Designing Water Cooled Inverter Filter Choke,

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  20°C incoming 55°C, maximal outcoming 1.5at max. pressure drop 1.5 m/s, max. speed 
Test voltage windingcooler, windingcore  4kV, 1minute, 60Hz 
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 knowhow for water cooling design.to my users.
Pc = K*Ptot/Nc
Qc = Kd*Ptot/4180/Nc
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
v = 10q/Apcs
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
α = 0.313v^{0.87}Dk^{0.13}
dTcw = Pc/α/Apw
Pc = 1000Ptot/Nc => Losses per cooler in W
α => Factor of convection in W/°K/cm^{2}
Apw => Contact surface coolerwater in cm^{2}
dTcw => Temperature drop between the cooler and water in the pipe in °K
dTcc = 2 Pc Lcp/(Ac+Apw)/λc
dTci = Pc Lci/Ac/λci
dTwi = Pc Lwi/Ac/λwi
dTw = Pc Lw/Aw/λw/16
dp = 0.01 K v^{2} 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 
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 = πDk^{2}/4 = 3.14x1.016^{2}/4 = 0.81cm^{2}
v = 10q/Apcs = 10x0.1/0.81 = 1.23 m/s
α = 3.13v^{0.87}Dk^{0.13 }= 0.313x1.23^{0.87}x1.016^{0.13}
= 0.374 W/°K/ cm^{2}
Lp = 12in = 30.48 cm
Apw = πDkLp = 3.14x1.016x30.48 = 97.23 cm^{2}
dTcw = Pc/α/Apw = 1500/0.374/97.23 = 41 °K
dp = 0.01 K v^{2} Lp/Dk = 0.01x0.070x1.23^{2}x30.48/1.016
= 0.03at per cooler
Lcp= 0.75"=1.9cm
Ac = 12"x3" = 1.06x 232 = 246 cm^{2 }( 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 in^{2} =
21.5 cm^{2}
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.
The cAlooler 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 Qfactor:at the max. tempeerature 165°C:
Qf = Lω/Irms^{2}/Ptot/Kd/(Nc/2) =
0.0005x376/600^{2}/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