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Designing Water Cooled Inverter Filter Choke,
0.5mH, 600Arms, 60Hz

General about inverter filter choke

Fig.1 illustrates the main circuit diagram of a three-phase inverter. The 3-phase mains Uin supplies the controlled rectifier R through the 3-phase commutation choke CC. The DC voltage Udc is regulated by the rectifier and smoothed with the capacitor C. The 3-phase AC voltage Uout is produced at the inverter outputs. The amplitude, frequency and form of this 3-phase AC voltage are regulated with the inverter and rectifier.

Fig. 1

The typical form of the inverter output voltage per phase is illustrated in Fig.2.

Fig. 2

At an inverter modulation frequency N*f, the output voltage Uout essentially consists of three components:

  • Fundamental frequency U with the frequency f (50Hz or 60Hz)
  • First harmonic U1=0.45*U with the frequency (N-1)*f.
  • Second harmonic U2=0.45*U with the frequency (N+1)*f

Accordingly, the current through the inverter filter choke Iout essentially comprises of 3 components:

  • Fundamental frequency I with the frequency f. This current is "impressed" and its amplitude depends on the inverter power.
  • First harmonic I1 with the frequency (N-1) * f. The amplitude of this current depends on the voltage U1, the modulation frequency N*f and the inductance L of the inverter filter choke:
    I1=U1/(2* *f*(N-1)*L).
  • Second harmonic I2 with the frequency (N+1) * f. The amplitude of this current depends on the voltage U2, the modulation frequency N*f and the inductance L of the inverter filter choke:
    I2=U2/(2* *f*(N+1)*L)

About input parameters

Inductance at 1000Hz and 900Apeak

0.5mH
Inverter frequency 60Hz
Modulation frequency (N=18) 1080Hz
RMS current at main frequency 600Arms at 60Hz
RMS currents at 1020Hz and 1140Hz 30Arms
Maximal total losses (warm) 10kW
Insulation clas H
Indirect water cooling 20C  incoming
55C, maximal outcoming
1.5at max. pressure drop
1.5 m/s, max. speed
Test voltage winding-cooler, winding-core 4kV, 1minute, 60Hz
   

 

About indirect water cooling

The indirect water cooling of the three phase inverter filter choke will be ralized with accordance to the construction in the Fig.3.

Fig.3

Note that the Small Chokes Program does not support water cooling. I selected this design example in order to transfer the know-how for water cooling design.to my users.

Power per cooler in W and kcal/s

  1. Pc = K*Ptot/Nc

  2. Qc = Kd*Ptot/4180/Nc

  • Kd => Factor of the losses distribution between cooler and air (0.8-0.9)
  • Ptot => Choke total losses in W
  • Nc => Number of coolers
  • Pc => Power per cooler in W
  • Qc => Power per cooler in kcal/s

Amount of water in l/s

q = Qc/dTw

  • Qc => Power per cooler in kcal/s

  • dTw => maximal temperature rise of water in K

  • q => Amount of water in l/s

Water speed in m/s

v = 10q/Apcs

 

  • q => Amount of water in l/s
  • Apcs => Cross section of pipe in cm2
  • v => Water speed in m/s (It must not be higher tan 1.5 m/s)

Equivalent pipe diameter

If the pipe cross section is rectangular then the equivalent pipe diameter can be calculated

Dk = 2ab/(a+b)

  • a, b => Sides of the rectangular pipe

  • Dk => Equivalent pipe diameter

Faktor of the convection between water and pipe in W/K/cm^2

α = 0.313v0.87Dk-0.13

  • V => Water speed in m/s
  • Dk => pipe diameter in m
  • Al => Factor of the convection in W/K/cm2

Temperature drop between cooler and water in K

dTcw = Pc/α/Apw

  • Pc = 1000Ptot/Nc => Losses per cooler in W

  • α => Factor of convection in W/K/cm2

  • Apw => Contact surface cooler-water in cm2

  • dTcw => Temperature drop between the cooler and water in the pipe in K

 Temperature drop between the pipe and the cooler surface in K

dTcc = 2 Pc Lcp/(Ac+Apw)/λc

  • Pc =  => Losses per cooler in W
  • Lcp => Equivalent distance cooler surface-pipe in cm
  • Apw => Contact surface cooler-water in cm2
  • Ac => Surface between the cooler and winding in  cm2 
  • λc=> Thermal conductivity  of cooler material (normally Al) 1.8 W/K/cm
  • dTcc = Temperature drop between the pipe and cooler surface in K

Temperature drop within the cooler insualtion in K

dTci = Pc Lci/Ac/λci

  • Pc => Losses per cooler in W
  • Lci => Thikness of the the cooler insulation cm
  • Ac => Surface between the cooler and winding in cm2
  • λci => Thermal conductivity  of cooler insulation  in W/K/cm
  • dTci = Temperature drop within the cooler insulation in K

