Richard Franklin
Member
Hello,
I have been tasked to survey a job performed by an Electrical Contractor which I have a voltage drop concern.
The installation is as follows:
The scope of the work was to install 18 Flat Screen CATV Monitors on a total of 14 floors in a commercial office building. The wiring method was 10/2 MC cable and EMT conduit. There are two branch circuit home runs back to a junction box for each floor. The wiring continues back thru a Riser on each floor to a 3-phase 4-wire 120/208 volt sub-panel designated for the power of this equipment. The branch circuit MC cables have a one-way distance of 825.0 feet for cable #1 and 1025.0 feet for cable #2. Each Monitor has a 120 volt duplex receptacle, using a 2 wire factory cord and connector terminating into a power supply reducing the voltage to 13.5 volts for the CATV Monitor. Every Monitor has a load of 0.5 ampere and each branch circuit has a load of 4.5 ampere.
Using the NEC Article 210.19 (A)(1) FPN No. 4 there is a 3% voltage drop on Branch circuits and 2% voltage drop on Feeders.
The voltage drop calculation formula that I have always been taught to use is as follows:
VD=2KIL divided by CM
VD=2 times K times I times L divided by CM
(OR)
CM=2KIL divided by VD
CM=2 times K times I times L divided by VD
VD-Voltage Drop
K-Constant for the conductor material (12 for Copper and 18 for Aluminum)
I-Amperage of the circuit
L-Length of the circuit
CM-Circular mills of the conductor (10AWG has a 10380 circular mill area)
For Single phase 120 volt or Single phase 277 volt the total distance is measured from the (LINE) voltage source, to the connected (LOAD) of the circuit and measured back to the (NEUTRAL) connection. On multi-phase circuits the distance is measured (ONE-WAY) from the (LINE) to the (LOAD).
With this in mind the allowed 3% voltage drop on a 120 volt circuit is 3.6 volts.
The distance of cable #1 is 825.0 feet times 2 = 1650.0 feet.
The distance of cable #2 is 1025.0 feet times 2 = 2050.0 feet.
Using the above formula for cable #1:
2 times 12 times 4.5 times 1650 : divided by 10380 = 17.17 volt (total drop)
120 volt - 17.17 volt = 102.83 volt (measured at the farthest outlet)
17.17 volt divided by 120 volt = .15 (15% voltage drop)
Using the above formula for cable #2:
2 times 12 times 4.5 times 2050 : divided by 10380 = 21.33 volt (total drop)
120 volt - 21.33 volt = 98.67 volts (measured at the farthest outlet)
21.33 volt divided by 120 volt = .18 (18% voltage drop)
I checked the voltage at each Monitor location using a True RMS meter and measured the following:
Monitor Load "not connected" - a steady 120.7 volts AC
Monitor Load "connected" - a variable of 0.5 to 2.0 volt drop
The Question is:
1) Why did I not see the voltage drop as calculated by the above Voltage
drop Calculation?
2)Could it be because the connected load is for electronic/solid state
equipment using a step down power supply?
3)I forwarded this survey to my Supervisors who in turn forwarded the
information to an Electrical Engineer in charge of the office building. The
reply I received back was that NEC Article 210.19 (A)(1) FPN No. 4 was not
a mandatory requirement. The installation by the contractor for this
wiring is correct and meets code requirements. I see in the Code Forum
that this is correct. I also see that in the Building Energy Code
ANSI/ASHRA/IESNA Standard 90.1-2004 -Chapter 8 that this is a
mandatory requirement for installations using the National Electrical Code.
Does anyone have any information pertaining to this requirement?
4)For the NEC Article 647 - Sensitive Electronic Equipment I do not see a
definition for this Article. Could CATV Monitors installations be required to
meet the requirements of this Article?
Thank you for you input in the above concerns.
I have been tasked to survey a job performed by an Electrical Contractor which I have a voltage drop concern.
The installation is as follows:
The scope of the work was to install 18 Flat Screen CATV Monitors on a total of 14 floors in a commercial office building. The wiring method was 10/2 MC cable and EMT conduit. There are two branch circuit home runs back to a junction box for each floor. The wiring continues back thru a Riser on each floor to a 3-phase 4-wire 120/208 volt sub-panel designated for the power of this equipment. The branch circuit MC cables have a one-way distance of 825.0 feet for cable #1 and 1025.0 feet for cable #2. Each Monitor has a 120 volt duplex receptacle, using a 2 wire factory cord and connector terminating into a power supply reducing the voltage to 13.5 volts for the CATV Monitor. Every Monitor has a load of 0.5 ampere and each branch circuit has a load of 4.5 ampere.
Using the NEC Article 210.19 (A)(1) FPN No. 4 there is a 3% voltage drop on Branch circuits and 2% voltage drop on Feeders.
The voltage drop calculation formula that I have always been taught to use is as follows:
VD=2KIL divided by CM
VD=2 times K times I times L divided by CM
(OR)
CM=2KIL divided by VD
CM=2 times K times I times L divided by VD
VD-Voltage Drop
K-Constant for the conductor material (12 for Copper and 18 for Aluminum)
I-Amperage of the circuit
L-Length of the circuit
CM-Circular mills of the conductor (10AWG has a 10380 circular mill area)
For Single phase 120 volt or Single phase 277 volt the total distance is measured from the (LINE) voltage source, to the connected (LOAD) of the circuit and measured back to the (NEUTRAL) connection. On multi-phase circuits the distance is measured (ONE-WAY) from the (LINE) to the (LOAD).
With this in mind the allowed 3% voltage drop on a 120 volt circuit is 3.6 volts.
The distance of cable #1 is 825.0 feet times 2 = 1650.0 feet.
The distance of cable #2 is 1025.0 feet times 2 = 2050.0 feet.
Using the above formula for cable #1:
2 times 12 times 4.5 times 1650 : divided by 10380 = 17.17 volt (total drop)
120 volt - 17.17 volt = 102.83 volt (measured at the farthest outlet)
17.17 volt divided by 120 volt = .15 (15% voltage drop)
Using the above formula for cable #2:
2 times 12 times 4.5 times 2050 : divided by 10380 = 21.33 volt (total drop)
120 volt - 21.33 volt = 98.67 volts (measured at the farthest outlet)
21.33 volt divided by 120 volt = .18 (18% voltage drop)
I checked the voltage at each Monitor location using a True RMS meter and measured the following:
Monitor Load "not connected" - a steady 120.7 volts AC
Monitor Load "connected" - a variable of 0.5 to 2.0 volt drop
The Question is:
1) Why did I not see the voltage drop as calculated by the above Voltage
drop Calculation?
2)Could it be because the connected load is for electronic/solid state
equipment using a step down power supply?
3)I forwarded this survey to my Supervisors who in turn forwarded the
information to an Electrical Engineer in charge of the office building. The
reply I received back was that NEC Article 210.19 (A)(1) FPN No. 4 was not
a mandatory requirement. The installation by the contractor for this
wiring is correct and meets code requirements. I see in the Code Forum
that this is correct. I also see that in the Building Energy Code
ANSI/ASHRA/IESNA Standard 90.1-2004 -Chapter 8 that this is a
mandatory requirement for installations using the National Electrical Code.
Does anyone have any information pertaining to this requirement?
4)For the NEC Article 647 - Sensitive Electronic Equipment I do not see a
definition for this Article. Could CATV Monitors installations be required to
meet the requirements of this Article?
Thank you for you input in the above concerns.
Looking for advice on the internet to take someone else to task?