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[mshields]

3 phase voltage drop calculation formula is (2 * L * R * I * .866)/1000

.866 is the square root of 3 divided by 2.

Which means you could write the formula (L * R * I * root3)/1000.

Why is the formula typically shown using .866

 

[mshields]

I just remembered - I think

I just remembered - I think

root3/2 is one of the trig signs for a power factor of .8?

 

[ggunn]

3 phase voltage drop calculation formula is (2 * L * R * I * .866)/1000

.866 is the square root of 3 divided by 2.

Which means you could write the formula (L * R * I * root3)/1000.

Why is the formula typically shown using .866

When I do voltage drop calculations the only formula I use is V = IR.

 

[mshields]

that does simplify things

that does simplify things

where R = 2LR/1000

where R is ohm's per thousand feet

and then I take it you just figure it for one leg such that for a three phase calc you would simply use 120 and get the same result. Yes, I see your point.

 

[Sahib]

The factor .866 is used to show that voltage drop in (balanced) 3phase lines is only .866 times that in 1 phase lines.

 

[topgone]

The factor .866 is used to show that voltage drop in (balanced) 3phase lines is only .866 times that in 1 phase lines.

How did you arrive at that figure?

 

[Ingenieur]

????

 

[Julius Right]

I think, a better formula [see IEEE 141 3.11.1 General mathematical formulas] is this:
V = IRcosf + IX sinf
where:
V is the voltage drop in circuit, line to neutral [V]
I is the current flowing in conductor [A]
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
The reason for using the line-to-neutral voltage is to permit the line-to-line voltage to be computed by multiplying by the following constants:
Single-phase 2
Three-phase 1.732

 

[topgone]

I think, a better formula [see IEEE 141 3.11.1 General mathematical formulas] is this:
V = IRcosf + IX sinf
where:
V is the voltage drop in circuit, line to neutral [V]
I is the current flowing in conductor [A]
R is the line resistance for one conductor, in ohms
X is the line reactance for one conductor, in ohms
f is the angle whose cosine is the load power factor
cosf is the load power factor, in decimals
sinf is the load reactive factor, in decimals
The reason for using the line-to-neutral voltage is to permit the line-to-line voltage to be computed by multiplying by the following constants:
Single-phase 2
Three-phase 1.732

Thanks for posting the above formula. I still would like to know where that 0.866 came from. i don't see it above!

 

[Julius Right]

Sorry, topgone, I already forgot the starting point!:ashamed1:

 

[Smart $]

Thanks for posting the above formula. I still would like to know where that 0.866 came from. i don't see it above!

It's one-half the square root of 3. I've not seen the Vd formula presented with the .866 factor, only the 1.732 (root of 3) factor as posted by Julius.

I think someone somewhere got the bright idea that half the root of three would give you the Line-to-Neutral voltage drop... but it does not. Line-to-neutral voltage drop is simply not multiplying by 2 or 1.732 (or in this case, .866).

 

[Sahib]

Thanks for posting the above formula. I still would like to know where that 0.866 came from. i don't see it above!

It is obtained by manipulation as done in the OP itself.

 

[Carultch]

3 phase voltage drop calculation formula is (2 * L * R * I * .866)/1000

.866 is the square root of 3 divided by 2.

Which means you could write the formula (L * R * I * root3)/1000.

Why is the formula typically shown using .866

It is more correct to show non-reducible square roots as expressions, as this calculates exactly what it is. As opposed to approximately what it is. Whether it is 2*0.866 or whether it is 1.73, both of these numbers only allow you three significant digits of calculation resolution. While this is usually good enough for our purposes, as circuit length is seldom known to more than 2 digits, if there ever is an application where it matters, this is a source of error. So it is useful to learn the original irrational number in its mathematical form.

One reason I can think that you might want to show it as 2*0.866 instead of 1.73, is that we are already accustomed to a factor of 2 in voltage drop formulas for DC & single phase. So instead of thinking that we swap this out for a 1.73, the thought is that we insert a factor of 0.866 and leave the 2 in its place.

 

[topgone]

It is obtained by manipulation as done in the OP itself.

See @smart$ post. You don't multiply with 0.866. It's just 2 or sqrt(3). But you can use it, it's your calculations anyway.:happyyes:

 

[Sahib]

See @smart$ post. You don't multiply with 0.866. It's just 2 or sqrt(3). But you can use it, it's your calculations anyway.:happyyes:

If you want to relate 3 phase line voltage drop with 1 phase line voltage drop, then the factor .866 appears.

 

[ggunn]

If you want to relate 3 phase line voltage drop with 1 phase line voltage drop, then the factor .866 appears.

Yes, but it is a number derived from other factors. "Simple" equations with derived quantities are easy (for me, anyway) to forget and/or get wrong. In keeping with the title of this thread, I always use Ohm's Law when calculating voltage drop.