Voltage Drop Calculators

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DetroitEE

Senior Member
Location
Detroit, MI
I was running the numbers on some of my feeders and the voltage drop calculator I usually use seemed to be giving me high numbers. I used another voltage drop calculator to verify my suspicion, and I was right, the voltage drop percentages were too high on the first one. I did a hand calc using the method described on this site, and it was consistant with the numbers to second one was giving me.

Here are my numbers:

208V 3 Phase
1 set of 500kcmil copper
250A load
75 degree conductors
steel conduit
200' run

From a hand calc:

VD = 1.73* 200' * 250A * 12.9 ohms (copper constant) * 1.12 (skin effect) / 500000 (500 kcmil area)


VD = 2.5 V

VD % = 2.5/208 = 1.2 %

Voltage drop calculator 1 is giving me 2.3% (which is wrong) and the second one is giving me 1.1%, which is very close.

This could be user error on my part, but it's not readily apparent to me. Please let me know if you guys see anything which would cause the first one to give me this number. I basically just want to know if the first calculator is programmed incorrectly, so that people will know not to rely one that particular one, and I would notify whoever made the calculator.
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
I think the program for your "usual" calculator is invalid. If you keep everything exactly as you describe, then change the conduit type, you get answers that vary from 1.6% to 2.3%. That low value is what I get when I leave the choice of conduit set to its default of "select conduit type." While the total inductive and capacitive reactance of a circuit can be influenced by the choice of conduit type, it won't have that much of an effect (i.e., 1.6% to 2.3% is a change of 30%).

I used "my usual calculator," and came up with 1.07%.
 

DetroitEE

Senior Member
Location
Detroit, MI
I think the program for your "usual" calculator is invalid. If you keep everything exactly as you describe, then change the conduit type, you get answers that vary from 1.6% to 2.3%. That low value is what I get when I leave the choice of conduit set to its default of "select conduit type." While the total inductive and capacitive reactance of a circuit can be influenced by the choice of conduit type, it won't have that much of an effect (i.e., 1.6% to 2.3% is a change of 30%).

I used "my usual calculator," and came up with 1.07%.
Great, thanks for the input. What calculator do you like to use?
 

charlie b

Moderator
Staff member
Location
Seattle, WA
Occupation
Electrical Engineer
What calculator do you like to use?
It is an Excel file that I downloaded from this site several years ago. For reasons I do not know, it was removed from this site, also several years ago. Since I did not create it and have never bothered to verify its accuracy (I only use it for a sanity check on VD numbers), and since I do not know whether its author claims a copyright on it, I do not feel it appropriate to share it with anyone.

 

GeorgeKoehl

Member
Location
Washington IL
Here are my numbers:

208V 3 Phase
1 set of 500kcmil copper
250A load
75 degree conductors
steel conduit
200' run

.
I ran those numbers through my ElectriCalc Pro and came up with a VD of 2.2 Volts and 1.1%

I checked the numbers again came up with the same answer.
very close to your figures.
 

dkarst

Senior Member
Location
Minnesota
Charlie, I tried this calulator:

http://www.electrician2.com/calculators/vd_calculator_initial.html

and for "208 volts 3-phase 3-wire" it indicates Evd 2.2; for '120/208 3-phase 4-wire" it shows Evd 1.3. Why would the neutral figure into the calculation and the answer be so different? Thanks.
Be a little careful of how this tool is changing the basis (i.e. line-neutral or line-line) when you select the two different configurations. The 2.6 volts times sqrt(3) = 4.5 volts but then they calculate the % change on different basis. If I get time later I will verify if there is a real computation problem or not.
 

dkarst

Senior Member
Location
Minnesota
Be a little careful of how this tool is changing the basis (i.e. line-neutral or line-line) when you select the two different configurations. The 2.6 volts times sqrt(3) = 4.5 volts but then they calculate the % change on different basis. If I get time later I will verify if there is a real computation problem or not.
I guess this reinforces the rule you should never type anything just before leaving the office :-? This on-line calculator seems to work fine but you have to input the conductor size correctly (which I did not). If you input the conductor size correctly and 208V, 3phase, 3-wire pull down, it gives the right answer of 1.1 % against the 208 V line voltage which is 2.2 volts which subtracted from 208 gives 205.8V.

