Transformer design
Transformer design
(OP)
I need to calculate undervoltage of exisitng transformer. rating : 75 MVA,34.5/13.8 KV, rated current at 13.8 KV =3137.8 A. The undervoltage is due to starting a compressor motor of 7.5 MW, 351 Ampere Induction motor on Direct on line start. The transformer is ruuning already with 50 MW load. The short circuit current capacity of transformer on 13.8 Kv side is 25.1 KA for 3 sec. Bus voltage on 13.8 KV is 13.95 KV.
Thanks in advance.
Thanks in advance.






RE: Transformer design
or Industrial power system-Shoalb Khan , item 9.7.2.6
RE: Transformer design
You should read thread238-221175: VOLTAGE DROP AT XFMR: VOLTAGE DROP AT XFMR, especially the files jghrist links to.
RE: Transformer design
RE: Transformer design
RE: Transformer design
see attached my calculation
RE: Transformer design
The required data could be [developing a little the modest list of odlanor]:
System short-circuit apparent power Ssys and system R/X-as jkristinn said.
If the maximum [usually] breaker rated current is 50 kA [rms] the Ssys=sqrt(3)*34.5*50= 3000 MVA [approx.]
Let’s say Ssys=3000 MVA at 34.5 kV and X/R=10.
Recalculating at 13.8 kV Zsys=13.8^2/3000= 0.06348 ohms and Xsys=sqrt[Zsys^2/(1+1/100)]= 0.063165 ohm and Rsys=0.0063165 ohms.
Now we shall appreciate the short-circuit impedance zk% and pcu% copper losses in the transformer. Let’s say zk=8% [in order to limit the short-circuit current at 25.1 kA if Sys=infinite.]and pcu=0.5%.
ZT=8%*13.8^2/75=0.203 ohm RT=0.5%*13.8^2/75=0.0127 ohm and XT= 0.2027
Also, let's say the steady state load power factor- before motor start- was 0.75 [in order to cover 75 MVA from 50 MW “declared” load and Smot at steady state].
Let's say the starting power factor of the motor was 0.15 [I think 0.4 is too high odlanor!
If we put as total active power-when starting- P=50+SQRT(3)*13.8*0.351*7*0.15= 58.8 MVA and total S=50/0.75+SQRT(3)*13.8*0.351*7=125.4 we get cos(fi)=P/S=0.469 then fi=1.0826 radians.
For System drop we use the IEEE Standard 141 VOLTAGE CONSIDERATION –the approximate formula: V=I*R*cos(fi)+I*X*sin(fi) and neglecting the supplementary angle introduced by transformer impedance fi=1.0826 radians.
DUsys=sqrt(3)* 125.4 /sqrt(3)/13.8*[(0.0063165 *cos(fi)+ 0.063165 *sin(fi)]=0.5334 kV
U2o=13.95-0.5334=13.42 kV
Using the way proposed by prc [See thread238-221175: VOLTAGE DROP AT XFMR VOLTAGE DROP AT XFMR)
according to IEC 60076-8 clause 7.4 The voltage drop equations :
A=S/U2^2*[XT*sin(fi)+RT*cos(fi)]
B=S/U2^2*[XT*cos(fi)-RT*sin(fi)]
DU2=U2o*(A-A^2+B^2/2...) –neglecting ....[of course!].
S=125.4 MVA U2o=13.42 kV
Recurrently-replacing U2 updated in A and B- we get finally:
A= 0.173807 B= 0.085796 DU2=11.45*(0.173807 -0.173807 ^2+0.085796 ^2/2)= 1.9765 kV.
Then total voltage drop at transformer terminals will be 13.8-11.45=2.35 kV[17%].
For a more accurate result we have to refer also to the tap position [if any].
Let's say between motor and the compressor is a clutch and the motor will start without any load [only J of the motor will be taken into consideration in order to calculate the starting time].
Let’s say the no of poles=8 freq=60 Hz rpm=3600/4=900 .
