Hello. Ive been doing the fire alarm submittals for our company for a few years now for Kentucky and Florida. I have recently started doing some for a sister company in Las Vegas and ive run into something they are asking for that no one else has in the other states. Ive done all the calculations for the NAC circutis of FACP, but the AHJ is aksing for the drops for the voice circuits too. I looked on the manufacturer site for the same style tools i use for their FACPS and they have every type of panel and model except for the voice amplifiers. I can not find any information in the datasheets of the speakers as far as amps or really anything other than they can be tapped at certain wattage and they need X volts to operate. So far the only information I know is the resistance of the wire for each circuit. Any ideas from anyone would be great
Voice Evac voltage Drop
- Thread starter Mocephus
- Start date
- Status
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
It looks to me like an AHJ is asking for something because one of their guys thought it would be a good idea without having any idea how to generate a result or interpret any value returned to them. It happens. Ask them how much voltage drop they are willing to accept and give them a number smaller than that. Prove me wrong.
Realistically, ask them for a sample calculation and do yours the same way without analyzing whether or not it makes any sense.
Last edited:
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Here's an idea that might fly with the AHJ: Calculate the round trip AC resistance of the wire from Table 9; call it I1. Call the impedance of the speaker I2. Percent voltage drop in the wiring is (I1 / (I1 + I2)) X 100%.
Here is what they sent me to calculate the loss. The odd part is when I use this formula unless im doing something wrong the circuit with the longer distance and more total used wattage is showing less loss. I attached the spreadsheet i was using to calculate it as well. The table for ohms on certain wire sized shows i have 9.78 for 16awg. i dont know i agree in those numbers but it is what they say i have to use
PLoss = 10 * Log [1 - ((2 * RL) / (2 * RL + (VLine squared / PRated))]
RL = (RRef / 1000) * D
With variables defined as follows:
D = length of wire used (in feet)
PLoss = power loss (in dB)
PRated = power driven on line from the amplifier (in watts)
RL = wire gauge resistance (in ohms)
RRef = wire resistance based on gauge of wire used (in ohms/ft.) 9.78
VLine = voltage on line (typically 25 volts or 70 volts) 25
Alternatively the distance may be calculated using a calculation similar to:
D = (61 / RRef) * (VLine squared / PRated)
So if someone want to use their brain and see if their numbers match mine id appreciate it
C3 16 AWG, 220 feet, 25 volt system, 13.75 watts on circuit
C4 16 AWG, 195 feet, 25 volt system, 9.5 watts on circuit
PLoss = 10 * Log [1 - ((2 * RL) / (2 * RL + (VLine squared / PRated))]
RL = (RRef / 1000) * D
With variables defined as follows:
D = length of wire used (in feet)
PLoss = power loss (in dB)
PRated = power driven on line from the amplifier (in watts)
RL = wire gauge resistance (in ohms)
RRef = wire resistance based on gauge of wire used (in ohms/ft.) 9.78
VLine = voltage on line (typically 25 volts or 70 volts) 25
Alternatively the distance may be calculated using a calculation similar to:
D = (61 / RRef) * (VLine squared / PRated)
So if someone want to use their brain and see if their numbers match mine id appreciate it
C3 16 AWG, 220 feet, 25 volt system, 13.75 watts on circuit
C4 16 AWG, 195 feet, 25 volt system, 9.5 watts on circuit
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
This is yet another example of AHJ requirements running amok, IMO.
Agree, but their claim is they have required it for years in Vegas so Im just trying to comply
ggunn
PE (Electrical), NABCEP certified
- Location
- Austin, TX, USA
- Occupation
- Consulting Electrical Engineer - Photovoltaic Systems
Get a power user of Excel to generate a spreadsheet with formulas where you can drop in the variables and have it spit out the answer.
So i did 1 circuit by hand and then put it all in excel and it came up with the same answer, but when i did the 2nd circuit which is longer and using more wattage the power loss was less is where i got confused
wwhitney
Senior Member
- Location
- Berkeley, CA
- Occupation
- Retired
Caveat: I have experience with formulas, not with this topic.
If I'm not mistaken, this is a formula for power transmitted, not for power loss. The 10 * and Log at the beginning are just putting the results on a logarithmic scale like dB, so let's ignore them.
The value inside the log is of the form (1 - A / (A+B)). Which is the same as B / (A + B).
If this is a simple series circuit of some wires and one device, then here A is the resistance of the wires (twice the one way resistance) and B is the impedance of the device. That is, from P = V2/R, we get R = V2/P.
So the argument of the Log is the ratio of the device impedance to the total circuit impedance. Which for a series circuit is the ratio of the power dissipated at the device to the power dissipated in the total circuit.
If you want power loss, I believe you'd omit the "1 minus" portion of the formula, so you'd be using the ratio of the wire resistance to the total circuit impedance.
Cheers, Wayne
Thanks Wayne. When i get rid of that 1- the total PLoss looks way more realistic. I was getting over 9 volts as a loss for each circuit now its less than 1.
- Status