Transformer design 2

[rajendrarohra1978]

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.

 

[odlanor]

search Google for: voltage drop due to motor starting
or Industrial power system-Shoalb Khan , item 9.7.2.6

 

[jkristinn]

To calculate the voltage drop (i.e. the regulation), you need the X/R ratio of the transformer impedance and the power factor of the total load (during motor starting).
You should read thread238-221175, especially the files jghrist links to.

 

[odlanor]

see attached

 

[odlanor]

see attached my calculation

 

[odlanor]

ERRATA
see attached my calculation

 

[7anoter4]

Since we don't know a lot of data we have to use our experience proposing these [as odlanor did].
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!]. Starting current will be 7 times the rated current of motor at rated voltage [Nevertheless the modern class D could be 4.5 kVA/HP=3 times, may be].
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)
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.

 

[meeko]

what program do i need to open odlanor's document(.m)? interesting thread so far!

 

[7anoter4]

I think I exaggerated a bit with the voltage drop on outer System. I had to presume the Utility assured the rated voltage for entire 75 MVA so only the supplementary current above the transformer rated has to be taken into consideration. Then Snew=125.4-75=50.4 MVA and DUsys=0.371637 kV. The voltage at transformer secondary terminals will be 11.63 kV and voltage drop from rated will be 15.7%.

 

[7anoter4]

CORRECTION:
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%.

 

[odlanor]

meeko
last file is .docx (word).

 

[odlanor]

I guess convenient for more clarify, showing my simplified calculation.
Odlanor


% Transformer design
% thread238-357593 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

 

[7anoter4]

I have to declare the p.u. method of calculation is very expeditious and accurate enough, odlanor. In Europe we waist our time trying to be accurate more than necessary.

 

[odlanor]

7anoter4, take it easy!
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 !!

 

[7anoter4]

odlanor, in my opinion the apparent power is not important since p.u. method is based on reactance ratio.