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Dear ;
As there's no answer till now, please find bellow the calculation of the Cables' Sections.
Assume that :
- ” Ku ≈ 70 % ” that means 50 of 70 AC are working in tha same time.
- The Methods of Installation :
– ” Horizontaly Perforated Cable Tray “
– ” Touching – Flat “
- ” Only 1 Cable Tray “
- The kind of Cable : ” Single Core Cable – Copper – PVC “
The total load current : Ib = 50 x 23.5 = 1175 A
The Reduction Factors will be defined accordingly to the following points :
- 3 or 4 cables per Phase → The Reduction Factor : 0.87
- The Ambiant Temperature is ” 55 °C ” in the air → The Reduction Factor : 0.61
So, the total Reduction Factors is : 0.87 x 0.61 = 0.5307
And the nominal current for needed cables is : 1175 / 0.5307 = 2214 A
Suppose that we will use ” Single Core Cable – 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 “, the number of cable per phase is : 2214 / 587 = 3.77
So, we will use :
- For each Phase : ” 4 Single Core Cables of 300 mm2 “
- For Neutral : ” 2 Single Core Cables of 300 mm2 “
Noting that :
- For Neutral : as all loads are ” 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 ” Drop Voltage “ to know if these cables are suitable or not.
Regards.
Dear ;
First, I'm surprised, because the ditributed Medieum Voltage that you should be used is normally defined by the ” Utility “, as the Utility defines accordingly to the consumed power from where the load should be electrified. In Syria for exp. The Utility defined as follow :
– The load up to ” 35kW ” will be connected directly to LV network ” 0.4kV “.
– The load up to ” 10MW ” should be connected to MV network ” 20kV “, that means the client will have a substation ” 20/0.4kV “.
– The load more than ” 10MW ” should be connected to HV network ” 66kV “, that means the client will have 2 substations ” 1st : 66/20kV ” & ” 2nd : 20/0.4kV “
But if you have the possibility to select one of 33kV or 11kV, please note that the transformer ” 33/0.415 kV ” is more expensive than ” 11/0.415 kV ” by around ” 15% “, and the Circuit Breakers ” 33 kV ” are also more expensive than ” 11 kV ” by around ” 3-4% “.
About the ” Step-Down ratio ” between HV and MV, please note that's approx ” 3.3 – 3.6 “, and some manufactures can produce transformers with more than this ration but the cost will be very expensive.
Regards.
Freeknow said:
Hi dears
I have 70 Unit of A/C unit 3phase 220 Current rated as in name plate is 23.5. My Qusi I want to total load for these units and feeder size, condsider ambeint tempreture is 55 degree.
With best
Dear ;
Please confirm the following points to be able to do the calculation :
– Supply : 3 Phases – 220V : is this value between 2 phases ?
– The rated current for each unit is : 23.5 A
– The Power Fcator ” Cos φ ” for unit ?
– Haw many units will be in working in the same time ? ( to define the Utilisation Factor – Ku )
– What kind of cables you use normally ” PVC or XLPE ” ? Cu or Alu ?
– What Methode of installation you will use ” Unperforated or Perforated cable tray, Ladder, … ” ?
– Are you sure that the ambiant temperature inside where the cables will be installed is ” 55 °C ” ?
Dear ;
As we know that the formula of motor speed ” r.p.m. ” ( without sleeping ) is :
r.p.m. = 60 x f / n
where :
– f : the frequency
– n : the nomber of each 2 poles
so, as in the most of case ” n ” for a motor is not changeable, that means to change the speed for a motor we can change the frequency, that we can do by using the ” Variable Speed Drive “. Noting that the frequency can be changed by the Variable Speed Drive from ” 0 ” to ” 100 Hz ” and in some V.S.D. we can achieve ” 500 Hz “. But we should be careful, because :
– It's possible that we have some Mechanical problems in the motor ( specially with the bearings ) if it runs on high speed for a long time ( more than his nominal speed ).
– In some case, we have a problem of heating.
Regards.
