SME
Power Systems · Problem type
Approximate voltage drop
Use this for “A line with R = 0.02 pu and X = 0.1 pu delivers P = 0.8 pu and Q = 0.6 pu at 1 pu. Find the sending voltage.”
Chart
This is an example. Type your own values to start a new problem.
ΔV (pu)
=
R (pu)
×
P (pu)
÷
V₂ (pu)
+
X (pu)
×
Q (pu)
÷
V₂ (pu)
V₁ (pu)
=
V₂ (pu)
+
ΔV (pu)
ΔVR
Drop from R
calculated · pu
ΔVX
Drop from X
calculated · pu
Formulas
ΔVR
=
R
×
P
V₂
0.016
=
0.02
×
0.8
1
ΔVX
=
X
×
Q
V₂
0.06
=
0.1
×
0.6
1
ΔV = ΔVR + ΔVX
0.076 = 0.016 + 0.06
V₁
=
V₂
+
ΔV
1.076 = 1 + 0.076
Assumptions
•
A short line with a small drop, so ΔV ≈ (RP + XQ) ÷ V₂, the part of the drop in phase with V₂.
•
The load is lagging or at unity power factor (Q ≥ 0); a leading load can raise the far end instead.
•
On a transmission line X is many times R, so the reactive power Q sets most of the drop.
Step-by-step
WE KNOW
R = 0.02 pu, X = 0.1 pu, P = 0.8 pu, Q = 0.6 pu, V₂ = 1 pu
FIND
drop from R (ΔVR), drop from X (ΔVX), voltage drop (ΔV), sending voltage (V₁)
Step 1 · Find drop from R (ΔVR)
Use ΔVR = R × P ÷ V₂
Real
power
drives
current
in
step
with
the
voltage,
and
R
drops
it
in
step
too.
ΔVR
=
0.02
×
0.8
1
ΔVR = 0.016 pu
Step 2 · Find drop from X (ΔVX)
Use ΔVX = X × Q ÷ V₂
Reactive
current
lags
90°,
and
X
turns
it
back
in
step
with
the
voltage.
ΔVX
=
0.1
×
0.6
1
ΔVX = 0.06 pu
Step 3 · Find voltage drop (ΔV)
Use ΔV = ΔVR + ΔVX
The two parts of the drop add.
ΔV = 0.016 + 0.06
ΔV = 0.076 pu
Step 4 · Find sending voltage (V₁)
Use V₁ = V₂ + ΔV
The sending end has to be higher than the far end by the drop.
V₁
=
1
+
0.076
V₁
=
1.076
pu
Check
Put the numbers back in. Both sides should match.
0.016
=
0.02
×
0.8
1
✓
0.06
=
0.1
×
0.6
1
✓
0.076 = 0.016 + 0.06 ✓
1.076 = 1 + 0.076 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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