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

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. 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

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

• Qc => Power per cooler in kcal/s

• dTw => maximal temperature rise of water in °K

• q => Amount of water in l/s

• 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)

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

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

• 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

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

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

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

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

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 = 3.58°K
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 = 3.13v^0.87Dk^-0.13   = 0.313x1.23^0.87x1.016^-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
dTcc = 2 Pc Lcp/(Ac+Apw)/λc = 2 x 1500 x 1.9 /(97+ 246)/1.8  = 9.2°K

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

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

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 = 42°K

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

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

Design page 1

Design page 1

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