Transformer Overload

[topgone]

TopGone...

While you seem to have no interest, others may be interested! So, here is the conclusion of my analysis: V-drop, across the transformer is in the order of 13%, not 23%!

Parameters used:

o Running-load is 150kVA (1/2 Xfmr Rating) @ 0.8 pf.

o Pump motor locked-rotor inrush is 600kVA @ 0.3 pf (4x motor rating).

o Resultant connected-load is ~ 730kVA @ 0.6 pf ( 2.4x Xfmr rating).

o Xfmr: kVA, 300; %Z, 4.75; Pcu, 4,000W @ FL; Pfe, 400W @ FL.

o % V-drop for running-load is ~ 2%.

o % V-Drop for combined loads is ~ 13%.

Regard, Phil

@Phil,
ThatMan ran his ETAP software and got 78% VD. I did my own calcs and got 77%. The thing is, you were using 750kVA as the transformer load. ThatMan posted the total load to be 900kVA (though this is a starting duty only).

 

[pv_n00b]

It does not seem like anyone has taken overload protection into consideration. If you put a 900kVA load on a 300kVA transformer, and therefore on the distribution system, some type of overload protection will most likely kick off and disconnect the power to the customer. Good luck with the fire pump then. Normally the utility would not build a system that allows a customer to load it's distribution system to failure.

 

[GoldDigger]

It does not seem like anyone has taken overload protection into consideration. If you put a 900kVA load on a 300kVA transformer, and therefore on the distribution system, some type of overload protection will most likely kick off and disconnect the power to the customer. Good luck with the fire pump then. Normally the utility would not build a system that allows a customer to load it's distribution system to failure.

AFAIK utilities routinely allow overloads which when continued long enough will destroy their transformers (allows in the sense of does not protect against them by OCPD.)
The primary side fusing then protects the distribution network when the transformer melts down and shorts out.
A high enough current to slowly destroy the transformer is generally not going to be high enough by itself to have any harmful effect on the upstream distribution equipment.

If you have one customer on a transformer the service conductors may or may not fail before the transformer.
If you have several customers on one transformer it is more likely that each set of service conductors will be able to serve as secondary OCPD for the transformer.

 

[topgone]

It does not seem like anyone has taken overload protection into consideration. If you put a 900kVA load on a 300kVA transformer, and therefore on the distribution system, some type of overload protection will most likely kick off and disconnect the power to the customer. Good luck with the fire pump then. Normally the utility would not build a system that allows a customer to load it's distribution system to failure.

Please spend more time backreading. Basically, the topic is about whether or not the fire pump can be started successfully, considering the size of the transformer is small (a large voltage drop will decrease the available starting torque on induction motors, hence stalling is a possibility). Unless you have a different definition of "overload", overloads for me are defined as persistent high load for a "long time" (not seconds and minutes when starting motors).

 

[iwire]

Please spend more time backreading. Basically, the topic is about whether or not the fire pump can be started successfully, considering the size of the transformer is small (a large voltage drop will decrease the available starting torque on induction motors, hence stalling is a possibility).

Respectfully I do not think that is the topic.

From the opening post

the code clearly states that the service must have adequate supply for the locked rotor current of the fire pump.

I take that to mean the issue is code compliance.

695.7 Voltage Drop.
(A) Starting. The voltage at the fire pump controller line
terminals shall not drop more than 15 percent below normal
(controller-rated voltage) under motor starting conditions.

Exception: This limitation shall not apply for emergency
run mechanical starting.

(B) Running. The voltage at the motor terminals shall not
drop more than 5 percent below the voltage rating of the
motor when the motor is operating at 115 percent of the fullload
current rating of the motor.

I had asked who owns the transformer, the OP said the utility. Which is why I responded with.

I do not see the issue.

In my opinion that NEC requirement stops at the point of service. If the customer owned and controlled portion is designed to meet the requirements of 695 I believe the NEC is satisfied.

 

[iwire]

The following code sections are likely more on point but still IMO does not apply to the utility transformer and conductors.

695.5 Transformers. Where the service or system voltage is
different from the utilization voltage of the fire pump motor,
transformer(s) protected by disconnecting means and overcurrent
protective devices shall be permitted to be installed
between the system supply and the fire pump controller in
accordance with 695.5(A) and (B), or with (C). Only transformers
covered in 695.5(C) shall be permitted to supply
loads not directly associated with the fire pump system.

(A) Size. Where a transformer supplies an electric motor
driven fire pump, it shall be rated at a minimum of 125 percent
of the sum of the fire pump motor(s) and pressure
maintenance pump(s) motor loads, and 100 percent of the
associated fire pump accessory equipment supplied by the
transformer.

(B) Overcurrent Protection. The primary overcurrent protective
device(s) shall be selected or set to carry indefinitely
the sum of the locked-rotor current of the fire pump motor(s)
and the pressure maintenance pump motor(s) and the full-load
current of the associated fire pump accessory equipment when
connected to this power supply. Secondary overcurrent protection
shall not be permitted. The requirement to carry the
locked-rotor currents indefinitely shall not apply to conductors
or devices other than overcurrent devices in the fire pump
motor circuit(s).

