Missing phase on a 3-phase heater load

[kpdci]

Hi there!

I've been asked the question as to what would happen if a phase is became disconnected on a wye connected heater load?

My first thought is that if C phase was missing for example, the output of the heater would simply produce a lower output. With the C phase disconnected, would that increase the current on the other two phases since there is no return on the C phase.

Am I thinking about that correctly? I tried looking up a reference on this but had no luck.

 

[gar]

160509-0851 EDT

Think about how to analyze the problem.

Assume a 4 wire wye source connected to a 4 terminal wye load, and the voltage drops to zero on one hot line relative to the neutral, then the only change is that one of three eqaul resistors is receiving no power. Thus, the heater is producing 2./3 of its normal power.

Now change the assumption to the same load being connected to the source, but with no neutral connection. With all three hot lines supplying power the heat power is the same as if the neutral was connected. With only two hot lines powered and the third line open, then the heater looks like two equal resistors in series, and a third resistor dangling unconnected to anything on one end, and connected at the other end to the midpoint of the two functioning resistors.

What is the power dissipation in the heater in this case compared to the fully operational heater. P = 2*R*V^2 where P is in watts, R is the resistance in ohms of each of the 3 resistors in the heater, and V is line-to-line voltage. Power in the properly working heater is P = 3*R*(V/1.732)^2. This ratio is 1/2, or less than if a neutral was present.

When the neutral and one hot line are missing, then the voltage across each resistor is less than when the neutral and two hots are present. As an example in a 208 system we have 104 V in one case and 120 V in the other.

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[kwired]

Gar is correct.

If it were delta connected heater elements and you lost one input line you simply lose 2/3's of the heating as you would only be supplying voltage across one of the three elements, but it would be normal operating voltage.

 

[Tony S]

Delta connected you would loose 1/3 power. One element has full line voltage the remaining two half line voltage.
Star connected it gets more interesting.

 

[Smart $]

Delta connected you would loose 1/3 power. One element has full line voltage the remaining two half line voltage.
...

The latter part of that is correct, but you'd be operating at half power. Half voltage amounts to approximately (~) quarter power. You'd have one element with the 1/3 normal total power and two elements with ~1/12th normal total power.

 

[Tony S]

Sorry it is 1/2 power

 

[kwired]

The latter part of that is correct, but you'd be operating at half power. Half voltage amounts to approximately (~) quarter power. You'd have one element with the 1/3 normal total power and two elements with ~1/12th normal total power.

Had to think about that for a little bit but I agree. Earlier when I said only one side of the delta I was only thinking of opening the opposite corner of the delta, in which case I would have been right if that happened, but if the corner remained intact and the supply conductor were opened further upstream what you described is what would happen.

 

[Tony S]

It’s an old college trick question and I walked in to the trap yet again.

 

[gar]

160510-2130 EDT

Some interesting observations:

Consider two separate three phase resistive loads, one wired delta and the other wye. Internally the resistances are all balanced. Both loads are designed for the same voltage and total power.

Internally the resistors for the delta are 3 times greater in resistance than for the wye connection. This results from the square root of 3 being squared. Thus, 3 is an exact value. As an exercise you can do the derivation.

If any external connection to the neutral terminal is open and if you measure the resistance between any two hot terminals the measured resistance is exactly the same for both the delta and wye configurations.

For the wye it is R + R = 2*R where R is the resistance of one wye resistor.

For delta we have 3*R in parallel with 6*R. This calculates to 3*R*6*R/(3*R + 6*R) = 18*R*R/9*R = 2*R.

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[kpdci]

Sorry for the late response but thank you for your response.

I'm still not seeing the answer to my question where sometimes I need it spelled out.

I realize the output power would be less but my actual question is what would happen to the amperage demand on the two legs still connected to the heater?

Given that each 3-phase leg is connected to a 1000 watt resistor load.

Given that a 3-phase calculation to determine the current on each leg is (I=P/E x 1.73) 3000 watts / 480 volts x 1.732 = 3.6 amps) correct?

If one of the legs opened up then one resistor would be out of the picture.
Now the power feeding the resistor bank is a single phase source across two 1000 watt resistors so the calculation would be (I = P/E) 2000/480 = 4.16 amps correct?

Am I overlooking something hear? Wouldn't more current be drawing back on two legs so the OCPD could feasibly open up, right?

 

[Smart $]

...
Am I overlooking something hear?...

Yes. The nominal condition has 277V across each element. When one leg of the supply is open, you only have 240V across the each of the two elements still heating. They will not be dissipating 1000W at reduced voltage. As the relationship goes, current will be reduced as well. Resistance will be slightly lower also, but not by as large a percentage as voltage and current. To estimate, use nominal resistance to determine current and power

 

[gar]

160513-1714 EDT

kpda:

You are not understanding the problem because you have failed to draw the circuit and analyze it.

From your description I believe you have a load consisting of three equal resistors each of 76.801 ohms wired in a wye configuration. This is a four terminal network. Connected to these four terminals are three hot wires and one neutral.

With a supply voltage of 480 V line-to-line there is a line-to-neutral voltage of about 277.13 V. The power dissipated in of the three resistors is 277.13*277.13/76.801 = 1000 W. The current from one hot line is 277.13/76.801 = 3.6084 A.

Open one hot supply line and the current in each of the other hot lines remains unchanged, and the power dissipated in each of the powered resistors remainds unchanged.

