SME
Power Systems · Problem type
Line losses
Use this for “A 33 kV line carries 10 MW at power factor 0.9 with 5 Ω per phase. Find the current and the loss.”
Chart
Pₗₒₛₛ(33) ≈ 0.5668
This is an example. Type your own values to start a new problem.
I (A)
=
P (MW)
÷
(
3
×
VL (kV)
×
pf
)
Pₗₒₛₛ (MW)
=
3
×
(
I (A)
)
2
×
R (Ω)
L (%)
=
Pₗₒₛₛ (MW)
÷
P (MW)
×
100%
Formulas
I
=
P
√
3
× VL × pf
0.1944
=
10
√
3
× 33 × 0.9
Pₗₒₛₛ
=
3
×
I
2
×
R
0.5668
=
3
×
0.1944
2
×
5
L
=
Pₗₒₛₛ
P
×
100
5.6684
=
0.5668
10
×
100
Formulas are worked in kA. The step-by-step shows the conversions.
Assumptions
•
A balanced three-phase line: VL is line to line, every value is rms, and R is the resistance of one conductor.
•
Each of the three conductors loses I²R, so the line loses 3I²R.
•
The voltage drop is small, so VL is the same at both ends; MW ÷ kV gives kA.
Step-by-step
WE KNOW
P = 10 MW, VL = 33 kV, pf = 0.9, R = 5 Ω
FIND
line current (I), power lost (Pₗₒₛₛ), loss as a percent (L)
Step 1 · Find line current (I)
Use I = P ÷ (√3 × VL × pf)
Three-phase
power
is
√
3
VL
I
times
the
power
factor,
so
divide
P
by
√
3
VL
pf.
I
=
10
√
3
× 33 × 0.9
I
=
10
51.4419
I
=
0.1944
kA
Step 2 · Find power lost (Pₗₒₛₛ)
Use Pₗₒₛₛ = 3 × I² × R
Each
conductor
heats
by
I
2
R
and
there
are
three;
kA²
×
Ω
gives
MW.
Pₗₒₛₛ
=
3
×
0.194394
2
×
5
Pₗₒₛₛ = 3 × 0.03779 × 5
Pₗₒₛₛ = 0.56684 MW
Step 3 · Find loss as a percent (L)
Use L = Pₗₒₛₛ ÷ P × 100
The loss as a share of the power delivered, in percent.
L
=
0.56684
10
×
100
L
=
0.05668
×
100
L
=
5.6684%
Step 4 · Convert the answers
Change each answer back to the unit you chose.
I = 0.1944 kA = 194.39403 A (1 kA = 1000 A)
Check
Put the numbers back in. Both sides should match.
0.194394
=
10
√
3
× 33 × 0.9
✓
0.56684
=
3
×
0.194394
2
×
5
✓
5.6684
=
0.56684
10
×
100
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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