Motor connections and currents

[draffy]

Can anyone help with the following questions. We use a 400V 3-phase power system.

Can anyone please give me the formula for calculating the line currents taken from the starter for a 3-phase motor if the same motor is connected in
(a) star/wye connection
(b) delta connection

The common formula is P = Root 3 x V x I x P.F. x Efficiency
But is that for a star or delta connection?

There are tables of typical motor currents available.
Are those currents for star or delta? e.g. 15 kW 400V 28A, 1.1kW 400V 2.5A

I have an ABB motor with 2.2 kW 380-440V star/wye 4.9A on the nameplate.
But the test sheet on the ABB web page says 400V delta 4.85A 2.2 kW; 400V star 6.3A 3kW
The nameplate says star - the test sheet says delta. How do I connect it?
I contacted ABB but am waiting for a reply.

I have seen statements that "small" motors are connected in star.
One said less than 4kW, another said less than 11kW.
Does this vary with manufacturers?
What happens if a small motor is connected in delta?

 

[aolalde]

draffy:

The internal connection of an induction three phase motor is defined by the manufacturer to fit the required magnetic flux at the voltage and frequency selected or for the starting type required.
Motors are built wye or delta connected, large or small.

The formula you gave applies to any three phase connection, but you should know what is the motor condition; at full load, partial load or starting.

I line = P*1000/(1.732*Vll*Eff*PF)

Note that P/ (Eff*PF) = kVA

Were:

I line = Current in amperes flowing on each line feeding the motor.
P= Power in kilowatts delivered by the motor. If the load changes, the power consumption changes too.
1.732 = the squared root of 3.
V ll = Line to line voltage in Volts.
Eff = Efficiency of the motor at the specific condition of load ( this characteristic changes with every motor changing condition)
PF = Power Factor of the motor at the specific condition of load ( this characteristic changes with every motor changing condition)
kVA = Total electric power in kiloVoltAmperes.



The starting condition is very special since the PF is regularly very low and at zero speed the power delivered is zero too, then all power consumed are kind of losses. When the motor starts turning its shaft, the impedance changes and the current gradually decays. The current calculated practically at this condition is the locked rotor current or inrush current.
The manufacturer provides the locked rotor kVAper kW (or kVAper HP) for each motor.
You could assume kVAlr = 8*kW (kVAlr = 6*HP)
were: kVAlr = kVA at inrush
kW (HP) = Nameplate nominal motor power.

Calculate; I lr = kVAlr *1000/(1.732*Vll)

Note that for a “wye start delta run” the motor winding is connected at start condition for higher voltage ( 1.732*Vplate) (wye) and only nameplate voltage is applied ( reduced voltage), then the connection is modified to the 100% Voltage ( delta) for continuous operation.

For the motor of your example (15 kW, 400 Volts, 28 amps, and assuming it allows wye-delta connection)

Ilr = 15*8*1000/(1.732*400)= 173.2 amperes (delta connected with 400 Volts at start)

If this motor is wye connected the nominal voltage becomes 400*1.732 = 692.8

Starting wye connected and applying 692.8 volts.

Ilr = 15*8*1000/(1.732*692.8)= 100 amperes

Since you will apply a reduced voltage ( 400 Volts) the current will be reduced proportional to the voltage drop

I lr = 100 * 400/692.8 = 57.73 Amperes.

The motor drives 173.2 amperes if started delta connected at 400 volts, when wye connected starts with 57.73 amperes at 400 volts line to line.

 

[jbartos]

Suggestions to draffy (Electrical) Feb 13, 2004 marked ///\\Can anyone help with the following questions. We use a 400V 3-phase power system.
///o.k.\Can anyone please give me the formula for calculating the line currents taken from the starter for a 3-phase motor if the same motor is connected in
(a) star/wye connection
///Iy=Pwatt,y/(sqrt3 x Vline-line x eff x pf)\\(b) delta connection
///Id=Pwatt,d/(sqrt3 x Vline-line x eff x pf)\\The common formula is P = Root 3 x V x I x P.F. x Efficiency
But is that for a star or delta connection?
///Yes, if properly applied. Specifically,
y-motor connection:
Pwatt,y=sqrt3 x Vline-line x Iline,y x pf x eff
d-motor connection:
Pwatt,d=sqrt3 x Vline-line x Iline,d x pf x eff
The difference is in Iline,y and Iline,d
Iline,y=(1/3) x Iline,d
since
the voltage across the y-connected motor winding will be Vline-neutral=(1/sqrt3) x Vline-line
and the current for d-connected motor winding will be
Id=Iline,d/sqrt3
Considering that the motor winding impedance Z is constant from y-connection to d-connection, then:
Ilind,d=sqrt3 x Id
Iline,y=Vline-neutral/Z
Id=Vline-line/Z=sqrt3 x Vline-neutral/Z=sqrt3 x Iline,y
hence,

Iline,d=sqrt3 x Id=sqrt3 x sqrt3 x Iline,y = 3 x Iline,y

Therefore,
Pwatt,d=3 x Pwatt,y
\\
There are tables of typical motor currents available.
///Usually, a manufacturer catalog provides tables with specific motor currents. The generalization is usually obtained by motors application engineers or designers.\\Are those currents for star or delta?
///This has to be properly stated in terms of the motor winding to avoid any ambiguity\\ e.g. 15 kW 400V 28A,
///This can be for d-connected motor or y-connected motor. The motor connection must be known, if needed, to avoid ambiguity.\\ 1.1kW 400V 2.5A
///The motor connection must be known, if needed, to avoid ambiguity.\\I have an ABB motor with 2.2 kW 380-440V star/wye 4.9A on the nameplate.
///This motor appear to be suitable for the motor terminal voltage range 380V to 440V and it is star/wye connected.\\But the test sheet on the ABB web page says 400V delta 4.85A 2.2 kW; 400V star 6.3A 3kW
///This is an inconsistent rating for the motor since the delta connection at 400V shows the smaller current, 4.85A with respect to the y-connection at 400V current, 6.3A.\\The nameplate says star - the test sheet says delta. How do I connect it?
///If the test sheet does not accompany the motor and overrides the nameplate data, then the motor nameplate information governs.\\I contacted ABB but am waiting for a reply.
///The confirmation from ABB is recommended because of the posted ambiguity.\\I have seen statements that "small" motors are connected in star.
///There are many statements as this one. The y-connected motor with three leads in the terminal box is someone simpler to produce, has smaller voltage stresses etc.\\One said less than 4kW, another said less than 11kW.
///This can be.\\Does this vary with manufacturers?
///Probably.\\What happens if a small motor is connected in delta?
///If provisions are made in the motor terminal box, it is understood that the manufacturer provided the provisions for y-connection and delta-connection. However, the y-connected motor should then have shaft load reduced to 1/3 of delta connection. However, the motor voltage rating is important. It is not possible to connect y-connection Vline-neutral voltage, e.g. designed for 220V, motor to 380V Vline-line motor since the motor winding would burn. Remember that the motor winding impedance is constant equal to Z.\\

 

[draffy]

Many thanks to jbartos

Your explanation was very good - just what I wanted.
You cleared up several items
- star is 1/3 delta current (for the same motor)
- a motor power rating is for either star or delta, not both
- the formula I gave applies to either star or delta

I work for consulting engineers and I have to predict motor full load currents in order to size cables before the motor is bought - and hence before I know whether it is star or delta.

Now I know that it doesn't matter. The manufacturer will make a motor with the typical power ratings and currents given in the tables - he will state on the nameplate which way it is to be connected to give those values, and maybe even have it already connected that way in the terminal box.

I have learned something today. Thanks.