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hi every one
imeed using and consttruction of auto transformer
and the method of cooling of power transformer
and the needs of cooling. last thig the information about
the PCBs
thank u
contact elyazel3000@hotmail.com
BoomBoom said:
Hello,
Do you know what is the accuracy error of energy meters ?
My electrical bill change a lot compare to my consumption behaviour.
Thanks.
Pl. refer to the relevant publication of Bureau of Indian Standards.
They have standardised the premissible error in different type of
meters in use. Pl. prolong the discussion.
Thank you for your replies and help. Also I managed to find some useful papers on the subject on the internet and I think I understand the P.U system now but still I have to apply it to a real life case to determine the short circuit current in a system.
However I would like to call on your help again and ask if anyone can tell me the imp/Km of standard 11kV, 33kV,66kV and 132kV. overhead transmission lines. Thank you.
Dear,
You can not use the same.
Reason:
Existing capacity : 37.5 KVA (Hope you forget to mention A)
To convert in KW : 37.5 x 0.8 = 30 KW
Whereas our new load is 62 KW
Hello,
You can use it in step up configuration, the only problem might be to work conected in parallel with other transfomers.
Some years ago we produced panels for Sony Outside Broadcast Vans .
We only used WAGO spring clamp terminals , terminating up to 25mm cables with great sucess ,easy to terminate cables and no problem with vibration .
We still use them today ( cost permitting) mainly the Weidmuller ones and would thoroughly recomend them.
Hello,
What you mentioned it is not exactally correct, as kWh is energy , so as you mentioned :
“Now the formula,which you know, KWH = Line Current * Line Voltage * Power factor can be applied.
The only unknown value ‘current’ can be calculated.”
You must not use de time, as P= dW/dt, the correct is kW = Line Current * 1,73*Line Voltage * Power factor * efficiency.
Hello,
The correct formulas were already mentioned, if you want to do it right you must consider power factor and the efficiency, allways
remember that when you read the kW in the motor plate, you are reading the output power.
In small motors the powerfactor and the efficiency are small, usually about 0,7 and 0,6.
Usually when the power factor and efficiency are unknow, people use 0,8 and 0,7 and some people forget about efficency ( very bad idea
for small motors ) and they do :
Pmec in kW
I = Pmec *10^3 / (1,73*400*0,8) = Pmec * 1/ (692 * 0,8 ) = Pmec * 1,8
Hello,
I dont know any book to suggest, but i can give you a simple example.
In a per unit system the values are refered to some values taken as bases of the system.
For example if we have many transformers conected in an energy distribution system, we can take
the values of one of them to make de bases of the per unit system :
T1 ( we take this one)
Sn=2500 kVA
Un=15 kV
T2
Sn=630 kVA
Un=15 kV
T3
Sn=500 kVA
Un=15 kV
T……
Bases :
Ub=U(t1)n = 15 kV
Sb=S(t1)n = 2500 kVA
Ib=2500*10^3 /(1,73*15*10^3) = 96,33 A
– Now lets make some calculations in per unit system
For example we Have :
S(T2) = (630 *10^3) / (2500*10^3) = 0,252 pu
In(T2) = 0,252 / 1 = 0,252 pu ( 1the denominator is because we have Un(T2)=15 kV=Ub= 1 pu)
S(T3) = (500 *10^3) / (2500*10^3) = 0,2 pu
In(T2) = 0,2 / 1 = 0,2 pu
So in pu :
Sn(T1) =1 pu , Un(T1)= 1 pu , In(T1)= 1 pu
Sn(T2)=0,252 pu, Un(T1)= 1 pu , In(T2)= 0,252 pu
Sn(T3)=0,2 pu, Un(T1)= 1 pu , In(T3)= 0,2 pu
We can see very easily wich equipment is above or under our network reference, we dont need to do calculations with
Ampere, Volt, VA, etc, it is also very usefull in some complex calculations.
If we want the real values we simple multiply the pu values by the bases :
Sn(T1) =1 * 2500 kVA = 2500 kVA , Un(T1)= 1 * 15 kV = 15 kV , In(T1)= 1 * 96,33 A = 96,33 A
Sn(T2)=0,252 * 2500 kVA = 630 kVA, Un(T1)= 1 * 15 kV = 15 kV , In(T2)= 0,252 * 96,33 A = 24,275 A
Sn(T3)=0,2 * 2500 kVA = 500 kVA, Un(T1)= 1 * 15 kV = 15 kV , In(T3)= 0,2 * 96,33 A = 19,266 A
the location of the earthing point is based on the earthing system your applying to your network (TN-C,TN-S, ,TT & IT)
TN-C
In this system the neutral and protective conductor are combined , the conductor is connected between the transformer star point and the equipment neutral /earth point . additional earthing point (rods) is not required on the equipment side
TN-S
in this system the earthing and nutral conductors between transformer and equipment are not combined ( separated ), they have to be connected back to the transformer star point, additional earthing point (rods) is not required on the equipment side
TT
in this system the only nutral conductors between transformer and equipment is connecetd , the exposed conductive part of the equipment shall be connected to local earth point by means of protective conductor.
IT
its the same as TT system except the the source neutral point is connected directly to earth ,its connected through low impedance earth electrode (earthing impedance)
there is also other compination such as TN-C-S
the protective cable size selection is based on BS ≤7671
- for phase conductor size S≤ 16mm2 the protective conductor size shall be equal to the phase conductor size
- for phase conductor size between 16<S≤35mm2 the protective conductor size shall be 16mm2
- for phase conductor size S>35mm2 the protective conductor shall be 1/2 phase conductor size
Regards
Amin
