{"id":10898,"date":"2026-08-29T23:43:09","date_gmt":"2026-08-29T15:43:09","guid":{"rendered":"https:\/\/q8bbzjnpm5.wpdns.site\/?p=10898"},"modified":"2026-08-29T23:43:09","modified_gmt":"2026-08-29T15:43:09","slug":"scientific-methods-for-selecting-transformers-based-on-the-power-of-electrical-equipment","status":"publish","type":"post","link":"https:\/\/subian-electric.com\/es\/news\/scientific-methods-for-selecting-transformers-based-on-the-power-of-electrical-equipment\/","title":{"rendered":"M\u00e9todos Cient\u00edficos para Seleccionar Transformadores Basados en la Potencia de Equipos El\u00e9ctricos"},"content":{"rendered":"

Una f\u00e1brica textil en Vietnam estaba a punto de poner en funcionamiento doce m\u00e1quinas de tejer con una potencia de 15 kW cada una. La conclusi\u00f3n del propietario era bastante simple: doce m\u00e1quinas multiplicadas por 15 kW dan 180 kW, por lo tanto, se necesita un transformador de 200 kVA. En ese momento, el contratista el\u00e9ctrico cancel\u00f3 el pedido. En realidad, cada m\u00e1quina consume alrededor de 18 kVA con un factor de potencia de 0.83; su corriente de arranque es hasta seis veces mayor que la corriente de funcionamiento; los compresores de aire que alimentan estas m\u00e1quinas a\u00f1aden 60 kVA; adem\u00e1s, ya se planea a\u00f1adir dos l\u00edneas a la f\u00e1brica. Al utilizar un enfoque cient\u00edficamente justificado, los c\u00e1lculos mostraron que se necesita un transformador de 500 kVA, que es m\u00e1s del doble del valor obtenido por un m\u00e9todo de c\u00e1lculo simple. Por eso, la selecci\u00f3n del transformador no debe basarse en la potencia del equipo.<\/p>\n

En este art\u00edculo, se presenta la selecci\u00f3n de transformadores cient\u00edficamente correcta, basada en los requisitos de potencia del equipo el\u00e9ctrico.<\/p>\n

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Razones para el fracaso de la suma de kW como m\u00e9todo de dimensionamiento<\/h2>\n

Utilizar la suma de los kW de placa de cada equipo para elegir un transformador que proporcione esa clasificaci\u00f3n de kVA es el error de dimensionamiento m\u00e1s com\u00fan que existe en la industria. Las siguientes son las razones cient\u00edficas de su fracaso:<\/p>\n

Se descuida el factor de potencia. La potencia real es mucho m\u00e1s baja que la potencia aparente de un motor, compresor o l\u00e1mpara. Por ejemplo, motores que utilizan 180 kW de potencia real, con un factor de potencia de 0.85, necesitar\u00edan 212 kVA. Esto muestra una diferencia del 18 por ciento que sobrecargar\u00eda un transformador de 200 kVA.
\nSe descuida la diversidad. En la pr\u00e1ctica, rara vez las m\u00e1quinas funcionan a m\u00e1xima capacidad al mismo tiempo. La suma de la potencia nominal de las m\u00e1quinas crea una brecha del 20-40 por ciento, lo que significa dinero desperdiciado en un transformador clasificado por encima de su valor requerido.
\nSe descuida la corriente de arranque. Los motores el\u00e9ctricos grandes pueden consumir hasta seis veces su corriente normal cuando arrancan. Esto significa que la tensi\u00f3n podr\u00eda colapsar y el rel\u00e9 apagarse.<\/p>\n

Un enfoque acad\u00e9mico es un m\u00e9todo de combinar todos los efectos y obtener el tama\u00f1o final del transformador no determinado por la corazonada, sino por la f\u00edsica.<\/p>\n

De kW de equipo a kVA de transformador<\/h2>\n

La ecuaci\u00f3n principal es kVA = kW \u00f7 factor de potencia. En casos donde est\u00e1n involucrados dispositivos y equipos trif\u00e1sicos, tambi\u00e9n se podr\u00eda calcular kVA utilizando la expresi\u00f3n kVA = (\u221a3 \u00d7 V \u00d7 I) \u00f7 1000. La placa de la m\u00e1quina tiene toda la informaci\u00f3n que necesitas: kW nominal, voltaje, corriente de carga completa y a veces un valor de eficiencia. Siempre debes usar kVA del bus del transformador en lugar de kVA para kVA de terminal del motor, ya que tomar\u00eda en consideraci\u00f3n cualquier p\u00e9rdida a lo largo de los alimentadores (p\u00e9rdidas de 2-5% a largas distancias).<\/p>\n