If you choose the 120/208 3phase 4-wire pull down, it gives 1.1% which against the line-neutral voltage of 120V is 1.3V which when subtracted from 120 gives you 118.7 volts. If you multiply the 118.7 * sqrt(3) = 205.6 volts. You can also multiply the line-neutral drop of 1.3 volts by sqrt(3) = 2.2 volts which is the line-line drop (same as above).
 
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mityeltu

Senior Member
Location
Tennessee
If you have access to it, IEEE 141 (Red Book) has several examples of voltage drop calculations. I believe Ugly's book alo has some nice examples from the NEC. (Ugly's book is alot cheaper than any IEEE book I've ever purchased)

I would not trust any of these calculators until I had verified a number of different calculations using standard methods that include both resistance and reactance. Otherwise you run the risk of installing the wrong size cable. This means either too small, which can be disasterous from a safety standpoint, or too big which is bad from a financial standpoint.

Ultimately, though, you as the engineer/designer are responsible for making sure your calculations are valid.
 

Howard Burger

Senior Member
thanks

thanks

dkarst, yes, I was wondering about your results...Thank you for the clarification and explanation about the difference between 3ph 3wire and 3ph 4wire calculator results. Perhaps for a rule of thumb, for 3ph I'll simply use the KID sqrt3/cm formula and go with the higher number as a hedge.
 

skeshesh

Senior Member
Location
Los Angeles, Ca
I tried out the first calculator (1st link) and got 1.2%. The note under the input field for conductor length states 'one-way distance' so I thought maybe the program always multiplies that number by 2 - I tried 100ft and there you have it. The language is unclear though so I dont know if this was just a conincidence with numbers or not. You're hand calculation does seem correct though.
 

DetroitEE

Senior Member
Location
Detroit, MI
I tried out the first calculator (1st link) and got 1.2%. The note under the input field for conductor length states 'one-way distance' so I thought maybe the program always multiplies that number by 2 - I tried 100ft and there you have it. The language is unclear though so I dont know if this was just a conincidence with numbers or not. You're hand calculation does seem correct though.
I think that's probably just a coincidence. 200' is the correct one way distance, and the correct voltage drop is around 1.1 or 1.2% for that distance, as confirmed by the hand calc and the second calculator.

I think there's definitely something wrong with the first one. It does give correct numbers for some calculations, but something about the parameters I put in for that particular feeder calc caused it to output an incorrect value.
 

jghrist

Senior Member
With this size conductor, you have to consider the reactance as well as the resistance. You also need to know the load power factor.

From NEC Table 9, r = 0.029 ohm/1000', x = 0.048 ohm/1000'

R = 0.0058 ohm, X = 0.0096 ohm

? = acos(pf)

phase-to-neutral voltage drop = I?(R?cos(?) + X?sin(?))

For pf = 100%, VD = 1.45V = 1.21%
For pf = 85%, VD = 2.5V = 2.08%
For pf = 50%, VD = 2.8V = 2.34%
 

dkarst

Senior Member
Location
Minnesota
With this size conductor, you have to consider the reactance as well as the resistance. You also need to know the load power factor.

From NEC Table 9, r = 0.029 ohm/1000', x = 0.048 ohm/1000'

R = 0.0058 ohm, X = 0.0096 ohm

? = acos(pf)

phase-to-neutral voltage drop = I?(R?cos(?) + X?sin(?))

For pf = 100%, VD = 1.45V = 1.21%
For pf = 85%, VD = 2.5V = 2.08%
For pf = 50%, VD = 2.8V = 2.34%
It is possible that the first on-line tool mentioned (that gives 2.2%) assumes some load power factor of say ~ 85% (inside the program and you don't see or choose). As noted by jghrist above (using IEEE-141 method), this is a more realistic answer with a more typical power factor... although 85% is on the low side... just a thought. It is also possible it is just lousy code :)
 
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