SpinUp Sample Exam Problems Discussion
#1
Posted 31 January 2012  03:34 AM
Thanks,
#2
Posted 01 February 2012  09:36 PM
Here is the math:Vavg=(1/pi) [(Integrate over 0 to pi (sinө dө)]=0.637Vm.
#3
Posted 01 February 2012  10:25 PM
Q. 156: 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
#4
Posted 02 February 2012  12:59 AM
#5
Posted 02 February 2012  01:12 PM
#6
Posted 02 February 2012  08:20 PM
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 210, 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, 02 February 2012  08:58 PM.
#7
Posted 02 February 2012  08:44 PM
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.
#8
Posted 02 February 2012  08:49 PM
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 Spinup or forum members will be appreciated.
#9
Posted 02 February 2012  10:08 PM
Thanks,
#10
Posted 02 February 2012  11:32 PM
Spinup prob 214:
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 Spinup 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.
#11
Posted 03 February 2012  12:19 AM
Spinup problem 215: In solution MVA_{L}=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.
S_{A} = 2 times 3 = 6
S_{B} = 3 times 4 = 12
Then I see S_{A} is equal to half of S_{B}
Since Gen B has a max load of 4MVA, therefore Gen A is half of 4 MVA which equals 2MVA
So MVA_{L} = S_{A} + S_{B }= 2MVA + 4 MVA = 6MVA
#12
Posted 03 February 2012  04:28 PM
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??
#13
Posted 03 February 2012  07:36 PM
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.
#14
Posted 03 February 2012  08:54 PM
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 10^{6} /.075 = 100 MVA
S short circuit for the conductors is = V^{2} / Z = (12 X10^{3})^{2}/ 15.312 = 9.4 MVA
The total short circuit is calculated with Trans and conductors similiar to series capacitors so = ((100 X 10^{6}) (9.4 X 10^{6}))/((100X10^{6}) +(9.4 X 10^{6})) = 8.6MVA
I_{sc} = S_{sc}/(sqroot 3 X V) = (8.6 x10 ^{6}) /((1.73)(12 x 10^{3}) = 414 A which is the same answer that is in the NCEES
#15
Posted 04 February 2012  08:27 PM
Should the answer is d? Would you let me know why the answer is c?
Thank you
#16
Posted 04 February 2012  11:11 PM
S_{A = }3x4 = 12
S_{B = }4x3 = 12
So Gen A should equal to Gen B.
S_{A = }S_{B = }3MVA (can not choose 4MVA since it's overload Gen A).
S_{T} = 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
Q_{c }= P [(tan(cos^{1}PF_{1})  tan(cos^{1}PF_{2})]
For NCEES sample # 502: From the formula above Q_{c} = 8.4KVAR. To improve the PF to .9 lagging we need to add capacitor in (Q<0). It means the reactive power Q_{c }= 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??
spinup proble 327 and 407, I got the correct answer, but the NCEES 503 confused me.
Thanks.
#17
Posted 05 February 2012  04:00 AM
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, 05 February 2012  04:01 AM.
#18
Posted 08 February 2012  04:12 PM
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 threephase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the shortcircuit 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,
(SscGens * SscTX)/(SscGens + SscTX)
From there, total fault current is calculated as (SscTot)/(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 baseS (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
(ZpuGen1 * ZpuGen2)/(ZpuGen1 + ZpuGen2) + ZpuTX = Ztotal
With this I calculate SFC = 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?
#19
Posted 08 February 2012  09:07 PM
I'm having some trouble understanding the solution for problem 302:
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 threephase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the shortcircuit 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,
(SscGens * SscTX)/(SscGens + SscTX)
From there, total fault current is calculated as (SscTot)/(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 baseS (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
(ZpuGen1 * ZpuGen2)/(ZpuGen1 + ZpuGen2) + ZpuTX = Ztotal
With this I calculate SFC = 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.
#20
Posted 08 February 2012  09:22 PM
I'm having some trouble understanding the solution for problem 302:
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 threephase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the shortcircuit 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,
(SscGens * SscTX)/(SscGens + SscTX)
From there, total fault current is calculated as (SscTot)/(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 baseS (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
(ZpuGen1 * ZpuGen2)/(ZpuGen1 + ZpuGen2) + ZpuTX = Ztotal
With this I calculate SFC = 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.
#21
Posted 08 February 2012  11:05 PM
#22
Posted 09 February 2012  03:04 PM
I've always used the perunit method since most of the time, they throw in some impedance in the line. Also, the Camera book describes looks like it describes the perunit process well enough, but then in the example problem (36.8), the solution uses the perunit method! That's where I got confused.
Still, using the perunit method I would think I should be getting the same answer. If anyone has any more input, please let me know!
#23
