Power And Distribution Transformers Sizing Calculations – Part Nine



Subject Of Previous Articles
Article
Glossary of Sizing Power and Distribution Transformers,
Resources used to calculate basic ratings of power and distribution transformers
the selection factors for the Power and Distribution Transformers
Applicable calculations procedures for sizing of power and distribution transformers
1-Applicable procedures for calculating power and distribution transformer ratios,
2-Applicable procedures for calculating power transformer efficiency,
3-Applicable procedures for calculating power transformer voltage regulation,
4-Special Cases In Transformers Sizing Calculations: Secondary Unit Substations
Special Cases In Transformers Sizing Calculations: Paralleled Transformers
 
Special Cases In Transformers Sizing Calculations: K-Factor Transformers
Special Cases In Transformers Sizing Calculations, K-Factor/Factor K Calculators
Note: I’d like from all of you to review our course “EP-3: Electrical Procurement – Transformers Courseto be more familiar with the contents of our new articles about the Power and Distribution Transformers sizing calculations.  

Today, we will explain other special cases for Power and Distribution Transformers sizing calculations; Transformers with Large Motor Loads .





Special Cases In Transformers Sizing Calculations
4- Transformers with Large Motor Loads



introduction



Many engineers when sizing a transformer for some loads they care only about selecting a transformer larger than the maximum demands they serve, or sometimes larger than the connected loads, so they do one of the following:

  1. Add up the kVA loads and match the size of the Transformer to this summation (∑KVA Loads).
  2. Add up the kVA loads, add a percentage like 20% of the total, then match the size of the Transformer to this summation (in this example, 1.20 X ∑KVA Loads).

Both above methods are non-professional, they can be used by non-specialist for the following reasons:

  1. The first method ignores the characteristics of the loads, for example it does not take account of inrush on Motor Loads, Lighting Loads, Harmonics and the like which can lead to premature failure of the actual Transformer or large voltage dips, on motor startup.
  2. The second method could unnecessarily increase the cost of the asset by over-compensating the required Transformer size.

Note: for best selecting and sizing of a transformer, please review all the articles listed in above.



Importance of Short Circuit and Motor Starting Calculations in Transformer Sizing
Many engineers didn’t know that Short Circuit Calculations together with Motor Starting Calculations are also used in Sizing Transformers and this what we will explain herein.

1- Short Circuit Calculations Contribution to Transformer sizing
  • Short Circuit condition brings down the voltage very dramatically. If a Short Circuit Test done for a transformer where full three phase fault happens at the secondary terminals of the transformer, the amount of voltage at the secondary terminals will be approaching zero, while the transformer is still requested to deliver short circuit power to the point of fault. If the upstream protection not operated to interrupt the fault quickly, the transformer will fail and damage.
  • Transformer damage (known as the transformer short-time withstand) is dependent on the amount of fault current and the impedance of the transformer. Usually this transformer short-time withstand could only last for a few seconds.
Motor Contribution to Short Circuit Capacity
 
When sizing the transformer for motor loads, the fault current contribution from the
motors will not be a consideration for sizing. However, the motor contribution must be considered when sizing all branch circuit fuses and circuit breakers. The interrupting capacity ratings of those devices must equal or exceed the total short circuit capacity available at the point of application.
 
 
 
Motor Starting Calculations Contribution to Transformer sizing
  • During start-up the motor absorbs a high current which causing a big voltage drop at the secondary of the transformer and a significant voltage drop supply network. This has an adverse influence on the operation of other loads and can result in the stalling and contactor drop leading to blackout of loads.
  • The magnitude of transient current involved in motor starting is however very much lower than the short circuit condition. But in effect, switching “on” to energize a large motor can be likened to a “soft short circuit”.
  • Reduction of voltage during start-up can be limited to an admissible range by application of a bigger transformer and larger cabling, but it increases the cost of installation. It is therefore better to reduce motor starting current to avoid an unnecessary oversizing of network elements, including transformer.



