has a time constant, t. When a 10-ohm resistor is placed in

prisoha [69]

Answer:

A) 30 mH

B ) 10-ohm

Explanation:

resistor = 10-ohm

Inductor = 30mH ( l )

L = inductance

R = resistance

r = internal resistance

values of the original Inductors

Note : inductor = constant time (t) case 1

inductor + 10-ohm resistor connected in series = constant time ( t/2) case2

inductor + 10-ohm resistor + 30 mH inductance in series = constant time (t) case3

<em>From the above cases</em>

case 1 = time constant ( t ) = L / R

case 2 = Req = R + r hence time constant t / 2 = L / R + r therefore

t = $\frac{2L}{R+r}$

case 3 = Leq = L + l , Req = R + r . constant time ( t )

hence Z = $\frac{L + l}{R + r}$ = t

A) Inductance

To calculate inductance equate case 1 to case 3

$\frac{L}{R}$ = $\frac{L + l}{R + r}$ = L / 10 = (L + 30) / ( 10 + 10 )

= 20 L = 10 L + 300 mH

10L = 300 m H

therefore L = 30 mH

B ) The internal resistance

equate case 1 to case 2

$\frac{L}{R}$ = $\frac{2L}{R + r}$

R + r = 2 R therefore ( r = R ) therefore internal resistance = 10-ohm

6 0

3 years ago