[Spir Georges GHALI]

Dear Mr. Ally ;

 

Please find below the relative information:

For ” 11 kV ” :

– 630kVA : Po : 1790W,     Pt(+70°C) : 6940W

– 800kVA : Po : 2470W,     Pt(+70°C) : 8370W

– 1000kVA : Po : 2940W,     Pt(+70°C) : 9800W

– 1250kVA : Po : 3520W,     Pt(+70°C) : 11800W

– 1600kVA : Po : 3890W,     Pt(+70°C) : 14400W

– 2000kVA : Po : 4830W,     Pt(+70°C) : 17100W

– 2500kVA : Po : 5990W,     Pt(+70°C) : 20700W

– 3200kVA : Po : 4600W,     Pt(+70°C) : 23400W

 

For ” 33 kV ” :

– 630kVA   : Po : 2100W,     Pt(+70°C) : 6750W

– 800kVA   : Po : 2580W,     Pt(+70°C) : 8370W

– 1000kVA : Po : 2800W,     Pt(+70°C) : 9280W

– 1250kVA : Po : 3000W,     Pt(+70°C) : 12350W

– 1600kVA : Po : 3600W,     Pt(+70°C) : 14600W

– 2000kVA : Po : 4600W,     Pt(+70°C) : 16200W

– 2500kVA : Po : 5780W,     Pt(+70°C) : 19800W

– 3200kVA : Po : 6620W,     Pt(+70°C) : 22500W

 

Where :

     – Po : the Losses without load

     – Pt(+70°C) : the losses with ful load at ” +70°C “

 

Noting that values are for ” Zucchini – Italy ” transformers, and there's some differences between manufactures.

 

Regards.

[Spir Georges GHALI]

Dear Mr. Ally ;

 

For ” 1 Ton – rotary type ” the nominal current is around ” 6 A “ ( and not  6 kW ).

 

Regards.

[Spir Georges GHALI]

Bhavik said:

Which type of transformer give highest secondary voltage?

1. D/D

2.D/S

3.S/S

4.S/D

---

Dear ;

 

If you mean the Voltage value between ” 2 phases “, certainly for the same transformer with ” S : Star connection ” we will have the highest voltage value, for the following reason :

Assume that we have a transformer has on secondary side ” 3 coils ” ( as we have 3 phases ) with ” 231 V ” as the voltage output for each coil, that means we have ” 6 connection points ” that can be connected as follow :

     – For ” D : Delta connection ” : we will have as output only ” 3 points ” ( only 3 phases ) where the voltage value between each 2 points is ” 231 V “.

     – For ” S : Star connection ” : we will have ” 4 points as output ( 3 phases + neutral ) where we will have ” 2 values ” of voltage output that are :

          – ” 231 V between each phase and the neutral

          – ” 231 x √¯3 = 400 V ” between each 2 phases.

 

Noting that :

    – For secondary : we can change the connection as we need a Neutral point or not.

    – For Primary : we can't do it, because it depends on the supply voltage value.

 

Regards.

[Spir Georges GHALI]

Ally Kanyondo said:

ABKHAN said:

Dear Alls,

can you tell me the right way to minimize the phase difference to install the single phase equipments, i.e. our company is a software house, they have 350 computers,each compoter has almost 350~500 watt supply and installed 20 nos. 2 ton air conditioners, these are all single phase supply 220V, 17 nos 4ton 3 phase air conditioners, and our transformer is 200kVA, is there is any formula to apply there i minimize the phase difference, and tell me it the 200kVA transformer is sufficient or not and how much capacity we of tranformer we required.

Regards,

AB KHAN

---

Dear Mr AB KHAN
Firstly Can you elaborate your quastion about minimizing phase difference.

secondly, to know whether your transformer is sufficient or not, you should calculate the total powe consumed by computers and air conditoners,

To calculate the total power do the following:

1. For computers, total active power(Pp)=350*500= 175kW

2.FOR AIR CONDIOTIONERS

 

1 TON=3.504kW, so

Total active Power for air condioners(Pa)= 3.504*(20*2*1.73+17*4)=480.7488kW,

NOTING THAT THE HIGHLTED NUMBER 1.73 USED TO CONVERT SINGLE PHASE POWER TO THREEPHASE.

