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Abdul Basit Khan said:
sir,
i am a user, and i need to define difference between 3 pole & 4 pole MCCB for 200kVA transformer & 275kVA generator in main distribution panel, if we use common neutral, what is the advantages and disadvantages.
Regards,
ABdul Basit Khan
Dear ;
The using of ” 3 Poles ” or ” 4 Poles ” MCCB either for transformers or generators, that means if the ” Neutral ” should be directly connected ” 3 poles ” or switched on & off with phases ” 4 Poles ” will be decided after knowing the ” Earthing System ” used in the project. So, if you know the Earthing System you can decide which MCCB should be used for your application.
Regarding the ” Earthing Systems “, we use as follow :
– TT, TN-S, IT (with distributed neutral) : the ” Neutral ” should be switched on & off with phases.
– IT (without distributed neutral ) : we don't need it as we haven't a neutral
– TN-C : as the Protection line & the Neutral are combined ” PEN “, this line should be directly and all ways connected.
Regards.
Vidhyut S. said:
We all know that capacitor supplies reactive power and Inductor consumes it.
How does this happen?
Why capacitor can't consume reactive power?
Dear ;
First, we know all the Capacitors are ” Loads “ that consume a reactive current, so, the Capacitors are not generator to supply any kind of power.
Assume that we have a pure inductive load and a Capacitor installed to the network, in this case we have the following currents :
– An Inductive current for the inductive load, where this current is ” leading ” by 90°
– A Reactive current for the capacitor load, where this current is ” lagging ” by 90°
so, the vectorial sum for these 2 currents is depending on the value of each one, if the value of the Inductive current is bigger than the Reactive current, the sum will be smaller than the inductive current, that seems as the capacitor supplies the reactive power to the inductive load.
Regards.
sureshkumar said:
Is there any MCB's for DC supply
Thanks.
Dear ;
Normally we can use the MCB's for ” DC supply ” but we should take into account the necessary information determinedby the manufacturers for this application, noting that the most important points are :
1 – System voltage : determined by the number of poles connected in series.
2 – Short Circuit Current : determined by the manufacturers depending on the ” L/R ” value
3 – Tripping caracteristics :
– The ” Thermal Protection ” : unchanged
– The ” Magnetic Protection ” : it will become less sensitive and should be derating by ” √2 “
Regards.
Avish said:
Can someone please describe to me the various factors affecting cable
sizing, and the step-by-step procedures how to do it, with examples if
possible. This has to do with one-size up cabling in a textile manufacturing plant, and to analuse what effect this will have on the cost, and also about the payback.Thanks.
Dear Avish ;
I answered many time about the section of cables calculation and how should be done, noting that the last one was before about one week where the name of Topic is ” Cable Size ” with an example about it.
In all case, the factors that affect the cable's size accordignly to ” IEC… ” are :
– The type of cable :
– Single Core or Multi Cores
– Cu or Alu
– PVC or XLPE
– The number of cables per phase
– The Method of Installation ” Unburied or Buried ” with the relevant ways for each one
– The numer of layers
– The ambient temperature
– The number of cables have been laid together.
– The Harmonics
After doing the necessary calculation and definingthe section of cable according to the above factors, we should take into consideration the following points :
– Calculate the percentage of ” Total Voltage Drop ” from the Source to the Load that should be compliance with the standard percentages, if not, the cable's section should be over sized.
– At the 1sf and the 2nd levels of the networks where the short circuit currents values are important, we should calculate the Thermal Stress to know if that cable can withstand it during the tripping time of the protection device, if not, the cable's section should be over sized.
Regards
Soufi said:
Did any one has selection table of cables, I mean for every size what is the current carrying capacity and the derating tables.
Dear Soufi ;
As there's many tables, please send me your e-mail to send them. noting that my e-mail is :
regards.
Dears ;
please find below an exsemple to define the section of a cable supplying a load.
Assume that we have the following load :
– Power : 450 kW
– Cos φ : 0.82
– Supply : 3Ph + N
– Voltage : 400/231V-50Hz
– Length from the panel : 125 m
– There's no Harmonics
– ku = 100%
So, the nominal current for this load is :
Ib = 450 x 103 / 1.73 x 400 x 0.82
Ib = 793 A
The Correction Factors ( or Reduction Factors ) will be defined accordingly to the Method of Installation and the kind of cable, assume that :
– Method of installation ” F ” :
- ” Horizontaly Perforated Cable Tray “
- ” Touching – Flat “
– The kind of Cable : ” Single Core Cable – Copper – PVC “
– 2 cables per Phase.
