Voltage drop for new feeder in an existing building

[Grouch1980]

Hi all,
If I have to add a new sub-panel on the 10th floor of an existing building, and let's say the new feeder to this sub-panel (originating from a distribution board ON the 10th floor) is 100 feet long, how does one know what volt drop percentage to use when calculating the wire size for this new feeder? The voltage drop from the cellar service up to the 10th floor distribution board may be at 2%, or 2.5%, or even already at the max of 3%. How would i know what percentage to use for the new feeder serving the sub-panel, and make sure I don't exceed 3% from the cellar service up the new sub-panel? It's an existing building, and let's say existing drawings and calculations don't exist (as sometimes is the case). Does the NEC cover this?

Thanks.

 

[don_resqcapt19]

Other than 2 or 3 very specific cases voltage drop is not a code requirement. There is an informational note that is not enforceable that says the total voltage drop to the finial outlet should not exceed 5%.

 

[Grouch1980]

Other than 2 or 3 very specific cases voltage drop is not a code requirement. There is an informational note that is not enforceable that says the total voltage drop to the finial outlet should not exceed 5%.

I know it's not a code requirement, however it is enforced by some of the energy codes, such as ASHRAE 90.1 where you can't exceed a 2% voltage drop for feeders. so if you're renovating the 10th floor of an existing building, and adding that sub-panel, how do you know what percentage to use for that new feeder serving the sub-panel? You wouldn't know what the existing drop is up to the existing distribution board on the 10th floor.

 

[Ingenieur]

I know it's not a code requirement, however it is enforced by some of the energy codes, such as ASHRAE 90.1 where you can't exceed a 2% voltage drop for feeders. so if you're renovating the 10th floor of an existing building, and adding that sub-panel, how do you know what percentage to use for that new feeder serving the sub-panel? You wouldn't know what the existing drop is up to the existing distribution board on the 10th floor.

2% for each feeder, not the sum, so 2%
they are concerned with heat load

at 100' v drop will not be an issue if ampacity is correct
what is the current?

you can measure the v drop ofbthe existing
measure v btm, v top, current simultaneously
delta v / i = feeder z
feeder ampacity x feeder z = v drop
or calc on the existing wire size

 

[infinity]

If you're in NYC voltage drop compensation is mandatory by NYC amendment.

Chapter 2

Wiring and Protection

ARTICLE 210
Branch Circuits

SECTION 210.19
Subsection 210.19(A)(1) – Add a new sentence at the end of the paragraph before the first Exception to
read as follows:
Conductors of branch circuits shall be sized to allow for a maximum voltage drop of 3 percent at the last
outlet supplying light, heat or power and the maximum voltage drop allowable for feeders and branch
circuit combined shall not exceed 5 percent.

 

[Grouch1980]

2% for each feeder, not the sum, so 2%
they are concerned with heat load

at 100' v drop will not be an issue if ampacity is correct
what is the current?

oh, I don't have a current for this particular case, this is more of a generic question, however you always come across this in renovations in existing buildings.

hmmm... i think the energy code means 2% for the entire run of feeders. So if you have main feeder A, feeder B, and then sub-feeder C... the sum cannot exceed 2% (it wouldn't be each feeder at 2%). After the 2%, your branch circuit can then be at 3%. Am I interpreting this wrong?

 

[Grouch1980]

If you're in NYC voltage drop compensation is mandatory by NYC amendment.

Nice! Thanks for that.

