voltage drop question
- Thread starter tawfiq
- Start date
- Status
Re: voltage drop question
I'm guessing 35 kilometers, and 100 millimeters squared. But I'm not sure how to calculate voltage drop on a line like that. I would assume inductance and capacitance plays a larger part in a line that long.
Steve
I'm guessing 35 kilometers, and 100 millimeters squared. But I'm not sure how to calculate voltage drop on a line like that. I would assume inductance and capacitance plays a larger part in a line that long.
Steve
Re: voltage drop question
Are you saying that you have 35 kilometers of line and 100 square millimeters of cross section?
Copper or aluminum?
The formula for resistance is:
R = Rho x L/A,
where Rho is the resistivity of the wire, L is the length, and A is the cross sectional area.
You must take into account the fact that Rho is typically given in Ohm-Centimeters and apply the various conversion factors.
Ed will have all this stuff handy, I am sure.
Once you have R, it is simply Ohm's law. The voltage has nothing to do with it though.
Are you saying that you have 35 kilometers of line and 100 square millimeters of cross section?
Copper or aluminum?
The formula for resistance is:
R = Rho x L/A,
where Rho is the resistivity of the wire, L is the length, and A is the cross sectional area.
You must take into account the fact that Rho is typically given in Ohm-Centimeters and apply the various conversion factors.
Ed will have all this stuff handy, I am sure.
Once you have R, it is simply Ohm's law. The voltage has nothing to do with it though.
- Location
- Massachusetts
Re: voltage drop question
Most times voltage drop is expressed as a percentage of the nominal system voltage.
60 volts of drop on a 120 volt circuit is 50% voltage drop. 60 volts of drop in a 13.8 Kv circuit is only about 0.004% voltage drop .
I don't know if I would say voltage has nothing to do with it.
Most times voltage drop is expressed as a percentage of the nominal system voltage.
60 volts of drop on a 120 volt circuit is 50% voltage drop. 60 volts of drop in a 13.8 Kv circuit is only about 0.004% voltage drop .
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
My first textbook on Power Systems Analysis (an oldie but still a goodie) tells me that there are three ways to model a transmission line, based on the length. A ?Short Transmission Line? is described as being under 80 kilometers, and that fits your situation. That is lucky for you, as the model for a ?Medium Transmission Line? involves a lumped shunt admittance at each end, and the model for a ?Long Transmission Line? involves differential equations.
The textbook formula for the ?voltage at the receiving end? (Vr), as a function of the ?voltage at the sending end? (Vs), the current (I), and the series impedance (Z) is as follows:
Vr = Vs - IZ
It should come as no surprise that this very much resembles the Ohm?s Law responses given by other Forum members. For longer lines, however, the formula would be quite different.
The tricky part is coming up with ?Z.? The properties given in NEC Table 8 will not serve this purpose. Since you are talking about a transmission line, you may have access to tables that list values for resistance and series impedance. Otherwise, it will be a bit of a chore, involving such things as the ?Geometric Mean Radius? of the individual conductors and the ?Geometric Mean Distances? between conductors.
[ January 14, 2005, 10:18 AM: Message edited by: charlie b ]
My first textbook on Power Systems Analysis (an oldie but still a goodie) tells me that there are three ways to model a transmission line, based on the length. A ?Short Transmission Line? is described as being under 80 kilometers, and that fits your situation. That is lucky for you, as the model for a ?Medium Transmission Line? involves a lumped shunt admittance at each end, and the model for a ?Long Transmission Line? involves differential equations.
The textbook formula for the ?voltage at the receiving end? (Vr), as a function of the ?voltage at the sending end? (Vs), the current (I), and the series impedance (Z) is as follows:
Vr = Vs - IZ
It should come as no surprise that this very much resembles the Ohm?s Law responses given by other Forum members. For longer lines, however, the formula would be quite different.
The tricky part is coming up with ?Z.? The properties given in NEC Table 8 will not serve this purpose. Since you are talking about a transmission line, you may have access to tables that list values for resistance and series impedance. Otherwise, it will be a bit of a chore, involving such things as the ?Geometric Mean Radius? of the individual conductors and the ?Geometric Mean Distances? between conductors.
