**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.

The typical form of the inverter output voltage per phase is illustrated in **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: - 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:

**I1=U1/(2* ¶*f*(N-1)*L)**.

**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 | |

Indirect water cooling | 20°C incoming 55°C, 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.

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**

**Pc = K*Ptot/Nc****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 cm
^{2} - 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/cm
^{2}

**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/cm
^{2} - Apw => Contact surface cooler-water in cm
^{2} - 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 cm
^{2} - Ac => Surface between the cooler and winding in cm
^{2} - 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 insulation in °K**

**dTci = Pc Lci/Ac/λci**

- Pc => Losses per cooler in W
- Lci => Thickness of the the cooler insulation cm
- Ac => Surface between the cooler and winding in cm
^{2} - λci => Thermal conductivity of cooler insulation in W/°K/cm
- dTci = Temperature drop within the cooler insulation in °K

**Temperature drop within the wire insulation in °K**

**dTwi = Pc Lwi/Ac/λwi**

- Pc => Losses per cooler in W
- Lwi => Thickness of the the wire insulation cm
- Ac => Surface between the cooler and winding in cm
^{2} - 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 cm
^{2} - 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 | 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.

**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

- 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 165°C:
- 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

Qf = Lω/Irms2/Ptot/Kd/(Nc/2) = 0.0005x376/6002/10000/3/0.9 >24 =2

**Design page 1**

**Design page 1**