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Transcript
風力發電期末報告
Short-Circuit Current of Wind Turbines With Doubly
Fed Induction Generator
報告人:廖仁呈
CCU
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Intorduction
• The short current contribution is important to know with respect to the coordination of
network protection, and the maximum currents that are allowed in a network.
• Most protection schemes that have been proposed are based on a socalled crowbar.
 This paper analyses the behavior of a crowbar-protected DFIG.
• It has received considerable attention in the past and some good approximate equations
to determine the maximum short-circuit current have been derived.
 First. We should find the equations.
CCU
2
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Induction Machine Response
• First, the maximum value of the short-circuit current of a conventional induction
machine is determined.
Fig.1 Equivalent circuit of induction machine for transient analysis.
2
With the Vs ,coupling factor
ks , kr

We can find the
is 
CCU
s
L's

 kr
L
and the leakage factor   1 m
Ls Lr
r
L's
3
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Induction Machine Response
• The short-circuit behavior of induction machines is strongly dependent on the
machine characteristics.
 Take into account machine characteristics with the two damping factor and
time constant, the is will becomes


2Vs t / T 's
is 
e
 1   e js t e t / T 'r
jX 's
CCU
4

Department of Electrical Engineering
National Chung Cheng University, Taiwan
Induction Machine Response
• The short-circuit current in this phase is then
the projection of the vector is on the a-phase,
i.e., its real part.
• Although the current vector does not reach
the maximum value exactly at t = T/2, the
current after half a period gives a good
approximation of the maximum current .
• The maximum current can thus be obtained by substituting t = T/2 in

2Vs T / 2T 's
is ,max 
e
 1   e js t e T / 2T 'r
jX 's

CCU

5
is ,max as
Department of Electrical Engineering
National Chung Cheng University, Taiwan
DFIG Protection
• In order to avoid breakdown of the converter switches, a crowbar is connected to the rotor
circuit.
 This can, for example,be done by connecting a set of resistors to the rotor winding via
bi-directional thyristors.
CCU
6
Department of Electrical Engineering
National Chung Cheng University, Taiwan
DFIG Short-circuit current
• When the bypass resistors are connected to protect the converter in case of a
fault, the transient time constant of the rotor becomes
• Taking into account these differences between an induction machine and a
doubly fed induction generator, is , max becomes

i 's ,max 
2Vs
X 's  Rcb
CCU
2
e
 t / T ' s

 1   e js t e  t / T 'r 
7
1.8Vs
X 's  Rcb
2
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Crowbar Resistance
• From i 's , max ,it can be observed that the maximum short-circuit current of the DFIG strongly
depends on the value of the crowbar resistance.
 On one hand, the resistance should be high to limit the short-circuit current.
 On the other hand, it should be low to avoid a too high voltage in the rotor circuit.
• An approximation of the maximum stator current is given by
CCU
8
is ,max
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Crowbar Resistance
• The voltage across the bypass resistors, and thus across the rotor and converter is
2Vr  Rcbir ,max
Rcb 
CCU
2Vr ,max X 's
3.2Vs  2Vr ,max
2
2
9
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Case study
• The simulation setup is shown in Fig. 4. The doubly fed induction generator is
directly connected to a voltage source.
• In order to simulate a short circuit, the voltage drops to zero.Simulations have been
done with three doubly fed induction generators with different rated powers:
3MW, 2.75 MW, and 660 kW.
CCU
10
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Case study
• In the first case, the calculated value is too low.
The maximum currents calculated from
i 's ,max
:5.9 p.u.
The simulated value is : 6.4 p.u.
CCU
11
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Case study
The maximum currents calculated from
i 's ,max
:7.8 p.u.
The simulated value is : 7.8 p.u.
CCU
12
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Case study
• In the last case it is too high.
The maximum currents calculated from
i 's ,max
:11.7 p.u.
The simulated value is : 10.1 p.u.
CCU
13
Department of Electrical Engineering
National Chung Cheng University, Taiwan
Conclusion
• The fault between the calculated and simulated values is less than 15%
in all cases.
• One of the goals of this paper is to derive an equation that requires only a limited
number of parameters.
 It’s the crowbar resistance.
• The maximum current decreases with increasing rated power.
CCU
14
Department of Electrical Engineering
National Chung Cheng University, Taiwan