Voltage drop calculations for CCVT's and Current TF's?

AI Thread Summary
Voltage drop calculations for cables used in instrumentation, particularly for Current Transformers (CT) and Potential Transformers (PT), are essential when dealing with long distances, such as 300 feet with #9 AWG cables. The resistance of the cable at 0.000729 ohms/ft can lead to significant voltage drop, potentially affecting measurements if it exceeds 5%. Users question whether voltage drop needs to be considered since the CT and PT outputs are for measurement rather than load applications. Calculating secondary voltage for CTs involves using the formula Vsec=Vpri*(Nsec/Npri), but there is uncertainty regarding the approach for CCVTs. Proper sizing of cables and understanding the impact of voltage drop on relay inputs is crucial for accurate readings.
this1ssteve
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I have not found a straightforward explanation on how to perform voltage drop calculations for cables used for instrumentations purposes. Suppose you have Current Transformer (CT) and Potential Transformer (PT) connections coming into a relay (IE SEL421). If we have a cable that runs from the yard equipment to the panels, let's say 300 Feet, using a #9 cable rated at 0.000729 ohms/ft, and we want the CT current to be 5A and the Potential coming into the relay to be at 67V, how do we determine this?
 
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this1ssteve said:
I have not found a straightforward explanation on how to perform voltage drop calculations for cables used for instrumentations purposes. Suppose you have Current Transformer (CT) and Potential Transformer (PT) connections coming into a relay (IE SEL421). If we have a cable that runs from the yard equipment to the panels, let's say 300 Feet, using a #9 cable rated at 0.000729 ohms/ft, and we want the CT current to be 5A and the Potential coming into the relay to be at 67V, how do we determine this?

Welcome to the PF.

What's a CCVT? And what's a TF?
 
this1ssteve said:
I have not found a straightforward explanation on how to perform voltage drop calculations for cables used for instrumentations purposes. Suppose you have Current Transformer (CT) and Potential Transformer (PT) connections coming into a relay (IE SEL421). If we have a cable that runs from the yard equipment to the panels, let's say 300 Feet, using a #9 cable rated at 0.000729 ohms/ft, and we want the CT current to be 5A and the Potential coming into the relay to be at 67V, how do we determine this?
I looked at the specs of the relay. Will the terminals even except a 9 AWG wire? Anyway, why are you needing to determine a voltage drop? Here's a snip from the relay's data sheet for the CT & PT inputs.

CT&PT input specs.jpg


If you don't get 67 volts from the PT when at the nominal line voltage, surely the relay software can provide a way to compensate I would think.

Regards
 
Berkeman:
-CCVT = Capacitive Coupling Voltage Transformer (used in Electrical Substations to step down huge potentials for metering purposes for protective relays)
-TF= Short for Transformer

Dlgoff:
So here is the scenario. We have a huge substation with very long runs between outdoor equipment and indoor relays. Let's say a protective relay needs to measure voltage and current on a line. It utilizes CCVT's (essentially to step down the huge voltages from the kilovolt level to 115V/67V) and CT's (Current transformers that are located on the bushings of a circuit breaker to step down the current from say 2000A to 5A) to be inputted into a protective relay in the control house. Well let's say the control house is located 1000 feet away and those cables are being ran with #9 AWG conductors from the outdoor equipment to the control house. With such a large run, there would be some sort of voltage drop to take into consideration. If there is too much loss (over 5% or so) you would need to size your cable to a bigger gauge. On a side note, the #9 conductor feeds into a panel, which has interconnecting wires that fit the size needed for the relay input.

I suppose my question is:
1) Would voltage drop be needed to be taken into consideration? It's not like the current or potential is being applied to a load, it is being used for measurement purposes, much like an ammeter or voltmeter on multimeter.
2) If so, how do we go about doing this? For CT's, I am not sure if I am doing the secondary voltage current correctly (Vsec=Vpri *(Nsec/Npri) ). For the CCVT's, I am clueless.
 
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