Exactly!

keep your head up. It gets better.

Billing demand charges are part of the total cost of purchasing electricity and should be included when calculating the overall average cost per kilowatt-hour since they contribute to the final price.

Thank you for the kind words, glad to hear it was helpful for you! Per unit seems much more harder and confusing at first but it is straight forward once you understand it.

which time stamp?

@ 11:12

@ I don’t get why you assume that all the current in the motor at no load is going to your stator resistance.

This is at 13:20

@@kinz8727 The reason why the total three-phase kVAR is not being divided by 3 at 11:12 is because the three-phase line values are being used in this formula. In the video at that time stamp, the magnetizing reactance (Xm) is being calculated by: Xm = VL²/Q-3ø Xm = (575 V)²/(49,513 VAR) Xm = 6.68 Ω Alternatively, you could also use single-phase values by dividing the three-phase kVAR by three, and the line voltage by √3: Xm = VL²/Q-3ø Xm = (VL/√3)²/(Q-3ø/3) Xm = (575 V/√3)² / ( 49,513 VAR / 3) Xm = 6.68 Ω It's the same result and either method may be used.

@@emmab-hq6ml part 4 and 5 will be released next week

Part 4 - kzbin.info/www/bejne/hanZd2xunqh_r6s Part 5 - kzbin.info/www/bejne/rqmwimWqn6qNmposi=fXtxWqwn-s1XRqqN

The USA uses mostly U.S. customary units (sometimes referred to as imperial units). This video for the PE exam which is a USA based licensing exam. I guess you could say we like our feet 🦶😀

would you mind clarifying? The period does not occur @ 4:00 in the video but the period shown in the video is correct.

glad you enjoyed it. I'm not sure if that particular video is on our KZbin channel it most likely follows this video in our online course at www.electricalpereview.com

My pleasure. Glad you found it helpful

It's a theoretical value that helps to simplify the calculation of inductance since current does not flow uniformly throughout the conductor due to the skin effect (en.wikipedia.org/wiki/Skin_effect).

It is unfortuante that this particular formula is not included in the NCEES Reference Handbook, however, you'll need to be familiar with it in order to solve two wattmeter method questions on the PE exam. For example, take a look at problem #49 in the official NCEES Practice Exam. It's a two wattmeter method problem that is solved using the formula in this video.

Glad you enjoyed it.

Current official passr rates from ncees.org: PE Exam - Electrical and Computer: Power 1st Time Pass: 59% Repeat Pass: 37%

@@electricalpereview Many thx，In China the rate is 10% around ，I‘m a instructor of Registered Electrical Engineer Exam in China。

Thanks for the kind words.

When a complex number is "negative" it really just means that it is ±180 degrees out of phase. For example: For voltage, just swap the positive and negative voltage polarity references (the "+" and "-" signs trade places), and for current, change the direction from forward to backward (or vice versa). Va(1) is positive due to KVL, since the voltage drop across Z(1) is smaller than Vp(1): Va(1) = Vp(1) - Vz1 Va(1) = Vp(1) - Ia(1)•Z(1) [see 7:14 for these steps]. Va(2) and Va(0) are "negative" due to where these are located on the circuit and their polarity direction. They appear across each sequence impedance Z(2) and Z(0), however, they are in the opposite polarity compared to the actual voltage drop across each of these sequence impedances (for a load, the voltage drop is positive where the current enters, and negative where the current exists). [See 8:47 for Va(2) and 11:28 for Va(0)]. The reason why these sequence voltages are "negative" is because during fault conditions they act as a temporary voltage source that delivers fault current. A voltage source has opposite polarity compared to a load, the voltage polarity is negative where the current enters and positive where the current leaves.

Thanks for the feedback. This is about an 8 year video, all of our new videos are very clear, but I keep the old ones up because people still find them helpful🙂

that's really neat, thanks for the share!

No need to since transformer 2's ratings were chosen as the new base values in this system. That makes the new base voltage in zone 2 equal to the transformer 2's primary rated voltage.

@@electricalpereview thanks

Agreed. Will be picking these shorts back up soon. Hard to fit it all in just 60 seconds 😀

Very normal. You can find more about our class for the NCEES Power PE Exam (and FE) at www.electricalpereview.com

On your calculator, make sure you are using the negative sign to the left of the enter button, not the minus sign below the multiplication sign.

@@electricalpereview yes, it works!! thank you so much

Failure mode philosophy for this would be metal fatigue

@@FemboyEngineer personally I've only ever seen timer auxiliary contacts or a timer bit from PLC used to bypass the autotransformer to apply full voltage once the motor is close to rated speed but I'm sure a centrifugal switch would work just the same but it would need to be already built into the motor which means you'd need to purchase a specialty motor as opposed to using a timer contact with any existing standard induction motor.

You are going to be spending a lot of time learning about either mechanical engineering or electrical power engineering. Choose the exam that most fits your job the most since what you will learn while you are studying will help you in the future.

If the gradient started at N = 3 (where the first $0 of the gradient starts) instead of N = 1 like in this video, then using the P given G formula (P/G) would transform the gradient to a single present worth value at N = 2. From there you could use a P given F formula (P/F) to move that single payment to the true N = 0 present worth value.

The circuit parameters solved for in the video are already referred to primary, no other step is needed since they were solved for using the open circuit and short circuit test values that were applied and measured at the primary terminals.

Thanks for the clarification. I missed that the primary is 120V and the secondary is 240V. You tend to think that the primary is always the high voltage side.

Our sister website, www.electricalfereview.com. It's a 100% free forever online course for the FE exam that I am currently working on that you can study from.

glad it helped!

Thanks for your comment, glad it helped!

Which video? Link it here and I can check it

check for the video titled "How to Solve Transmission Line Inductance and Reactance Problems (Electrical Power PE Exam)". i can't link it because it says unavailable@@electricalpereview

inductance for 3 phase@@electricalpereview

@@kozaTG I was only able to locate single-phase transmission line inductance and inductive reactance: kzbin.info/www/bejne/ZpamiKyvbsSXb68 Can you tell me where you are seeing the link for three-phase that will not open?

yeah that's what i mean where's the 3 phase one?@@electricalpereview

Solving for just the rotor resistance R2, and not the rotor slip resistance R2/s

Glad to hear it