Power And Distribution Transformers Sizing Calculations – Part Five



In Article “Power and Distribution Transformers sizing calculations – part One”, we indicate that the contents of our articles for Power and Distribution Transformers sizing calculations will include the following points:

  • Glossary of Sizing Power and Distribution Transformers,
  • Power and distribution transformer components,
  • Power and distribution transformer classification: construction and application,
  • Three-phase power and distribution transformer connections,
  • Power and Distribution Transformers sizing calculations.

The following points were explained before (or will be explained) in our course “EP-3:Electrical Procurement – Transformers Course”:

  • Power and distribution transformer components,
  • Power and distribution transformer classification: construction and application,
  • Three-phase power and distribution transformer connections,

So, we will not go through these points here, we will focus only on the following two points:

  •  Glossary of Sizing Power and Distribution Transformers,
  • Power and Distribution Transformers sizing calculations.
And we already explained the Glossary of Sizing Power and Distribution Transformers in Article “Power and Distribution Transformers sizing calculations – part One”.

Also, in ArticlePower and Distribution Transformers Sizing Calculations – Part Two” ,we indicate that Our study for the Power and Distribution Transformers sizing calculations will include the explanations of the following points:

  • Resources used to calculate basic ratings of power and distribution transformers,
  • Selection Factors,
  • Calculations procedures For Sizing of Power and Distribution Transformers,
  • Special cases.

And we explained in this article the Resources used to calculate basic ratings of power and distribution transformers while we explained the selection factors for the Power and Distribution Transformers in article “Power and Distribution Transformers Sizing Calculations – Part Three

In article “Power And Distribution Transformers Sizing Calculations – Part Four”, we indicted that the accurate sizing calculations of Power and Distribution Transformers will include the following:

  • Applicable calculations procedures for sizing of power and distribution transformers,
  • Applicable procedures for calculating power and distribution transformer ratios,
  • Applicable procedures for calculating power transformer efficiency,
  • Applicable procedures for calculating power transformer voltage regulation.
And we explained the Applicable calculations procedures for sizing of Power and Distribution Transformers in this article.

Today we will explain the following:

  • Applicable procedures for calculating power and distribution transformer ratios,
  • Applicable procedures for calculating power transformer efficiency,
  • Applicable procedures for calculating power transformer voltage regulation,
  • Special Cases.
 
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.

 

 
2- Applicable Procedures For Calculating Power And Distribution Transformer Ratios.
 

  

 
Step# 1:
 
Determine the turns ratio.
 
Np/Ns = Ep/Es = a         (a: transformer turns ratio)
 
Step# 2:
 
Determine the volts-per-turn ratios.
 
volts-per-turn (primary) = volts-per-turn (secondary)
 
Ep/Np = Es/Ns
 
Then:
Ep = (Np/Ns) x Es
Es = (Ns/Np) x Ep
 
And:
Np = (Ep/Es) x Ns
Ns = (Es/Ep) x Np
 
Step# 3:
 
Determine the ampere-turns relationships.
 
ampere-turns (primary) = ampere-turns (secondary)
Ip x Np = Is X Ns
 
Then:
Ip = (Ns/Np) x Is
Is = (Np/Ns) x Ip
 
And:
Np = (Is/Ip) x Np
Ns = (Ip/Is) x Np
 
Step# 4:
 
Determine the volt-ampere relationships.
 
VAin = VAout
Ep x Ip = Es x Is (for single-phase transformers)
3 x Ep x Ip = 3 x Es x Is (for three-phase transformers)
 
Or
 
kVAin = kVAout
kVp x Ip = kVs x Is (for single-phase transformers)
3 x kVp x Ip = 3 x kVs x Is (for three-phase transformers)
 
Where:
Ep or KVp : Primary Voltage,
Es or KVs: Secondary Voltage,
Ip: Primary Current,
Is: Secondary Current,
Np: Primary Turns Number,
Ns: Secondary Turns Number.
 

 

 
3- Applicable Procedures for Calculating
Power Transformer Efficiency.
 

  

 
Note:
 
Efficiency is calculated by dividing the output real power by the input real power. The efficiency will vary depending on the amount of load because the copper losses of a transformer vary with load.
 
Step# 1:
 
Determine the transformer kVA, total copper losses, and total iron losses.
 
Step# 2:
 
Calculate the output power of the transformer at the given power factor.
 
