AC 3 Phase Symmetrical Components Star Load Fault Questions - Would you like to check my answers?

Discussion in 'Homework Help' started by CasterSkux, Oct 7, 2014.

1. CasterSkux Thread Starter New Member

Oct 7, 2014
2
0
Hello there

I'm doing an Online Electrical Engineering course and we're up to the last component of the course. We're currently doing Asymmetrical 3 phase, because it's online we haven't had much practice with these maths concepts compared to say university.

Here are the questions with my answers:

1. In a 3 phase circuit with resistance earthed and neutral and earth fault in phase “A” which produces a fault current of 800 amps.

Load currents may be taken as zero.

What will be the positive, negative and zero sequence currents in this system?

I have Ia = 800 for Phase A, Ib = 0 and Ic = 0 and I assume this is a Phase-to-earth fault

According to this source https://www.selinc.com/WorkArea/DownloadAsset.aspx?id=100688, when Phase B and Phase C is zero (as suggested by "Load currents may be taken as zero"): the zero,
positive and negative sequence are all 1/3Ia.

Therefore Ia = 800A and I0, I1, I2 = 267A. Is this right?

2. In a three phase 4 wire system with phase to neutral voltage of 230V, a balanced set of resistive loads of 8 ohms are connected between each phase and neutral.

If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?

Because it's a 4 wire system with a neutral, the voltage difference through the resistors will utilize the phase to neutral voltage, and since they're both 8 ohms, the current magnitude will be the same.

Therefore |Ia| = |Ib| = V/R = 230/8 = 28.75 Amps

I turned Phase A and Phase B into their phasor values respectively to:
Ia = 28.75 arg(0)
Ib = 28.75 arg(240)
and because C is blown:
Ic = 0

I've calculated the Asymmetrical components using a calculator (under 012 Polar):
The printscreen is here:

I think Neutral current flows by adding Ia + Ib utilizing Kirchoff's current law, this equates to 28.75 arg(60), but I doubt I use this in my phase diagram.

Is this the right answer/approach?

3. In the circuit above in problem 2, if there is no neutral wire used and Line to Line voltage of 400Volts is given, and connected in Star configuration

Note (only 3 wire system now)
.

If the fuse cartridge of phase “C” blows, what will be the currents in the corresponding phases and the symmetrical components of the load currents?
Please note: (This is the known condition, when a motor is subjected to loss of a phase. Which is also known as “single-phasing or poling”).

Because there is no neutral, the potential difference is 400V, and the star load becomes a series resistance of 8+8 = 16 ohms as Phase C is gone. The current is 400/16 = 25 amps.

Using Kirchhoff's voltage law, we can find the individual voltage drops of each resistor: 25*8 = 200V for both Ia and Ib resistors: this is consistent with the content in the bottom of http://www.allaboutcircuits.com/vol_2/chpt_10/5.html

I don't know the angle direction for each however, and I'm assuming Ic is = 0 because it's blown to begin with. I've been told it's 180degrees apart between Ia and Ib, but don't know how I can see it? anyone to clarify?

I'd really appreciate the feedback, I don't have a strong theoretical Elec Eng background compared to people who has been to University haha

Cheers!

2. CasterSkux Thread Starter New Member

Oct 7, 2014
2
0
Hello again,

I'm happy with my answer for Question 3:
Ia = 25 arg (0)
Ib = 25 arg (180)

3. MrAl Distinguished Member

Jun 17, 2014
2,579
523
Hi,

I did not check your answers, but if you go about it in a general way you should be able to prove your results.

The only difference between the three phase circuit and single phase is that the 'other' two phases have a different phase angle, and that means that instead of using single numbers for the source voltages we have to use complex numbers. Everything else is the same though, so all the rules for single phase circuits will apply just the same. Since we have three voltages it also helps to use superposition, and that should lead to the solution for any problem setup.

For example, with three phases v1,v2,v3, and three resistances R1,R2,R3 with one resistor going to each voltage source and the three tied to a common floating node, the voltages are first converted to complex:
V1=v1*(cos(A1),sin(A1))
V2=v2*(cos(A2),sin(A2))
V3=v3*(cos(A3),sin(A3))

and then we would proceed the same way as if the sources where just DC. For example, the voltage response from source V1 at the floating node would be:
V1*R23/(R23+R1)
where R23 is R2 in parallel with R3, or written out a little more:
v1*(1,0)*R23/(R23+R1)
where (1,0) is (cos(A1),cos(A1)) the complex representation of the unit source.

Similarly, the response for the second source would be:
v2*(-1/2,sqrt(3)/2)*R13/(R13+R2)

and the complex part here came from V2 having the angle A2 of 120 degrees. After including the part for V3 you can then combine the results and that would give you the voltage at the floating node regardless what the voltage amplitudes were or the resistances were, so the system can be balanced or extremely unbalanced it doesnt matter anymore.

Not sure if this helps for what you wanted, but if you do it this way you should at least be able to check your answers.

BTW just to be clear, the two element notation used above:
(-1/2,sqrt(3)/2)

could be written out also as:
-1/2+j*sqrt(3)/2

so for source V2 we get:
v2*(-1/2+j*sqrt(3)/2)*R13/(R13+R2)

Last edited: Oct 8, 2014