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Spin-Up Sample Exam Problems Discussion

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Please ask for explanation of Spin-up sample exam problems. Hope forum members will help to understand the problems and their solutions.

Thanks,

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Q. 1-56: In electrical engineering, average value of sine wave is not zero rather it is 63.7% of peak value. Its always averaged over half cycle (not over full cycle).

Here is the math:Vavg=(1/pi) [(Integrate over 0 to pi (sinө dө)]=0.637Vm.

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Q. 1-56: In electrical engineering, average value of sine wave is not zero rather it is 63.7% of peak value. Its always averaged over half cycle (not over full cycle).

Here is the math:Vavg=(1/pi) [(Integrate over 0 to pi (sinө dө)]=0.637Vm.

In Camara's reference book in Table 27.2 , it says the average of a Sine Wave is 0. Half wave is .637

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The integral of a sine wave over a full period (360 degrees) is = 0

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I noticed that question last night. The question states "the sine wave shown," which is a full cycle. For the full period shown in the diagram, the average is 0.

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In electrical engineering, we used to have “Form Factor” of sine wave which is defined as ratio of RMS value to average value. If average is zero, what will happen?

My understanding as we learned from college:

In case of symmetrical alternating signal, the average value is zero. Hence in that case, the average value is obtained by adding or integrating the instantaneous values of signal over one half –cycle only.

Additional:

In the same book, problem 2-10, answer described as “The average magnitude of sine wave is the peak *(2/pi) (i.e., 63.7% of peak value). What is the difference between average value and average magnitude?

Edited by Insaf

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IWhat is the difference between average value and average magnitude?

The average value takes into account that for half the period the value is negative and it therefore subtracts. The magnitude is always positive and would therefore be the output of a full wave rectifier.

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Spin-up prob 2-14:

Slack or swing bus where voltage magnitude and angle are specified (generally angle set to zero). Normally there can only be one slack or swing bus in the system and it is generally chosen from among the voltage controlled buses.

In this problem slack and swing buses are described as different, but its not. It looks voltage controlled buses referred as generator’s bus (major types of buses: Load Buses, Voltage Controlled buses and Swing / Slack bus). According to the definition of the buses, slack or swing bus (one of the voltage controlled bus chosen as slack or swing bus) keeps voltage constant.

Any explanation from Spin-up or forum members will be appreciated.

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Spin-up problem 2-15: In solution MVAL=6, from where it is coming?

Thanks,

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Spin-up prob 2-14:

Slack or swing bus where voltage magnitude and angle are specified (generally angle set to zero). Normally there can only be one slack or swing bus in the system and it is generally chosen from among the voltage controlled buses.

In this problem slack and swing buses are described as different, but its not. It looks voltage controlled buses referred as generator’s bus (major types of buses: Load Buses, Voltage Controlled buses and Swing / Slack bus). According to the definition of the buses, slack or swing bus (one of the voltage controlled bus chosen as slack or swing bus) keeps voltage constant.

Any explanation from Spin-up or forum members will be appreciated.

The generator bus is the answer. It is known as the voltage controlled bus. My book does not say the slack and swing buses are described differently. I see the slack bus as a possible answer or swing bus as another possible answer. I think the answers are just trying to confuse you.

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Spin-up problem 2-15: In solution MVAL=6, from where it is coming?

Thanks,

For these type of problems I use the trial and error method. I multiply the generator MVA by the other generators impedance.

SA = 2 times 3 = 6

SB = 3 times 4 = 12

Then I see SA is equal to half of SB

Since Gen B has a max load of 4MVA, therefore Gen A is half of 4 MVA which equals 2MVA

So MVAL = SA + SB = 2MVA + 4 MVA = 6MVA

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I like this solution better. The solution from the test takes more more time to calculate. I tried to use this method to solve one of the sample from the NCEES and the answer is easier to understand.

Thanks Ivory.

There is a couple problems calculating the fault current by using MVA method. Usually, we need to choose the base MVA, base Voltage, then calc the total fault Z and fault MVA. Then using fault MVA and fault Z to calc the fault current. But Spinup using different way to solve it. It confuse me since I use SpinUp method to solve one of the problem from NCEES (from 500 series - the one with 100MVA base, 20mile transmisstion line, 7.5KVA Transf) and it did not work. Any idea??

