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Design: 150 kVA single phase transformer

Short answer

Core diameter 233.9 mm, limb 198.9×124.0 mm; window 132.6×331.6 mm; yoke depth 198.9 mm, height 205.5 mm. Overall frame 742.6×565.4×198.9 mm. Leakage reactance calculation is not fully provided.

Design a 150 kVA, 50 Hz, 6600/400 V, 1-phase, core type oil immersed natural cooled distribution transformer. Given that: Maximum flux density in core = 1.35 Wb/m²; Current density = 2.75 A/mm²; Core type = cruciform two stepped; Window space factor = 0.27; Stacking factor = 0.9; Ratio of window height to width = 2.5; Take hot rolled steel sheet and area of yoke is 20% greater than area of core; Axial depth of LV winding = 268 mm; Axial depth of HV winding = 276 mm; Inside diameter of LV winding = 255 mm; Radial depth of LV winding = 14 mm; Radial depth of HV winding = 18 mm; Width of insulation between LV and HV = 11 mm; Outside diameter of LV winding = 293 mm; Inside diameter of HV winding = 314 mm; Outside diameter of HV winding = 351 mm. Calculate: i) Dimensions of the core, window and yoke ii) Overall dimensions of the frame iii) Leakage reactance of the transformer iv) Draw overall dimension of the transformer

Standard method (A.K. Sawhney, A Course in Electrical Machine Design): Et=KQE_t = K\sqrt{Q}, net core area ratio for the core type, output equation for the window, yoke depth = width of largest stamping. In a single-phase core type transformer half of each winding is on each limb, so the ampere-turns per limb are IHVTHV/2I_{HV}T_{HV}/2. KK is not given; KK = 0.75 is taken (single-phase core type). Winding data are given: LV 255/293 mm (inside/outside diameter), HV 314/351 mm, radial depths b1b_1 = 14 mm, b2b_2 = 18 mm, gap aa = 11 mm. Axial length taken as the mean of 268 and 276 mm, LcL_c = 272 mm.

i) Dimensions of core, window and yoke

Voltage per turn, Et=KQE_t = K\sqrt{Q} (QQ in kVA):

Et=0.75150=9.186 Vϕm=Et4.44f=9.1864.44×50=0.04138 WbAi=ϕmBm=0.041381.35=0.03065 m2=306.5 cm2\begin{aligned} E_t &= 0.75\sqrt{150} = 9.186\ \text{V}\\ \phi_m &= \frac{E_t}{4.44f} = \frac{9.186}{4.44\times50} = 0.04138\ \text{Wb}\\ A_i &= \frac{\phi_m}{B_m} = \frac{0.04138}{1.35} = 0.03065\ \text{m}^2 = 306.5\ \text{cm}^2 \end{aligned}

For a cruciform (two-stepped) core with stacking factor 0.9, Ai=0.56d2A_i = 0.56d^2 (standard ratio, Sawhney):

d=Ai0.56=0.030650.56=0.2339 m=233.9 mma=0.85d=198.9 mmb=0.53d=124.0 mm\begin{aligned} d &= \sqrt{\frac{A_i}{0.56}} = \sqrt{\frac{0.03065}{0.56}} = 0.2339\ \text{m} = 233.9\ \text{mm}\\ a &= 0.85d = 198.9\ \text{mm}\\ b &= 0.53d = 124.0\ \text{mm} \end{aligned}

Gross core area Agi=Ai/0.9=0.03405A_{gi} = A_i/0.9 = 0.03405 m².

Output equation (single phase): Q=2.22fBmδKwAwAi×10−3Q = 2.22fB_m\delta K_wA_wA_i\times10^{-3}

Aw=Q2.22fBmδKwAi×10−3=1502.22×50×1.35×2.75×106×0.27×0.03065×10−3=0.04399 m2Ww=Aw2.5=132.6 mm,Hw=2.5Ww=331.6 mmD=d+Ww=366.6 mm (distance between adjacent limb centres)\begin{aligned} A_w &= \frac{Q}{2.22fB_m\delta K_wA_i\times10^{-3}}\\ &= \frac{150}{2.22\times50\times1.35\times2.75\times10^6\times0.27\times0.03065\times10^{-3}}\\ &= 0.04399\ \text{m}^2\\ W_w &= \sqrt{\frac{A_w}{2.5}} = 132.6\ \text{mm},\quad H_w = 2.5W_w = 331.6\ \text{mm}\\ D &= d + W_w = 366.6\ \text{mm (distance between adjacent limb centres)} \end{aligned}

Check: dd = 234 mm < LV inside diameter 255 mm; HV outside diameter 351 mm < DD = 367 mm; coil height 276 mm < HwH_w = 332 mm. The given windings fit.