Para configuraciones donde solo se conoce la potencia del motor en caballos de fuerza, se debe calcular utilizando la conversi\u00f3n 1 hp \u2248 0.746 kW y dividir el n\u00famero por el valor de eficiencia del motor (generalmente entre 0.90 y 0.95). En este caso, un motor de 100 hp consume 100 \u00d7 0.746 \u00f7 0.93 \u00f7 0.85, resultando en aproximadamente 94 kVA mientras est\u00e1 en operaci\u00f3n y de cinco a ocho veces m\u00e1s al arrancar.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Tipo de equipo<\/th>\nFactor de potencia t\u00edpico<\/th>\nDiversidad t\u00edpica<\/th>\nkVA por 100 kW de placa de identificaci\u00f3n<\/th>\n<\/tr>\n<\/thead>\n
Motores de inducci\u00f3n (general)<\/td>\n0.80\u20130.88<\/td>\n0.70\u20130.85<\/td>\n114\u2013125<\/td>\n<\/tr>\n
Motores impulsados por VFD<\/td>\n95 (entrada)<\/td>\n0.70\u20130.80<\/td>\n105<\/td>\n<\/tr>\n
Calefacci\u00f3n resistiva \/ hornos<\/td>\n1.00<\/td>\n0.60\u20130.80<\/td>\n100<\/td>\n<\/tr>\n
Iluminaci\u00f3n LED<\/td>\n0.95\u20130.99<\/td>\n0.80\u20130.90<\/td>\n101\u2013105<\/td>\n<\/tr>\n
M\u00e1quinas de soldadura<\/td>\n0.45\u20130.70<\/td>\n0.30\u20130.50<\/td>\n143\u2013222<\/td>\n<\/tr>\n
Compresores (de tornillo)<\/td>\n0.82\u20130.90<\/td>\n0.60\u20130.80<\/td>\n111\u2013122<\/td>\n<\/tr>\n
Enfriadores HVAC<\/td>\n0.80\u20130.88<\/td>\n0.70\u20130.90<\/td>\n114\u2013125<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Nota sobre la fila de soldadura: un bajo factor de potencia y una diversidad muy baja significan que el transformador debe dimensionarse para una alta relaci\u00f3n kVA por kW, aunque la demanda simult\u00e1nea real sea peque\u00f1a. Esta es la raz\u00f3n por la que las reglas gen\u00e9ricas fallan: el m\u00e9todo cient\u00edfico siempre trabaja a partir de las caracter\u00edsticas el\u00e9ctricas reales del equipo.<\/p>\n

Factores de demanda, diversidad y coincidencia<\/h2>\n

Los factores de demanda, diversidad y coincidencia explican la misma relaci\u00f3n entre c\u00f3mo act\u00faan las cargas en el bus del transformador.<\/p>\n

El factor de demanda se define como la demanda m\u00e1xima dividida por la carga conectada para el grupo y generalmente es menor que uno porque una carga tiende a funcionar por debajo de su capacidad total.<\/p>\n

El factor de diversidad se conoce como la demanda m\u00e1xima total dividida por la demanda coincidente m\u00e1xima y generalmente es mayor que uno.<\/p>\n

El factor de coincidencia es similar al factor de diversidad en que rige la cantidad real de carga que funciona al mismo tiempo y generalmente est\u00e1 entre 0.5 y 0.9.<\/p>\n

Los ingenieros suelen combinar uno de estos factores con la cantidad de kVA porque conduce a un ejemplo claro de c\u00f3mo encontrar la cantidad total de demanda de dise\u00f1o en t\u00e9rminos de ahorro en p\u00e9rdidas y compra de transformadores.
\nPor ejemplo, con una carga conectada de 500 kVA, el valor del factor de demanda es 0.9 y el del factor de coincidencia es 0.8, por lo que la demanda de dise\u00f1o resulta ser 500 x 0.9 x 0.8 = 360 kVA.<\/p>\n

El problema del arranque del motor<\/h2>\n

La corriente de arranque puede ser conocida como el asesino silencioso cuando se trata de transformadores peque\u00f1os. Para un motor de 150 kW, el arranque directo requiere una corriente de arranque que puede ser de 5 a 7 veces la corriente de carga completa, que dura de 3 a 10 segundos. Esto debe ocurrir mientras el transformador proporciona kVA de arranque y el resto de su carga de corriente, adem\u00e1s de intentar limitar las ca\u00eddas de voltaje en los terminales del motor a aproximadamente 10 a 15% para que el contactor no se desconecte.<\/p>\n

