per unit

[LMAO]

I was just wondering, when you have a 7.5MVA transformer, 60kV:12kV, 7.5% impedance; how do you find the actual impedance? You need to have the base voltage as well as base KVA to find the base impedance; What should you use as your base voltage? high or low side (60k or 12k)?
thanks

 

[gar]

100815-1830 EST

If your load is on the low voltage side, then I would assume you are concerned with viewing the impedance from that side. And from the high side if the load is on the high side. You can view it either way depending upon what problem you want to solve.

.

 

[cadpoint]

Take a look at this definition through this link:

Percentage Impedance (Z%)

I thought it is on target; enjoy…

 

[dkarst]

One of the primary advantages of the per unit system is the p.u. equivalent impedance of any transformer is the same referred to either primary or secondary side. Since the base voltage of each side is different, the actual impedance viewed from each side vary. I would recommend getting a copy of a power system analysis text such as Grainger and Stevenson (or something more modern if you prefer) to see this in more detail.

The general equation that applies is: Quantity in per unit (p.u.) = Actual Quantity/Base Value of Quantity, just solve this for actual quantity and you have it.

 

[BJ Conner]

I was just wondering, when you have a 7.5MVA transformer, 60kV:12kV, 7.5% impedance; how do you find the actual impedance? You need to have the base voltage as well as base KVA to find the base impedance; What should you use as your base voltage? high or low side (60k or 12k)?
thanks

Use either one. Impedance changes with the square of the turns ratio.

Percent Impedance is defined as the percentage of primary voltage require to produce full load current in the secondary if the terminals are shorted.
If you put 7.5 % of 60KV on the high side (.075x60,000= 4,500 volts) 360 amps flow in the low side (with the terminals shorted). The high side current would be (12 KV/60KV)*360= 72 Amps.
You can figure the impedance for either side.
For the Primary

Z=E/(I*Sqr(3))= 4,500/(72*1.732)=36.08 Ohms

For the Secondary

Z= 900/(360*1.732)= 1.44 Ohms

Note the two differ by yhe square of the voltage ratio.
1.44* ( Vprimary/Vsecondary)^2=36.08

 

[dkarst]

BJ Conner is correct. If you want to stay more in "per-unit land", you apply the relationship Zbase = Vb^2/Sbase, where Vb is the line voltage and Sbase is total 3-phase rated voltamperes. Since you normally know the line voltage and rated VA, this is an important relationship and doesn't require you to wonder if there is a sqrt(3) somewhere.

So on the high voltage side Zbase = 60kV^2/7.5 MVA = 480 ohms. The actual impedance value is calculated using my previous post Zactual = Zbase * Zp.u = 480 * 0.075 = 36 ohms.

Same thing on low side: Zbase on that side is 12kV^2/7.5MVA = 19.2 ohms for base impedance. Multiply by 0.075 and you get 1.44 ohms.

 

[dkarst]

...you apply the relationship Zbase = Vb^2/Sbase, where Vb is the line voltage and Sbase is total 3-phase rated voltamperes.

A more straightforward equation to apply without worrying about exponents may be Zbase = ((Vb kV)^2)/Sbase-MVA where Vb-kV is the base voltage IN KILOVOLTS and Sbase-MVA is the base VA IN MEGAVOLTAMPERES.

So in this case on the high side Zbase is simply = 60^2/7.5 = 480 ohms since the 60 is already given in kV and the base VA is given in MVA.