0
\$\begingroup\$

I have several resistive heating element and I want to draw the maximum power from a 400 V, 3 phase, 16 A socket.

If I design the heating elements to connect between phase and null = 230 V RMS * 16 A, I get 3 * 3680 W = 11 kW max Power out of the socket.

If I design the heating elements to connect between two phases I get 400 V RMS, but I can draw less than 16 A * 400 V. How much less?

Is it 12 A per element between two phases?

In which case I can draw 14,4 kW out of the socket.

\$\endgroup\$
  • \$\begingroup\$ Can you explain why you can draw less? \$\endgroup\$ – Marcus Müller Jan 15 '17 at 17:37
1
\$\begingroup\$

You are absolutely right that you can draw less current through each resistor when connecting phase to phase. The actual current for each resistive element will be 9.23 A not the 11 A you calculated.

The current flowing in each line is limited by the wiring for each phase, which is rated at 16 A continuous. No matter how you connect the load resistances, there can only be a 16 A line current in each phase wiring run.
You have two choices ...Delta or Wye connected loads.

enter image description here

In the case of Delta connected circuit the line current (that flowing through the wiring) is 1.732 * the individual resistor current when at least two resistive loads are connected to that line. The converse being with a 16 A line current limit, you can only get 16/1.732 --> 9.23 A flowing in each resistor. This limits the total W to 3.693 kW per resistor and 11 kW total.

In the case of Wye connected load resistors the line current is limited by the wiring to 16 A, the voltage across each resistor becomes 230 V --> 3680 W so total power is 11 kW.

Net result is that you can only dissipate the same total watts in either Delta or Wye connected loads.

\$\endgroup\$
0
\$\begingroup\$

So this is what your run-off-the-mill 3-phase heater looks like

schematic

simulate this circuit – Schematic created using CircuitLab

$$R_1=R_2=R_3=R$$

Notice that at the center point, your neutral wire could be connected — but due to the 120° phase offset summing to 360°, that point is at a constant 0V, anyways, so this is, strictly speaking, optional¹.

Now, apply the Y-Δ transform:

schematic

simulate this circuit

For symmetry reasons, we can deduct that \$R_{12}=R_{23}=R_{31}=R_\Lambda\$.

$$R_\Lambda = \frac{RR+RR+RR}{R}= \frac{3R^2}{R} = 3R$$

The important part here is that *we can't tell whether it's Y or Δ from the outside**! In other words, if your circuit breaker trips for \$R\$ in Y-configuration, it'll trip for \$3R\$ in Δ-configuration.

Since the power going into the two circuits can't be different, it's impossible that you find one to give you more heating than the other.


¹ I was actually at a friend's place where one of the three phases in her multi-apartment building was dead – so she had no light in the kitchen. Unless she turned on the heater – which led to the ungrounded center point of the Y-configured heater getting shifted, and thus, the dead P3 getting seeing enough voltage to turn on an incandescent light bulb dimly. Scary stuff.

\$\endgroup\$
0
\$\begingroup\$

Well the fuse is 16A. This is for each wire/phase in the supply. So if a phase current exceeds 16A for a period of time it will trip.

What power can you draw from that? It doesn't depend on your wiring, it can be calculated as a wye or delta but the result is the same: 11kW.

If you calculate it as a wye/star, you have 230VAC of voltage and 16A of current times x3. P = 230V * 16A * 3 = 11.0kW

If you calculate it as delta, you have 400VAC of voltage and line-to-line current of I_ll=16A * sqrt(3) = 16A * 1.732 = 27.713A. P = 400V * 16A * sqrt(3) = 11.0kW.

These are for purely resistive loads. If you have a phase-shift between current and voltage you will have a reactive part aswell. The fuse are tripping on the sum of the resistive and reactive current, so having a reactive part (lagging or leading) will reduce the amount of real power that can be delivered. The phase shift is cos(phi) and is often between 0.7-1.0 lagging, depending on the application.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.