Posted 09 February 2012  04:14 PM
#24
Posted 11 February 2012  05:03 PM
I'm having some trouble understanding the solution for problem 302:
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 threephase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the shortcircuit 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,
(SscGens * SscTX)/(SscGens + SscTX)
From there, total fault current is calculated as (SscTot)/(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 baseS (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
(ZpuGen1 * ZpuGen2)/(ZpuGen1 + ZpuGen2) + ZpuTX = Ztotal
With this I calculate SFC = 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. SpinUp 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
#25
Posted 12 February 2012  01:43 PM
I'm having some trouble understanding the solution for problem 302:
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 threephase distribution system has a voltage of 480V on the secondary side. Use the MVA method to find the shortcircuit 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,
(SscGens * SscTX)/(SscGens + SscTX)
From there, total fault current is calculated as (SscTot)/(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 baseS (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
(ZpuGen1 * ZpuGen2)/(ZpuGen1 + ZpuGen2) + ZpuTX = Ztotal
With this I calculate SFC = 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. SpinUp 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
#26
Posted 13 February 2012  02:19 PM
#27
Posted 25 February 2012  08:43 PM
I believe for calculating "daily load factor" 24 hours should be used instead of 6 hours. Any comments about the spinup solution will be appreciated.
Note: This answer could be right if there is any "quarter day load factor".
Thanks,
#28
Posted 25 February 2012  11:27 PM
Spin up problem 265:
I believe for calculating "daily load factor" 24 hours should be used instead of 6 hours. Any comments about the spinup 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.
#29
Posted 26 February 2012  12:23 AM
#30
Posted 26 February 2012  12:25 AM
Why the solutions methodologies are different ?
Thanks,
#31
Posted 26 February 2012  02:34 PM
Spin up problem 269 and 301:
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.
#32
Posted 05 March 2012  05:08 PM
The peak load is 500KW and the avg is 2500MW/(6hrs) = 416.67KW
My solution thus is 416.67/500= 83%.
#33
Posted 18 March 2012  02:36 AM
Thanks,
#34
Posted 18 March 2012  01:14 PM
Spinup prob 378: Why MVA_{L}= 12, any feedback will be appreciated ?
Thanks,
S_{A} = 6 X 2 = 12
S_{B} = 8 X 3 = 24
When in parallel, S_{B} = 2S_{A}
The max load of Trans B is 8MVA which limits Trans A to 4MVA. Therefore the total Max load is 4MVA + 8MVA = 12 MVA
#35
Posted 30 March 2012  02:12 AM
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
#36
Posted 30 March 2012  02:16 AM
#37
Posted 01 March 2014  07:40 PM
I had a few questions:
1) Spin Ups Problem 180: When a generator is overloaded, the frequency will drop...but when will a generator trip offline?
2) Spin Ups Problem 24: Can anyone help explain how to differentiate between buck and boost autotransformers?
3) Spin Ups Problem 215: 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 01000V, medium voltage is 1kV to 36kV, and anything higher is high voltage.
Thank you.
#38
Posted 03 March 2014  08:30 PM
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
#39
Posted 04 March 2014  02:23 AM
For problem 24, 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 stepup and boost were the same and stepdown and buck were the same but the answer for this problem is stepup, 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?
#40
Posted 04 March 2014  11:32 AM
#41
Posted 06 April 2014  02:09 AM
For problem 19 in exam 5, why is 2 x 10^2 used rather than 4 x 10^ 7?
I have seen two equations for inductance per phase, (2 x 10^7) ln (GMD/GMR) and (4 x 10^7) ln (GMD/GMR). Can anyone clarify when each should be used? Thank you.
#42
Posted 07 April 2014  01:43 AM
Is (2 x 10^7) ln (GMD/GMR) used when per phase inductance is asked for? And (4 x 10^7) ln (GMD/GMR) for total inductance?
#43
Posted 07 April 2014  01:50 AM
Do some searching on 3 Phase Inductance vs Single Phase Inductance Calculations.
Also tagged with one or more of these keywords: spinup, spinup, sample, exam, problems, solution, discussion, pe power
PE Exam Prep Forum →
Civil Engineering PE Exam →
Books I used to pass the 8hr examStarted by ggguy23 , 19 Aug 2014 pe, exam, books, construction and 2 more... 



PE Exam Prep Forum →
Civil Engineering PE Exam →
Transportation →
EET helped me PASS the NCEES PE ExamStarted by OBPassed2014 , 10 Jun 2014 eet, pe, ncees, exam, seismic and 4 more... 



Exam Discussion →
PE Exam Results →
APR 2014 →
CA civil first time pass rateStarted by mytchell , 26 May 2014 CA, civil, pass, rate, first and 4 more... 



Exam Discussion →
PE Exam Results →
APR 2014 →
NEVADA Results in...Started by apprentice , 22 May 2014 PE Power 



Vendor Forums →
Vendor Giveaways / eb.com Deals →
Engineering Reference Manual Giveaway & 20% Off Instant Promo CodeStarted by PPI , 03 Feb 2014 free, engineering, exam and 4 more... 


0 user(s) are reading this topic
0 members, 0 guests, 0 anonymous users