[ January 14, 2005, 10:18 AM: Message edited by: charlie b ]
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
That?s why we are all here: to learn. You can look at Table 8, and discover that 100 mm2 is just slightly smaller than a 4/0.
Re: voltage drop question
At what length does a transmission line start to act like a piece of coax or twin line at higher frequencies? By my calculations the wavelength at 60 Hz is near infinity. Well, it is pretty long.
I would think this could be modeled as a lumped xmission line and simulated on the computer.
At what length does a transmission line start to act like a piece of coax or twin line at higher frequencies? By my calculations the wavelength at 60 Hz is near infinity. Well, it is pretty long.
I would think this could be modeled as a lumped xmission line and simulated on the computer.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
I have not dealt with higher frequencies than 60 hz. The textbook to which I referred is ?Elements of Power System Analysis,? by William D. Stevenson, Jr., and my copy is copyrighted 1982. It draws the line between ?short? and ?medium? at 80 km (50 miles), and the line between ?medium? and ?long? at 240 km (150 miles). The computer simulation to which you refer is certain to be able to take into account all series and shunt impedance values, so I am sure it is the best approach.
I have not dealt with higher frequencies than 60 hz. The textbook to which I referred is ?Elements of Power System Analysis,? by William D. Stevenson, Jr., and my copy is copyrighted 1982. It draws the line between ?short? and ?medium? at 80 km (50 miles), and the line between ?medium? and ?long? at 240 km (150 miles). The computer simulation to which you refer is certain to be able to take into account all series and shunt impedance values, so I am sure it is the best approach.
jtester
Senior Member
- Location
- Las Cruces N.M.
Re: voltage drop question
Usually transmission lines are 69 kv and up. I would think at 13.8 kv distribution modeling is accurate. There are relatively simple equations to calculate circuit impedance based on line configuration, phase spacing, phase-neutral spacing, etc. These must be used even at distribution voltages. Resistance by itself is inappropriate.
Jim T
Usually transmission lines are 69 kv and up. I would think at 13.8 kv distribution modeling is accurate. There are relatively simple equations to calculate circuit impedance based on line configuration, phase spacing, phase-neutral spacing, etc. These must be used even at distribution voltages. Resistance by itself is inappropriate.
Jim T
Re: voltage drop question
Rattus:
I come up with about 500 mi for the wavelength at 60Hz. If the transmission line starts getting to be a significant fraction of that (like maybe more than 10%) you have to start taking into account the variation of voltage and current along the line.
So Charlie's 85 Km (which would be about 38 mi) turns out to be about 8% of the wavelength. The posters 35Km is only 3% of the wavelength. So we can ignore any effects the length has on Vdrop.
Steve
Steve
Rattus:
I come up with about 500 mi for the wavelength at 60Hz. If the transmission line starts getting to be a significant fraction of that (like maybe more than 10%) you have to start taking into account the variation of voltage and current along the line.
So Charlie's 85 Km (which would be about 38 mi) turns out to be about 8% of the wavelength. The posters 35Km is only 3% of the wavelength. So we can ignore any effects the length has on Vdrop.
Steve
Steve
Re: voltage drop question
The formula for caculating Voltage Drop is
VD = I(RCosΘ + XsinΘ) where
I = amps
R = line resistance at a given Temperature for the
given length.
X = Reactance at the given length
The problem is that X varies with the distance between the phase conductors. AS Charlie said
the distance is Geometric Mean Distances between conductors or GMD. IF the spacing in feet between the 3 conductors is A,B and C then GMD is equal to the cube root(A x B x C).
X = Xa + Xd where Xa is the reactance at 1 ft spacing and Xd is the reactance at the GMD spacing.
Xa = 0.2794 x (f/60)xlog 1/GMR) where f = frequency and GMR = 0.7788 x r (r = radius of the conductor).
So tawfiq
you need to determine GMD, R, X and the PF.
