SME
Power Systems
Power flow
Use this for “Two buses at 1 pu are joined by X = 0.5 pu. Find P at δ = 30°.”
Chart
Local maximum at (90, 2)
P(30) = 1
This is an example. Type your own values to start a new problem.
Pₘₐₓ (pu)
=
V₁ (pu)
×
V₂ (pu)
÷
X (pu)
P (pu)
=
Pₘₐₓ (pu)
×
sin(
δ
°)
PMW (MW)
=
P (pu)
×
Sbase (MVA)
Formulas
Pₘₐₓ
=
V₁
×
V₂
X
2
=
1
×
1
0.5
P
=
Pₘₐₓ
×
sin(
δ
)
1 = 2 × sin(30°)
PMW
=
P
×
Sbase
100 = 1 × 100
Assumptions
•
The line is lossless (R ≪ X), so the power sent is the power received: P = V₁V₂ sin δ ÷ X.
•
V₁ leads V₂ by the power angle δ, so real power flows from bus 1 to bus 2.
•
P peaks at δ = 90°: past it the line can’t carry more, and the buses fall out of step.
•
Per-unit values are fractions of the base; P in MW is P in pu times Sbase in MVA.
Step-by-step
WE KNOW
V₁ = 1 pu, V₂ = 1 pu, X = 0.5 pu, δ = 30°, Sbase = 100 MVA
FIND
maximum power (Pₘₐₓ), real power (P), power in megawatts (PMW)
Step 1 · Find maximum power (Pₘₐₓ)
Use Pₘₐₓ = V₁ × V₂ ÷ X
At
δ
=
90°
sin
δ
is
1,
so
the
most
the
line
carries
is
V₁
V₂
X
.
Pₘₐₓ
=
1
×
1
0.5
Pₘₐₓ = 2 pu
Step 2 · Find real power (P)
Use P = Pₘₐₓ × sin(δ)
The power follows the sine of the angle between the two bus voltages.
P
=
2
×
sin(30°)
P
=
2
×
0.5
P
=
1
pu
Step 3 · Find power in megawatts (PMW)
Use PMW = P × Sbase
One pu of power is Sbase, so multiply the per-unit power by the base in MVA.
PMW = 1 × 100
PMW = 100 MW
Check
Put the numbers back in. Both sides should match.
2
=
1
×
1
0.5
✓
1 = 2 × sin(30°) ✓
100 = 1 × 100 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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