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ikram shehzad said:
on 1000 kw load our power factor is 0.85. how much kvar power is required to improve power factor up to 0.98. please tell me the formula to improve power factor by capacitors. thanks.
Dear Ikram ;
We have a very simple formula to calculate the Q ( kVAR ) that's :
Q = P ( tg φ1 – tg φ2 ) where :
– Q : is the power of Capacitors in ” kVAR “
– P : is the real power load in ” kW “
– tg φ1 : tangent φ1 ( before compensation ) ( calculated from Cos φ1 )
– tg φ2 : tangent φ2 ( after compensation ) ( calculated from Cos φ2 )
So, in your case :
– Cos φ1 = 0.85 → tg φ1 = 0.619
– Cos φ2 = 0.98 → tg φ2 = 0.203
Q = 1000 ( 0.619 – 0.203 ) = 416 kVAR
Noting that you should add at least 10-15% from the calculated Q, because if you have a drop voltage, then the real reactive power of capacitors is less than mentioned on.
Regards
ikram shehzad said:
dear all,
i have 20 no,s 400 watt,s helogen light,s. these lights are 50 feet far from the panel. can anybody provide me the exact formula to get exact cable size for it. if anybody have any formula or website or any information about this then please give me. thats my mail address.
m.ikram786@yahoo.com , m.ikram555@gmail.com
thank,s.
Dear Ikram ;
First, you should know the value of the voltage, then by the total power of lifgts and the voltage's value you can calculate the current of this loas that we call ” Ib “.
Then, you should know the type of installtion's methode, the ambiant temperature, by which you will calculate the total factor coorection ” k “, then calculate Iz = Ib / k, so, by this value and from the tables of ” IEC 60364-5-52 ” you can do your choice of the section of cable, but it's not the end, becuase you should calculate the total value of the ” drop voltage ” from the source to this load, where the IEC defines that the maximum value of the total drop voltage for luminars is ” 3-5% “, so, if with the choiced cable we have more, we should over-domensionned the section of the cable.
Regards.
Spir Georges GHALI said:
farang said:
I have a oil filled 3 phase transformer 315 kva on a pole pripary voltage is 33 kv secondary is 220/380.
What would be the rated output in amps per phase. Here everyone brings in 3 phase to thier pannel there are hardly any 3 phase loads, maybe a pool pump motor or two. The rest are 1 phase A/C and lighting loads.
Our ampient temperature is 35C
What can I safely draw in current per phase on this transformer?
Thanks
farang
Dear Farang ;
Each transformer hase an ” S : Apparent Power ” in kVA, either for single phase with neutral or for 3 phases with or witout neutral (depending of the connection).
The formula for Single phase + neutral is : S = V x I where :
– S : apparent power in kVA
– V : the nominal voltage between phase & neutral without load.
– I : the nominal current for the transformer
and for 3 phase + neutral, the formula is : S = 1.73 x U x I where :
– U : the nominal voltage between 2 phases without load
So. using one of these formula accordingly the the kind of transformer, we can calculat the nominal current for any one, noting that the ” I – nominal current ” for 3 phases with or without neutral means the value of current for each phase, in this case, when we have single phase loads we will distribute them to the 3 phases of the transformer trying to have approximetly an equivalant installed power on each phase.
Normally, we try to have the maximum load on any transformer not more than 80-85% of the apparent power.
Regards.
farang said:
I have a oil filled 3 phase transformer 315 kva on a pole pripary voltage is 33 kv secondary is 220/380.
What would be the rated output in amps per phase. Here everyone brings in 3 phase to thier pannel there are hardly any 3 phase loads, maybe a pool pump motor or two. The rest are 1 phase A/C and lighting loads.
Our ampient temperature is 35C
What can I safely draw in current per phase on this transformer?
Thanks
farang
Dear Farang ;
Each transformer hase an ” S : Apparent Power ” in kVA, either for single phase with neutral or for 3 phases with or witout neutral (depending of the connection).
The formula for Single phase + neutral is : S = V x I where :
– S : apparent power in kVA
– V : the nominal voltage between phase & neutral without load.
