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Formulas of Power Engineering
Cross-section
for direct - and single phase alternating current
(current given)
q =
1 x I x l
(mm²)
κ x u
for three-phase current
q =
1.732 x I x cosφ x l
(mm²)
κ x u
for direct - and single phase alternating current
(capacity given)
q =
2 x I x P
(mm²)
κ x u x U
for three-phase current
q =
l x P
(mm²)
κ x u x U
Voltage drop
For low voltage cable network of normal operation, a voltage drop of 3-5% is adviseable.
Exeption: higher values (7%) can be permitted in case of network extension or in short circuit.
for direct current (current given)
u =
2 x I x l
(V)
κ x q
single phase alternating current (current given)
u =
2 x I x cosφ x l
(V)
κ x q
for three-phase current
(current given)
u =
1.732 x I x cosφ x l
(V)
κ x q
for direct current (capacity given)
u =
2 x l x P
(V)
κ x q x U
single phase alternating current (capacity given)
u =
2 x l x P
(V)
κ x q x U
for three-phase current (capacity given)
u =
l x P
(V)
κ x q x U
u
=
Voltage drop (V)
U
=
Operating voltage (V)
P
=
Power (W)
RW
=
Effective resistance (Ω/km)
L
=
Inductance (mH/km)
ωL
=
Inductive resistance (Ω/km)
ω = 2 π f at 50 Hz = 314
q
=
Cross-section (mm²)
I
=
Working current (A)
l
=
Length of line (m)
κ
=
Electrical conductivity of conductors
(m/Ω x mm²)
κ Copper: 58
κ Aluminum: 33
Nominal voltage
The nominal voltage is to be expressed with two values of alternating current U0/U (in V).
U0
=
voltage between conducter and ground or metallic covering (shield, armoring, concentric conductor)
U
=
voltage between two outer conductors
U0
=
U/√3 for three-phase current systems
U0
=
U/2 for single phase and direct current systems
U0/U0
=
1 outer conductor is grounded for single phase and direct current systems
Nominal current
I in A
Active current
IW = I x cosφ
Reactive current
I0 = I x sin φ
Active power
S = U x I for single phase current
S = 1.732 x U x I for three-phase current
Apparent power
P = U x I x cosφ for single phase current
P = 1.732 x U x I x cos φ for three-phase current
P = U x I for direct current
Reactive power
Q = U x I x sin φ for single phase current
Q = 1.732 x U x I x sin φ for three-phase current
(Voltampere reactive) Q = P x tan φ
Phase angle
φ is a phase angle between voltage and current
cos φ
=
1.0
0.9
0.8
0.7
0.6
0.5
sin φ
=
0
0.44
0.6
0.71
0.8
0.87
Insulation resistance
RInsu =
SInsu
x In
Da
x 10-8 (MΩ x km)
l
d
Specific insulation resistance
RS =
R x 2π x l x 108
In
Da
di
Da
=
Outer diameter over insulation
d
=
Conductor diameter
di
=
Inner diameter of insulation
l
=
Length of the line
SInsu
=
Spec. resistance of insulation materials (Ω x cm)
Mutual capacity
for single conductor, three-cond. and H-cable
CB =
ξr x 10³
(nF/km)
18 In
Da
d
Inductance
single-phase
0.4 x ( In
Da
+ 0.25) mH/km
r
three-phase
0.2 x ( In
Da
+ 0.25) mH/km
r
Da
=
Distance mid to mid of both conductors
r
=
Conductor radius
ξr
=
Dielectric constant
0.25
=
Factor for low frequency
Ground capacitance
EC = 0.6 x CB
Charging current
(only for three-phase)
ILad = U x 2 π f x CB x 10-6 (A/km per conductor at 50Hz)
Charging power
PLad = ILad x U
Leakage and loss factor
G = tanδ x ωC(S)
tanδ = G/ωC
ω
=
2 π f
C
=
Capacity
tanδ
=
Loss factor
S
=
Siemens = 1/1Ω
Dielectric loss
DV = U² U x 2 π f x CB x tan x 10-6 (W/km)
It should be noted that for the current load of the insulated cables and wires of selected cross-section, the power ratings are to be considered.
To estimate the voltage drop of insulated cables and wires with large cross-sections of single and three-phase overhead line, the active resistance as well as indictive resistance must be considered.
Formula for single phase:
U = 2 x l x I x (RW x cosφ + ωLx sinφ) x 10-3 (V)
Formula for three-phase
U = 1.732 x l x I x (RW x cosφ + ωLx sinφ) x 10-3 (V)
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