Cable Sizing Calculation for Star Delta Starter

[SSKLC]

Hi,
I am trying to size cables from Star Delta Starter to Motor

During Star connection,
Load Current in Cable from SWGR – Motor Term. U1, V1, W1 = 1/3 (33%) of actual FLC
Load Current in Cable from SWGR – Motor Term. U2, V2, W2 = 1/3 (33%) of actual FLC
Starting Current in Cable from SWGR – Motor Term. U1, V1, W1 = 1/3 (33%) of actual Starting Current
Starting Current in Cable from SWGR – Motor Term. U2, V2, W2 = 1/3 (33%) of actual Starting Current

During Delta Connection,
Load Current in Cable from SWGR – Motor Term. U1, V1, W1 = 1/√3 (58%) of FLC
Load Current in Cable from SWGR – Motor Term. U2, V2, W2 = = 1/√3 (58%) of FLC
(Note that reclosing current transient during star to delta transition is ignored as present only for very small duration)

For Star Connection,
Running VD = √3 x (33% of FLC) x (2xL) x (RcosØ + XsinØ) / (No.of runs X V)
Starting VD = √3 x (33% X starting current Mulitplier x FLC) x (2xL) x (RcosØ + XsinØ) / (No.of runs X V)

For Delta Connection,
Running VD = √3 x (58% of FLC) x (2xL) x (RcosØ + XsinØ) / (No.of runs X V)

Based on above, worst case cable size to be selected for cables between Star Delta Starter to Motor.
Please advise if I am missing anything.

 

[waross]

Too much information.

I am trying to size cables from Star Delta Starter to Motor

Six leads, two cables per phase, each cable ampacity 58% of nameplate delta running current.
Correction in bold.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[SSKLC]

Thanks Waross

Yes, I have considered 58% current of nameplate current for cable sizing.
However, I am confused about voltage drop formula in delta...as delta is formed inside MCC.
Whereas, voltage drop is being measured inside delta winding.

 

[waross]

Consider one phase. You have a conductor from the MCC, the phase winding and a conductor returning to the MCC.
These are all in series and there is no other connection to other circuits in the field.q
The current is 58% of nameplate current and the voltage drop on each phase is 58% of full load current times the total impedance of both conductors.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

"..I am confused about voltage drop formula in delta...as delta is formed inside MCC. Whereas, voltage drop is being measured inside delta winding..".
I have the following opinion for your consideration.
1. There are six conductors from the Starter (in the MCC) to the motor. The voltage drop (VD) of each conductor is calculated (impedance of the length of the conductor x current). For small conductors R >> X, and load pf ignored. Therefore VD is about R x I. Attention: Reference BS7671, the data is tabulated in mV/A/m instead of Z/m.
Che Kuan Yau (Singapore)

 

[waross]

The Canadian Electrical Code includes voltage drop tables with the following note:
Inductive reactance has not been included
because it is a function of conductor size and spacing.
The CEC gives current, conductor size in AWG and distance in meters.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

" ..The current is 50% of nameplate current and the voltage drop on each phase is (58% of full load current)squared times the total impedance of both conductors. Example. Full load current = 100 Amps. Total impedance of conductors = .005 Ohms. Voltage drop = 58 Amps x 58 Amps x .005 Ohms = 17.11 Volts...".
I am confused. Voltage drop = 58 Amps x 58 Amps x .005 Ohms = 17.11 V. does not make sense. 58A x 58A x 0.005 Ohm = 16.82 W. Where I[sup]2[/sup]x R = Power (W), not volt (V).
Che Kuan Yau (Singapore)

 

[waross]

I stand corrected.
Need more coffee.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

"...I am confused about voltage drop formula in delta...as delta is formed inside MCC. Whereas, voltage drop is being measured inside delta winding".
I have the work out for your consideration.
1. Assumption: MCC voltage 400V; Motor rated 400V, full-load (in delta) 100A and conductor route length 20m (from MCC to the Motor). Take VD 3% at the Motor terminals during running at full-load in Delta.
2. MCC voltage 400V. Motor terminal voltage is expected > 0.97 x 400V = 388V. The VD is expected to be < 0.03 x 400V = 12V. There are two conductors/phase from the MCC to the Motor. Each 20m conductor VD is expected < 12V/2 = 6V.
3. Reference BS 7671 PVC Ambient 30 [sup]o[/sup] C, Max temp 70 [sup]o[/sup] C, ... 16 mm[sup]2[/sup] rated 62A, 2.4 mV/A/m*. For 20m, the VD is (2.4* x 58A x 20m)/1000 = 2.784V ; which is < 6V .
3. With this conductor size, it would be in general fine with other installation derating factors and length say slightly longer than 20m.
Che Kuan Yau (Singapore)

 

[SSKLC]

Thanks for detailed response Che !

However, I am still having doubt.

1) The cable from MCC to motor and from Motor back to MCC, both are in series with motor winding before formation of delta in MCC.
Considering this, why voltage drop should get half? Instead 58% current of nameplate delta will flow through both the cables, so, Isn't voltage drop should get double ? Refer Page 11 of attached file.

2) As per your calculation, Voltage drop is calculated I*R. Is this Voltage drop measured as Phase to neutral or Phase to Phase ?
Considering Voltage drop will be mearured as Phase to Phase, Is I*R voltage drop measured is correct otr it should be √3x58%xIxR.

Any advise on above is highly appreciated.

