At actual application we rarely know the resistance of our
equipment. What usually specified by manufacturer is only a power consumption
at nominal power supply (can be 24 VDC or 48 VDC). So, how to get the equation
to calculate the voltage drop across some distance with using the power
consumption data?
From previous illustration in the part 1, we could express the voltage drop
(Delta V) as below:
Delta V = V – VR2
Delta V = V – I*R2
Because I=V/RTOT then,
Delta V = V – ( (V/RTOT) * R2)
We take out the same V so that we could get the following
equation,
Delta V = V* (1 – (R2/RTOT) )
Since the RTOT = R1+R2+R3 then,
Delta V = V* (1 – (R2 / (R1+R2+R3) ) )
Since R1 = R2 = Length of cable (L) * Resistance of cable
(Rcable) then,
Delta V = V* (1 – (R2 / ((L*Rcable) + R2 + (L*Rcable)) )
Delta V = V* (1 – (R2 / ((2*L*Rcable) + R2)
Unfortunately we don’t know the R2 of the equipment (but we
only know the power consumption), then we should replace this in term of power
consumption (P).
R2 = P/I2
While I = P/V, then
R2 = P / (P/V)2
R2 = V2 / P
Now we substitute this R2 value to voltage drop equation
above.
Delta V = V* (1 – ( (V2 / P) / ((2*L*Rcable) + (V2 / P) )
Delta V = V* ( ((2*L*Rcable) + (V2/P) – (V2/P)) /
(2*L*Rcable) + (V2/P) )
Delta V = V* ((2*L*Rcable) / (2*L*Rcable) + (V2/P))
While (2*L*Rcable) + (V2/P) = (2*L*Rcable) + R2 = RTOT
Then we could simplify the above equation to be:
Delta V = V/RTOT * (2*L*Rcable)
Delta V = I * (2*L*Rcable)
Where,
Delta V = Voltage drop across some distance (Volt)
I = electric current (Ampere) = P/V
L = Length of cable (km or m)
Rcable = OHM/km
or OHM/m
Usually Rcable represents in OHM/km rather than in OHM/m.
And length of cable represents in m. In this case the voltage drop equation
will be:
Delta V = (I * 2*L*Rcable) / 1000
Let’s take some example.
We have a solenoid valve put at distance 250 meter from the
system cabinet (power supply putted). This solenoid valve has a power
consumption 5 watt@24 VDC. We need to check what cable size that can be used
with the requirement of voltage drop must not exceed 5% at solenoid valve
terminal. The following example cable data will be used:
Cable Cross Sectional Area

Rcable

1.0 mmsq

20 OHM/km

1.5 mmsq

15 OHM/km

2.5 mmsq

10 OHM/km

Calculate the electric current:
I = P/V = 5 watt / 24 V = 0.2083 A
Calculate the voltage drop by using formula:
Delta V = (I * 2*L*Rcable) / 1000 with the preliminary
Rcable using the 1.0 mmsq properties.
Delta V = (0.2083*2*250*20)/1000
Delta V = 2.083 Volt
Delta V = 2.083 Volt / 24 Volt *100%
Delta V = 8.67%
It exceeds our requirement which is 5%. So using 1.0 mmsq
cable isn’t sufficient for our application. Let’s try with 1.5 mmsq cable.
Delta V = (0.2083*2*250*15)/1000
Delta V = 1.56 Volt
Delta V = 1.56Volt / 24 Volt *100%
Delta V = 6.5%
By using 1.5 mmsq cables it still exceed the requirement.
Let’s try again with 2.5 mmsq cable, looks like this cable will suit for this
application.
Delta V = (0.2083*2*250*10)/1000
Delta V = 1.04 Volt
Delta V = 1.04Volt / 24 Volt *100%
Delta V = 4.34% which is less than 5% requirement.
Yes, this 2.5 mmsq cable is suitable for our applications.
Thus this cable size will be selected for this application.
salam wisdu,
ReplyDeletethank you for sharingg :)
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