Diplomarbeit - von Werner Schuster

Niederspannungsaufschließung
ΔU
A
A−1
=
kV 1 kV 2
P 1 =110,5kW P 2 =110,5kW
3 ⋅ I B ⋅l1
⋅ cosϕ
3 ⋅ 318,
26A
⋅ 60m
⋅1
=
= 5,01 V
γ 70°
C ⋅ A
m
27,
5 ⋅ 240mm²
Ωmm²
ΔU
A−1
5,
01V
Δu
A − 1 = ⋅100%
= ⋅100%
= 1,25 %
U
400V
ΔU1− 2
P2
⋅l
2
=
γ ⋅ A⋅U
=
27,
5
110 ⋅10³
kW ⋅ 38m
= 1,58 V
m
⋅ 240mm²
⋅ 400V
Ωmm²
ΔU1−2
1,
58V
Δu
1−
2 = ⋅100%
= ⋅100%
= 0,40 %
U
400V
Δ U
U A −2
= ΔU
A−1+
Δ 1−2
Δ U
u A −2
= Δu
A−1
+ Δ 1−2
ΔU
Δu
U
B−2
B − 2
Δ 2−1
B l 3 =98m l 2 =38m
=5,01 V + 1,58 V = 6,59 V
=1,25 % + 0,40 % = 1,65 %
kV 2 kV 1
P 1 =110,5kW P 2 =110,5kW
3 ⋅ I B ⋅l
3 ⋅ cosϕ
3 ⋅318,
26A
⋅ 98m
⋅1
=
=
= 8,19 V
γ 70°
C ⋅ A
m
27,
5 ⋅ 240mm²
Ωmm²
ΔU
B−2
8,
19V
= ⋅100%
= ⋅100%
= 2,03 %
U
400V
P2
⋅l
2 110,
5 ⋅10³
kW ⋅ 38m
= =
γ ⋅ A⋅U
m
27,
5 ⋅ 240mm²
⋅ 400V
Ωmm²
ΔU
2−1
1,
58V
Δu
2−
1 = ⋅100%
= ⋅100%
= 0,40 %
U
400V
Δ U
U B −1
= ΔU
B−2
+ Δ 2−1
Δ U
l 1 =60m l 2 =38m
u B −1
= Δu
B−2
+ Δ 2−1
=8,19 V + 1,58 V = 9,77 V
=2,03 % + 0,40 % = 2,43 %
= 1,58 V
Schriebl Seite 160