10 second rating of resistor in a neutral grounding transfomer?

[gauthamweb]

Hi,

I'm trying to calculate the rating of resistor used in neutral grounding transformer. I know the capacitance to ground values, and hence I know how to calculate the size of the resistor in ohms in accordance to that.

However, after choosing a resistor, I am asked to specify the 10 second rating. Please tell me if the value of current I get in the secondary of the neutral transformer during a single line to ground fault is the 10 second rating of resistor itself or if ten times (usually) that value becomes its '10 second rating'?

 

[jumper]

Since you are a student I assume this a homework problem.

We will help, but we will not do your work.

What is your answer and how did you derive it?

 

[Ingenieur]

Go snoop around the Post Glover site
read this http://www.postglover.com/wp-content/uploads/downloads/2014/04/NG112-06_Technical_Information.pdf

you are on the right track
Current is constant for the ratings
time changes

we spec continuous in case protection fails
can't afford to lose the grounding path

 

[topgone]

Go snoop around the Post Glover site
read this http://www.postglover.com/wp-content/uploads/downloads/2014/04/NG112-06_Technical_Information.pdf

you are on the right track
Current is constant for the ratings
time changes

we spec continuous in case protection fails
can't afford to lose the grounding path

Tip:
Try researching for the time required before a certain type of metal melts. You only need to design your resistor to withstand that fault current level and duration (10 seconds), else your resistor will be a very large equipment!

 

[Ingenieur]

Tip:
Try researching for the time required before a certain type of metal melts. You only need to design your resistor to withstand that fault current level and duration (10 seconds), else your resistor will be a very large equipment!

It depends whether you want to trip power, ie, ground fault protective relaying or you want annunciation so you can continue to operate while you isolate and rectify the fault

the IEEE has ratings from 10 sec all the way to essentially continuous
1 min, 10 min, etc

• Steady-State Rating
This rating applies where the NGR is expectedto be operating under ground fault conditions formore than an average of 90 days per year and/or it is desirable to keep the temperature risebelow 385 ̊C.


We use 25 A on 12470 systems and 15 A 1000, 600 and 480
we use 3 levels of instantaneous tripping with 0.25 sec on the back up for coordination
with 2 points of interruption

we still want the ngr to be rated continuous in case something hangs up
unlikely but this is a life safety system

 

[topgone]

It depends whether you want to trip power, ie, ground fault protective relaying or you want annunciation so you can continue to operate while you isolate and rectify the fault

the IEEE has ratings from 10 sec all the way to essentially continuous
1 min, 10 min, etc

• Steady-State Rating
This rating applies where the NGR is expectedto be operating under ground fault conditions formore than an average of 90 days per year and/or it is desirable to keep the temperature risebelow 385 ̊C.


We use 25 A on 12470 systems and 15 A 1000, 600 and 480
we use 3 levels of instantaneous tripping with 0.25 sec on the back up for coordination
with 2 points of interruption

we still want the ngr to be rated continuous in case something hangs up
unlikely but this is a life safety system

Still, given those voluminous reference you have, you don't get to know how to compute for the short-time ratings! Yes?

 

[Ingenieur]

Still, given those voluminous reference you have, you don't get to know how to compute for the short-time ratings! Yes?

In my sleep
determine total C leakage current, usually <2A in my systems
determine the gnd fault frame potential you are designing for
I like to shoot for 40 vac
I allow for a ground ckt of 2.7 Ohm, we set our monitors at 3
40/2.7 = 15 A
must be at least twice the C current

for a 480 vac system the V across the ngr is 277 (480/sqrt3)
277/15 = 18.5 Ohm

ngr rating
15 A
480 we spec at least ph V for transients
18.5 Ohm
continuous
power 15^2 x 18.5 = 4.2 kW

the current rating is the same regardless of the time spec'ed
the time spec is determined by the application, eg, the engineer
just more wire, thicker, more insulators and heat sinks

protection
zero seq on the feeder set at 6 A or less
i on the ngr leg set at 7.5 A or less
v across the ngr with a small td set at 7.5 x 18.5 vac ~ 138 vac or less

the zero seq trips the feeder
the other 2 trip the main in sequence

 

[gauthamweb]

Since you are a student I assume this a homework problem.

We will help, but we will not do your work.

What is your answer and how did you derive it?

I had values of various capacitances in the generator system, using which I found Xc. That turned out to be 8187 ohm (one phase to ground). We have an 18 kV system in the generator. I chose the Line to neutral voltage times 1.1 margin, and approximated the primary of the neutral transformer to 12 kV. I kept a 240 volt secondary. I know I need a resistor that's about Xc/3. So that way, on the 240 volt side 3R*(12K/240)= 8187 ohm. So my resistor is 1.1 ohm.

What is my 10 second rating? I'm guessing that the 240 Volt secondary squared divided by the value of 1.1 is my 10 second rating.

