Ask question

Ask question

All categories

3 years ago

12

A 3-phase induction motor with 4 poles is being driven at 45 Hz and is running in its normal operating range. When connected to

a certain load, the motor rotates at a speed of 1070 rpm and outputs 800W a. What is the slip and developed torque? b. What is the new slip and motor speed if the developed torque is reduced to 4 Nm?

1 answer:

Dennis_Churaev [7]3 years ago

5 0

Answer:

a) The slip for the given conditions is 0.2074 or 20.74% and the developed torque is 7.14 Nm.

b) The new slip after reducing the torque to 4 Nm is 0.1162 or 11.62% and the new motor speed is 1193.13 rpm.

Explanation:

In order to find the slip we can use the definition as the relative difference between Synchronous speed and the rotor speed and that the developed torque is the ratio between output power and rotor angular velocity.

Slip.

We can find the synchronous speed u sing the following formula

$N_s =\cfrac{120f}p$

Where f stands for the frequency and p the number of poles, so we have

$N_s = \cfrac{120(45)}{4}\\N_s =1350 \,rpm$

Replacing on the slip formula

$S = \cfrac{N_s-N_r}{N_s}$

we get

$S = \cfrac{1350-1070}{1350}\\ S = 0.2074$

Thus the slip for the given conditions is 0.2074 or 20.74%.

Developed torque.

We can find the angular velocity in radians per second

$\omega_r =1070 \cfrac{rev}{min} \times \cfrac{1 \, min}{60 \, s}\times \cfrac{2\pi rad}{1 \, rev}\\\omega_r =112.05 \, \cfrac{rad}{s}$

Thus we can replace on the torque formula

$\tau = \cfrac{P}{\omega_r}\\\tau=\cfrac{800 W}{112.05 \, \cfrac{rad}{s}}$

We get

$\tau = 7.14\, N m$

The developed torque is 7.14 Nm.

Slip for the reduced torque.

The torque is proportional to the slip, so we can write

$\cfrac{\tau_1}{\tau_2}= \cfrac{S_1}{S_2}$

Thus solving for the new slip $S_2$ we have:

$S_2 = S_1 \cfrac{\tau_2}{\tau_1}$

Replacing the values obtained on the previous part we have

$S_2 = 0.2074 \cfrac{4 Nm}{7.14 Nm}\\ S_2=0.1162$

So the new slip after reducing the torque to 4 Nm is 0.1162 or 11.62%

Motor speed.

We can use the slip definition

$S_2 =\cfrac{N_s-N_{r_2}}{N_s}$

Solving for the motor speed we have

$S_2N_s =N_s-N_{r_2}$

$N_{r_2}=N_s-S_2N_s \\N_{r_2}=N_s(1-S_2)$

Replacing values we have

$N_{r_2}=1350 \, rpm(1-0.1162)\\N_{r_2}=1193.13 rpm$

The new motor speed is 1193.13 rpm.

You might be interested in

garri49 [273]

Concentrating solar power (CSP) plants use mirrors to concentrate the sun's energy to drive traditional steam turbines or engines that create electricity. The thermal energy concentrated in a CSP plant can be stored and used to produce electricity when it is needed, day or night. Today, roughly 1,815 megawatts (MWac) of CSP plants are in operation in the United States.

Parabolic Trough
Parabolic trough systems use curved mirrors to focus the sun’s energy onto a receiver tube that runs down the center of a trough. In the receiver tube, a high-temperature heat transfer fluid (such as a synthetic oil) absorbs the sun’s energy, reaching temperatures of 750°F or higher, and passes through a heat exchanger to heat water and produce steam. The steam drives a conventional steam turbine power system to generate electricity. A typical solar collector field contains hundreds of parallel rows of troughs connected as a series of loops, which are placed on a north-south axis so the troughs can track the sun from east to west. Individual collector modules are typically 15-20 feet tall and 300-450 feet long.

Compact Linear Fresnel Reflector
CLFR uses the principles of curved-mirror trough systems, but with long parallel rows of lower-cost flat mirrors. These modular reflectors focus the sun's energy onto elevated receivers, which consist of a system of tubes through which water flows. The concentrated sunlight boils the water, generating high-pressure steam for direct use in power generation and industrial steam applications.

3 0

3 years ago

Read 2 more answers

State two reasons why industrial files must be always be used fitted upon the individual

NikAS [45]

Explanation:

dnndndndndndndndndnndmfnfnf

4 0

3 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago

Consider a car manufacturing firm, with more than 100 facilities worldwide. What would be the positive aspects of having high nu

liberstina [14]

More work getting done and faster

7 0

3 years ago

Read 2 more answers

What are some of the main causes of accidents?

Ainat [17]

Answer

Being drunk, texting, being distracted

Explanation:

Hope this helped.

3 0

4 years ago