Fire Pump Conductor & Conduit Sizing Question?

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Re: Fire Pump Conductor & Conduit Sizing Question?

Originally posted by bob:
Iwire
Compute the VD using 267 amps, 208 v, 420 ft
1 350 per phase and see if you get the same results as I show, please.
Click to expand...
Using those perimeters I (actually the program) comes up with 3.64% or 7.57 volts of drop.

I do not understand why we would be looking into the voltage drop at FLA when we need to consider starting conditions.
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Is anyone able to do the formula that will size the conductors appropiately?

75HP Fire Pump = 211A x 125% = 263.75A
1/3HP Jockey Pump = 2.4A x 125% = 3A
Total Amps for Conductor Sizing = 266.75(267A)

Table 310-16, 75 Degree Column, is 300kcmil @ 285A. Annex C, Table C.1 shows 3, Type RHH current carrying conductors in a 2-1/2" EMT Conduit.

Voltage Drop - Vd=1.73KxLxI/Cm or Vd=1.73(12.9)x420'x285A/300,000 = 8.9 Volts Dropped or 4.3% total VD, which is within the 15% VD referred to in section 695.7.

Even though the disconnect switch, fuse, and circuit breakers are sized according to locked-rotor currents, the feeder conductors to the fire pump and associated equipment are required to have an ampacity not less than 125% of the full-load current (FLC) rating of the fire pump motor(s) and pressure maintenance pump motor(s), + 100% of associated accessory equipment.

BTW iwire, the engineer has been on vacation for the past 2 weeks, and he returns tomorrow. We are trying to have this situation resolved by the time he returns. Thanks for the assistance.
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Originally posted by tnt8197:
Even though the disconnect switch, fuse, and circuit breakers are sized according to locked-rotor currents, the feeder conductors to the fire pump and associated equipment are required to have an ampacity not less than 125% of the full-load current (FLC) rating of the fire pump motor(s) and pressure maintenance pump motor(s), + 100% of associated accessory equipment.
Click to expand...
See that is a problem you seem to think the conductors are a separate issue.

If you cam find a way to meet this requirment...

695.7 Voltage Drop.
The voltage at the controller line terminals shall not drop more than 15 percent below normal (controller-rated voltage) under motor starting conditions........
Click to expand...
...by sizing the conductors based on only the FLC I will be very surprised.

Just how do you expect to maintain less than 15% VD at at the controller line terminals under motor starting conditions with VD calculations based on the FLC and not the LRA?

It can't be done that way. :D
 
Re: Fire Pump Conductor & Conduit Sizing Question?

tnt,
Look at Table 430.251(B) in the the 2005 code. You will have to use 1200A in computing the 15% starting voltage drop. That is the current under motor starting conditons.
Don
 
Re: Fire Pump Conductor & Conduit Sizing Question?

iwire,

For the approximate fourth time within this post, I am asking once again. Can you show all the code references & formulas with how you come up with your interpertations of the code? Your rhetoric means nothing to me, unless they are backed up by the above-referenced facts. Thanks. :roll:
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Tnt8197, Iwire and Don have given you the applicable code references and Bob has explained the formulas, what else are you looking for?

The bottom line is, 695.7, specifies that the voltage drop not exceed 15% under motor starting conditions, so LRA is the number to be used.

Roger
 
Re: Fire Pump Conductor & Conduit Sizing Question?

tnt,
What do these words "under motor starting conditions", from 695.7, mean to you?
Don
 
Re: Fire Pump Conductor & Conduit Sizing Question?

iwire,
For the approximate fourth time within this post, I am asking once again. Can you show all the code references & formulas with how you come up with your interpretation of the code? Your rhetoric means nothing to me, unless they are backed up by the above-referenced facts.
Click to expand...
TNT
You can only hope to attain the knowledge of Don and Iwire in your life time. I wish I knew on half of what they can quote for the NEC.

The NEC is not an engineering manual. It is a set of rules established over time to provide a safe electrical installation. You will not find a statement that LRA and motor starting current are the same thing. You obtain the information thru education and association with other talented individuals such as those on this forum.
You have entered 285 amps but the post said 267 amps. I am going to use the 285 figure.

