[Spir Georges GHALI]

Dear Mr. Mill

Refer to this topic, I want to draw your attention that according to the IE… definitions, we have 2 kinds of faults can happen in any Electrical Network that are :

• Overcurrent Faults
• Earth Faults

Where the “ Overcurrent Faults ” are divided as follow :

1- Over-Load Fault where his percentage is between “ 15-20 % ” to “ 50-80 % ” of “ ILoad ”.

2- Short Circuit Fault where his percentage is more than “ ≈150 % ” of “ I ” and may be achieved to “ 10 time ” or even more.

Also the “ Short Circuit Faults ” are classified depending on his kind as following :

1. 3 Phases Short Circuit with or without Neutral ( possibility : ≈ 5-10% ) : the flown current is the biggest one.
2. 2 Phases Short Circuit with or without Neutral ( possibility : ≈ 10-20% ) : the flown current is “ ≈ 86% of 3 phases short circuit current ”.
3. 1 Phase & Neutral Short Circuit ( possibility : ≈ 70-80% ) : the flown current is the smallest one comparing with 1 or 2.

Rem. : it’s not necessary that a Short Circuit Fault be connected also to the Earth.

 

By the way, the figure no. 2 in the Topic is not relative to the 2^nd fault “ Phase to Phase Fault ” ( it’s the same as no. 3 ).

[Spir Georges GHALI]

Dear Mr. Christopher ;

 

Please note that the equation mentioned in my reply is general, and the complete one is :

 

( Vp / Vs ) = ( Is / Ip ) = ( Np / Ns )

where :

Np : the number of turns of Primary coil.

Ns : the number of turns of secondary coil.

[Spir Georges GHALI]

Dear Mr. Dikran

 

Most of Generators’ assemblers define the Apparent Power ” kVA ” with ” P.F. = 0.80 “, but we should always read the  Generator’s Name Plate to know the exact PF value be defined by the assembler.

[Spir Georges GHALI]

@Christopher said:
My first engineering job out of school is designing electrical panels for industrial process equipment.  My present learning experience involves transformers.  According to my calculations, I have a total of 3.75 A being drawn on the secondary side of a transformer.  The transformer is 480 VAC primary/120 VAC secondary and is rated 1kVA.  So this 3.75 A would result in a 15 A draw on the primary side, would it not?  And assuming a purely resistive load, that would equal 15 A x 480 VAC = 7.2kVA, no?  There were originally 4 A fuses on each of the two lines leading to the transformer primary side.  I figured my 15A x 1.2 safety factor / 2 lines = 9 A, so I thought the fuses should be 10 A rather than 4.  But I’m thinking my real problem is that my transformer is too small.  Where is my thought process breaking down here?  Thank you in advance for setting me straight.

Dear ;

As I understood of you is you have the following :

1- Transformer :  S = 1 kVA,  Vp = 480 V,  Vs = 120 V, where :

• i.      “ Vp ” : the value of Primary’s Voltage
• ii.      “ Vs ” : the value of Secondary’s Voltage

2- The current of loads is “ 3.75 A ”.

 

According to the Transformer’s specifications the maximum current that can be drawn from Secondary side is :

Is max = S / Vs = 1000 / 120 = 8.333 A

So, the Power of the Transformed is enough to supply this load.

 

By the way, and sorry to say that your calculation of the load’s current from Primary side is not correct, because that current can be calculated as follow :

( Vp / Vs ) = ( Is / Ip )

480 / 120 = 3.75 / Ip

Then :  Ip = 3.75 ( 120 / 480 )     è     Ip = 0.9375 A

That means the Fuses ” 4A ” at primary size doesn’t blow.

In reality, the real drawn current of Primary side is “ 0.9375  +  Transformer Losses ”. 

 

Rem. : if the Voltage’s values of Primary & Secondary are different, the calculation should be modified.

[Spir Georges GHALI]

Dear Mr. Mill ;

 

Refer to this topic, I would mention to the following points :

– In the table and all equations you mention only to the “ small Single Phase Generators ”, where are the 3 Phases Generators ?

– The table mentioned to the small powers of Single Phase Generators and the relevant Loads in Watts, but those loads’ values are theoretical, because :

• Most of actual loads aren’t 100% resistive, that means the PF is smaller than 1 ( exp. if we have a load “ P : 800W, PF : 0.70 ”, that means the minimum apparent power that we need is “ Smin : 800/0.70 = 1140VA ”, so, 1kVA generator can’t afford this load of 800W ).
• Normally, the name plate of any generator mention to a “ PF value ”, and that value guides us to know to the maximum active power “ P : W  or  kW ” can be consumed of it. Noting that you mentioned to this point in “ Standby Power Rating ” paragraph.

