volt drop formula impasse 1

[g3942]

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?

 

[dpc]

Unless your circuit impedance Z is purely resistive, the voltage and current cannot be in phase.

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

 

[IRstuff]

And, you apparently have a reactance X, specified, so V and I can't be in phase any more, particularly since R and X are close in value.

TTFN

FAQ731-376

 

[g3942]

Well it certainly can't be in phase on both sides of the impedance, but can be in phase on either the source or load side. I just picked the load side.

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.

 

[IRstuff]

But, what you say, and what you're doing appear to be two different things. Your OP states that you're calculating the problem from different sides, the so-called "exact" equation appears to be from the supply side, while your circuit analysis approach is presumably from the load side.

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

FAQ731-376

 

[dpc]

The current and voltage at the *receiving* end can be in phase if the load is purely resistive. In fact, they have to be.

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

 

[magoo2]

I'm not sure what the real issue is. The approximate equation for voltage drop is after all "approximate" and different than the values calculated by the exact equation and the difference in voltage magnitudes. It seems to be that you're somehow expecting the two approaches to yield identical results. It ain't gonna happen!

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%.

 

[g3942]

ok. I was trying to take the results and force a solution into the complex equation V=IZ, but I did not have all the information I needed. To make it work, I was having to force an angle for the current because I could not legitimately jump from load to source across the conductor impedance.

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

 

[7anoter4]

First of all the load seems to be a simply resistance [a big boiler for instance of 1500*120/1000= 180 kw] . Then the rated voltage of this item may be 120 V. So I=1500 A only if VD=120 V. Rload=120/1500=0.08 ohm
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.

 

[dpc]

A star for 7anoter4 - I didn't check the math, but I'm impressed he got the phasor diagram into his post!

Cheers,

Dave

"The more the universe seems comprehensible, the more it also seems pointless." -- Steven Weinberg

 

[magoo2]

Let me try it from the basic equations.

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.

 

[7anoter4]

I agree with magoo2 up to this point:
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!

 

[7anoter4]

This error is due to the fact that here the current angle is referred to ES instead of ER and the actual
value is not 0 but 0.092090976 rad [about 5.27 degrees].If you substitute for this one you will get the accurate value.

 

[g3942]

The IEEE reference was IEEE Std 141-1993 (Red Book):

 

[7anoter4]

I agree the theoretical voltage drop is actually as indicated in IEEE 141.But I don't understand why one needs this. What do you need is "eR" since you have the load reference to this voltage –rated current and rated lag angle. This formula is valid also for voltage drop across a transformer or other dipole.