Dynamic action of short circuit currents. Electrodynamic resistance of electrical devices. Electrodynamic and thermal action of short circuit currents Thermal and dynamic action of short circuit currents
When electric current flows through conductors, the conductors heat up. When a conductor is heated by a load current, part of the generated heat is dissipated into the environment, and the degree of dissipation depends on the cooling conditions.
When a short-circuit current flows, the temperature of the conductors increases significantly, since the currents during a short-circuit increase sharply, and the duration of the short-circuit is short, so the heat released in the conductor does not have time to be transferred to the environment and almost all of it goes to heating the conductor. Heating of the conductor during a short circuit can reach dangerous values, leading to melting or charring of the insulation, deformation and melting of live parts, etc.
The criterion for the thermal resistance of conductors is the permissible temperatures of heating them by short-circuit currents (x permissible, °C).
A conductor or device is considered thermally resistant if its heating temperature during a short circuit does not exceed permissible values. The thermal resistance condition in the general case looks like this, °C:
x con? x extra (4.1.)
where x con is the final value of the conductor temperature in short circuit mode.
It is recommended to quantitatively assess the degree of thermal impact of short-circuit current on conductors and electrical devices using the Joule integral
where i Kt is the total short-circuit current at an arbitrary time t, A; t off - estimated short circuit duration, s.
The Joule integral is a complex function that depends on the parameters of energy sources, the configuration of the original design circuit, the electrical distance of the fault location from the sources and other factors. For approximate calculations of the Joule integral V k in circuits that are located at a significant distance from power sources, you can use the formula, kA 2 * s,
where is the effective value of the periodic component of the short-circuit current at the moment t = 0 from the equivalent source, kA; - equivalent decay time constant of the aperiodic component of the short-circuit current, s; t off - estimated short circuit duration, s.
The most difficult case is the case of determining the Joule integral during a short circuit near generators or synchronous compensators. But in educational design, here too you can use formula (4.1.3.), since the resulting value of Vk will be somewhat overestimated, and the conductors and devices selected in powerful connections (generator, communication transformer, etc.) for long-term conditions and electrodynamic resistance, have significant reserves of thermal resistance. Based on the above considerations, in formula (4.1.3.) as T a.eq we can take the largest of the values of T a of those sources that feed the short circuit site, if there were several of them, since this leads to an increase in the calculated Joule integral and not gives errors when testing devices for thermal resistance.
When determining the Joule integral, it is necessary to determine t off quite accurately. According to the PUE, the estimated duration of the short circuit t off is the sum of the operating time of the main relay protection of this circuit (t pz) taking into account the action of automatic reclosure and the total time of switching off the circuit breaker (t trip.v), which is indicated in the catalog data of the circuit breakers, s,
t off = t pz + t off (4.4.)
For generator circuits with Р nomG? 60 MW PUE is recommended to take t off = 4 s, i.e. according to the duration of backup protection.
Manufacturers in catalogs provide the values of the guaranteed root-mean-square thermal current (t ter, kA) and the permissible time of its flow (t ter, s) for electrical devices (switches, disconnectors, current transformers, etc.).
In this case, the condition for the thermal resistance of devices in short-circuit mode looks like this: kA 2 * s,
B to? t ter (4.5.)
When checking the thermal resistance of a conductor having a standard cross-section q std, mm 2, the condition must be met
q std? q min (4.6.)
The PUE stipulates a number of cases when it is permissible not to check conductors and devices for thermal resistance during a short circuit. This applies to overhead power line wires, devices and circuit conductors protected by fuses, etc.
Electrodynamic action of currents short circuit.
During short circuits, as a result of the occurrence of short circuit shock current in busbars and other switchgear structures, electrodynamic forces arise, which, in turn, create a bending moment and mechanical stress in the metal. The latter must be less than the maximum permissible stresses for a given metal
According to the literature, the permissible design stress for aluminum is 80 MPa.