Calculated motor rotor diameter will be 1.5 m and the ideal length 0.9 m then the mass of the rotor will be 12900 kgf. In this case GD^2=kgf*dia^2/4=12900*1.5^2/4=7256 kg.m^2 [approx.6*7256=43540 lbs*ft^2].Let’s say no load torque will act against the motor so the accelerating torque will be approx Tacc=2*Tfl. [Tfl = Php 5,252 / ωmax wmax=2*pi()*rpm
Tfl=7.5*746*5252/2/pi()/900 =5196.4 lbs.ft
tstart=wk^2*(ωmax -0)/Tacc/308=5196.4*2*pi()*900/2/5196.4/308 =9.18 sec.
The transformer has to withstand 9.18 sec 125.4 MVA [125.4/75=167.2%].I think it is not a problem.
RE: Transformer design
RE: Transformer design
RE: Transformer design
I have to apologize-my OFFICE 2010 it is still a problem for me.
DUsys=0.23154 kV and the secondary voltage will be 11.81 kV. Then total VD=(13.8-11.81)/13.8*100=14.4%.
RE: Transformer design
last file is .docx (word).
RE: Transformer design
Odlanor
% Transformer design
% thread238-357593: Transformer design jan 2014
% rajendrarohra1978 (Electrical)
% 8 Jan 14 5:51
% I need to calculate undervoltage of exisitng transformer. rating : 75 MVA,34.5/13.8 KV, rated current at 13.8 KV =3137.8 A.
% The undervoltage is due to starting a compressor motor of 7.5 MW, 351 Ampere Induction motor on Direct on line start.
% The transformer is runing already with 50 MW load. The short circuit current capacity of transformer on 13.8 Kv side
% is 25.1 KA for 3 sec. Bus voltage on 13.8 KV is 13.95 KV.
% Thanks in advance.
%%
%% initial data existing or assumed
MVAb = 10; % potencia de base ; kVb = 13.8;
kVh = 34.5 ; % kV
kVl = 13.8; %kV
Xt = 6; % Transf impedance in per-cent , base 75 MVA,34.5/13.8 KV
fpsm = 0.40; % motor starting power factor, ask motor vendor
Pm = 7; %MW
fpload = 0.90 ; % load power factor
fprm = fpload; % motor running power factor
Pload= 50.0; %MW
Imr= 351.0; %A motor current running
Ims_Imr = 6.5 ; % ratio Ims/Imr , Ims=locked-rotor motor current in A, ask motor vendor
Iccl = 25.1; % kA short-circuit 13.8kV bus
kVbus = 13.95; % kV
%%
%%preliminary check
MVAlbus = 1.732*kVl*Iccl; % MVA shortcitcuit power 13.8kV bus
MVAload = Pload/fpload;
MVAmr = Pm/fprm; % MVA motor full load
MVAms = Ims_Imr * MVAmr; % MVA motor locked rotor
%%
%%per-unit method (neglect resistance component) - MVAb = 10; kVb = 13.8;
Xsys = (kVbus/kVb)^2 * MVAb / MVAlbus; % pu ;
Xms = (kVbus/kVb)^2 * MVAb / MVAms; % pu; % starting motor
Xload = (kVbus/kVb)^2 * MVAb / MVAload; % pu; %
%%
%% voltage drops at 13.8kV bus
Xsload = (Xms * Xload) / ( Xms + Xload); % Xms // Xload total load during motor starting
vd = Xsys / ( Xsys + Xsload); % 0.1503 pu or 15% VOLTAGE DROPS AT 13.8kVbus
RE: Transformer design
RE: Transformer design
My drop voltage is 15.0% bases 10MVA-13.8kV
your drop voltage is 14.5% bases 125.4MVA-13.95kV
My drop voltage in your bases is 15.0 * (125.4/10) * (13.8/13.95)=>184% ??!!
I am lost !!
RE: Transformer design