NCEES 2009 Power question 525

[td2010]

I get confuse of the reply for this problem from thread

http://engineerboards.com/index.php?showto...;hl=NCEES+Power

IF the rating of the input of the transformer is 50KVA, should the output rating also be 50KVA ?

Consider the transformer as a black box, what make the output power is greater than the input power ?

Thanks

 

[Flyer_PE]

Both the primary and secondary windings are rated for 50 kVA. However, when you connect the transformer as an auto-transformer, you're taking advantage of the current rating of the low voltage winding on a higher voltage circuit. The rating in this configuration gets much larger since you're not really passing all of the input power through the primary winding.

The output power will not be greater than the input. All you are doing is increasing the rating of the transformer.

 

[td2010]

Both the primary and secondary windings are rated for 50 kVA. However, when you connect the transformer as an auto-transformer, you're taking advantage of the current rating of the low voltage winding on a higher voltage circuit. The rating in this configuration gets much larger since you're not really passing all of the input power through the primary winding.
The output power will not be greater than the input. All you are doing is increasing the rating of the transformer.

anyone has good source (reference or website ) for studying auto transformer ? how it actually connect and rated...etc. I am still confuse. :{ Thanks

 

[mull982]

Both the primary and secondary windings are rated for 50 kVA. However, when you connect the transformer as an auto-transformer, you're taking advantage of the current rating of the low voltage winding on a higher voltage circuit. The rating in this configuration gets much larger since you're not really passing all of the input power through the primary winding.
The output power will not be greater than the input. All you are doing is increasing the rating of the transformer.

anyone has good source (reference or website ) for studying auto transformer ? how it actually connect and rated...etc. I am still confuse. :{ Thanks

I dont understand why the solution adds the currents in the bottom transformer winding instead of subtract them. Since the currents are flowing in different directions in this winding shouldn't the difference be used for calculating this winding rating?

I also think the wrong current is used on the 2420V side. They calculate the secondary current at 120V rather then 2420V. Shouldn't secondary current be 50kVA/2420=20.6A?

 

[cableguy]

Some tidbits about autotransformers... to solve this problem, you just need one small formula:

KVA Rating AutoXFMR = KVA(nameplate) x (1 + Ncommon/Nseries)

So in this instance, it'd be = 50 x (1 + 2300/120) = 1008.33 KVA

For an autotransformer circuit, if you're working with currents, if it's a step up, Icommon = I1 - I2, arrow pointing down. If it's a step down, Icommon = I2 - I1, arrow pointing up. You can get I2 and I1 by dividing the single phase VA rating by the winding voltage (since each winding is rated at that KVA). Since this is a single phase transformer, I1 = 50kVA / 2300, I2 = 50kVA / 120 - but note that the actual voltage on the output is 2420.

It looks to me like in the solution to this problem, they have one of the arrows drawn in the wrong spot. Icommon (which is where they have their arrow) should not equal to 21.74... The way they have it drawn, if you apply Kirchoff's current law, the input line from the 2300 volt source would be over 400 amps. It "should" be that the 21.74 is the input from the 2300 volt line (which would be 50 kVA, the rating of the winding), and then the common winding would have just under 400 amps flowing "up".

 

[mull982]

Some tidbits about autotransformers... to solve this problem, you just need one small formula:
KVA Rating AutoXFMR = KVA(nameplate) x (1 + Ncommon/Nseries)

So in this instance, it'd be = 50 x (1 + 2300/120) = 1008.33 KVA

For an autotransformer circuit, if you're working with currents, if it's a step up, Icommon = I1 - I2, arrow pointing down. If it's a step down, Icommon = I2 - I1, arrow pointing up. You can get I2 and I1 by dividing the single phase VA rating by the winding voltage (since each winding is rated at that KVA). Since this is a single phase transformer, I1 = 50kVA / 2300, I2 = 50kVA / 120 - but note that the actual voltage on the output is 2420.

It looks to me like in the solution to this problem, they have one of the arrows drawn in the wrong spot. Icommon (which is where they have their arrow) should not equal to 21.74... The way they have it drawn, if you apply Kirchoff's current law, the input line from the 2300 volt source would be over 400 amps. It "should" be that the 21.74 is the input from the 2300 volt line (which would be 50 kVA, the rating of the winding), and then the common winding would have just under 400 amps flowing "up".

I dont understand why the current on the secondary though is calculated at 120V. Doesnt the current on the secondary have to flow all the way from the - to + through both windings and therefore at 2400V. Similar to a step down transformer when you calculate the current on the primary the current flows all the way from + to - on the primary through both windings and you use the total input voltage not just the partial voltage of the one winding.

I understand the arrow convention on both but I dont understand why the secondary current is calcualted like it is for the step-up. Why is this different then how the primary current is calculated on a step down?

 

[dora]