The electrodynamic force of the shock current of a short circuit in a three-phase short circuit is determined by the force of interaction between the conductors when the shock current flows through them.
where is the shock current at points K1, K2, kA,
Distance between insulators of one phase mm,
Distance between conductors of adjacent phases, mm
For KSO-366 cameras: mm; mm
Let's calculate the interaction force between AT 15x3 busbars on the 10 kV side using formula (62):
Let's consider the tire as a uniformly loaded beam and calculate the bending moment created by the shock current
where is the interaction force, N
Distance between tires, m
Bending moment
To determine the mechanical stress in the metal, it is necessary to calculate the moment of resistance, taking into account the location of the tires. Tires can be placed either flat or on edge.
Figure 2.5.1.1. Tire placement flat
Figure 2.5.1.2 Arrangement of tires on edge
In my course project the tires are placed flat. In this case, the moment of resistance is determined by the formula
where is the moment of resistance,
Tire width, cm,
Tire thickness, cm
Let's determine the design voltage in the tires:
From the conditions we see that AT brand tires (15x3) are tested for electrodynamic resistance. Similarly, we will check rectangular buses of the AT brand (15x3) on the 0.4 kV side.
Let's calculate the interaction force between AT busbars (15x3) on the 0.4 kV side, (63)
Let's calculate the bending moment created by the shock current (64):
Let's determine the design voltage in the tires (62):
From the test we see that AT brand tires (15x3) are tested for electrodynamic resistance.
Thermal effect of short circuit currents
During short circuits, live parts, including cables, can heat up to a temperature significantly higher than during normal operation.
When checking for thermal resistance, the cross-section of the cable or busbars is checked using the formula:
where VC is the thermal impulse,
st - coefficient depending on the material of the conductor, taken according to the PUE: st = 85 for aluminum conductors; st = 88 for copper cores
Let us first determine the thermal impulse:
VK = ·t off, (68)
where I pk is the periodic component current, I pk = I pk1 = kA = 2350 A
t off - shutdown time in case of short circuit,
t off = t off.. + t on, (69)
where t is off - switch operation time; s, t off =0.2s,
t z - protection response time; s, t s = 1.1s
t off = 0.2 + 1.1 = 1.3s
Let us determine the thermal impulse for the overhead line and busbars on the 10 kV side (68):
V k1 = 1.3= 7179250
Let's determine the minimum cross-section of ASBG brand CL (3x16) (67):
F min == 31.52 mm²
According to the condition of testing for thermal resistance, the selected cross-section of the CL brand
ASBG (3x16) must be greater than or equal to the minimum design cross-section
F min S extra (70)
31.52 mm² 16 mm²
From the condition we see that the selected cross-section of the ASBG brand (3x16) cable line does not pass, we re-select it to a larger cross-section of the ASBG brand (3x35):
30.72 mm² 35 mm²
From the condition we see that the selected section of the ASBG brand cable line (3x35) passes
Let's determine the minimum cross-section of an AT 15x3 tire (66):
F min == 31.52 mm²
We check condition (70):
31.52 mm² 45 mm²
From the condition we see that the rectangular busbars on the 10 kV side of the AT brand (15x3) pass
We will carry out the test on the 0.4 kV side by comparing temperatures; for this we will draw up table 2.5.2.1 parameters of current-carrying parts
Table 2.5.2.1 Parameters of live parts
To check CL AAB 2 (4x25) for thermal resistance on the low side, we will clarify the heating temperature in normal operating mode because The heating current does not coincide with the long-term permissible current.
