Easy way to calculate transformer and motor details
Transformer details:
- Transformer size
- Transformer secondary voltage (v2)
- % Impedance (z)
- Allowable voltage drop in system
Motor details for one and more motors:
- Motor full load current (ILM)
- Locked rotor current multiplier (Min)
- Locked rotor current multiplier (Max)
- Locked rotor current (L1) (Min)
- Locked rotor current (L2) (Max)
- Motor inrush KVA at starting (IrsM)
- How many times motor starts per hour
and Calculations:
- Full load current of transformer (ILX)
- Short circuit current at transformer secondary (Isc)
- Maximum KVA of transformer at Isc (Q1)
- Motor inrush KVA At starting (IrsM)
- Voltage drop at transformer secondary due to motor inrush (Vd)
- Transformer secondary voltage
Results you get:
- Motor Full Load Current
- Voltage Drop at Transformer Secondary Due to Motor inRush
- Transformer Size:(20% Added due to motor starts more than once/Hour)
Calculation formulas used:
- ILX = KVA / (1.732 x v2)
- ILM = (Kw x 1000) / (1.732 x Vm x P.F)
- Isc = ILx / z
- Q1 = (v2 x ISC x ILM x 1.732) / 1000
- IrsM = (Vm x L2 x 1.732) / 1000
- Vd = IrsM / Q1
| Software: | Calculate size of transformer and voltage due to starting of large size motor |
| Version: | 10.6.13 |
| Developer: | Jignesh Parmar |
| Size: | 40 KB |
| Price: | Free |
| Download: | Right here | Video Courses | Membership | Download Updates |
how to size proper transformer for pump motor
Whether this calculation holds good for heavy starts as for ballmills.
Rule of thumb large motors should be ___ of the KVA of XFMR?
I need the size of the Generator and Transformer for 400 kw motor
IEEE Standards provide correct formula. Please refer and accordingly revise Excel sheet tabulated formulas. Regards.
Motor KW is 300KW
Motor KVA at 0.8 PF is 300/0.8 = 375 KVA ,not 240 KVA given in the sheet
Also Motor KW is out put KW ,not input KW which needs to consider efficiency ,say 90% for 300KW motor
Input KW = 375/0.9 = 417 KVA
motor full load current = 417000/ 1.732×460 = 522 Amp ,not 470 Agiven ,so starting current and volt drop also will increase proportionately also.
The sheet needs such major corrections , CS NAMBISAN
Excellent example. Could you please modify it to include 75 HP motor with pf of 0.8 lag and starting pf of 0.3. and constant load 20 KVA, pf = 0.8 lag
Thanks
What is the rating of transformer which can feed motor 1500 HP , 4160 V and 60 HZ ?
Ganesh, I agree with you, It should include the KVA of the loads connected to the transformer before starting the motor. Including the loads KVA to the starting KVA of the motor may end up with the transformer size is incorrect (VD% is above the limit. Comment of Jignesh Parmar is highly appreciated.
Please note that in thi volatge drop calculation existing load on transformer is not considered. only motor starting KVA is considered for voltage drop calculation. Kindly tell me how to consider other conneted load in this calcullation.
Kw=KVA*p.f
KVA=Kw/p.f
To calculate sizing of Transformer 30% future expansion should be considered.
Kw=KVA*p.f
KVA=Kw/p.f
In the excel sheet he multiply the KW by Power Factor but actual KVA= KW/Pf
motor kVA from kW and cos phi will be rather E11/E13 than E11/E13
isn’t it ?
Yes, it should be
This calculation has two mistakes.
1. You have not considered the existing load in the transformer during motor starting, as if we are starting the motor alone without any other load. Normally transformers will be loaded 50% of its rating. So atleast you have to consider 30 to 40 % load in the transfomer and add that to the motor starting KVA. Here it is around 300 to 400 KVA. IF you add that the transformer selection will be wrong.
2. Motor starting KVA can be calculated directly from starting KVA/HP, converting KW to HP, which will be 400HP*4.5 = 1800KVA, which is correct way of doing. In that case your figure is wrong. In both cases the transformer will not be suitable.
Usually in LV MCC, motor KW is restricted to 160KW the maximum due to the above reasons.
otor starting KVA can be calculated directly from starting KVA/HP, converting KW to HP, which will be 400HP*4.5 = 1800KVA, which is correct way of doing. In that case your figure is wrong. In both cases the transformer will not be suitable.
where from u get this.??
Yup they are right it is not downlodable. i am trying since two months and for your information when we registered ourself then only we are able to give comments here. pls do the needful
Muchas gracias.
Calculation is good but i understand that it does not include starting method of motor??? or does it include this effect in form of motor code. This is one gray area I am not clear about. Can you please confirm the same please….
Thank you
J
Not downloadable !
the excel spreadsheet is not in downloadable format. so i have not been able to download the table even though i can see it in pictorial form.
Spreadsheet is downloadable and link works. Did you register at EEP?
Please check your calculations otherwise give password so we can correct your mistakes (E13 is incorrect as kVA increases with a lower PF not decreases.
dear sir, Please correct the calculation at E14 (conversion from Kw to KVA). OR can u give me the password for this X-cel sheet