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Current and Power calculations for unbalanced loads on a 3 phase system
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Aaron
Thu, 03 Sep 09 13:29:23 +0000
I have 3 resistance heaters. One is rated for 880 watts at 220 VAC and
the other two are each rated for 400 watts at 220 VAC. I am trying to
calculate the theoretical currents for each phase if I wire the heaters
in a delta configuration across a 3 phase supply. Can anybody point help?

Thanks,
Aaron
6 replies folowing
Sean Breheny
Thu, 03 Sep 09 16:34:40 +0000
Hi Aaron,

I am assuming that your three phase system itself is balanced (i.e.,
the magnitude of the voltages on each phase with respect to ground is
the same), but only your load is unbalanced. I am also assuming
negligible cable voltage drop.

Then, the phase to ground voltage on each phase is:
A) Vpg at 0 deg
B) Vpg at 120 deg
C) Vpg at 240 deg

The voltage between phases A and B is (in phasor form):
Vpg*(cos(0 deg)+j*sin(0 deg) - Vpg*(cos(120 deg)+j*sin(120 deg)
which is
(Vpg+j*0)-(-0.5*Vpg+j*Vpg*sqrt(3)/2)=1.5*Vpg-j*Vpg*sqrt(3)/2.
The magnitude of this is
Vpg*sqrt(1.5^2+(sqrt(3)/2)^2)=Vpg*sqrt(2.25+3/4)=Vpg*sqrt(3)
So, Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
Similarly, Vbc=Vpg*(cos(120 deg)+j*sin(120 deg)-Vpg*(cos(240 deg)+j*sin(240 deg)
which is (-0.5*Vpg+j*Vpg*sqrt(3)/2)-(-0.5*Vpg-j*Vpg*sqrt(3)/2)=(0+j*Vpg(sqrt(3))
Vca is then Vpg*(cos(240 deg)+j*sin(240 deg)-Vpg*(cos(0 deg)+j*sin(0 deg)
which is (-0.5*Vpg-j*Vpg*sqrt(3)/2)-(Vpg+j*0)=(-1.5*Vpg-j*Vpg*sqrt(3)/2)

Summarizing:
Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
Vbc=(0+j*Vpg(sqrt(3))
Vca=(-1.5*Vpg-j*Vpg*sqrt(3)/2)

Note that in all three cases the phase to phase voltage magnitude is
Vpg*sqrt(3).

Now, computing the load currents is easy (I am going to assume resistive loads):

Load D is connected from A to B with resistance Rd.
Load E is from B to C with resistance Re.
Load F is from C to A with resistance Rf.

The current out of phase A is Vab/Rd-Vca/Rf.
Current out of phase B is -Vab/Rd+Vbc/Re
Current coming out of phase C is Vca/Rf-Vbc/Re

Ia= (1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd - (-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf
Ib= -(1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd + (0+j*Vpg(sqrt(3))/Re
Ic=(-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf - (0+j*Vpg(sqrt(3))/Re

Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))
Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))
Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))

If the R's are equal, this reduces to:
Ia=Vpg*3/R
Ib=Vpg*(-1.5/R+j*3*sqrt(3)/(2*R))
Ic=Vpg*(-1.5/R-j*3*sqrt(3)/(2*R))

The magnitude of each of these is Vpg*3/R.

Now, for your case. Assuming resistive loads, and assuming you are
using a 208VAC 3-phase system (208VAC phase to phase voltage, 120V
phase to ground):

800Watts at 220VAC is 60.5 ohms
400Watts at 220VAC is 121 ohms.

Say that Rd=60.5, and Re=Rf=121 ohms (the 800W heater is connected
between A and B and the 400W heaters are between B and C and C and A).

Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))=120*(1.5*(1/60.5+1/121)+j*(0.866)*(1/121-1/60.5))
Ia=4.46-j*0.86 Amps Magnitude=4.54 Amps

Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))=120*(-1.5/60.5+j*0.866*(1/60.5+2/121))
Ib= -2.98+j*3.44 Amps Magnitude=4.55 Amps

Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))=120*(-1.5/121-j*0.866*(3/121))
Ic= -1.49-j*2.58 Amps Magnitude=2.98 Amps

Notice that these all sum to zero when added as complex numbers (which
they must as there is no neutral or ground connection).

All of this can be done much more compactly and easily in matrix form,
but I wanted to spell it all out for you in case you were not familiar
with matrix algebra.

