All categories

3 years ago

What is the electrical consumption in KVA of a motor powered by a 3-phase, 60 Hz, 460 VAC supply that continuously draws 17 A

Physics

1 answer:

MrRa [10]3 years ago

8 0

Answer:

15.34 kVA

Explanation:

A motor is a device that converts electrical energy into mechanical energy. It takes in electrical energy at the input and produce torque (motion) at the output.

The power consumption for a three phase motor is the product of voltage and current and √3. The √3 is because it is a three phase supply.

Hence Power (P) =√3 × voltage (V) × current (I)

P = √3 × V × I

Given that voltage (V) = 460 V, current (I) = 17 A. Hence:

P = √3 × V × I = √3 × 460 × 17 = 13544.64 VA

But 1000 VA = 1 kVA. Hence:

$P=13544.64\ VA*\frac{1\ kVA}{1000\ VA}=13.54\ kVA$

You might be interested in

How do terrestrial and giant planets differ? List as many ways as you can think of.

Leni [432]

Answer: The differences between terrestrial planets and the giant planets are s follows-

The inner planets namely Mercury, Venus, Earth and Mars are the terrestrial planets, whereas the outer planets namely Jupiter, Saturn, Uranus and Neptune are known as the outer planets.
Inner planets are composed mainly of silicate materials as well as some metals, whereas the giant planets are comprised of water (in different states) and gases such as Hydrogen and Helium.
The density of the inner planets are more in comparison to the outer planets, where earth has the highest of about 5.5 gm/cm³ and Saturn has the lowest of about 0.7 gm/cm³.
Due to the location of the inner planets near to the sun, they have high boiling point, whereas outer planets are much far from the sun so they have a low boiling point.

5 0

4 years ago

How easy is it to die in quick sand (I know the answer to this one too i just want to see who understands)

matrenka [14]

There have been 580 cases!! Wow

8 0

3 years ago

Read 2 more answers

A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is

puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

$\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}$

$\dfrac{1}{u}=\dfrac{103}{4080}$

$u =\dfrac{4080}{103}$

The magnification is

$m = \dfrac{-v}{u}$

$m=\dfrac{-240\times103}{4080}$

$m = -6.05$

Hence, The magnification is -6.05.

6 0

3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$