Then the total active power(P)=Pp+pa=480.7488kW+175kW=655.7488kW

Assume the diversity factor is 0.7

Then the diversified power=0.7*P=0.7*655.7488=460kW

Assuming also the power factor is 0.85

The total Apparent Power(S)=P/0.85=540kVA.

From there you can know know whether your chosen tr is right or not.

In your calculations you should not forget to calculate the total power for lighting then add to Pp and Pa.

Regards.

---

Dear Mr. Ally ;

 

Refer to your above calculation I have the following points :

– You applied the Diversity Fcator ” Kc “, but you forgeten to apply the ” Utilisation Factor – Ku “, as there's certainly some computers that are switched-off.

– The Electric Power for ” AC 1 ton ” is not in all case ” 3.504 kW “, because it depends on the Compressor type ” Piston  or  Rotary ” ( Exp. 1 ton / Rotary type  → ≈ 6 Amp. ).

– In this case, it will be better to do al follow :

          – First, prepare the ” Loads Distribution Table “.

          – Calculate the Current for each Phase ” IL1, IL2, IL3 “.

          – Calculate the Active Power ” P ” accordignly to the bigest current's value.

          – Multiply the ” P ” by ” Ku  &  Kc ” to have the real need Active Power.

          – Assume ( if not knowing ) the Power Factor value, and then caculate the need Apparent Power.

 

Regards.

[Spir Georges GHALI]

Ally Kanyondo said:

Dear Mr Ghali,

Apart fom those mentioned above,

What about the losses of transformer with rating above 500kVA, i.e with the following ranges

750kVA-2000kVA

 

 

Regards.

---

Dera Mr. Ally ;

 

Please define the Working Voltages, are the same as above or different ?

 

Regards.

[Spir Georges GHALI]

Dear ;

 

First, all equipments that contain a ” Reactance ” like ” Motors, Transformers, … ” and are working on ” AC power supply ” need absolutly the Reactive power, and this reactive power is depending on the specifications of these equipments. Exp. if there's 2 motors have the same value of the active power but for the 1st ” Cos φ = 0.80 ” and for the 2nd ” Cos φ = 0.85 “, the 1st motor needs as reactive power more than the 2nd motor. Noting that the Reactive power doesn't transfer to any kind of work, for exp. only the active power of a motor will be transferred to a mechanical work.

 

The most disadvatages of a big reactive power are :

     – The invoice of the consumed reactive energy will be important.

     – A big Jule's losses.

     – The Voltage Drop will be also big specially in the transmission lines as their distances are important.

     – For a source ” P² = S² – Q² “, so, as the Reactive power increase, the available active power decrease.

 

For these reasons, and as we can't do any thing for the consumed reactive power of equipments, we try to improve always the value of ” Cos φ ” for the networks, when doing, the advantages are :

     – Decrease the amount of the invoices.

     – Decrease the Jule's losses.

     – Decrease the Voltage Drop specially on the Transmission lines.

     – More available of Active power for a source.

[Spir Georges GHALI]

Dear All ;

 

Regarding the using of any Calculation Software for LV networks, I have some points that are :

A – As the study of any LV electric network should contain the specifications of all components, the goal of this kind of Software is only to help the Engineers to make easier the calculation instead to do it manually and spend a lot of time, so, once we do the calculation by any Software, and have all results as : Section of Cables, Nominal Currents' values for loads, Short Circuit Currents' values, Earth Fault's value, Thermal Stress values for cables, ext.…, we can too easily define the specifications of all Circuit Breakers that should be installed at any point of network, but certainly without using the following advantages :

–          ” Cascading  or  Back-up ” as that's the results of many testes for many combinations of C.B.

–          ” Discrimination  or  Selectivity “, but we can do it if we have a good idea about the tripping curves of C.B.

    So, it's not necessary to use the same Circuit Breakers that are chosen by the Software.

 

B – To use any Calculation Software, we should know very well the designing method of LV network for many reasons like :

– As any Software contains the default parameters where most of them should be modified as ” Method of Installation, Ambient Temperature, Kind of Cables, Ku, Kc, Voltage Drop Percentages, ext., so, the using of the default parameters is not convenient all time and for all networks.

– Any Software hasn't all what we need, and in some time it stops the calculation, where we should do some modification to continue.