So, the Reduction Factor : 0.91
– The Ambiant Temperature is ” 45 °C ” in the air
So, the Reduction Factor : 0.79
And the total Reduction Factors is : 0.91 x 0.79 = 0.718
And the nominal current for needed cables is : 793 / 0.718 = 1105 A
We assume that we will use ” Single Core Cable – Cu – PVC – 300 mm2 “, and from table
” A.52-10 ” of ” IEC 364-5-52 “, we find that the Current Carrying Capcity for ” 300mm2 ” for ” Touching – Flat / Methode F ” is ” 587 A “, and the number of cable per phase is :
1105 / 587 = 1.88So, we will use :
– For each Phase : ” 2 Single Core Cables of 300 mm2 “
– For Neutral : ” 1 Single Core Cable of 300 mm2 “
Noting that :
– for Neutral : the load is ” 3 phases “, and we haven't the 3rd Harmonic.
– If ” Ku ” is different, the calculation should be re-done.
After that, you should calculate the ” total Voltage Drop “ from the source till the load to decide if that cables are suitable or not.
To calculate the ” Voltage Drop “, we use the following formula :
Δ U = 1.73 . I ( R . Cos φ + X . Sin φ )
1 – For section : from the panel to the load :
As the cable is ” Singl core, 300 mm2, Touching-Flate “, So :
R = ρ . L / S → 18.51 x 1.23 x 125 / 300 x 2 → 4.74 mΩ
X = λ . L → 0.13 x 125 / 2 → 8.12 mΩ
Cos φ = 0.82 → Sin φ = 0.57
Δ U = 1.73 x 793 ( 0.00474 x 0.82 + 0.00812 x 0.57 ) → 11.68 V
Δ U = 2.92%
2 – For section : from the source till the load :
To be able to calculate this section we should have the following information :
– The section of cable if the panel is supplying directly from the source, or the sections of cables if there's many sections.
– The length of cable or lenghts of cables.
– The kind of cable or cables.
– The nominal for this section or the nominal currents for these sections.
– The Cos φ for this section or for these sections.
– The Method of installation for each section.
By using the same formula, we caculate the percentage of Voltage Drop for each section, then we sum these perecentages to have the ” Total Voltage Drop ” that should be compliance with the standard percentage, if not, the section of cable should be oversized.
By the way, if the load is a motor, we should calculate the Total Voltage Drop 2 times that :
– After running ( nominal current for the motor ) ( as before )
– During running ( current & Cos φ for the motor during running )
And the Total Drop Voltage during running shouldn't be exceeded 10-12%
jaymez83 said:
@6, if your going to specify a percentage for volt drop you should let them know that this volt drop is the total from the intake point to the final circuit not just from a sub-board.
Dear Jaymez83 ;
I think that we do a changing of infortmation at the engineering level, so, that means an engineer should know that the ” total Voltage Drop ” for any point is from the Source to that point.
by the way, I want to drow your attention that when calculating the Voltage Drop it's better that we don't use the formula ” mV x A x m ” for the following reasons :
– The value ” mV ” is for a specific value of Cos φ that is not the case, and also for Sin φ
– I'm not sure if ” mV ” contains ” X . Sin φ ” as for a long cable the value of ” X ” is important.
Dear Ahmad ;
It will be better to define the Standard that your are using in your country, if you use the ” IEC… “, please find below the method to define the Cable's Section.
1- Calculate the Load's Current that we called ” Ib “.
2- Accordingly to :
– The methode oh installation of cable ” Barried or Unbarried ” and the type of installation for each one.
– The number of the circuits laid togother.
– The ambiant temperature.
we will define the ” Correction Factors – k1, k2, … ” from the table of ” IEC 60364-5-52 “
3- We calculate the Total Correction Factor ” k ” by multiplying all correction factors ” k = k1 x k2 x … “
4- we calculate the nominal current for the cable in the working condition ” I’z ” by the formula : I’z = Ib / k “
5- From the ” IEC 60364-5-52 ” tables that mentioned to the Nominal Current of each section in the Standard Conditions ” Iz ” depending on :
– The ” methode oh installation “
– The kind of cable ” Cu or Alu ; Single Core or Multi Cores ; PVC or XLPE “
we will define the suitable cable's section where ” Iz ≥ I’z “.
5- Then we calculate the Voltage Drop by the Formula : ” Δ V = 1.73 x I ( R . Cosφ + X . Sinφ ) “, where : R = ρ . L / S ; X = λ . L
6- This value of the Voltage Drop should be smaller or equal than the value defined by ” IEC… ” ( 3% for Lighting ; 5% for Machines ), if it's not the case, then the Cable's Section should be oversized.
7- In the 1st & 2nd levels of the network where the values of the short circuit currents are important, it will be better to be sure that the cable can withstand the short circuit current during the tripping time of the protection device, where we use the formula : ” I² . t ≤ K² . S² “,
if it's not the case, then the section of cable should be oversized.
