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# Sdhabik

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Project Engineer

## Previous Fields

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Electrical & Instrumentation
EIT
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Electrical

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1. ## Graffeo: Ind. Motor

Still waiting for someone to response on this.
2. ## Open Wye - Open Delta

Thank you Zach for the attachment, and happy to say that is what i followed. I believe Open wye must be the same approach like open delta that you have mentioned. I wonder if some one agree with my above answers so that i can know i am in right track .
3. ## NEC - Capacitors with motors

My understanding may be incorrect but i see the 460.9 statement as following: "When a capacitor is connected on the load side of the motor overload device, line current will be reduced due to an improved power factor, which must be taken into account when selecting the rating of a motor overload device. But this effect of capacitor mentioned above shall not be considered while determining the motor circuit conductor rating. This is for determining motor overload device only. "
4. ## NEC Code questions

If no other adjustments or correction factors are given use 75degC. You can see in NEC'17_Annex D Example D7 doing the same .
5. ## Graffeo: Ind. Motor

I tried to solve Graffeo Ex38 by Wilid method but i got different answers than one in the Graffeo. I think i am missing something with my calculation. Please let me know who have done this problem with Wildi method. For those who don't know the question: "A 120V, 6OHz, 6 pole, delta connected, three phase induction motor has the following characteristics. At standstill: Z stator = 0.2+j 0.25 ohm; Z rotor = 0.3+j 0.35 ohm. Find Pmax. The slip at which Pmax occurs and the Torque N- m at that point." Book ans: Pmax= 16.88 KW; Slip=0.278; Torque=186.05 N-m My ans: Pmax=8.687 KW; Slip=0.476; Torque=69 N-m
6. ## Graffeo: Ind. Motor

Sorry for the late response. Thank you @ellen3720 for your kind suggestion. I do have Wildi book. I just want to know if Graffeo Pmax came from same logic as mentioned in Wildi. Yes definitely i will follow Wildi.
7. ## Voltage drop

While doing VD problems, my instinct will say to use (5+15j) * 30 < -36.86 = 474.34volts which i think is the most accurate one. And choose the most near one option. I know the difference is too much but no other way. If got extra time in exam to check then will apply VD = I (RCosφ + XSinφ) at the end. I wish in actual exam we don't have to see both answers as option choice and hope the answer is near to our calculation. What i conclude to do is if the question asks for NEC. use NEC: VD = I (RCosφ + XSinφ). If not then use Ohms law.
8. ## Engineering Pro Guides Question 54

It depends on which phase to phase fault occurred. IA=0 if there is Phase B to Phase C fault. Here in this problem phase A to phase B fault occurs so condition will be IC=0.
9. ## NEC - Capacitors with motors

I appreciate your views on this. Here is mine. First i thought it should be 72A because of the statement in NEC 460.8 "The ampacity of conductors that connect a capacitor to the terminals of a motor or to motor circuit conductors shall not be less than one-third the ampacity of the motor circuit conductors and in no case less than 135 percent of the rated current of the capacitor." So back to question thinking that only conductor from capacitor to motor can be 1.35*60 = 81A rated, but the question is asking for feeder ckt conductor, it will be 72A as it is. But when i peek at NEC 2011handbook i saw this additional statement that change my mind to 81 A. It states that "The capacitor circuit conductors and disconnecting means must have an ampacity not less than 135 percent of the rated current of the capacitor. The reason is that all capacitors are manufactured with a tolerance of zero percent to 15 percent, so a 100-kVAr capacitor may actually draw a current equivalent to that of a 115-kVAr capacitor. In addition, the current drawn by a capacitor varies directly with the line voltage, and any variation in the line voltage from a pure sine wave form causes the capacitor to draw an increased current. Considering these several factors, the increased current can amount to 135 percent of the rated current of the capacitor." So then i thought the capacitor mentioned in the question may draw 81A from the source, 72A feeder may not be sufficient. Thats why my answer changed to 81A. I may not be correct. These are just my views.
10. ## Open Wye - Open Delta

Question: "A three-phase transformer bank is to handle 500 kVA and have a 34.5/11-kV voltage ratio. Find the rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) if the transformer bank is connected to open wye—open delta." My ans: HV: 19.9KV LV: 11 KV a: 1.81 S: 288KVA FYI, i only have question no solution. Question seems so challenging cuz i have not practice problems regarding the title. So somebody please teach me. Are my answers correct?
11. ## Calculate apparent power when real power given in PU

I really appreciate your explanation. Thank you @Zach Stone, P.E. for enlighten us. The above statement elucidate my confusion. Sorry guys for making it more complicated from my side. However, learned a good lesson from the discussion. Thank you all.
12. ## CI Test 4- problem 5

"what would be the advantage if you were to reconnect a conventional voltage transformer as an autotransformer ? " As you can see the question is mentioning to reconnect the same transformer as auto-transformer. If you reconnect the same transformer there is no logic to get less wingdings. So the answer is only first one not the second. I don't see typo here.
13. ## NEC - Capacitors with motors

Found somewhere in the board, may be useful for discussion. "If a motor with 60A needs a feeder ckt of 1.25*60=72A and a capacitor rated for 60A is connected somewhere near motor terminals ( A hypothetical case for understanding purpose only), will the motor feeder ckt conductors now required be 1.35*60 = 81A rated or 72 A rated. ( reference Articles 460.8.A and 430.22), if yes why, if no why?"
14. ## CI Test 4- problem 5

Let say a conventional transformer is rated 10kVA. If we use this same 10kVA transformer and connect it to auto transformer connection, its rating can goes max. say 60 kVA (depends on connection). Hence it can handle more power. But if you are comparing 10 kVA conventional transformer with another 10 kVA auto transformer, then for same kVA rating, auto transformer will require less wingdings or less copper. Hope it makes clear to you.
15. ## NCEES (version 2018) #122

Interval #2 Area; A2 = (25kW x 15min)+ (1/2 x 25kW x 15min) = 562.5 kW min Interval #2 time interval; t2: 15 min. Avg power in interval 2 = A2/t2 = 562.5 kW min / 15 min = 37.5 kW Does it make sense now?
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