Force
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1.1.1 ABILITY TO WITHSTAND SHORT-CIRCUIT
The ability to withstand short circuit of power transformer is done by First: By calculating the asymmetric short circuit current Second: Determining the axial forces and radial force Third: Calculating the stresses on various assembled part Fourth: Compiling the design value for input variables Fifth: Determining the limits of stress Sixth: Comparison of calculated value and permitted value
REFERENCES: The calculation of axial forces is done by residual ampere turn method (m.a. waters) And the following references are included 1. “The short circuit strength of power transformer” M. WATERS. 2. Sollergren,B .”Calculation of short circuit in transformers.” Electra, Report no. 67, 1979, pp.29-75. 3. Mcnutt, W.J. “shortcircuit characteristics of transformers.” IEEE SUMMER MEETING, Portland, July 1976, Paper No. ch1 159-3/76, pp. 30-37. 4. “Transformer engineering” Design and practice by: S.V. Kul Karni & S.A. Khaparde.
ENERGYPAC ENGINEERING LTD.
1.1.1.1 CALCULATION HYPOTHESIS
These stresses calculations are based on following hypothesis:
1. Square cross section with homogeneous ampere-turns distribution 2. Windings are concentric & regular radial gaps are maintained. 3. the final axial length(shrunk) of hv and lv winding is fully matched 1.1.1.2 PERMISSIBLE LIMITS Following parametric limits are permissible as per IEEMA and the other limits ( critical resistance to clloapse is calculated ): a) Hoop Stress kg/cm2
=1250
b) Compessive Pressure in radial spacers kg/cm2 c) Bending Stress on Clamping Ring (Permawood Ring) MT/cm2
1.1.2 PEAK ASYMETRICAL SHORT CIRCUIT CURRENT:
ENERGYPAC ENGINEERING LTD.
=500 =1.1
1.1.2.1 MEANING OF ABREVIATIONS: Iph = RATED PHASE CURRENT ZT = PER UNIT IMPEDANCE OF TRANSFORMER ZS = PER UNIT IMPEDANCE OF TRANSFORMER K1 = K x √2 (IEC 60076-5 1976 AND IT’S AMMENDMENT) EZ = TOTAL PER UNIT IMPEDANCE %X = PERCENT RECTANCE OF TRANSFORMER %R= PERCENT RESISTANCE OF TRANSFORMER
1.1.2.2 CALCULATIONS:
ZT
=
12.216
ZS(HV) = RATED MVAFAULT MVA = 5010000= 0.005 ; EZ(HV) = 0.12216 + 0.005 = 0.12716 ZS(LV) = RATED MVAFAULT MVA = 501000= 0.05 0.17216
%X%R
=
12.210.4044
=
; EZ(LV) = 0.12216 + 0.05 =
30.19 ; K1 = K x √2 = 1.8 x √2 = 2.55
ISC(HV) = IphHV× K1 EZ(HV) = 157.83× 2.55 0.12716 = 3165 ISC(LV) =IphLV× K1 EZ(LV) = 874.44× 2.55 0.17216 = 12957
1.1.3 AXIAL IMBALANCE FORCE (as per IEEMP (M.A Waters)): 1.1.3.1 MEANING OF ABREVIATIONS: Dm = MEAN DIA OF WINDING
ENERGYPAC ENGINEERING LTD.
