Star Delta

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Star Delta Calculation

Eng / Ahmed Ramadan Page(1) of (2)

I have a question about star delta calculation. I will try and write all the information I have related to my question along with my calculations. But before I proceed I will like to make it clear that I will not be doing any installation or designing and this is just primarily for my own understanding. Main Supply Voltage = 400V/3Phase/50Hz Motor Power = 18.5KW (Condenser Pump) (Continuous Duty) Ambient Temperature = 40C Starting Method = Star Delta Cable Installation Method = Conduit Distance from the panel to starter = 1meter Distance from the starter to motor = 20meter I want to calculate the following, a. MCCB b. Cable size from main supply panel to starter and from starter to motor c. Size of Main contactor, Delta Contactor and Star Contactor d. Size of overload My calculation Full Load Current = 18.5KW/1.732*400*0.85*0.94 (0.85 = p.f and 0.94 = efficiency) FLC = 33.42A (FLC I have obtained from ABB Motor Catalogue is 34A) MCCB = 1.5*FLC = 1.5*33.42A = 50.13A (I am confused on how to select the best MCCB rating. I mean like NEC states, if I am right to select MCCB at 2.5*FLC. What about IEC or BS7671?) I am trying to follow the BS7671 Cable size From Panel to starter. Cable size = 16mm (Table 4D2A 16mm cable ampere rating = 62A) Applying derating factor at 40C we have 62*0.87 = 53.94A This is sufficient in my opinion. Keeping in mind I have a 50A MCCB installed. Voltage drop Starting current in star delta = 3*FLC = 3*33.42 = 100.26

Star Delta Calculation

Eng / Ahmed Ramadan Page(2) of (2)

Vd (Running) = 2.4*1*33.42/1000 = 0.08V (OK) Vd (Starting) = 2.4*1*100.26/1000 = 0.24V (OK) Cable Size from starter to motor I will have 6 cables going from the starter to motor. Each set of cable need to be rated at 58% of the FLC. Therefore Cable size = 0.58*33.42 = 19.38A ( From table 4D2A I have selected 4mm which has a 27A rated capacity) Applying derating factor at 40C = 23.49A which is good Voltage drop Vd = 9.5*20*33.42/1000=3.68V (Which is good) Main Contactor = 0.58*FLC = 0.58*33.42 = 19.38 Delta contactor = 0.58*FLC = 0.58*33.42 = 19.38 Star Contactor = 0.33*FLC = 0.33*33.42 = 11.02A Overload = 0.58*FLC = 0.58*33.42 = 19.38A I know there are a lot of calculations but can someone check them for me. Are they right? If not can you please highlight my mistakes and if its totally wrong can you please explain the calculation or point me to a good literature. Many thanks Regards, NEC 430.52 mentions that the maximum rating of the Inverse time breaker should be 250%. So you can choose a breaker rating whose rating is less than 250% and more than 100% of the motor rated current.The breaker is there to provide protection against Short circuits.To protect the motor from over load Over load relays are provided The breaker selected should be able to withstand the current developed during the change over from star to delta which is about 300 % for a few seconds. It is better to go through the selection charts of a good manufacturers in Switching devices as they will provide the details for selecting contactors and breakers for different ratings of motors.

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