Una de las tres soluciones discutidas aqu\u00ed ayuda a resolver problemas de dimensionamiento. Los arrancadores suaves ayudan a reducir las corrientes de arranque a entre 2 y 3 veces la corriente de carga completa. Los VFD reducen la corriente de arranque a aproximadamente 1.2 a 1.5 veces. El arranque estrella-tri\u00e1ngulo lo reduce a 33% del arranque directo, pero este m\u00e9todo tiene un par reducido. A menudo es posible reducir el tama\u00f1o en una clasificaci\u00f3n est\u00e1ndar si se puede controlar la corriente de arranque. Si el control no es posible y si la instalaci\u00f3n tiene motores de velocidad fija grandes, entonces el tama\u00f1o del transformador debe basarse en la corriente de arranque en lugar de en cualquier carga en funcionamiento.<\/p>\n\n\n\n\n\n\n\n\n
M\u00e9todo de Arranque<\/th>\nCorriente de Arranque T\u00edpica (% de FLC)<\/th>\nImpacto en el Dimensionamiento del Transformador<\/th>\n<\/tr>\n<\/thead>\n
Direct-on-line (DOL)<\/td>\n500\u2013700%<\/td>\nSignificant \u2014 may add one rating step<\/td>\n<\/tr>\n
Star-delta<\/td>\n~170\u2013230%<\/td>\nModerate<\/td>\n<\/tr>\n
Soft starter<\/td>\n200\u2013300%<\/td>\nModerate<\/td>\n<\/tr>\n
VFD<\/td>\n120\u2013150%<\/td>\nMinimal<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The engineering rule of thumb: if the largest motor exceeds 25% of the transformer rating, verify the starting scenario explicitly with a voltage-drop calculation. If it exceeds 50%, expect to either bump the rating or change the starting method \u2014 the arithmetic is in IEC 60076-7 and IEEE C57.12.00 guidance on cyclic and short-time loading.<\/p>\n

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Worked Examples Across Industries<\/h2>\n

Example A – Textiles Mill. Connected load = 12 looms (15 kW each (PF = 0.83)) + 3 compressors (30 kW each (PF = 0.85)) + 40 kW lighting (PF = 0.95) + 20 kW HVAC. Running kVA = 12 * 18.1 + 3 * 35.3 + 42 + 21 = ~ 345 kVA. Demand factor = 0.9 and coincidence (CF) = 0.8 -> 345 * 0.9 * 0.8 = 248 kVA. With a growth of 15%, the power rating required = 285 kVA or standard rating of 315 kVA. Largest single load is 35 kVA (about 11% of the rating) thus no restrictions start-up.<\/p>\n

Example B – Food Processing Plant. Cooling compressor (100 kW) (6\u00d7 start\/ PF = 0.85), machinery (60 kW), CIP pump (30 kW) and lighting (25 kW). Running kVA = 118 + 71 + 35 + 26 = 250 kVA and coincidence factor (CF) = 0.75 -> the power required = 188 kVA. During start-up, the cooling compressor momentarily has a power demand of about 6 x 118 = 708 kVA for 5 seconds. Design condition: 188 + (708 – 118) = ~ 778 kVA momentarily. 315 kVA unit could not withstand that, the practical solution is 630 kVA unit and\/or soft starting.<\/p>\n\n\n\n\n\n\n\n\n
Scenario<\/th>\nNaive kW Sum<\/th>\nScientific kVA Result<\/th>\nDifference<\/th>\n<\/tr>\n<\/thead>\n
Textile mill (Example A)<\/td>\n270 kVA<\/td>\n315 kVA<\/td>\n+17%<\/td>\n<\/tr>\n
Food plant (Example B)<\/td>\n250 kVA<\/td>\n630 kVA<\/td>\n+152%<\/td>\n<\/tr>\n
Warehouse lighting + HVAC<\/td>\n150 kVA<\/td>\n160 kVA<\/td>\n+7%<\/td>\n<\/tr>\n
Machine shop with 50 kW lathes<\/td>\n300 kVA<\/td>\n400 kVA<\/td>\n+33%<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The pattern is clear: the more motors and the larger the largest motor, the bigger the gap between naive and scientific sizing. The scientific method is not about inflating every number \u2014 it is about finding the true constraint, which is sometimes capacity and sometimes starting current.<\/p>\n