Isc = SHORT CIRCUIT CURRENT N= TURN OF WINDING H = HEIGHT OF WINDING Ka = CONSTANT (SEE ANEX-A)
1.1.3.2 CALCULATIONS:
Fa(hv) = 4π2 ×Ka×N(hv)×Isc(hv)2×DmH×10-7 Newton =4π2 ×0.1387×1334×31652×9122115×10-7 =4.2×106 Newton
Fa(lv) = 4π2 ×Ka×N(lv)×Isc(lv)2×DmH×10-7 Newton =4π2 ×0.1387×241×129572×6762115×10-7 =1.7×106 Newton 1.1.3.3 CONCLUSION:
The result axial imbalance force for winding hv and lv are listed below: winding
CALCULATED VALUE
Fa(hv)
4.2×106 Newton
Fa(lv)
1.7×106 Newton
1.1.4 AXIAL COMPRESSIVE FORCE (as per IEEMP (M.A Waters)): 1.1.4.1 MEANING OF ABREVIATIONS: KVA = RATING OF THE JOB %Z = SHORT CIRCUIT IMPEDANCE HW= HEIGHT OF WINDING
1.1.4.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
Fc
=
34×KVAHw×%Z×9964×10-3 Newton
=34×50,000211.5×.1221×9964×10-3 Newton =0.655×106 Newton
Fc(HV) = 13×Fc= 0.218×106 Newton Fc(LV) = 23×Fc= 0.437×106 Newton 1.1.4.3 CONCLUSION:
The result axial compressive force for winding hv and lv are listed below: winding
CALCULATED VALUE
Fa(hv)
0.218×106 Newton
Fa(lv)
0.437×106 Newton
1.1.5 AXIAL COMPRESSIVE STRESS : 1.1.5.1 MEANING OF ABREVIATIONS: FC= AXIAL COMPRESSIVE FORCE OF WINDING
n = NO OF CONDUCTOR RADIALLY t= THICKNESS OF CONDUCTOR w= WIDTH OF KEYBLOCK s = NO OF BLOCK/CIRCLE
1.1.5.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
Qc
=
Fcn×t×w×sNewton /mm2
Qc(HV) = 0.218×1062×2.55×40×24 N /mm2 =44.52 N/mm2 Qc(LV) =0.437×10612×2.7×40×16 N /mm2 =21 N/mm2 1.1.5.3 CONCLUSION:
The result required no of vertical support for winding hv and lv are listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
Qc(HV)
44.52 N/mm2
20 N/mm2
Qc(LV)
21 N/mm2
20 N/mm2
comment
1.1.6 BENDING STRESS ON CLAMPING RING: 1.1.6.1 MEANING OF ABREVIATIONS: BS= BENDING STRESS FC= TOTAL AXIAL FORCE OF WINDING (IN TON)
n = NO OF JACKING POINTS t= THICKNESS OF PERMAWOOD RING b= WIDTH OF PERMAWOOD RING Dout = O/D OF PERMAWOOD RING Din= I/D OF PERMAWOOD RING
1.1.6.2 CALCULATIONS:
FC = Fa-13 Fc = (4.2×106-0.218×106) N = 3.982×106 N=399.63 TON
ENERGYPAC ENGINEERING LTD.
b =(Dout-Din)2cm= (115-60.0)2cm =27.5 BS(MAX)
cm
6×π×Fc×D8×t2×n2×b
=
MT/cm2
=
6×π×399.63×10-
3×1158×62×102×27.5 MT/cm2=0.00109 MT/cm2 1.1.6.3 CONCLUSION:
The result of bending stress on clamping ring listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
BS(MAX
0.00109 MT/cm2
1.1 MT/cm2
comment
1.1.7 COMPRESSIVE STRESS ON WINDINGS: 1.1.7.1 MEANING OF ABREVIATIONS: PI= COMPRESSIVE STRESS ON INNER WINDING POUT= COMPRESSIVE STRESS ONOUTER WINDING FC= AXIAL COMPRESSIVE FORCE OF WINDING( HV & LV) Fa= AXIAL IMBALANCE FORCE OF WINDING FI = MAX COMPRESSIVE FORCE ON INNER SPACERS FOUT = MAX COMPRESSIVE FORCE ON OUTER SPACERS
Asin= TOTAL SUPPORTED AREA OF INNERWINDING SPACERS Asout = TOTAL SUPPORTED AREA OF OUTER WINDING SPACERS w= WIDTH OF KEYBLOCK b= LENGTH OF KEYBLOCK n = NO OF BLOCK/CIRCLE
1.1.7.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
FI = Fa+13 Fc = (4.2×106+0.218×106) N = 4.418×106 N = 443.4 ×103 kg FOUT = Fa+23 Fc = (4.2×106+0.437×106) N = 465.37 ×103 kg Asin = w×b×n= 4×9.6×16 cm2= 614.4 cm2 Asout = w×b×n= 4×8×24 cm2= 768 cm2
continued………………………….
PI =FiAin= 443.4 ×103 614.4=721.67 kgcm2 Pout=FoutAout= 465.37 ×103 768=605.95kgcm2 1.1.7.3 CONCLUSION:
The result of compressive stress on windings for winding hv and lv are listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
PI(lv)
721.67 kg/cm2
500 kg/cm2
Pout(hv)
605.95 kg/cm2
500 kg/cm2
comment
ENERGYPAC ENGINEERING LTD.
