92 APPENDIX 1 DERIVATION OF TRANSFORMER EQUATIONS The necessary equations for the formation of optimum design problem of a transformer is derived below. OUTPUT / CONSTRAINT EQUATION Single Phase Transformer The voltage induced in a transformer winding with T turns and excited by a source having a frequency f Hz is given by Voltage per turn Et E 4.44 f m T (A1.1) The window in a single phase transformer contains one primary and one secondary winding. Total copper area in window Ac = Copper area of primary winding + Copper area of secondary winding = primary turns x area of primary conductors + secondary turns x area of secondary conductors = T p a p Ts a s 93 Taking the current density secondary windings, a p I p / and a s to be same in both primary and Is / Ip Total conductor area in window, Ac T p I Ts s (T p I p Ts I s ) / 2 AT as T p I p Ts I s (A1.2) AT if magnetizing mmf is neglected. The window space factor K w is defined as the ratio of copper area in window to total area of window, or Kw Ac Aw Conductor area in window, Ac K w Aw From Equations (A1.2) and (A1.3), or AT 2 AT (A1.3) K w Aw Kw Aw 2 (A1.4) Rating of a single phase transformer in kVA, Q Vp I p 10 3 E p I p 10 3 (asVp E p ) Et T p I p 10 3 Et AT 10 3 K A K A Et w w 10 3 4.44 f m w w 10 3 2 2 2.22 f m Kw Aw 10 3 (A1.5) 94 But m Bm A i Q 2.22 f Bm K w Aw Ai 10 3 kVA (A1.6) Three Phase Transformer In this case, each window contains two primary and two secondary windings. Proceeding as in the case of single phase transformer, Total conductor area in each window, Ac = 2(a pTp +a sTs ) = 2(IpTp + IsTs ) / 4 AT (A1.7) Total conductor area is also equal to K w Aw 4 AT or AT K w Aw Kw Aw 4 (A1.8) 95 Rating of a three phase transformer in kVA Q 3Vp I p 10 3 3 E p I p 10 3 3 Et T p I p 10 3 3 Et AT 10 3 K A 3 4.44 f m w w 4 3.33 f m Kw Aw 10 3 10 3 (A1.9) 3.33 f Bm Kw Aw Ai 10 3 kVA (A1.10) As the transformer is expected to deliver its rated kVA, Equations (A1.6) and (A1.10) are taken as the constraint equations for single phase and three phase transformers respectively. CONDITIONS FOR OPTIMUM DESIGN Minimum Cost / Volume / Weight Consider a single phase transformer. Its kVA output is Q = 2.22fBm K w A A x10-3 w i Assuming that the flux and current densities are constant, the product Ac A is constant for a transformer of given rating. Let this product i Ac Ai M2 (A1.11) 96 The optimum design problem is, therefore, that of determining the minimum value of total cost. Now, r where Ai Ac Bm Ai and AT r 2 Bm 2 (A1.12) is a function of r only, as Bm and are constant. Thus from Equations (A1.11) and (A1.12), Ai Ac 2Bm Ai Ac r or m AT and m M and Ac M Let CT be the total cost, CT C Cc i ci Gi cc Gc ci gi Al i i cc g c Ac Lmt Substituting for A and Ac , i CT c g l M i ii cc g c Lmt M 97 Differentiating CT w.r.t , dCT d 1 2 cc g c Lmt M 1 c glM 2 i ii For minimum cost, dCT d ci gi li cc g c Lmt or 3 2 0 1 ci gi li Ai cc g c Lmt Ac or c G or Ci i i cc Gc Cc (A1.13) Hence, for minimum total cost, the cost of iron must be equal to the cost of copper or conductor. Similar conditions apply to other quantities, namely, for minimum volume of transformer, Volume of iron = Volume of conductor i.e. Ui Uc (A1.14) and for minimum weight of transformer, Weight of iron = Weight of conductor i.e. Gi Gc (A1.15) 98 The above condition also holds good for a three phase transformer. Minimum Full Load Loss The main component of full load loss is i) Full load iron loss, P and i ii) Full load copper loss, Pc . The efficiency will be maximum for minimum full load loss. Total full load loss, PT Pi Pc (A1.16) At any fraction x of full load, the total losses are P i x2 Pc . If Q is the output at full load, the output at fraction x of full load is xQ . Efficiency at output, xQ , x xQ xQ P x2 Pc i This efficiency is maximum when d x 0 dx 2 d x ( xQ Pi x Pc ) Q xQ (Q 2 x Pc ) dx ( xQ Pi x2 Pc )2 For maximum efficiency, ( xQ Pi x2 Pc ) x (Q 2 x Pc ) xQ Pi x2 Pc xQ 2 x2 Pc 99 or Pi x2 Pc For full load (i.e. x =1), Pi Pc Hence for minimum full load loss, the condition is Full load iron loss = Full load copper loss i.e. Pi Pc (A1.17) The above condition is applicable to both single and three phase transformers.