Yoke:

Yoke area 20% more than core area (hot-rolled steel, to reduce yoke loss and magnetising current). Yoke depth = width of largest stamping:

Ay=1.2Ai=0.03678 m2,Agy=Ay0.9=0.04087 m2Dy=a=198.9 mmHy=AgyDy=0.040870.1989=205.5 mm\begin{aligned} A_y &= 1.2A_i = 0.03678\ \text{m}^2,\quad A_{gy} = \frac{A_y}{0.9} = 0.04087\ \text{m}^2\\ D_y &= a = 198.9\ \text{mm}\\ H_y &= \frac{A_{gy}}{D_y} = \frac{0.04087}{0.1989} = 205.5\ \text{mm} \end{aligned}

ii) Overall dimensions of frame

H=Hw+2Hy=331.6+2(205.5)=742.6 mmW=D+a=366.6+198.9=565.4 mmDepth=Dy=a=198.9 mm\begin{aligned} H &= H_w + 2H_y = 331.6 + 2(205.5) = 742.6\ \text{mm}\\ W &= D + a = 366.6 + 198.9 = 565.4\ \text{mm}\\ \text{Depth} &= D_y = a = 198.9\ \text{mm} \end{aligned}

iii) Leakage reactance

Currents: ILV=QVLV=375.00I_{LV} = \frac{Q}{V_{LV}} = 375.00 A; IHV=22.727I_{HV} = 22.727 A.

TLV=VLVEt=400.09.186=43.55≈44Et (revised)=400.044=9.091 VTHV=TLVVHVVLV=726.0≈726aLV=ILVδ=136.36 mm2,aHV=IHVδHV=8.264 mm2\begin{aligned} T_{LV} &= \frac{V_{LV}}{E_t} = \frac{400.0}{9.186} = 43.55 \approx 44\\ E_t\ (\text{revised}) &= \frac{400.0}{44} = 9.091\ \text{V}\\ T_{HV} &= T_{LV}\frac{V_{HV}}{V_{LV}} = 726.0 \approx 726\\ a_{LV} &= \frac{I_{LV}}{\delta} = 136.36\ \text{mm}^2,\quad a_{HV} = \frac{I_{HV}}{\delta_{HV}} = 8.264\ \text{mm}^2 \end{aligned}

Radial depths: b1=14b_1 = 14 mm (LV), b2=18b_2 = 18 mm (HV), duct a=11a = 11 mm.

Mean lengths of turn:

Lmt,LV=π255.0+293.02=0.8608 mLmt,HV=π314.0+351.02=1.0446 mLmt=π255.0+351.02=0.9519 m (both windings, for reactance)\begin{aligned} L_{mt,LV} &= \pi\frac{255.0 + 293.0}{2} = 0.8608\ \text{m}\\ L_{mt,HV} &= \pi\frac{314.0 + 351.0}{2} = 1.0446\ \text{m}\\ L_{mt} &= \pi\frac{255.0 + 351.0}{2} = 0.9519\ \text{m (both windings, for reactance)} \end{aligned}

Leakage reactance (per unit), with ATAT per limb =8250= 8250 and axial winding length Lc=272.0L_c = 272.0 mm:

εx=2πfμ0ATLc⋅LmtEt(a+b1+b23)=2π(50)(4π×10−7)82500.2720⋅0.95199.091(11+14+183)×10−3=0.0272 p.u.XHV=εxVHVIHV=0.0272×6600.022.73=7.889 Ω (referred to HV)XLV=XHV(TLVTHV)2=0.02898 Ω\begin{aligned} \varepsilon_x &= 2\pi f\mu_0\frac{AT}{L_c}\cdot\frac{L_{mt}}{E_t}\left(a + \frac{b_1+b_2}{3}\right)\\ &= 2\pi(50)(4\pi\times10^{-7})\frac{8250}{0.2720}\cdot\frac{0.9519}{9.091}\left(11 + \frac{14+18}{3}\right)\times10^{-3}\\ &= 0.0272\ \text{p.u.}\\ X_{HV} &= \varepsilon_x\frac{V_{HV}}{I_{HV}} = 0.0272\times\frac{6600.0}{22.73} = 7.889\ \Omega\ \text{(referred to HV)}\\ X_{LV} &= X_{HV}\left(\frac{T_{LV}}{T_{HV}}\right)^2 = 0.02898\ \Omega \end{aligned}

iv) Overall dimensions (sketch)

 +-----------------------+  ---
 |  yoke, Hy = 206 mm    |   ^
 +-----+-----------+-----+   |
 |limb |  window   |limb |   H = 743 mm
 |d=234| 133 x 332 |     |   |
 +-----+-----------+-----+   |
 |        yoke           |   v
 +-----------------------+  ---
 |<--------- W --------->|
 W = 565 mm, D = 367 mm (limb centres)
 Depth = 199 mm

Answer: dd = 234 mm; window 133 × 332 mm; yoke 206 × 199 mm; frame 743 × 565 × 199 mm; leakage reactance ≈ 7.89 Ω referred to HV (2.72%).

✓ Answer checked by Claude Sonnet 5.5

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