Complex Imaginary Test 3 Problem 43

[Wildsoldier PE]

I think there is an error in this problem what do you all think?

 

[DK PE]

There wasn't anything mentioned on winding configuration, correct? That is what can make this type of problem interesting.

 

[rick.conner]

Agree with you. they obviously took 8118va and divided by 1.732. not sure why??

 

[Wildsoldier PE]

There is a bunch of problems in this tests that complex imaginary didn't use the square of 3 and they have fix a few......every time you have a 3 phase voltage no matter if it is delta or wye to get the line current you always multiply the voltage by square of 3 to get what i called the "3phase voltage"....looks like the author of this books have a little confusion on when square of 3 needs to apply.

DK Pe

They didn't mention anything on the winding configuration so i had to assume that the current is the full load current that flows thru the line.

 

[Insaf]

Let consider secondary side is Delta, so phase voltage = 62.5 V and phase current= 75/sqrt(3)=43.3 A. Now KVA (1-phase) rating = 62.5*43.3=2.706 kVA.

Go another way, suppose secondary side is Y-connected, so line to neutral voltage = 62.5/sqrt(3)=36.08 and phase current=line current=75 A.

KVA (1-phase)=36.08*75=2.706 kVA.

In both cases, 3 phase KVA=3*2.706=8.11 KVA and it indicates that transformer configuration is not necessary in this case. Simply "sqrt(3)*62.5*75" will give the desired result 8.11 kVA.

Bottom line is that Complex Imaginary choose wrong answer.

Thanks

 

[DK PE]

Try this version: Say the winding configuration is Y - Δ and each transformer has the 4:1 turns ratio as given (assume you are using three 1Φ transformers to make a 3Φ bank). Assume the input line voltage is 250V, therefore primary winding voltage is 250/√3 = 144V. This gets converted to a 144/4 = 36V on the Δ secondary winding (which is the line voltage on that side. Then assume that the load current is the 75A as given (this makes sense, it would be unusual in practice to be given a winding current, you would more likely know the load current). Now the total power is *poof* ... P = √3 * 36 *75 = 4.7 kVA.

Yea, I know this took a lot of assumptions that were missing from the problem statement... but IMHO it is a more relevant problem...

 

[vdubEE]

Try this version: Say the winding configuration is Y - Δ and each transformer has the 4:1 turns ratio as given (assume you are using three 1Φ transformers to make a 3Φ bank). Assume the input line voltage is 250V, therefore primary winding voltage is 250/√3 = 144V. This gets converted to a 144/4 = 36V on the Δ secondary winding (which is the line voltage on that side. Then assume that the load current is the 75A as given (this makes sense, it would be unusual in practice to be given a winding current, you would more likely know the load current). Now the total power is *poof* ... P = √3 * 36 *75 = 4.7 kVA. Yea, I know this took a lot of assumptions that were missing from the problem statement... but IMHO it is a more relevant problem...

The only problem is they came to the 4.7kVA answer without multiplying by square root of 3. They got to that answer by multiplying the secondary voltage by the secondary current only. Seems they stated that it was an ideal 3-phase transformer but used single phase transformer calculation instead.

It seems that Complex Imaginary might be working on fixing all the errors in their materials since they are not responding to any emails about possible errors in their practice exams. It is kind of frustrating since it would be nice to know before the test this coming Friday.