Spin-Up Sample Exam Problems Discussion

[Power12]

hmmm..... I get the same answer as Maulinite when doing it the PU method.

 

[mauldinite]

Ok, further reading clarified that you treat the S-values as you would capacitors. Thanks for helping me out everyone!

I've always used the per-unit method since most of the time, they throw in some impedance in the line. Also, the Camera book describes looks like it describes the per-unit process well enough, but then in the example problem (36.8), the solution uses the per-unit method! That's where I got confused.

Still, using the per-unit method I would think I should be getting the same answer. If anyone has any more input, please let me know!

 

[mauldinite]

Woops. Wrote that last post a little too fast. I meant that the Camera book describes the MVA method well enough.

 

[spinup]

I'm having some trouble understanding the solution for problem 3-02:

The diagram shows 2 gens in parallel, G1 is 175KVA, 2%; and G2 is 275KVA, 4%. These are tied to a transformer that is 400KVA, 5%, with a fault on the low side of the transformer. The question is as follows:

"Using the provided diagram, the three-phase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the short-circuit current at the fault."

The method used in the solution is to calculate the MVA fault contribution for each generator and add them. This value is taken in parallel with the transformer fault MVA (Srated/Zpu). In other words,

(Ssc-Gens * Ssc-TX)/(Ssc-Gens + Ssc-TX)

From there, total fault current is calculated as (Ssc-Tot)/(Vrated*sqrt(3)). Their answer is 6,375 A.

That goes against the method I've known as the MVA method (from the Camara PPI book). With that method, I would select a base-S (I used 400KVA). With that base I would calculate the new Zpu values for each generator and the transformer. Since the gens are in parallel, I then have

(Zpu-Gen1 * Zpu-Gen2)/(Zpu-Gen1 + Zpu-Gen2) + Zpu-TX = Ztotal

With this I calculate S-FC = Sbase/Ztotal = 7.062 MVA.

From there, I get Isc = Ssc/(Vrated * sqrt(3)) = 8495 A.

My biggest hangup is that the Spinup solution takes the generator power contributions in series and then it takes the total contribution in parallel with the max of the transformer. Can anyone steer me in the right direction here?

It is important that everyone understand that the answer should be the same no matter which method is used to come up with the solution. You should use the method that you are most comfortable with on the exam. Spin-Up will be providing this question as the "Question for the Week" with the solution done both ways. The question and both solution methods will be posted tomorrow (Sunday).

www.spinupexams.com

Joan

 

[spinup]

I'm having some trouble understanding the solution for problem 3-02:

The diagram shows 2 gens in parallel, G1 is 175KVA, 2%; and G2 is 275KVA, 4%. These are tied to a transformer that is 400KVA, 5%, with a fault on the low side of the transformer. The question is as follows:

"Using the provided diagram, the three-phase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the short-circuit current at the fault."

The method used in the solution is to calculate the MVA fault contribution for each generator and add them. This value is taken in parallel with the transformer fault MVA (Srated/Zpu). In other words,

(Ssc-Gens * Ssc-TX)/(Ssc-Gens + Ssc-TX)

From there, total fault current is calculated as (Ssc-Tot)/(Vrated*sqrt(3)). Their answer is 6,375 A.

That goes against the method I've known as the MVA method (from the Camara PPI book). With that method, I would select a base-S (I used 400KVA). With that base I would calculate the new Zpu values for each generator and the transformer. Since the gens are in parallel, I then have

(Zpu-Gen1 * Zpu-Gen2)/(Zpu-Gen1 + Zpu-Gen2) + Zpu-TX = Ztotal

With this I calculate S-FC = Sbase/Ztotal = 7.062 MVA.

From there, I get Isc = Ssc/(Vrated * sqrt(3)) = 8495 A.

My biggest hangup is that the Spinup solution takes the generator power contributions in series and then it takes the total contribution in parallel with the max of the transformer. Can anyone steer me in the right direction here?

It is important that everyone understand that the answer should be the same no matter which method is used to come up with the solution. You should use the method that you are most comfortable with on the exam. Spin-Up will be providing this question as the "Question for the Week" with the solution done both ways. The question and both solution methods will be posted tomorrow (Sunday).

www.spinupexams.com

Joan

QFTW has been uploaded.

www.spinupexams.com

Joan

 

[mauldinite]

Ahhhhh, was swapping Sbase new and Sbase old the whole time! Thanks Spin-Up!

 

[Insaf]

Spin up problem 2-65:

I believe for calculating "daily load factor" 24 hours should be used instead of 6 hours. Any comments about the spin-up solution will be appreciated.

Note: This answer could be right if there is any "quarter day load factor".

Thanks,

 

[cali78]

Spin up problem 2-65:

I believe for calculating "daily load factor" 24 hours should be used instead of 6 hours. Any comments about the spin-up solution will be appreciated.

Note: This answer could be right if there is any "quarter day load factor".

Thanks,

Load Factor is also a function of availability. Since the generation was only avail for 6 hours for a given day and used at half capacity for 5 hours and zero capacity for the other hour than the Load Factor is

((.5 hrs use )(5 hr avail) + (0)(1 hr avail))/ ( 6 hrs of Max Capacity) = 42%

The load factor for the generator is a measure of the output of the generator compared to the maximum output the generator could "produce". Since the generator is only avail for use to produce power for 6 hrs daily, then one needs to use 6 hrs. If the generator was avail for 24 hours then 24 hrs would of been used. The problem was a little tricky.

