--- title: "Output (kVA) equation of single phase transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:output-equation-single-phase-transformer" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-single-phase-transformer site: BCT Engineering --- # Output (kVA) equation of single phase transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2079 Bhadra, 6 marks; 2074 Asoj, 6 marks; 2071 Chaitra, 8 marks (asked 3 times) - Video: [2.2. Output equation of single phase transformer](https://www.youtube.com/watch?v=636pQP1osRM) — Nisha Prakash - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-single-phase-transformer - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Derive an expression for kVA output of a single phase transformer from design point of view. ## Short answer The kVA rating is Q = 2.22 f Bₘ δ K_w A_w Aᵢ × 10⁻³. It comes from the EMF per turn 4.44fBₘAᵢ, the copper area K_wA_w = 2(AT)/δ, and the relation AT = V/(4.44fBₘAᵢ), yielding the product of core and window areas. ## Answer The output equation relates the kVA rating of a transformer to its main dimensions (net core area $A_i$ and window area $A_w$) through the chosen magnetic loading $B_m$ and electric loading $\delta$. It is the starting point of transformer design. #### Notation - $f$ = frequency (Hz), $B_m$ = maximum flux density (Wb/m²), $\phi_m = B_mA_i$ - $A_i$ = net iron area of limb (m²), $A_w$ = window area (m²) - $K_w$ = window space factor = copper area / window area - $\delta$ = current density (A/m²), $T$ = turns, $I$ = current - Subscripts 1, 2 for primary and secondary #### Step 1: Voltage EMF per turn and terminal voltage (neglecting drops): $$ E_t = 4.44 f\phi_m = 4.44fB_mA_i, \qquad V \approx E = 4.44fB_mA_iT $$ #### Step 2: Window (copper) area In a single-phase core type transformer the window holds one primary and one secondary winding (half of each on each limb, but a full set passes through the window). $$ \begin{aligned} a_1 &= \frac{I_1}{\delta},\quad a_2 = \frac{I_2}{\delta}\\ \text{Copper area in window} &= T_1a_1 + T_2a_2 = \frac{T_1I_1 + T_2I_2}{\delta} = \frac{2AT}{\delta} \end{aligned} $$ since $T_1I_1 = T_2I_2 = AT$. This copper area is also $K_wA_w$, so $$ K_wA_w = \frac{2AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{2} $$ #### Step 3: Output $$ \begin{aligned} Q &= V_1I_1\times10^{-3} = 4.44fB_mA_iT_1I_1\times10^{-3}\\ &= 4.44fB_mA_i \cdot \frac{K_wA_w\delta}{2}\times10^{-3}\\ Q &= 2.22\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned} $$ #### Remarks - The same equation applies to a single-phase shell type transformer, because there too the window carries one full set of windings. - For a given rating, higher $B_m$ or $\delta$ gives a smaller core and window, but increases iron or copper losses and temperature rise. - In design, $E_t = K\sqrt{Q}$ is used first to get $A_i$; the output equation then gives $A_w$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The EMF induced per turn in a transformer is: A. 4.44 f Bm Ai B. 2.22 f Bm Ai C. 4.44 Bm Ai D. 4.44 f Bm Answer: A. Et = 4.44 f φm with φm = Bm Ai. (Key point: EMF) 2. The copper area in the window of a single-phase transformer is: A. AT/δ B. 2 AT/δ C. AT δ D. AT/(2δ) Answer: B. T1a1 + T2a2 = (T1I1 + T2I2)/δ = 2AT/δ, which equals Kw Aw. (Key point: Copper area) 3. From the window space factor, the primary ampere-turns are: A. Kw Aw δ B. 2 Kw Aw δ C. Kw Aw δ/2 D. Kw Aw/δ Answer: C. Kw Aw = 2AT/δ, so AT = Kw Aw δ/2. (Key point: Ampere-turns) 4. The output equation is: A. Q = 4.44 f Bm δ Kw Aw Ai x 10^-3 B. Q = 1.11 f Bm δ Kw Aw Ai x 10^-3 C. Q = 2.22 f Bm δ Kw Aw Ai x 10^-3 kVA D. Q = 2.22 f Bm Aw x 10^-3 Answer: C. Q = V1 I1 = 4.44 f Bm Ai x (Kw Aw δ/2) x 10^-3. 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