Sean


On Thu, Sep 3, 2009 at 9:29 AM, Aaron wrote:

> I have 3 resistance heaters.  One is rated for 880 watts at 220 VAC and
> the other two are each rated for 400 watts at 220 VAC.  I am trying to
> calculate the theoretical currents for each phase if I wire the heaters
> in a delta configuration across a 3 phase supply.  Can anybody point help?
>
> Thanks,
> Aaron
> --
> > > >

dicsEE
Fri, 04 Sep 09 15:38:11 +0000
 
Aaron
Fri, 04 Sep 09 15:38:11 +0000
Thanks Sean!

I am in the process of digesting all that and working through it myself
with the help of some of my old textbooks...

Aaron

Sean Breheny wrote:

> Hi Aaron,
>
> I am assuming that your three phase system itself is balanced (i.e.,
> the magnitude of the voltages on each phase with respect to ground is
> the same), but only your load is unbalanced. I am also assuming
> negligible cable voltage drop.
>
> Then, the phase to ground voltage on each phase is:
> A) Vpg at 0 deg
> B) Vpg at 120 deg
> C) Vpg at 240 deg
>
> The voltage between phases A and B is (in phasor form):
> Vpg*(cos(0 deg)+j*sin(0 deg) - Vpg*(cos(120 deg)+j*sin(120 deg)
> which is
> (Vpg+j*0)-(-0.5*Vpg+j*Vpg*sqrt(3)/2)=1.5*Vpg-j*Vpg*sqrt(3)/2.
> The magnitude of this is
> Vpg*sqrt(1.5^2+(sqrt(3)/2)^2)=Vpg*sqrt(2.25+3/4)=Vpg*sqrt(3)
> So, Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
> Similarly, Vbc=Vpg*(cos(120 deg)+j*sin(120 deg)-Vpg*(cos(240 deg)+j*sin(240 deg)
> which is (-0.5*Vpg+j*Vpg*sqrt(3)/2)-(-0.5*Vpg-j*Vpg*sqrt(3)/2)=(0+j*Vpg(sqrt(3))
> Vca is then Vpg*(cos(240 deg)+j*sin(240 deg)-Vpg*(cos(0 deg)+j*sin(0 deg)
> which is (-0.5*Vpg-j*Vpg*sqrt(3)/2)-(Vpg+j*0)=(-1.5*Vpg-j*Vpg*sqrt(3)/2)
>
> Summarizing:
> Vab=1.5*Vpg-j*Vpg*sqrt(3)/2
> Vbc=(0+j*Vpg(sqrt(3))
> Vca=(-1.5*Vpg-j*Vpg*sqrt(3)/2)
>
> Note that in all three cases the phase to phase voltage magnitude is
> Vpg*sqrt(3).
>
> Now, computing the load currents is easy (I am going to assume resistive loads):
>
> Load D is connected from A to B with resistance Rd.
> Load E is from B to C with resistance Re.
> Load F is from C to A with resistance Rf.
>
> The current out of phase A is Vab/Rd-Vca/Rf.
> Current out of phase B is -Vab/Rd+Vbc/Re
> Current coming out of phase C is Vca/Rf-Vbc/Re
>
> Ia= (1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd - (-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf
> Ib= -(1.5*Vpg-j*Vpg*sqrt(3)/2)/Rd + (0+j*Vpg(sqrt(3))/Re
> Ic=(-1.5*Vpg-j*Vpg*sqrt(3)/2)/Rf - (0+j*Vpg(sqrt(3))/Re
>
> Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))
> Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))
> Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))
>
> If the R's are equal, this reduces to:
> Ia=Vpg*3/R
> Ib=Vpg*(-1.5/R+j*3*sqrt(3)/(2*R))
> Ic=Vpg*(-1.5/R-j*3*sqrt(3)/(2*R))
>
> The magnitude of each of these is Vpg*3/R.
>
> Now, for your case. Assuming resistive loads, and assuming you are
> using a 208VAC 3-phase system (208VAC phase to phase voltage, 120V
> phase to ground):
>
> 800Watts at 220VAC is 60.5 ohms
> 400Watts at 220VAC is 121 ohms.
>
> Say that Rd=60.5, and Re=Rf=121 ohms (the 800W heater is connected
> between A and B and the 400W heaters are between B and C and C and A).
>
> Ia=Vpg*(1.5(1/Rd+1/Rf)+j*(sqrt(3)/2)*(1/Rf-1/Rd))=120*(1.5*(1/60.5+1/121)+j*(0.866)*(1/121-1/60.5))
> Ia=4.46-j*0.86 Amps Magnitude=4.54 Amps
>
> Ib=Vpg*(1.5*(-1/Rd)+j*(sqrt(3)/2)*(1/Rd+2/Re))=120*(-1.5/60.5+j*0.866*(1/60.5+2/121))
> Ib= -2.98+j*3.44 Amps Magnitude=4.55 Amps
>
> Ic=Vpg*(1.5*(-1/Rf)+j*(sqrt(3)/2)*(-1/Rf-2/Re))=120*(-1.5/121-j*0.866*(3/121))
> Ic= -1.49-j*2.58 Amps Magnitude=2.98 Amps
>
> Notice that these all sum to zero when added as complex numbers (which
> they must as there is no neutral or ground connection).
>
> All of this can be done much more compactly and easily in matrix form,
> but I wanted to spell it all out for you in case you were not familiar
> with matrix algebra.
>
> Sean
>
>
> On Thu, Sep 3, 2009 at 9:29 AM, Aaron wrote:
>
>> I have 3 resistance heaters. One is rated for 880 watts at 220 VAC and
>> the other two are each rated for 400 watts at 220 VAC. I am trying to
>> calculate the theoretical currents for each phase if I wire the heaters
>> in a delta configuration across a 3 phase supply. Can anybody point help?
>>
>> Thanks,
>> Aaron
>> --
>> >> >> >>
>>
>
>
Sean Breheny
Fri, 04 Sep 09 15:53:26 +0000
Sure, you're welcome.

