· Code Requirement for Voltage Drop Calculations?
NFPA 72 2013 Edition Section 7.2.1 - "Where documentation is required by the enforcing authority, the following list shall represent the minimum documentation required for all fire alarm and emergency communication systems, including new systems and additions or alternations to existing systems.", Within this list, you will find #7, Battery Calculations and #8, Voltage Drop Calculations.
Keep in mind almost all fire alarm control units are 24
volts DC. Also note that there are a few
fire alarm control panels that are 12 volt DC.
Now these panels are typically combination fire and burglar system. Just remember that the calculations for NAC voltage
drops are the same for these systems, however the cut off voltage for a 12 volt
system will be roughly half that of a 24 volt system.
· What is the reason for voltage drop calculations?
NAC voltage drop calculations are critical for determining
if your notification appliances will in fact work with the provided head
end equipment. (This is of course based
on the installation contractor installing the system per plans and noted wiring distances). If you perform your NAC voltage drop
calculations correctly at the time of design, you will know exactly how many
remote power supplies and NAC circuits are needed as well as the wall space
requirements and 120 VAC connections required by the electrical contractor. Keep in mind that this is a requirement of
NFPA 72.
· Voltage Drop Calculation Methods
There are primarily two methods to perform a NAC voltage
drop calculation. These methods are
better known as “Point to Point (PTP)” and “End of Line (EOL)”.
· Point to
Point Voltage Drop requires much more math than the
“EOL” method. However, the extra work
pays off as this method is more accurate.
- Designers generally use this method with a spreadsheet as the math can become tedious
- This is the method typically used by panel manufacturers within their own calculation programs
- Since it is less conservative of the “EOL” method, it allows for more devices on a circuit.
- There are cases of a 30% difference between the PTP and EOL methods
- Can be done easily by hand or with a calculator
- Results are less accurate that provide lots of head room for future expansion
Starting Voltage and Cut-Off Voltage
UL (Underwriters Laboratories) 864, 9^{th} Edition
Standards for Fire Alarm Control Panels:
- All panels must have a demonstrated 20.4 VDC panel cut off voltage.
You may be asking, “Where did they come up with 20.4 VDC on a
24 Volt system?”
It is actually quite simple.
20.4 VDC is 85% of 24 VDC. Or like we stated earlier, there are a few 12
VDC systems floating around. In their
case, the demonstrated voltage shall be 10.2 VDC. Once again, 10.2 VDC is 85% of 12 VDC.
Now, above we mentioned a term “Cut-Off Voltage”. All fire alarm control units (FACU) have and
internal voltage drop. The voltage at
the actual NAC output terminal is always less than 20.4/10.2 volts at cut-off.
- This amount of drop varies with every panel. The variance can be as much as .5 VDC to 2.5 VDC.
Keep in mind that this value is not often found within the
panel documentation. I have found that
the easiest way to obtain this value is to contact the panel manufacture and
get it in writing.
You now may be asking yourself, “Why is it so critical that I
get this value from the manufacturer and not just use the 20.4/10.2 value
figured from the 85% set forth by UL 864 9^{th} Edition?”
In order for your NAC voltage drop calculations to be
precise and as accurate as possible based on the facts and information
provided, you must use the specific panel/power supply terminal cut-off value.
How to
determine your NAC Voltage Drop using the End of Line method:
Step #1
Add up the total current draw for your entire notification
appliance circuit. This is based on the
manufacture, type (strobe only, horn/strobe, mini horn, audible level, wall,
ceiling, etc.) Make sure to consult with
the appliance documentation to get these figures.
Step #2
Add up the total wire length for the run and multiple it by
2 (if class B). The 2 represents the number
of conductors used in the run.
Step #3
Multiply the total wire length times the wire resistance
value per foot for a total circuit wire resistance. The wire resistance per foot can be found in
Table 8 “Conductor Properties” in Chapter 9 of the National electrical Code.
Step #4
Using Ohm’s Law we know that Current (I) x Resistance (R) =
Voltage (E). Simply take the total
current found in step #1 and multiply it by the resistance found in Step
#3. This will give you the volts
dropped.
Step #5
Subtract the volts dropped from the panel/power terminal
cut-off voltage to obtain the voltage that will be supplied to the last
appliance on the circuit. This value MUST exceed 16 volts.
Keep in mind that this method is not as accurate as the
Point to Point method. This method
assumes that the voltage drop at each appliance will be the same when in
reality, they are not.
Below is an
example of an End of Line voltage drop calculation:
Diagram notes:
- · We will assume that the terminal cut-off voltage is .5 volts below the 20.4 VDC giving us a voltage of 19.9.
- · Use the wire lengths shown in the diagram
- · V1=85mA / V2=75mA / V3=115mA / V4=100mA
- · The circuit is using #12 AWG wire
- · Use the Table 8 of the NEC (National Electric Code) provided previously in this document
Using
the Diagram and notes above, can you provide the voltage drop for this circuit
using the End of Line method? Give it a
try and when you are ready move on to the next page where we will break it down
for you.
