Calculate Phase Voltage On Unbalanced Load if Neutral Disconnected

[blueboxcat]

Homework Statement

An Unbalanced 3-phase star connected load has the details as below.
Line Voltage=400V
Voltage R-N = 230V 0°
Voltage B-N = 230V +120°
Voltage W-N= 230V -120°

Impedance R-N = 100+J36
Impedance B-N = 102+J0
Impedance W-N = 50+J0


A.) Determine the neutral current(im confident in this)

B.) Determine the voltage between the star point and neutral wire if the neutral wire(Vn-o) becomes disconnected. (Im not sure how to do this)

Homework Equations

The Attempt at a Solution

I calculated neutral current.

converted all impedances to polar to calculate current

Ir=Vph/Zr=230V 0° / 106.28ohm 19.8° = 2.1641A -19.8°
Ib=230V 120° / 102ohm 0° = 2.25A 120°
Ic=230V 240° / 50ohm 240° = 4.6A 240°

Converted these to Rectangular hen Added them together to determine neutral current

Ir=2.16A -19.8°=2.0361-J0.733
Ib=2.25A 120° =-1.125+J1.95
Iy=4.60A 240° =-2.3-J3.98

(2.0361-J0.733)+(-1.125+J1.95)+(-2.3-J3.98) = -1.3889-J2.7631 = 3.09A -116.69°

Now i do not know how to approach question b.) Determine the voltage between the star point and neutral wire if the neutral wire(Vn-o) becomes disconnected.

 

[gneill]

Now i do not know how to approach question b.) Determine the voltage between the star point and neutral wire if the neutral wire(Vn-o) becomes disconnected.

Draw the schematic without the neutral wire. If you take the junction of the voltage sources as the reference node (Neutral), there's one other node and it happens to be the one you're interested in. Write the nodal equation and solve for the potential of that node.

 

[blueboxcat]

Im having trouble understanding your explanation, if i make the star point the reference node, where is this other node? are you able to demonstrate to me how to solve it?

 

[gneill]

Im having trouble understanding your explanation, if i make the star point the reference node, where is this other node? are you able to demonstrate to me how to solve it?

Sorry, we can't provide solutions to homework, only help, hints, clarifications, etc.

You have star-connected sources and star-connected loads. In the 4-wire system the Neutral wire joins the node where the sources come together to the node where the loads come together. Here you are asked to remove that Neutral wire and determine the potential difference between these now separate nodes. Looks like an ideal case for nodal analysis.

Here's your circuit with the neutral wire still in place:

 

[blueboxcat]

Hey, I am not confident in node voltage analysis as I am only in my first month of power engineering. Though i do intend to look into it. I found a similar problem in a book at the library and found the solution, are you able to confirm if this is correct or not?

Vn-o= ((Vr x Yr)+(Vy x Yy)+(Vb x Yb)) / (Yr + Yy + Yb)

Yr = 1/100+j36 = 1/106.28 , 19.8° = 9.41 , -19.8°
Yb = 1/102+J0 = 1/102, 0° = 0.009804 , 0°
Yw = 1/50+ J0 = 1/50 , 0° = 0.02 , 0°

Vn-o = ((230 , 0° x 9.41 , -19.8) + (230 , -120° x 0.02) + (230 , 120° x 0.009804 , 0°)) / ((8.85-J3.19)+(0.009804) + (.02)

Vn-o = (2161.7652 , -19.88°) / (9.4356 , -19.76)

Vn-o = 229.11V -12°

Im doubting whether my answer is correct because it is incredibly close to the supply phase voltage(230v) . But I am thinking this could also be a coincidence. Are you able to tell me if I am wrong and if so, where i have went wrong? Thank you

 

[gneill]

Hey, I am not confident in node voltage analysis as I am only in my first month of power engineering. Though i do intend to look into it. I found a similar problem in a book at the library and found the solution, are you able to confirm if this is correct or not?

Vn-o= ((Vr x Yr)+(Vy x Yy)+(Vb x Yb)) / (Yr + Yy + Yb)

The above formula is correct. Do you understand where it comes from (how to derive it)? You won't have the library at your disposal at exam time...

Yr = 1/100+j36 = 1/106.28 , 19.8° = 9.41 , -19.8°
Yb = 1/102+J0 = 1/102, 0° = 0.009804 , 0°
Yw = 1/50+ J0 = 1/50 , 0° = 0.02 , 0°

Recheck the value I've indicated in red; Looks waaay too big.

Vn-o = ((230 , 0° x 9.41 , -19.8) + (230 , -120° x 0.02) + (230 , 120° x 0.009804 , 0°)) / ((8.85-J3.19)+(0.009804) + (.02)

Vn-o = (2161.7652 , -19.88°) / (9.4356 , -19.76)

Vn-o = 229.11V -12°

Im doubting whether my answer is correct because it is incredibly close to the supply phase voltage(230v) . But I am thinking this could also be a coincidence. Are you able to tell me if I am wrong and if so, where i have went wrong? Thank you

Fix the value I've indicated and redo your math. The result should be less than 100V.

I'd suggest looking up and reviewing the nodal analysis technique. It's usefulness will come up again and again in electrical systems.