– In some time we should modify even some chosen components like the Cables' Section to use some available Circuit Breakers or vise-versa.

– The default parameters are not convenient for all countries like ” the Transformers' parameters “, for that we should know what the country's utility defined as Transformers' Parameters and modify them to have the correct results.

[Spir Georges GHALI]

Abhilash said:

Dear All,

Kindly support me in the below matter for Type Test  IEC61439

I want to type test (ASTA) 250A SMDB with 4 outoings as per IEC 61439. Our enclosure is IP54 . I came to know that the Incomer MCCB 250A will not withstand the fulload current while temperature rise test since the enclosure is IP54 . 

If any of you have knowledge in this topic  kindly help me friends.

Abhilash

Dubai

---

Dear ;

 

Certainly, you know before that this SMDB should be ” Totally Type Tested ” and ” IP 54 “, so, did you use any Software to design it ( like : Sis-Power  or  Rapsoide  from Schneider Electric, XLPro2 from Legrand ) ? if yes, how this panel is not ” Totally Type Tested  &  IP 54 ” ? Because this Software helps to design the Distribution Panels to be ” Totally Type Tested ” accordignly to your specifications and installed equipments where also ” IP .. ” should be defined. Noting that this Software will choose the suitable metallic enclosure after calculation to Total Losses Powers for all installed equipments and compare it with the Heating dissipation of the enclosure. But if not ?????

 

By the way, normally we don't ” Dearate ” the Electric Panel, but in your case we optimize the circulated currents ( deacrese ) till the temperature rise to be constant value. Noting that in practice if the variation of temperature after ” 8 hours ” doesn't exceed” 1 K/h ” we can accept the ” Temperature Rise Test “.

 

Regards.

[Spir Georges GHALI]

Ldscp said:

Trying to get approximate numbers for core/iron/parasitic losses in large (50-500 kVA) distribution transformers that might serve one light-industrial building or complex, taking 13,800V to 480V 3-phase.The application is very low duty-cycle, so unloaded losses will dominate.

Have spent quite a bit of time browsing the Web; most postings either talk about standards or promote advanced tech that may not be in service yet.

Is there a general guideline our group might use?

Thanks very much for all advice.

 

Mike

---

Dear Mr. Mike ;

 

As I don't use these special Voltages, I can give you an idea about that I used for some special applications where the Voltages are ” 13.2 KV  &  420 V ” and the manufacter is ” Zucchini – Italy “.

– 100kVA : Po : 380W,     Pt(+70°C) : 1800W

– 160kVA : Po : 480W,     Pt(+70°C) : 2550W

– 250kVA : Po : 750W,     Pt(+70°C) : 3120W

– 315kVA : Po : 1050W,     Pt(+70°C) : 4050W

– 400kVA : Po : 1320W,     Pt(+70°C) : 5000W

– 500kVA : Po : 1630W,     Pt(+70°C) : 5960W

Where :

     – Po : the Losses without load

     – Pt(+70°C) : the losses with ful load at ” +70°C “

 

Regards.

[Spir Georges GHALI]

Dear Mr. Ally ;

 

Regarding the highlighted points, please find below the explanation :

– Ssc : 500 MVA : The IEC defined the maximum value of Short Circuit Power for each level, exp. for ” 11kV  &  20kV ” this value is till now ” 500MVA “, but if in your country the Ssc is smaller, you can either use this value or the defined value in your country. Some persons prefer do not calculate the RMV & ZMV, and they consider that the Ssc value is too big ( infinitif value )( making the calcultaion more easier ).

 

– RMV / ZMV : The value of this ratio have been also defined by IEC as the value of ” XMV ” is too bigger than ” RMV “, noting that in the past this ratio was ” 0.2 ” but as the MV networks is composed now of thousands of KM it's better to taken into consideration ” 0.1 “

 

– The value ” 1.05  & Cmax ( 1.05 ) : it's the voltage's flactuation of the medium voltage network.

 

– The value of ” X ” : as this value is depending on the Cable's Length, the IEC defined this value accordingly to the kind of cable and the method of installation, that means if the kind of cable is different than ” Touching – Flat ” the X will be also different. for exp. for Multi cores cable ” X = 0.08 mΩ ” wahtever the installation method.

