In Camara's reference book in Table 27.2 , it says the average of a Sine Wave is 0. Half wave is .637Q. 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.
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.IWhat is the difference between average value and average magnitude?
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.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.
For these type of problems I use the trial and error method. I multiply the generator MVA by the other generators impedance.Spin-up problem 2-15: In solution MVAL=6, from where it is coming?
Thanks,
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 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 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??
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?
The MVA method does not include converting to per unit. There must be an error in Camara's book.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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