SIZING TRANSFORMERS WITH LARGE MOTOR LOADS (PART II)

LESSONS WE NEED TO LEARN

SIZING TRANSFORMERS

WITH LARGE MOTOR LOADS

(PART II)

By Doods A. Amora, PEE
(December, 2006)

(Last of Two Parts)

6.0: VOLTAGE SAGS IN LARGE MOTOR STARTING

The voltage dip at the transformer terminals is proportional to the motor load required in start-up. As discussed earlier, the voltage drop can be expressed as a percentage of the inrush motor load compared to the maximum capability of the transformer.

% Voltage Drop = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)

EXAMPLE:

Now, supposing a lone 2,000 hp, 4.16 kV, Code Letter J motor (Starting kVA = 7.6 times its rated hp) is connected to the 7,500 kVA transformer described above, would it not be that the transformer is too large for the motor load? Probably yes, probably not! Let us see why.

A) IF THE MOTOR CONTROLLER IS ACROSS-THE-LINE (FULL-VOLTAGE CONTROLLER):

Starting kVA of 2,000 hp Motor (SkVA = 7.6):
Starting kVA of 2,000 hp motor = 2,000 kVA x 7.6 or 15,200 kVA

Three-Phase Short Circuit Capacity of the Transformer:
3Φ SC kVA of 7,500 kVA Trafo = 7,500 kVA/0.08 or 93,750 kVA

Note: In this case, the 7.5 MVA transformer has a maximum of 93.75 MVA
short circuit capability.

The voltage drop on motor inrush will be:

VD at Trafo Terminals = 15,200/(15,200 + 93,750) = 0.1395 or, 13.95%
VD at Trafo Terminals = 580 v
Trafo Terminal Volts During Motor Start-Up = 3,580 v

The transformer output voltage will drop from 4,160v to 3,580v (4,160 x 0.8605 = 3,580 v) as against a minimum requirement of 3,600v for a 4,000v motor. Thus, we can see that the transformer even at 7.5 MVA is small for a 2,000 hp motor!

Therefore, the transformer must be sized to a short circuit capability equal to or greater than 15,200 kVA times 10, or, 152,000 SC kVA in order to have a voltage drop of 10% or less.

The next higher standard size transformer at 10,000 kVA 8.0% impedance would have a short circuit output capability of 125,000 kVA which is still not sufficient. Or a 12.5 MVA transformer could be the choice!

In this particular application, the ratio of the selected standard size transformer kVA to motor kVA is 12,500 kVA/2,000 kVA = 6.25. Thus the transformer rating is 625% larger than the rating of the motor! Fantastic!

Note that the voltage sag in reality will even be bigger (actual voltage smaller) because the source can never be infinite in fault duty plus in addition, the voltage drop at the distribution cables from the transformer to the motor end. The total voltage sag is the sum of the sag in the secondary unit substation transformer and the secondary circuit. In the case of very large motors of several hundred to a few thousand horsepower, the impedance of the supply system (or the ‘available fault duty’) should be considered.


B) IF THE MOTOR CONTROLLER IS AN AUTOTRANSFORMER REDUCED VOLTAGE STARTER:

Now therefore, here comes the wisdom in using reduced voltage starting scheme and the choice of Code Letter specification of a motor. Normally in practice, large motors need to be equipped with Auto-Transformer RVS controllers while the motor Code Letter specification can be chosen.

Let us now see the effect with a 2,000 hp Code Letter E motor (Starting kVA = 4.8 times rating) is controlled by an Auto-transformer RVS. Establish further that the Auto-transformer RVS shall be at 65% tap.

From the table below, the line current at 65% tap draws 46% of the full voltage starting current.


Table 2: MOTOR STARTING METHODS FOR SQUIRREL CAGE INDUCTION MOTORS:
(Table re-configured for easy reference) in this blog:

 Full Voltage : 100%: Line Starting Current or kVA

 Reduced Voltage (RVS Autotransformer Type):

80% Tap: 68% of Line Starting Current or kVA of 100% of Full Voltage Starting
65% Tap: 46% of Line Starting Current or kVA of 100% of Full Voltage Starting

 Reduced In-Rush:

Wye-Delta: 33% of Line Starting Current or kVA of 100% of Full Voltage Starting

NOTE: In addition to methods listed above, users should consider solid-state soft-start motor controllers and/or adjustable speed drives.


In this example:
Motor Starting kVA = 46% of 2,000 kVA x 4.8 or 4,416 kVA
Motor Starting kVA can therefore be placed at 4,416 kVA.

The voltage drop at transformer terminals on motor inrush will be:

VD at Trafo Terminals = 4,416/(4,416 + 93,750) = 0.045 or, 4.5%
VD at Trafo Terminals = 187.2 v
Trafo Terminal Volts During Motor Start-Up = 3,972.8 v

The transformer output voltage will drop from 4,160v to 3,972.8 (4,160 x 0.955 = 3,972 v) against a minimum requirement of 3,600v for a 4,000v motor.

