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visweswar rao said:
HAI IAM VISSU,
IAM WORKING AS AN ELECTRICAL ENGINEER IN SPINNING MILL.
RECENTLY APSPDCL POWER BILLING ON KVAH INSTEAD OF KWH.
SO WE HAVE TO MAINTAIN PF 0.99 TO UNITY OR LEADING PF.
IF WE MAINTAIN LEADING POWER FACTOR , ANY DISADVANTAGES ARE THERE?
PLS LET ME KNOW.
Dear ;
Certainly, you will some advantages when the Power Factor will be high, but be carful, because if you have a capacitive PF, then it's possible that the level of voltage will be higher that will be danger for some sensitive equipments,so, for that reason the ideal value for PF is : 0.93-0.96 but not more, because if you try to achieve 0.99 then certainly during switching-off for 1 big step ( 50 or 60kVAR ) you will have a capacitive power factor.
Regards.
sagar said:
effect of harmonics on industrial crane or other indusrial load such as vfd operated motor while using welding m/c in shop
Dear ;
The high level of harmonics have a negative effects on the network's components like : Cables, Capacitors, Circuit Breakers, Transformers, and Generators, as well as on the equipments like : Motors, Electronic Machines, ext.
In general, the effects on the motors are : Pulsating & Unstable Torque, some Vibration, Noising, quick Ageing, and the motor's Losses will be bigger.
Regards.
jj said:
Hi Guys, I'm new here. Please help me out. I hv a 1 no. Exhaust Air Fan of 25kW with Full Load Current of 43.5A -400v. What is the required power/current rating should I supply? (i.e. isolator/socket outlet rating)
Can any of you guys out there, show me how can I calculate this and what is the formula?
Thanks.
Dear ;
If you want to install an Isolater or Socket, the nimonal current should be ” 63A “, But if you want to protect the fan's motor, it's better to use a ” Motor Circuit Breaker ” that should be adjusted around 43A.
Regards.
therealabdo said:
Dear Engineer;
Actually i was working on installing a power factor correcting on a factory and i heard an info i just want to confirm it:
We all know that power factor correcting panels reduces the Kvar. and you get some discounts on the bill
What I heard is that with each 1 Kvar u SAVE, 10 Watta increases in ur consumption. Which means reducing Kvar will increase ur Watta which results a higher bill..but with the discounts u gain …the bill will be approximately the same…What do u think?
Dear ;
Adding to what said before, you will have certainly a benefit and the bill, even with the losses of the capacitors, will be less, because the in Syria if the power factor is less than 0.90 you will pay some penalities plus the value of KVARh.
Regards.
nigel said:
can someone calculate the amps for these motor
1 5hp/3ph/ac drive motor 415volts/50Hz
2 2hp/3ph/ac conveyer drive motor 415v/50Hz
3 3hp/3ph/ac pump motor 415v/50Hz
4 5hp/3ph/ac pump motor 415v/50Hz
5 10hp/3ph/acdrive motor 415v/50Hz
6 2hp/3ph/ac pump motor 415v/50Hz
7 4kw/200v/dc conveyer feed motor
8 5kva/3ph/ac 415v/50Hz
9 .5hp/1ph/ac fuelpump motor
10 1hp/1ph/ac water feed motor
11 2hp/3ph/ac conveyor drive motor
12 10 hp/3ph/ac cold roomcompressor motor
Dear ;
To be able to calculate the nominal current for a motor, you should know also the Power Fcator ” Cos φ ” for this motor, and if it's Single Phase or 3Phases motor, then we can apply the formula :
– For Single pahse : I = P / V x Cos φ
– For 3 phases : I = P / 1.73 x U x Cos φ
where :
– P : Motor's Power in ” kW “, but if we have it in ” HP ” we can take each HP = 0.742 kW
– V : Voltage between phase & neutral
– U : Voltage between 2 phases
But we should also take into consideration another factor that's ” The Efficiency ” of the motor, So the final value of the motor's current is :
In = I / Efficiency
Regards.
ayekyawmyint said:
Why to use shant reactor?
Dear ;
If you mean the shunt reactor installed in serise withe capacitors, the role of this reactor is to protect the capacitor by preventing the harmonic to go in the capacitor that called ” Detuned Capacitos “, because the reactor with capacitor in serise connection are composed a ” Serise Resonance Circuit ” where it hase a resonance frequency. So, when we will install the detuned capacitors in a polluated network, we should know all harmonics that we have in this network, then choicing of the detuned capacitors will be done accordingly to the following condition :
The frequency resonance of the detuned capacitor should be absolutly smallest than the smal harmonic frequency
But if you mean the reactor of a filter, this reactor with the capacitor in serise connection composed a ” Filter ” to eliminate a special harmonic, that means if we have in a network the 5th harmonic with a big value of current and we decide to eliminate it, we should install a Filter where his resonance frequency should be approximatly 250Hz. In reality, the resonance frequency for this filter is 240 Hz, because if we use exactly the same value we will have a big short circuit current for this frequency.