 

[Ingenieur]

I had values of various capacitances in the generator system, using which I found Xc. That turned out to be 8187 ohm (one phase to ground). We have an 18 kV system in the generator. I chose the Line to neutral voltage times 1.1 margin, and approximated the primary of the neutral transformer to 12 kV. I kept a 240 volt secondary. I know I need a resistor that's about Xc/3. So that way, on the 240 volt side 3R*(12K/240)= 8187 ohm. So my resistor is 1.1 ohm.

What is my 10 second rating? I'm guessing that the 240 Volt secondary squared divided by the value of 1.1 is my 10 second rating.

what is the configuration of the grounding xfmr?
138/240 Y ?

what current will flow thru the R?

 

[gauthamweb]

what is the configuration of the grounding xfmr?
138/240 Y ?

what current will flow thru the R?

The grounding xfmr is a single phase transformer. I'm taking a 12kV primary to 240 volt secondary transformer. The current flowing through it will be 188 A during a single phase to ground fault.

 

[Ingenieur]

The grounding xfmr is a single phase transformer. I'm taking a 12kV primary to at240 volt secondary transformer. The current flowing through it will be 188 A during a single phase to ground fault.

The ngr needs rated for 188? A for 10 sec

or 240/1.1 = 218 A

 

[gauthamweb]

The ngr needs rated for 188? A for 10 sec

or 240/1.1 = 218 A

Sorry that's 218... I took the voltage that would appear if 18/sqrt.3 reflected on the secondary instead of 12 kV reflecting on secondary. We'll assume the entire 240 V appears and hence let's take 218 A.

 

[topgone]

I had values of various capacitances in the generator system, using which I found Xc. That turned out to be 8187 ohm (one phase to ground). We have an 18 kV system in the generator. I chose the Line to neutral voltage times 1.1 margin, and approximated the primary of the neutral transformer to 12 kV. I kept a 240 volt secondary. I know I need a resistor that's about Xc/3. So that way, on the 240 volt side 3R*(12K/240)= 8187 ohm. So my resistor is 1.1 ohm.

What is my 10 second rating? I'm guessing that the 240 Volt secondary squared divided by the value of 1.1 is my 10 second rating.

Your transformer rating is 75 kVA. If you design the transformer to withstand a 10-sec fault duration = 7 kVA(overload factor = 10.5)
Here:
phase amps = 12 kV/8187 ohms = 1.4A;
neutral amps = 3 x 1.4 = 4.2A;
kVA rating = 4.2 *18 kV = 75 kVA
10-sec rating = 75/10.5 = 7 kVA

 

[Ingenieur]

Your transformer rating is 75 kVA. If you design the transformer to withstand a 10-sec fault duration = 7 kVA(overload factor = 10.5)
Here:
phase amps = 12 kV/8187 ohms = 1.4A;
neutral amps = 3 x 1.4 = 4.2A;
kVA rating = 4.2 *18 kV = 75 kVA
10-sec rating = 75/10.5 = 7 kVA

???
1.4 primary C coupling current
he designed for a factor of 3 or 4.2

but it only sees 12kv not full phase v
it is located between the gen X0 and earth
4.2 x 12000 = 50.4 kva

Seconday
12000/240 x 4.2 = 210 A 218 without the rounding
or 240 V x 210 A = 50.4 A

xfmr is 50.4 kva
ngr has to be rated to carry 210 A for 10 sec

where does the 'overload factor' come from?
he wanted the ngr rating
ngr ratings are based on current and time for a given voltage
IEEE standards determine time rating class

nrg
12/18 kv
220 A
10 sec IEEE class

Xfmr
12000/240
1 ph
50 kva

 

[Ingenieur]

What is impressive is they are teaching this subject matter

in the last 30+ years power engineering fell out of vogue
telecom, computers, signals, integrated electronics were the focus

now that the current crop of power engineers are waddling towards the rocking chair there is going to be a shortage...at a time when most needed

infrastructure is nearing end of life cycle
new topology of distributed generation
Effeciency driven tech like facts/hvdc
environmental issues, getting more out of existing row's, cleaner generation

it's good to see but we need programs in the US too

 

[Ingenieur]

My previous link
pg 12 shows op's exact setup
pg 11 shows ngr spec: V, I and desired time rating

The neutral grounding resistor is rated as follows:

• Voltage: Line-to-neutral voltage of the system to which it is connected.
• Initial Current: The initial current which will flow through the resistor with rated voltage applied.
• Time: The “on time” for which the resistor can operate without exceeding the allowable temperature rise.

