Voltage Drop Calculation for Single Phase

The formulae are:

R = Z * Cos Φ

X = Z * Sin Φ

Therefore Z can only be equal to R÷Cos Φ and X÷Sin Φ

which are obvious from your impedance triangle also.

So, there is something wrong with the formulae you are quoting. Where did you come across them?

General Voltage Drop Formula is V= I*Z

and Voltage drop formula used for Single phase calculation is V = 2*I*(RCosΦ+XSinΦ)*L

L is Length

RCosΦ+XSinΦ must be impedance..... Am i rit???? or is there any mistake in my observation

Welcome to CR4 bvishnukumar89 (somehow your id has reverted to Guest, please mail Admin)

V = I*Z is applicable to all loads, single phase too. i am unable to understand the other formula you quote. Doesn't look right.

If L = Length, then the Resistance of a conductor = ρ * L ÷ A where ρ is resistivity and A is cross-sectional area.

no the actual impedance there is refered as impedance per KM so L factor is used

OOPs.

R*CosΦ= Z*{CosΦ}² and X*SinΦ = Z*{sinΦ}², therefore,

R*CosΦ+X*SinΦ= Z*(SinΦ²+CosΦ²) = Z*1 = Z

Nothing electrical about this

Your complex looking formula is nothing but V = I * Z after all.

1. KVSRIDHAR has given you the algebra.
2. If you prefer geometry, draw a perpendicular (to the Z side of the triangle) to the corner which is the join of the R and X sides.
3. The length from the corner with Φ in it to that point where new perpendicular cuts Z is RcosΦ. N.B. The new triangle has a right angle where perpendicular cuts Z, that being the meaning and purpose of a perpendicular.
4. The length from corner at other end of Z to the new perpendicular is XsinΦ.
5. Their sum is length Z = RcosΦ + XsinΦ.
6. N.B. The angles where the new perpendicular splits the right angle (90 degrees) at the RX corner are 90 -Φ and Φ. A "right angle" triangle has 3 angles which sum to 180 degrees.

Good one. GA to you.

A small modification perhaps to #6?

in all Δs, the three angles total 180°. In a right-triangle, the two angles other than the right-angle therefore total 90°

kvsridhar writing that there was something "not right" about your formula set me thinking....

1. He was right - the formula supposes that you can get R and X values for each leg of the loop of cable (run length L) which comprises parallel "go" and "return"conductors, each L long - then multiply by 2 for the loop voltage drop.
2. National standards for wiring give the volt drops per amp per metre RUN at 50 Hz say, for the very good reason that the loop inductance of two parallel conductors depends very much on their diameter and centre to centre distance. For three phase, symmetry is important, else one gets unequal inductances in each phase. When runs are kilometres, at high current, this can cause trouble due to unbalanced voltage and current.
3. Halving the values for a loop, as if inductance can be isolated [without considering the magnetic field of the return circuit] for one conductor,then doubling the value for a loop is not sound thinking.
4. Even with direct current, the effect of the magnetic field of the return conductor is not zero, though negligible.
   
5. If the formula you gave was from an educational course, please feel free to criticise your teachers for encouraging wrong thinking. Point it out to them!
   

My uneasiness arose from (a) inductance would depend on the coupling, for example the non-inductive shunt in the local test lab is created by winding the two parts in opposition so that the inductance cancels itself. Or the cable may not be a straight run, may get coiled somewhere.

and (b) capacitance would depend on the dielectric, the proximity of the conductors etc..

Anyway, i never had to worry about this in my career, so i am no expert for sure

Thanks for the correction on the angles and the good diagrams. You may not have had to deal with power cabling much, but your fundamental knowledge obviously alerts you when there is a source of error!

I wish you a good day.

For the voltage drop formula, see the section 9.4.3.1 (Voltage drop in short cables), page 210 of the 'Handbook of Electrical Engineering For Practitioners in the Oil, Gas and Petrochemical Industry' by Alan L. Sheldrake. In Figure 9.1, the voltage drop is AC = AB + BC = IR cosØ + IX sinØ = I x (R cosØ + X sinØ).

For single phase, the total voltage drop is 2 times because of the return path and for three phase, it is multiplied by √3.

Single phase voltage drop: 2 x I x (R cosØ + X sinØ) x L

For three phase voltage drop: 1.732 x I x (Rcos Ø + X sinØ) x L

Where,

I - Current through the conductor

R - Resistance (in ohm) of the conductor per KM (or 1000 feet)

L - Length of conductor in KM (or 1000 feet)

X - Reactance (in ohm) of the conductor per KM (or 1000 feet)

Ø – Load power factor angle

Reference thread:

http://cr4.globalspec.com/thread/51446/voltage-drop

- MS

A GA as usual msamad, excellent link, thanks.

Hope to get a little more knowledge from you on this.

1. Why is the expression (R*Cosφ+X*Sinφ) used in place of Z ?

2. Is X only X_L or X_C or a combination of both ?

3. What do cable manufacturers give in their catalogues ... L and C per some length ?

Would appreciate if you could clarify, thanks.

Thank you very much.

This is what CR4 is all about, i have gained unexpected knowledge from what started as an innocous thread by someone else. May your tribe increase, 67model and msamad !

Yes Mr. Sridhar, I always say I learned more from CR4 than what I contributed. It is a great engineering forum for learning from lot of talented people across the world.

- MS