I never learned polyphase power in school - I've had to pick it up
along the way. The way I computed this is definitely valid, but it may
not be the usual way to do it. I know that there are rules of thumb
and shortcuts for delta and wye configurations, etc.

To check my answers, I used LTSpice, a free circuit simulator from
Linear Tech. I would recommend that you try using that to help you
figure this stuff out. Nothing beats a quick way to check your
answers! You can get it at:


http://www.linear.com/designtools/software/ltspice.jsp

Sean


On Fri, Sep 4, 2009 at 11:38 AM, Aaron wrote:

> Thanks Sean!
>
> I am in the process of digesting all that and working through it myself
> with the help of some of my old textbooks...
>
> Aaron
>
Dwayne Reid
Fri, 04 Sep 09 16:36:16 +0000
At 07:29 AM 9/3/2009, Aaron wrote:
>I have 3 resistance heaters. One is rated for 880 watts at 220 VAC and
>the other two are each rated for 400 watts at 220 VAC. I am trying to
>calculate the theoretical currents for each phase if I wire the heaters
>in a delta configuration across a 3 phase supply. Can anybody point help?


Easy.

Lets call the 3 phase supply lines A, B, C.

Find the current through each heater when connected across the
supply. Call those currents Iab, Iac, Ibc where Iab is the current
of the load connected from lines A to B.

Note that 220Vac is NOT a standard voltage in North America - its
208Vac. That will reduce the heater current (and output power) from
your rated 220Vac. You are going to have to figure out the
resistance of each heater (Eg. R= (220V)^2 / 800W), then figure out
the current at 208Vac. Or: find the current at 220V, then divide
that by the ratio of 220V / 208V. Whichever is easier for you.

Add Iab to Iac, then divide that result by 2, then multiply that by
SqRoot (3) (about 1.73). That amount is the current in conductor
A. In other words, (Iab + Iac) * 0.866 .
Add Iab to Ibc, then divide that result by 2, then multiply that by
SqRoot (3) (about 1.73). That amount is the current in conductor
B. In other words, (Iab + Ibc) * 0.866 .
Add Iac to Ibc, then divide that result by 2, then multiply that by
SqRoot (3) (about 1.73). That amount is the current in conductor
C. In other words, (Iac + Ibc) * 0.866 .

I hope this helps.

dwayne

Dwayne Reid
Trinity Electronics Systems Ltd Edmonton, AB, CANADA
(780) 489-3199 voice (780) 487-6397 fax
www.trinity-electronics.com
Custom Electronics Design and Manufacturing
Sean Breheny
Fri, 04 Sep 09 18:11:59 +0000
Yes, that works, too. I should have realized in advance when I did my
calculation that the resistors cannot change the phase relationships
so the result can be calculated more easily, without complex numbers.
My method would work for general impedances but that's not needed
here.