1.1.8 CRITICAL RESISTANCE TO COLLAPSE: 1.1.8.1 MEANING OF ABREVIATIONS: E= MODULUS OF ELASTICITY OF COPPER= m= TOTAL NO OF PARALLEL CONDUCTORS Dm= MEAN DIA OF WINDING( HV & LV) Iph/cond= PHASE CURRENT PER CONDUCTOR OF THE WINDING t = THICKNESS OF THE CONDUCTOR FOUT = MAX COMPRESSIVE FORCE ON OUTER SPACERS
As= TOTAL SUPPORTED AREA OF SPACERS J = CURRENT DENSITY WCR= CRITICAL RESISTANCE TO COLLAPSE
1.1.8.2 CALCULATIONS:
WCR =1.05×E×m×Iph/cond2t×Dm×J2+4500×As×t3×JIph/cond WCR(LV) =1.05×1.13×106×12×72.920.27×67.6×249.62+4500×614.4×0.273×249.672.9kg cm2 =66.543×103+ 186.325×103 kgcm2 = 252.868×103kgcm2
ENERGYPAC ENGINEERING LTD.
WCR(HV =1.05×1.13×106×2×78.920.22×91.2×3072+4500×768×0.223×30778.9kgcm2 =7.818 ×103+143×103kgcm2 =151×103kgcm2 the total compressive force of the inner and outer spacers must be less than these critical resistance to collapse 1.1.8.3 CONCLUSION:
The result of compressive stress on windings for winding hv and lv are listed below: winding
CALCULATED VALUE
PI(lv)
721.67 kg/cm2
252.868×103 kg/cm2
Pout(hv)
605.95 kg/cm2
151×103 kg/cm2
wcr
comment
1.1.9 HOOP STRESS: 1.1.9.1 MEANING OF ABREVIATIONS:
ENERGYPAC ENGINEERING LTD.
hoop = HOOP STREES Rdc= DC RESISTANCE OF WINDING IN CM HW= WINDING HEIGHT K(CU) = 0.03 × K x √22.55
2
= 0.03
1.1.9.2 CALCULATIONS:
hoop(HV)
=K(CU)
0.127162×211.5
hoop(LV)
= 1029.22
= K(CU)
0.172162×211.5
×Rdc_hv×Iph_hv2EZ(hv)
2×Hw
0.03×4.71×157.832
Kg/cm2
×Rdc_lv×Iph_lv2EZ(lv) 2×Hw
= 339.58
=
=
0.03
×0.0928×874.442
Kg/cm2
1.1.9.2 CONCLUSION:
The result of hoop stress for hv and lv are listed below:
winding
Permissible value
HV
Calculated value 1029.22 Kg/cm2
LV
339.58 Kg/cm2
1250 Kg/cm2
comment
1250 Kg/cm2
1.1.10 REQUIRED NO OF SUPPORT FOR RADIAL FORCE: 1.1.10.1 MEANING OF ABREVIATIONS: Dm = MEAN DIA OF WINDING
ENERGYPAC ENGINEERING LTD.
t = THICKNESS OF CONDUCTOR OF THE RELATED WINDING EO = MODULUS OF ELASTICITY OF COPPER =1.136 ×106
Ns= REQUIRED NO OF VERTICAL SUPPORT FOR WINDING 1.1.10.1 CALCULATIONS:
Ns(hv) =Dm t×12Xhoop(HV) EO =1100.29×12X1029.22 1.136 ×106 = 39.55 ≅40
Ns(lv) =Dm t×12Xhoop(lV) EO =67.600.27×12X339.58 1.136 ×106 = 14.99 ≅15
1.1.10.2 CONCLUSION:
The result required no of vertical support for winding hv and lv are listed below:
winding HV
Required supports Available supports 40
LV
15
comment
ANNEXTURE-A: CALCULATION OF CONSTANT Ka: 15 mm.
HV TAP
ENERGYPAC ENGINEERING LTD.
746
LV
HV M
2115
1490
=H
T = 2x746+623
HV TAP
746
d1 =118 d2 =212
Kt=ATS OF TAPATS OF MAIN+ATS OF TAP=5041334+504=0.274 X1 = 15mm
; d1x1=11815=7.86
X2=H-T4+X1= 2115-14924+15= 170.75 mm
; d2x2=212170.75=1.241
Ka= 1-Kt1+d1x12 + Kt21+d2x22 = 0.09162+0.0471=0.1387 (FOR LV & HV)
ENERGYPAC ENGINEERING LTD.