Voltage Drop - Vd=1.73KxLxI/Cm or Vd=1.73(12.9)x420'x285A/300,000 = 8.9 Volts Dropped or 4.3% total VD, which is within the 15% VD referred to in section 695.7.
Click to expand...
Your formula appears correct except that you used FLA instead of the starting current. LRA and starting amps are the same number. LRA is approx 6 x FLA. If you can not except this then do a google search on motor starting and find out for your self. Using your formula
Vd=1.73(12.9)x420'x285Ax6/300,000 = 53.4 Volts drop.
Another way of making the line to neutral VD caculation is

VD = IR Cos Theta + IX Sin Theta where

Theta = the power factor angle
I = FLA or LRA depending what you want.
R = wire resistance
X = wire reactance
Making an assumption that the PF = 0.85 the PF angle is 31.8 degrees.
Cos 31.8 = 0.85 and Sin 31.5 = 0.52
R of 350kcm = 0.039 / 1000ft table 9
X of 350kcm = 0.05 / 1000 ft table 9
I = 285 amps

VD=(285 x 0.85 x 0.039 x 420)/1000
+ (285 x 0.52 x 0.05 x 420)/1000 =4.0+3.1= 7.1 vd
Line to neutral. For Line to line 7.1 x 1.73 = 12.3 volts.
Using the same formula with LRA or starting amps
amps?= 1200 amps
VD = 16.7 + 13 = 29.7 or 30 volts L to N
VD L to L = 30 x 1.73 = 52 volts which is the same as using your formula.

Vd=1.73(12.9)x420'x285Ax6/300,000 = 53.4 Volts drop.
Click to expand...
In reality the motor starting PF is about 0.40.
VD would be about 31 volts L to N and 54 L to L.

TNT
Note that your formula results of 53.4 volts is the same voltage and my formula Line to Line voltage of 52 volts.

[ July 17, 2005, 04:06 PM: Message edited by: bob ]
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Originally posted by tnt8197:
Your rhetoric means nothing to me, unless they are backed up by the above-referenced facts. Thanks. :)

You thought you where right and wanted some back up for this debate your in, looks like you will be eating some crow. ;)

695.7 is about as cut and dry as code sections come.

The fire pump controller is located near the fire pump as required by 695.12(A), correct?

The feeder to the fire pump controller is 420' from the supply, correct?

When the fire pump controller closes the 420' long conductors feeding the controller will be carrying the locked rotor current of the fire pump motor, correct?

In order to prevent more than 15 percent voltage drop at the controller line terminals as required by 695.7 the conductors will need to be sized based on the LRC, correct?

Have great day, it's been a pleasure. :p :roll:
Click to expand...
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Than per the LRA the voltage drop calculation should be as follows:

Vd = 1.73x12.9x420x1222/350000 = 32.7V or 15.7% VD

Vd - 1.73x12.9x420x1222/400000 = 28.6V or 13.7% VD

I am not claiming to be right or wrong in this situation, but am requesting that the calculations and formulas be confirmed. Thanks!
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Is the voltage drop through the service transformer considered in the 15% maximum drop on motor starting? This thread has been looking at only the drop in the cables, not the total drop at the controller.

I assume the Code intent is to prevent contactor coil drop out or motor stalling during the starting voltage dip. Oversized conductors might be used to compensate for excessive service transformer voltage drop.

This would be unusual because the relative size of fire pumps and service transformers usually means the transformer voltage drop is insignificant. But it could happen, especially on a retrofit.
 
Re: Fire Pump Conductor & Conduit Sizing Question?

Vd = 1.73x12.9x420x1222/350000 = 32.7V or 15.7% VD
Vd - 1.73x12.9x420x1222/400000 = 28.6V or 13.7% VD
Click to expand...
How did you get 1222 amps for the LRA?
6 x 267 = 1602 amps.

[ July 18, 2005, 01:58 PM: Message edited by: bob ]
 
Re: Fire Pump Conductor & Conduit Sizing Question?

[/QUOTE]How did you get 1222 amps for the LRA?
6 x 267 = 1602 amps. [/QB][/QUOTE]

tnt,
Look at Table 430.251(B) in the the 2005 code. You will have to use 1200A in computing the 15% starting voltage drop. That is the current under motor starting conditons.
Don

Plus the 22 LRA for the Jockey pump. Thanks for the assistance.
 
Re: Fire Pump Conductor & Conduit Sizing Question?

350 mcm
VD= IR Cos A + IX Sin A
VD = (1222x0.039x0.85 + 1222x0.05x0.52)420/1000
VD = (40.5 + 31.8)x .42 = 30.3 v Line to Neutral
VD L to L = 30.3 x 1.73 = 52.5 volts 25%

400 mcm
VD =(1222x0.035x0.85 + 1222x0.049x0.52)420/1000
VD = (36.3 + 31.1)420/1000 = 28.3 Line to Neutral
VD L to L = 28.3 x 1.73 = 49 volts 23%

RC
You hav a good point regarding the transformer voltage drop. We have not taken that into account. We do not have the TR information but the VD can be significant..

[ July 18, 2005, 11:27 PM: Message edited by: bob ]
 
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