Power networks and the per-unit system

Từ vựng

English	Ý nghĩa
per-unit	giá trị chuẩn hóa

English

Speaking in per-unit

Choose $S_{base}$ once and $V_{base}$ per voltage level; then $Z_{base} = V_{base}^{2}/S_{base}$ and every impedance becomes a pure number.
A transformer's nameplate 10% IS its per-unit impedance: 10% of rated voltage drops inside at rated current.

One-line diagram of the fault study annotated in per-unit

Example

100 MVA, 11/132 kV transformer (Z = 10%) feeding a 40 Ω line: $Z_{base}(132) = 174.2\ \Omega$, line $= 0.230$ pu, chain $= 0.330$ pu, fault $= 3.03$ pu — and back in units, 1327 A at 132 kV or 15,926 A at 11 kV, one number serving both levels.

{exercise:uni-ee0951-fault}

Luyện tập

On a 100 MVA base at 132 kV, the impedance base is about —

Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay

174 Ω
1.32 Ω
1320 Ω
17.4 Ω

Luyện tập

A transformer nameplate says Z = 10%. This means —

Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay

at rated current, 10% of rated voltage drops inside it
it loses 10% of its power
it is 90% efficient
its no-load current is 10%

English

Changing bases

$Z_{new} = Z_{old} \times (S_{new}/S_{old}) \times (V_{old}/V_{new})^{2}$ — and when the voltage bases already match the transformer ratio, only the $S$ ratio survives.
An 8% transformer on 60 MVA becomes $0.08 \times 100/60 = 0.133$ pu on a 100 MVA study; a generator at 0.25 pu on 80 MVA becomes 0.313.

{exercise:uni-ee0952-rebase}

Luyện tập

An 8% transformer impedance on 60 MVA, used in a 100 MVA study (same voltage bases), is —

Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay