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bogdanovich [222]
3 years ago
12

A three-phase, 480 Volt, 120 horsepower, 50 Hertz four-pole induction motor delivers rated output power at a slip of 4%. Determi

ne the synchronous speed, the actual motor speed and the slip speed?
Engineering
2 answers:
Bezzdna [24]3 years ago
8 0

Answer:

Explanation:

Given:

Number of poles, np = 4 poles

E = 480 V

Frequency, f = 50 Hz

Slip, s = 4%

A.

Speed, Vs = 120 x Frequency (Hz)/number of poles

= 120 × 50/4

= 1500 rpm

B.

Actual motor speed, V = Vs × (1 - s)

V = 1500 × (1 - 0.04)

= 1440 rpm

C.

Slip speed, s = Vs - V

= 1500 - 1440

= 60 rpm

kkurt [141]3 years ago
4 0

Answer:

a) n_{s} = 750\,rpm, b) n_{r} = 720\,rpm, c) n_{p} = 30\,rpm

Explanation:

a) Synchronous speed is:

n_{s} = f\cdot \frac{60}{n_{poles}}

n_{s} = (50\,hz)\cdot \left(\frac{60}{4}\right)

n_{s} = 750\,rpm

b) Actual motor speed is:

S = \frac{n_{s}-n_{r}}{n_{s}} \times 100\%

n_{r} = \left(1-\frac{S}{100}\right)\cdot n_{s}

n_{r} = \left( 1 - \frac{4}{100}  \right)\cdot 750\,rpm

n_{r} = 720\,rpm

c) Slip speed is:

n_{p} = n_{s} - n_{r}

n_{p} = 750\,rpm - 720\,rpm

n_{p} = 30\,rpm

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A structural component in the shape of a flat plate 25.0 mm thick is to be fabricated from a metal alloy for which the yield str
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Answer:

The critical length of surface flaw = 6.176 mm

Explanation:

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Design stress. =0.3 × yield strength

= 0.3 × 545

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Dimensionless parameter. Y = 1.3

The critical length of surface flaw is given by

= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2

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5 0
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Read 2 more answers
The size of Carvins Cove water reservoir is 3.2 billion gallons. Approximately, 11 cfs of water is continuous withdrawn from thi
Zolol [24]

Answer:

471 days

Explanation:

Capacity of Carvins Cove water reservoir = 3.2 billion gallons i.e. 3.2 x 10˄9 gallons

As,  

1 gallon = 0.133 cubic feet (cf)

Therefore,  

Capacity of Carvins Cove water reservoir in cf  = 3.2 x 10˄9 x 0.133

                                                                         = 4.28 x 10˄8

 

Applying Mass balance i.e

Accumulation = Mass In - Mass out   (Eq. 01)

Here  

Mass In = 0.5 cfs

Mass out = 11 cfs

Putting values in (Eq. 01)

Accumulation  = 0.5 - 11

                         = - 10.5 cfs

 

Negative accumulation shows that reservoir is depleting i.e. at a rate of 10.5 cubic feet per second.

Converting depletion of reservoir in cubic feet per hour = 10.5 x 3600

                                                                                       = 37,800

 

Converting depletion of reservoir in cubic feet per day = 37, 800 x 24

                                                                                         = 907,200  

 

i.e. 907,200 cubic feet volume is being depleted in days = 1 day

1 cubic feet volume is being depleted in days = 1/907,200 day

4.28 x 10˄8 cubic feet volume will deplete in days  = (4.28 x 10˄8) x                    1/907,200

                                                                                 = 471 Days.

 

Hence in case of continuous drought reservoir will last for 471 days before dry-up.

8 0
2 years ago
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