n= 0 +(additional n - 0) · () 2 ; (71)
n=15+(65-15) () 2 = 15.69C
Let us determine the thermal equivalent for normal operation according to the schedule in Fig. 3.13 literature
An=0.12 10 4 A 2 S/mmI
Let us determine the actual flow time of the short circuit current
t real = t in + t in, (72)
where t is off - switch operation time; With,
t z - protection response time; With
t act = 0.2+1.1=1.3s
Let us determine the reduced time of the aperiodic component of the short circuit current
t pr.a = 0.003 · ", (73)
where "=; because Ipko= Ipk, means “=1
t pr.a = 0.003·1= 0.003 s
Let us determine the reduced time of the periodic component of the short circuit current according to Figure 3.12 in the literature: t pr.p = 0.85 s
Let us determine the total reduced time:
t pr = t pr + t pr.p (74)
t pr = 0.003+0.85 = 0.853 s
Let us determine the thermal equivalent for a short circuit:
A k = A n +, (75)
A k = 0.12 · 10 4 += 0.205 · 10 4 A 2 s/mmI,
therefore, the heating temperature is 30C
The following condition must be met:
The condition is met, therefore, the CL passes through thermal resistance.
Let's check the tires for thermal resistance:
To check a rectangular tire of the AT brand (15x3) for thermal resistance on the low side, we will clarify the heating temperature in normal operating mode because The heating current does not coincide with the long-term permissible current (71):
n=25+(88-25) () 2 = 48.69C
Let us determine the thermal equivalent for normal operation according to the schedule in Fig. 3.13 literature, An=0.38 10 4 A 2 C/mmI
Let us determine the thermal equivalent for a short circuit (75):
A k = 0.38 10 4 += 0.76 10 4 A 2 s/mmI,
therefore, the heating temperature is 110C
Condition (76) must be satisfied:
The condition is met, therefore, AT brand tires (15x3) are thermally resistant.
The ability of devices, conductors and insulators to withstand electrodynamic and thermal effects that occur when the largest short-circuit currents pass through them is called electrodynamic and thermal resistance, respectively.
In the case of a short circuit with sufficient accuracy for practice, the heating process can be assumed to be adiabatic:
Where i k(t) is a function characterizing the change in short-circuit current over time; R J is the resistance of the conductor at a given temperature J; C J is the specific heat capacity of the conductor at a given temperature; G- mass of the conductor.
Considering that the resistance of the conductor and its specific heat capacity are functions of temperature:
,
Where r 0 and c 0 - resistivity and heat capacity of the conductor at the initial temperature J H =0 °C; a and b are temperature coefficients of resistance and heat capacity; S, l, g - area cross section, length and density of the conductor.
Separating the variables and integrating within the required limits, we obtain the equation
which allows you to determine the final temperature of the conductor Jc when heated by a short-circuit current from the initial temperature J n. However, the analytical solution of this equation is difficult, and therefore, for common conductor materials, the dependences of the values of the second integral on the final temperature (at J n =0), which are presented in Fig. 2.8.
Rice. 2.8. Curves for determining the heating temperature of live parts during a short circuit
First integral depending on short-circuit current and tripping time t off, is called the quadratic current pulse V. Its approximate value can be expressed in terms of the effective values of the total current and its components
Where 
effective value of the total short-circuit current at the moment of time t; I P, t- effective value of the periodic component; I a, t– aperiodic component.
Thus, the pulse of the quadratic short-circuit current is equal to the sum of the pulses from the periodic B n and aperiodic B but a component.
The impulse from the periodic component can be determined by the graphic-analytical method by replacing the smooth curve with a step one with ordinates corresponding to the average values of the squares of effective current values for each time interval:
In cases where the fault location is remote from the generators or it is necessary to roughly (overestimate) the impulse from the periodic component, it can be assumed that the periodic component does not attenuate, i.e. 
.
The pulse from the aperiodic component of the short-circuit current is equal to:
When we find
Then the final temperature of the conductor will be equal to
.
In Fig. 2.8 we plot along the ordinate axis J n and along the corresponding curve (point A) we find A n. Adding to A n (on the x-axis) value B/S 2, we get A n and the corresponding conductor temperature J k (point b on the curve).
The final temperature during a short circuit should not be higher than permissible under the condition of maintaining insulation or under the condition of mechanical strength (for bare conductors).
