A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100 MVA and 437.38 A respectively. The line-to-line load voltage in kV is _______
V = 0.9 pu. and cos ϕ = 0.8 lag.
Base power = 100 MVA
Base current = 437.38 A
P.U. value = Actual value / base value
Now, the base voltage \(= \frac{{MVA\;Base}}{\sqrt3\times{Base\;current}}\)
\({V_{Base}} = \frac{{100 \times {{10}^6}}}{\sqrt3\times{437.38}} = 132\;kV\)
Now, Vactual = 0.9 pu × 132 kV
⇒ Vactual = 118.8 kV
\( ⇒ {V_{L - L}} = {{{V_{actual}}}}\)
VL-L = 118.80 kV
Important Questions from Per Unit System
The relation between the old and new per unit impedance values is given by:
A synchronous generator is rated at 40 MVA, 14.6 kV and 50 Hz. The base impedance of the generator will be
- The per unit impedance Z (Pu) in 3 - phase system is -
- A synchronous generator is rated at 40 MVA, 10 kV and 50 Hz. The base impedance of the generator will be:
The per unit impedance of a line is X p.u. If base voltage is tripled and base MVA is doubled, the new per unit impedance is:
