Calculate TC Size & Voltage Drop due to starting of Large Motor
June 5, 2013 12 Comments
Calculate TC Size & Voltage Drop due to starting of Large Motor
- Calculate Voltage drop in Transformer ,1000KVA,11/0.480KV,impedance 5.75%, due to starting of 300KW,460V,0.8 Power Factor, Motor code D(kva/hp).Motor Start 2 times per Hour and The allowable Voltage drop at Transformer Secondary terminal is 10%.
Motor current / Torque:
- Motor Full Load Current= (Kwx1000)/(1.732x Volt (L-L)x P.F)
- Motor Full Load Current=300×1000/1.732x460x0.8= 471 Amp.
- Motor Locked Rotor Current =Multiplier x Motor Full Load Current
|
Locked Rotor Current (Kva/Hp) |
||
|
Motor Code |
Min |
Max |
|
A |
3.15 |
|
|
B |
3.16 |
3.55 |
|
C |
3.56 |
4 |
|
D |
4.1 |
4.5 |
|
E |
4.6 |
5 |
|
F |
5.1 |
5.6 |
|
G |
5.7 |
6.3 |
|
H |
6.4 |
7.1 |
|
J |
7.2 |
8 |
|
K |
8.1 |
9 |
|
L |
9.1 |
10 |
|
M |
10.1 |
11.2 |
|
N |
11.3 |
12.5 |
|
P |
12.6 |
14 |
|
R |
14.1 |
16 |
|
S |
16.1 |
18 |
|
T |
18.1 |
20 |
|
U |
20.1 |
22.4 |
|
V |
22.5 |
|
- Min Motor Locked Rotor Current (L1)=4.10×471=1930 Amp
- Max Motor Locked Rotor Current(L2) =4.50×471=2118 Amp
- Motor inrush Kva at Starting (Irsm)=Volt x locked Rotor Multiplier x Full Load Currentx1.732 / 1000
- Motor inrush Kva at Starting (Irsm)=460 x 4.5x471x1.732 / 1000=1688 Kva
Transformer:
- Transformer Full Load Current= Kva/(1.732xVolt)
- Transformer Full Load Current=1000/(1.732×480)=1203 Amp.
- Short Circuit Current at TC Secondary (Isc) =Transformer Full Load Current / Impedance.
- Short Circuit Current at TC Secondary= 1203/(5.75/100)= 20919 Amp
- Maximum Kva of TC at rated Short Circuit Current (Q1) = (Volt x Iscx1.732)/1000.
- Maximum Kva of TC at rated Short Circuit Current (Q1)=480x20919x1.732/1000= 17391 Kva.
- Voltage Drop at Transformer secondary due to Motor Inrush (Vd)= (Irsm) / Q1
- Voltage Drop at Transformer secondary due to Motor Inrush (Vd) =1688/17391 =10%
- Voltage Drop at Transformer Secondary is 10% which is within permissible Limit.
- Motor Full Load Current<=65% of Transformer Full Load Current
- 471 Amp <=65%x1203 amp = 471 Amp<= 781 Amp
- Here Voltage Drop is within Limit and Motor Full Load Current<=TC Full Load Current.
- Size of Transformer is Adequate.
JIGNESH PARMAR
Sir,
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Pls provide relevant info for buy the unlocked version of earth mat design .
thanks
Nathan.
Electronic Energy meters are based on current pulses and calibrated for standard PF (0.8 PF) and standard voltage so by improving PF we may reduce Electricity bills.
Hi Jignesh bhai,
Thanks for your wondeful notes on electrical systems.
can you please check and clarify the below point.
■Short Circuit Current at TC Secondary= 1203/5.75= 20919 Amp
which you have posted under
Calculate TC Size & Voltage Drop due to starting of Large Motor
Typing Error…S.C Current= Full Load current/%Z =1203/(5.75/100)=20919Amp.
Thanks for highlighting error…
Dear Sir,
As per excel sheet, we have selected 3.5Cx120 Sq.mm Al. Cable for 125Kw motor (DOL)(I,e 415VAC, 50HZ) and its starting current 600%.
And now our cable vendor telling it will not meet , bcz starting current is more.
kindly advise on this.
Regards,
Umesh MM
9986006728
It is totally depend upon
(1) S.C capacity of your system as well as Cable.
(2) allowable starting Voltage Drop
(3) No of run of cable (Cable current capacity and distance)
All above this will decide size of cable
Hi Jignesh,
My query is, generally we check the voltage dip of the largest starting motor only in the motor terminals to ensure that motor starts and its torque overcomes the load torque with a min. possible voltage i.e 85% of rated motor voltage.
Which is more essential to check, whether voltage drop at transformer terminal or voltage dip at the largest motor terminal ?
Pl advice.
Rgds,
Ganesh
Hi Jignesh ,
Motor locked current and starting current of the motor both are same . If same than it’s also depends upon the selection of startor of the motor .
I am waiting your reply .
Regards
Shiv
hello sir I would like to know what can be the voltage dip formula for dg sizing in case of starting the largest motor in system?
Good morning sir:
Considering that the operation of the motors will be by using VFDs, but in an abnormal situation I must start a motor directly, in this case it can be accepted that the voltage drop at start, in the secondary of the transformer due to the magnetizing current is more than 10%? What is the maximum voltage drop you recommend?
Regards
Gustavo
why is 65% chosen as basis?
Dear Sir,
Please guide me on transformer sizing. Electric loads as follows:
1. motor loads of 125kW, 155kW, 315kW,
2. Lighting load of 20kW
3. Residential 130kW