volt drop formula impasse
volt drop formula impasse
(OP)
I hope this doesn't repeat but I'm not sure if my 1st post went through. I had a typo in the 1st one anyway.
I am trying to compare voltage drop calculations and I am going in circles.
I am trying to compare the IEEE exact VD formula (for line-neutral):
VD_exact = Vs + IRcos@ + IXsin@ - sqrt[Vs^2 - (IXcos@ - IRsin@)^2]
with a calculation to get the source and load voltage using vectors for voltage, current, and impedance:
Vs = Vl + IZ
VD = |Vs| - |Vl| = |Vl + Iz| - |Vl|
I can't get the two numbers to match.
example:
given:
R = 0.0204 ohms/kft
X = 0.0390 ohms/kft
single phase length = 100 ft
I = 1500 + j0 amps
so:
R = 0.0204/1000*100 = 0.00204 ohms
X = 0.0390/1000*100 = 0.00390 ohms
let load power factor = 100% => @=0
VD_exact = 120 + 1500*0.00204*1 + 1500*0.0039*0 - sqrt[(120^2 - (1500*0.0039*1 - 1500*0.00204*0)^2] = 123.06 - 119.8573 = 3.2027 volts
Since it is single phase, VD = VD_exact * 2 = 6.405 volts, which is close to a couple of simpler formulas. This gives me Vl = 113.5946 volts
Now use AC circuit anaylsis, and start at the load end. Since I have already said that the current and voltage at the load were in phase, I get:
Vl = 113.5946 + j0 volts
I = 1500 + j0 amps
Z = 2*(0.00204 + j0.00390) ohms
so IZ = 6.12 + j11.7 volts
VD = |113.5946 + j0 + 6.12 + j11.7| - |113.5946 + j0| = 6.69 volts /= 6.405 volts
What in the world am I overlooking?
I am trying to compare voltage drop calculations and I am going in circles.
I am trying to compare the IEEE exact VD formula (for line-neutral):
VD_exact = Vs + IRcos@ + IXsin@ - sqrt[Vs^2 - (IXcos@ - IRsin@)^2]
with a calculation to get the source and load voltage using vectors for voltage, current, and impedance:
Vs = Vl + IZ
VD = |Vs| - |Vl| = |Vl + Iz| - |Vl|
I can't get the two numbers to match.
example:
given:
R = 0.0204 ohms/kft
X = 0.0390 ohms/kft
single phase length = 100 ft
I = 1500 + j0 amps
so:
R = 0.0204/1000*100 = 0.00204 ohms
X = 0.0390/1000*100 = 0.00390 ohms
let load power factor = 100% => @=0
VD_exact = 120 + 1500*0.00204*1 + 1500*0.0039*0 - sqrt[(120^2 - (1500*0.0039*1 - 1500*0.00204*0)^2] = 123.06 - 119.8573 = 3.2027 volts
Since it is single phase, VD = VD_exact * 2 = 6.405 volts, which is close to a couple of simpler formulas. This gives me Vl = 113.5946 volts
Now use AC circuit anaylsis, and start at the load end. Since I have already said that the current and voltage at the load were in phase, I get:
Vl = 113.5946 + j0 volts
I = 1500 + j0 amps
Z = 2*(0.00204 + j0.00390) ohms
so IZ = 6.12 + j11.7 volts
VD = |113.5946 + j0 + 6.12 + j11.7| - |113.5946 + j0| = 6.69 volts /= 6.405 volts
What in the world am I overlooking?





RE: volt drop formula impasse
Also, for an exact solution, bear in mind that (in general), the voltage drop changes the current drawn at the receiving end, which changes the voltage drop which changes the current, etc. So iteration is generally necessary.
"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg
RE: volt drop formula impasse
TTFN
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RE: volt drop formula impasse
Suppose before reaching the source side, the current (with a phase angle of zero) saw a capacitive reactance that was equal to the inductive reactance of the impedance in question. Then we are back to a zero phase angle at the load side and the voltage is in phase with the current.
Neither solution is addressing the change in current draw so I should be able to find the same answer. The exact solution is based on the same V=IZ formula as the other, but I can't seem to see where they are diverging.
RE: volt drop formula impasse
Yet, in your statement of the exact solution, you've set the phase angles to 0 on the supply side, which appears to contradict your statement that you're treating the load side as being in phase.
TTFN
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RE: volt drop formula impasse
But the source voltage and current cannot possibly be in phase if there is any reactance in the circuit impedance (neglecting the special case of a resonant circuit).
You can't just decide that the voltage and current at either end are in phase - that's part of the computation based on the impedance of the circuit and the load power factor.
"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg
RE: volt drop formula impasse
From my perspective, they are close enough.
The approximate expression is a lot easier to calculate, but it is "approximate". There was an IEEE paper published in the 80s or so on this subject and it reported a maximum error between the two methods of 5 or 6%.
