11. PHYSICAL FUNDAMENTALS (JBS READY)
11.10. ABILITY TO WITHSTAND SHORT CIRCUIT CURRENTS (JBS READY)
IEC 60076-5 Power transformers – Part 5: Ability to withstand short circuit identifies the requirements for transformers to sustain without damage the effects of overcurrents originated by external short circuits.
For three-phase transformers with two separate windings, the r.m.s. value of the symmetrical short- circuit current is calculated as:
)
(
3
s scZ
Z
U
I
+
⋅
=
kA (40)Where Zs is the short-circuit impedance of the system.
S U Z
2 s
s = In ohms per phase (equivalent star connection) (41)
⊗
t1 t12 t2
hw
Where
Us is the rated voltage of the system, in kilovolts (kV);
S is the short-circuit apparent power of the system, in megavoltamperes (MVA). U and Z are defined as follows:
U is the rated voltage Ur of the winding under consideration, in kilovolts (kV)
Z is the short-circuit impedance of the transformer referred to the winding under consideration; it is calculated as follows:
r 2 r z S 100 U u Z ⋅ ⋅
= In ohms per phase (equivalent
star connection) (42) where
us is the measured short-circuit impedance at rated current and frequency at reference temperature, as a percentage;
Sr is the rated power of the transformer, in megavoltamperes (MVA).
The duration of the symmetrical short-circuit current must be limited to avoid excessive overheating. According to IEC 60076-5 the duration shall be 2s, unless otherwise agreed between purchaser and supplier. During this time the temperature in the windings shall not exceed certain values given in IEC 60076-5, when calculated by means of formulas given in the standard.
Another aspect of the short circuit topic is the mechanical forces in the windings due to overcurrents. Unlike temperatures that need some time to rise, the mechanical forces follow the current instantaneously. The first cycles of a short-circuit current are asymmetric with respect to the time axis. The degree of asymmetry depends on point of the sinusoidal voltage curve where the short-circuit occurs. The degree of asymmetry is a statistical variable with its maximum when the short-circuit occurs in the range where the voltage passes through zero and its minimum in the range where the voltage has its peak value. This will be demonstrated in the following.
R X ) t sin( Uˆ ω +α i(t Figure 11-12
Consider the circuit in Figure 6-12. Assume that the switch makes at the instant t = 0, thus simulating a short-circuit. The current i(t) is expressed by the following equation:
(
sin( t)
e sin( )) Iˆ ) t (i = ⋅ ω +α−ϕ − −tτ ⋅ α−ϕ A (43) in which Z Uˆ Iˆ= A (44)that is the peak value of the symmetrical, stationary short-circuit current Z = R + jωL [Ω]
α = switching angle of the voltage u(t) at the instant of the short circuit [rad] ϕ = phase angle of the circuit impedance (=arctg ωL/R) [rad]
τ = L/R = tgϕ/ω [s] (the time constant of the circuit) ω = 2·π·f [s-1]
Transformer handbook. Draft. Rev. 02Q Page 138 of 197 The current starts from zero and consists of two components:
• an alternating steady-state component of fundamental frequency, • an unidirectional component decreasing exponentially with time.
The first peak value of the short-circuit current determines the maximum force that acts on the windings. The maximum value of this current depends on the X/R ratio and the switching angle α. It occurs in almost all cases when α = 0 (or π).
In power systems X >>R, which means that ϕ ≈ π/2.
Equation (43) can then with reasonable approximation be written as: ) e ) 2 t (sin( Iˆ ) t (i = ⋅ ω −π + −tτ A (45)
The first current peak closely corresponds to the time when ωt = π.
Figure 11-13 max sc sc 2 k I I k Iˆ i= ⋅ = ⋅ ⋅ = A (46)
where the asymmetry factor k is
X R e 1
k = + −π (47)
Figure 11-13 shows an example on how the short-circuit current elapses. In this example X/R = 8, k = 1,68.
The peak factor is k·√2 and its dependence of X/R is shown in Figure 11-14. The peak factor value in the example in Figure 11-13 is 2,38. The value of the first peak of the short circuit current is the r.m.s. value of the stationary short circuit current multiplied with the peak factor.
In a real case with a transformer installed in a network:
X = the sum of the network and the transformer reactance [Ω] R = the sum of the network and the transformer resistance [Ω] X/R and the peak factor increase with increasing transformer power rating.
-1 -0,5 0 0,5 1 1,5 2 0 20 40 60 80 t (ms) Is c p .u .
Figure 11-14 Mechanical forces in the windings
Current flows through the winding conductors, which are situated in the magnetic leakage field. The conductors are then subject to mechanical forces.
These forces are not static. They are pulsating. Each time the current passes through zero the forces are also zero. At normal load current the forces are small. They increase with the square of the current, so during the high overcurrents that arise if a short-circuit in the system occur, the forces must be given attention when designing the transformer. The short circuit current may amount to 10 – 20 times the rated current of the transformer, which means that the forces in the windings may be 100 – 400 times larger at a short circuit than in normal service.
The forces cause large movements in the windings. These movements are invisible for the human eye. But rapid film recordings played in slow motion show the size of the movements and illustrate the violence of the forces.
B I
F = I Х B
Figure 11-15
When considering the ability of a design to withstand it is usual to split the forces into radial and axial components as indicated in Figure 11-15, which shows an upper part of an outer winding. The radial force is directed outwards and causes a tensile force in the winding conductors. In the corresponding inner winding the radial component is directed inwards. That may cause buckling of the winding if it has not been made robust enough.
The axial forces are caused by the radial component of the magnetic field at the ends of the windings. These forces may lead to tilting of the conductors between the axial spacers in the winding. The force on each turn or disc adds together. The sum of the forces is balanced at the other end of the winding. The whole winding is subject to a strong axial pressure. See Figure 11-16. 1,4 1,6 1,8 2,0 2,2 2,4 2,6 2,8 3,0
1
10
100
X/R
k
√
2
Transformer handbook. Draft. Rev. 02Q Page 140 of 197
A B
A Axial current forces B Resulting axial pressure Figure 11-16
The ability to withstand tilting of the conductors depends on the diameter of the winding, the distance between the spacers around the circumference and the dimensions of the conductor. In case of axial “openings” in the winding, that is one or more places along the height of the winding where there are no ampere-turns, there will be axial forces directed towards the yokes. The framework keeping the core and windings together must be designed to be able to withstand such forces.
A more comprehensive treatment of the short-circuit topic is given in the book “SHORT-CIRCUIT DUTY OF POWER TRANSFORMERS – THE ABB APPROACH” by Giorgio Bertagnolli, issued by ABB. The book is available from ABB at request.