Thus, we can see that the 7.5 MVA transformer now becomes sufficient for a 2,000 hp motor with a different code letter and with the use of an RVS. Note also that if the auto-transformer RVS is at 80% tap, the voltage drop as seen by the motor would still satisfy the 10% VD requirements.


7.0: MULTIPLE MOTORS IN A TRANSFORMER

What if there are smaller motors plus other loads among a few large ones in a single transformer source?

EXAMPLE:

Assume that in the 7.5 MVA Substation transformer described above, there are 2,150 kVA smaller 460 v motors in the system through a number of transformers downstream the 4,160 v bus of the substation. Other load, load growth included is 750 kVA. Assume also that there are 800 hp 4.16 kV Code Letter E motors directly connected at the 4.16 kv bus. How many 800 hp motors can be energized into the system complying with the 10% voltage sag requirements? Assume that RVS’s for the 800 hp motors are at 65% tap and the motors shall not be started at one time.

If there are several motors on one transformer:

TRAFO KVA = (Maximum Demand kVA of Small Group Motors) + (Maximum Demand kVA of Other Loads Including Load Growth) + (kVA Ratings of all Large Motors) + (Additional Trafo kVA capacity necessary to accommodate the inrush current of the largest motor)

Thus:

TRAFO KVA = 7,500 kVA
Max Demand of Small Group Motors = 2,150 kVA

Other Loads Including Load Growth = 750 kVA

kVA of Large Motors = KVA of LM (Unknown)
Additional Trafo kVA for 800 hp Start-Up = Unknown

Solving for the Starting kVA for 800 hp Code Letter E motor
with RVS at 65% tap:

800 hp Starting kVA = (0.46 x 800 x 4.8)
800 hp Starting kVA = 1766 kVA

Solving for the Additional Trafo kVA Needed for 800 hp Start-Up:

Trafo SC kVA Needed for one 800 hp Start-Up to maintain a 10% VD = 1,766 x10
Trafo SC kVA for one 800 hp Start-Up to maintain a 10% VD = 17,660 kVA

Additional Trafo kVA Needed for one 800 hp Start-Up = SC kVA x %IZ of Trafo
Additional Trafo kVA for one 800 hp Start-Up = 17,660 x 0.08 = 1,413 kVA

Solving for the Number of 800 hp Motors
that can be placed into the system:

Recalling: TRAFO KVA = (Max Demand kVA of Small Group Motors) + (Max Demand kVA of Other Loads Including Load Growth) + (kVA Ratings of all Large Motors) + (Additional Trafo kVA capacity necessary to accommodate the inrush current of the largest motor)

7,500 kVA = 2,150 kVA + 750 kVA + KVALM + 1,413 kVA

KVA Large Motors = 7,500 – 2,150 – 750 - 1,413
KVA Large Motors = 3,187 kVA

Number of 800 hp motors = 3,187/800
Number of 800 hp motors = 3.984, say 4

The transformer selected will be capable of running and starting all motors
provided that only one large motor is started at any one time. Additional capacity will be required for motors starting simultaneously.


8.0: BACK TO THE BOARD PROBLEM

A 250 hp motor driving a pump is served by 3 x 100 kVA 13.8kV- 480v transformer bank. Other data are as follows:

 Transformer Impedance: 3.7% IZ
 Short Circuit Capacity of Source: 200 MVA
 Motor Starting Current: 8 times of FLC
 Length of Cable from Transformer to Motor Controller: 50 ft

Solve for:

a) Size of Cable & Conduit
b) Branch Circuit Breaker & Controller Size if Auto-Transformer type RVS
c) Determine the voltage dip during motor start-up using full voltage starter
d) Determine the voltage dip during motor start-up using Auto-Transformer Reduced Voltage Starter tapped at 65%.

Solution:

a) Solving for the Cable Size & Conduit
Motor Full Load Current = 302 A (From NEC Table)
Size of Cable = 1.25 x 302 or 377 A
USE: 500 MCM THW in 4”Φ Conduit

b) Solving for the Branch Circuit Breaker & RVS Controller Size
Branch Circuit Breaker = 2.0 x 302 A or 604 A
USE: 600AT/600AF, 3P, 600V MCCB
USE: Auto-Transformer RVS NEMA Size 5

c) Solving for the Voltage Dip Using Full Voltage Starter:

SC kVA at Transformer Terminal

Let: kVA Base = 300 kVA
V Base = 480 v
I Base = 361 A

X @ 13.8 kV = kVA Base / SC KVA@ 13.8 kV Source
X @ 13.8 kV = 300 / 200,000
X @ 13.8 kV = 0.00150 pu

