IEEE 141 Method for Voltage Drop Derivation

[enterprisenx]

Hello,

Could someone possibly explain how the "Actual Voltage Drop" formula (page 98 of IEEE 141-1993) is derived from the phasor diagram on the previous page?

Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

I realize it does not take into account temperature variances and more detailed evaluation of magnetic fields. However, in a basic voltage drop, I would have approached it differently.

The IEEE equation assumes we know the current. If that is true and we know the value of R and X of the cable based on 1000 ft., then we can calculate voltage drop without even needing to know the source voltage, es.
Zcable = R + X

VdropL-N = I * Zcable
Vdrop_1phase = I * Zcable * 2
Vdrop_3phase = I * Zcable * sqrt(3)

If one uses the longer version the standard provides, I find it more confusing when you assume there is unit power factor (ie: cos(theta)= 1 ...sin(theta)=0)

You will be left with the following equation:
Vdrop = es + IR + sqrt(es^2 - IX)
If you have unit power factor, there should be no reactance (X).

Thank you all for taking the time to explain or further disprove the IEEE 141 method for cable voltage drop.

 

[Smart $]

Hello,

Could someone possibly explain how the "Actual Voltage Drop" formula (page 98 of IEEE 141-1993) is derived from the phasor diagram on the previous page?

Hmmm... kinda hard to see that phasor diagram from here

Any chance you can scan and post it?

The "old" one looked like this...

Note the dimension marked "ERROR". I believe this is the reason for the current formula.

The old formula: VD = IRcosθ + IXsinθ

The old formula error resulted from calculations using triangular methods. The current formula resolves that problem by using a circular method.

 

[enterprisenx]

IEEE 141 Method for Voltage Drop Derivation

Hello,
The phasor diagram you linked to is what I am referring to... However, the error component is not really my concern for my questions. I accept there is an error component because there are other cable and environmental characteristics neither equation takes into account.


The simpler equation:
V = I Rcos(theta) + I Xsin(theta) is confusing

Z = sqrt(R^2 + X^2) or
Z = R + jX = R + X i i, j = sqrt(-1)

In Polar Form:
R = Z cos(theta)
X = Z sin(theta)

But the simpler formula is stating V = I * Z where Z = Rcos(theta) + Xsin(theta). But by polar definitions, Z = R/cos(theta) or Z = X/sin(theta).


The more complex equation
Vd = es + IRcos(theta) + IX sin(theta) + sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

seems odd to incorporate the source voltage since that is unneccessary to determine voltage drop since V = IZ. All you need is I and Z.

The voltage drop formula states cos(theta) is load power factor. However, the equations to determine cable voltage drop are basically neglecting the actual load and just looking at the Z of the cable. So in effect, it is just a source and a "resistor" in a simple circuit where the "resistor" is the cable with R and X values.

The longer equation again seems invalid since as mentioned before, if power factor goes to 1 (ie: cos(theta) = 1), you still don't cancel out the reactance which should cancel out with unit power factor.

Thank you for providing further clarification.

 

[wirenut1980]

At unity power factor, theta = 0, and X = 0.

Using your more complex equation, which is different than what you typed:
Vd = es + IRcos(theta) + IX sin(theta) - sqrt(es^2 - (IXcos(theta) - IRsin(theta))^2)

Vd = es + IRcos(0) + IXsin(0) - sqrt(es^2 - I(0)cos(0) - IRsin(0))^2)

Vd = es + IR - es

Vd = IR

 

[jghrist]

The voltage drop is not I?Z because by voltage drop, we mean the difference in voltage magnitudes.
The imaginary part (vertical projection in the phasor diagram) of I?Z is:
VDI = I?X?cos?-I?R?sin?
The real part (horizontal projection in the phasor diagram) of I?Z is:
VDR = I?R?cos?+I?X?sin?
The real part of es is:
esR = sqrt(es?-VDI?) which is the second part of the IEEE-141 equation.
The receiving end voltage is
er = esR-VDR
the voltage drop is
VD = es-er = es - esR + VDR
VD = es - sqrt(es?-VDI?) + I?R?cos?+I?X?sin?
VD = es + I?R?cos?+I?X?sin? - sqrt(es?-(I?X?cos?-I?R?sin?)?)

 

[enterprisenx]

IEEE 141 Method for Voltage Drop Derivation

How is the X (reactance) canceled out in the longer equation if there is unity power factor where theta = 0 deg; cos(theta)= 1 ; sin(theta) = 0.

The VDI component of the sqrt (es - VDI) would still have IX * cos(0) ; IX remaining.

Or do I need to be looking at in such a way X still exists from the cable even if the load power factor is 1?

For example, a cable feeding an incandescent light bulb which the light bulb is expected to have near unit power factor. However, the cable still has its own R and X.

Thanks again for your comments on this.

 

[jghrist]

The X does not cancel out. It is part of the impedance regardless of the load power factor. With unity power factor, the circuit reactance mostly causes a voltage angle change. In the simplified equation, there is no voltage drop caused by I?X because the magnitude of es is assumed to be equal to the real part of es. If there were no circuit resistance, then es, er, and I?X vectors would form a right triangle.

 

[enterprisenx]

IEEE 141 Method for Voltage Drop Derivation

Good - That's what I was concluding.

Can you tell me where / why I am going wrong below....

Le't say the following is given:
cos(theta) = 0.8 ; sin(theta) = 0.6
i = sqrt(-1)
I = 1.86 A (current through cable)
es = 120 V
Resistance of Cable with length taken into account is 0.1729 ohms
Reactance of Cable with length taken into account is 0.0181 ohms

Proceeding...

I = 1.86 (0.8 + 0.6 * i) = 1.488 + 1.116 * i

Z of cable = R + X * i = 0.1729 + 0.0181 * i

V = I * Z = (1.488 + 1.116 * i)(0.1729 + 0.0181 * i)

V = 0.2370756 + 0.2198892 * i

abs(V) = 0.323 V

V (single phase) = 0.323 * 2 = 0.646 V <----<---

When I use the longer IEEE method, I get V single phase 0.555 V.

Thanks again as it is appreciated. I really want to understand this.

 

[jghrist]

I = 1.86 (0.8 + 0.6 * i) = 1.488 + 1.116 * i

Should be I = 1.86 (0.8 - 0.6 * i) = 1.488 - 1.116 * i

V = I * Z = (1.488 + 1.116 * i)(0.1729 + 0.0181 * i)
V = 0.2370756 + 0.2198892 * i
abs(V) = 0.323 V

Should be
V = I * Z = (1.488 - 1.116 * i)(0.1729 + 0.0181 * i)
V = 0.27747 - 0.16602 * i
abs(V) = 0.323 V

This is the magnitude of the vector voltage between es and er, but it is not the difference between the magnitude of es and the magnitude of er.

 

[enterprisenx]

IEEE 141 Method for Voltage Drop Derivation

Thank you so much for taking the time to explain this. I think I have a better handle on it now.

Based on what I calculated, is there a purpose or example of wanting to know the magnitude of the vector "es,er" since it is not the voltage drop?

 

[jghrist]

Thank you so much for taking the time to explain this. I think I have a better handle on it now.

Based on what I calculated, is there a purpose or example of wanting to know the magnitude of the vector "es,er" since it is not the voltage drop?

Not that I know of, except maybe to better understand the problem.