# per unit impedance

Where is my mistake?

Question: Three single-phase two-winding transformers, each rated 25 MVA, 54.2/5.42 kV, are connected to form a three-phase Y-$$\\Delta\$$ bank with a balanced Y-connected resistive load of 0.6 $$\\Omega\$$ per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the load resistance $$\R_L\$$ in ohms refered to the high-voltage side and the per-unit value of this load resistance on the chosen base.

My approach: Since secondary is $$\\triangle\$$ the line voltage is 5.42kV. $$a=\frac{54.2}{5.42}=10$$ $$Z_b=\frac{V_L^2}{P_{3-\phi}}=\frac{(94 * 10^3)^2}{75 * 10^6}=117.81 \Omega$$ $$R_P=a^2*R_S=100*0.6=60 \Omega$$ $$Z_{p.u.}=\frac{60}{117.81}= 0.51 p.u.$$

The solution manual: Determine the load resistance referred to the high-voltage side. $$Z_{L(high)}=Z_L(\frac{V_{AB}}{V_{high}})$$ $$=(0.6)(\frac{94 kV}{5.42 kV})^2$$ $$=180.47 \Omega$$ Thus, the load resistance refered to the high-voltage side is $$\\boxed{180.47\Omega}\$$.

Source: Power Systems Analysis Publisher : McGraw-Hill Education; 1st edition (January 1, 1994) by John Grainger & William Stevenson ISBN-10 : 0070612935 ISBN-13 : 978-0070612938

• Please write the question and explanation into the post instead of using an image. Images can't be found by search engines, and create difficulties for blind users. Jun 1 at 13:34
• I agree with @ThePhoton, take the time to make a nice, clean post with MathJax. I edited it for you to get you started down the right path. ;-) Jun 2 at 1:38

The load resistance referred to the high-voltage side is just, $$R_P=R_L*N^2=0.6\Omega*10^2=60\Omega$$

In the per-unit system we have the luxury of picking one, and just one MVA base for our entire system. The question tells you that it has been selected for you and is 75 MVA three-phase.

Once you pick the voltage base of 1 bus, all of the rest of the buses are constrained by either direct connection or by transformation. So, all other buses connected to this 1st bus by transmission lines will be at the exact same voltage base. For buses connected to this 1st bus via transformation they will have a voltage base equal to the voltage base of this 1st bus divided by turns ratio.

Your 1st bus has voltage base of 94kV ph-ph and power base of 75 MVA three-phase.

So, $$Z_{base}=\frac{94^2}{75}=117.81\Omega\text{ for first bus}$$

Your 2nd bus has voltage base $$\=\frac{94kV}{\frac{54.2}{5.42}}=9.4kV\$$

So, $$Z_{base}=\frac{9.4^2}{75}=1.178\Omega\text{ for second bus}$$

and, $$R_L= \frac{0.6\Omega}{1.1781\Omega}=0.509 \Omega\text{ in p.u.}$$

As a check, $$R_P= \frac{60\Omega}{117.81\Omega}=0.509 \Omega\text{ in p.u.}$$

A major reason we really like the per-unit system is that it gets rid of transformations.

Note: The solution manual is wrong, they chose the voltage base (94kV) for the numerator in their turns ratio square calculation. Probably a sleep deprived or otherwise distracted grad student.

i) ON THE LOW TENSION SIDE: The base for low tension side is 75MVA, 5.42KV Actual Value = 0.6 ohm Base Value = [(5.42)(5.42)]/(75) = 0.3916 ohm Per unit value of RL = (0.6 / 0.3916) = 1.52

ii) ON THE HIGH TENSION SIDE: The base for high tension side is 75MVA, 94KV Actual Value = 0.6[(9494)/(5.425.42)] = 180 ohm Base Value = (94*94)/75 = 117.81 ohm Per unit value of RL = 180/117.81 = 1.52

• no, the base on the low tension side is not 5.42 kV. The problem specifies the high tension side base as 94 kV. Once that is set, the low tension side voltage base is constrained by the transformer turns ratio. You can only pick 1 votage base, the rest are determined by actual transformer ratios. Jun 17 at 17:36
• uotechnology.edu.iq/dep-electromechanic/typicall/… Please refer this link problem 1-4 and let me know still Iam wrong or what… Jun 18 at 6:26
• Hi @abarna, yes your answer is still wrong. See note 1 on page 8 of the document you referenced in your comment above. Once the 94 kV was specified as base on HV side, the turns ratio (10) requires the LV voltage base to be $\frac{94}{10}=9.4 kV$. Jun 18 at 13:43
• Sorry for my mistake, I got the point! Jun 18 at 13:55
• No problem @Abarna. You should either correct or delete your answer. Jun 20 at 17:41