THE SUDDEN SHORT-CIRCUIT AT TERMINALS

Sudden short-circuit at the terminals of large induction motors produces severe problems in the corresponding local power grid. [16]

Large induction machines with cage rotors are skin-effect influenced so the double cage representation is required. The space-phasor model for double-cage IMs is obtained from (13.66) with V_r = 0 and eliminating the stator core loss equation, with the rotor core loss equation now representing the second rotor cage. In rotor coordinates,

Vs = R i_s s + Ψ + ω Ψp s j _r s; Ψ = Ψ_{s m} + L i ; _sl s Ψ_m = L i_m m

0 = R i_r1 r1 + Ψp r1; im = is +ir1 +i ; r1 Ψ_r1 = Ψ_m + L i_rl1 r1 (13.89) 0 = R i_r2 r2 + Ψp r2; Ψ_r2 = Ψ_m + L_rl2ir2

To simplify the solution, we may consider that the speed remains constant (eventually with initial slip S₀ = 0).

Eliminating ir1 and ir2 from (13.89) yields

Ψ_s ( )p = L p i p( ) _s ( ) (13.90)

The operational inductance L(p) is

L p( )= Ls (1(1++T ‘p 1T’p 10 )()( ++T”pT “p0 )) (13.91)

The transient and subtransient inductances L_s’, L_s” are defined as for synchronous machines.

L_s = limL p_s→₀ ( )

To short-circuit the machine model, – V e_s ^jγ must be applied to the terminals and thus Equation (13.89) with (13.90) yields (in Laplace form):

^jγ

Denoting ^{R R} ct (13.97)

Vs ejγ 1 1−e− α+ ω( j )t +

i ts( )= − jω1 Ls ( 1 )  L ‘1s − L1s e− T’t −e− α+ ω( j 1 )t 

(13.100)

+ 1 − 1 e− T”t −e− α+ ω( j 1 )t 

 L “s L ‘s  



Now the current in phase a, i_a(t), is i_a ( )t = Re i t[( _s ( )+i_s⁰( )t e) ^jω1t ]=

V 2 1 e−αt sin γ −  ω11L ‘s − ω11Ls e− T’t + ω11L “s − ω11L ‘s e− T”t sin(ω + γ1t ) ω1L “s 

(13.101)

A few remarks are in order.

• The subtransient component, related to α, T”, L_s” reflects the decay of the leakage flux towards the rotor surface (upper cage)

• The transient component, related to T’, Ls’, reflects the decay of leakage flux pertaining to the lower cage

• The final value of short-circuit current is zero

The torque expression first requires the stator flux solution.

j 1

We may neglect α and add Ψ_s ( )t with Ψ_s⁰ to obtain Ψ_st ( )t .

Ψst ( )t = V ejsω1jγ e− α+ ω( j 1)t (13.106)

Now, from (13.100), (13.105), and (13.102), the torque is

Te = 3V2 ωp11  ω11L ‘s − ω11Ls e− T’t + ω11L “s − ω11L ‘s e− T”t sin(ω1t) (13.107) Neglecting the damping components in (13.107), the peak torque is

Tepeak ≈ − 3p V^ω1₁ 2 ^ω₁1L “_s (13.108)

This result is analogous to that of peak torque at start up.

Typical parameters for a 30 kW motors are [17, 18]

T’ = 50.70ms, T” = 3.222 ms, T₀’ = 590 ms, T₀” = 5.244 ms, 1/α = 20.35ms, L_m = 30.65 mH, L_sl = 1 mH (L_s = 31.65 mH), R_s = 0.0868 Ω Analyzing, after acquisition, the sudden short-circuit current, the time constants (T₀’, T₀”, T’, T”) may be determined by curve-fitting methods.

Using power electronics, the IM may be separated very quickly from the power grid and then short-circuited at various initial voltages and frequency. [17] Such tests may be an alternative to standstill frequency response tests in determining the machine parameters.

Finally, we should note that the torque and current peaks at sudden shortcircuit are not (in general) larger than for direct connection to the grid at any speed. It should be stressed that when direct connection to the grid is applied at various initial speeds, the torque and current peaks are about the same. Reconnection, after a short turn-off period before the speed and the rotor current have decreased, notably produces larger transients if the residual voltage at stator terminals is in phase opposition with the supply voltage at the reconnection instant.

The question arises: which is the most severe transient? So far apparently there is no definite answer to this question, but the most severe transient up to now is reported in [18].