Great video! However It's important to clarify that the current calculation considers a single phase motor. You need to multiply the voltage by √3 for 3 phase motors. Single phase: A = 1000 ⋅ kVA / V Three phase: A = 1000 ⋅ kVA / (√3 × V)
Thank you for the clarification. So if you measure each phase of this motor at fully loaded conditions the reading will be approximately 59.2 amps correct?
Just a little correction on your triangle. Power factor is the cos(theta) of the angle (we call theta) you have labeled as 0.7. 0.7 (the power factor) is the result of that calculation, not the angle itself. Just a little clarification, since you can’t work out the equations otherwise. Thanks for the video!
The best way to learn this stuff is by picking up the following book... - Electronics Fundamentals: A Systems Approach 😉 If you're not good at Algebra and Trigonometry then don't waste your time. Personally, I recommend at least Pre-Calculus due to all the logarithmic equations when it comes to Capacitor/Inductor calculations. Power Factor is the ratio between [True Power] & [Apparent Power]. Ideally, you want a Power Factor of (1) which means there is no Reactive Power hence [True Power] & [Apparent Power] are equal thus no wasted energy 😉 Power Triangle --------------------------- - True Power = Adjacent Side - Reactive Power = Opposite Side (wasted energy) - Apparent Power = Hypotenuse - Cosφ = [ Adjacent Side / Hypotenuse ] - Cosφ = [ True Power (W) / Apparent Power (V·A) ] - Power Factor = Cosφ If for some reason you need the phase angle... - φ = Cos⁻¹ [ True Power (W) / Apparent Power (V·A) ]
Thanks for sharing. I wanted to draw attention to one thing. In your calculation you basically applied the power factor to the Power output by dividing the power output by the pf. While mathematically the end result will be the same, I think it is important to recognize that theoretically the pf is more like a ‘knockdown’ of the electrical input power rather. That means, strictly from the engineering point of view and dir the purposes of understanding, we ought to multiply the input power rather by pf rather than dividing the output by the pf.
is the Power factor given on the name plate ? Because it seems like you gave yourself .7 equals the power factor. But I didn't see where you solved for that.
Good morning, I have a VFD in which three phase induction motor is attached to and I have a capacitor bank to correct power factor lag. I want to know if it's a good idea to implement both power factor correction and a VFD?
As long as your project has the budget for it haha but usually with VFD's and its ability to soft start the motor it helps the motor not lose power which overall helps the power factor. So I would say if you need a VFD and still using capacitor banks then either the sizing of the motor, Tranformer, Cable or VFD is sized wrong or them some complexed loads which is go require a nice budget lol
7460w / 240v = 31amps .. vs your calculated 59.2amps Is this motor really wasting 28 amps ... That seems pretty extreme. So in this case, the electric company would be charging you nearly double the power that you're actually using. Is this common? Or is this example a particularly bad efficiency?
An induction motor has 40kVA input power, a power factor of 82 percent, and an efficiency of 92 percent. What is the horsepower? How do you solve this?
Thanks for your tutorial it is very helpful, if there is no info about the motor how can find efficiency, voltage source current resistance power voltage required Vin Vout, maximum and minimum voltage and so on. Can you help us please. Many thanks
Earned a subscription from me!!! Thank you so much for the simplicity and straightforward explanation 👊🏻
Great video! However It's important to clarify that the current calculation considers a single phase motor. You need to multiply the voltage by √3 for 3 phase motors.
Single phase: A = 1000 ⋅ kVA / V
Three phase: A = 1000 ⋅ kVA / (√3 × V)
√3 = tan(60°) 🥸
This was a good video for a quick refresh :) Thanks!
Thank you for the clarification. So if you measure each phase of this motor at fully loaded conditions the reading will be approximately 59.2 amps correct?
Just a little correction on your triangle. Power factor is the cos(theta) of the angle (we call theta) you have labeled as 0.7. 0.7 (the power factor) is the result of that calculation, not the angle itself. Just a little clarification, since you can’t work out the equations otherwise. Thanks for the video!
you are right. should be cos-1 70 in order to get the angle.
The best way to learn this stuff is by picking up the following book...
- Electronics Fundamentals: A Systems Approach 😉
If you're not good at Algebra and Trigonometry then don't waste your time. Personally, I recommend at least Pre-Calculus due to all the logarithmic equations when it comes to Capacitor/Inductor calculations.
Power Factor is the ratio between [True Power] & [Apparent Power]. Ideally, you want a Power Factor of (1) which means there is no Reactive Power hence [True Power] & [Apparent Power] are equal thus no wasted energy 😉
Power Triangle
---------------------------
- True Power = Adjacent Side
- Reactive Power = Opposite Side (wasted energy)
- Apparent Power = Hypotenuse
- Cosφ = [ Adjacent Side / Hypotenuse ]
- Cosφ = [ True Power (W) / Apparent Power (V·A) ]
- Power Factor = Cosφ
If for some reason you need the phase angle...
- φ = Cos⁻¹ [ True Power (W) / Apparent Power (V·A) ]
Thanks for sharing. I wanted to draw attention to one thing.
In your calculation you basically applied the power factor to the Power output by dividing the power output by the pf.
While mathematically the end result will be the same, I think it is important to recognize that theoretically the pf is more like a ‘knockdown’ of the electrical input power rather.
That means, strictly from the engineering point of view and dir the purposes of understanding, we ought to multiply the input power rather by pf rather than dividing the output by the pf.
H0w can I take this course? I'm currently at
SOROHT SHAW
Would like to know the PF for a VFD driven Induction Motor
is the Power factor given on the name plate ? Because it seems like you gave yourself .7 equals the power factor. But I didn't see where you solved for that.
Good morning, I have a VFD in which three phase induction motor is attached to and I have a capacitor bank to correct power factor lag. I want to know if it's a good idea to implement both power factor correction and a VFD?
As long as your project has the budget for it haha but usually with VFD's and its ability to soft start the motor it helps the motor not lose power which overall helps the power factor. So I would say if you need a VFD and still using capacitor banks then either the sizing of the motor, Tranformer, Cable or VFD is sized wrong or them some complexed loads which is go require a nice budget lol
7460w / 240v = 31amps .. vs your calculated 59.2amps
Is this motor really wasting 28 amps ... That seems pretty extreme. So in this case, the electric company would be charging you nearly double the power that you're actually using. Is this common? Or is this example a particularly bad efficiency?
An induction motor has 40kVA input power, a power factor of 82 percent, and an efficiency of 92 percent. What is the horsepower?
How do you solve this?
thank you for the video.
Nice IKEA Chair, I have one at home. I am from the UAE
Thanks for your tutorial it is very helpful, if there is no info about the motor how can find efficiency, voltage source current resistance power voltage required Vin Vout, maximum and minimum voltage and so on.
Can you help us please.
Many thanks
Thank you, now the calculation makes sense
Can’t you just look at the name plate and look for the FLA to get the current ?
Yup but when you’re in school they ask stuff like this.
But with all due respect, it was very helpful. Thank you.
The coffee peace was so pointless lol
Switching to facecam is no bueno
VA= V•I•√3
I really wish you would shorten your intro and get to the point. People are on UA-cam so that they can cut the crap and get the info they need.