3 Choice of the designing and calculation of windings
3.1 Calculation of LV windings.
Determine a voltage of one coil after a rounding of calculated number of the coils:
Vturn
=
=5.922 V.
Acting induction in the rod:
Вleg
=
= 1.514 T.
3.1.1 Calculation of cylindrical LV windings.
3.1.1.1 Assume approximate number of layers of the winding nlayer
Assume nlayer = 3.
3.1.1.2 Approximate number of coils in the layer:
Wlayer
=
= 13turns
Assume Wlayer =13 coils.
3.1.1.3 Approximate height of coil with insulation:
hturn
=
= 0.01985m,
where L – height of the LV winding on item 2.8.
3.1.1.4 Cross section of winding wire collects from one conductor because the cross section of winding conductor is less than 75 mm2 (on item 2.14.).
3.1.1.5 Number of elementary conductors: nel = 2,
3.1.1.6 Approximate cross section of elementary conductors:
Sel
=
= 6.255·10-5
m2.
3.1.1.7 Approximate height of elementary conductor, from which is composed a coil of a winding with insulating:
h´wire
=
= 0.017 m.
3.1.1.8 Approximate height of elementary conductor without insulation:
hw = h´wire – δiz = 0.017 – 0.5·10-3 = 0.0165 m,
where δiz = 0,5 mm - two-sided thickness of insulation (tabl. 5.1, р. 211 [1]).
3.1.1.9 On the base hw = 0.0165 m and necessary cross section Sel = 0.6255 mm2 (when was need Sel =0.671 mm2) under the tabl. 5.2, р. 212 [1] choose an aluminium conductor trademark АПБ a =4 mm and b =17 mm.
Characteristic of a coil shall write down as:
АПБ×1×
.
Accommodation of conductors is shown in figure. 3.1.
Figure 3.1 – Determination of height of the coil of LV winding
Figure 3.2 – Cylindrical, multilayer from rectangular wire of the coil of LV winding.
3.1.1.10 Calculation height of the coil:
htl = nel·b1 =2·17.5·10-3 = 35·10-3 m.
3.1.1.11 Calculation the number of the coils in the layer:
Wlay
=
=13.858
I assume that Wlayer =13.
3.1.1.12 Number of the layers:
n1
=
= 2.814
assume n1 = 3 layers and number of coils W1= 39.
3.1.1.13 Real cross section of the coil:
Swire1 = nel·Selement·10-6 = 2·67.1·10-6=134.2·10-6 m2.
3.1.1.14 Real density of the current:
Δ1
=
=1.721·106 A/m2
it is in an allowable limit 1,2-2,5 А/m2 (tabl. 5.7, р. 257 [1]).
3.1.1.15 axial size of coil:
hturn1 = nel·b1·10-3 = 2·18·10-3 =35·10-3 m,
Where b1 = 18 mm – the height of elementary conductor with insulation.
3.1.1.16 Final axial size of a winding:
L1 = hturn1·(Wlayer +1) + 0.01 = 35·10-3·(13 + 1) + 0.01 = 0.5 m.
3.1.1.17 Radial size of the winding (under the fig. 6.2 and 6.3, р. 268[1]).
a1 = (nlayer · a1 + a11)·10-3,
where a1=45 mm the width of elementary conductor with insulation;
a11=0 mm for winding without channel
a1 = (nlayer · a1 + a11)·10-3 = (3·4.5 + 0)·10-3 = 0.014 m,
3.1.1.18 Internal diameter of the winding:
D1´ = dn + 2·a01·10-3 = 0.16 + 2*4·10-3 = 0.16001 m.
3.1.1.19 Outside diameter of a winding:
D1´´ = D1´ + 2·a1 = 0.16001 + 2·0.014= 0.187 m.
3.1.1.20 Surface of cooling of the LV winding at absence of the cooling channel:
S01=3·КЗ·π·(D1´ + D1´´)·L = 3·0.75·3.14·(0.16 +0.187)· 0.485 = 1.226 m2,
where с = 3 – the number of active rods;
КЗ = 0.75 – coefficient, which takes into account closing a part of a surface of the winding by the lath and others insulation details.