Reviewing Some Motor basics
But now how we can size the transformers supplying large motor loads? Before answering the above question, we need to review the following information about motors:

  1. Inrush Current,
  2. Motor starting KVA and Locked-Rotor Current,
  3. Voltage and Frequency Variation.
1- Inrush Current
It is the initial input current drawn by an electrical device when first turned on.  It usually applies to:

  • Inductive loads such as transformers, Inductors and electric motors.
  • AC/DC power supplies that use a simple rectifier/capacitor input stage. 

Inrush current is higher than the normal operating current or what is called “steady state” current.

An example of an electric motor inrush current is shown in Figure-1.  It shows the peak current for the first half cycle as being close to 30 amps and then decaying over subsequent half cycles as the motor spools up.



Fig.1: Electric Motor Inrush Current
Another example of the inrush current of a transformer is shown in Figure-2. It shows the current decay to excitation current (steady state current) that keep the transformer energized after the peak inrush current dissipated.


Fig.2: Transformer Inrush Current


2- Motor starting KVA and Locked-Rotor Current
 
Locked-rotor current is the steady-state current of a motor with the rotor locked and with rated voltage applied at rated frequency. NEMA has designated a set of code letters to define locked-rotor KVA/HP. This code letter appears on the nameplate of all AC squirrel-cage induction motors. KVA per horsepower is calculated as follows:
 
For three-phase motors:
KVA/HP = √3 x current (in amperes) x volts / (1000 x Hp)
 
For single-phase motors:
KVA/Hp = current (in amperes) x volts / (1000 x Hp)
 
 
 
 
CODE LETTER
KVA per HP
A
0 - 3.15
B
3.15 – 3.55
C
3.55 – 4.0
D
4.0 – 4.5
E
4.5 – 5.0
F
5.0 – 5.6
G
5.6 – 6.3
H
6.3 – 7.1
J
7.1 – 8.0
K
8.0 – 9.0
L
9.0 - 10.0
M
10.0 - 11.2
N
11.2 - 12.5
P
12.5 - 14.0
R
14.0 - 16.0
S
16.0 - 18.0
T
18.0 - 20.0
U
20.0 - 22.4
V
22.4 - and up
Table-1: NEMA code letter designations (starting KVA’s)
 
 
The locked-rotor kilovolt-ampere-per-horsepower range includes the lower figure up to, but not including, the higher figure. For example, 3.14 is letter “A” and 3.15 is letter “B”.
By manipulating the preceding equation for KVA/Hp for three-phase motors the following equation can be derived for calculating locked-rotor current:
 
LRA = (1000 x Hp x Locked-Rotor KVA/Hp) / (√3 x Volts)
This equation can then be used to determine approximate starting current of any particular motor.
Starting KVA = (1000 x Hp x Locked-Rotor KVA/Hp)
Starting KVA is also known as “Locked Rotor kVA” or sometimes “Locked Rotor Amperes”.
 
for example, the approximate starting current for a 7 1/2 Hp, 230 volt motor with a locked-rotor KVA code letter G would be:
 
LRA = (1000 x 7.5 x 6.0) / (√3 x230) = 113 Amps
Starting KVA = (1000 x 7.5 x 6.0) = 45 KVA
For more information about Motor starting KVA and Locked-Rotor Current, please review our article “Motor Selection Procedures – Part One”.
3- Voltage and Frequency Variation
  • all motors are designed to operate successfully with limited voltage and frequency variations. However, voltage variation with rated frequency must be limited to ±10% and frequency variations with rated voltage must be limited to ±5%. The combined variation of voltage and frequency must be limited to the arithmetic sum of 10%.
  • Variations are expressed as deviation from motor nameplate values, not necessarily system nominal values. The allowable ±10% voltage variation is based upon the assumption that horsepower will not exceed nameplate rating and that motor temperature may increase.
  • Therefore, the voltage drop on inrush should not be allowed to drop more than -10% of the rated voltage. This means 208v for 230v or 414v for 460 volt or 2.07 kV for 2.3 kV and 3.6 kV for 4.0 kV motors. It means that a 4 kV motor can still operate satisfactorily at 3,600 V.
 So, Effect of variations in voltage:
An increase or decrease in voltage may result in increased heating at rated horsepower load. Under extended operation this may accelerate insulation deterioration and shorten motor insulation life. The affected motors may trip off as provided for by its protection – or if not, the motor burns.


