UTC Fire and Security DET-TRONICS 95-8533, Инструкция по эксплуатации

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7.1 3-2 95-8533
The Eagle Quantum Premier system utilizes a power
supply that provides an isolated 24 vdc battery
backed-up power to the fire protection devices as
described in NFPA 72. More than one power supply
may be used in a system to provide power to different
sets of equipment as part of the system.
The power supply wiring may consist of one or more
daisy-chained wire segments providing power to the
devices. For each of the daisy-chained wire
segments, the installer must calculate the voltage
drops that occur across the devices in order to
determine the gauge of the wire that will be installed.
A power supply wiring diagram should contain
information describing wire distances and current
draws associated with all devices connected to the
wire segment. A typical power supply wiring
recommendation is that the voltage drop from the
power source to the end device should not exceed
ten percent. Using 24 vdc as a reference, the
maximum voltage drop should not exceed 2.4 vdc. A
wire gauge must be selected to ensure that the end
device has at least 21.6 vdc or higher.
In order to calculate the power supply voltage for the
end device, calculate the voltage drops that occur
due to each wire segment between the devices. This
involves determining the total current draw and the
two conductor wire resistance per each wire segment.
Example:
Can 18 AWG wire be used to power three devices
from the 24 vdc power supply? Refer to the figure
below for wiring and device current draw information
along with voltage drop calculations.
Answer: If the Authority Having Jurisdiction (AHJ)
requires a voltage loss of 10% or less, only 16 AWG
wire could be used, since the end device would
require 21.4 vdc. If there is no local requirement, then
18 AWG wire could be used to provide power to the
devices.
Device 1
65 mA
Current Draw
Device 2
65 mA
Current Draw
Device 3
565 mA
Current Draw
24 vdc Power
Supply
18 AWG Single Wire Resistance: R = 0.6385 Ohms per 100 ft
2 Conductor Resistance: CR = 2 • R
Device 1 Voltage = Supply Voltage – (Voltage Drop)
= 24 – (I • CR)
= 24 – (0.695 • 0.6385)
= 23.55 vdc
Device 2 Voltage = Device 1 Voltage – (Voltage Drop)
= 23.55 – (I • CR)
= 23.55 – (0.630 • 1.9155)
= 22.35 vdc
Device 3 Voltage = Device 2 Voltage – (Voltage Drop)
= 22.35 – (I • CR)
= 22.35 – (0.565 • 1.9155)
= 21.27 vdc
50 ft
0.6385 Ohms
150 ft
1.9155 Ohms
150 ft
1.9155 Ohms
Total Current
695 mA
Total Current
630 mA
=
Device 2
+
Device 3
Total Current
565 mA
=
Device 3