The ability to withstand short circuit of power transformer is done by First: By calculating the asymmetric short circuit current Second: Determining the axial forces and radial force Third: Calculating the stresses on various assembled part Fourth: Compiling the design value for input variables Fifth: Determining the limits of stress Sixth: Comparison of calculated value and permitted value
REFERENCES: The calculation of axial forces is done by residual ampere turn method (m.a. waters) And the following references are included 1. “The short circuit strength of power transformer” M. WATERS. 2. Sollergren,B .”Calculation of short circuit in transformers.” Electra, Report no. 67, 1979, pp.29-75. 3. Mcnutt, W.J. “shortcircuit characteristics of transformers.” IEEE SUMMER MEETING, Portland, July 1976, Paper No. ch1 159-3/76, pp. 30-37. 4. “Transformer engineering” Design and practice by: S.V. Kul Karni & S.A. Khaparde.
ENERGYPAC ENGINEERING LTD.
1.1.1.1 CALCULATION HYPOTHESIS
These stresses calculations are based on following hypothesis:
1. Square cross section with homogeneous ampere-turns distribution 2. Windings are concentric & regular radial gaps are maintained. 3. the final axial length(shrunk) of hv and lv winding is fully matched 1.1.1.2 PERMISSIBLE LIMITS Following parametric limits are permissible as per IEEMA and the other limits ( critical resistance to clloapse is calculated ): a) Hoop Stress kg/cm2
=1250
b) Compessive Pressure in radial spacers kg/cm2 c) Bending Stress on Clamping Ring (Permawood Ring) MT/cm2
1.1.2 PEAK ASYMETRICAL SHORT CIRCUIT CURRENT:
ENERGYPAC ENGINEERING LTD.
=500 =1.1
1.1.2.1 MEANING OF ABREVIATIONS: Iph = RATED PHASE CURRENT ZT = PER UNIT IMPEDANCE OF TRANSFORMER ZS = PER UNIT IMPEDANCE OF TRANSFORMER K1 = K x √2 (IEC 60076-5 1976 AND IT’S AMMENDMENT) EZ = TOTAL PER UNIT IMPEDANCE %X = PERCENT RECTANCE OF TRANSFORMER %R= PERCENT RESISTANCE OF TRANSFORMER
1.1.2.2 CALCULATIONS:
ZT
=
12.216
ZS(HV) = RATED MVAFAULT MVA = 5010000= 0.005 ; EZ(HV) = 0.12216 + 0.005 = 0.12716 ZS(LV) = RATED MVAFAULT MVA = 501000= 0.05 0.17216
%X%R
=
12.210.4044
=
; EZ(LV) = 0.12216 + 0.05 =
30.19 ; K1 = K x √2 = 1.8 x √2 = 2.55
ISC(HV) = IphHV× K1 EZ(HV) = 157.83× 2.55 0.12716 = 3165 ISC(LV) =IphLV× K1 EZ(LV) = 874.44× 2.55 0.17216 = 12957
1.1.3 AXIAL IMBALANCE FORCE (as per IEEMP (M.A Waters)): 1.1.3.1 MEANING OF ABREVIATIONS: Dm = MEAN DIA OF WINDING
ENERGYPAC ENGINEERING LTD.