 

– The value ” Cmin ” : as there's a Voltage Drop when a short circuit happen that will be in the most of cases around 5%, and we want to calculate the minimum value of the short circuit current, we should take it into consideration. For me, when I calculate the minimum short circuit current for the final circuits in the network I prefer to take  it into consideration even smaller than ” 0.95 ” ( 0.80-0.85 ).

 

Regards.

[Spir Georges GHALI]

duclampv said:

could you explain to me the conditons of 2 transformers operate parallel?

---

Dear ;

 

To connect 2 Transformers on parallel, we should take into consideration the following points for the Transformers and the Panelboards and the installed Circuit Breakers :

 

1- Transformers :

– The specifications ” Power, Usc, Wtr, Vector Group, Primary & Secondary connection, V1, V2, … ” of these 2 transformers should be the same.

– It's better that are produced by the same manufactures.

– The ” Tap-Off ” changer ( primary side ) of both should be connected to the same point.

– We should pay attention of the connection of LV sides ” L1(Tr1)  →  L1(Tr2),  L2(Tr1)  → L2(Tr2),  L3(Tr1)  →  L3(Tr2),  N(Tr1)  →  N(Tr2) “

– If we use Cables or Busbars between the transformers and the MDB it's be better to be the same kind and the same length.

– As there's no fine adjustment for the Output Voltage of transformers, it's better that are connected to the same Feeder of Medium Voltage.

 

2- Panelboards :

– If the Panelboards are ” Totally Type Tested Assemblies “, the test's Short Circuit Withstand Current for the distribution's Busbars should be ” > 2 x Isc3max “ ( depending on the values of ” Isc3max ” in the inatallation palces ).

 

3- Circuit Breakers :

The most important points about the installed Circuit Breakers are :

 

A- The Breaking Capacities ” B.C. ” :

As the value of the maximum short circuit current ” Isc3max ” is same for the 2 transformers, so :

   – For main C.B. of MDB : Equal or Bigger than the maximum short circuit current for 1 transformer ” B.C.  ≥  Isc3max “.

   – For other installed C.B. : Equal or Bigger than the maximum short circuit current for 2 transformer ” B.C.  ≥  2 x Isc3max “ ( 2 transformers are connected ).

 

B- The Magnetic Protection ” Im ” :

As the value of the minimum short circuit current ” Isc1min ” is same for the 2 transformers, and also the earth fault current ” If ” is same for both, and as the Magnetic Protection ” Im ” should trip due to ” Short Circuit  &  Earth ” faults if thers's no Earth Fault Protection, so :

   – For all C.B.  : The ” Im ” should be Smaller than the smallest value of ” Isc1min  or  If ” for 1 transformer ( as only 1 transformer is connected ).

 

But if there's also an Earth Faults Protection that should trip due to an ” Earth ” faults, the adjustments will be :

   – For all C.B.  : The ” Im ” should be Smaller than ” Isc1min ” for 1 transformer ( as only 1 transformer is connected ).

   – For all C.B.  : The ” Ig ” should be Smaller than ” If ” for 1 transformer ( as only 1  transformer is connected ).

Rem. : The values of ” Isc1min  &  If ” should be calculated for each C.B.

 

Regards.

[Spir Georges GHALI]

Dear Mr. Ally ;

 

Certainly you can.

Please find below another important points :

– The ” Loop Impedance “ for a fault at any point in the network should be calculated from the source to this point, that means : MV, Tr, and all cables till the point of fault.

– If the used Earthing System is different than ” TN-C “, in this case, we should calculate ” Isc1min  &  If ” specialy if the ” N  &  PE ” have different sections that means ” Isc1min ≠ If “.

– If you use ” IT  system ” : it's better to define first the ” Fault's Loop ” then calculate the Loop Impedance, as there's many method to realize the ” IT “.

 

I's at your disposal for any inquiry.

 

Regards.

[Spir Georges GHALI]

ally said:

Dear Ghali

 

Let me add to the above question,

 

Assume the ratings are as follows:-

1. Transformer rating is 1000kVA, 11kV/0.4kv

2. Type of earthing system is TN-C

3. Cable length is 35m.

 

Help us to calculte the remaining items(short circuits).

 

I real appreciate your solutions.

