The voltage shall not drop more than 10% at the terminals of a running motor as per NEMA requirements.
[10% less than motor rated voltage.]This includes the common drop and the individual cable drop in steady state regime.
You have to calculate the voltage drop up to motor terminal for a steady state load and to add the voltage drop up to the last common MCC in the case of the start of the biggest motor [or which presents the biggest current at start].
There is a short way-not so accurate and conservative but efficient-taking the maximum starting current.
If the transformer apparent power is 5 MVA at 4.15 kV the maximum motor power supplied from could be 1/4=1025 HP.
Iftrf=5/sqrt(3)/4.15=0.6956 kA transformer rated current.
Pmot=1025*.746=764.65 kw
Motor rated voltage=4000 V. [as per catalogue].
If the induction motor is squirrel cage normal rotor take it as DOL starting.
Irated=0.764.65/sqrt(3)/4/0.8=0.138 Ka [0.8=pf*eff=0.9*0.89] motor rated current.
Istart=8*0.138=1.1037 kA at full rated voltage.
The total voltage drop up to motor terminals take it as 20%, so you can adjust the Istart at 80% Istart=1.1037*0.8=0.883 kA.
The starting pf of the motor could be 0.2 then the Imotorstart=0.883*(0.2+j*0.98)=0.1766+j*0.865 Ka.
Let's say the transformer load in steady state is 5 MVA pf=0.8.
Itrfsteady=0.6956*(0.8+j*0.6)=0.5565+j*0.4174 kA
Imotorsteady=0.138*(0.9+j*0.436)=0.1242+j*0.06 kA
You can extract the starting motor steady state current in order to find the remaining running load current.
and the remaining load current will be: Itrfremain=0.432+j*0.357 KA.
During the start of the biggest [proposed] motor : Itrfst=Itrfremain+Imotst=0.6086+j*1.222 KA.
Let's calculate the voltage drop through the transformer when this motor will start.
In order to find the transformer resistance let's say the transformer copper losses =40 kw then Rtrf%=40/1000/5*100=0.8%.
The transformer reactance will be sqrt(5.6^2-0.8^2)=5.54%.
Rtrf=0.8*4.16^2/5/100=0.0277 ohm transformer resistance
Xtrf=5.54/100*4.16^2/5=0.19175 ohm transformer reactance.
The voltage drop across the transformer will be
Volt=sqrt(3)*(Itrfstreal*Rtrf+Itrfstim*Xtrf)=0.4388 kV
The voltage at transformer terminals will be 3.72 kV that means 3.72/4=93% of motor rated voltage.
You have to chose the right cable feeder so that at motor starting terminal the voltage will not be less than 0.8*4=3.2 kV.
and for other running motor not less than 10%[4*.9=3.6kV].
If the cable is in acceptable limits than this could be the maximum power motor you may connect to this transformer.
If not a smaller motor should be taken into consideration.
If the utility supply voltage could be less than rated by 5% and your transformer has not possibility to raise the voltage back,
you have to take it into consideration as a supplementary drop [but only for steady state regime].
