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Hello Akhila,
These will help you to understand smart grid…
1. http://ge.ecomagination.com/smartgrid/#/landing_page
2. http://www2.schneider-electric.com/sites/corporate/en/group/energy-challenge/smart-grid.page
Rick said:
I have been trying to get the load shedding guide but the “system” states I am registered already. How can I get a copy?
Hello Rick,
If the system said you are registered then it means the 1st step (registration) was successful: no need to try to register again, you should get the mail to access the survey & download within 48 or 72 hours maximum.
Have you received this mail now? Please check your spam folder too.
If you don't get this invitation after 72 hours and it's not in your spam, tell us again by mail at engineering@electrical-equipment.com
Thank You!
Hello,
Which kind of software you are talking about? Which features do you want? Please specify so that our community members will be able to help you.
You may want to look at this article published in the blog.
Thank You!
SeRU said:
I'm sorry, I have request by Catalan, you can translate by google traductor.
Asimetria
En condicions normals les tensions d'alimentació són asimètriques i les càrregues equilibrades.
Són dissimètriques i desequilibrades en cas d'avaria (trencament de l'aïllament) i interrupció de fases. A més, amb càrregues monofásicas, l'equilibri pot ser sol de tipus estadístic.
És necessari afrontar l'estudi de la xarxa trifàsica fins i tot en les condicions anòmales d'avaria per a dimensionar les proteccions.
Es pot recórrer a un sistema d'equacions derivat del principi de Kirchhoff, per a utilitzar formules dels sistemes equilibrats, i a pesar que es pot comprendre mitjançant l'aportació dels components de la instal·lació, és útil la teoria dels components
asimètrics.
Es pot demostrar que qualsevol trio de vectors pot ser descompost en tres trios.
Sobre la base s'obté que cada sistema trifàsic comunament asimètric i equilibrat pugui descompondre's en tres sistemes trifàsic que es reconduiran a l'estudi separat de tres circuits monofásicos corresponents, respectivament, a la seqüència directa, a la seqüència negativa, a la seqüència zero.
La normativa EN50160 defineix, relativament als sistemes elèctrics en BT, que “en condicions de normal exercici
per a cada període d'una setmana, el 95% dels valors mitjos eficaços, calculats cada 10 minuts, de la component de seqüència negativa de la tensió d'alimentació ha de ser compresa en l’ interval entre 0 i 2% de la component de seqüència directa”.
En algunes regions amb instal·lació usuàries de connexions amb línies parcialment monofásica o bifásica, poden haver desequilibris fins a un 3% als terminals d'alimentació trifàsic.
L'instrument permet la mesura i registre dels següents paràmetres que defineixen la percentual de l'asimetria sobre les tensions d'un sistema elèctric
On:
Ei = seqüència de la terna negativa
Ed = seqüència de la terna directa
E0 = seqüència de la terna zero
Here is the English translation proposed by Google translator:
AsymmetryUnder normal conditions the supply voltages are balanced and asymmetrical loads.
They are dissimetriques and unbalanced in case of failure (insulation break) and interruption phases. In addition, monophasic charges, the balance may be one type statistic.
It is necessary to tackle the study of three-phase network even in abnormal conditions of failure for dimension protection.
You can use a system equations derived from Kirchhoff principle, to use the formulas balanced systems, and although one can understand by providing components of the installation, the theory is useful components asymmetric.
It can be shown that any trio of vectors can be decomposed into three trios.
Obtained on the basis that each commonly asymmetrical three-phase balanced system can be broken into three three-phase systems that redirect the study of three separate circuits monofásicos corresponding respectively to the direct sequence, the negative sequence , zero sequence.
The regulations EN50160 defines the electrical systems in relatively BT, “In normal conditions of exercise for each period of a week, 95% of the average effective values, calculated every 10 minutes, the negative sequence component of voltage Power must be understood in the interval between 0 and 2% of the component direct sequence.”
In some regions with users installing connections with lines monophasic or biphasic partially, imbalances may have up to 3% by three-phase power supply terminals.
The instrument can measure and record the following parameters define the percentage of asymmetry about the tensions of a system Electrical
Where:
Ei = negative sequence of the triplet
Ed = direct sequence of the triplet
E0 = zero sequence of the triplet
Hello Anil,
Here are the links of National Grid and Grid Code:
1. http://www.nationalgrid.com/uk/Electricity/Codes/gridcode/ (National Grid)
2. http://www.nationalgrid.com/NR/rdonlyres/67374C36-1635-42E8-A2B8-B7B8B9AF2408/44739/Z_CompleteGridCode_I4R5.pdf (Complete Grid Code)
Thanks!
Hello,
Can you please tell us what point you want to make? Do you have any question regarding Power Protection?