IEEE-32-1972 Standards
IEEE-32 is the standard used for rating and testing neutral grounding resistors. The most important parameters to consider from the IEEE-32 are: the allowable temperature rises of the element for different “on” times; the applied potential tests; the dielectric tests, and the resistance tolerance tests that are required. Post Glover Neutral Grounding Resistors are designated and built to pass all these rigorous tests.
• Time Rating
IEEE Standard 32 specifies standing time ratings for Neutral Grounding Resistors (NGRs)with permissible temperature rises above 30 ̊C ambient as shown in Table 3. Time ratings indicate the time the grounding resistor can operate under fault conditions without exceeding the temperature rises.
• 10-Second Rating
This rating is applied on NGRs that are used with a protective relay to prevent damage to both. the NGR and the protected equipment. The relay must clear the fault within 10 seconds.

a 10 sec rated ngr is allowed a 760 C rise from ambient 30 C

 

[topgone]

My previous link
pg 12 shows op's exact setup
pg 11 shows ngr spec: V, I and desired time rating

The neutral grounding resistor is rated as follows:

• Voltage: Line-to-neutral voltage of the system to which it is connected.
• Initial Current: The initial current which will flow through the resistor with rated voltage applied.
• Time: The “on time” for which the resistor can operate without exceeding the allowable temperature rise.

IEEE-32-1972 Standards
IEEE-32 is the standard used for rating and testing neutral grounding resistors. The most important parameters to consider from the IEEE-32 are: the allowable temperature rises of the element for different “on” times; the applied potential tests; the dielectric tests, and the resistance tolerance tests that are required. Post Glover Neutral Grounding Resistors are designated and built to pass all these rigorous tests.
• Time Rating
IEEE Standard 32 specifies standing time ratings for Neutral Grounding Resistors (NGRs)with permissible temperature rises above 30 ̊C ambient as shown in Table 3. Time ratings indicate the time the grounding resistor can operate under fault conditions without exceeding the temperature rises.
• 10-Second Rating
This rating is applied on NGRs that are used with a protective relay to prevent damage to both. the NGR and the protected equipment. The relay must clear the fault within 10 seconds.

a 10 sec rated ngr is allowed a 760 C rise from ambient 30 C

Google is your friend. Try harder, so your professor will give you an "A".

 

[Ingenieur]

Google is your friend. Try harder, so your professor will give you an "A".

That is from an ngr manufactuers manual
the link of which I posted
what you previously posted is nonsensical and does not give the ngr current and undersizes the xfmr by a factor of 10.5

Originally Posted by topgone
Your transformer rating is 75 kVA. If you design the transformer to withstand a 10-sec fault duration = 7 kVA(overload factor = 10.5)
Here:
phase amps = 12 kV/8187 ohms = 1.4A;
neutral amps = 3 x 1.4 = 4.2A;
kVA rating = 4.2 *18 kV = 75 kVA
10-sec rating = 75/10.5 = 7 kVA

the 18kv should be 18/sqrt3, the xfmr and ngr are ph-gnd
the 10.5 is silly
the ngr is on the 240 secondary
so he needs a 7 kva ngr at 240 vac???

perhaps YOU should read the link

according to you
ngr I VALUE = 7000/240 = 29 A

If the primary C current is 1.4 in the secondary it is 12000/240 x 1.4 = 70A
you charging I is greater than the ngr lol 70 >> 29

don't help the kid flunk

 

[topgone]

That is from an ngr manufactuers manual
the link of which I posted
what you previously posted is nonsensical and does not give the ngr current and undersizes the xfmr by a factor of 10.5

Originally Posted by topgone
Your transformer rating is 75 kVA. If you design the transformer to withstand a 10-sec fault duration = 7 kVA(overload factor = 10.5)
Here:
phase amps = 12 kV/8187 ohms = 1.4A;
neutral amps = 3 x 1.4 = 4.2A;
kVA rating = 4.2 *18 kV = 75 kVA
10-sec rating = 75/10.5 = 7 kVA

the 18kv should be 18/sqrt3, the xfmr and ngr are ph-gnd
the 10.5 is silly
the ngr is on the 240 secondary
so he needs a 7 kva ngr at 240 vac???

perhaps YOU should read the link

according to you
ngr I VALUE = 7000/240 = 29 A

If the primary C current is 1.4 in the secondary it is 12000/240 x 1.4 = 70A
you charging I is greater than the ngr lol 70 >> 29

don't help the kid flunk

Tip:
Charging current adds up in the neutral! You have three lines, so In = 3 x Ic.
You'll never flunk if you think harder!

 

[Ingenieur]

Tip:
Charging current adds up in the neutral! You have three lines, so In = 3 x Ic.
You'll never flunk if you think harder!

you'll never pass if you keep thinking like that
he gave us the total reactance Xc

by you calculations a 7kva will do
but charging kva is 17kva
that xfmr won't last too long
using your numbers charging kva is 50
you 7 kva will last even less

keep trying
he asked the ngr current value, not the xfmr kva