## Voltage drop Equation

## Voltage drop Equation

(OP)

1. In voltage drop calculations the Voltage drop equation is IR Cos (phi) + IX Sin (phi). What is phi here?

Phi - Power factor angle of the load

(or)

Phi - Internal Impedance angle of the transformer (arc tan (X/R)) where X/R is the X/R of the transformer.

2. For example, In cable size adequacy check, Phi is the load power factor angle, not the arc tan (X/R) of the cable. Here why not the cable angle is considered?

3. Is voltage drop at full load due to %Z is same as the voltage regulation? what is the difference between as it appears to be the same? How power factor affects the voltage regulation?

Phi - Power factor angle of the load

(or)

Phi - Internal Impedance angle of the transformer (arc tan (X/R)) where X/R is the X/R of the transformer.

2. For example, In cable size adequacy check, Phi is the load power factor angle, not the arc tan (X/R) of the cable. Here why not the cable angle is considered?

3. Is voltage drop at full load due to %Z is same as the voltage regulation? what is the difference between as it appears to be the same? How power factor affects the voltage regulation?

## RE: Voltage drop Equation

2. I am guessing the cables are close enough together that their positive sequence reactance is lower and I think you are being conservative if you just assume the cable voltage drop is inline with the load. Vload = Vsource * (Zload)/(Zload + Zcable). Zload + Zcable is the most when they are in phase

## RE: Voltage drop Equation

At normal loading the load impedance is in series with the transformer impedance. With typical loads this is mostly resistive.

Transformer regulation is the voltage drop at full load with a typical power factor.

The Canadian Electrical Code has voltage drop cables that take the reactance of the cables into account.

Bill

--------------------

"Why not the best?"

Jimmy Carter

## RE: Voltage drop Equation

2. Cable angle is considered in the R and X.

3. Voltage regulation is a particular form for expressing voltage drop for a transformer. Regulation is defined as the change in output (secondary) voltage which occurs when the load is

reduced from rated kVA to zero, with the applied (primary) voltage maintained constant and is expressed as a percentage of the full-load secondary voltage.

Reg (pu) = R⋅cos(phi) + X⋅sin(phi) + [X⋅cos(phi) − R⋅sin(phi)]²/2, where R and X are per unit resistance and reactance. Reg % is 100 x Reg (pu).

## RE: Voltage drop Equation

http://www.electrician2.com/calculators/vd_calcula...

## RE: Voltage drop Equation

The resistive voltage drop that occurs due to cable resistance is IR Cos (phi) Where Phi is cable impedance angle and the reactance voltage drop that occurs due to the cable reactance is IX cos (phi). My question is why this phi is excluded in the voltage drop equation instead only load power factor angle(theta) is considered. Also, this phi (cable impedance angle) is independent of load.

Does not including cable impedance angle in the voltage drop calculation means it is less accurate?

I'm not questioning the well-established equation but it is rather my inability to understand why only load power factor (theta) is considered and why not also phi?

## RE: Voltage drop Equation

If we should measure the power factor at high voltage terminals the R and X of the transformer would be taken into consideration. However, in across transformer voltage drop calculation only load power factor is required.

From the attached sketch the voltage drop will be:

DV=|Vp|-|V's| it is the subtraction from absolute values.

Vp=DVp+DVs+V's=Zp*Ip+Z's*I's+V's

V's=Vload*wp/ws where wp=number of turns of the high voltage and ws=number of turns of the low voltage winding.[ wp/ws=~VHV/VLV transformer rated voltages].

I's=I'load [I'load=Iload*ws/wp] =Iload*[(cos(φ)-jsin(φ)]*Ns/Np Ip=Io+I's Zp=Rp+jXp Z’s=R’s+jX’s

[Since Io depends on Vp in the short-circuit case Vp~0 then Io~0. Zsc=[(Rp+Rs')+j(Xp+Xs')]

However, even in the case Vp=Vrated we may neglect Io as it is less than 3-5%.

An "accurate" calculation taking Io into account will increase the voltage drop by 1.5-2% only.]

So Ip=I's=I'load==Iload*[(cos(φ)-jsin(φ)]*Ns/Np=V's/(R'load+jX'load).Q.E.D.

## RE: Voltage drop Equation

## RE: Voltage drop Equation

In my opinion, the associated angles of the power factor of the load and the power factor of the cable are taking in consideration for the voltage drop calculation as follow:

1) The power factor of the load drive the angle of the current with respect voltage at the load.2) The Voltage drop is also equal to the cable resistive voltage drop

(same direction as the current)plus the voltage drop of the cable inductance (90^{o}with respect to the current).Hope this help.

## RE: Voltage drop Equation

What it is odd in IEEE 141/1993 formula it is the fact we need to know Es and cos(fi).

Actually you may know Es and cos(alpha)[a] or Er and cos(fi)[ [of the actual load].

In case you know Es and cos(alpha) then:

Er^2=OG^2+FG^2

OG=Es-AE-AG

AE=I*R*cos(a); AG=I*X*sin(a)

FG=I*X*cos(a)-I*R*sin(a)

Er^2=[Es-I*R*cos(a)-I*X*sin(a)]^2+[I*X*cos(a)-I*R*sin(a)]^2

Er=sqrt([Es-I*R*cos(a)-I*X*sin(a)]^2+[I*X*cos(a)-I*R*sin(a)]^2) and VD=Es-Er.

VD=Es-sqrt([Es-I*R*cos(a)-I*X*sin(a)]^2+[I*X*cos(a)-I*R*sin(a)]^2)

In case you know Er and cos(fi) then:

Es=sqrt{[Er+I*R*cos(fi)+I*X*sin(fi)]^2+[I*X*cos(fi)-I*R*sin(fi)]^2} and VD=Es-Er.

In order to get alpha[a] if you know Er and cos(fi) we use the formula:

a=asin((Er*sin(fi)+I*X)/Es)

We get the same VD in both cases.