RE: volt drop formula impasse
I had the current angle relative to the load voltage, but without the load impedance (or the load voltage angle), I can't find the actual current vector. I think.
The trigonometric exact formula is still handy in that you can find the voltage magnitudes even with one unknown remaining.
thanks for the input
RE: volt drop formula impasse
The resistance R and reactance X are the parameter of 2*500MCM cable[it seems to be].
Second: I don't know which IEEE standard is stated the VD_exact relation but it seems to me it is wrong.
The correct relation is:
VD=-I*R*cos(@)-I*X*sin(@))+SQRT[VS^2-(I*R*sin(@)-I*X*cos(@))^2]
From the attached sketch using a simply trigonometry you'll get as following:
VS^2=OE^2+CE^2
OE=VD+BE
BE=I*R*cos(@)+I*X*sin(@)
CE=CD-DE
CD=I*X*cos(@)
DE=I*R*sin(@)
From here you'll get :
VD^2+2*VD*(I*R*cos(@)+I*X*sin(@))+[I^2*(R^2+X^2)-VS^2]=0
If VD=x 2*b=2**(I*R*cos(@)+I*X*sin(@)) and c=[I^2*(R^2+X^2)-VS^2]
You'll get :
x^2+2*b*x+c=0
x=-b+sqrt(b^2-a*c)
So VD_exact=-I*R*cos(@)-I*X*sin(@))+SQRT[VS^2-(I*R*sin(@)-I*X*cos(@))^2]
If I=1500 A then VD_exact is 113.308 V. But for VD=113.308 I=1416.35 A
As explained dpc you have to reiterate [to put the new I1=1416.35 A in the VD_exact equation and recalculate then I2] and repeat this process until ( In-In+1)/In<.001.
Then you'll get I= 1421.11A and VD_exact= 113.69 V.
The complex calculation may be simplified starting from I calculation:
If VS is the reference VS=120+j0 V
Ztotal=R+Rload+j*X
I=VS/Ztotal
I=120/(0.00408+.08+j*0.0078) I=120*(0.00408+.08-j*0.0078)/((0.00408+.08)^2+0.0078^2)
Iactiv= 120*(0.00408+.08)/((0.00408+.08)^2+0.0078^2)=1415.034 A
Ireactiv= -120*(0.0078)/((0.00408+.08)^2+0.0078^2)=-131.27 A
|I|=SQRT(1415.034^2+131.27^2)= 1421.11 A
VD=VS-Z*I [in complex ,of course]
VD=120-(0.00408+j*.0078)*(1415.034-j*131.27) and you'll get:
VD=113.20-11.57j and |VD|=113.79 approx VD_exact.
RE: volt drop formula impasse
Cheers,
Dave
"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg
RE: volt drop formula impasse
If we assume a lagging power factor, we can write:
I = I cos - j I sin
ES = I (R + j X) + ER
= (I cos - j I sin ) (R + j X) + ER
= I R cos + I X sin + j (I X cos - I R sin ) + ER
Rearranging
ER = ES – ( I R cos + I X sin ) - j (I X cos - I R sin )
Ignoring the imaginary term, you get the approximate calculation, or
ER = ES – ( I R cos + I X sin )
where the drop is the term in parentheses, which agrees with IEEE and most other sources.
If you stay with the exact approach, you get
ER = ES – sqrt[( I R cos + I X sin )2 + (I X cos - I R sin )2]
which is different than what g4932 reported as an exact equation from IEEE. It is also different than what 7anoter4 reports. I think he got a sign wrong.
Hope this helps.
RE: volt drop formula impasse
ER = ES – ( I R cos + I X sin ) - j (I X cos - I R sin )
But if we shall develop the calculation we shall get as following:
ERactiv=ES-( I R cos + I X sin )
ERreact=-(I X cos - I R sin )
ER=SQRT(ES^2-2*ES*( I R cos + I X sin )+( I R cos + I X sin )^2+(I X cos - I R sin )^2)
( I R cos + I X sin )^2=(I R cos )^2+2*I^2*R*X*cos *sin + (I X sin )^2
(I X cos - I R sin )^2=(I X cos )^2-2*I^2*R*X*cos *sin +(I R sin )^2
And finally we got:
ER=SQRT(ES^2-2*ES*( I R cos + I X sin )+ I ^2(R^2 +X^2))
If we shall take the O.P. example:
ES= 120 V R=0.00408 ohm X=0.0078 ohm and I= 1421.11A
ER= 114.73 0.83% error from 113.79 as per last result.
That means these two relations are equivalent. But you have to know I value a priori for the second!
RE: volt drop formula impasse
value is not 0 but 0.092090976 rad [about 5.27 degrees].If you substitute for this one you will get the accurate value.
RE: volt drop formula impasse
RE: volt drop formula impasse