Preparing for a power system technical interview can be stressful. Depending on the role you apply for, technical questions will be asked to assess your overall knowledge of power systems analysis, power devices, and system protection. In this article, I will discuss ten questions you will likely see in a power systems technical interview. Answering these questions correctly will show the interview panel your understanding of important power system concepts and may increase your chance of being hired. However, the key points in the sample answers are not exhaustive. And, you should NEVER remember the answers mechanically, but ALWAYS understand the concepts being asked and describe them in your own words. Your answers should be relevant and concise. Don't explain every detail (this is very important!), but you may ask the interviewer if your answer fully addresses their questions. Be ready to defend your answer and elaborate if needed.
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Question 1: What is the AC power flow? What is the "DC" power flow? What are the pros and cons?
While these are technically three questions, they are highly related. Power flow is arguably the most important concept of power system engineering. Before interviewing for a power engineering job, you should know what power flow is and be able to compare its variations.
Sample answer:
The objective of the AC power flow is to
- determine the voltage and current levels at each bus in a power system, and
- calculate the active and reactive power flow in each transmission line/transformer under a given set of operating conditions, such as the load and generation levels and the network topology.
The "DC" power flow is a simplified version of the AC power flow. The classic DC power flow assumes the following:
- Reactive power is negligible (Q = 0).
- The bus voltages are fixed at their nominal values, i.e., 1 p.u.
- The angle difference across any line is small.
Compared to the AC power flow, the "DC" version is faster, non-iterative, and guarantees convergence. On the other hand, the AC version is more accurate and a better representation of physical reality.
Question 2: What are the typical bus types in the AC power flow analysis? Explain them.
As a power system engineer who regularly performs power system analysis, it is important to understand different types of buses and their modeling assumptions to be able to interpret power flow results.
Sample answer:
Generally speaking, there are three types of buses in the power flow analysis. They are the PQ bus, PV bus, and slack/swing bus.
- PQ bus: These buses represent the loads in the system. For PQ buses, the real and reactive power drawn from the bus is known. The voltage magnitude and angle at these buses are solved in the power flow.
- PV bus: These buses represent the generators in the system. For PV buses, the real power injection (from the generator) and voltage magnitude at the bus are known. The reactive power output and the voltage angle at the bus are solved in the power flow.
- Slack or swing bus: This is the bus at which the system's real and reactive power balance is maintained. For the slack bus, the voltage magnitude and angle (normally set to 0 as the reference) are fixed.
The above answer should give you an A. But if you really want to impress the interviewer, you can talk about the PV-PQ bus conversion logic. In short, A PV bus will become a PQ bus if the reactive power regulation of the generator is exhausted.
Question 3: What are the critical inputs to solve a power flow?
You will unlikely need to solve a power flow by hand in an interview. However, you may be asked to describe the critical inputs to run a power flow to show your understanding of the concept.
Sample answer:
You need the following information to solve a classic AC power flow:
- The real and reactive power drawn from each PQ bus.
- The real power output of the generator(s) and the voltage magnitude at each PV bus.
- The voltage magnitude and the angle at the slack bus.
- Series resistance (R), reactance (X), and charging susceptance (B) of each transmission line, and reactance of each transformer.
- The network topology.
Question 4: What is a transmission line's Surge Impedance Loading (SIL)?
Surge Impedance Loading (SIL) is important in transmission design and operations. It helps power engineers and operators estimate the voltage profile and determine if reactive compensation is needed.
Sample answer:
- Surge Impedance Loading (SIL) is the real power flow level (in MW) that makes a transmission line act like a resistor – the net reactive power injection or consumption across the line is zero.
- When the active power flowing in the line is less than its SIL, the line produces reactive power.
- When the active power flowing in the line is greater than its SIL, the line consumes reactive power.
Question 5: How can you correct a low voltage problem in real-time system operations?
