Voltage drop calculation across cable

Hi i'm more towards instrumentation and i need help with this calculation. Kindly please help step by step.

We have 4 flame detectors and 8 gas detectors in this one loop connected in series.

Each gas detector load is :

8.0 watts peak @ 24vdc

Each flame detector load is :

17 watts at 24 Vdc with EOL resistor installed and heater on maximum

Voltage at Unit Control Panel : 24VDC.
Cable length from PLC panel to package : 380m
Cable core size : 21Cx2.5mm2
According to cable spec: Maximum conductor resistance for 20C 2.5mm2 cable is 7.41ohm/km

What is the voltage drop?

Reason for this calculation is this power loop is having lower voltage than expected during commissioning. Hence a voltage drop calculation is recommended to be done .
 
We have 4 flame detectors and 8 gas detectors in this one loop connected in series.
Are you saying your 24VDC power supply is connected in series to each flame detector and gas detector? Are you sure it's not connected in parallel (+ to +, - to - at each device)? I've never heard of connecting power in series (24V supply + to first detector +, first detector - to second detector +, ..., last detector - to 24V supply -).

Or are you talking about a 4-20mA loop and each device is loop powered (which it seems impossible given the very high power requirements you stated)?

Each gas detector load is :

8.0 watts peak @ 24vdc

Each flame detector load is :

17 watts at 24 Vdc with EOL resistor installed and heater on maximum

Voltage at Unit Control Panel : 24VDC.
Cable length from PLC panel to package : 380m
Cable core size : 21Cx2.5mm2
According to cable spec: Maximum conductor resistance for 20C 2.5mm2 cable is 7.41ohm/km

What is the voltage drop?
Power = Voltage x Current (P = I x V), so Current = Power / Voltage (I = P / V)

Voltage = Current x Resistance (V = I x R)

The conductor resistance rating is in ohms/km, there's 1000m in 1 km, so your cable resistance is 0.00741 ohm/m. At 380m, the maximum cable resistance is 2.8158 ohms.

Calculate max current for each detector, add them up, multiply by the max cable resistance. That's your maximum voltage drop due to cabling.

Also, make sure your 24VDC power supply is capable of supplying the required power/current for all of the devices. Overloading a power supply can reduce the output voltage significantly (and may even damage the power supply over time).
 
May i know if this calculation is correct? By right the voltage drop should be less than 2V. I'm getting way higher values.

Also may i know when do we use this formula? >> Vd=(2ILR/1000)? Is it applicable here?
Gas Detector

8Watts @24VDC
P=IV
I=P/V
I=8/24
=0.3A
Therefore, 8 gas detectors
8x3A=2.4A

Flame detector
17W @ 24VDC
I=17/24
=0.7A

Therefore, 5 flame detectors
5X0.7 A=3.5A

Voltage drop across 8 gas detectors:

Vd=IxRxL
Vd=2.4Ax0.00741ohm/m x 380m
=6.75V

Voltage drop across 5 flame detectors:
Vd=0.7Ax0.00741ohm/m x 380m
=1.97V

Vd across cable: 6.75+1.97V
=8.72V
 
Are you saying your 24VDC power supply is connected in series to each flame detector and gas detector? Are you sure it's not connected in parallel (+ to +, - to - at each device)? I've never heard of connecting power in series (24V supply + to first detector +, first detector - to second detector +, ..., last detector - to 24V supply -).

Or are you talking about a 4-20mA loop and each device is loop powered (which it seems impossible given the very high power requirements you stated)?


Power = Voltage x Current (P = I x V), so Current = Power / Voltage (I = P / V)

Voltage = Current x Resistance (V = I x R)

The conductor resistance rating is in ohms/km, there's 1000m in 1 km, so your cable resistance is 0.00741 ohm/m. At 380m, the maximum cable resistance is 2.8158 ohms.

Calculate max current for each detector, add them up, multiply by the max cable resistance. That's your maximum voltage drop due to cabling.

Also, make sure your 24VDC power supply is capable of supplying the required power/current for all of the devices. Overloading a power supply can reduce the output voltage significantly (and may even damage the power supply over time).
May i know if this calculation is correct? According to vendor the voltage drop should be less than 2V. I'm getting way higher values.
In this attachment, it shows the the unit found voltage for power loop 1 and 2 at FGS devices gas detectors and flame detectors inside the package which is too low. (FYI)

Also may i know when do we use this formula? >> Vd=(2ILR/1000)? Is it applicable here?
Gas Detector

8Watts @24VDC
P=IV
I=P/V
I=8/24
=0.3A
Therefore, 8 gas detectors
8x3A=2.4A

Flame detector
17W @ 24VDC
I=17/24
=0.7A

Therefore, 5 flame detectors
5X0.7 A=3.5A

Voltage drop across 8 gas detectors:

Vd=IxRxL
Vd=2.4Ax0.00741ohm/m x 380m
=6.75V

Voltage drop across 5 flame detectors:
Vd=0.7Ax0.00741ohm/m x 380m
=1.97V

Vd across cable: 6.75+1.97V
=8.72V
 

Attachments

Thibhika,

That is a lot of current to be pulling through a small wire such an extremely long distance - Yes there is going to be a significant voltage drop. (Still not sure how exactly its wired, is everything on 1 or is there 2 loops? is there 4 or 5 flame detectors) Heres 8 gas detectors on one 380m conductor.
1641788169357.png

Do you have any spare wires ran between the PLC and the package that you could use? To either split up the load or to use in parallel with the existing conductor.
 
May i know if this calculation is correct? According to vendor the voltage drop should be less than 2V. I'm getting way higher values.
No, I don't think it's correct. There's one important clarification in the cable length, in that you must consider the voltage drop on both both the positive conductor and the negative conductor, resulting in multiplying your length by 2 to get the total length of the power wires (more on this below). You also need to be aware that this calculation is a worst-case calculation, since you are using the maximum power draw ratings for your gas and flame detectors. Any measurements done under normal operation (i.e. not maximum power draw) may be far less than the calculated maximum voltage drop.

Secondly, your calculation of the flame detectors is incorrect. Do you have 4 or 5 flame detectors? In your OP, you state 4, but now you use 5. Also, in your calculation here, you're only calculating for a single flame detector (note that you should be using 3.5A, not 0.7A):
Flame detector
17W @ 24VDC
I=17/24
=0.7A

Therefore, 5 flame detectors
5X0.7 A=3.5A

Voltage drop across 5 flame detectors:
Vd=0.7Ax0.00741ohm/m x 380m
=1.97V
Also may i know when do we use this formula? >> Vd=(2ILR/1000)? Is it applicable here?
Yes, I believe this formula is correct and applicable here, assuming the following:
- L is the length of the cable in meters
- R is the resistance/km of each conductor in the cabling

Therefore, the 2LR/1000 component of the formula is used to calculate the total resistance of the cabling. The factor of 2 is there since you need to include the positive conductor run from the supply to the device and the negative conductor run back from the device to the supply. The 1000 is there to convert from km to m.

As MattyIce stated, you are pulling a lot of current through a wire that has significant resistance, which results in a large voltage drop. Therefore, you must change either the wiring topology to result in less current per cable or use lower resistance (i.e. lower gauge / thicker) cabling.
 
Top