Temperature drop within the wire insualtion in K

dTwi = Pc Lwi/Ac/λwi

  • Pc => Losses per cooler in W
  • Lwi => Thikness of the the wire insulation cm
  • Ac => Surface between the cooler and winding in cm2
  • λwi => Thermal conductivity  of wire insulation  in W/K/cm
  • dTwi = Temperature drop within the wire insulation in K

Temperature drop within the winding in K

dTw = Pc Lw/Aw/λw/16

  • Pc =  => Losses per cooler in W
  • Lw => Average turn length in cm
  • Aw => Winding cross section (all turns) in cm2
  • λw=> Thermal conductivity  of winding material (normally Cu) 3.5 W/K/cm
  • dTw = Temperature drop within the winding in K

Water pressure drop in the pipe in at

dp = 0.01 K v2 Lt/Dk

  • Lp => Pipe length in cm

  • Dk => Pipe diameter in cm

  • K => Factor
     
    Dk (m) 0.0.004 0.006 0.006 0.007 0.008 0.009 0.010 0.011 0.012
    K 0.094 0.087 0.083 0.080 0.075 0.073 0.070 0.068 0.066

 Provisional "core selection"

Bofore you start coke design you need approximately to know how big has to be the cooler. Due to the fact that the cooler fits tightly over the leg along the window hight its cooling surface to the winding is apppoximately equal with the leg surface along the window hight.

Pc = Kd Ptot/Nc = 0.9x10000/6 = 1500 W
Qc = Pc/4180 = 0.358 kcal

dTw = 3.58K
q = Qc/dTw = 0.358/3.58 = 0.1 l/s

Dk = 0.4 " = 0.1.016 cm
Apcs = πDk2/4 = 3.14x1.0162/4 = 0.81cm2 
v = 10q/Apcs = 10x0.1/0.81 = 1.23 m/s

 α = 3.13v0.87Dk-0.13   = 0.313x1.230.87x1.016
-0.13 =  0.374  W/K/ cm2  
 
Lp = 12in = 30.48 cm
Apw = πDkLp = 3.14x1.016x30.48 = 97.23 cm2
dTcw = Pc/α/Apw = 1500/0.374/97.23 = 41 K

dp = 0.01 K v2 Lp/Dk = 0.01x0.070x1.232x30.48/1.016 =  0.03at per cooler

Lcp= 0.75"=1.9cm
Ac = 12"x3" = 1.06x 232 =  246 cm2 ( Leg Width x Height), Factor 1.06 due to the oval form
λc=1.8 W/K/cm

d
Tcc = 2 Pc Lcp/(Ac+Apw)/λc = 2 x 1500 x 1.9 /(97+ 246)/1.8  = 9.2K

λci = 0.015 W/K/cm (Huntsman XB 2710+XB 271, 20kV/mm, Tmax=120C)
Lci = 0.1cm (Insulation thicknes of the coated coole))
dTci = Pc Lci/Ac/λci = 1500x0.1/0.015/246 = 40K

λwi = 0.004 W/K/cm
Lwi = 0.01 cm
dTwi = Pc Lwi/Ac/λwi = 1500x0.01/0.004/246 = 15K

Lw = 13" = 33cm
Aw= 40 x 0.46" x 0.182 = 3.34 in2 = 21.5 cm2
λw = 3.5 W/K/cm

dTw = Pc Lw/Aw/λw/16 = 1500 x 33/ 21.5/3.5/16 = 42K

Winding teperature rise will be 2+41+40+15 +9.2 + 42 =149.2K. Due to the fact that the winding losses are lower than 3000W (view the design results) the temperature rise will be approx. 20% lower ( 125K instead 149K)

This provisional thermal design of the cooler shows that the core with the leg sizes 3" x 12" is, from thermal point of view, big enough.

General design rules

  • The cAl-ooler size is H=12", W=3" T=1.2". In order to avoid any gaps between the cooler and the winding the contact form has to be oval (see the Fig3). The cooler has to be coated with 1.5mm resin from Huntsman XB 2710+XB2711. This resign has very high thermal conductivity (0.015W/K/cm) and relative high break down voltage (>20kV/mm).

  • Due to the fact that the water cooled are small the amperturns are very high and the gap size very big. For a "good" gap of approx <0.1" the choke has to be made by more than 10 gaps per leg and with the high induction

  • The criterion of the design is Q-factor:at the max. tempeerature 165C:
     Qf = Lω/Irms2/Ptot/Kd/(Nc/2) =
    0.0005x376/6002/10000/3/0.9 >24 =2

  • The insulation in the gaps and aroun the gaps has to be clas H.

  • From thermal point of view the optimal construction is one leyer winding with Cu rectangular wire

 

 

Design page 1

 

Design page 1

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