11 kw Ingersoll Rand Motor

a 11 kw,could help me with the turns, pitch, and wire size

Dear afritech:

To determining of coil pitch, no of turn, wire size, you should inform the following data such as :

1. Core Dimensions , (Length and Inner Diameter) .. in M

2. No of Slot

3. Voltage rating

4. Speed or no of poles

5. Freq

6. Cos phi

The following calculation has just informed you how to determining of coil pitch, no of turn, and wire size,

all of parameters require are assumption only such as the Bav (flux density in the core, No of slot, Core dimensions (Diameter and Length are correct) however you must recheck with the existing data, especially for Bav, you can refer to the literatures for machine design.

You can calculate by your self……

Assumption Data:

No of Slot = 36

No of Pole = 4

No of Phase = 3

Freq = 50 Hz

Cos ф = 0.85

Voltage = 380 volts

Core Dimensions : D = 30 cm, L/D = max 0.65 or L = 20 cm

Thus : No of Slot per Pole per Phase

( SPP ) Slot / pole / ph = (36/ 12) = 3 slot

Slot / ph (g ) `` = 36 / 3 = 12

Full Pitch Coil will be:

Coil pitch for full pitch = Slot / pole = 36 / 4 = 9

Its mean the coil span will be = 1 – 10 (for full pitch)

Slot Angle ( in Electric degrees)

Slot angle (α ) = ∏P / Slot

∏ = 180 ( North to South )

Slot Angle (α) =( 180 x 4 ) / 36 = 20 deg elect

Normally to reduce of harmonic the pitch winding will be used a chording factor.

Possibility of chording factor for SPP = 3

Coil pitch (kp) = 2/3 with pitch factor = 0.866

Coil pitch (kp) = Cos ( α / 2)

Thus, the coil pitch will be = 1800 elct – α

= 180⁰ elct – 20⁰elct = 160⁰elct

= 160⁰elct / 20⁰elct = 8 slot

Coil span = 1 - 9

Distribution factor (kd) = (Sin (mα)/2) / m Sin (α/2)

m = SPP = Slot/pole/ph = 3

=( Sin (3x20)/2) / 3 Sin (20/2) = 0.5 / 0.52095 = 0.9598

Winding Factor (kw) = kp x kd = 0.866 x 0.9598 = 0.8312

Basic formulae:

Co

=

(11 kw x Bav x ac x Cos φ xη)x 10‾³

D2L = KW / ( Co x N )

KW = Motor Out Put

E = 4.44 x kw x f x T1 x Øm

Ø per pole = ( Bav x ∏ x D x L ) / P

Bav = Flux density

D = diameter of core

L = Length of core

P = pole

Let say :

D = 0.3 M and L/D max = 0.65, L = 0.2 m

Bav = 0.489 Wb / M2 for class F

Ø per pole = ( 0.489 WB/M2 x ∏ x 0.3M x 0.2M ) / 4

= 0.0922 / 4 = 0.02305 WB

E = 380 VOLTS

F = 50 HZ

kw = 0.8312 ( winding factor)

T1 = E1 / ( 4.44 x kw x f x Øm )

T1 = 380 / ( 4.44 x 0.8312 x 50 x 0.02305) = 380 / 4.25 = app 89 turn

Slot / phase (g ) = 36 / 3 = 12

If Winding is double layer

No of conductor in the slot =( T1 x 2 ) / g = ( 89 x 2 ) / 12 = app 15 turn

Synchronous Speed = 1500 RPM

Let say for the cooling with speed = 1500 RPM, Current Carrying Capacity for 1mm2 = 5 Amps

The cross section of conductor will be :

If Cos ф = 0.85

Current IFL = 11 kW / ( √3 x 0.85 x 380 ) = 39.3 Amps app = app 20 Amps

Wire size to accommodate of IFL with 1 mm2 = 5 Amps

Wire size = 20 amps / 5 amps = 4 mm2

Wire Cross Section Area (q ) = (1 / 4) x ∏ D²

Wire Diameter ( D² ) = 4 q / ∏ = (4 x 4 mm2) / 3.14 = 5.1 mm2

Wire Diameter (D) = √5.1 mm² = app 2.25 mm

Rgds

sis