Regards.
saraswatapalit said:
Please help by providing a detail note on generator transformer sizing. What are the factors that governs this rating?
thanks in advance
best regard
saraswata
Dear ;
For exp. you have a factory for which you should define the sizes of Transforme & Genrator, first, you should know the following :
– The total installed power in this factory
– The loads that are running in normal working ” P “
– The kind of loads ” Single Phase or 3 Phases ” and how the ditribution will be done for the single phase loads
– The average value of Cos φ for the factory ( we can calculate it if we know the Cos φ for each load )
– After that we should know an idea if there's some extention and the total power of it
Then the needed apparent power is : S1 = P( total ) / Cos φ, and ad the needed apparent power should be less the power of the transformer by ” 15-20% “, we can say that the apparent power of the transformer is : S ( tr ) = S1 / 0.8
So accordingly to the value of calculated ” S ” we can do our choice of the real transformer from the standard rating that are : 25, 50, 100, 160, 200, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600, 2000, 2500, 3200, 4000, 6300 kVA
For the generator, we should know the vital loads that should be always in running, once we know, we do the same procedure to define the power of generator, noting that once the generator runs it should be loaded at teast 45% of his power.
Regards.
ghulam mujtaba said:
HI i am mujtaba want to know that how can we calculate size of cable as per requirement of load
Dear ;
First, by knowing the load's power ” P “, the kind of load ” Single Phase or 3 Phases “, and the Cos φ, we can claculate the nominal current ” Ib ” of load by applying the formula :
– For single phase : Ib = P / V x Cos φ
– For 3 phases : Ib = P / 1.73 x U x Cos φ
where :
– V : voltage between phase & neutral
– U : voltage between 2 phases
Then you should decide or know the methode of installation of cables as we have many methode ( better to refer to IEC 345-5-52 ), and the ambiant temperature, so, we can find the correction factors and the totat ” k “, so, we calculate the nominal current for cable in standard conditions ” Iz = Ib / k “, and by this value and the table of ” IEC 364-5-52 ” we can do the choice of cable accordingly to the kind of cable ” Single or Multi “, ” Cu or Alu “, ” PVC or XLPE “.
After knowing the section, we should calculate the percentage of drop voltage of the cable, if the percentage is equal or less the standard values, the cable is OK, if not, we should ove-size the section of cable. But be carful that the standard values of Drop Voltage percentages are from Source to the load, that means all cables from the source ( transformer or generator ) to the load.
Regards.
vijayakumar said:
dears
how to maintain correct powerfactor battery charging unit factory.
Dear ;
First, we should know the value of Cos φ for the factory that we called Cos φ1, and the value of the running power ” P ” in normal working factory, then define accordingly to the local low, the value of Cos φ2 that we should achieve, and then calculate the needed capacitors ” Q ” by applying the following formula :
Q = P ( tg φ1 – tg φ2 )
where :
– Q : the power in ” kVAR ” for needed capacitors
– P : the running power in ” kW “
– tg φ1 : the tangent φ1 ( we know it from Cos φ1 )
– tg φ2 : the tangent φ2 ( we know it from Cos φ1 )
then accordingly to the value of ” Q ” and the kind of loads, we can do the combination of capacitors.
But be carful, you should know if the factory's network is polluated or not, if yes, you should know the percentage of this polluation, then we can do the choice of the kind of capacitors as we have : Standard Capacitors, Overrated Capacitors, and the Detuned Capacitors.
Regards.
excalibur said:
I have a project. It consists of a lot motors, my problem is how can i know the right ampacity of cable wires and breaker to be use? can you share me some tables?
Dear ;
First, you should know from the name plate the following specifications for each motor :
– Powe ” kW or HP ” , Single Phase or 3 phases “, Cos φ , methods of running as we have ” Direct on Line , Star-Delta , Auto-transformer , Soft Starter , Variable Speed Drive ” , the distance between the motor & the supply's panel , and the methode of installation of cables with the ambiant temperature.
By these info, you can calculate the nominal current of the motor ” In ” then accordingly to the methode on running, you can estimate the insush current. Then accordingly to the methods of installation you can define the correction factors and the total ” K “, then you calculate the value ” Iz ” that's :
Iz = In / k
Refering to the tables of ” IEC 364-5-52 ” and the value of ” Iz “, we can know the section of cable, after that we should calculate the percenrages of drop voltage of cable in normal working & in running time, if the values of drop votltage from the source till the motor are equal or less the standard percentages the section of cable is OK,but if not, we should over-size the section. Noting that the standard persentages of drop voltage is :
– in normal working ( after running ) : 5 %, can be exeeded to 6 or 7 % in some special case
– in running time : 10-12 % ( not more )
but I prefer to draw your attention that these percentages are from source to the load ( not from the supply's panel to the motor ).
