Voltage Drop for 3 phase Induction motor

[XinY0503]

I am doing the calculation for voltage drop at motor terminal but only get 3% voltage drop and seems wrong. Can anyone help me please?
The motor LRA is 2714A, FLC 405A, 3000HP, 4KV 3 Phase 60 Hz. But there is no transformer, it connected directly to the 4160V bus line.
Tha cable is 3 conductor copper 350MCM 5000V and length is 200M.

Appericate your help in advance!
Thanks

 

[RRaghunath]

How about the source impedance? You need to account that as well in voltage drop calculation.
Voltage drop is estimated for starting condition and for running condition.
The limits are 15% at the busbar and 20% at the motor terminals, for starting condition.
You may find some freely downloadable excel sheets for estimating voltage drop on the internet. However, to verify the correctness rests on you.

 

[7anoter4]

In my opinion, the voltage drop has to be 5.6%.
If the source it is a transformer of 50 MVA [63 A maximum rated] 11% short-circuit voltage [0.038 ohm]
The cable [3*350 MCM copper as per Kerite catalogue 0.04+j0.033 ohm/kft ]200 m Zcbl=0.034 ohm
Induction motor of 3000 HP [2255 kW per Siemens catalogue] Irat 360 A [for 4.16 kV] Ilock/Irat=5.5 Istartrat=1980 A
Zmotor=4160/sqrt(3)/1980=1.213 ohm
Ztotal=1.285
Istart=4160/sqrt(3)/1.285=1869A
DV=sqrt(3)*Istart*(Zsys+Zcable)=sqrt(3)*1869*(0.038+0.034)=233.1V=5.6%

 

[XinY0503]

Thanks RRaghunath, I think you are right and I include the Station Transfomer in my calculation and get a reasonable number.
But i just have another question for the 3 phase ST, the rating is 17.8/23.7/29.7 MVA @65 Degree, Z_HX = Z_HY = 8.3% @17.8MVA, Z_XY = 16.6% @17.8MVA.
Which value transfomr KVA I should use in my calculation?
For the percentage Impedance, as this is a three winding transformer and based on the circuit model, Z% = Z_XY+Z_HX//Z_HY =20.45%. Then should I use 17.8MVA for transofmer KVA or...?

This are the two equation I used for calculation.

Thank you!

 

[RRaghunath]

The motor is 4kV rated, so you have to use that winding related impedances and kVA.
If you are using pu method, you will be having a Base MVA.

 

[7anoter4]

In my opinion if the VX=4.1 kV then ZH-L=8.3/100*4.1^2/17.8=0.0784 ohm viewed from 4.1 kV. Istart=1812 A. That means Zsystem=0.0784 ohm and voltage drop at motor start will be 8.5%

 

[XinY0503]

Hi 7anoter4,Thanks for the information and calculation. I completed the calcualtion for voltage drop including the source transformer and cable and get a resonable number.
The reason I am calulating the voltage drop is because I want to know whether the cable is capable of powering the motor and fan as the 3000HP is the new motor and we are trying to using the old cable to avoid more cost.

I just had another question and would be appericated if you can help. I was wondering is acceleration time vs current curve doesn't consider load torque?
Because based on the Torque vs RPM curve for motor and load, i was able to calculate the acceleration time. The acceleration time of 80% voltage with load toruqe considered is 44s and 32s for without load torque. And when I check the acceleration time VS current curve, it seems like the t is around 32s which means the acceleration curve is for without load torque condition. But most source i found online indicate this curve is with load torque considered. I am a bit confused.

 

[7anoter4]

Calculated using data of 2255 kW/1200 rpm [6 poles] Siemens production for load inertia of 1600 kg.m^2
[approx. 6 times motor inertia] the starting time for 100% voltage supply is 21 sec and at 80% it is 33 sec.

 

[7anoter4]

Sorry: 6 poles ;1600 kg.m^2