Figure 6.2 – Pipes radiator with direct pipes
Нbmin = АР + С1 + С2 = 0.71 + 0.085 + 0.10 = 0.895 m,
where С1 = 0,085 m and С2 = 0,10 m – distance from axes of flanges up to bottom and top border of section of a tank (tabl.9.9, р. 442 [1]).
6.2.6 Surface of convection of smooth wall of the tank:
SКgl = Нbmin·[2·(А – В) + π·В] = 0.895·[2·(1.004 – 0.414) + π·0.414] = 2.221m2.
6.2.7 Approximate surface of emanation of a tank with radiators:
Si = К·SКgl = 1.5· 2,221 = 3.331m2,
where К = 1.35–1.5 – factor taking into account equipment of a tank (for tank with pipes);
6.2.8 Approximate necessary surface of convection for the given meaning Θbv:
S´conversion
=
S´conversion
=
= 9.9119 m2
Surface of convection sum from:
surface of smooth wall of the tank SКgl;
surface of tank’s cover,
SК.КР
= 0.5·[(A
– B)·(B
+ 0.16) +
]
SК.КР
= 0.5·
= 0.299 m2.
6.2.9 Surface of convection of radiators:
ΣSК.Р = S´conversion – SКgl – SК.КР = 9.9119 – 2.221– 0.299 = 6.599 m2.
6.2.10 Surface of convection of radiators, reduced to the surface of smooth wall of the tank:
SК.Р = SТР·Кf + SККp = 2.135·1.344 + 0.34= 3.209 m2,
where Кf = 1.4*0.96 – by tabl. 9.6, р. 432 [1].
6.2.11 Necessary number of radiators:
nR
=
= 2.056
I accept nR =3 radiators. Arrangement of radiators is on fig.6.3.
6.2.12 Surface of convection of tank:
SК = ΣSКР + SКgl + SК.КР = 9.628 + 3.209 + 0.299 = 12.148m2,
where ΣПКР = nR·ПКР – total surface of convection of radiators.
6.2.13 Surface of emanation is determined after performance of the sketch of accommodation of radiators on walls of a tank (fig.6.3):
П´I = РI·НB,
П´I = =4.089·0.895=3.66m2.
Figure 6.3 – Sketch of accommodation of radiators on walls of a tank
6.3 Calculation of excess of temperature of windings and oil of the transformer.
6.3.1 Average excess of temperature of tank’s walls in comparison with temperature of air:
Θbvcp
=
=
=
40.397 оС,
where К = 1.05÷1.10, assume К = 1.05.
6.3.2. Average excess of temperature of oil near a wall of the tank:
=nr·(Sтр·Кf+Sкk)=3·(2.135·1+0.34)
= 7.425 m2
,
where Кf=1.
ΘМbcp
= 0.165·К1·
;
ΘМbcp
= 0.165·1·
=
5 оС,
where К1 = 1.0 – for natural cooling.
6.3.3 Excess of average temperature of oil in comparison with temperature of air:
ΘМvcp = ΘМbcp + Θbvcp = 5 + 33.831 = 38.831 оС.
6.3.4 Excess of temperature of oil in top layers in comparison with temperature of air:
ΘМvcp = ΘМbcp + ΘδvcpВ = 5 + 33.831= 38.831 < 60 оC.
.
6.3.5 Excess of temperature of the windings in comparison with temperature of air:
LV winding:
ΘОv1 = ΘОМСР1 + Θbvcp + ΘМbcp = 19.112 + 33.381 + 5
= 57.943 оС < 65 оС;
HV winding:
ΘОv2 = ΘОМСР2 + Θbvcp + ΘМbcp = 21.308 + 33.831 + 5 =
= 60.138 оС < 65 оС.
7 Definition of weight of constructional materials and oil of the transformer
7.1 Weight of an active part of the transformer:
Gap = 1.2·(Gwire + Gsteel) = 1.2·(57.549+431.755) = 587.165 kg
where, Gwire=GM1+GM1*0.05+GM2+GM2*0.05+Swire1+Swire2=23.075+23.075*0.05+ 32.503+32.503*0.05+1.34*10-4+8.81*10-6 = 57.549 kg – weight of winding’s wire with lead-in;
Gsteel = 431.755 kg –weight of steel of magnetic circuit – by p. 5.1.16.
7.2 Weight of tank of the transformer:
Gt = γst·St·δb = 7650·2.976·5·10-3 = 113.851kg
where St = [2·(А – В) + π·В]·Нb + [2·В· (2·С + 0.25·π·В) = 2.976 m2 – general surface of tank, cover and bottom of the transformer.
Thickness of wall of the tank is definition by the tabl.7.1, р.39[2] δ=5 mm.
7.3 Weight of pipes of the cooling system (for the tank with radiators):
Gstr = nr·G1stp = 3·34.14 = 102.42 kg,
where Gstp=34.14 kg – weight of steel of the radiators (by tabl. 9.9, р.442[1]).
7.4 Volume of active part of the transformer:
Vap
=
= 0.112 m3,
where γap – average density of active part of the transformer (5 – 5.5)·103 kg/m3 (by р.39[2]).
7.5 Volume of tank of the transformer:
Vt = (2·С + 0.25·π·В) ·В·Нb = (2·0.293 + 0.25·π·0.414) 0.414·0.895 = 0.338 m3.
7.6 Weight of oil in the tank:
Got = 0.9· (Vt – Vap) ·103 = 0.9· (0.338 – 0.112) ·103 = 203.775 kg.
7.7 Weight of oil in pipes (for the tank with radiators):
Gor = nr·Golr = 3·24 = 72 kg,
where Golr = 24 кg – weight of oil in one radiator (by tabl. 9.9, р. 442 [1]).
7.8 General weight of oil in the transformer:
Goiltrans = 1.05·(Got + Gor) = 1.05·(203.775 + 72) = 289.564 kg.
7.9 Weight of insulating cardboard:
Gcard = Kcard·Gwire = 0.25·57.549= 14.387 kg,
where Kcard = 0.25 – use factor of cardboard (by tabl. 7.2, р. 40 [2]).
7.10 General weight of transformer:
Gtotal=1.2·(Gap+Gt+Goiltrans+Gcard+Gstr)
Gtr=1.2·(587.165+113.851 +289.564+14.387+102.42)=1328.866kg.