nstitte of Mechatronics and nformation Systems for internal se only Transformers Exercise A three-phase transformer has the following nominal ratings rated power S 60 kva, Calclate rated voltage ratio 15750 x,5 / V, 1 ± winding connection Dyn, rated freqency f 50 Hz, rated no- crrent i0 0, 75, rated core losses P 0 900 W, rated short-circit voltage k 6, rated winding losses P k 600 W. 1. ratings of the spply needed to perform the nominal no- test,. parameters of the magnetising branch of the Γ type eqivalent circit,. ratings of the spply needed to perform the nominal short-circit test, 4. parameters of the branch of the Γ type eqivalent circit, 5. relative vales of the Γ type eqivalent circit parameters, 6. relative, percent vales of the voltage drops on the short circit elements of the Γ type eqivalent circit, 7. relative, percent vales of voltage drops nder and the voltage on the LV winding, when energised from the HV winding on tap 0 with rated voltage 15750 V and rated freqency f f 50 Hz, for the following s 1 a. 0, 8 ind. b. 0,5 0, 8 ind. c. and cosϕ 1 d. 0, 6 ind. e. 0, 8 cap.. k X k Fe 1ph ph E ph X m Edited by Witold Kbiak, Ph.D., M.Sc., Eng. 1
nstitte of Mechatronics and nformation Systems for internal se only Soltion 1. The no- test shold be exected energising the LV winding of the transformer with nominal voltage, at nominal freqency. Therefore the spply shold be able to provide line voltage V ominal no- line crrent 0 () i0 S i0 60 10 0,75 6,80 A 100 100 100 0 () Apparent power of the spply sorce i0 i0 0,75 S0 () 0() S 60 10 4,75 kva 100 100 100. Active component of the no- phase crrent (nominal vale) P0 P0 900 0 acph() 1,99 A ph (Y) eactive component of the no- phase crrent (nominal vale) 6,8 1,99 6,695 A mph () 0ph() 0acph() 0()(Y) 0acph() esistance of the magnetising branch (nominal vale) ph Fe () 177,8 Ω P0 P0 900 eactance of the magnetising branch (nominal vale) ph Xm () 4,49 Ω 6,695 mph() mph() mph (). The short-circit test shold be exected energising the HV winding of the transformer with decreased voltage, at nominal freqency, to obtain the nominal crrent the LV winding short-circited. Therefore the spply shold be able to provide line crrent S 60 10,09 A 15750 k (1) ominal no- line short circit voltage k 6 k (1) 15750 945 V 100 100 Apparent power of the spply sorce k k 6 Sk (1) k(1) S 60 10 7,80 kva 100 100 100 4. Active component of the short-circit phase voltage (nominal vale) 600 kph (1) 157,5 V,09 1ph Edited by Witold Kbiak, Ph.D., M.Sc., Eng.
nstitte of Mechatronics and nformation Systems for internal se only eactive component of the short-circit phase voltage (nominal vale) 945 157,5 91,8 V kxph (1) kph(1) kph(1) k(1)( ) kph(1) esistance of the branch (nominal vale) 600 k (1) 11,8 Ω,09 1ph eactance of the branch (nominal vale) X kxph(1) kxph(1) kxph(1) 91,8 69,90 Ω 1ph,09 k (1) 5. The relative vales shold be calclated as ratios to appropriate phase impedances. Therefore in case of the magnetising branch it shold be ph Z ph 0,540 Ω S ph S 60 10 ominal relative vale of the magnetising branch resistance Fe() 177,8 rfe 700 r.. Z 0,54 ph ominal relative vale of the magnetising reactance Xm() 4,49 xm 15,8 r.. Z 0,54 n case of the branch the vales shold be related to 1ph 15750 Z1 ph 1181Ω S 1ph S 60 10 ph ominal relative vale of the winding s resistance k(1) 11,8 rk 0,01001 r.. Z 1181 1ph ominal relative vale of the leakage reactance Xk(1) 69,9 xk 0,05919 r.. Z 1181 1ph 6. elative percent vale of the active component of the short-circit phase voltage (nominal vale) k(1) k 100 100 100 100 S 600 100 1 60 10 k Edited by Witold Kbiak, Ph.D., M.Sc., Eng.
nstitte of Mechatronics and nformation Systems for internal se only t shold be noted, that the relative percent vale of the winding resistance is eqal kph(1) 1ph rk 100 100 100 100 Z 1ph 1ph 1ph 1ph S 1ph 600 k 100 1 60 10 elative percent vale of the reactive component of the short-circit phase voltage (nominal vale) 6 1 5,916 kx k k 7. The voltage drop on the transformer nder indctance is given by the exact formla ( cosϕ + sin ϕ ) + 0,005 β ( cosϕ ϕ ) β k kx kx k sin where β The voltage drop on the transformer nder capacitance is given by the exact formla ( cosϕ ϕ ) + 0,005 β ( cosϕ + ϕ ) β k kx kx k sin sally it is acceptable to se a simplified eqation ( k cosϕ ± kx ϕ ) (the sign in the formla depends on the type of ). a. 0, 8 ind. ( k cos ϕ + kx ϕ ) 1 (1 0,8 + 5,916 0,6) 4, 50 4,5 1 8,6 V b. 0,5 0, 8 ind. ( k cosϕ + kx sin ϕ ) 0,5 (1 0,8 + 5,916 0,6), 175,175 1 91, V c. and cosϕ 1 ( k cos ϕ + kx ϕ ) 1 (1 1+ 5,916 0) 1, 000 1 1 96,0 V 100 100 Edited by Witold Kbiak, Ph.D., M.Sc., Eng. 4
nstitte of Mechatronics and nformation Systems for internal se only d. 0, 6 ind. ( k cosϕ + kx ϕ ) 1 (1 0,6 + 5,916 0,8) 5, 5, 1 78,7 V e. 0, 8 cap. ( k cosϕ kx sin ϕ ) 1 (1 0,8 5,916 0,6), 750,75 410,8 V Edited by Witold Kbiak, Ph.D., M.Sc., Eng. 5