Standard Ratings and the Sizing Table<\/h2>\n

After computing design demand, select the next standard IEC 60076-1 rating. Standard kVA values include 25, 50, 100, 160, 250, 315, 400, 500, 630, 800, 1000, 1250, 1600, 2000, 2500, and 3150. Custom ratings exist but cost 15\u201330% more and lengthen lead time because cores and tanks are non-standard.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
Equipment Connected (kVA, diversified)<\/th>\nAdd Growth (15%)<\/th>\nClasificaci\u00f3n Recomendada<\/th>\nTypical Price Range<\/th>\n<\/tr>\n<\/thead>\n
Up to 85<\/td>\n98<\/td>\n100<\/td>\n$2,500\u2013$6,000<\/td>\n<\/tr>\n
86\u2013135<\/td>\n155<\/td>\n160<\/td>\n$3,500\u2013$8,000<\/td>\n<\/tr>\n
136\u2013215<\/td>\n247<\/td>\n250<\/td>\n$4,500\u2013$11,000<\/td>\n<\/tr>\n
216\u2013270<\/td>\n310<\/td>\n315<\/td>\n$6,000\u2013$13,000<\/td>\n<\/tr>\n
271\u2013430<\/td>\n495<\/td>\n500<\/td>\n$8,000\u2013$16,000<\/td>\n<\/tr>\n
431\u2013545<\/td>\n627<\/td>\n630<\/td>\n$10,000\u2013$20,000<\/td>\n<\/tr>\n
546\u2013870<\/td>\n1,000<\/td>\n1,000<\/td>\n$15,000\u2013$32,000<\/td>\n<\/tr>\n
871\u20131,090<\/td>\n1,254<\/td>\n1,250<\/td>\n$20,000\u2013$40,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Prices are indicative FOB China for oil-immersed units and vary by specification, brand, and region. The cost of stepping up one rating \u2014 say from 500 to 630 kVA \u2014 is typically 15\u201325% of the unit price, while an undersized transformer’s failure costs 100% plus downtime. The headroom is cheap.<\/p>\n

Supporting Specifications: Voltage, Losses, Enclosure<\/h2>\n

Once kVA is established, the rest of the parameters are determined scientifically instead of with arbitrary guesswork. From the customer\u2019s system design, voltage ratio is determined: the primary voltage is either 11, 22, 33, or 35 kV (i.e. the voltage at which electricity enters customer\u2019s premises), and secondary voltage equals either 400, 415, or 480 V (utilization voltage). Vector group is based on the earthing system pattern and should be Dyn11 for grounded star low-voltage use, and normally Yd11 (from where the voltage is increased, low-voltage current less than 5% in the former system) or Yd1 for the respective high-voltage case. Impedance (uk%) is used to obtain fault coordination: 4% (for smaller distribution transformers), from 5% to 6% (for power rating 630-1250 kVA), and higher than that for power rating over 1250 kVA or as it is required by grid codes.<\/p>\n

Losses are the dominant cost over a period of 30 years. It would be crucial to state guaranteed no-load losses and guaranteed load losses in accordance with IEC 60076-1 standards and compare the total costs of ownership against competitors. Thus, for instance, a transformer of 1,000 kVA with amorphous core that saves 600 W of no-load losses saves quite an amount of electricity (almost $2,000 in 15 years at rate of 0.1 USD per one kWh) that makes it reasonable to pay for the difference. Further, enclosure should be chosen according to the conditions of environment (IP rating): IP54 for oil-immersed in the open air, IP21 to IP44 for dry type indoors, and IP56 for washdown applications.<\/p>\n