Isc = SHORT CIRCUIT CURRENT N= TURN OF WINDING H = HEIGHT OF WINDING Ka = CONSTANT (SEE ANEX-A)
1.1.3.2 CALCULATIONS:
Fa(hv) = 4π2 ×Ka×N(hv)×Isc(hv)2×DmH×10-7 Newton =4π2 ×0.1387×1334×31652×9122115×10-7 =4.2×106 Newton
Fa(lv) = 4π2 ×Ka×N(lv)×Isc(lv)2×DmH×10-7 Newton =4π2 ×0.1387×241×129572×6762115×10-7 =1.7×106 Newton 1.1.3.3 CONCLUSION:
The result axial imbalance force for winding hv and lv are listed below: winding
CALCULATED VALUE
Fa(hv)
4.2×106 Newton
Fa(lv)
1.7×106 Newton
1.1.4 AXIAL COMPRESSIVE FORCE (as per IEEMP (M.A Waters)): 1.1.4.1 MEANING OF ABREVIATIONS: KVA = RATING OF THE JOB %Z = SHORT CIRCUIT IMPEDANCE HW= HEIGHT OF WINDING
1.1.4.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
Fc
=
34×KVAHw×%Z×9964×10-3 Newton
=34×50,000211.5×.1221×9964×10-3 Newton =0.655×106 Newton
Fc(HV) = 13×Fc= 0.218×106 Newton Fc(LV) = 23×Fc= 0.437×106 Newton 1.1.4.3 CONCLUSION:
The result axial compressive force for winding hv and lv are listed below: winding
CALCULATED VALUE
Fa(hv)
0.218×106 Newton
Fa(lv)
0.437×106 Newton
1.1.5 AXIAL COMPRESSIVE STRESS : 1.1.5.1 MEANING OF ABREVIATIONS: FC= AXIAL COMPRESSIVE FORCE OF WINDING
n = NO OF CONDUCTOR RADIALLY t= THICKNESS OF CONDUCTOR w= WIDTH OF KEYBLOCK s = NO OF BLOCK/CIRCLE
1.1.5.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
Qc
=
Fcn×t×w×sNewton /mm2
Qc(HV) = 0.218×1062×2.55×40×24 N /mm2 =44.52 N/mm2 Qc(LV) =0.437×10612×2.7×40×16 N /mm2 =21 N/mm2 1.1.5.3 CONCLUSION:
The result required no of vertical support for winding hv and lv are listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
Qc(HV)
44.52 N/mm2
20 N/mm2
Qc(LV)
21 N/mm2
20 N/mm2
comment
1.1.6 BENDING STRESS ON CLAMPING RING: 1.1.6.1 MEANING OF ABREVIATIONS: BS= BENDING STRESS FC= TOTAL AXIAL FORCE OF WINDING (IN TON)
n = NO OF JACKING POINTS t= THICKNESS OF PERMAWOOD RING b= WIDTH OF PERMAWOOD RING Dout = O/D OF PERMAWOOD RING Din= I/D OF PERMAWOOD RING
1.1.6.2 CALCULATIONS:
FC = Fa-13 Fc = (4.2×106-0.218×106) N = 3.982×106 N=399.63 TON
ENERGYPAC ENGINEERING LTD.
b =(Dout-Din)2cm= (115-60.0)2cm =27.5 BS(MAX)
cm
6×π×Fc×D8×t2×n2×b
=
MT/cm2
=
6×π×399.63×10-
3×1158×62×102×27.5 MT/cm2=0.00109 MT/cm2 1.1.6.3 CONCLUSION:
The result of bending stress on clamping ring listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
BS(MAX
0.00109 MT/cm2
1.1 MT/cm2
comment
1.1.7 COMPRESSIVE STRESS ON WINDINGS: 1.1.7.1 MEANING OF ABREVIATIONS: PI= COMPRESSIVE STRESS ON INNER WINDING POUT= COMPRESSIVE STRESS ONOUTER WINDING FC= AXIAL COMPRESSIVE FORCE OF WINDING( HV & LV) Fa= AXIAL IMBALANCE FORCE OF WINDING FI = MAX COMPRESSIVE FORCE ON INNER SPACERS FOUT = MAX COMPRESSIVE FORCE ON OUTER SPACERS
Asin= TOTAL SUPPORTED AREA OF INNERWINDING SPACERS Asout = TOTAL SUPPORTED AREA OF OUTER WINDING SPACERS w= WIDTH OF KEYBLOCK b= LENGTH OF KEYBLOCK n = NO OF BLOCK/CIRCLE
1.1.7.2
ENERGYPAC ENGINEERING LTD.
CALCULATIONS:
FI = Fa+13 Fc = (4.2×106+0.218×106) N = 4.418×106 N = 443.4 ×103 kg FOUT = Fa+23 Fc = (4.2×106+0.437×106) N = 465.37 ×103 kg Asin = w×b×n= 4×9.6×16 cm2= 614.4 cm2 Asout = w×b×n= 4×8×24 cm2= 768 cm2
continued………………………….
PI =FiAin= 443.4 ×103 614.4=721.67 kgcm2 Pout=FoutAout= 465.37 ×103 768=605.95kgcm2 1.1.7.3 CONCLUSION:
The result of compressive stress on windings for winding hv and lv are listed below: winding
CALCULATED VALUE
PERMITTED LIMIT
PI(lv)
721.67 kg/cm2
500 kg/cm2
Pout(hv)
605.95 kg/cm2
500 kg/cm2
comment
ENERGYPAC ENGINEERING LTD.