Dear ;
First, all equipments that contain a ” Reactance ” like ” Motors, Transformers, … ” and are working on ” AC power supply ” need absolutly the Reactive power, and this reactive power is depending on the specifications of these equipments. Exp. if there's 2 motors have the same value of the active power but for the 1st ” Cos φ = 0.80 ” and for the 2nd ” Cos φ = 0.85 “, the 1st motor needs as reactive power more than the 2nd motor. Noting that the Reactive power doesn't transfer to any kind of work, for exp. only the active power of a motor will be transferred to a mechanical work.
The most disadvatages of a big reactive power are :
– The invoice of the consumed reactive energy will be important.
– A big Jule's losses.
– The Voltage Drop will be also big specially in the transmission lines as their distances are important.
– For a source ” P² = S² – Q² “, so, as the Reactive power increase, the available active power decrease.
For these reasons, and as we can't do any thing for the consumed reactive power of equipments, we try to improve always the value of ” Cos φ ” for the networks, when doing, the advantages are :
– Decrease the amount of the invoices.
– Decrease the Jule's losses.
– Decrease the Voltage Drop specially on the Transmission lines.
– More available of Active power for a source.
mkomaravelly said:
Dear,
We have got a requirement from customer side, more over its in Agricultural Area. Customer is preferring 3 Phase 10kVA Rating Transformer for the above mentioned hp.
Instead of 10kVA, we wanted to supply him 16kVA for other issues.
So,please reply me.
Thank You.
Dear ;
Refer to your answer, please note that the procedure to define the Power of transformer is as follow :
– The motor's power in ” kW ” is : 7.5 ( HP ) x 0742 = 5.5 kW
– The needed power assuming that ” Cos φ = 0.85 ” = 5.5 / 0.85 ≈ 6.5 kVA ( if the Cos φ is different, please re-do the calculation )
So, 10 kVA if suitable with around 50% plus than the needed power, but certainly with the standard percentages of ” Drop Voltage “.
Regards.
mkomaravelly said:
Dear All,
I have a requirement for Transformer..
7.5hp Motor with 3 Phase, 415Volts, For this requirement I need a Transformer.
Can any body suggest me required 3 Phase Transformer Capacity for the above said requirement.
Thank you.
Dear ;
Fisrt, this motor is a small & standard motor, and you can run it by using any ” 3 Phases + Neutral / 415/239V “, so, why you need a transformer ?
Please, explain more about the kind of supply that you have.
Dears ;
Adding to what said before, we should be careful of the following points :
– It's possible that the Power factors ” Cos φ ” aren't for all ” 0.80 “, so, the nominal current will be changed.
– When we know the nominal current of a motor, we can select the cable's section either from the tables of ” IEC 364-5-52 ” ( best ) or from the tables of the manufactures, but these values are in the standard conditions, so, we should integrate the Correction factors accordingly to : the kind of cable, Method of installation, Ambient temperature, and the no. of cables in the same conduit.
– After the correct slection of cable's section, and as the cable will electrify a motor, we should calculate the ” Drop Voltage ” in 2 cases :
– During the nominal working ( after running ) where the current is the ” nominal current ” for the motor.
– During the running where the current is the ” Inrush current “.
if these calculated values are suitable, we can install & use the selected cable, but if not, we should over-size the section of cable.
– Then we will select the CB, noting that this CB is responsible to protect only the cable ( not the motor ).
nagaraju said:
Dear Engineers,
Greetings.
I want to be know about synchroniziation with full fledged details,regarding example of generators and transmission lines also
thanking you.
Dear ;
We use the Synchronization to be able to connect 2 sources on parallel to the same bus-bar, so, to do it we should realize the following conditions :
– The Value of ” Voltage ” for both should be exaclty the same.
– The Value of ” Frequency ” for both should be exactly the same.
– The direction of ” Rotation ” of pahses should be the same.
– There's no any ” Phase shift ” between them at the moment of connection.
– It will be better that :
– The powers are the same.
– The specifications are the same.
– The manufacture for both is the same ( to be sure that they have the same specifications ).
There's now special equipments that can do it automatically, or another can provide a signal once these conditions are realized, so, we can connect them manually but in this case it should be done immediately after the signal.