– I think we should also mention that the maximum loads of any Standby Generator should not exceed “ 80  to  85% ” of its Rated Apparent Power “ S : kVA ”.

– It’s mentioned in “ Prim Power Rating ” paragraph that for this type a 10% of the rated apparent power is allowed as an Overload only one time each 12 hours, but it’s not mentioned the duration of this allowed overload ???

[Spir Georges GHALI]

@Arunraje03 said:
Dear aLL,

Need your HeLp.!!

I want run a pump of 0.5HP only 5mins with time gap of 4hours.

Suppose at beginning of the day say at 8am I will manually start the motor it should be on only for 5mins. after 4hour (i.e. at 12pm) it should be ON automatically only for 5mins and again (at 4pm) it should be ON again for 5mins.

This cycle should repeat everyday.

Can anybody help me in this regards please provide me detail Bill-of-Material required and connection diagram.

Regards.

Arunraje.

Daer ;

You can easily do the sequential run & off by using a ” Daily Timer ” produce by ” Schneider Electric, hager, ABB, Legrand, … “, and as the running time is only ” 5 min ” you should have :

– Digital Daily Timer

– With reserve 

but be careful that the auxiliary contact of timer should energize the coil of suitable contactor that will run & stop the motor, as the nominal current of the auxiliary contact is small ” ≈ 6A ” and not for AC3

[Spir Georges GHALI]

@zulfadhli said:
Sir Georges,

 

Let talk about power generation,

I had done some visit to Power Generation Control Room during my university year. I donno if this is concept is wrong or correct. The Technician said that, the generator produce active and reactive power. But, production of reactive power is limited to provide extra life span to the generator. In addition, he said that, therefore on the transmission line system, utility company design reactor to supply reactive power into the transmission line to recap the usage of reactive power in the transmission and improve power quality in term of power correction.

dear Mr. zulfahdli ;

As we know all, any Electrical Power Generation Center generates ” Apparent Power ” in ” kva “, that will be divided into ” Active  &  Reactive Powers ” depending on the Power Factor’s value of the loads, where the resistive loads consume only ” Active Power “, and the ” Inductive & Capacitive Loads ” consume both kind of power ” Active & Reactive ” depending on the Power Factor’s value.

Beside , in most of countries, the Utilities oblige the end users of important values of Electric Energy to install the convenient Capacitors to improve the Power Factor’s Value of there sites till a predefined value ( ≥0.93 ). Also, the utilities use ” Capacitors and/or Synchronous Motors ” to improve the Power Factor value in the Transformation & Generation Centrals.

[Spir Georges GHALI]

@legendeer said:
Ok, below are some technical specifications:

 

Type                                        Oil Immersed

Rated frequency                         50Hz                         

Rated power (ONAN)                  1500Kva                           

Tap changer                             Yes

High Voltage Winding connection    Delta

Low Voltage Winding connection    Star

Vector Group                            DYN 11

 

The full technical spec is too long, based on this, any cost input?

 

Thanks for your help

Dear ;

The budgetary price is around ” 20500 Euro “, of the following transformer :

– Manufactur : Elvim – Greece

– Power : 1600 kVA

– Voltage : 20 / 0.4 kV

– With Tap Changer

– Cooling : ONAN

– Connection : Dyn11

Regards. 

[Spir Georges GHALI]

Dear Mr. Schonek

First, you’re right, the symbol of Reactive Power according of “ IEC 60027-1 ” is “ var ”.

It’s also right that the Reactive Power is mathematical concept, but as it’s measured that means it’s a consumption of a kind of Power.

[Spir Georges GHALI]

@legendeer said:

Hi,   I have a query that i hope some of you could help with. I’ve been erecting transmission and distribution substations for awhile now, and am now moving into supplying distribution transformers. I’m trying to get a price range for 33/.433 distribution transformers, anyone with experience on this? Just a rough figure for me to think about. Make/model doesn’t matter. Thanks alot for any input!

Dear ;

To be able to give you an idea about the unit price, you should complete the specifications’ inquiry like :

– The type of Transformer ” Immersed  or  Dry “

– The Transformer’s power ” kVA “

– The type of connection ( exp. Dyn11 )

– The type of cooling

– With or without Protection like ” Bukelz, … “

[Spir Georges GHALI]

@zulfadhli said:

@Spir Georges GHALI said:
Dear Mr. Mill ;

As I remember, that I mentioned before in other Topic to the same remarks that are :

1- The Reactive Energy doesn’t oscillate between “ Source & Load ” ( as mentioned in this Topic ).

2- The reactive Energy is a “ Consumed Energy ” by the installed loads either Inductive or Capacitive, and for that the Utilities install the Reactive Energy Counters “ kVAR meters ”, because if it’s oscillate that means it doesn’t be consumed and no need for these counters.

