Voltage drop calculation for resid. service

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jrob

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I'm trying to figure out the voltage drop for a 120/240, single phase, 125 amp service. The distance from the power source is approx. 125 ft, and the wire gage is 4/0 aluminum in PVC conduit. I've already done the calculations, but it just doesn't seem right. If anyone could help me out with this I would appreciate it.
VD ( single phase ) = 2 x (K x Q ) x I x D / CM

.63 ( single phase ) = 2 x ( 21.2 x .2 ) x 125 x 125 / 211600
 
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Where did that factor "Q" come from? My answer is 3.13 volts, and it differs from yours by that factor of 0.2.
 
The voltage drop wil be dependant on the actual load on the conductors, not the capacity of the service. Will the calculated load actually be 125-amps?
 
I get 3.1 volts also, figuring the 125 amp load as 125 amps. The actual connected load is likely to be considerably less that 125 amp so the VD would drop in proportion to the load. Bottom line, your 4/0 is quite adequate for what you have described.
 
haskindm said:
The voltage drop wil be dependant on the actual load on the conductors, not the capacity of the service. Will the calculated load actually be 125-amps?

For sizing purposes, there would be no practical reason to base the voltage drop calculation on the current load instead of using the service size.
 
kingpb said:
For sizing purposes, there would be no practical reason to base the voltage drop calculation on the current load instead of using the service size.

Of course there is a practical reason.... it's called money.

kingpb you must work with our tax dollars....no regard for costs. :grin:
 
I got the Q from a formula in Mike Holts electrical exam book. It's an adjustment factor for AC circuits with conductors 2/0 or bigger. It says to divide the ac ohms to neutral impedance listed in ch. 9 table 9 by the dc resistance listed in ch. 9 table 8.
 
If my actuall load were 125 amps can you tell me approx. how many feet my power source could be and still be within the 3% NEC reccomendation. If you could give me the formula I would appreciate it.
 
See if this is in the ball park:

3% Vd = 3.75v

CM x Vd / (2KI) = Distance

211600 x 3.75 = 793,500

Divide by 2KI = 5300

Distance = 149 ft (one way)

I used the the CM posted and did not do a Q adjustment.
 
iwire said:
Of course there is a practical reason.... it's called money.

kingpb you must work with our tax dollars....no regard for costs. :grin:

Better, power plants! Design is based on reliability, not low cost. Cost of a new plant is built into your utility rates. Need a new plant? They just raise your rates!

It's a vicious cycle, but it pays well:grin:
 
yea i agree with IWIRE but how do you really get a practical amp reading on a house? at what given times of the day? would you use all nameplate ratings of appliances and recep/general lighting loads? or is it easier to just go by the main ocpd?
 
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jrob said:
I'm trying to figure out the voltage drop for a 120/240, single phase, 125 amp service. The distance from the power source is approx. 125 ft, and the wire gage is 4/0 aluminum in PVC conduit...

This is utility service and they serve the approximate connected load, if this were a feeder we feed the calculated/connected load. The utility company is not as thorough with their calculation. I am curious though, this service should not be your responsibility it usually is the responsibility of the utility company.
Take a voltage reading at the main with everything off then turn everything on to create the largest load possible, even plug stuff in, and take another voltage reading (note your amperes as well) see what your actual VD is. You may be able to ask the utility company to consider changing their service. Can they serve creating a greater than 3% VD, or 5% at the farthest outlet?
 
Geraldselectric said:
See if this is in the ball park:

3% Vd = 3.75v

CM x Vd / (2KI) = Distance

211600 x 3.75 = 793,500

Divide by 2KI = 5300

Distance = 149 ft (one way)

I used the the CM posted and did not do a Q adjustment.

Just to clarify, Vd should be 7.2v NOT 3.75v = 1523520

Max Distance would be 288 ft.
 
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