3.1.1.21 approximate density of heat flow of the LV winding:
q1
=
= 972.314 Wt/m2.
It is in permissible limit from 800 to 1400 Wt/m2.
3.2 Calculation of HV windings.
3.2.1 The number of coils of HV winding by the rated voltage
W2
= W1·
= 585
Assume 585 turns.
3.2.2 Approximate density of current in the windings:
Δ2 = 2·Δavd – Δ1 = 2·1.846– 1.721= 1.971 MА/m2.
3.2.3 Approximate cross section of the windings’ coil:
P2
=
=7.81 .10-6
m2.
Determination of sizes of the coil of the HV winding is shown in figure. 3.3.
Figure 3.3 – Determination of sizes of the coil of the HV winding
3.3 Calculation of multi-layer cylindrical HV winding from rectangular wire.
Choose an aluminum conductor trademark АПБ d2 = 3.35 mm and d`2 =3.85 mm. Characteristic of a coil shall write down as:
АПБ×1×
,
I choose wire with cross section Swire2=8.81 mm2.
3.3.1 Complete cross section of the coil:
Swire21= nv2· Swire2·10-6 = 1·8.81·10-6 = 8.81*10-6 m.
3.3.2 The current density:
Δ2
=
= 1.971 MА/mm2,
it is in an allowable limit 1.2-2.5MА/m2 (tabl. 5.7, р. 257 [1]).
3.3.3 Calculation the number of coils in the layer:
Wsl2
=
= 128.87 coils;
where L2 = L1 = 0.5 m;
assume the number of coils in the layer 129.
3.3.4 The number of layers in the winding into accounts the number of coils for regulation of voltage:
nsl2
= 1.05·
= 4.762
assume nsl2 =5 layers.
3.3.5 The number of coils of regulation steps:
WP
=
= 14.625 turns,
assume 15 turns.
Location of regulation coils and circuit of switch is shown on fig. 3.5.
3.3.6 The working voltage of two layers:
UМ.Sl
= 2·Wsl2
·Vturn
= 2·129·5.922 = 1.528 V,
at this voltage under the tabl. 4.7, р.190 [1] choose the number of layers cable’s paper, projection of insulation 16 mm from each winding’s end. Paper-bakelite cylindrical rings by thickness 4 mm is placed to end faces of each layer above and below.
Common thickness of insulation δМSL = 4×0.12 = 0.48 mm.
3.3.7 The radial size of winding:
a2 = [d2`·nSL2 + δМSL ·(nSL2 – 1) + а´22 nк]·10-3
where а´22 = 0 winding without channel;
a2= [3.85·5 + 0.48·(5 – 1) + 0]·10-3 = 0.02117 m.
Let assume that a12 =0.027 m, because of the power losses are need to decrease.
Figure 3.4 - HV winding of transformer
а) Arrangement of regulation branch in outside layer of winding;
b) Circuit of switch branchings.
3.3.8 inside the diameter of winding:
D´2 = D´´1 + 2·a12 = 0.187 + 2·27·10-3= 0.241 m.
3.3.9 Outside diameter of winding:
D´´2 = D´2 + 2· a2 = 0.241 + 2·0.021 = 0.283 m.
The insulation distance between external windings of the next rods:
а22 = а´´22·10-3 = 10·10-3 = 0.01 m,
Where а´´22 = 10 mm (tabl. 4.5, р. 184 [1]).
3.3.10 A surface of cooling of the HV winding:
S02 = c·n·k·π·(D´2 + D´´2)·L2 = 3·1·1·3.14·(0.241 +0.283)·0.5=2.471m2,
where c = 3 – the number of rods of magnetic system;
п=1 and k=1 – under the tabl. 3.1, р.17 [2].
3.3.11 Approximate density of heat flow of the HV winding:
q2 = 0.55·PSCc·103/S02 = 0.55·2.65·103/2.471 = 589.849 Wt/m2 <1400 Wt/m2