– I : the nominal current for the transformer
and for 3 phase + neutral, the formula is : S = 1.73 x V x I
So. using one of these formula accordingly the the kind of transformer, we can calculat the nominal current for any one, noting that the ” I – nominal current ” for 3 phases with or without neutral means the value of current for each phase, in this case, when we have single phase loads we will distribute them to the 3 phases of the transformer trying to have approximetly an equivalant installed power on each phase.
Normally, we try to have the maximum load on any transformer not more than 80-85% of the apparent power.
Regards.
KIAN KUMAR said:
dear sir!
how to work the MCB ? & wha is the principal of the MCB?
how to work the RCCB ? &what is the principal of the RCCB?
Regards
G.Kiran kumar
Banglore
Dear Kian ;
MCB :
the MCB's are Circuit Breakers that are reposnibles to protect any elctric circuit against ” Oveload & Short Circuit ” problems, where are composed of : External body, External connctions, External Handle, Thermal protection, Magnetic protection, Moving & Fixed contacts, Mechansim, Arc shut chamber, and the intsrnal connections.
– The Thermal Protection si reponsible to protect against ” Overload “, where the standard values ” In ” are ” 0.5, 1, 2, 3, 4, 6, 8, 10, 13, 16, 20, 25, 32, 40, 50, 63, 80, 100, & 125A, noting the 8A & 13A aren't used now.
– The Magnetic Protection is reponsible to protect against ” Short Circuit “, where the standard values are : B, C, D, K, Z, noting taht :
– The most using values are : B, C, & D
– B : 3-5 In
– C : 5-10 In
– D : 10-20 In
– By the external Handle and the Mechanism we can close & open the contacts, that means close & open the electric circuit.
– The MCB's can withstand a high values of short circuit currents accordignly the their ” Icu ” values, where the standard of these values are : 1500, 3000, 4500, 6000, 10000, 15000, 20000, & 25000A.
– The relative standard of the MCB's is ” IEC 60898 ” that defines all specifications, Conditions, & the sequences of testing.
RCCB :
the ” RCCB ” is ” Residual Current Circuit Breaker ” that are composed of : External body, External connctions, External Handle, Differencial protection, Moving & Fixed contacts, Mechansim, and the intsrnal connections. These RCCB's are sreponsibles to detect any small difference between Phase & Neutral currents in 2 poles RCCB or between the Phases & Neutral currents in 4 poles RCCB, we call this small current an ” Earth Lekage Current “. Noting that there's no any other protection in the RCCB that means there's no thermal & magnetic protections inside.
– The standard values of the differencial protection : 6, 12, 30, 100, 300, 500 mA, noting that to protect the pepoles we use ” 30mA “, to protecto against fire we use ” 300 or 500mA “, the using of ” 6 & 12mA ” is for a very special applications.
– The standard nominal current are : 25, 40, 63, 80, 100, 125A
– The relative international standard is ” IEC 61008 ” that defines all specifications, Conditions, & the sequences of testing.
Regards.
tukang listrik said:
hello there, i have a problem and need some solutions from all fellows here.
are there any formulas that can show the relationship between the area of cable, its purity of the material, and the current capacity? until now i mostly find just the table of the ampacity, but don't know how to calculate it.
Dear Tukang ;
First, the calculation methode of the nominal current for any cable, accordingly the the section & material, is very complicated, so, it's better to refer not the tables of manufacturers but to the table of ” IEC 60364-5-52 “ where you can find the nominal currents for each section accordingly to the methods of installation and the type of cable either for Cu or Alu. Noting that these values are refered to the standard installation conditions, so, you should know : how the cable will be isntalled, the ambiant temperature around the cable, the number of circuits, then you can calculate the Correction Fcator, by which you can calculate the current of the cable accordingly to the installation conditions.
By the way, there's now a new version of this standard.
Regards.
electricalexpert65 said:
Ali said:
what is the approperiate setting for O/L shall it be 115% of the FLA or of the actual current of the motor or the same as FLA of the motor
some one says 115% of FLA and others says 100% FLA and other says 125%FLA
i heared many things so i'm confusing
Not more than the rated full load current of the motor.