Moderator
Hello,
These sites explain shunt reactors and its use…
1. http://www.qualitypower.co.in/shunt-reactor.php
3. http://www.arevausitr.com/pdf/TECH%20NEWS_SHUNT_71665.pdf
I hope, this answers your question…
Regards,
Fierro said:
1º – Debes decidir a que voltaje vas a trabajar los motores: Por ejemplo 480 voltios 3 Fases 60 ciclos.
2º – Debes hacer una lista con los siguientes datos por cada motor: Por ejemplo
Motor Nº HP Fases Voltios Amperios nominales RPM Factor de servicio
1 5 3 480 7.3 1800 1.15
2 20 3 480 26 1800 1.15
3 50 3 480 63 1200 1.15
4 75 3 480 91 1800 1.15
5 1 3 480 1.7 1800 1.15
6 0.5 3 480 0.8 1800 1.15
7 125 3 480 145 1800 1.15
3º – Para escoger el calibre del cable debes tener en cuenta:
Los amperios nominales del motor
La longitud del cable
La corriente necesaria arrancar la máquina que el motor mueve (Rotor bloqueado), en algunas maquinas la corriente puede llegar a 8 veces la corriente nominal del motor- El tiempor necesario para alcanzar la velocidad nominal del equipo que el motor mueve.
Por ejemplo: Supongamos un motor de 125 HP , 3F- 480 Voltios 1800 RPM , 145 amperios de ccorriente nominal, que mueve un un ventilador centrífugo – El Motor esta a 100 metros del arrancador.
Un ventilador centrifugo toma aproxximadamente 6 segundos para alcanzar lal velocidad nominal, durante los primeros 2 segundos la corriente en el motor es mas o menos 6 veces la corriente nominal del motor (690 amperios).
Es necesario conocer si el arrancador es directo (A travez de la linea), Estrella delta, Fase partida, por autotransformador, “SoftStart”, Variador de velocidad (Invrter)
Si el arrancador es directo en general el calibre del cable (Conductor, wire) es 1.5 veces la corriente nominal del motor pero para una distancia no mayor de 15 metros, para distancias mayores es necesario calcular la regulación (Caida de voltaje) en el extremo del cable (En bornes del motor) la caida de voltaje no debe ser mayor del 3% (Para 480 voltios 14 voltios de caida) durante los dos primeros segundos.
Si el arrancador es estrella delta la corriente es el 58% (de los calculos anteriores)
Si el arrancador es Softstart la corriente de arranque puede estar entre el 130 y 170% de la nominal. (Es necesario tener en cuenta que algunas maquinas requieren un par de arranque muy elevador y es posible que el arrancador softstart o el variador de velocidad no permitan el arranque del equipo
4º Para escoger el interruptor se tiene en cuenta el tipo de arrancador, el tiempo que dura el arranque, el calibre del cable, la corriente de cortocircuito en bornes de entrada al interruptor, las caracteristicas del interruptor en cuanto a tiempo de disparo por sobrecorriente y por cortocircuito, nuemero de arranques por hora (Es necesario tener en cuenta que los fabricantes de los motores establecen el numero de arranques por hora, Tener en cuenta que si el motor permite 6 arranques por hora no quiere decir que cada en 12 minutos puedo hacer 6 arranques, lo que quiere decir es que que el segundo arranqwue se hace 10 minutos después del primero y asi sucesivamente- Sin embargo lo mejor es colocar un sensor de temperatura dentro de los devanados del motor para que prohiba el arranque en caso de que la temperatura de los devandos sobrepase lo permitido).
CONCLUSION:
Si no tienes experiencia en este tipo de trabajo lo mas recomendable es que te asesores de un ingeniero con experiencia o una firma igualmente experimentada.
Escribió Israel Henao
Para el motor de 125 HP el interruptor puede estar entre 275 y 350 amperios.
Para el caso anterior el interruptor
Hello, I am posting the English translation (by Google), so that everyone will understand the post:
1 – You decide that you will be working voltage motors: for example 480 volts 3 Phase 60 cycles.
2 – You can list the following information for each engine: For example
No. Phase Motor HP Volts Amps RPM rated service factor
1 5 3 480 7.3 1800 1.15
2 20 3 480 26 1800 1.15
3 50 3 480 63 1200 1.15
4 75 3 480 91 1800 1.15
5 1 3 480 1.7 1800 1.15
6 0.5 3 480 0.8 1800 1.15
7 125 3 480 145 01.15 1800
3 – To choose the wire size to keep in mind:
Rated motor amps
Cable length
The current required to start the machine moves the motor (rotor locked), some current machines can reach 8 times rated current of the motor-The tiempor necessary to achieve the rated speed of the computer that the engine moves.
