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Temperature rise and tubes for 500 kVA transformer

Short answer

Plain tank rise 79.7 °C. Required tube area 9.698 m². With 50 mm diameter, 1.5 m long tubes, 42 needed. Arrangement: 14 tubes on each long side, 7 on each short side in a single row.

For a 500 kVA, 50 Hz, 6600/400, single phase core type, oil immersed, natural cooled power transformer, the design parameters are: Constant for output voltage per turn (K) = 0.8; Resistivity of copper = 0.021 Ω-mm²/m; Maximum flux density in core (Bm) = 1.5 Wb/m²; Current density = 2.75 A/mm²; Core type = Cruciform; Window space factor (Kw) = 0.27; Stacking factor (Ki) = 0.9; Ratio of height to width of windows (Hw/Ww) = 2.5; Ratio of yoke height to width (Hy/Dy) = 1; Axial depth of LV winding = 402 mm; Axial depth of HV winding = 377.5 mm; Inside diameter of LV winding = 310 mm; Outside diameter of LV winding = 348 mm; Inside diameter of HV winding = 360 mm; Outside diameter of HV winding = 418 mm; Taking iron loss = 1460 W, Copper loss = 3865 W at full load, Height of tank = 1.6 m, Length of tank = 1.05 m, Width of tank = 0.62 m, find the temperature rise. If the mean rise of oil is not to rise 35°C, find the necessary number of tubes and show its arrangement.

Only the losses and tank size are needed for this part; the core and winding data are used in the full design of the same transformer.

Basis

  • A plain tank wall loses heat by radiation (≈ 6 W/m²°C) and natural convection (≈ 6.5 W/m²°C): total 12.5 W/m²°C.
  • Tubes screen each other, so they add almost nothing by radiation, but they improve convection by about 35%: 6.5×1.35≈8.86.5\times1.35 \approx 8.8 W/m²°C.
  • Top and bottom of the tank are neglected.

Temperature rise with a plain tank

Total loss P=Pi+Pc=1460+3865=5325P = P_i + P_c = 1460 + 3865 = 5325 W.

St=2(Lt+Wt)Ht=2(1.05+0.62)×1.6=5.344 m2θ=P12.5St=532512.5×5.344=79.7 ∘C\begin{aligned} S_t &= 2(L_t + W_t)H_t = 2(1.05 + 0.62)\times1.6 = 5.344\ \text{m}^2\\ \theta &= \frac{P}{12.5S_t} = \frac{5325}{12.5\times5.344} = 79.7\ ^\circ\text{C} \end{aligned}

This is far above the permitted mean oil rise of 35 °C, so cooling tubes are needed.

Number of tubes

Let the total tube area be AtA_t:

P=θ (12.5St+8.8At)At=18.8(532535−12.5×5.344)=18.8(152.14−66.80)=9.698 m2\begin{aligned} P &= \theta\,(12.5S_t + 8.8A_t)\\ A_t &= \frac{1}{8.8}\left(\frac{5325}{35} - 12.5\times5.344\right) = \frac{1}{8.8}(152.14 - 66.80) = 9.698\ \text{m}^2 \end{aligned}

Assume tubes of 50 mm diameter and 1.5 m mean length (tank height 1.6 m):

Area of one tube=π×0.05×1.5=0.2356 m2nt=9.6980.2356=41.16≈42 tubes\begin{aligned} \text{Area of one tube} &= \pi\times0.05\times1.5 = 0.2356\ \text{m}^2\\ n_t &= \frac{9.698}{0.2356} = 41.16 \approx 42\ \text{tubes} \end{aligned}

Check: 35(12.5×5.344+8.8×42×0.2356)=538635(12.5\times5.344 + 8.8\times42\times0.2356) = 5386 W ≥ 5325 W.

Arrangement of tubes: with 50 mm tubes at 75 mm centre-to-centre spacing, one row round the tank can hold about 1050/75=141050/75 = 14 tubes on each long side and 620/75≈8620/75 \approx 8 on each short side. So 42 tubes are placed in a single row: 14 on each long side and 7 on each short side (2×14+2×7=422\times14 + 2\times7 = 42).

        o o o o o o o o o o o o o o    (14)
      +-------------------------------+
   o  |                               |  o
   o  |                               |  o
   o  |        TANK (top view)        |  o
(7)o  |      1.05 m  x  0.62 m        |  o (7)
   o  |                               |  o
   o  |                               |  o
   o  |                               |  o
      +-------------------------------+
        o o o o o o o o o o o o o o    (14)
   o = cooling tube, 50 mm dia, 75 mm pitch

Answer: temperature rise with plain tank ≈ 79.7 °C; 42 tubes of 50 mm diameter, 1.5 m long, keep the mean oil rise within 35 °C.

✓ Answer checked by Claude Sonnet 5.5

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