Brands and Price Ranges by Rating<\/h2>\n

Brand choice is the commercial layer on top of the technical calculation. For the same scientific specification, prices diverge by a factor of two to three depending on brand positioning and documentation quality.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n
Marca<\/th>\nFuerza<\/th>\n250 kVA Price<\/th>\n630 kVA Price<\/th>\n1,250 kVA Price<\/th>\n<\/tr>\n<\/thead>\n
ABB<\/td>\nGlobal service, full range<\/td>\n$12,000\u2013$20,000<\/td>\n$22,000\u2013$38,000<\/td>\n$38,000\u2013$60,000<\/td>\n<\/tr>\n
Siemens<\/td>\nEfficiency, digital monitoring<\/td>\n$12,000\u2013$22,000<\/td>\n$24,000\u2013$40,000<\/td>\n$40,000\u2013$65,000<\/td>\n<\/tr>\n
Schneider Electric<\/td>\nDry-type and LV integration<\/td>\n$11,000\u2013$18,000<\/td>\n$20,000\u2013$34,000<\/td>\n$35,000\u2013$55,000<\/td>\n<\/tr>\n
Hitachi Energy<\/td>\nGrid assets, power transformers<\/td>\n$12,000\u2013$21,000<\/td>\n$23,000\u2013$38,000<\/td>\n$39,000\u2013$62,000<\/td>\n<\/tr>\n
Eaton<\/td>\nNorth America compliance<\/td>\n$10,000\u2013$17,000<\/td>\n$18,000\u2013$32,000<\/td>\n$32,000\u2013$52,000<\/td>\n<\/tr>\n
CG Power \/ TBEA<\/td>\nVolume pricing<\/td>\n$7,000\u2013$12,000<\/td>\n$13,000\u2013$22,000<\/td>\n$24,000\u2013$40,000<\/td>\n<\/tr>\n
Jiangsu Subian Electric Power<\/td>\nIEC 60076, test reports, copper<\/td>\n$6,000\u2013$11,000<\/td>\n$11,000\u2013$20,000<\/td>\n$20,000\u2013$36,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The global companies talked about above achieved their status through years of work and contribution for the energy sector and other critical areas. However, the same approach will not work for industrial plants, warehouses, and renewable energy installations, where you are already protected by scientific specifications. Jiangsu Subian Electric Power produces IEC 60076-certified transformers in the range of 10 kVA to 100 MVA and provides copper-wound transformers as a standard product, and all the items come with a report proving that the losses and parameters you needed for the transformer correspond to the item delivered. In the case of transformers where the calculations show some specific rating, it is easy to assess the level of work done by suppliers. Rather than asking customers to trust what is written in brochures, Subian shares proofs of its claims.<\/p>\n

The Complete Selection Formula and Checklist<\/h2>\n

In summary, the calculation for sizing transformers based on the equipment-power selection formula is:<\/p>\n

The design kVA set = Sum (load_kVA \u00d7 demand factor) \u00d7 coincidence factor \u00d7 (1 + growth margin)<\/p>\n

and then the two criteria of achievable loads: sustained load \u2264 rating (oil\/dry) in 80\u201390%, and starting transient takes place within the short-duration capability limit of the unit as per IEC 60076-7. The checklist will consist of the following operations:<\/p>\n

Compile the list of machines with kW, PF, and full load current parameters.
\nConvert to kVA per apparatus and classify by feeder.
\nApply demand and coincidence packages per group.
\nVerify the maximum starting demand of the motor against the rating.
\nAdd 10\u201320% margin of growth and round to a standard rating.
\nSpecify voltage, vector group, impedance, losses, and enclosure.
\nCompare bidders based on their performances and TCO requirements.<\/p>\n

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Preguntas Frecuentes<\/h2>\n

How do I size a transformer from the equipment’s kW rating?<\/h3>\n

The kW of each load should be divided by its power factor so that you may convert it to kVA, sum them up to get the total, and then you must apply the demand factor (around 0.9 for most machines) and the coincidence factor (between 0.7 and 0.85), make sure to account for about 10-20 percent growth in total and round it to the next standard rating. For instance, 180 kW at PF of 0.85 equals to 212 kVA, applying coincidence of 0.85 and 15 percent growth, we get the design demand of approximately 207 kVA which implies that we have to use a 250 kVA machine.<\/p>\n

What is the rule of thumb for transformer sizing from load?<\/h3>\n

A common first-pass rule is that the transformer should be rated 1.2\u20131.5\u00d7 the expected maximum kW demand, depending on power factor and future growth. It is a screening tool only. The scientific method always wins where motors are large, power factors are low, or harmonics are present \u2014 use the formula above, not the rule, for the final decision.<\/p>\n

Why is the transformer rating bigger than the sum of my equipment?<\/h3>\n

Three reasons: power factor (kVA > kW), starting current (a 100 kW motor can demand 600+ kVA for seconds), and future growth (which you always add). If your equipment runs at 0.8 PF with a large DOL motor, a rating 25\u201360% above the naive kW sum is normal \u2014 the food-plant example in this guide needed 152% more.<\/p>\n

Can I oversize a transformer for future expansion?<\/h3>\n

Yes, and it is usually wise \u2014 but only within reason. A 10\u201320% margin costs 15\u201325% more per step up and slightly raises no-load losses, which run 24\/7. Oversizing beyond about 40% wastes capital and energy unless you are certain the load will arrive within 2\u20133 years. The scientific answer: match the margin to the actual expansion plan.<\/p>\n

How much does it cost to upgrade the transformer rating one step?<\/h3>\n

Moving from 500 to 630 kVA typically adds $2,000\u2013$4,000 (about 15\u201325%) for the unit itself; moving from 1,000 to 1,250 kVA adds $4,000\u2013$8,000. Against that, an undersized transformer that fails in service costs $8,000\u2013$30,000 in replacement plus production downtime. The upgrade premium is the cheapest insurance in the project.<\/p>\n

Referencias<\/h2>\n