1.1.8 CRITICAL RESISTANCE TO COLLAPSE: 1.1.8.1 MEANING OF ABREVIATIONS: E= MODULUS OF ELASTICITY OF COPPER= m= TOTAL NO OF PARALLEL CONDUCTORS Dm= MEAN DIA OF WINDING( HV & LV) Iph/cond= PHASE CURRENT PER CONDUCTOR OF THE WINDING t = THICKNESS OF THE CONDUCTOR FOUT = MAX COMPRESSIVE FORCE ON OUTER SPACERS
As= TOTAL SUPPORTED AREA OF SPACERS J = CURRENT DENSITY WCR= CRITICAL RESISTANCE TO COLLAPSE
1.1.8.2 CALCULATIONS:
WCR =1.05×E×m×Iph/cond2t×Dm×J2+4500×As×t3×JIph/cond WCR(LV) =1.05×1.13×106×12×72.920.27×67.6×249.62+4500×614.4×0.273×249.672.9kg cm2 =66.543×103+ 186.325×103 kgcm2 = 252.868×103kgcm2
ENERGYPAC ENGINEERING LTD.
WCR(HV =1.05×1.13×106×2×78.920.22×91.2×3072+4500×768×0.223×30778.9kgcm2 =7.818 ×103+143×103kgcm2 =151×103kgcm2 the total compressive force of the inner and outer spacers must be less than these critical resistance to collapse 1.1.8.3 CONCLUSION:
The result of compressive stress on windings for winding hv and lv are listed below: winding
CALCULATED VALUE
PI(lv)
721.67 kg/cm2
252.868×103 kg/cm2
Pout(hv)
605.95 kg/cm2
151×103 kg/cm2
wcr
comment
1.1.9 HOOP STRESS: 1.1.9.1 MEANING OF ABREVIATIONS:
ENERGYPAC ENGINEERING LTD.
hoop = HOOP STREES Rdc= DC RESISTANCE OF WINDING IN CM HW= WINDING HEIGHT K(CU) = 0.03 × K x √22.55
2
= 0.03
1.1.9.2 CALCULATIONS:
hoop(HV)
=K(CU)
0.127162×211.5
hoop(LV)
= 1029.22
= K(CU)
0.172162×211.5
×Rdc_hv×Iph_hv2EZ(hv)
2×Hw
0.03×4.71×157.832
Kg/cm2
×Rdc_lv×Iph_lv2EZ(lv) 2×Hw
= 339.58
=
=
0.03
×0.0928×874.442
Kg/cm2
1.1.9.2 CONCLUSION:
The result of hoop stress for hv and lv are listed below:
winding
Permissible value
HV
Calculated value 1029.22 Kg/cm2
LV
339.58 Kg/cm2
1250 Kg/cm2
comment
1250 Kg/cm2
1.1.10 REQUIRED NO OF SUPPORT FOR RADIAL FORCE: 1.1.10.1 MEANING OF ABREVIATIONS: Dm = MEAN DIA OF WINDING
ENERGYPAC ENGINEERING LTD.
t = THICKNESS OF CONDUCTOR OF THE RELATED WINDING EO = MODULUS OF ELASTICITY OF COPPER =1.136 ×106
Ns= REQUIRED NO OF VERTICAL SUPPORT FOR WINDING 1.1.10.1 CALCULATIONS:
Ns(hv) =Dm t×12Xhoop(HV) EO =1100.29×12X1029.22 1.136 ×106 = 39.55 ≅40
Ns(lv) =Dm t×12Xhoop(lV) EO =67.600.27×12X339.58 1.136 ×106 = 14.99 ≅15
1.1.10.2 CONCLUSION:
The result required no of vertical support for winding hv and lv are listed below:
winding HV
Required supports Available supports 40
LV
15
comment
ANNEXTURE-A: CALCULATION OF CONSTANT Ka: 15 mm.
HV TAP
ENERGYPAC ENGINEERING LTD.
746
LV
HV M
2115
1490
=H
T = 2x746+623
HV TAP
746
d1 =118 d2 =212
Kt=ATS OF TAPATS OF MAIN+ATS OF TAP=5041334+504=0.274 X1 = 15mm
; d1x1=11815=7.86
X2=H-T4+X1= 2115-14924+15= 170.75 mm
; d2x2=212170.75=1.241
Ka= 1-Kt1+d1x12 + Kt21+d2x22 = 0.09162+0.0471=0.1387 (FOR LV & HV)
ENERGYPAC ENGINEERING LTD.