Low voltage could occur in real-time operations for various reasons. Low voltages, if not corrected appropriately, could lead to voltage collapse and uncontrolled loss of load. As a power engineer, you should know how to mitigate a voltage issue and discuss their trade-offs if needed.
Sample answer:
Various actions can be taken to correct a low-voltage problem.
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- Switch in shunt capacitors. (increase Q injection)
- Switch out shunt reactors. (reduce Q consumption)
- Stop pumping hydro. (reduce load)
- Turn on generators in the low-voltage area. (increase Q injection)
- Increase the terminal voltage set point of generators in the low voltage area. (increase Q injection if they still have Q regulation available)
- Switch in series capacitor if applicable. (reduce Q consumption across the line)
- Return outaged transmission lines if applicable. (lower overall system impedance and reduce Q consumption across the line)
- Load management, including shedding firm load (a last resort).
The interviewer will likely not expect you to answer all the options above, but you should at least give a couple of options and discuss their trade-offs.
Question 6: How can you correct a high voltage problem in real-time system operations?
Compared to low voltage issues, high voltage, if not handled appropriately, could be equally harmful, especially to lines and generators. As a power engineer, you should know how to mitigate a voltage issue and their trade-offs. You can pretty much take the options in the question above and reverse them to get the correct answer for this one.
Sample answer:
Various actions can be taken to correct a high-voltage problem.
- Switch out shunt capacitors.
- Switch in shunt reactors.
- Start pumping hydro and charging battery storage.
- Turn off generators in the high-voltage area. (only applicable in a "load pocket," why?)
- Decrease the terminal voltage set point of generators in the high voltage area.
- Switch out the series capacitor if applicable
- Switch out lightly loaded transmission lines if applicable (study first)
Again, the interviewer will likely not expect you to answer all the options above, but you should at least give a couple of options and discuss their trade-offs.
Question 7: What is the difference between phase voltage and line voltage? What is the relationship between the phase and line voltages in a balanced three-phase system?
These two questions are highly related, so I again lump them together. One important difference between the high-voltage power system and the residential electricity supply is that the former uses a three-phase system. As a power system engineer, it is critical to differentiate the two systems and understand their mathematical and physical relationships.
Sample answer:
- Line voltage is the voltage between two phases, e.g., phase A and B, or phase B and C, whereas the phase voltage is the voltage between the given phase and neutral (ground).
- In a balanced three-phase system, the line voltage is √3 times the phase voltage.
Question 8: What is an Area Control Error (ACE)?
You should never work in the control room if you don't understand ACE. You may not remember the exact ACE equation (although you should), but you must know the ACE components and understand why ACE is important in real-time operations.
Sample answer:
- Area Control Error (ACE) is the difference between the scheduled and actual interchange in MW, considering the frequency bias and meter error contribution.
- Holding the frequency component constant, ACE indicates whether the generation and load in the system are balanced.
Question 9: What is the power factor?
Accurately describing the power factor shows a candidate's understanding of AC circuits and the power triangle. Each electric utility company has minimum power factor criteria to ensure the transmission system's efficient usage.
Sample answer:
- In power systems, the power factor is the ratio of the real power and the apparent power drawn by the load at a bus.
- The power factor can range from 0 to 1. A power factor of 0 means no real power component, whereas a 1 power factor means no reactive power component.
- The higher the power factor, the more efficient the transmission system is utilized, as it is better to compensate for reactive power locally.
Question 10: Why do the three protection zones overlap?
System protection is a broad topic. As a power system engineer, you should know basic protection schemes and their design philosophy. The three-zone design is one of the fundamental protection designs you should know.
Sample answer:
- The main reason for overlapping protection zones is to provide redundancy so that secondary/tertiary relays can back up the primary ones if they fail to operate.
- For example, if the primary (Zone 1) relay fails to detect and clear the fault, Zone 2 relays will kick in with a delay and clear the fault.
I hope the above questions and sample answers are helpful to you. If you are interviewing for a position in the energy market, the following article summarizes 10 popular questions/concepts you know.