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I like this solution better. The solution from the test takes more more time to calculate. I tried to use this method to solve one of the sample from the NCEES and the answer is easier to understand.

Thanks Ivory.

There is a couple problems calculating the fault current by using MVA method. Usually, we need to choose the base MVA, base Voltage, then calc the total fault Z and fault MVA. Then using fault MVA and fault Z to calc the fault current. But Spinup using different way to solve it. It confuse me since I use SpinUp method to solve one of the problem from NCEES (from 500 series - the one with 100MVA base, 20mile transmisstion line, 7.5KVA Transf) and it did not work. Any idea??

There is two ways to solve these questions, either PU or MVA. Show us the way you solved it with your calculations. We could then point out where you went wrong. Both methods will give the same answer.

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I like this solution better. The solution from the test takes more more time to calculate. I tried to use this method to solve one of the sample from the NCEES and the answer is easier to understand.

Thanks Ivory.

There is a couple problems calculating the fault current by using MVA method. Usually, we need to choose the base MVA, base Voltage, then calc the total fault Z and fault MVA. Then using fault MVA and fault Z to calc the fault current. But Spinup using different way to solve it. It confuse me since I use SpinUp method to solve one of the problem from NCEES (from 500 series - the one with 100MVA base, 20mile transmisstion line, 7.5KVA Transf) and it did not work. Any idea??

There is two ways to solve these questions, either PU or MVA. Show us the way you solved it with your calculations. We could then point out where you went wrong. Both methods will give the same answer.

I agree with EEVA, both methods should get the same answer. During the exam if I have enough time I will do it both ways to check myself, to make sure I solved the fault problem correctly.

Bethy, you must be talking about NCEES prob # 513. The way to solve this problem using the method from spinup is

S short circuit for the transformer is = S/Z = 7.5 X 106 /.075 = 100 MVA

S short circuit for the conductors is = V2 / Z = (12 X103)2/ 15.312 = 9.4 MVA

The total short circuit is calculated with Trans and conductors similiar to series capacitors so = ((100 X 106) (9.4 X 106))/((100X106) +(9.4 X 106)) = 8.6MVA

Isc = Ssc/(sqroot 3 X V) = (8.6 x10 6) /((1.73)(12 x 103) = 414 A which is the same answer that is in the NCEES

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Spin-up Problem 3-4:

Should the answer is d? Would you let me know why the answer is c?

Thank you

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I used the same method.

SA = 3x4 = 12

SB = 4x3 = 12

So Gen A should equal to Gen B.

SA = SB = 3MVA (can not choose 4MVA since it's overload Gen A).

ST = 6MVA (answer C)

Thanks to EEVA and Power12 for the MVA fault method. I did the calc wrong on the series capacitors. But I have another question and need your help to understand problem with power factor correction (lagging and leading reactive power). I got confuse every time I try to solve it. I used the formula below to calc the KVAR added

Qc = P [(tan(cos-1PF1) - tan(cos-1PF2)]

For NCEES sample # 502: From the formula above Qc = 8.4KVAR. To improve the PF to .9 lagging we need to add capacitor in (Q<0). It means the reactive power Qc = -8.4KVAR (negative).

Then from #503: With my understanding the answer should have the negative sign (-) 3370KVAR. But the answer is positive KVAR (Q>0).

Am I missing something here?? The equestion asking "the total reactive power supplied by a 3 phase capacitor....Do we need to find the magnitude of the reactive power? What is the key point here??

Any easy way (rule of thumb) to recognize when it should be negative or positive KVAR??

spin-up proble 3-27 and 4-07, I got the correct answer, but the NCEES 503 confused me.

Thanks.

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Regarding problem 503:

You calculated perfectly (-) 3370 KVAR. The var supplied by capacitor is always negative and this question asking only to find the magnitude of KVAR required.

Please note that in question # 502, you have to decide what type of var (capacitive or inductive) is required to correct the power factor.

Thanks,

Edited by Insaf

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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?

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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?

Mauldinite,

You are looking at it backwards. You handle the parallel power sources as additive and the series as capacitors.

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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?

Mauldinite,

You are looking at it backwards. You handle the parallel power sources as additive and the series as capacitors.

The MVA method does not include converting to per unit. There must be an error in Camara's book.

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

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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!

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

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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

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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

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