 

[waross]

With the delta formed at the motor, the single conductor from the MCC powers two phases.
Those phases are out of phase with each other.
As a result, the resultant current is less than twice the current in each phase.
And yes,there is a voltage drop in both conductors so your calculations will use the sum of the length of both conductors.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

I may had failed to understand your questions. I try to explain as following, for your consideration.
1. There are two conductors per phase from the MCC to the Motor. Each conductor having 58A. As there are two conductors, the VD per conductor is 1/2 of total.
2. The VD of the conductor is (I x R ). Where R is the resistance of the conductor. It has nothing to do with whether Phase to neutral or Phase to Phase ?
3. The VD between the MCC and the motor is measured between Line to Line. E.g. 400V at MCC, 388V at the motor terminals. That is VD of 3%.
Che Kuan Yau (Singapore)

 

[SSKLC]

Che,

My question is we are measuring voltage line to line, so, considering 3Phase system, factor of √3 should be applicable.
so, when calculating voltage drop formula should be √3x58%xIxR.
This is where I am confused.

 

[che12345]

"..My question is we are measuring voltage line to line, so, considering 3Phase system, factor of √3 should be applicable. so, when calculating voltage drop formula should be √3x58%xIxR".
1. In the example we are considering:
a) Voltage measured at the MCC , between L1 and L2 = 400V.
b) Voltage measured at the Motor terminal between U1 and U2 = 388V.
c) The conductor Voltage drop between starter thermal O/L terminal 2 and Motor terminal U1 = 6V . The same 6V VD between D contactor terminal 2 and Motor terminal V2.
2. Current
a) in MCC to the MCCB = 100A.
b) from thermal O/L terminal 2 to Motor terminal U1 = 58A.
c) from D contactor terminal 2 to Motor terminal V2 = 58A.
3. Delta configuration for CW rotation.
L1 to Motor U1 and V2
L2 to Motor V1 and W2
l3 to Motor W1 and U2
Che Kuan Yau (Singapore)

 

[waross]

√3 is applicable for three phase circuits.
When the delta is formed at the MCC you effectively have three single phase circuits.
You have used 1/√3 (0.57735) to calculate the current in the individual windings.
You do not use √3 a second time to calculate voltage drop.

Cable Sizing Calculation for Star Delta Starter

I am sure that Mr Che will agree that we have both been remiss in not mentioning a second factor when it comes to cable sizing in this instance;
A star delta arrangement requires 6 conductors from the MCC to the motor.
In many installations, the conductors are contained in a single raceway or cable.
When 4 to 6 current carrying conductors are present and in contact, some codes (CEC, NEC, Others?) call for derating of the ampacity to 80% of listed ampacity.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear all, I repeat my opinion submitted on 18 May.
1. "...3. Reference BS 7671 PVC Ambient 30 o C, Max temp 70 o C, ... 16 mm2 rated 62A, 2.4 mV/A/m*. For 20m, the VD is (2.4* x 58A x 20m)/1000 = 2.784V ; which is < 6V . With this conductor size, it would be in general fine with other installation derating factors and length say slightly longer than 20m..."
2. BTW: 16 mm[sup]2[/sup] = 6 AWG approx.
Che Kuan Yau (Singapore)

 

[waross]

In our example we have based our calculations on 100 Amps full load current.
With the delta formed in the MCC, the full load current in each conductor will be 58 Amps.
Per the CEC, the ampacity must be 125% of the rated current.
125% of 58 Amps = 72.5 Amps ampacity.
If there are two raceways or cables with three conductors in each raceway or cable, the ampacity should be 58 Amps for a full load current of 100 Amps.
If there is one raceway or cable with six conductors, then the ampacity should be 90.63 Amps to allow for 80% derating per North American codes.
The cable ampacity should be based on 75C temperature rated insulation.
This is a code issue and may not be the same world wide.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear All, let us have some fun and coffee.
1. NEC Art 430-22 ".. single motor . General. Branch-circuit conductor that supply a single motor...... shall have an ampacity of not less than 125 percent of the motor's full-load current rating ....".
2. Further down, " For a Wye-start, delta-run connected motors, the selection of branch-circuit conductors on the line-side of the controller shall be based on the motor full-load current. The selection of the conductors between the controller and the motor shall be based on 58 percent of the motor full-load current".
What is your take?
Che Kuan Yau (Singapore)

 

[waross]

2. Further down, " For a Wye-start, delta-run connected motors, the selection of branch-circuit conductors on the line-side of the controller shall be based on the motor full-load current. The selection of the conductors between the controller and the motor shall be based on 58 percent of the motor full-load current".

There is no such rule in the CEC.
In the CEC,the 125% applies to the conductors feeding the motor.
The notes on the rules make this clear.
Under the CEC, the following would apply.
Conductor ampacity at 75C = 100 Amps.
4 to 6 conductors in raceway or cable, derate to 80% The ampacity is now 80 Amps.
Motor supply conductors at 125% of motor full load current. 80 Amps /125% = 64 Amps.
So for a motor full load current of 64 Amps and six conductors in a raceway or cable a cable with an ampacity of 100 Amps is required.
I have no idea as to the code requirements of our original poster.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[che12345]

Dear All. It is great fun and time for more coffee.
The CEC is apparently NOT in harmonization with that of the NEC !? It is a great fun or a laughing matter, to put it mildly.
Che Kuan Yau (Singapore)