For example: Suppose a 125 HP engine, 3F-480 Volts 1800 RPM, 145 amps nominal ccorriente, which drives a centrifugal fan – the engine is 100 meters from the starter.
A centrifugal fan aproxximadamente takes 6 seconds to reach full speed lal, during the first 2 seconds the motor current is about 6 times the motor rated current (690 amps).
You need to know if the starter is direct (THROUGH line), Star Delta, Phase departure, autotransformer, “SoftStart” Variable Speed (Invrter)
If the starter is usually direct gauge of wire (conductor, wire) is 1.5 times the rated current of the motor but for a distance not exceeding 15 meters, for greater distances need to calculate the adjustment (voltage drop) at the end cable (at the motor terminals) voltage drop should not exceed 3% (480 volts to 14 volts of drop) during the first two seconds.
If the star delta starter is current is 58% (from previous calculations)
If the starter is soft start starting current may be between 130 and 170% of nominal. (It should be noted that some machines require a very high starting torque and may softstart starter or drive controller does not allow the computer boots
4 – To select the switch takes into account the type of starter, the duration of the boot, wire size, the short circuit current at the input terminals to the switch, the characteristics of the switch in terms of trip time and overcurrent short circuit, nuemero of starts per hour (must be noted that engine manufacturers set the number of starts per hour, Keep in mind that if the engine allows 6 starts per hour does not mean that every 12 minutes I can do 6 starts, what it means is that the latter is arranqwue 10 minutes after the first and so on “But it's best to place a temperature sensor inside the motor windings to forbid the start if the Devandas temperature exceeds permitted).
CONCLUSION:
If you are inexperienced in this type of work it is more advisable advisers you an experienced engineer or firm also experienced.
Israel Henao wrote
For the 125 HP motor switch may be between 275 and 350 amps.
For the former case the switch
Fierro said:
Para conectar en paralelo dos o mas transformadores se requiere:
1º- Que la impedancia de los transformadores sea igual
2º- Que el voltaje de los transformadores sea igual
3º- Que la secuencia sea igual
4º- Que las barras o coductores que unen los secundarios de los transformadores tengan impedancia igual.
Al colocar transformadores en paralelo la corriente de cortocircuito a la salida de los transformadores colocados en paralelo será mayor, si por ejemplo conectamos en paralelo dos transformadores de 500 KVA con salida a 480 voltios y una impedancia del 5% la corriente maxima de cortocircuito de un transformador es 12 ka la corriente de cortocircuito de los dos transformadores será de 24 kA.
Al colocar en paralelo dos transformmadores la perdidas en hierro y cobre serán la suma de las de cada transformador.
El consumo de la carga no varía ; Si la carga exige 400 kva serán los mismos 400 kva con ransformador que con dos, pero las perdidas en el hierro seran el doble.
Escribio Israel Henao
Hello, I am posting an English translation (by Google) so that everyone will understand:
To connect in parallel two or more transformers are required:
1 – That the impedance of the transformer is equal
2 – That the transformer voltage is equal
3 – That the sequence is equal
4 – That the bars or coductores joining the side of processors have the same impedance.
By placing transformers in parallel to the short circuit current laid out parallel processors will be higher, if example connect in parallel two 500 KVA transformers with output at 480 volts and an impedance of 5% of the maximum current short circuit of a transformer is 12 kA short circuit current of the two transformers is 24 kA.
By placing in parallel two transformmadores the iron and copper losses be the sum of each transformer.
The consumption of the load remains unchanged, if the load requires 400 kva will be same 400 kva with ransformers than two, but lost in the iron will double.
You can check these pages:
1. http://en.wikipedia.org/wiki/Electronvolt
2. http://en.wikipedia.org/wiki/Joule
Please tell me if this does not answer your question.
Thank You!
Moderator
Hello,
Can you please give specific title to your topic instead of 'Help me please…'? The new specific title will help our readers to make understand the content which would get more replies.
Thank You!
Moderator
Hello,
Can you please give specific title to your topic instead of 'Help me plssss…'? The new specific title will help our readers to make understand the content which would get more replies.
Thank You!
Moderator
Hello,
Can you please give specific title to your topic instead of 'Pls help…'? The new specific title will help our readers to make understand the content which would get more replies.
Thank You!
Regards,
Moderator
I found some links related to your question:
1. CATHODIC PROTECTION, P E FRANCIS, http://www.npl.co.uk/upload/pdf/cathodic_protection_in_practise.pdf
2. Process of Cathodic Protection, Courtesy of CP Design Center by MESA Products, Inc., http://www.cpdesigncenter.com/public/technical/cp_process.htm
3. Patent application title: Process for Cathodic Protection of Electrode Materials,
Read more: http://www.faqs.org/patents/app/20080283417
