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3 years ago

A 60-Hz 3-phase induction motor is required to drive a load at approximately 850 rpm. How many poles should the motor have

Engineering

1 answer:

Anastaziya [24]3 years ago

7 0

Answer:

8 poles

Explanation:

Given

$Frequency, f = 60Hz$

$Speed, n_s = 850rpm$

Required

Calculate the number of poles the motor should have?

To solve this question, we make use of synchronous angular velocity formula

$n_s = \frac{120f}{P}$

Where

$P = Number\ of\ Poles$

Substitute values for $n_s$ and f in $n_s = \frac{120f}{P}$

$850 = \frac{120 * 60}{P}$

$850 = \frac{7200}{P}$

Multiply both sides by P

$P * 850 = \frac{7200}{P} * P$

$P * 850 = 7200$

Divide both sides by 850

$\frac{P * 850}{850} = \frac{7200}{850}$

$P = \frac{7200}{850}$

$P = 8.47058823529$

However, the number of poles must be an integer.

So:

$P = 8$

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3 years ago

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

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Answer:

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3 years ago

engineering uses data from pareto charts to analyse motor faults caused during their production. Explain one advantage of using

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3 years ago

Please define the coefficient of thermal expansion?

Vikki [24]

Answer:

The coefficient of thermal expansion tells us how much a material can expand due to heat.

Explanation:

Thermal expansion occurs when a material is subjected to heat and changes it's shape, area and volume as a result of that heat. How much that material changes is dependent on it's coefficient of thermal expansion.

Different materials have different coefficients of thermal expansion (i.e. It is a material property and differs from one material to the next). It is important to understand how materials behave when heated, especially for engineering applications when a change in dimension might pose a problem or risk (eg. building large structures).

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3 years ago

Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes th

Svet_ta [14]

Answer:

The time required to elute the two species is 53.3727 min

Explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

$Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621$

The mathematical expression for the time to elute the two species is:

$\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}$

Here

ReB = 1.5

$t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min$

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3 years ago