Voltage drop

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Re: Voltage drop

john,

I'm curious as to how you and I can come up with such different voltage numbers, so I ran the calculations with a different formula. I used VD=I*R (instead of the spreadsheet I have set up). Using 20' of wire (10' there and 10' back) at a resistance of .00193/ft., and the cumulative total of current at each point, I came up with similar results to my spread sheet. What formula are you using?

Looking at the 14th and 15th fixture in your calculations, you have a VD listed as .51 volts and I have it listed as .03 volts. If VD=I*R is valid, the current for 1 lamp (load between 14 and 15 is .08333 amps and the resistance in the wire is (from table 8 in the NEC) .00193 ohms/ft (times 20' of wire, 10' there and 10' return) we get VD=.0833*.0386=.00322 volts dropped between the 14th and 15th fixture.

If you look at the 1st fixture, you get VD=12.5(cumulative amperage of the 15 lamps)*.0386=.4825 volts dropped between the breaker and the 1st lamp. I have this listed as .48 and you have it listed as .2, one of us is using the wrong formula.
 
Re: Voltage drop

Stiff,

The current is 0.833A per lamp, not 0.0833A.

Now, the method you are using is not exact because none of the lamps see 120V. Close enough, but not exact. Using this approximation though, I see about 116V at the end of the line.

An easier approach is to take the average current times the total resistance.

Vd = 0.00386*150*12.5/2 = 3.6V at lamp 15. Still not exact, but close enough.
 
Re: Voltage drop

Originally posted by rattus:
Stiff,

The current is 0.833A per lamp, not 0.0833A.

Now, the method you are using is not exact because none of the lamps see 120V. Close enough, but not exact. Using this approximation though, I see about 116V at the end of the line.

An easier approach is to take the average current times the total resistance.

Vd = 0.00386*150*12.5/2 = 3.6V at lamp 15. Still not exact, but close enough.
Click to expand...
Thanks for catching that Rattus. The posted voltages still look correct (I checked my spread sheet and I had .8333 in it). The end voltage I had posted was 116.4 volts.
 
Re: Voltage drop

Rattus,

How could you set up a spread sheet to accurately calculate the voltage drop? Every time I've tried it (using the voltage the lamp sees) the spread sheet comes back with a circular reference error (since the voltage is dependent on the current and the current is dependent on the voltage).
 
Re: Voltage drop

Originally posted by hardworkingstiff:
Rattus,

How could you set up a spread sheet to accurately calculate the voltage drop? Every time I've tried it (using the voltage the lamp sees) the spread sheet comes back with a circular reference error (since the voltage is dependent on the current and the current is dependent on the voltage).
Click to expand...
Stiff, I would set up three columns.

Column A would be the current per section;

In A15 you would enter 0.833, then in A14 you would enter =A15 + 0.833, then copy that formula into A13 thru A1.

Column B would be the voltage drop per section:

In B1 you would enter =.00386*A1
Now copy this formula to B2 thru B15

Column C would be the cumulative voltage drop:

In C1 enter B1.
In C2 enter =C1 + B2
Copy this formula to C3 thru C15.

Check it out.
 
Re: Voltage drop

This is an interesting problem. I entered the load as 100 watts at 120 volts and the distance given. At each lamp the current increased due to the load staying at 100 watts as the voltage dropped. I then used a constant current of 0.83 amps in place of the 100 watts and the voltage at the last location was 116 v & 3.36% vd.
 
Re: Voltage drop

Rattus,

If I understand what you are saying, the 0.833 entered in A15 would only be accurate at 120 volts. If the voltage indeed had been reduced to 116 volts then A15 would be 0.8056. This is where it gets tricky.
 
Re: Voltage drop

Here is my spreadsheet results. Each row represents one section.

Current Drop Cum. Drop
12.495 0.48231 0.482
11.662 0.45015 0.932
10.829 0.41800 1.350
9.996 0.38585 1.736
9.163 0.35369 2.090
8.330 0.32154 2.412
7.497 0.28938 2.701
6.664 0.25723 2.958
5.831 0.22508 3.183
4.998 0.19292 3.376
4.165 0.16077 3.537
3.332 0.12862 3.666
2.499 0.09646 3.762
1.666 0.06431 3.826
0.833 0.03215 3.858
 
Re: Voltage drop

If you round your voltage drop to two digits to the right of the decimal, it's exactly what I posted earlier. The problem is although it is close, it still is based on 120 volts at each lamp to get the current rating for the voltage drop (like you mentioned earlier). Every time I try to set up a spread sheet the interrelation between current, voltage, and voltage drop won't allow the spread sheet to calculate it because of the circular reference.
 
Re: Voltage drop

Originally posted by hardworkingstiff:
Rattus,

If I understand what you are saying, the 0.833 entered in A15 would only be accurate at 120 volts. If the voltage indeed had been reduced to 116 volts then A15 would be 0.8056. This is where it gets tricky.
Click to expand...
Stiff,

This is indeed an approximation, but we do not need to compute the drop exactly. With the assumptions we are making, we compute a result greater than actual.

One could solve 15 loop equations or write a repeating fraction, but these methods are quite tedious. Furthermore, the resistance of each lamp changes somewhat as the voltage changes. Furthermore, we have not considered the tolerance of each individual lamp.

It is important to know when it is acceptable to use an approximation. For example, multiply the average current by the total resistance.

Vd15 = 6.6A * .00193 Ohms/ft * 300 ft = 3.8V

Take the sum of Column A and divide by 15 to get the average current.
 
Re: Voltage drop

Stiff, here are the formulas:

A2+0.833 0.0386*A1 B1
A3+0.833 0.0386*A2 C1+B2
A4+0.833 0.0386*A3 C2+B3
A5+0.833 0.0386*A4 C3+B4
A6+0.833 0.0386*A5 C4+B5
A7+0.833 0.0386*A6 C5+B6
A8+0.833 0.0386*A7 C6+B7
A9+0.833 0.0386*A8 C7+B8
A10+0.833 0.0386*A9 C8+B9
A11+0.833 0.0386*A10 C9+B10
A12+0.833 0.0386*A11 C10+B11
A13+0.833 0.0386*A12 C11+B12
A14+0.833 0.0386*A13 C12+B13
A15+0.833 0.0386*A14 C13+B14
0.833 0.0386*A15 C14+B15

Be sure to insert "=" in front of each entry.
 
Re: Voltage drop

OK men,

This ought to settle it. I simulated this problem with PSPICE and here is what I got.

($N_0001) 120.0000 ($N_0002) 119.5300

($N_0003) 119.0900 ($N_0004) 118.6800

($N_0005) 118.3000 ($N_0006) 117.9600

($N_0007) 117.6500 ($N_0008) 117.3600

($N_0009) 117.1100 ($N_0010) 116.9000

($N_0011) 116.7100 ($N_0012) 116.5500

($N_0013) 116.4300 ($N_0014) 116.3300

($N_0015) 116.2700 ($N_0016) 116.2400

My approximation came to a drop of 3.8V. Close enough!

[ November 21, 2005, 10:13 PM: Message edited by: rattus ]
 
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