Question

If an equipment has a PU impedance of 0.9 to a base of 20 MVA, 30 KV. The PU impedance to base of 50 MVA and 11 KV???

Answer 1

Given :

Per unit impedance of 20 MVA base = 0.9

The potential of 20 MVA base = 30KV

The potential of 50 MVA base = 11KV

To find :

The per unit impedance of the 50MVA base = ?

Solution :

• The relation between impedance and per-unit impedance is given by the formula

              $R=R_{pu} /times \frac{V_{base} ^{2} }{S_{base} }$

              $R=R_{pu} \times \frac{V_{base} ^{2}(old) }{S_{base} (old)}\rightarrow Equation(1)$

              $R=R_{pu} \times \frac{V_{base} ^{2}(new) }{S_{base} (new)}\rightarrow Equation(2)$

• By equating equation (1) & (2)

              $R_{pu} \times \frac{V_{base} ^{2}(old) }{S_{base} (old)}=R_{pu} \times \frac{V_{base} ^{2}(new) }{S_{base} (new)}$      

              $0.9\times \frac{30^{2}}{20}=R_{pu} \times \frac{11^{2}}{50}$

              $R_{pu} =16.73$

The per unit impedance is 16.73.