X T = %IZ/100 x kVA Base /Trafo kVA
X T = 3.7/100 x 300/300
X T = 0.037 pu

X EQ = 0.00150 + 0.037
X EQ = 0.0385 pu

I SC = IBASE / X EQ
I SC = 361 A / 0.0385
I SC = 9,377 A

Three-Phase Short Circuit Capacity of the Transformer:
KVA SC = √3 x 0.48 x 9,377
KVA SC = 7,796 kVA

Motor Starting kVA:

Motor Starting kVA = 250 hp x 8
Motor Starting kVA = 2,000 kVA

The voltage drop on motor inrush at the transformer terminals will be:

VD at Trafo Terminals = 2,000/(2,000 + 7,796) = 0.2042 or, 20.42%
VD at Trafo Terminals = 98 v
Trafo Terminal Volts During Motor Start-Up = 382 v

Voltage Drop at the Cables:

Impedance ZC of 500 MCM THW Cable in Metallic Conduit= 0.00575 Ω per 100 ft (From IEEE Table)

Impedance of C1: 50 ft, from transformer to Motor Controller:
ZC = 0.00575 Ω/100 ft x 50 ft
ZC = 0.002875 Ω

Voltage Drop along the Cable C:
VDC = √3 x (IMOTOR LINE START) x ZC1
VDC = √3 x (302 x 8) x 0.002875
VDC = 12.0 v

Total Voltage Drop at Motor Controller:
98.0 + 12.0 = 110 v

Voltage at Motor Controller Line Side Terminals during Start-Up = 370 v
Voltage Requirements during Start Up = 414 v for 460 v motors

The set-up can not satisfy voltage dip requirements. Not even 3 x 250 kVA (750 kVA). In fact a 3 x 333 kVA (1,000 kVA) set-up at 5.2%IZ can barely satisfy the requirement!

There could be two remedies for the situation. One is to change the motor controller from full voltage to auto-transformer RVS type. The other alternative is to change the Transformer from 300 kVA (3 x 100 kVA) to 1,000 kVA (3 x 333 kVA).

d) Solving for the Voltage Dip Using Auto-Transformer RVS:

If RVS Auto-Transformer Tap to 65%:
Motor Starting kVA = 0.46 x 250 hp x 8
Motor Starting kVA = 920 kVA

The voltage drop on motor inrush at the transformer terminals will be:

VD at Trafo Terminals = 920/(920 + 7,796) = 0.1055 or, 10.55%
VD at Trafo Terminals = 50.64 v
Trafo Terminal Volts During Motor Start-Up = 429.36 v

Voltage Drop along the Cable C1:
VDC1 = √3 x (IMOTOR LINE START) x ZC1
VDC1 = √3 x (302 x 8 x 0.46) x 0.002875
VDC1 = 5.534 v

Total Voltage Drop at Motor Controller:
50.64 + 5.534 = 56.174 v

Voltage at Motor Controller Line Side Terminals during Start-Up = 423.826 v
Voltage Requirements during Start Up = 414 v for 460 v motors

Therefore, the 3 x 100 kVA transformer need not be replaced if its Auto-Transformer RVS is tapped at 65%. Note that if the RVS is tapped at 80%, the transformer needed must be 750 kVA (3 x 250 kVA)!

9.0: THE REAL ENGINEERING ECONOMICS

While first costs are very important to clients & plant owners; correct engineering practice should not be sidestepped in selecting best systems. More often than not, too much cheap engineering is in fact, costly. By making the first cost not objectionable, it does not mean that the engineer shall use cheap, undersized and substandard materials or equipment. Applying correct Value Engineering, the design must fundamentally conform to good engineering practice, codes and standards. This fundamental or minimum requisite must not be compromised under the cloak of costs.

It is important for the Industrial Power Systems Designer to understand that First Cost is not enough in determining the economics of the project. Total Life Cycle Cost of the power system depends on equipment purchase price & quality, construction & installation costs, operating costs including losses, outage costs, repair costs, useful lifetime of the equipment and administrative costs.

In the USA & other advance countries, the insurance companies play lead roles in the safety and soundness in any installations. That’s why they are strongly behind the National Electrical Code and the National Fire Protection Code. In the Philippines, if it’s “faulty electrical wirings”, the insurance companies pay without much fuss, and nobody seems to ask how standards were complied.

But alas, more often than not, what design engineers consider as “best solutions” may not correspond to what owners perceive. In the name of cost, the ‘economical solutions’ that clients want usually end up into sickly systems that are vulnerable to failure or worse, violations to the Code and Correct Engineering Practice for which we engineers are supposed to profess. Note that there is a common misguided thought in electrical engineering design economics. For instance, why use an expensive RVS when Across-the-Line Starter will do? Or why a 1,000 kVA transformer for a 250 hp motor? In this case, we say that firstly, the engineer must know his codes & standards. The standards must first be satisfied before any value engineering can be focused. Sensitive to safety & system operability in the ultimate analysis, it is the engineer's responsibility to find solutions that satisfy both parties.

Doods A. Amora
December, 2006