Sizing Wire

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JonathanWaldner

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Just wondering what size wire do i need if my run is 4000 feet at 480 from my switch wanting to step it up to 7200 an then from 7200 down again to 208/120 what size wire will i need
 
LarryFine

LarryFine

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Welcome to the forum.

You left out the most important info: calculated load current (and intended capacity).
 
infinity

infinity

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So you're going from 480 step up to 7200 and then back down to 208Y/120, 400 amps at the end with a total length of run 4000'?
 
roger

roger

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What type of wire do you plan on using for the 7200? Is it underground or overhead?
 
roger

roger

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It's not as simple as just sizing. You will need to use a 15KV type cable. Is it going to be in conduit or direct burial? Have you ever made MV terminations? It might be worth consulting with a local EE or EC that has done this type of work.
 
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JonathanWaldner

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Electrician
roger said:
It's not as simple as just sizing. You will need to use a 15KV type cable. Is it going to be in conduit or direct burial? Have you ever made MV terminations? It might be worth consulting with a local EE or EC that has done this type of work.
Click to expand...
Direct burial and in a poly type conduit
 
Jpflex

Jpflex

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JonathanWaldner said:
Just wondering what size wire do i need if my run is 4000 feet at 480 from my switch wanting to step it up to 7200 an then from 7200 down again to 208/120 what size wire will i need
Click to expand...
You did not say if this was a single phase or 3 phase service

Assuming this is a 3 phase service I calculated as small as a # 10 AWG conductor for the 4,000 foot feeder span between the two transforners (ONLY THIS 4000 ft section run)

It may seem too small but with such as high 7,200 volts your output current will be so low 11.554 I amperes estimated

144 Vd = 1.732 x 12.9 (copper) x 11.555i (amperes) x 4,000 (one way distance) / wire circular mill

Solve for CM = 7,171.42
#10 AWG = 10,380 cm circular mills size



The maxim voltage drop on the 4,000 foot run will be 144 volts to be within a 2% max voltage drop for the feeder/ link between transformers

For a 3 phase system you may need 144.1 kVA transformers minimum. See picture working from second transformer up its:

400 I amperes output secondary to service disconnect at 208/120 E

Second transformer primary
primary 11.555 i ampere at 7,200 E volts

First transformer feeds second transformer primary at 7,200 E at 11.555 i amperes

First transformer primary sees 173.25 i amperes at 480 volts



What i do not know is why calculating voltage drop using chapter 9 table 8 wire resistance yields a different VD value other than 144 volts dropped?

First voltage drop method = 144 volts
1.732 (3 phase) x 12.9 (copper) x 4,000’ (feet) / circular mill = 144 VD

Second voltage drop method
I X R
11.555i x 1.21 (ohms per 1,000 ft) x 4000’ x 2 (two way distance) / 1,000 = 111.85 i VD

Clearly both methods are acceptable for calculating voltage drop but why do they always yield inconsistent results? 144 Vd does not equal 111.85 VD?
 

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J

JonathanWaldner

Member
Location
Cut Bank
Occupation
Electrician
Jpflex said:
You did not say if this was a single phase or 3 phase service

Assuming this is a 3 phase service I calculated as small as a # 10 AWG conductor for the 4,000 foot feeder span between the two transforners (ONLY THIS 4000 ft section run)

It may seem too small but with such as high 7,200 volts your output current will be so low 11.554 I amperes estimated

144 Vd = 1.732 x 12.9 (copper) x 11.555i (amperes) x 4,000 (one way distance) / wire circular mill

Solve for CM = 7,171.42
#10 AWG = 10,380 cm circular mills size



The maxim voltage drop on the 4,000 foot run will be 144 volts to be within a 2% max voltage drop for the feeder/ link between transformers

For a 3 phase system you may need 144.1 kVA transformers minimum. See picture working from second transformer up its:

400 I amperes output secondary to service disconnect at 208/120 E

Second transformer primary
primary 11.555 i ampere at 7,200 E volts

First transformer feeds second transformer primary at 7,200 E at 11.555 i amperes

First transformer primary sees 173.25 i amperes at 480 volts



What i do not know is why calculating voltage drop using chapter 9 table 8 wire resistance yields a different VD value other than 144 volts dropped?

First voltage drop method = 144 volts
1.732 (3 phase) x 12.9 (copper) x 4,000’ (feet) / circular mill = 144 VD

Second voltage drop method
I X R
11.555i x 1.21 (ohms per 1,000 ft) x 4000’ x 2 (two way distance) / 1,000 = 111.85 i VD

Clearly both methods are acceptable for calculating voltage drop but why do they always yield inconsistent results? 144 Vd does not equal 111.85 VD?
Click to expand...
It is a 3 phase run
 
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