 

Your are a real MENTOR most of in the forum.

 

Regards.

---

Dear Mr. Ally ;

 

Thanks a lot for your complement.

Please find hereafter the manual Calculation of the important Short Circuit Currents ” …max  &  …min “.

To calculate the Short Circuit Currents, we should calculate the Impedances ” R  &  X ” for : Medium Voltage – MV, Transformer – Tr, and the Cables between the Output of Tr and the Input of MDB that we do as follow :

 

– Medium Voltage :

We have the following information :

– The value of the Short Circuit Power on ” 11 kV ” is : ” Scs ≈ 500 MVA “

– The Voltages : 11 / 0.4 kV

– RMV / ZMV = 0.1

So :

 ZMV = U² / Scs   →   ( 1.05 x 400 )² / 500 x 10⁶   →   ZMV = 0.352 mΩ

Rem. : as we should calculate the values of ” R  &  X ” from LV side of transformer, we can use directly ” 400V “, or we use ” 11kV ” but in this case and after calculation we should multiply the result by the ” SQR of the Transformer Ratio – ( 11/0.4 )²  “.

RMV = 0.1 x ZMV    →                                                                      RMV = 0.035 mΩ

XMV² = ( ZMV )² – ( RMV )²    →    ( 0.352 )² – ( 0.035 )²    →         XMV = 0.35 mΩ 

 

– Transformer :

As the Transformer's power is ” Sn = 1000 kVA   →   In = 1445 A ” and I haven't the other information, I will assume that :

– Usc = 6 % ( percentage of Short Circuit Voltage )

– Wtr = 15 kW ( full load Copper Loses )

So :

ZTR = Usc x U² / Sn   →   0.06 x ( 400 )² / 1000 x 10³   →               ZTR = 9.6 mΩ

RTR = Wtr / 3 x In²   →   15000 / 3 x ( 1445 )²   →                          RTR = 2.39 mΩ

XTR² = ( ZTR )² – ( RTR )²    →   ( 9.6 )² – ( 2.39 )²    →                XTR = 9.29 mΩ

Rem. : if the values of ” Usc  &  Wtr ” are different, the calculation should be re-done.

 

– Cables :

As the previous cables have been defined accordingly to the Loads' Current ” 1175 A “, but now :

– They should be defined accordingly to the transformer's nominal current ” 1445 A “

– The number of cables per phase will be more than 4

– These cables will be installed on 2 Cable Trays.

So, The Reduction Factors are :  0.81   &   0.61

By doing the same previous calculation, we obtain the cables' sections in the same working conditions that are :

– For each Phase        : ” 5 Single Core Cables of  300 mm² “

Regarding the following points :

– All loads are ” 3 Phases “.

– Assuming that we haven't the 3^rd harmonic.

– The used Earthing System is ” TN-C “.

We can use the following cables :

– For  PEN                   : ” 1 Single Core Cables of  300 mm² “

Calculation of ” X ” for Cables :

As the Installation's Method is ” Touching – Flat “, the ” X  per 1m = 0.13 mΩ/m “, so :

Xca( Ph ) = 0.13 x 35 / 5    →                                                                Xca( Ph ) = 0.91 mΩ

Xca( PEN ) = 0.13 x 35    →                                                                   Xca( PEN ) = 4.55 mΩ

Calculation of ” R ” for Cables :

As the Resistivity ” ρ “of Copper is different between the ” Cold ( 20 °C )  or  Hot ( 70 °C ) ” states of cables, we should calculate ” R ” in these 2 states, that we do as follow :

In Cold state ” 20 °C ” :

Rca(20)( Ph ) = ρ x L / S = 18.51 x 35 / 5 x 300    →                              Rca(20)( Ph ) = 0.432 mΩ

Rca(20)( PEN ) = ρ x L / S = 18.51 x 35 / 300    →                                 Rca(20)( PEN ) = 2.159 mΩ

In Hot state ” 70 °C ” :

Rca(70)( Ph ) = 1.23 x ρ x L / S = 1.23 x 18.51 x 35 / 5 x 300    →          Rca(70)( Ph ) = 0.531 mΩ

Rca(70)( PEN ) = 1.23 x ρ x L / S = 1.23 x 18.51 x 35 / 300    →             Rca(70)( PEN ) = 2.656 mΩ