Assume that we want to connect 2 generators on parallel, where they have the same specifications mentioned above, in this case, at least one of them should have the possibility ” manual or Electrical ” to adjust : the Voltage and the Frequency ( by changing the r.p.m. of the engine ) that will be used to adjust these values to be the same as the 1st source. The procedure to connect them is as follow :
– We run the 1st generator, close the relative CB, so the Synchronization equipments will measure the ” Voltage & frequency ” for it.
– We run the 2nd generator, so, the Synchronization equipments will measure the ” Voltage & the frequency ” for it, and in the same time the ” Shift phase ” apparatus will measure the ” Shift angle ” between them.
– We adjust the ” Voltage ” value for the 2nd to be exactly as the 1st.
– We adjust the ” frequency ” value for the 2nd to be exactly as the 1st.
( these adjustments can be done also automatically by the Synchronization equipments if the 2nd generator is equipped ).
– If the connection will be done automatically, the Synchronization equipments will do it once the conditions are realized, but if will be done manually, we should monitor the ” Shift phase ” apparatus and once the value of the shift angle is ” 0 ” we close immediatly the CB for the 2nd generator.
If the powers of these 2 generatos aren't the same,in this case we should use also the ” Load Shearing Apparatus “.
Dear Sir, how r u? me working in one pump station,for making SLD i need to calculate the total load,f i have cable size like 3c, 185mm2 ,then which proper size of mccb we will use? i need some proper formul/tableshOW I WILL calculade load?And how i will calculated real power,reactive power?waiting for prompt response. Thanks
Dear ;
First, I thing that you are talking about the disgning of the network for A Pumping Station, so, to be able to define the necessary procedure, let me the following points :
– The power supply voltage for the motors is ” MV or LV ” ?
– Are you talking about an extention for this station ? or it's totally new one.
– An idea about : the No. of motors and the avearage power.
– The power supply is ” Transformer or Generator or both “
nael.samman said:
There is another way, if you can;t modify in internal connections
you can use a capaitor with the following arrangement
1- the connection should DELTA
2- the delta poles are A,B,C
3- L1 is connected to A, N is connected to B, the capacitor is connected to A and C
this should do the trick
Regarding the capacitor rating, frankly i don't know the calculations, maybe Mr.Spir (My mentor) can help with this.
Dear Mr. Nael ;
We use this solution for some small & special application ( not 15kW ) where thers's no possibility to have ” 3 phase + neutral “, noting that if there's inside a motor the torque will ne be stable.
ajay123 said:
i have 15Kw very old motor 1983 make,i want run this motor with AC drives.
i want know that if any effect on motor winding of peak voltage generated by drive and what's solution of this.
Dear ;
Normally, there's no problem to use AC drive with this motor, noting that u used for motors even old than this motor, but you don't mentioned to the following points :
– If it's an ” AC motor – Squirrel Cage ” or another kind.
– What's the working voltage for it ?
– If it has been rewinded.
– Certainly, it's 3 phases motor.
Dear ;
First, you mentioned that the school has a Single Phase ” 120/240 V “, and that is not possible, because a Single Phase supply meanse that you have only 1 value 120V or 240V, and you haven't both in the same time, but if you have both in the same time that means you have ” 3 Phase + Neutral “.
So, if you have a Single Phase ” 240V “, you have the possibility that Mr. Nael Samman mentioned, but if it's ” 120V ” and you have the same possibility you should use a transformer ” 120/240V ” with the necessary power. If you use 1 of these 2 cases you should be careful because after re-connecting the cooker to be supplied by 1 phase + neutral, the nominal current of the cooker will be 3 times than his initial value.
But if the cooker can't be re-connected, the school should absolutly have ” 3 phases + neutral “.
VNR said:
Hey,
I have two different server racks connected with two different phases vis different MCBs. Source DB is same. Recently we have experienced tripping of MCBs of both racks with in One minute time delay. Input supply is from latest version of MGE UPS systems.
Not able to find out the reason. Can some one explain the reason?
Dear ;
Please let me know if this One minute time delay is between the 2 MCB's, or both MCB's trip at the same time but this delay is more than a previous recorded tripping time.