Pout = kVA x 1000 x power factor (p.f.)     in watts
 
Step# 3:
 
Calculate the required input power.
 
Pin = Pout + copper losses + iron losses
 
Step# 4:
 
Calculate the percent efficiency.
 
% efficiency = (Pout/Pin) x 100
 

  

 
4- Applicable Procedures for Calculating
Power Transformer Voltage Regulation
 

  

 
Note:
 
As a transformer becomes loaded, the voltage at the secondary terminals of the transformer decreases. This decrease in voltage is caused by the voltage drop across the internal impedance of the transformer. The higher the transformer impedance, the higher the voltage drop.
 
Step# 1:
 
Determine the voltage (E full-load) of the transformer at the secondary terminals under full-load conditions.
 
Step# 2:
 
Calculate the percent voltage regulation.
 
% Voltage Regulation = [(E no-load – E full-load)/(E no-load)] x 100
 
 where E no-load = the transformer’s rated secondary voltage
 

  

 
4- Special Cases In Transformers Sizing Calculations
 

 

 
 
 
 
Some factors must be taken in consideration in some cases when sizing the power and distribution transformers these are the special cases which will be as follows:
 
  1. Secondary Unit Substations,
  2. Paralleled Transformers,
  3. K-Factor Transformers,
  4. Transformers with Large Motor Loads.
 

 

 
1- Secondary Unit Substations
 

  

 
What is the Secondary Unit Substations?
 
A secondary unit substation, sometimes called a power center, is a close-coupled assembly consisting of three-phase power transformers, enclosed high voltage incoming line sections, and enclosed secondary low voltage outgoing sections encompassing the following electrical ratings:
 
  • Transformer kVA: 112.5 thru 2500 kVA (self-cooled rating), liquid-filled, dry-type, or cast coil
  • Primary Voltages: 2.4 kV thru 34.5 kV
  • Secondary Voltages: 208, 240, 480, or 600 Volt (maximum)
 
As a result of locating power transformers and their close-coupled secondary switchboards as close as possible to the areas of load concentration, the secondary distribution cables or busways are kept to minimum lengths. This concept has obvious advantages such as:
 
  • Reduced power losses.
  • Improved voltage regulation.
  • Improved service continuity.
  • Reduced exposure to low voltage faults.
  • Increased flexibility.
  • Minimum installed cost.
  • Efficient space utilization.
 
Components of Secondary Unit Substations:
 
The major components of a unit substation are:
 
  • primary switchgear, for example, power fuses, medium voltage circuit breakers, or interrupters,
  • the transformer (liquid-filled or dry type),
  • The secondary distribution section (main and feeder low voltage power circuit breakers).
 
 

  

 
Types of Secondary Unit Substations
 
There are two main types of Secondary Unit Substation as follows:
 
  1. Single-Ended Substations
  2. Double-Ended Substations
 
1- Single-Ended Substations:
 
Figure-1 shows the one-line diagram and physical layout of a secondary unit substation that uses a radial system arrangement. This type of radial substation arrangement is called a single-ended substation because there is only one incoming line section at the one end (west) of the unit (prepackaged) assembly.
 
 
 
Fig-1: Single-Ended Substations
 
 
2- Double-Ended Substations:
 
Figure-2 shows the one-line diagram and physical layout of a secondary unit substation that uses a secondary selective system arrangement. This type of arrangement, the secondary selective system, overcomes the major disadvantage of the radial system in that it provides duplicate paths of supply to the secondary bus of each load center. This selective system has two step-down transformers, each with its own incoming primary feeder. The secondary bus associated with each transformer is connected through a tie breaker. Normally, the system is operated with the tie connection open, that is, as two separate radial systems operating independently of each other.
 
 
 
  
Fig-2: Double-Ended Substations
 
With the loss of one of the primary feeders and/or transformers, the main secondary breaker for that circuit can be opened and the tie breaker closed, allowing the one remaining primary feeder and transformer to energize all of the secondary bus. The service to one-half of the load is momentarily interrupted during this transition period.
 

 
 

 
Factors Must Be Taken into Consideration when Sizing Unit Substations
 
The selection of kVA and impedance ratings of unit substation transformers is very critical to the levels of available fault current on the secondary main bus side of the substation. This available fault current on the secondary side of the transformer, assuming that there are no other fault current sources on the secondary side, is called The Transformer Let-Through Current.
 