 

Calculation of Short Circuit Currents :

To calculate the ” Loop Impedances ” of Short Circuits, it will be better to integrate all results in a table like the following :

	Component	X : mΩ	R : mΩ
1	Medium Voltage : 500MVA – 11/0.4kV	0.35	0.035
2	Transformer : 1000kVA – 6% – 15kW	9.29	2.39
3	Cables :		20 °C	70 °C
	Phase : 5 x 300mm2	0.91	0.432	0.531
	PEN    : 1 x 300mm2	4.55	2.159	2.656

 

Calculation of the maximum Short Circuit Current ” Isc3max ” :

We assume in this case that we have a short circuit between ” 3 Phases ” at the input of MDB but the cable are in the ” Cold ” state :

Isc3max = Cmax x Uo / 1.73 x Zsc      ,      where :

– Cmax = 1.05

– Uo = 400V

Zsc² = ( 0.35 + 9.29 + 0.91 )² + ( 0.035 + 2.39 + 0.432 )²     →                         Zsc = 10.93 mΩ

So :

Isc3max = 1.05 x 400 / 1.73 x 10.93 10^-3      →   Isc3max = 22.21 x 10³ A      →    22.21 kA

The Breaking Capacity of the Circuit Breaker should be  ” BC  ≥  23 kA “

Rem. : The ” Isc3max ” at the Transformer's Output is ” 24.42 kA “.

 

Calculation of the minimum Short Circuit Currents ” Isc1min   &  If ” :

As we have ” PEN ” that means the ” Isc1min  &  If ” have the same value, so, we will calculate only one.

We assume in this case that we have a short circuit between ” 1 Phase & PEN ” at the input of MDB but the cable are in the ” Hot ” state :

Isc1min = If = Cmin x Vo / Zsc      ,      where :

– Cmin = 0.95

– Vo = 231 V

Zsc² = ( 0.35 + 9.29 + 0.91 + 4.55 )² + ( 0.035 + 2.39 + 0.531 + 2.656 )²   →     Zsc = 16.109 mΩ

So :

Isc1min = If = 0.95 x 231 / 16.109 x 10^-3     →   Isc1min = If = 13.62 x 10³ A      →   13620 A

The Magnetic Protection should be adjusted at ” Im = 10600 A “ or smaller accordingly to the protection unit adjustments.

 

[Spir Georges GHALI]

jatin333 said:

use Variable freq.Drive for converting 1-phase to three phase power

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Dear Mr. Jatin333

 

Your solution can be used in apseical cases, and please note the following :

– For most of manufatures of Variable Speed Drives, the maximum power of a VSD where the Input is ” Ph + N ” and the Output is ” 3 Phases ” is ” 3 kW “.

– The output of these VSD is only ” 3 Phases ” without neutral.

– The protection of the Output currents is dedicated for motors, that means the currents' values of 3 phases should be the same, so, if the power is suitable ” ≤ 3kW ” we should disable this protection as we don't know if the load has the same current of 3 phases.

– A lot of VSD Fonctions should also be disabled.

 

Regards.

[Spir Georges GHALI]

Dear ;

 

I think that you posted another topic for Cables' Section Calculation where I explained the procedure to define the suitable sections of Cables to electrify the  70 AC, the cables are :

– For each Phase : ” 4 x Single Core Cable of  300mm² “

– For Neutral       : ” 2 x Single Core Cable of  300mm² “

 

About the Circuit Breaker size :

we do as follow to define the specifications of the CB

– The Nominal Current for CB :

          In = 1250A

– The Thermal Protection :

          Ir = 1200A ( as the cables can carry ” 1246A ” with the installation conditions )

– The Magnetic Protection :

          Im : we should calculate : the minimum Short Circuit Current at the end of cable ” Isc1min “, and the Earth Fault at the end of cable, then the value of ” Im ” will be smaller by ” 20% ” than the smallest one.

– The Breaking Capacity :

          BC: we should calculate the maximum Short Circuit Current at the beginning of cable ” Isc3max “, then the value of ” BC ” will be equal or bigger than the calculated value.

 

But as I don't know : the Power of Transformer, the Earting system, and the Length of cable, I can't calculate the 2 last specifications.

 

Regards.

[Author]

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