Wires and Voltage Ratings Calculation _ Utilities _ Notifier

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    08/05/13 Wires and Voltage Ratings Calculation | Utilities | Notifier

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    Wires and Voltage Ratings Calculation

    CABLES DIMENSIONING

    This is a method to help on calculations relative to the electric cables dimensioning in the direct current operating

    systems.

    From the following formulas it is possible to obtain:

    which voltage drop will be in relation to the line length, the current draw, the section of the cable usedhow much current can be drawn in the end of a line depending on its length, the section of the cable used

    and the voltage drop accepted

    which cable section must be used to have a certain voltage drop in relation to the length and the current

    draw of the load

    Each conductor, depending on its section, gives a certain resistance to the passage of current. The following is a

    table that summarizes the specific resistance values for the copper cables and the sections more usually used.

    Cable section in mm2Specific resistance

    0,22 0,090

    0,50 0,0351,00 0,018

    1,50 0,012

    2,00 0,009

    2,50 0,007

    3,00 0,006

    Definitions

    V. drop: Voltage drop on the load (difference between the output tension from the supplier and the minimum

    operating tension of the connected device)

    Length: Line length ( distance from the supplierx 2 ) in metres

    I. load: Max. Current drawn from the device connected in the end of the line ( in mA )

    Section: Section of the cable used ( which corresponds to the specific resistance as shown in the table

    above) in mm2

    Calculation of the VOLTAGE DROP

    (Specific resistance x Line length) x I.load

    V. drop = = Volt

    1000

    EXAMPLEA sounder that draws max. 2A must be installed on a 100 metres line. The line is installed using a cable with a section

    of 1 mm2.

    Which voltage drop will be in the end of the line?

    Solution: ( 0,018 x 200 ) x 2000 / 1000 = 7,2 Volt

    Calculation of the maximum current allowed in the end of the line

    V. drop max allowed x 1000

    I.load max = = mA

    Specific resistance x Line length

    EXAMPLE

    Is it possible to install a 24 V sounder that draws 2A max. in the end of a line 100 metres long using a cable with the

    section of 1 mm2 taking into consideration that the sounder operates with a minimum of 13 V ?

    Solution:

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    ( 24 - 13 ) x 1000 / 0,018 x 200 = 3055 mA = 3 A .... so the answer is : YES.

    And if the sounder operates with a minimum of 18 V ?

    Solution:

    ( 24 - 18 ) x 1000 / 0,018 x 200 = 1666 mA = 1,6 A .... so the answe is : NO.

    Dimensioning of the CONDUCTORS SECTIONS that supply a sensor or a call point

    Together with the choice of the battery capacity and the definition of the current suppliable by the supplier, the

    correct dimensioning of the cables sections allows the devices to have optimal tension values according with the

    regulations established by the manufacturer.

    Supplying a sensor or a call point with a lower tension than the minimum value stated means for the device a

    situation of instability, low efficiency, poor immunity to the noises.

    The following data must be considered:

    Example

    Voltage at

    SourceVs

    For a correct and preferable result consider a

    critical situation as mains loss24 V

    Minimum

    voltage on

    sensor

    Vc Pointed out by the manufacturer data 13,5 V

    Sensor current Ic

    Refer to manufacturer data sheet or measured by

    a multimeter (in milliampere)

    NOTE: In the case of NOT self-powered devices

    must be used the higher absorption in the

    different conditions: Service, Stand-By, Alarm

    10 mA

    Line length LSegment of cable between the source and the

    sensor (in metres)50 m

    Calculated the data, they can be insert in an easy formula to obtain the measure of the smallest cable sec tion that is

    able to guarantee the optimal functioning: the minimum section, measured in mm2, must be equal or bigger than:

    Section = ( 2L x Ic x 0,038 ) : [ ( VS - Vc ) x 1000 ]

    EXAMPLE

    Section = (100 m x 10 mA x 0,038 ) : ( 24 V - 13,5 V) x 1000 ) = 0,0036 mm2

    The c able for the installation must have the c onductors sec tion equal or bigger than 0,04 mm2

    The method described can be used to dimension with a good approximation a more complex tension network, with

    ramifications.

    In this case it is enough to identify all the connection points from the source on, to effect the calculation for each

    single segment between two points and then summing all the results.

    IMPORTANT NOTE

    Take always into account the minimum sections given by the bodies of regulations that happen to be bigger than

    those obtained from the calculations.