 

[Ingenieur]

oh, I don't have a current for this particular case, this is more of a generic question, however you always come across this in renovations in existing buildings.

hmmm... i think the energy code means 2% for the entire run of feeders. So if you have main feeder A, feeder B, and then sub-feeder C... the sum cannot exceed 2% (it wouldn't be each feeder at 2%). After the 2%, your branch circuit can then be at 3%. Am I interpreting this wrong?

just looked at 90.1, correct
it is total feeders 2%
branch 3%
same as nec guidelines

the only way to do it is measure or calc from as-builts

a 200 A feeder sized for 160 A load, 3 ph
v drop / sys v x 100 = (1.732 x L/1000 x Z x i)/ sys v x 100
assume max 2%
0.02 = (1.732 x L/1000 x Z x i)/sys v
L = 11.55/(Z x i) x sys v

max for 208/3 at 160 A with 3/0 (200 A) Cu, steel cnd
max one-way L = 11.55/(0.094 x 160) x 208 = 160'
1% 80', etc

 

[Grouch1980]

just looked at 90.1, correct
it is total feeders 2%
branch 3%
same as nec guidelines

the only way to do it is measure or calc from as-builts

a 200 A feeder sized for 160 A load, 3 ph
v drop / sys v x 100 = (1.732 x L/1000 x Z x i)/ sys v x 100
assume max 2%
0.02 = (1.732 x L/1000 x Z x i)/sys v
L = 11.55/(Z x i) x sys v

max for 208/3 at 160 A with 3/0 (200 A) Cu, steel cnd
max one-way L = 11.55/(0.094 x 160) x 208 = 160'
1% 80', etc

Got it. So there's no rule of thumb / formula, other than to actually take a field measurement of the voltage at the point where the new connection is being made, or find as-builts / design drawings and calculate it.

 

[ramsy]

max (2%) for 208/3 at 160 A with 3/0 (200 A) Cu, steel cnd
max one-way L = 11.55/(0.094 x 160) x 208 = 160'
1% 80', etc

That is Slick -

Start new thread for "Conduit fill", & "Point-to-Point SCA"

2" EMT holds four 3/0 + #4 EGC

4,393 Tot. SCA for 150kva w/ STD 1.2 pct.Z

Now Knuckle heads are ready to roll

 

[Ingenieur]

Got it. So there's no rule of thumb / formula, other than to actually take a field measurement of the voltage at the point where the new connection is being made, or find as-builts / design drawings and calculate it.

you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
v load 480
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
if loading is 80% (cond sized 1.25 x load)
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits

 

[Grouch1980]

you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
v load 480
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
if loading is 80% (cond sized 1.25 x load)
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits

Thanks. ..... "the closer the i is to the rated ampacity the better"... why is that? i is your calculated demand amps.

 

[ramsy]

you need 2 v readings
supply and load at the same time (2 guys, do each ph and avg)
take i readings at same time, should be steady, avg 3 phases
feeder Z = (v supply - v load) / i Ohms

max v drop = Z x cond ampacity (x 80% depending on application)

eg
v supply 485
v load 480
i = 100
Z = 5/100 = 0.05 Ohm

if 250 kcmil rated ampacity = 255
v drop = 0.05 x 255 = 13 or 2.7%
if loading is 80% (cond sized 1.25 x load)
v drop = 0.05 x 0.8 x 255 = 10 v or 2.1%

the closer the i is to the rated ampacity the better
>30% of rated ampacity with good measurements yeilds good results

you can also remove power and actually measure a loop by shorting one end
or calc off as-buits

you need 2 v readings..
feeder Z = (v supply - v load) / I "CLamp meter"

Field measurements with Logged impedance (Z) can be assumed constant so Voltage Drop is solved with any Load.

While (Z) and temperature will vary with load, any conservative error is on the side of caution.

Z = VD/I

If Logged Feeder VD = 5v with 100A Load, then Z = 0.05

Now Solve Voltage Drop for any amperage,

VD = Z*I

If 250 kcmil rated ampacity = 255 then 80% = 204A

0.05*204 = 10.2v of 485v or VD = 2.1%

Where actual termination temperature is needed, Temp must be logged.

 

[Ingenieur]

Thanks. ..... "the closer the i is to the rated ampacity the better"... why is that? i is your calculated demand amps.

the higher the i the higher the v drop
so the lower the instrument messurement error

i is the measured amps when tested

 

[Grouch1980]

the higher the i the higher the v drop
so the lower the instrument messurement error

i is the measured amps when tested

got it, thanks.