Measuring a Delta connected load with a wattmeter

[Pitt123]

How are the calculations done in a digital power meter to calculate a delta connected load?

I know for a wye connected load there is a CT and PT on each phase as well as a PT lead on ground/neutral and it simply takes the L-N voltage on one phase multiples it by the current on each phase and then adds up the three phases for the total cpower (assuming pf is 1 here)

For a delta connected load however does the powermeter somehow use the L-L voltages since there really isn't a L-N voltage on a delta system? Or is a L-N voltage on each phase calculated from each L-L voltage and the calcualted L-N voltage used in the calculation?

 

[erickench]

You can connect the PTs L-L and then the CT primaries to that same L-L. It would have to be correct. The product of the L-L voltage and L-L current would give you the VAs for that phase. I do not think the L-N voltage is calculated from the L-L voltage. The meter merely measures the voltage and current L-L just as it would L-N.

 

[david luchini]

Usually on a delta system you would have the PTs measure the line to line voltage, and use 2 CT to measure 2 of the line currents. Blondell's Theorem is used to get the 3 phase power (P=Vab*Ia + Vcb*Ic).

 

[Pitt123]

You can connect the PTs L-L and then the CT primaries to that same L-L. It would have to be correct. The product of the L-L voltage and L-L current would give you the VAs for that phase. I do not think the L-N voltage is calculated from the L-L voltage. The meter merely measures the voltage and current L-L just as it would L-N.

What are you refering to when you are talking about L-L currents? Are the line currents before the primary of the delta considered L-L currents, or aren't they simply considered Line currents? Each line current splits inside the delta to the other two phases so how are they considered L-L currents?

 

[Pitt123]

Usually on a delta system you would have the PTs measure the line to line voltage, and use 2 CT to measure 2 of the line currents. Blondell's Theorem is used to get the 3 phase power (P=Vab*Ia + Vcb*Ic).

So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?

 

[T.M.Haja Sahib]

So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?

No matter star or delta,balanced or unbalanced loads with neutral or without neutral,take the three RYB power lines to the loads and connect them to your watt meter per instructions in it.It is calculated in the same way as in two element watt meter.

 

[erickench]

What are you refering to when you are talking about L-L currents? Are the line currents before the primary of the delta considered L-L currents, or aren't they simply considered Line currents? Each line current splits inside the delta to the other two phases so how are they considered L-L currents?

I'm referring to the currents that flow from A to B.

 

[david luchini]

So with a wye connected load we measure line current and L-N current on each phase where as with a delta load only two of the line currents are measured as you described above?

I think you mean L-N voltage above. But yes, in a 4 wire wye system a meter would measure three line currents and three L-N voltages to get the total power: P= Van*Ia + Vbn*Ib + Vcn*Ic.

And in a 3 wire delta system a meter would measure two line currents and the L-L voltages to get the total power: P=Vab*Ia + Vcb*Ic.

 

[Pitt123]

So lets say that the 3-phase meter was set up to measure a wye load however the load being measured was actually a delta load and there were unbalanced currents?

I've seen this before and the current readings were taken so I'm curious how this works in this case?

 

[david luchini]

So lets say that the 3-phase meter was set up to measure a wye load however the load being measured was actually a delta load and there were unbalanced currents?

I've seen this before and the current readings were taken so I'm curious how this works in this case?

It doesn't matter if the load is delta or wye, it matters if the system is 3 wire or 4 wire. In a 3 wire system, you need two current measurements, in a 4 wire system, you need three current measurements.

Imagine if you had a 120/208V wye system, with three wye connected loads (we'll assume resistive loads) so that the load currents were 17.333A, 19.067A, & 20.800A. You can see that the line currents are the same as the load currents. So solving for total power:

P=(17.333*120)+(19.067*120)*(20.800*120)=6864W total.