Conductor thermal resistance condition:
The thermal resistance of devices is usually characterized by the rated thermal resistance current I ter at a certain duration of its passage, called the nominal thermal resistance time t ter. To check the device for thermal resistance, compare the value of the thermal impulse normalized by the manufacturer with the calculated value. The condition for the thermal resistance of the apparatus is formulated as:
The method for calculating the thermal and dynamic resistance of conductors and devices is given in more detail in the guidelines for calculating short circuit currents and selecting electrical equipment RD 153–34.0–20.527–98
Electrodynamic force of interaction between two parallel conductors (Fig. 1) of arbitrary cross-section, flowing currents i 1 and i 2, determined by the formula
F=2.04k f i 1 i 2 · l/a 10 -8, kg ,
Where i 1 and i 2 – instantaneous values of currents in conductors, a ; l– length of parallel conductors, cm; a– distance between the axes of the conductors, cm; k f - shape coefficient.
The interaction force between two parallel conductors is evenly distributed along their length. In practical calculations this uniformly distributed force is replaced by the resultant force F, applied to the conductors in the middle of their length.
When the direction of currents in conductors is the same, they attract, and when they are in different directions, they repel.
Shape factor k f depends on the cross-sectional shape of the conductors and their relative position. For round and tubular conductors k f =1; for conductors of other cross-sectional shapes: k f =1 in cases where the cross-section of the conductors is small and their length is large compared to the distance between them and it can be assumed that the entire current is concentrated in the axis of the conductor. Yes, they accept k f =1 when determining the forces of interaction between the m/y phases of busbar structures of switchgears, regardless of the cross-sectional shape of the busbars, because the distance between busbars of different phases in switchgears is quite large and amounts to several hundred millimeters or more.
If the distance between conductors (buses) of rectangular, box-shaped and other sections is small, then k f ≠1.
The force acting on a current-carrying conductor is determined as the result of its interaction with currents in the conductors of the other two phases, while the middle phase conductor is in the most severe conditions. The greatest specific force on the middle phase conductor can be determined from the expression, N/m,
f=√3·10 -7 ·k f ·I 2 m/a,
where I m is the current amplitude in phase, A; a – distance between adjacent phases, m.
Coefficient √3 takes into account phase displacements of currents in conductors.
The interaction of conductors increases significantly in the short-circuit mode, when the total short-circuit current reaches its highest value - shock. When assessing phase interaction, it is necessary to consider two-phase and three-phase short circuits.
To determine the specific force during a three-phase short circuit in a system of conductors, use the expression
f (3) =√3·10 -7 ·k f · i ( 3)2 y /a,
Where i (3) y– shock current of three-phase short circuit, A.
In the case of a two-phase short circuit, the influence of the third (undamaged) phase is negligible, taking into account that ׀i 1׀= ׀i 2 |=|i (2)2 y |. Hence,
f (2) =2·10 -7 ·k f · i ( 2)2 y /a,
Where i ( 2) y – shock current of a two-phase short circuit, A.
Considering that the interphase force with a three-phase short circuit is greater than with a two-phase one. Therefore, the design type of short circuit when assessing electrodynamic forces is considered three-phase.
To prevent mechanical damage under the influence of forces arising in conductors when short-circuit currents flow through them, all elements of the current-carrying structure must have sufficient electrodynamic resistance.
Electrodynamic resistance is usually understood as the ability of devices or conductors to withstand mechanical forces arising during the flow of short-circuit currents, without deformations that prevent their further normal operation.
Thermal effect of short-circuit currents. When a short-circuit current flows, the temperature of the conductor increases. The duration of the short circuit process is usually short (within a few seconds), so the heat released in the conductor does not have time to be transferred to the environment and is almost entirely used to heat the conductor. A conductor or device should be considered thermally resistant if its temperature during a short circuit does not exceed permissible values.
