[Ally Kanyondo]

Smithy said:

I think you need to be a little clearer in what you are asking and trying to achieve.

Are you trying to place lightning dods on the highest points of the building to attract lightning to set locations from which you then provide heavy conductor connections directly to earth so that you control the path of lightning currents rather than allowing them to pass anywhere they like through the building structure. This is lightning protection and is protection for the structure. It is suaually accomplished using air terminals on the highest part of the building with one or more down conductor connections into a buried earth grid. You can get some guidance on the coverage of air terminals from standards such as AS1768 or equivalent other British, intenational or American standards. It may also be worth talking to lightning protection experts such as Erico as there are many types of lightning conductors, other than simple rods, which provide improved areas of coverage.

or

Are you trying to connect lightning devices to electrical systems which may have components outdorrs and subject to lightning strike in order to conduct the high impulse voltage to ground to pretect internal electrical equipment from the excessive voltage spike. This is lightning arresting.

---

Dear Smithy

Thanx for the clarification, my problem is about lightning protection as the heading states. I know it is accomplished by using air terminals.

My problem is to know if there is a formula to calculate the maximum area one air teminal should cover, Assume the building has four point of the same level of which i can put air terminal, should i put air terminal in all points i must make some calculations to know how many i should connect.

Also give me the way i should contact this expert ERICO.

Regards

[Ally Kanyondo]

Devilina said:

aah now thats u good answer!!! i thot so tooo!! :) thnxxx ally!!

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Your welcome devilina, Dont hesitate to post other quistions.

 

 

Regards

[Ally Kanyondo]

Devilina said:

why do we use the symbol of a coil ..not a resistor while drawing a diagram of DC shunt machine  ?
(as given in this figure)  

                                                       ? 

Dear Devlina.

The coil is used to show the resistance of the coil(Field Resistance) and to distinguish from resistance of the load R.

 

Kind regards.

 

---

[Ally Kanyondo]

kusha02 said:

I think this site will help you ,
 

http://www.siemens.hu/htm/ajanlataink/trf/download/geafolproduct_english.pdf

 

I see Ghail help you but one more isnt anything :) 

---

Dear Mr. Kusha02

 

Thanks for your completement

 

Regards.

[Ally Kanyondo]

Spir Georges GHALI said:

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.]

 

Dear Mr. Ghali

 

Thank you very mch for your clarification.

 

Regards.

---

[Ally Kanyondo]

Spir Georges GHALI said:

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 said:

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.

 

Dear Mr Ghali

Thanks for your above comments. i didnt know if the rotary type have the power of 6kW for 1 ton.

I am taking it as a note. I always thankfull for your advice. Please don forget to work on my previous post about the formula for calculating max area of lightning protection. 

 

Regards

 

---

[Ally Kanyondo]

Spir Georges GHALI said:

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

 

 Dear Mr Ghali

 

For the above range, take the working voltage as 11kV and 33kV(distribution levels  in my country)

 

Regards.

 

---

Dera Mr. Ally ;

 

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

 

Regards.

---

[Ally Kanyondo]

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.

[Ally Kanyondo]

Hi,

Erickench is right, you should arrange you loads in the distribution panel or distribution board diagonal in order to ensure load balancing.

See the table below.

	R	B	Y	N
LOAD 1	X
LOAD 2		X
LOAD 3			X
LOAD 4	X
LOAD 5		X
LOAD 6			X
LOAD 7	X

 

Where R,B and Y are phases and X is the total load of each load.

From the that you can play with you loads in order to balance the phase differences.

 

Regards.

[Ally Kanyondo]

avnish khare said:

hi, i want to discuss abt most efficient electrical software,which is mostly used thes days power system network.plz suggest me to learn about the any most efficient software

---

Just use solutionselectrical software. The best and easiest software to use.

For more info go to http://www.solutionselectricalsoftware.com

 

 

 

Regards.

[Ally Kanyondo]

Hi,

 

I thank you shoul also consider the cost of buying that transformer compared to buying the new bell that satisfy the voltage levels

used in hong kong.

 

 

Regards.

[Ally Kanyondo]

Hi engineers,

 

I recommend the use of circuit breakers instead of fuses due to the following two major reasons:-

1. On heavy short circuits You cannot achieve DISCRIMINATION btn fuses.

2. The current-Time characteristics of a fuse can not be co-related with that of the protected apparatus

 

 

 

Regards

[Ally Kanyondo]

Hamdullah said:

Dear HVAC !

      What comes the value 1.73  ?   Here metioned ( max for 1Ø and for 3Ø devide it with 1.73 )

---

The value of 1.73 comes from the square root of 3.  That means √3=1.73205

 

 

Regards

[Ally Kanyondo]

Dear Mr. Thomas

 

Is the transformer breakdown appears more frequently during rain season only?.

 

If ts so, try to check the cable terminations on both sides primary and secondary side.

 

Somewhere water is leaking.

 

 

 

 

Regards.

[Ally Kanyondo]

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.

[Author]

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