Q3Electrical Machine Design

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Q.3. Calculate the core and window area required for a 1000 kVA, 6000/400V, 50 Hz single phase core type transformer. Assume a maximum flux density of 1.25 Wb/m², a current density of 2.5 A/mm², voltage per turn = 30V, and window space factor = 0.32.

Answer 1

Step 1 — Core (net iron) area from the given voltage per turn:

A_{i} = \frac{E _{t}}{4.44 f B _{m}} = \frac{30}{4.44 \times 50 \times 1.25} = \frac{30}{277.5} \approx 0.1081 m^{2} = 1081 cm^{2}

Step 2 — Number of turns on each winding (used to size conductors, and as intermediate context for the window calculation): primary (HV, 6000V) turns N1 = 6000/30 = 200 turns; secondary (LV, 400V) turns N2 = 400/30 ≈ 13.3, rounded to a practical integer (13 or 14 turns) in the actual detailed winding design, though for the purpose of this main-dimensions calculation the exact turns count does not affect the core/window area result.

Step 3 — Window area from the kVA output equation: the standard single-phase transformer output equation relating kVA rating to the core and window areas is:

Q (kVA) = 2.22 f B_{m} δ K_{w} A_{i} A_{w} \times 1 0^{3}

where δ is the current density (A/mm²), Kw is the window space factor, and Ai, Aw are in m². Rearranging to solve for the window area Aw with Q=1000 kVA, f=50 Hz, Bm=1.25 Wb/m², δ=2.5 A/mm², Kw=0.32, and Ai=0.1081 m²:

A_{w} = \frac{Q}{2.22 f B _{m} δ K _{w} A _{i} \times 1 0 ^{3}} = \frac{1000}{2.22 \times 50 \times 1.25 \times 2.5 \times 0.32 \times 0.1081 \times 1000}

A_{w} \approx \frac{1000}{12000} \approx 0.0833 m^{2} = 833 cm^{2}

Interpretation: the required net core (iron) cross-sectional area of approximately 1081 cm² establishes the size of the core limb needed to carry the specified flux at the design flux density, while the required window area of approximately 833 cm² establishes the space needed within the core window to accommodate both the primary and secondary windings (including their necessary insulation clearances) while keeping the current density at the specified 2.5 A/mm², given that only a fraction (Kw=0.32) of the total window area is actually occupied by copper conductor cross-section, the remainder being insulation, clearances, and cooling ducts. Both of these calculated areas, together with the assumed stacking factor for translating net iron area to gross/physical core dimensions, form the essential starting point from which the detailed core leg dimensions, window height/width, and overall transformer frame size are subsequently determined in the complete transformer design process.

Cross-check using the aspect ratio D²L-style relation: it is instructive to note that the product Ai×Aw (sometimes called the output constant product) directly determines the kVA rating for fixed Bm, δ, Kw, and frequency, in a manner exactly analogous to how the D²L product determines the output of a rotating machine — here Ai×Aw ≈ 1081×833 ≈ 900,700 cm⁴, and this single combined product is preserved even if the designer later chooses a different split between a taller, narrower window (larger Aw, requiring correspondingly smaller Ai for the same product) versus a shorter, wider one, provided the specified flux density, current density and space factor are held fixed. This is precisely why transformer design, especially at the main-dimensions stage, is often described as determining the required Ai-Aw product first, and only subsequently apportioning it between the specific core cross-section shape (Stage 2 of the design) and the specific winding/window layout (Stage 3), mirroring the analogous two-stage process (first D²L, then the specific split into D and L) used in rotating-machine design elsewhere in this paper.

Q3Electrical Machine Design

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