Additionally, for information about power systems engineering career options, please refer to my two other articles below:
Thanks for reading.
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When exactly does the generator turn into a motor?
When its prime mover stops applying torque to the shaft.
The only difference betwen a generator and a motor is direction of power flow. If you shut off the fuel to one of the diesels the electrical machine will commence being a motor and spin the diesel at synchronous speed, drawing power from the other machine.. At least until some protective relay trips it.
The old British term "Dynamo" was to me a more intuitive name than motor or generator.
When there is noe more reactive export - only import?
There's a naming convention.
For the scenaio you described , Reactive current circulates between the machines.
Have you ever played with the voltage adjust? There's a point where current meter indicates a minimum, either raising or lowering the voltage adjust from there causes current to rise. At that point your machines are matched and any reactive power is going out to the external load.
So let's pretend for a moment you have no external load, just two machines in parallel.
When they are matched the current will be zero(assuming engine throttles are also matched).
Raise the voltage on either machine and current increases - reactive current is circulating between the machines.
By convention, the overexcited machine is said to be exporting and the underexcited one importing. Also by convention, the overexcited one is said to be "in the lag" and the underexcited one "in the lead".
Lead and lag refers to the phase of current WRT terminal voltage. It'll be 90 deg ahead or behind.
Leading current AIDS the machine's internal field, raising terminal volts
Lagging current OPPOSES it lowering terminal volts
For our simple case of two machines circulating only reactive current,
one is in the lead and one is in the lag
leading machine(underexcited) is accepting(importing) reactive power and lagging machine(overexcited) is supplying(exporting) it.
The point at which the machine changes from exporting to importing would be the point where circulating current falls to zero, ie machines matched.
So the best answer to your question would be "When reactive current is [STRIKE]zero[/STRIKE]" edit er, make that
minimum but i thought it needed that lead-in. Sorry if I'm being too long winded.
Does that help any?
It's best to start with minimum system, just two machines and grasp how reactive current helps or hinders the internal field.
Then add an external load , it just offsets things by however much reactive current is drawn by the load. So instead of current going through zero as you change voltage it'd go through a minimum.
Since you're on a boat, I encourage you to experiment with your machines.
I had the luxury of two 450kw generators and freedom to do these things, there's nothing like grabbing that knob and getting the feel of how much knob gives how much circulating current. I could easily cause 300 amps of circulating current, therer wasn't much impedance between my machines.
So, you asked initially to effect "is one generator applying 10 volts to the other one'
well almost. It's applying enough current to raise the other one's voltage by ten volts(?)
I hope i answered the right questions - sometimes i wander.
old jim
I'm just left wondering why old jim did not answer sooner.
I feel like I'm maybe making a nuisance of myself here by overposting, so i stay on EE and Nuke threads mostly. And i wait for others to have their say usually somebody else gives a better answer than i could have.
(I think this is due to transformers and some measurement inaccuracies of the AVR sensing voltage - correct me if I'm wrong)
i think you're exactly right on that one.
So, secondly i think you misunderstood my question.
Quite likely, i do that a lot. It's called 'Asperger's Syndrome".
Anyways, my first initial question in this thread is that when a generator is importing VAR, is that current doing any work that makes the prime mover's work on the shaft easier?
No. VARS transmit no work.
There's two different ways to show that.
1. . Recall VAR current is 90 deg out of phase.
Watts is Volts X Amps X cosine of angle between them, and since cos(90deg) = zero that product is also zero - no watts so no torque. That's a math solution.
2. . If you draw two sinewaves 90 deg out of phase, label one Volts and the other Amps,
now quadrant by quadrant multiply them to get 'instantaneous' watts. You'll see by symmetry that there's equal positive and negative products so the area under the resulting curve is zero. That's a graphic solution.
Now shift them back in phase and repeat - all area is now positive (because negative volts times negative amps = positive watts).
Observe if you shift them 180 out all area is negative, which means you swapped from generator to motor.