Suppose this same wye load was only connected to a delta source. Meaning no connection to the neutral point of the three resistors. Now open one hot line and the load between the two connected hot lines is 2*76.801 = 153.60 ohms. These two resistors have 480 V applied across the pair. Current is the same in each hot line and is 480/153.6 = 3.125 A. Power dissipated is 480*3.125 = 1500 W. Exactly 1/2 of the original 3000 W.

I haven't proofread or checked this.

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[JFletcher]

gar, if the heating element were circular, with the 3 phases wired 120* apart, delta, and you remove one wire, what would be the outcome? say the element is wired to 480V 3Ø, has 48Ω resistance between connection points for 10A draw per phase-- I calculate 4.8kW on each third of the element for 14.4kW.

Now, if I lift one conductor, I would have one part of the element still at 48Ω between connections, the other side of the loop would be 96Ω. Still have 480V line-line. One third the element is still at 10A, and the 2/3 (96Ω part) would draw 5A, for 15A total, yes? Would the two conductors draw 7.5A each vs the 10A if I had all three phases? If so, then would I have 15A x 480V= 7200W, or 50% normal power?

I think that follows what you wrote about winding up with half the power, tho my math seems wrong as theres no accounting for three phase vs single phase power, and iirc, if a 3Ø motor single phases, it runs at 57.7% power.

 

[kwired]

gar, if the heating element were circular, with the 3 phases wired 120* apart, delta, and you remove one wire, what would be the outcome? say the element is wired to 480V 3Ø, has 48Ω resistance between connection points for 10A draw per phase-- I calculate 4.8kW on each third of the element for 14.4kW.

Now, if I lift one conductor, I would have one part of the element still at 48Ω between connections, the other side of the loop would be 96Ω. Still have 480V line-line. One third the element is still at 10A, and the 2/3 (96Ω part) would draw 5A, for 15A total, yes? Would the two conductors draw 7.5A each vs the 10A if I had all three phases? If so, then would I have 15A x 480V= 7200W, or 50% normal power?

I think that follows what you wrote about winding up with half the power, tho my math seems wrong as theres no accounting for three phase vs single phase power, and iirc, if a 3Ø motor single phases, it runs at 57.7% power.

motors are not a fixed resistor and are a different animal.

If the motor is not already running when the "phase" is lost it won't even develop any torque to start, otherwise high torque loads already running will usually stall, high inertia loads will often keep running in a single phase condition but will have other complications if you don't detect the single phasing and shut them down.

 

[gar]

160514-1132 EDT

JFletcher:

if the heating element were circular, with the 3 phases wired 120* apart, delta, and you remove one wire, what would be the outcome? say the element is wired to 480V 3Ø, has 48Ω resistance between connection points for 10A draw per phase-- I calculate 4.8kW on each third of the element for 14.4kW.

I believe you are saying that you have a delta configuration of 3 equal resistors. Each resistor is 48 ohns.

If you apply 3 phase power to this load from a a 480 V line-to-line source, then within each resistor there is a 10 A current flow. That is from 480/48 = 10 A. You are correct on 4800 W dissipated in each of the three resistors. The three phase line current is, however, the vector sum of two of these currents, or 17.32 A. Note that 277.14 V * 17.32 A = 4800 W.

When you open one three phase supply line, then your analysis is correct that the line current is 10 + 5 = 15 A. That same current flows in both wires, not half of the current in each wire. You have one single closed path. The current anywhere in that circuit must be the same. Total power is now 15*480 = 7200 W or 1/2 of the fully powered heater.

kwired gave you an adequate comment on the motor. You need a definition from someone for the assumptions to come up with a 57.7% value.

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[GoldDigger]

Well kpdci, what you are missing is that with the supply voltages changed you may not have 1000W individual elements anymore.
In the wye you simply have two 1000W elements fed instead of three. Exact the same current as originally in the two connected wires.
In the delta you have a 1000W element in parallel with two 250W elements in series. The resistance of the element stays, for all purposes, unchanged. The wattage does not.
Also keep in mind that once you go unbalanced the current in the two lines feeding a delta load is no longer in phase with the line to neutral voltage.

If you feed a wye load from the ungrounded conductors only and leave the center point open the current drawn from two hots only will not be same as when you have the neutral connected.

 

[domnic]

HELP Three phase heater

HELP Three phase heater

Could someone please draw a diagram of a three phase delta and a wye heater with one element open say 480 v 3000 watt . amps and watts and ohms with all good elements good and with one element open. THANK YOU.

 

[GoldDigger]

Could someone please draw a diagram of a three phase delta and a wye heater with one element open say 480 v 3000 watt . amps and watts and ohms with all good elements good and with one element open. THANK YOU.

And there we have the root of some of the confusion. From the start most have interpreted the question to apply to a good set of resistors and opening one of the three ungrounded conductors feeding it.
You are now introducing the question of opening one of the load resistors while all three lines, and the neutral if used, are intact.

 

[gar]

160515-1358 EDT

If one resistor (element) of a delta configuration heater opens, and all three or four supply lines are operational as expected, then it is quite obvious that the load power is reduced to 2/3 of a fully working heater.

For any or all of the combinations discussed the reader should be able to easily draw their own diagram and see what voltages are across what resistors. Therefore be able to determine the power dissipation by themselves.

.

 

[domnic]

heater

heater

160515-1358 EDT

If one resistor (element) of a delta configuration heater opens, and all three or four supply lines are operational as expected, then it is quite obvious that the load power is reduced to 2/3 of a fully working heater.

For any or all of the combinations discussed the reader should be able to easily draw their own diagram and see what voltages are across what resistors. Therefore be able to determine the power dissipation by themselves.

.

How do you have four supply lines to a delta configuration ?