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hodyreva [135]
3 years ago
15

What is the difference between an induced and a permanent magnet?

Physics
1 answer:
MissTica3 years ago
8 0
INDUCTION MOTOR:-

Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control


Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation

Power density:-Induction produced by squirrel cage rotor inherently limits power density

Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos

Cost:-Relatively modest initial cost; higher operating costs

PERMANENT MAGNET MORTOR:-

speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range

Reliability:-Lower operating temperatures reduces wear and tear, maintenance • Extends bearing and insulation life • Robust construction for years of trouble-free operation in harsh environments.

power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.

Accuracy:-Without feedback, can be difficult to locate and position to the pinpoint accuracy of servomotors

<span>Cost:-Exhibit higher efficiency, so their energy use is smaller and full return on their initial purchase cost is realized more quickly</span>
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Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

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We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

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7 0
3 years ago
when a circular plate of metal is heated in an oven, its radius increases at .03 cm/min, at what rate is the area increasing whe
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Answer:

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We have given rate of change of radius \frac{dr}{dt}=0.03cm/min

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So \frac{dA}{dt}=2\pi r\frac{dr}{dt}

Puting the value of r and \frac{dr}{dt}

\frac{dA}{dt}=2\times 3.14\times 52\times 0.03=9.796cm^2/min

So rate of change of area will be 9.796cm^2/min

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