voltage drop calculation

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karthik

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In general when we do voltage drop calculatio for

1. 1 phase AC system we use the following formula
VD = 2xIxRxL/1000

2. 3 phase AC system we use the following formula
VD = 1.732xIxRxL/1000

I want to make sure in both the cases do we have to consider only run of the cable irrespective of the number of runs.

Plz clarify
 
Re: voltage drop calculation

Originally posted by karthik: I want to make sure in both the cases do we have to consider only run of the cable irrespective of the number of runs.
Click to expand...
If by "number of runs" you mean that you have two or more parallel conductors for each phase and neutral, then you do have to take them into account. The way you take them into account is to reduce the value of "R," or resistance. You initially get "R" from a table, based on the size and type of conductor. If you have two parallel runs, then you divide that value by 2, before you use "R" in the equation. If you have three parallel runs, then you divide that value by 3.

Any number of equal resistors that are placed in parallel will give you a net total resistance of the resistance of one resistor divided by the number of resistors.
 
Re: voltage drop calculation

Bryan: That version of the equation only works for single phase. Also, the value of "R" you use in your equation is not the same as the value of "R" in the OP's version of the equation. So it is not necessarily "easier."
 
Re: voltage drop calculation

this is the formula i've used. sometimes this is easier.

VD = (2*K*L*I/CMA)*F

K is a constant, 12.5 for copper 18 for aluminum.
F is a factor, .866 for 3 phase, and 1 for single phase.

CMA is the circular mils
Line Size-CMA
4/0 - 211,600
2/0 - 133,100
#2 - 66,360
#4 - 41,740
#6 - 26,240
#12 - 6,530

[ January 06, 2006, 02:10 PM: Message edited by: dave_asdf ]
 
Re: voltage drop calculation

"K" is only an approximate value if the conductor size is not know. I've seen values of "K" from as low as 11.5 to as high as 13. (Usually based on temperature differences) The Table 8 values average right at 12.9 using R x CM / 1000 = K.

[ January 06, 2006, 02:11 PM: Message edited by: bphgravity ]
 
Re: voltage drop calculation

Also, the value of "R" you use in your equation is not the same as the value of "R" in the OP's version of the equation.
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Charlie
Why is the value of R different.
 
Re: voltage drop calculation

Both versions of the equation are valid. The values of "R" in the two versions are different because in the OP's version, "R" is resistance per unit length (i.e., ohms per 1000 feet), as taken from the tables. However, in Bryan's version, "R" is resistance, in ohms. In order to get the actual resistance, you need to know the resistance per unit length and the length. That puts you back to the OP's version of the equation.
 
Re: voltage drop calculation

OK. They both end up with the same value of R. The methods of getting it are different.
 
Re: voltage drop calculation

This may prove both VD=I/R and VD=KIL/CM

Measured _______________ Proof-1999 NEC Tbl 8
Nominal V Et= 120.0 vac ---> 14 AWG CM= 4110
Load Volt Er= 114.0 vac ---> Total feet L= 127.5
Load Amps I= 15.0 Amps ---> Kcu@75c K= 12.9
V-drop% VD%= 5.0 %.... ---> Ohm@75c R= 3.14
VD= Et-Er VD= 6.0 vac.. ---> KIL/CM VD= 6.0 vac
========== ========= =================
Z=VD/I -----> 0.4 ohm ----> (L/1000)*R= 0.4 ohm

[ January 06, 2006, 05:01 PM: Message edited by: ramsy ]
 
Re: voltage drop calculation

I like the formulas in the OP.

But I don't really get the question:

I want to make sure in both the cases do we have to consider only run of the cable irrespective of the number of runs.
Click to expand...
He could be asking about paralell conductors or seperate circuits.
 
Re: voltage drop calculation

You can find an accurate value for K for any size conductor. It's different for every conductor size and temperature dependant, like Bryan said.

But using a value like 12.8 or 12.9 will not yield the actuall resistance in most cases. You'll get an error from the difference between the generalized value for K and the actual value for the conductor.

Personally, I'll never understand how using a generalized K value became common practice.

But then the error's small too.
 
Re: voltage drop calculation

I believe the OP was wanting to know why 3 phase equation didn't use the multiplier "2" for two way distance (to load and back).
 
Re: voltage drop calculation

1. 1 phase AC system we use the following formula
VD = 2xIxRxL/1000

2. 3 phase AC system we use the following formula
VD = 1.732xIxRxL/1000

To the formula mentioned above i have put the number of runs in the denimenator wheather itis 1 phase or 3 phase.

Is that correct..........
 
Re: voltage drop calculation

Sorry to misunderstand your original question, :eek:
Yes, the number of parallel runs would be in the denominator of the equation for both single phase and 3 phase
 
Re: voltage drop calculation

Originally posted by ccjersey: I believe the OP was wanting to know why 3 phase equation didn't use the multiplier "2" for two way distance (to load and back).
Click to expand...
If that had been the question (and the OP has now explained that it was not), then the answer would be that the factor of 1.732 takes into account the fact that current leaving (to the load) on one of the three wires can return (to the source) on either or both of the other two wires.
 
Re: voltage drop calculation

Originally posted by charlie b:
If that had been the question (and the OP has now explained that it was not), then the answer would be that the factor of 1.732 takes into account the fact that current leaving (to the load) on one of the three wires can return (to the source) on either or both of the other two wires.
Click to expand...
Charlie.Correct me if iam wrong.You mean to say that for a 3 phase system with the factor of 1.732 will take care of the return length from load to source as we normally consider for a 1 phase system with the factor of 2.Is my understanding is correct.
 
Re: voltage drop calculation

The magic number, 1.732, arises because any two line currents in a balanced system are 120 degrees apart. Line to line voltage drop is affected accordingly. In a single phase system the currents are 180 degrees apart and we use the multiplier, "2".

The symbol "R" denotes a fixed resistance. It is confusing to use it for resistivity.

Bulk resistivity is denoted by the Greek letter, Rho, and its units are Ohm-centimeters or perhaps Ohm-meters. Then for example,

R (Ohms) = Rho (Ohm-cm) x length (cm) / area (cm^2)
 
Re: voltage drop calculation

Originally posted by karthik: You mean to say that for a 3 phase system with the factor of 1.732 will take care of the return length from load to source as we normally consider for a 1 phase system with the factor of 2.
Click to expand...
That is correct.

[ January 10, 2006, 10:22 AM: Message edited by: charlie b ]
 
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