Now imagine a 120/208V wye system where you had three (resistive) loads of 2080W, 2288W, & 2496W connected in a delta connection. You can see that your load currents would be Iab=10A<30, Ibc=11A<-90 and Ica=12A<-210. The meter, however, is measuring the line current, not the load current, so the meter would see currents of:

Ia=Iab-Ica=19.078<-3.005
Ib=Ibc-Iab=18.193<-118.424
Ic=Ica-Ibc=19.925<-238.563

So the meter multiplies these three line currents times the three line voltages (Van=120<0, Vbn=120<-120, Vcn=120<-240) and sums them to get the total power.

 

[Pitt123]

It doesn't matter if the load is delta or wye, it matters if the system is 3 wire or 4 wire. In a 3 wire system, you need two current measurements, in a 4 wire system, you need three current measurements.

Imagine if you had a 120/208V wye system, with three wye connected loads (we'll assume resistive loads) so that the load currents were 17.333A, 19.067A, & 20.800A. You can see that the line currents are the same as the load currents. So solving for total power:

P=(17.333*120)+(19.067*120)*(20.800*120)=6864W total.

Now imagine a 120/208V wye system where you had three (resistive) loads of 2080W, 2288W, & 2496W connected in a delta connection. You can see that your load currents would be Iab=10A<30, Ibc=11A<-90 and Ica=12A<-210. The meter, however, is measuring the line current, not the load current, so the meter would see currents of:

Ia=Iab-Ica=19.078<-3.005
Ib=Ibc-Iab=18.193<-118.424
Ic=Ica-Ibc=19.925<-238.563

So the meter multiplies these three line currents times the three line voltages (Van=120<0, Vbn=120<-120, Vcn=120<-240) and sums them to get the total power.

Great example! That makes sense now. Thanks

I guess I was thinking that the line currents measured in a delta were actually 1.73x greater than the phase currents inside the delta so this would lead to a higher power reading that wasn't correct but I guess its sort of a tradeoff because although the line currents being measured are 1.73x greater than the phase currents inside the delta the L-N voltage being measured is 1.73x less than the phase voltage being measured inside the delta. So to some degree its a tradeoff with the multiplier.

So with your example above you showed measurment with 3 current measurments although you could have gotten the same results with only 2 current measurments?

 

[Pitt123]

one additional follow up question to above.

I've seen with meters when measuring in wye mode it lets you see the power on each individual line whereas when in the delta mode it only shows the total power. Why is that?

 

[david luchini]

So with your example above you showed measurment with 3 current measurments although you could have gotten the same results with only 2 current measurments?

Don't forget that in the example we have a delta connected load on a wye system. It doesn't matter if the load is wye connected or delta connected, a meter on a wye system will always have 3 current measurements.

one additional follow up question to above.

I've seen with meters when measuring in wye mode it lets you see the power on each individual line whereas when in the delta mode it only shows the total power. Why is that?

That relates to the way the meter measures power per Blondell's Theorem as mentioned before.

For a 4 wire wye system a meter would measure three line currents and three L-N voltages to get the total power: P= Van*Ia + Vbn*Ib + Vcn*Ic. You can see that it measures the power on each line, (Van*Ia, Vbn*Ib, Vcn*Ic) so it can show you the power on each line individually as well as the total power.

But in a 3 wire delta system a meter would measure two line currents and the L-L voltages to get the total power: P=Vab*Ia + Vcb*Ic. As you can see there are only two measurements which are summed together to give you the total power. Therefore, the meter cannot give you the power for each individual line.

 

[david luchini]

So with your example above you showed measurment with 3 current measurments although you could have gotten the same results with only 2 current measurments?

If we had a 3 wire delta system, with the same loads as in the example above, and the meter measured the two line currents Ia=19.078<-3.005 and Ic=19.925<-238.563, and it measured Vab=208<30 and Vbc=208<-90 (therefore Vcb=208<90) then the meter calculates total power as P=Vab*Ia + Vcb+Ic:

P=(208<30)(19.078<-3.005) + (208<90)(19.925<-238.568) = (208)(19.078)(cos(33.005)) + (208)(19.925)(cos(328.568)) =3328+3536 =6864 Watts total. You can see that the two element meter on the delta system gives the same power measurement as the three element meter on the wye system for the same delta connected load.