General Rule For Sizing Transformers With Large Motor Loads

We must determine the voltage dip caused by the motor inrush on start-up.
  1. If the voltage dip within the allowable percentage of the system voltage. Then no oversizing of the transformer is required.
  2. If the voltage dip exceeds the operating requirement of the system, then the transformer must provide extra kVA.
Note:

The NEMA specification for a standard motor requires the motor to
be capable of operating at plus or minus 10% of nameplate voltage. Therefore, the
voltage dip must not exceed 10% of nameplate voltage of the motor.
 
 


Voltage Dip Calculation
As explained before, during start-up the motor absorbs a high current which causing a big voltage drop at the secondary of the transformer and a significant voltage drop supply network. This has an adverse influence on the operation of other loads and can result in the stalling and contactor drop leading to blackout of loads.
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 (Motor Starting kVA) compared to the maximum capability of the transformer (Motor Starting kVA + Short Circuit kVA).


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

Note: A voltage sag (U.S. English) equal to voltage dip (British English).



Case#1: A Single Three Phase Motor On Single Transformer
 
 
The following calculations will determine the extra kVA capacity required for a three
phase transformer that is used to feed a single three phase motor that is started with full voltage applied to its terminals, or, "across-the-line."
The following precautions should be followed:
  1. When one transformer is used to operate one motor, the running amperes of the motor should not exceed 65% of the transformer’s full load ampere rating.
  2. If any motor is started more than once per hour, add 20% to that motor's minimum kVA rating to compensate for heat losses within the transformer.
  
Example:
 
A 7,500 kVA, 8.0% IZ, 69-4.16 kV transformer loaded with a single 2,000 hp, 4.16 kV, Code Letter J motor. Check if this transformer size is adequate or not?
 
 
Answer:

a) If the motor controller is across-the-line (full-voltage controller):
 
1- from table-1 in above, Starting kVA of 2,000 hp Motor code letter J= 2,000 kVA x 7.6 = 15,200 kVA
 
2- Three-Phase Short Circuit Capacity of the 7,500 kVA transformer = transformer KVA X 100 / transformer impedance= 7,500 KVA x 100 /8 = 93,750 kVA
 
Note: In this case, the 7.5 MVA transformer has a maximum of 93.75 MVA short circuit capability.
 
3- The voltage drop on motor inrush will be:
 
 % Voltage dip = (Motor Starting kVA) x 100 /(Motor Starting kVA + Short Circuit kVA)
 
%VD at transformer Terminals = 15,200 x 100/(15,200 + 93,750) = 13.95% from the secondary voltage 4.16 KV
So, VD at transformer Terminals = 13.95% x 4.16 KV = 580 V
transformer Terminal Volts During Motor Start-Up = 4160 – 580 = 3,580 V
 
   
Result:
The transformer output voltage will drop more than 10%, so this 7.5 MVA transformer is small for a 2,000 hp motor!
 
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 will give %VD at transformer Terminals 10.84% . so, 12.5 MVA transformer could be the choice which will have a short circuit output capability of 156,250 kVA which will give %VD at transformer Terminals 8.865%.
 
 


Case#2: Multiple Motors On A Single Transformer

The minimum transformer kVA is given by transformer manufacturers so that a transformer may be sized properly for multiple motors.
If there are five motors on one transformer, add the minimum kVA ratings and then add transformer capacity as necessary to accommodate the inrush current of the largest motor.
KVA transformer min = kVA ratings for all motors except largest + transformer KVA as necessary to accommodate the inrush current of the largest motor (as in above Example)
 
 
The following precautions should be followed:
  1. If several motors are being operated from one transformer, avoid having all motors start at the same time. If this is impractical, then size the transformer so that the total running current does not exceed 65% of the transformer’s full load ampere rating.
  2. Also, if any motor is started more than once per hour, add 20% to that motor's minimum kVA rating to compensate for heat losses within the transformer.


This was the last article in Power and Distribution Transformers Sizing Calculations. Please keep following for other articles in MEP Equipment Sizing Calculations.


  


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