Now it's true that VARS exchange energy between source and load for fractions of a cycle but they average out to zero. That's the positive and negative areas you saw in the graphical solution.. In a single phase machine that would make a pulsating torque but it would average zero.
Also I think that even though the current is supposed to 90 degrees leading the voltage, it still must be somewhat lower (89? 89.9?)due to heat losses in the windings providing som real power (?)
I believe you could say that. But don't get hung up on that thought. Practically, there'll always be a resistive and a reactive component to load current so it'll have some angle between -90 and +90 deg.. For learning we break load current into ideal components at zero and 90 degrees and study them separately.
And, IF a generator is importing huge amounts of reactive current from another generator (all protection functions are disabled, machine will eventually burn or pole slip will occur etc), still with prime mover input -- will that current make it easier for the prime mover to turn the shaft?
Hmmmm there's a possibliliy we're not on same track here. Try this thought:
That imported reactive current does not transmit any torque to the shaft so it niether helps nor hinders the prime mover.
What it does do is this: it adds to the flux that's made by the rotating field of the underexcited machine. Now - if that underexcited machine has so little excitation that its field can no longer handle the torque of its prime mover, it will accelerate above synchronous speed and try to run as an induction generator. That can be mechanically pretty violent so we don't want it to happen.
But the reason the machine sped up is not because the other one helped it, instead because its own field got too weak to hold on to the torque from its engine.
Let me lift a phrase fom an old post last January
If as a kid you ever played with magnets, put one on top of a table and one underneath and moved them from below, you know magnets can transmit force.
Of course so can electromagnets. It shouldn't be much of astretch to realize they can transmit torque, too.
Rotor of a synchronous machine is an electromagnet that physically rotates. A stator with external voltage applied is also an electromagnet and its field rotates because of the precession of three phase currents around windings. In a motor the stator field drags the rotor along, in a ngenerator the rotor drags the stator field around.
We had a long discussion with a cool guy named Bassalisk about this last January, you might check these two threads. He had some good drawings and some very good thoughts.
https://www.physicsforums.com/showthread.php?t=
https://www.physicsforums.com/showthread.php?t=
another question:
when reactive imported current is opposing the excitation current (in the rotor, right?), does that mean that the excitation current is canceled out?
Well - it's flux we should be thinking of, and magneto-motive force MMF.
Flux is MMF divided by (properties of the iron magnetiic path ~ reluctance)
and MMF is amp-turns.
The amp-turns of the reactive armature current oppose the amp-turns of the field
so it's the MMFs that cancel each other out.
Both currents still flow but the end result is not unlike they canceled one another. That's called Armature Reaction and is a big part of those two links i gave.
Or does the AVR somehow turning down the excitation by some kind of detection of something?
The AVR measures reactive current and modifies the excitation to make the machines share vars gracefully. Often they measure phase B current and add that to phase AC voltage, note they're already 90 degrees out of phase so it's a dirt simple way to get var correction. It'll be called "Reactive Droop". Typical setting is 5% meaning full reactive amps would lower voltage by 5%. That way a machine that's hogging vars will be cut back by its AVR. Or a slacker will be kicked up a little.
If machines are tied together by transformers you'd subtract transformer impedance from that 5% setting. Note that with a high impedance transformer , > the droop you want, you might have to reverse the AVR droop setting to make it look like there's 5% impedance between machine and bus. Then it'll be callaed a "line loss compensator" or "reactive rise" or something the manufacturer found to his liking. Be careful with that adjustment - it chenges the reactive compensation from negative into positive feedback so the AVR gets real touchy, might need more damping.
Sigh it's endless, but really fun. I envy you that setup - have a good time !
I sense you want an intuitive feel for these machines. Until tha sinks in that i can't make myself believe the formulas, afterward they come easily.. I hope i have helped you
Wow - an engineer on a boat in China. Just like "Sand Pebbles" , one of my favorite movies.
old jim