--- title: "Output equation of three phase transformer" subject: "Electric Machine Design" chapter: 3 key: "semester_5/electric_machine_design:output-equation-three-phase-transformer" source: https://www.bctengineer.com/q/semester_5/electric_machine_design/output-equation-three-phase-transformer site: BCT Engineering --- # Output equation of three phase transformer Electric Machine Design · Chapter 3: Transformer Design - Asked in: 2081 Bhadra, 6 marks; 2072 Chaitra, 6 marks (asked 2 times) - Video: [2.3. Output equation of three phase transformers](https://www.youtube.com/watch?v=vpJ0jt7a3R4) — 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-three-phase-transformer - Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machine_design/chapter-3 ## Question Derive an output equation for the three phase core type transformer. ## Short answer For a three-phase core type transformer, the kVA rating is Q = 3.33 f B_m δ K_w A_w A_i × 10⁻³, with f in Hz, B_m in Wb/m², δ in A/m², K_w window space factor, A_w window area in m² and A_i net core area in m². The equation combines voltage per phase E_t = 4.44f B_m A_i and the window ampere-turn relation AT = K_w A_w δ/4. ## Answer The output equation gives the kVA rating of a three-phase core type transformer in terms of the net core area $A_i$, window area $A_w$, flux density $B_m$ and current density $\delta$. ```text +-------+-------+-------+ | | | | [L|H] W1 [H|L|H] W2 [H|L] | R | Y | B | +-------+-------+-------+ 3 limbs, 2 windows; each window holds LV + HV of two different phases ``` #### Step 1: Voltage per phase $$ E_t = 4.44f\phi_m = 4.44fB_mA_i, \qquad V_p \approx 4.44fB_mA_iT_p $$ #### Step 2: Copper in a window A three-phase core type transformer has three limbs and two windows. Each window holds one coil side from each of two adjacent limbs. Each coil side carries the full cross-section of one primary and one secondary winding: $$ \text{Copper area in a window} = 2(a_pT_p + a_sT_s) $$ With $a_p = I_p/\delta$, $a_s = I_s/\delta$ and $I_pT_p = I_sT_s = AT$: $$ K_wA_w = 2\left(\frac{AT}{\delta} + \frac{AT}{\delta}\right) = \frac{4AT}{\delta} \;\Rightarrow\; AT = \frac{K_wA_w\delta}{4} $$ #### Step 3: Output $$ \begin{aligned} Q &= 3V_pI_p\times10^{-3} = 3\times4.44fB_mA_iT_pI_p\times10^{-3}\\ &= 3\times4.44fB_mA_i\times\frac{K_wA_w\delta}{4}\times10^{-3}\\ Q &= 3.33\,f\,B_m\,\delta\,K_w\,A_w\,A_i\times10^{-3}\ \text{kVA} \end{aligned} $$ where $Q$ is the total three-phase rating in kVA. #### Use in design - First, $E_t = K\sqrt{Q}$ (with $K \approx 0.45$ for distribution and 0.6–0.7 for power transformers) gives $\phi_m$ and $A_i = \phi_m/B_m$. - Then the output equation gives $A_w$, from which window width and height are found with a chosen $H_w/W_w$ ratio (2 to 4). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In a 3-phase core type transformer, the copper area in a window is: A. a_pT_p + a_sT_s B. 2(a_pT_p + a_sT_s) C. 3(a_pT_p + a_sT_s) D. (a_pT_p + a_sT_s)/2 Answer: B. Each window holds one primary and one secondary of two limbs, giving twice the single-limb copper. (Key point: Copper in window) 2. From Kw Aw = 4AT/delta, the ampere-turns are: A. Kw Aw delta B. Kw Aw delta / 2 C. Kw Aw delta / 3 D. Kw Aw delta / 4 Answer: D. AT = Kw Aw delta / 4. (Key point: Ampere-turns result) 3. The resulting output equation is: A. Q = 2.22 f Bm delta Kw Aw Ai x10^-3 kVA B. Q = 3.33 f Bm delta Kw Aw Ai x10^-3 kVA C. Q = 4.44 f Bm delta Kw Aw Ai x10^-3 kVA D. Q = 1.11 f Bm delta Kw Aw Ai x10^-3 kVA Answer: B. 3 x 4.44 / 4 = 3.33. (Key point: Final output equation) 4. After finding Ai from Et = K sqrt(Q), what does the output equation give? A. Number of turns B. Window area Aw C. Core loss D. Tank size Answer: B. Aw is found, then window width and height using Hw/Ww of 2 to 4. 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