Dear Ali ;
First, we should the numonal current of the motor ” In ” from the name plate, them we adjust the O/L relay eactly as the ” In ” value, because :
For each thermal relay we have a Tripping Curve, by which we can find that there's no any tripping order if the motor current is equal to In, but if it's more than In the thernal relay needs some time ” accordignly the the tripping curve ” to stop the motor. that means if you adjust the thermal relay at 115% of In, your motor will run on overload for a time bigger than the time if adjusted on 100% of In, so, if the motor can accept it you can do it, but normally we don't do it. In some application, we need to run the motor with an overload ( we should know it ) for some time without tripping ( normally smal time ), in this case, we adjust at 100% of In and from the tripping curve we can know the tripping time at this overload, so we can decide if it will be OK or we should re-adjust, but the more important is the protection of the motor.
By the way, I want to draw your attention that there's many Tripping curves accordingly of the class of thermal relay ” Class 5, 10, 20 ” and for the Electronic Thermal Relays there's more, so, accordingly to the kind of application ” Standard or Severe ” we do the choice of the kind of tripping curve.
Regards.
paule said:
Is there any value per standard to use as reference for Total Harmonic in the system. What is the value we can say it is safe for a sensitive instrument.
Need to educate me, how the hamonic and distortion is appearing in the electrical system?please need your expert advice, thanks
Dear Paule ;
First, the harmonics currents are generating by the Elctronics Loads like : Computers, Printers, Faxes machines, Variable Speed Drives, DC Drives, Electrnic transformers for lighting, ext. These loads generates the Harmonics Currents, and all distortion of currents we called ” THD I – Total Harmonic Distortion of Current ” that affacte the voltage sine wave ( becuase we have the Harmonic Impedances ), so, we have that we called the distortion of Voltage ” THD U or THD V “. Noting that the most importatnt value is ” THD U or THD V “.
The ” IEC 61000-2-2,3,… ” defines the maximim percantages value of Voltage Distrotion of each harmonic for ” HV, MV, and LV networks, for exp. for LV networks : for n=3, the maximum is ” 5 % “, for ” n=5 ” the max is ” 6 % “, and also this IEC… defines these percantages for Odd & Even Harmonics. In France, the EDF defines the max percantage of ” THD V ” on the LV networks by ” 5 % “.
Regards.
amit kulkarni said:
hi can u help me @ transformer (3 phase) losses any formulae or thumb rules.if the transformer is 50 % loaded there are some losses but how to calculate? how we to calculate % of transformer loading?
Dear Amit ;
there are 2 kinds of losses : Iron Losse & Cupper losse, the iron loose is not affected by the load, but the cupper load si affected by the load.
Normally, on the name plate of any transformer there's a percentage value ” i % ” that's the percentage current of the nominal current for the iron losse, so, if we have it we can calculate the iron losse. and also on the name plate there is a value of cupper losse on full load in ” watt “, by whici we can calculate the ” R ” of the transformer then accordingly the load's current and the value of R we can calculate the cupper losse, notint taht the formula by which we calculate the ” R ” of any transformer is : R = Pcu/3 . In2 ; where :
– Pcu : is the full load cupper losse
– In2 : is the square on the transformer nominal current
Regards.
Dear ;
Adding to the parameters of Mr. Reihan, I want to add the following :
As it's a new factory, you have 2 possibilities :
– If the factory is just running, you can mesure the Harmonics that you have, then will decide whcich harminocs should be eliminated by Filters.
– If the factory is not running yet, you should know for each harmonics generator products the ” Harmonics Spectrum ” that you can obtain from the products' producers, then you will have a good idea about the hamonics & their levels once the factory runs, then will decide whcich harminocs should be eliminated by Filters
in all cas, you should absolutly know the harmonics and their levels to decide what you should be done, because in some cases where the levels of some harmonics are low, we can do nothing for them as the costs of filters wre heigh or veru heigh depending on the current level's of each hamonics shold be eliminated.
regards.
Spir Georges GHALI