 

[Insaf]

Thanks Cali78!!

 

[Insaf]

Spin up problem 2-69 and 3-01:

Why the solutions methodologies are different ?

Thanks,

 

[Ivory]

Spin up problem 2-69 and 3-01:

Why the solutions methodologies are different ?

Thanks,

Careful on those. They are different type of questions, therefore different solution methodologies. The sawtooths are the same in the picture, but one of them are asking for average and the other average "magnitude". They are different questions. I used to mix them up and not pay attention to that word "magnitude" which changes things.

On the real exam I am going to read all the questions slowly, to make sure I do not get misled and provide the wrong answer.

 

[EEmarcus]

Back to question 2-65 the question asks for load factor. My formula for this is LF=(Avg Load)/(Peak Load). But the answer provided is a formula for capacity factor. CF= (Avg Load for a period)/(Output Capacity of Plant). This is according to my reference book. Aren't these two things different?

The peak load is 500KW and the avg is 2500MW/(6hrs) = 416.67KW

My solution thus is 416.67/500= 83%.

 

[Insaf]

Spin-up prob 3-78: Why MVAL= 12, any feedback will be appreciated ?

Thanks,

 

[Ivory]

Spin-up prob 3-78: Why MVAL= 12, any feedback will be appreciated ?

Thanks,

SA = 6 X 2 = 12

SB = 8 X 3 = 24

When in parallel, SB = 2SA

The max load of Trans B is 8MVA which limits Trans A to 4MVA. Therefore the total Max load is 4MVA + 8MVA = 12 MVA

 

[BH_Cubed]

Problem 5-53

I'm not quite sure how where the -38.25<-27.8 is coming from in the second line of the solution. It seems like they multiplied the current 100<-36.9 by the impedance 0.51<9.1 then by the 750/1000 for the length of the cable. Shouldn't you multiply the impedance by the 750/1000 first because the impedance is per unit length? When I do this, I get a different answer. I calculated Z=(.51<9.1)*(750/1000) = 0.38<9.1 . Then IZ = (100<-36.9)*(0.38<9.1) = 37.98<-27.8. If you subtract that from 480 you get 442V at the load not 422V. I think the solution may be incorrect. or maybe I'm doing something wrong? 442V is answer C

 

[BH_Cubed]

oh way nevermind, i'm missing the root 3. duh

:beat:

 

[GreenNGold]

I had a few questions:

1) Spin Ups Problem 1-80: When a generator is overloaded, the frequency will drop...but when will a generator trip offline?

2) Spin Ups Problem 2-4: Can anyone help explain how to differentiate between buck and boost autotransformers?

3) Spin Ups Problem 2-15: Are the equations for parallel transformers & parallel generators the same?

KVA1 = ((KVA1/Z1%) / ((KVA1/Z1%) + (KVA2/Z2%))) * KVAL

KVA2 = ((KVA2/Z2%) / ((KVA1/Z1%) + (KVA2/Z2%))) * KVAL

I used the two above equations and solved for MVAL for each case. I got MVAL = 9 for case 1 and MVAL = 6 for case 2. Then I used 6MVA as my load because anything higher would overload generator B, plugged it back into this equation KVA2 = ((KVA2/Z2%) / ((KVA1/Z1%) + (KVA2/Z2%))) * KVAL and got MVA2 = 4 MVA. Does the way I did this make sense?

4) What are the ranges for low voltage, medium voltage, and high voltage? I googled it and it says Low voltage is 0-1000V, medium voltage is 1kV to 36kV, and anything higher is high voltage.

Thank you.

 

[Ship Wreck PE]

A generator is normally protected with a overcurrent breaker located on the generator, and when the current exceeds the set amount for a given time, it will trip out the circuit.
Most newer generators are programmable and they will not transfer until the frequency, voltage, and phases are all correct, They can be programmed to drop out at low voltage, high voltage, low frequency, high frequency, Phase loss, or an overcurrent condition, But they can be programmed to not do all of this also.

A buck transformer lowers the voltage and the boost transformer raises the voltage, and in the spinup workbook the buck transformer is noted by the transformer diagram being wider than the boost..


LV = Voltage levels that are less than or equal to 1 kV

MV = Voltage levels that are greater than 1 kV, but less than or equal to 69 kV

HV = Voltage levels that are greater than 69 kV, but less than or equal to 230 kV

EHV = Voltage levels that are greater than 230 kV, but less than or equal to 800 kV

UHV = Voltage levels that are greater than 800 kV

 

[GreenNGold]

For problem 2-4, isn't it a buck because of the location of the dots? I don't think I completely understand what you mean by the buck transformer being wider than the boost. I always thought step-up and boost were the same and step-down and buck were the same but the answer for this problem is step-up, buck.

I assume the transformer is a step up because the load is connected across the series and common winding, and it is a buck because the currents will cancel due to the dot convention?

 

[Ship Wreck PE]

I sold my spinup book so I really don't recall the problems.