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LuckyWell [14K]

3 years ago

10

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic

field of the stator is rotating

1 answer:

rusak2 [61]3 years ago

4 0

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

$n_s=\frac{120f}{p}=\frac{120*60}{4}=1800\ rpm$

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

$n=n_s-sn_s\\\\n=1800-0.05(1800)\\\\n=1800-90\\\\n=1710\ rpm$

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\\\\f_r=0.04*60\\\\f_r=2.4\ Hz$

4) At standstill, the motor speed is zero hence the slip = 1:

$s=\frac{n_s-n}{n_s}\\ \\n=0\\\\s=\frac{n_s-0}{n_s}\\\\s=1$

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\\\\f_r=1*60\\\\f_r=60\ Hz$

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2 years ago

A three-point bending test is performed on a glass specimen having a rectangular cross section of height d 5 mm (0.2 in.) and wi

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3 years ago

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

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Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

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We are to determine the maximum temperature in the wall

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Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

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$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

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From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

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Substituting the values we get

T = 167 ° C

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3 years ago

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Answer:

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Explanation:

Given data :

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3 years ago