Transformer Question

Professional Engineer & PE Exam Forum

Help Support Professional Engineer & PE Exam Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

a4u2fear

Well-known member
Joined
Jan 5, 2018
Messages
156
Reaction score
60
A Delta-Delta transformer, 10:1, Primary side line V = 1732, Primary side line I = 17.3, 0.94 lagging PF, ABC phase sequence

PRIMARY SIDE:

VAB=VL=1732<0

PF angle = cos-1(.94) = -20 deg

IL=17.3<(0-20)=17.3<-20 deg

IAB=(IL/1.73)<(+30deg) = 10<(-20+30)=10<+10 deg

Is what I have written correct?  It disagrees with a solution I have which I think is wrong.

 
A Delta-Delta transformer, 10:1, Primary side line V = 1732, Primary side line I = 17.3, 0.94 lagging PF, ABC phase sequence

PRIMARY SIDE:

VAB=VL=1732<0

PF angle = cos-1(.94) = -20 deg

IL=17.3<(0-20)=17.3<-20 deg

IAB=(IL/1.73)<(+30deg) = 10<(-20+30)=10<+10 deg

Is what I have written correct?  It disagrees with a solution I have which I think is wrong.
Does the problem have a picture associated with it and can you post the solution?

 
i didn't post my full solution with my question because i thought its primary calculations were wrong.

why did they calculate it as if it were a WYE system then convert to delta?  That's why my angles don't match

 
i didn't post my full solution with my question because i thought its primary calculations were wrong.

why did they calculate it as if it were a WYE system then convert to delta?  That's why my angles don't match
Understood, but know that I have the problem I can work it out and also know that the values I'm using are correct.   :thumbs:

So to answer your question, converting things to a wye is a common approach for solving Delta's but is not required in this problem either as long as you keep track of angles correctly. 

An issue with problems like this is the assumption of what is angle "0".  When not provided with an absolute point of reference and you have to guess... its best to put the Phase value to "0" instead of the Line value.  This is in part because in the coil, Phase Voltage and Phase current are in phase (if unity power factor) so it makes the maths easier.  

Doing so, remember that I_P leads I_L in a delta by 30 Degrees plus whatever lag / lead associated with power factor. 

  • V_L = V_AB = 1732   0 V
  • I_P = (I_L / 1.73) ∠ (0 - PF) = (17.3 / 1.73) ∠ (0 - 20) =      10  -20 A
  • I_L = I_L (ang[I_P] - 30) = 17.3 (-20 - 30) =      17.3 (-50) A
The Conversions is similar transformer system (delta / delta, Wye / Wye) have no phase angle shift between the Primary and Secondary.  As such, its just a transformation ratio.

  • V_l = V_ab = V_AB / Ration = (1732 / 10)  0 V   =  173.2  0 V 
  • I_p = (I_P * 10) 0 = 10 * 10   -20 V =      100  -20 A
  • I_l = (I_L * 10) 0 = 17.3 * 10 -50 =      173 (-50) A


So what happened with your solution?  You put the Phase Voltage at Zero Degrees, and then put the Phase Current at -30 degrees (when you assigned I_L as angle (0-20).  You basically made the Line Voltage and Phase Voltage equal phase angle which shouldn't be.  Phase Current and Phase Voltage are in phase with each other (barring power factor leading or lagging) so by making the Phase Voltage and Line Current equal with each other, your angles became messed up. 

Their long winded maths of converting to wye helps keep track of angles, but adds multiple steps more.  

(Note, i don't like their nomenclature.  I'm used to Uppercase on the Low side, lower case on the High side... but at least they defined it in the problem.)

 
Last edited by a moderator:
No problem.  Hopefully it makes sense and helps!

For what its worth, it took me a while to figure it out too.  Phase angle is the last thing I care about and being that its a relative term anyway based on what you initially assign to zero... blah. 

In the real world no one cares about phase angle unless your doing studies... just get through the test and go back to blissful ignorance!   

 

Latest posts

Back
Top