I suggest you do a google search and find a cable MFG. Get them to send you a engineering manual
with this information. You can look most of this information up in the manual. Good luck.
Corrected X to read reactance per rattus correction.
[ January 14, 2005, 11:47 PM: Message edited by: bob ]
The formula for caculating Voltage Drop is
VD = I(RCosΘ + XsinΘ) where
I = amps
R = line resistance at a given Temperature for the
given length.
X = Reactance at the given length
The problem is that X varies with the distance between the phase conductors. AS Charlie said
the distance is Geometric Mean Distances between conductors or GMD. IF the spacing in feet between the 3 conductors is A,B and C then GMD is equal to the cube root(A x B x C).
X = Xa + Xd where Xa is the reactance at 1 ft spacing and Xd is the reactance at the GMD spacing.
Xa = 0.2794 x (f/60)xlog 1/GMR) where f = frequency and GMR = 0.7788 x r (r = radius of the conductor).
So tawfiq
you need to determine GMD, R, X and the PF.
I suggest you do a google search and find a cable MFG. Get them to send you a engineering manual
with this information. You can look most of this information up in the manual. Good luck.
Corrected X to read reactance per rattus correction.
[ January 14, 2005, 11:47 PM: Message edited by: bob ]
Re: voltage drop question
Bob, I think you mis-spoke when you said "Xa is the inductance". I think you meant to say, "Inductive reactance".
It appears that the inductance per unit lengh and capacitance per unit length are buried in this formula.
Just for grins, I wonder if anyone ever thinks about the characteristic impedance of a power transmission line?
I see now that this problem is more than one of simple resistance.
Bob, I think you mis-spoke when you said "Xa is the inductance". I think you meant to say, "Inductive reactance".
It appears that the inductance per unit lengh and capacitance per unit length are buried in this formula.
Just for grins, I wonder if anyone ever thinks about the characteristic impedance of a power transmission line?
I see now that this problem is more than one of simple resistance.
Re: voltage drop question
Rattus
Thanks for the correction.
I'm not sure what you mean. The R and X are usually given in units per 1000 ft or per mile.
I have a table that lists Xd as ohms per mile.
[ January 14, 2005, 11:49 PM: Message edited by: bob ]
Rattus
Thanks for the correction.
Capacitance is not a factor in the equation.
I'm not sure what you mean. The R and X are usually given in units per 1000 ft or per mile.
I have a table that lists Xd as ohms per mile.
[ January 14, 2005, 11:49 PM: Message edited by: bob ]
Re: voltage drop question
Bob, The classic xmission line is described as having series resistance and inductance plus shunt capacitance--equally distributed along its length. I would guess that with the line spacings involved, the capacitance either has little effect, or is buried in the formula.
With computer simulation and a good model, these calculations can be made in a few seconds--after you find all your mistakes.
Bob, The classic xmission line is described as having series resistance and inductance plus shunt capacitance--equally distributed along its length. I would guess that with the line spacings involved, the capacitance either has little effect, or is buried in the formula.
With computer simulation and a good model, these calculations can be made in a few seconds--after you find all your mistakes.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
Several of us have posted similar formulas, and you can use any version you wish. Table 8 is at the back of the NEC. There are also tables that give resistance and impedance for various types of conductors. Using a value from the tables might give you a ?rough estimate? of your voltage drop. But your results will not be accurate, until you calculate the impedance that applies to the configuration of the line. As you can see from Bob's post, the math is not trivial. You should look for help from a member of your company who has done this type of calculation, or from a vendor who has the related software readily available.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
That bring us back to the short, medium, and long line models I mentioned earlier. This line fits into the short line model. The short line model treats shunt impedance as being negligible, regardless of the geometry of the conductor spacing. All it considers in the series inductance, and it lumps that into a single element.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
Re: voltage drop question
I don?t, but then I don?t design such things. You?ll have to ask someone who does. I do, however, remember how to calculate the length, taking into account the sag. The line will follow the curve of a ?hyperbolic cosine.?
- Status