Dear ;
Normally, we use the formula ” I² x t ≤ K² x S² ” to know if the cable can withstand the short circuit current during the tripping time of the protection device, and as ” I² x t ” should be equal or smaller than ” K² x S² “, we have 2 possibilities :
– Either, by knowing the value of the short circuit current and by using the ” Time / Current Curve ” we can define the value of ” t ” then do the calculation to know if ” I² x t ” is equal or smaller than ” K² x S² ” then decide if the section cable is enough or should be oversized.
– Or, by knowing the value of the short circuit current and the section of cable & the value of K, we can do the calculation to find the value of ” t “, then compare the value with the tripping time of the protection device that ca be obtained by ” Time / Current Curve “, if the ” t ” calculated value is bigger than the tripping time, the cable's section is enough, if not the cable's section should be oversized.
Dear ;
As we know all that according to the Standard ” IEC 61439-1 ” ( and also the old version ” IEC 60439-1 ” ) the Assembly Manufacturer is responsible to declare the value of ” Ue ” that should be combined with the rated current for correct functioning of the main and auxiliary circuits.
I think in your case, it will be better to ask the Assembly Manufacturer about the reason for this low value.
Dears ;
As we have many types of ” Earthing System “, and each one of them has advantages & disadvantages, so, Mr. Laurent mentioned to the advantages of ” IT system ” that are :
– Too limited value of the 1st fault's current that means long life of equipments.
– The continuity of service : very very important in some applications like Hospitals.
– By using the ” IMD “, we will have an idea about the insulation's level of the networks that means : optimizing of maintenance.
– Ext…
These advantages can't be realized by another system.
Certainly this system is expensive and in some cases ( depending on realisation ) we spend some time to locate the place of the fault, but we need it in some applications. As we have, for 3 phases IT, 2 main types ” IT without distribution of Neutral & IT with distribution of Neutral ” ( depending on the loads' type ), so by choosing the correct one and using the correct ” IMD equipments ” ( depending on the cost ), the IT will be reliable and very usefull.
Certainly, I'm for The ” IT Earthing System “, especially in some places like :
– Hospitals ” Operation Rooms ” as we can't accept the tripping of electricity once an earth fault occurs.
– In the areas where the explosion's risk is high.
– In some applications where the continuity of service is very important.
By the way, the fault's location management is not easy, but if we install at the beginning the necessary products of ” IMD ” it will be more easier, because if we have many outputs after the Isolation Transformer, we can use the ” IMD ” equipments that's composed of ” one Injection equipment ” and ” Detectors equipments with Torroids ” where the number of them is suitable for the outputs' number, so, this combination can detect and locate the output where the fault occurs.
journeyman said:
Transformer capacity: 100KVA/13.8KV/231/133Y
Transformer LV overcurrent protection Device: 400 Amp,Ceramic knife blade fuse
Secondary Service LV cable size bet. xtfmr LV bushing and 400 amp fuse box(approx.6 Mtr).: 4×300 mm2,alum cbl.
Secondary Service LV cable size bet.. 400 amp fuse box and secondary service cndctr.(approx.10 Mtr.): 4x300mm2,alum cbl.
Secondary service LV cbl size bet.4×300 mm2 cbl and load.(Approx.150 mtr length,Direct joint ): 120mm2 alum cbl.
Secondary service LV cbl size bet.120mm2 and load: 70mm2,alum cbl.
Secondary overcurrent protection device bet.120mm2 cbl and load:150 Amp c.breaker.
If the cause is overheating due to overload what will happen to the cable (70 mm2) at the upstream side of 150 amp protective device?
Dear ;
Excuse-me, but some things are not correct :
First :
You have a transformer ” 100 kVA ” with ” 231/133 V ” as output voltages, so, the nominal currant ” In ” for this transformer from LV side is :
– In : 100 x 10³ / 1.73 x 231 = 250 A
So, how this transformer is protected by ” Fuse : 400A ” ?
Second :
You mentioned to many outputs ” Cables' Sections & Fuses “, but you don't mentioned to the Loads' Powers !, because if the sum of the Fuses' currents is the total currents loads, this transformer will be certainly overloaded.
Third :
You defined for some outputs 2 cables' sections ” 300mm² & 120 mm² ” , ” 120 mm² & 70 mm² ” ?
Please clarify more precisely .
Regards.
Bhavik said:
how can improve power factor of open ended cable?
1.increase capacitance
2. reduce capacitance
3. increase resistance
4. increase IR
Dear Mr. Bhavik ;
If the end of cable is ” Open ” as you said, that means there no load's currents, and if there's no load, there's no consumed reactive power, so, for what we install the capacitors ? and what's the Power Factor that should be improved ?
Regards.