End of Line
Voltage Drop Calculation Break Down:
Step #1
Add up the total current for all four notification
appliances within the circuit. We know
that V1 = 85 mA, V2 = 75 mA, V3 = 115 mA and V4 = 100 mA. The total of all four of these is 375 mA.
Step #2
Add up the total wire length and multiply it by 2. We know based on the diagram that the first
section is 200 feet, the second section is 150 feet, the third section is 25
feet and the final section is 70 feet.
This totals up to 445 feet x 2 = 890
Total Feet
Step #3
We know from the Table 8 “Conductor Properties” we have a
value of 1.98 Ohms/1000 feet of #12 AWG stranded uncoated wire. To find our resistance for our circuit simply
take the total wire length (890 Feet) and divide it by 1000. This gives us a value of .89. Now take .89 and multiply it by the 1.98
value found in the NEC Table. (.89 x
1.98 = 1.7622 Ohms of Resistance)
Step #4
Using Ohm’s Law we know that Current (I) x Resistance (R) =
Voltage (E). Take the total current
found in Step #1 (.375) and multiply it by the total found in Step #3
(1.7622). .375 x 1.7622 = .660825 volts dropped.
Step #5
Finally we need to subtract the .660825 volts from our
terminal cut-off voltage. We know from
the previous diagram and notes that we have a terminal cut-off voltage of 19.9
volts. 19.9 volts - .660825 = 19.239 Volts at the last appliance.
Point to
Point Voltage Drop Calculation Break Down:
Diagram notes:
- · We will assume that the terminal cut-off voltage is .5 volts below the 20.4 VDC giving us a voltage of 19.9.
- · Use the wire lengths shown in the diagram
- · V1=85mA / V2=75mA / V3=115mA / V4=100mA
- · The circuit is using #12 AWG wire
- · Use the Table 8 of the NEC provided previously in this document
Calculation Breakdown:
To perform a point to point voltage drop calculation is
basically the same as the End of Line method however; we are going to do a
breakdown of each path/appliance.
Calculation #1
- · First wire run section resistance multiplied by the total current for appliance V1, V2, V3 and V4
- · Subtract the total from the terminal cut-off voltage to get the voltage drop for V1.
Calculation #2
- · Second wire run section resistance multiplied by the total current for appliances V2, V3 and V4.
- · Subtract the total of V1 from the terminal cut-off voltage to get the voltage drop of V2
Calculation #3
- · Third wire run section resistance multiplied by the total current for appliances V3 and V4.
- · Subtract the total of V2 from the terminal cut-off voltage to get the voltage drop of V3
Calculation #4
- · Fourth wire run section resistance multiplied by the total current for appliances V4.
- · Subtract the total of V3 from the terminal cut-off voltage to get the voltage drop of V4
If this last value
is greater than 16 volts, the circuit should work.
Using
the Diagram and notes above, can you provide the voltage drop for this circuit
using the Point to Point method? Give it a
try and when you are ready move on to the next page where we will break it down
for you.
Calculation # 1
- · 200 Feet x 2 = 400 Feet. 400 / 1000 = .4 x 1.98 = .792 Ohms (From the FACU to V1)
- · .792 x .375 (Current of all Appliances) = .297 volts dropped @ V1
- · 19.9 (Terminal Cut-Off Voltage) - .297 = 19.603 VDC @ V1
Calculation #2
- · 150 Feet x 2 = 300 Feet. 300 / 1000 = .3 x 1.98 = .594 Ohms (From the FACU to V1)
- · .594 x .290 (Current of Appliances V2-V4) = .17226 volts dropped @ V2
- · 19.603 (Terminal Cut-Off Voltage) - .17226 = 19.43074 VDC @ V2
Calculation # 3
- · 25 Feet x 2 = 50 Feet. 50 / 1000 = .05 x 1.98 = .099 Ohms (From the FACU to V1)
- · .099 x .215 (Current of Appliances V3-V4) = .021285 volts dropped @ V3
- · 19.43074 (Terminal Cut-Off Voltage) - .021285 = 19.40946 VDC @ V3
- · 70 Feet x 2 = 140 Feet. 140 / 1000 = .14 x 1.98 = .2772 Ohms (From the FACU to V1)
- · .2772 x .200 (Current of Appliance V4) = .05544 volts dropped @ V4
- · 19.40946 (Terminal Cut-Off Voltage) - .05544 = 19.35402 VDC @ V4
Both of these
calculations are commonly used and widely accepted by your local AHJs. As you can see the PTP method came up with a
total voltage drop of 19.35402 while the EOL method came up with 19.239. Remember both of these examples used the same
parameters. I personally recommend using
the Point to Point method purely based on its accuracy.
This comment has been removed by the author.
ReplyDeleteHi Kyle. Not sure if anyone has already informed you, but in the Point to Point calc above there's a math error.
ReplyDelete.594 x .290 (Current of Appliances V2-V4) = .16356 volts dropped @ V2
The actual number is .594 x .290 = 0.17226
So all the values from that point on are slightly off. In the end the difference is only hundredths or thousandths of a difference. So no biggie. But since this is a teaching website I thought you should know.
corrected. Thank you so much for your feedback!
Delete