  

 
Procedures For Calculating Transformer Let-Through Current
 
Transformer impedance dictates how much fault current a transformer can supply to a fault for a given kVA size. The higher the impedance, the lower the available fault current. Sometimes transformer impedances are used to limit the let-through current (sometimes called the infinite bus fault current calculations).
By calculating the transformer let-through current, you can estimate the available fault current at the secondary terminals of a transformer. Transformer let-through current can be calculated using the following steps:
 
Step#1:
 
Calculate rated secondary current for the transformer.
 
Isec = kVA/(1.732 x kV)
 
Step#2:
 
Determine % Z from Table-1 for single-phase transformers and Table-2 for three-phase transformers.
 
% Z =
 
Note: Convert %Z into decimal form
Example: 5% = 0.05
 
Step#3: 
 
Calculate the available short circuit current (Transformer Let-Through Current).
 
ISCA (ILT ) = Isec/Z
 
 

  

 
Impedance (%Z)
 
  • The impedance (%Z) of the transformer, which is based on the self-cooled (OA) rating of the transformer, is the main factor in limiting the magnitude of the fault current available on the low voltage section of the system, especially at points close to the substation secondary bus.
  • Because of the importance of the transformer impedance in limiting the available fault current, unit substation transformers are designed on purpose to have impedances values of at least 5.0%, with a typical value being 5.75%. This higher impedance value is obtained by increasing the leakage reactance of the transformer windings, rather than increasing the winding resistance. Higher-resistance windings would increase the heat losses in the transformer, which is not desirable.
 
 
Tables of Standard Impedances for Single-Phase and Three-Phase Transformers
 
  • When selecting a transformer, standard impedance values should be used.
  • Table-1 lists the standard impedance values for single-phase transformers.
  • Table-2 lists standard impedance values for three-phase transformers.
 
 
 
TYPICAL SINGLE-PHASE POWER TRANSFORMER IMPEDANCES
RATING
kVA
PRIMARY NOMINAL
VOLTAGE
kV
IMPEDANCE
%
3
2.4
2.2
13.8
2.8
10
2.4
2.2
13.8
2.4
34.5
5.5
25
2.4
2.5
13.8
2.3
34.5
5.5
50
2.4
2.4
13.8
2.5
34.5
5.5
69.0
6.5
100
2.4
5.5
69.0
6.5
167
2.4
3.7
13.8
3.8
34.5
5.5
69.0
6.5
 
Table-1: Standard Impedances for Single-Phase Power Transformers
 
TYPICAL THREE-PHASE POWER TRANSFORMER IMPEDANCES
SELF COOLED
NOMINAL HIGH VOLTAGE WINDING RATING
RATING
MVA
UP TO
UP TO
13.8 KV
UP TO
34.5 KV
UP TO
69 KV
UP TO
115 KV
UP TO
230 KV
1
5.0
5.5
6.0
 
 
5
6.5
6.5
7.5
9.0
 
10
10.0
10.0
10.0
10.0
 
15
13.0
13.0
10.5
10.0
 
30
 
12.5
10.5
10.0
 
60
 
 
11.0
12.5
15.0
90
 
 
11.0
15.0
15.0
120
 
 
 
 
17.0
 
Table-2: Standard Impedances for Three-Phase Power Transformers
 

  

 
Example:
 
Referring to Figure 56, what is the transformer let-through current (ILT) at the secondary bus? What is ILT if the impedance of the transformer is specified at 5% versus 5.75%? What is ILT if the impedance of the transformer is specified at 7% versus 5.75%?
 
 
Answer:
 
 1. ILT = IFLA-sec/Z = kVA/[( 3 x kVsec) x Z]
 
• Where Z is expressed as a decimal
 
2. ILT @ 5.75% = 2500/[( 3 x .48) x .0575] = 3007/0.0575 = 52.3 kA
 
3. ILT @ 5% = 3007/0.05 = 60.1 kA
 
4. ILT @ 7% = 3007/0.07 = 43.0 kA
 

 
In the next article, we will continue discussing other special cases for Power and Distribution Transformers sizing calculations which are:

  • Paralleled Transformers,
  • K-Factor Transformers,
  • Transformers with Large Motor Loads.
So, please keep following.

 

 


 

Post a Comment

Leave a comment to help all for better understanding