Regards.
jatin333 said:
what is % impedance of transformer? how we calculate %impedence of transformer,
what is purpose of % impedence?
Dear Mr. Jatin ;
In some case called the impedance of transformer that is given in ” % “, but the sientific name is ” Usc or Ucc ” that means the percentage of primary voltage ( HV ) that should be applied to have the nimonal current ” In ” on short circuited secondary.
Normally, each transformer hase a Resistance ” Rtr ” and an Inductance ” Xtr “, where we can calculate by the following formula :
– Ztr = Usc x U²/Sn
– Rtr = Ptr / 3 x In²
– Xtr² = Ztr² – Rtr²
Noting that :
– Ztr : the impedance of transformer
– Usc : the voltage in percentage of transformer
– U : the voltage between 2 phases
– In : the nominal current for the transformer's secondary
– Ptr : the total loss at full load
As desiegner, we should calculate the ” Rtr & Xtr ” to the transformer ten add themm to the values ” R & X ” of cables to calculate the different values of short circuit currents.
Regards.
David Knight said:
Hi all
Please can someone help me, i am trying to calc the PF factor of my heavyduty machines.
I know the voltage and the amps but how do i calc the PF factor??
Please help its driving me made
David Knight said:
Hi all
Please can someone help me, i am trying to calc the PF factor of my heavyduty machines.
I know the voltage and the amps but how do i calc the PF factor??
Please help its driving me made
Dear Davis ;
First, if you know the Voltage, the current of the machine, the kind of machine ” Sigle phase or 3 phases “, and the power of the machine, you can do a simple calculation to have the value of Cos φ, where :
– For Single phase machine : P = V x I x Cos φ
– For 3 phases machine : P = 1.73 x U x I x Cosφ
Noting that :
– V : value of voltage betwwen phase & neutral
– U : value of voltage between 2 phases.
– I : the current of the machine
but, if you don't know the power of the machine, so you should either measure directly the value of Cos φ.
regards.
jatin333 said:
how tansformers are paralleled and why?
Dear Mr. Jatin ;
Certainly, there ia a lot of advantages that were been explained before, we use this solution for exp. if a factory hase 1 transformer 1000 kVA that's suitable for the loads, and there's a new extention where the total of loads needs more than 1000 kVA, we have 2 solution, chang the actual transformer or add another one in parallel where the conditions were been mentioned before. But the most important point to know that after the installation of the 2nd transformer, the maximum short circuit current on the basbar will be doubled, that means we should re-view the value of the breaking capacities of installed circuit breakers if are suitable to the new valur of short circuit cureent or not, if not the circuit breakers should be changed.
Regards.
asegoma said:
Panelboard
Dear Asegoma ;
First at all, the choice of the CB ( circuit breaker ) should be correctly done accordingly the some points that are :
– The nominal ” In ” current of the CB.
– The correct value of the thermal protection ” Ir “
– The ambiant temperature around the CB, because most of CB are designed accordingly to +30 or +40°C; that means if ambiant temperature around the CB will be more you should absolutly derate the CB accordingly to the ” Derating Temperature ” done by the manufacturs.
– The connection should be done accordingly to the manufacturs instructions specialy for the big currents.
– The connection points should be tighted accordingly to the manufatures instructions.
– The section of cables should be well calculated accordingly to the circulated current.
Regards.
neighbor said:
Can You recommend any sofware?
Dear Neighbor ;
If you need a software by which you can calculate any Low Voltage Networke, from my experiance by using many of them, I suggest the software done by ” Schneider Electric ” that called ” Ecodial … ” where there's many version and the very good one in ” Ecodial 3.38 International ” & the new version ” Ecodial 4.1 Int “.
The advatages of this software after calculation are :
– You will have all values like : Ib for loads, Isc3max, Isc1max, Isc1min, If at any points of the installation.
– You will have the average Cos φ1 for all loads before compensation.
– You can add the capacitors bank, and the software calculate the the needed power for these capacitors to achieve Cos φ2 after compemsation.
– You have all values of the viltage drop at any points of network.
– The needed section of cables that should be used for all loads, where the software can calculate also the Thermal Stress of cables.
– The software can do the choice of Circuit Breakers to realize the ” Discrimination ” & ” Cascading “.
– You will have the values od adjustement for all circuit breakers that should be sone if they are adjustables.
– The software is able to do the choice for the Contactors, Motor Circuit Breakers, Variable Speed Drives, & the Soft Starters.
Rem. : The version 3.38 is Free on the Website of Schneider Electric.
Regards