As for 220Vac, that is a standard (well, it's more like 230V) for
North America, but that's for single phase power. 208Vac is the
closest for 3-phase. I think the original poster meant that he has
heaters which are intended for 220Vac and he is going to run them from
208. That is a fairly standard thing to do. Most industrial buildings
obtain 120V for normal outlets by connecting them in a wye
configuration with the center connected to neutral and the phases at
208V relative to each other.

Note that the tolerances on three phase power seem (in my experience)
to be wider than typical 120V circuits. I've seen 480V be as high as
510 and as low as 460 (about 6% off). Under normal circumstances, I've
never seen 120V be more than 125V or less than 115V (4% off).

Also, the OP may know this but all power distribution voltages and
currents (residential or industrial) are stated as RMS values. This
means that a 120V circuit has 120V RMS from hot to neutral, which is
actually 170V peak from hot to neutral.

Sean


On Fri, Sep 4, 2009 at 12:36 PM, Dwayne Reid wrote:

> At 07:29 AM 9/3/2009, Aaron wrote:
>>I have 3 resistance heaters.  One is rated for 880 watts at 220 VAC and
>>the other two are each rated for 400 watts at 220 VAC.  I am trying to
>>calculate the theoretical currents for each phase if I wire the heaters
>>in a delta configuration across a 3 phase supply.  Can anybody point help?
>
> Easy.
>
> Lets call the 3 phase supply lines A, B, C.
>
> Find the current through each heater when connected across the
> supply.  Call those currents Iab, Iac, Ibc where Iab is the current
> of the load connected from lines A to B.
>
> Note that 220Vac is NOT a standard voltage in North America - its
> 208Vac.  That will reduce the heater current (and output power) from
> your rated 220Vac.  You are going to have to figure out the
> resistance of each heater (Eg. R= (220V)^2 / 800W), then figure out
> the current at 208Vac.  Or: find the current at 220V, then divide
> that by the ratio of 220V / 208V.  Whichever is easier for you.
>
> Add Iab to Iac, then divide that result by 2, then multiply that by
> SqRoot (3) (about 1.73).  That amount is the current in conductor
> A.  In other words, (Iab + Iac) * 0.866 .
> Add Iab to Ibc, then divide that result by 2, then multiply that by
> SqRoot (3) (about 1.73).  That amount is the current in conductor
> B.  In other words, (Iab + Ibc) * 0.866 .
> Add Iac to Ibc, then divide that result by 2, then multiply that by
> SqRoot (3) (about 1.73).  That amount is the current in conductor
> C.  In other words, (Iac + Ibc) * 0.866 .
>
> I hope this helps.
>
> dwayne
>
> --
> Dwayne Reid  
> Trinity Electronics Systems Ltd    Edmonton, AB, CANADA
> (780) 489-3199 voice          (780) 487-6397 fax
> www.trinity-electronics.com
> Custom Electronics Design and Manufacturing
>
> --
> > > >

Herbert Graf
Fri, 04 Sep 09 19:34:06 +0000
On Fri, 2009-09-04 at 14:11 -0400, Sean Breheny wrote:
> Note that the tolerances on three phase power seem (in my experience)
> to be wider than typical 120V circuits. I've seen 480V be as high as
> 510 and as low as 460 (about 6% off). Under normal circumstances, I've
> never seen 120V be more than 125V or less than 115V (4% off).


The apartment building I grew up in (which was built in the 60s)
supplied each unit with 2 phases of a 3 phase source. Caused an issue
when getting replacement elements for the oven since most are 220V yet
we only had 208V supplied.

Another consequence is our "120V" outlets varied widely, I vividly
remember my meter on one outlet reading 107V and on another (which was
on the second phase) reading 125V.

This became quite an issue when we upgraded our window air conditioners,
the phase with 107V couldn't cut it, the AC when starting up would trip
it's internal overload (or if not that the fuse would eventually blow).

That building was also my only exposure to alu wiring. I didn't like it,
the alu is "tougher" then copper so to bend 14 gauge you had to pretty
much use pliers, copper I can do easily with my fingers.

TTYL