(C) Feeder Source. Where a feeder source is provided in
accordance with 695.3(C), transformers supplying the fire
pump system shall be permitted to supply other loads. All
other loads shall be calculated in accordance with Article
220, including demand factors as applicable.

(1) Size. Transformers shall be rated at a minimum of
125 percent of the sum of the fire pump motor(s) and
pressure maintenance pump(s) motor loads, and 100 percent
of the remaining load supplied by the transformer.

(2) Overcurrent Protection. The transformer size, the feeder
size, and the overcurrent protective device(s) shall be coordinated
such that overcurrent protection is provided for the
transformer in accordance with 450.3 and for the feeder in
accordance with 215.3, and such that the overcurrent protective
device(s) is selected or set to carry indefinitely the sum of
the locked-rotor current of the fire pump motor(s), the pressure
maintenance pump motor(s), the full-load current of the associated
fire pump accessory equipment, and 100 percent of the
remaining loads supplied by the transformer. The requirement
to carry the locked-rotor currents indefinitely shall not apply to
conductors or devices other than overcurrent devices in the
fire pump motor circuit(s).

90.2 Scope.
(A) Covered. This Code covers the installation of electrical
conductors, equipment, and raceways; signaling and communications
conductors, equipment, and raceways; and optical
fiber cables and raceways for the following:

(1) Public and private premises, including buildings, structures,
mobile homes, recreational vehicles, and floating
buildings

(B) Not Covered. This Code does not cover the following:

(5) Installations under the exclusive control of an electric
utility where such installations

a. Consist of service drops or service laterals, and associated
metering, or

 

[Phil Corso]

TopGone…

Percent Voltage Regulation vs Percent Impedance
Wow… Fantastic… Quite a revelation… calculate No-load to On-load V-drop, knowing just %-Imp! Think of all the time I wasted during my 60+year career doing it the old-fashioned way (w/o computer), i.e., using the oft-forgotten PVR method (Percent-Voltage-Regulation), i.e., knowing, Load-Factor (ratio of Connected-load to Xfmr-capacity), Pf (Power-factor of connected-load) as well as Pcu and Pfe (FL Copper and Iron-losses of the Xfmr)!

About Computers
My friend, typical of large numbers of today’s EE’s you have fallen prey to the theory that the computer is all powerful! You don’t need a brain! Have you forgotten the old saw… Garbage-In, Garbage-Out!

Now a word about the parameters I used.
The OP asked, “What will happen to a 300-kVA transformer if 900-kVA of load is running?” Furthermore, he said the motor was 150 Hp (~150-kVA)! I thought, if the xfmr is rated 300, then the over-load is 600-kVA! Thus, he must mean that the motor has an inrush of 600-Kva (4xMotor rating), paralleled with an already running 300-kVA! Hmm, I thought… he’s an engineer! Obviously, he made a mistake… who would add a 600-kVA load to a xfmr already operating at 100% of capacity? Maybe he meant a 5x inrush of the 150-Hp motor?

Regards, Phil Corso

Ps: Suggested reading to all… PVR! Or you could contact me for the Abridged version!

 

[mayanees]

Fire Pump xfmr sizing

Fire Pump xfmr sizing

TM,
What I have done in this situation is to model the system in SKM using the reported Utility contribution for the primary of the transformer, xfmr impedance, and accurate secondary cable lengths. Then apply the locked-rotor current to the pump motor and determine the starting voltage drop. It's then a pass/fail based on the NEC 795 15% voltage drop criteria. .. and 5% while running, but the starting current is typically the limiter.

 

[topgone]

TopGone…

Percent Voltage Regulation vs Percent Impedance
Wow… Fantastic… Quite a revelation… calculate No-load to On-load V-drop, knowing just %-Imp! Think of all the time I wasted during my 60+year career doing it the old-fashioned way (w/o computer), i.e., using the oft-forgotten PVR method (Percent-Voltage-Regulation), i.e., knowing, Load-Factor (ratio of Connected-load to Xfmr-capacity), Pf (Power-factor of connected-load) as well as Pcu and Pfe (FL Copper and Iron-losses of the Xfmr)!

About Computers
My friend, typical of large numbers of today’s EE’s you have fallen prey to the theory that the computer is all powerful! You don’t need a brain! Have you forgotten the old saw… Garbage-In, Garbage-Out!

Now a word about the parameters I used.
The OP asked, “What will happen to a 300-kVA transformer if 900-kVA of load is running?” Furthermore, he said the motor was 150 Hp (~150-kVA)! I thought, if the xfmr is rated 300, then the over-load is 600-kVA! Thus, he must mean that the motor has an inrush of 600-Kva (4xMotor rating), paralleled with an already running 300-kVA! Hmm, I thought… he’s an engineer! Obviously, he made a mistake… who would add a 600-kVA load to a xfmr already operating at 100% of capacity? Maybe he meant a 5x inrush of the 150-Hp motor?

