Delta-Delta 3 phase transformer HNC help

1. May 12, 2015

craig clarke

<< Thread moved from the technical engineering forums to HH >>

I'm currently studying for a HNC in electrical engineering and am currently stuck on a question on one of my assignments. The question is:

A Delta-delta, 3 phase transformer steps down the system line voltage of 132kV to a local power distribution level of 11kV. The total connected load is 21 MW at 0.86 pf lagging. Calculate:

a) The apparent power taken by the load
b) The current in the HV lines
c) The current in the LV lines
d) The currents in the primary and secondary windings of the transformer
e) The load carried by each phase winding of the transformer

a) Apparent power = 21MW/cos(0.86 = 32MW

b) Current in HV line = 32MW/√3x132kV = 139.9A
c) Current in the LV line = 32MW/√3x11kV = 1679.5A
d) Current in primary winding = Primary current/√3 = 80.25A
Current in secondary winding = Secondary current/ √3 = 969A

Could someone please check these answers for me and i'm not sure how to calculate the load carried by each phase winding but am I correct in thinking that each winding have the same answer?

Last edited by a moderator: May 12, 2015
2. May 13, 2015

Staff: Mentor

Delta-Delta 3 phase transformer HNC help

Hi craig clarke. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

If you'd like some answers checked then you need to include your working. That's the requirement when seeking homework help at Physics Forms. Otherwise, helpers would have to work the problem themselves in its entirety.

Where the load is balanced, phase currents and voltages are all identical.

Last edited by a moderator: May 7, 2017
3. May 13, 2015

craig clarke

I have included my calculations and answers of a,b,c and d I'm stuck on quest e. I'm not sure on what formula to use. I want to know what formula to use for calculating the load carried by each phase winding and if I've answered the other questions correctly.

Thanks

4. May 13, 2015

Staff: Mentor

In (e) the intended answer might be its kVA.

Last edited: May 13, 2015
5. May 13, 2015

Staff: Mentor

Isn't 0.86 already cos(Φ)? Why take the cos of a cos?

6. May 13, 2015

craig clarke

PF=cos(φ

where φ is the phase angle between the voltage and current, the power factor is 0.86 lagging, so φ=0.86

The PF is the cosine of the phase angle which is 0.86?

7. May 13, 2015

craig clarke

Will the equation for this calculation be

Has it is a delta transformer and will be balanced?

kilovolt-amps = 3 × amps × volts / 1000

8. May 13, 2015

Staff: Mentor

The power factor is indeed the cosine of the phase angle. Here it's given to you as PF = 0.86. So 0.86 = cos(φ).
That would make φ = cos-1(0.86). Or if you prefer, φ = arccos(0.86).

9. May 13, 2015

Staff: Mentor

winding voltage x winding current /1000, I think, for kVA