The heating temperature of the conductor during a short circuit can be determined in the following way. During a short circuit in time dt a certain amount of heat is released in the conductor
dQ=I 2 k , t r θ dt,
Where Ik, t– effective value of the total short-circuit current at the moment t KZ; rθ– active resistance of the conductor at a given temperature θ :
rθ=ρ 0 (1+αθ)l/q,
here ρ 0 is the specific active resistance of the conductor at θ=0 0; l– length of the conductor; q– its cross-section; α - temperature coefficient of resistance.
Almost all the heat goes to heating the conductor
dQ=Gc θ dθ,
Where G – conductor mass; c θ– specific heat capacity of the conductor material at temperature θ.
The heating process during a short circuit is determined by the equation
I 2 k , t r θ dt= Gc θ dθ.
When choosing electrical devices, it is usually not necessary to determine the temperature of live parts, since the manufacturer, based on special tests and calculations, guarantees the time and rms current of thermal resistance. In other words, the catalogs provide the value of the guaranteed pulse of the rms short-circuit current, which the device can withstand without damage preventing further normal operation. The condition for testing thermal resistance in this case is as follows:
B to ≤I 2 ter t ter,
Where B to– calculated pulse of the quadratic short-circuit current, determined according to the method described above; I ter and t ter – respectively, the root mean square current of thermal resistance and the time of its flow (nominal value).
The effects of short circuit currents are checked
1) for dynamic stability - devices and conductors protected by fuses with inserts for rated currents up to 60 A inclusive; electrical equipment protected by current-limiting fuses for high rated currents should be checked for dynamic stability based on the highest instantaneous value of the short-circuit current passed by the fuse.
For thermal stability - devices and conductors protected by fuses for any rated currents,
2) conductors in circuits to individual electrical receivers, including workshop transformers with a total power of up to 1000 kVA and with primary voltage up to 20 kV inclusive, if the necessary redundancy is provided in the electrical part, in which the disconnection of these receivers does not cause disruption to the production process, if damage to the conductors cannot cause an explosion and if replacing damaged conductors is without any particular difficulties.
3) conductors in circuits to individual electrical receivers and bleaching distribution points for non-essential purposes, provided that their damage during a short circuit cannot cause an explosion;
In short circuit mode, current-carrying elements of an electrical installation (busbars, cables, etc.) for a short short circuit time t(seconds or fractions of a second) are heated by short-circuit current. from some initial temperature θ n to temperature θ max. Short-circuit currents many times higher than the normal mode currents, therefore, despite the short duration of the short circuit, the temperature of the conductors increases sharply and θ max becomes much greater than θ N (Fig. 6.1). Determining the temperature θ max and comparing it with the short-term permissible θ max additional is a task thermal calculations for short circuit mode
Fig.6.1 Heating of the conductor in short circuit mode
A little time t short circuit allows you to perform thermal calculations during short circuit. without taking into account heat transfer to the environment during this time. Let us consider the heating of the conductor by the periodic component of the short-circuit current, leaving aside for now the additional heating of its aperiodic component of the short-circuit current. Such a separate consideration of the two components of the short-circuit current. it is possible that this directly follows from the expression for the effective short circuit current I short circuit. :
I 2 k.z = I 2 p t + I 2 at (6.1)
where I at is the value of the aperiodic component, and I p t – the periodic component.
The energy spent on heating the conductor with current t p t is expressed by Lenz's law. Then the initial expression for heating the conductor looks like:
i 2 n t R np t = C m θ (6.2)
where R np is the resistance of the conductor, C is the specific heat capacity of the conductor material, m is the weight of the conductor.
Due to the fact that the current changes over the short-circuit time. and the heat capacity and resistance of the conductor are a function of temperature, the original heating equation is differential:
i 2 p t ρ o (1+ αθ) dt = s lγc o(1+ β θ)d θ (6.3)
where i p t is the instantaneous value of the periodic component of the short circuit.