3- The Capacitors and Inductive Loads consume Reactive Energy, and they don’t store this energy and swinging back and forth between source and load ( as mentioned in this Topic ).

4- The Transformers’ Power are rated in “ Apparent Power – kVA ” because we don’t know the type of loads that will be supplied by them, as the Apparent power will be divided into “ Active Power  &  Reactive Power ” according to the “ Power Factor value ” of the supplied loads.

I agree with your explanation.

1. Reactive Power is indeed consume by inductive load and produced by Capacitive load. 

2. Utility Company does install kVAR meter as in located in Industry Area, as they install / use a lot of inductive machinery.

 

Dear Mr. Zulfadhli

For remark No. 1 : OK for the 1st part that Inductive Loads consume Active Energy & Reactive Energy, but please note that :

1- The Capacitors doesn’t produce any power.

1- The Capacitors are ” Loads “ consume a lot of Reactive Energy and too small of Active Energy.

2- The Current’s Vector of Capacitive Loads is ” Leading “, and the Current’s Vector of Inductive Loads is ” Lagging “, therefore the Sum of these 2 currents is ” Vector Adding “.

[Spir Georges GHALI]

Dear Mr. Mill ;

As I remember, that I mentioned before in other Topic to the same remarks that are :

1- The Reactive Energy doesn’t oscillate between “ Source & Load ” ( as mentioned in this Topic ).

2- The reactive Energy is a “ Consumed Energy ” by the installed loads either Inductive or Capacitive, and for that the Utilities install the Reactive Energy Counters “ kVAR meters ”, because if it’s oscillate that means it doesn’t be consumed and no need for these counters.

3- The Capacitors and Inductive Loads consume Reactive Energy, and they don’t store this energy and swinging back and forth between source and load ( as mentioned in this Topic ).

4- The Transformers’ Power are rated in “ Apparent Power – kVA ” because we don’t know the type of loads that will be supplied by them, as the Apparent power will be divided into “ Active Power  &  Reactive Power ” according to the “ Power Factor value ” of the supplied loads.

[Spir Georges GHALI]

@guest said:
how much ampere should i use for my 15hp, 220v, 3 phase motor?

Dear;

refer to your information about the motor, and to decide the nominal current should be used, we should calculate the nominal current for this motor as follow :

You mentioned to the Motor’s power in ” HP : 15 “, the supply voltage ” 220V “, and it’s ” 3 Phases Motor “, to be able to calculate the nominal current, I assumed that the Voltage is between 2 phases, and the Power Factor for this Motor is ” Cos φ = 0.80 “, so :

I = P / 1.732 x U x Cos φ 

I = 15 x 745 / 1.732 x 220 x 0.805

I = 36.6 A

 In this case, if you will use a Switch, his nominal current should be at least ” 40A “, bur be careful by using only Switch  there’s no any protection for your motor, but if you want to use ” Contactor + Thermal Relay ” you can use : 40A Contactor, and  Thermal Relay can be adjusted at ” 36 A “.

 

By the way, if the Voltage value between 2 phases is different, the calculation should be redone to have the relevant value of Nominal Current, and the same for ” Cos  φ “

[Spir Georges GHALI]

@guest said:
Power factor correction and harmonic mitigation go hand in hand and should be tackeled in totality

Dear ;

 

You’re 100% right, as I say always, before installing and even calculating the Capacitors’s Power, we should know all about the existed Harmonics Currents in the Network, then decide if we can install the Capacitors without any Network’s treatment and in this case what Type of Capacitors should be used, or we should do the Harmonics’ treatment before than calculate and install the Capacitors.   

[Spir Georges GHALI]

@surinder said:
dear sir,

         I am facing problem in a 400 kva 11000/440v transformer. Its powerfactor drops to 0.2 in night at noload.

I have connected 150 KVAR capacitor bank in night by a timer. But it improved to 0.6 only. In day time powerfactor is ok around 0.95. So pl. suggest what i do now.

Thanks

Surinder

 

Dear ;

 

As we know all, the Power Factor of any transformer without loads is a small or too small value depending on Transformer’s No-load current named ” iο ” as it’s almost Inductive Current, therefore, to improve the Power Factor for any transformer we use the following equation to calculate the Capacitors’ power should be installed :

Qο = Sn x iο

where :

• Qο : the Capacitors’ Power ” kVAR “
• Sn : the Transformer’s Power ” kVA “
• iο = the Transformer’s No-load Current ” % “

Refer to your information, I think that the value of No-load Current of this transformer is big enough, and may be you need more than ” 150 kVAR “, but if not, it will better to check if the installed Capacitors work correctly and the supply voltage is connected to all steps.

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