Regards, Phil Corso

Ps: Suggested reading to all… PVR! Or you could contact me for the Abridged version!

Phil,
You're stretching the figures beyond what the OP was asking, IMO. That additional 600kVA you're mentioning isn't an additional load on a 300kVA. Let's do some quick calcs so things get clear.

• A 150HP motor is equivalent to 112 kW and will be equal to 130kVA (@ 0.86PF).
• Assume your fire pump motor draws 6 times its FLA or basically draws 6 times its kVA.
• At starting, your pump will draw = 6 x 130 kVA = 780 SkVA
• Assume further that the 300kVA transformer is loaded at 40% = 300 x 0.4 = 120kVA
• The total kVA load of the 300kVa transformer during fire pump starting is = 780 + 120 = 900kVA

As you can see, I am arbitrarily choosing round numbers so everyone can see why 900kVA is within the ball park.

 

[Phil Corso]

TopGone…
Combining your values, i.e., 120kVA-Run+780kVA-Start, yields a 13.9 % drop! Nowhere near 23%! Now, what did your Computer calculate?
Phil

 

[Julius Right]

In my opinion, the size of the transformer should be as per NEC Art. 695.5 (A):
(A) Size. Where a transformer supplies an electric motor driven fire pump, it shall be rated at a minimum of 125 percent of the sum of the fire pump motor(s) and pressure maintenance pump(s) motor loads, and 100 percent of the associated fire pump accessory equipment supplied by the transformer.
The overcurrent protection shall be selected or set as per 695.4 Continuity of Power.
(B) Connection Through Disconnecting Means and Overcurrent Device.
If we shall take instead of full load current of fire pump motor the rated transformer current –which is more than full load current of fire pump motor- any transformer could withstand
2 minute 6*Irated , 10 minute 3*Irated and 24*Irated for up to 10 seconds.

 

[Phil Corso]

Phil,

• A 150HP motor is equivalent to 112 kW and will be equal to 130kVA (@ 0.86PF).

• 
  
  Your conversion of HPout to KVAin is quite removed from an oldy, but goody, Rule-of-Thumb! Here is my conversion!:
  
  o HPout = 112 kWout (ok).
  
  o kWin = kWout/eff'y! If 92%, then, kWin = 112/0.92 = 122 kW.
  
  o kVAin = kWin/Pf = 122/0.846 ~ 144!
  
  Phil

 

[Phil Corso]

TopGone, ThatMan…

The error being made by you, and others, i.e., is the use, presumably, of the formula, V-Drop = Vs - Lf*I*Z, which is incorrect! Why? Because %Z is fixed… determined for full-load conditions. On the other hand, the %-VR method adjusts for all load conditions, as I had explained earlier! Its equation modifies both copper and iron-losses proportionally to Load-Factor, and, just as important, Load-PF!

Let me know if additional retail is required!

Regards, Phil Corso

 

[kwired]

I think this can possibly go beyond NEC and into other fire protection codes. If the utility can not or will not provide necessary power to run that pump under locked rotor conditions it may not be considered a reliable enough power source and some other on site powered fire pump could possibly be required.

 

[iwire]

I think this can possibly go beyond NEC and into other fire protection codes. If the utility can not or will not provide necessary power to run that pump under locked rotor conditions it may not be considered a reliable enough power source and some other on site powered fire pump could possibly be required.

Here is the thing.

The fire pump motor cannot read the NEC, it will start.

 

[kwired]

Here is the thing.

The fire pump motor cannot read the NEC, it will start.

If it is in locked rotor condition did it start, or was it just an attemp to start, or even a failure after it had been running? I don't know the other codes that apply to fire pump installations, but would guess the reliability of whatever is powering them (could even be other then electric power sources) is addressed in those codes.

 

[Phil Corso]

I have a meeting with the client about this on Tuesday, and I would like to at least say I explored this scenario. Your thoughts?

ThatMan...

I am interested in the outcome of your meeting, if you are permitted to reveal it!

Regards, Phil Corso

 

[charlie b]

That being the case I do not see the issue.

The issue is that if the utility has to upsize their transformer, there might be a cost to the owner for that. In addition, there might be costs to the owner to upgrade the incoming service conductors, and perhaps even the main service panel.

 

[Phil Corso]

The issue is that if the utility has to upsize their transformer, there might be a cost to the owner for that. In addition, there might be costs to the owner to upgrade the incoming service conductors, and perhaps even the main service panel.

Charlie...

Not necessarily! Using parameters suggested by TopGone, and my guesstimate regarding Xfmr losses, the maximum V-Drop is ~ 16%. Why not recalculate using the existing Xfmr’s parameters!

Regards, Phil Corso

 

[iwire]

The issue is that if the utility has to upsize their transformer, there might be a cost to the owner for that. In addition, there might be costs to the owner to upgrade the incoming service conductors, and perhaps even the main service panel.

You missed the point.

In my opinion the NEC requirements do not apply to the utility owned and controlled equipment.