ρ o (1+ αθ) - conductor resistance at temperature θ o C, ohm
c o(1+ β θ) - specific heat capacity of the conductor material at θ o C , Tue. s/y o C
ρ o and c o– specific resistance and heat capacity at 0 o C
α and β temperature coefficients of change ρ and c
s l – conductor volume, cm 3; γ – beat conductor material weight, g/cm 3
Having separated the variables and rearranged the coefficients, we rewrite the equation as follows:
Dt = To d θ (6.4)
Where To = γ
During the short circuit time t, the temperature of the conductor rises from the initial value θ n to θ max of the final value, therefore both sides of the equation should be integrated within the specified limits:
The law of change in the value i p t over time is quite complex, therefore the integration of this function is carried out by replacing areas (integrals). Fig.6.2. illustrates this method.∞
Fig.6.2 Graph for determining the fictitious time of the periodic component.
On the graph in Fig. 6.2, the area of the OABC corresponding to the short-circuit time. t equal to the heat from the short-circuit current. during t, those.
area OABC = dt
The same amount of heat could be generated by a steady (constant) short-circuit current. I 2 ∞ but for a different time t fp. This time can be found by constructing a rectangle ODEF of equal area. For determining t FP at a known time t according to the calculated short-circuit current curves. dependency built t fп =f (λ) (Fig. 6.3), and λ = I” / I∞. Thus the integral can be calculated as:
t
fp (6.6)
Fig.6.3 Curves for determining fictitious time
The heat generated by the aperiodic component of the short-circuit current. i and t is determined by the equation similar to equation 6.6:
t f.a. (6.7)
where t f.a. – time during which the steady-state short-circuit current will release the same amount of heat as the aperiodic component of the short-circuit current. during the short circuit t.
The aperiodic component decays with the time constant of the circuit to the short-circuit point. T a: i a t =√2 I" o e - t / Ta (6.8)
Where I" o – known (equation 5.9) effective value of the subtransient component of the short-circuit current. at a time equal to 0. This function is easily integrated and as a result the value of the fictitious time of the aperiodic component:
t f.a = T a λ 2, (6.9)
where λ = I"o/
Full dummy time t f = t fp + t f.a
Integrating the right side of equation 6.5 is complex and leads to a cumbersome expression for determining the desired temperature θ max. Based on this expression, the calculated curves were constructed under the assumption that the initial temperature of the conductor θ n =0. The order in which curves are used follows from their construction. First, find the initial temperature of the conductor at the moment of short circuit. θ n. :
θ n = θ environment + (θ additional - θ environment) I 2 slave / I 2 additional (6.10)
where θ of the medium is the calculated temperature of the medium
θ additional – long-term permissible temperature of the conductor
I slave– operating current through the conductor
I additional– permissible current through the conductor
The values of θ additional are given in the reference tables for selecting buses and cables. The θ environment is taken to be the maximum possible during operation (for example +40 o C). Having determined the initial temperature, find from the curves (Figure 6.4) the value of the corresponding abscissa a n. Then t f is calculated and the abscissa is determined: a k = a n + t f. The value of θ max is determined by the value of a k. Next, the value of θ max is compared with θ max. for a given type of conductor material.
Fig.6.4 Curves for determining the heating temperature of conductors during short circuits.
Due to the fact that the duration of the short circuit is short (does not exceed a few seconds), temperatures that are significantly higher than the permissible temperatures during prolonged heating are taken as θ maxdop. It is taken into account that the insulation of the conductors is capable of withstanding θ max. without compromising further work.
For bare conductors (switchboard busbars), θ maxdop is taken from the conditions of the mechanical strength of the material. For example, for bare copper busbars θ maxdop = 300 o C.
Conductors protected by fuses need not be checked for thermal stability, as well as conductors protected by current-limiting switches and circuit breakers, without a specially introduced time delay when triggered.
Selective circuit breakers (circuit machines with an adjustable time delay when switching off a short circuit) are checked for thermal resistance according to the following conditions:
I 2 ¥ t f< (I 2 t) доп. ,
where I ¥ - steady-state short-circuit current; t f – fictitious short-circuit time;
(I 2 t) add. – thermal stability according to technical specifications(reference data).
ELECTRODYNAMIC ACTION CURRENT K, W,
When current flows i in the circuit in the latter an electrodynamic force arises F, tending to deform the contour (Fig. 6.5). At a constant current value, the increment in field energy W when the contour is deformed in the direction X equal to the work done by the electromagnetic force F over the same period of time.
dW = Fdx(6.11)
Where X- coordinate of the direction of the force.
Equation 6.11 is called Maxwell's equation.
Rice. 6.5 The action of electrodynamic forces on a current-carrying circuit.
Magnetic energy W in a circuit with inductance L and electric shock i is determined by the well-known expression:
F = (6.13)
With two circuits with inductances L 1 and L 2 and according to currents i 1 and i 2 and mutual inductance M energy magnetic field W is determined by the expression:
W= L 1 i 2 1 + L 2 i 2 2 + M i 1 i 2(6.14)
Electrodynamic force tending to change the relative position of rigid circuits ( L 1 = const; L 2 = const) is equal to:
F = i 1 i 2 (6.15)
Mutual inductance (H) of two parallel conductors located in the same plane at a distance much less than their length.
M= 2l(ln - 1)10 -7 Hn (6.16)
dM/dx = dM/da = (2l/a) 10 -7 (6.17)
And F = (2i 1 i 2 l / a) 10 -7 N (6.18)
This formula is used to determine the force of interaction between the busbars of distribution devices during the passage of short circuit currents.
When calculating the mechanical strength of busbars in short-circuit mode, we proceed from the assumption that the busbar of each phase is a multi-span beam, freely lying on rigid supports and under the influence of a uniformly distributed load. Distribution board busbars. satisfy the requirements of electrodynamic stability if the value of the maximum design voltage in the bus is less than or equal to the maximum permissible voltage, i.e. σ calc. ≤ σ add.
a) reducing the magnitude of the short circuit current;
b) increasing the distance between tire axles;
c) reducing the span length between support insulators;
d) changes in tire section size.
The maximum voltage in the bus when the tires are located flat is determined by the ratios:
When the number of spans is more than two
σ calc. = (1.06 K f i 2 r L 2 / a h 2 b) * 10 -10, kPa (6.19.)
with a number of spans equal to two
σ calc. = (1.33 K f i 2 r L 2 / a h 2 b) * 10 -10, kPa (6.20)
When the tires are located according to Fig. 6.6 a. maximum voltage
in the bus is equal to: .
σ calc. = (1.06 K f i 2 r L 2 / a h b 2) * 10 -10, kPa (6.21)
with the number of spans equal to two,
σ calc. = (1.33 K f i 2 r L 2 / a h b 2) * 10 -10, kPa (6.22)
where i r is the total short-circuit shock current;
a - distance between phase axes, cm, usually a = 6......7 cm
L - span length, cm, usually L = 60 cm;
h - tire height, cm;
b - tire thickness, cm;
Kf - tire shape coefficient, determined from the curves presented in Fig. 6.7
Rice. 6.6 Arrangement of single-span buses
Rice. 6.7 Dependence of the tire form factor on the relative position and configuration.
Circuit breakers are checked for electrodynamic stability against short-circuit shock current. until the switch turns off. In addition to dynamic resistance, selective (generator) circuit breakers are also tested for ultimate breaking capacity.
The maximum breaking capacity is determined by the permissible current value at the moment the contacts diverge. Dynamic resistance test condition:
i beats calc.< i уд. доп. ;
for breaking capacity:
I t calc.< I t доп,
where i beat. calc. – calculated short-circuit shock current. for the point selected for the purpose of checking the machine; i beats add. – permissible value of short-circuit shock current. machine; I t calculated – calculated effective value of the short-circuit current. at the moment of divergence of arc extinguishing contacts (corresponding to the time setting); I t additional, is the permissible effective value of the circuit breaker current at the moment of divergence of the arc extinguishing contacts.