como hacer un match entre dos columnas en excel

Como hacer un match entre dos columnas en Excel?Instrucciones1. 1Introduce o importa el primer grupo de datos en lacolumnaA de la hoja de clculo de Excel.2. 2Introduce o importa la segundacolumnade datos en lacolumnaC de la misma hoja de clculo.1. 3Deja lacolumnaB vaca.2. 4Teclea la siguiente frmula en la celda B1: "=IF(ISERROR(MATCH(A1,$C$1:$C$#,0)),"",A1)", donde # es el nmero de la fila del ltimo valor de lacolumnaC.3. 5Selecciona las celdas de lacolumnaB, empezando en B1, y continuando hacia abajo hasta llegar al nmero de fila que se corresponde con la ltima fila en lascolumnasA y B.4. 6Abre el men "Editar",haz clicen "Rellenar" y despus en "Abajo", y habrs terminado de comparar las celdas. LacolumnaB se rellenar con los nmeros duplicados en lascolumnasA y C.

Cmo comparar hojas de clculo en Excel 2007

Comparar las hojas de clculo de Microsoft Excel es til cuando se deben analizar las similitudes y diferencias de los datos. Mediante el uso de la herramienta "Side by Side View" ("vista lado a lado") en Excel 2007, puedes comparar tablas, grficos y otros datos, a partir de hojas de clculo independientes, dentro de la misma ventana de Excel. Adems, puedes sincronizar el desplazamiento de cada hoja de clculo, lo que permite comparar los datos a travs de la ventana de Excel. Utiliza la opcin "Side by Side View" para analizar los datos de la hoja de clculo de Excel y reducir los errores al hacer comparaciones visuales.InstruccionesComparar hojas de clculo en el mismo libro1. 1

Jupiterimages/Comstock/Getty ImagesHaz clic en la pestaa "Ver" y haz clic en "Nueva ventana" de la "ventana" del grupo.2. 2

Michael Blann/Lifesize/Getty ImagesHaz clic en el botn "Side by Side View" de la "ventana" del grupo.3. 3

PhotoObjects.net/PhotoObjects.net/Getty ImagesHaz clic en las pestaas en la parte inferior de cada hoja de clculo para mostrar las hojas de clculo que deseas comparar.4. 4

Hemera Technologies/AbleStock.com/Getty ImagesHaz clic en la opcin "Desplazamiento sincrnico" de la "Ventana" del grupo para desplazarte entre ambas hojas de clculo al mismo tiempo.

Comparar hojas de libros diferentes1. 1

Jupiterimages/Creatas/Getty ImagesAbre los dos libros que contienen las hojas de clculo que deseas comparar.2. 2

Jupiterimages/Photos.com/Getty ImagesHaz clic en la pestaa "Ver" y luego en "Nueva ventana" de la "ventana" del grupo.3. 3

Jupiterimages/Polka Dot/Getty ImagesHaz clic en el botn "Side by Side View" de la "ventana" del grupo.4. 4

Jupiterimages/Photos.com/Getty ImagesHaz clic en las pestaas en la parte inferior de cada hoja de clculo para mostrar las hojas de clculo que deseas comparar.5. 5

Ablestock.com/AbleStock.com/Getty ImagesHaz clic en la opcin "Desplazamiento sincrnico" de la "ventana" del grupo para desplazarte entre ambas hojas de clculo al mismo tiempo.

Como comparar dos listas en Excel?

Otra tarea bastante comn en Excel es comparar entre dos listas. El objetivo es encontrar qu diferencias hay entre dos listas (que personas de una lista no aparecen en la otra, por ejemplo). Este tipo de comparacin se puede hacer de varias maneras. En esta nota mostrar cmo hacerlo usandoConditional Formatting(Data ---> Conditional Formatting).Supongamos esta hoja

Para sealar los faltantes en cada lista, daremos un fondo de color azul a los nombre de la lista 1 que no aparecen en la lista 2 y un fondo de color rosa a los nombre de la lista 2 que no aparecen en la lista 1.Procedemos de la siguiente manera:1. seleccionamos el rango de los nombres en la lista 1 (A2:A10)2. en la barra del men pulsamos Format ---> Conditional Formating3. seleccionamos Formula Is4. en la ventanilla escribimos la formula=COUNTIF($B$2:$B$10,A2)=0(prestar atencin al signo $ en la frmula)5. apretar el botn Format, seleccionar Pattern y elegir el color azul.

Ahora hacemos lo mismo con la segunda lista (el rango ser B2:B10) y utilizamos la frmula=COUNTIF($A$2:$A$10,B2)=0Apretamos OK

Ahora podemos ver claramente quien falta en cada lista.Otra tcnica es utilizar la frmula =MATCH, lo que mostrar en alguna futura nota.1. Hi,

In a AC power quadrant diagram, how to understand the physical meaningsof powers that falls in each quadrant?

I: +P, +QII: -P, +QII: -P, -QIV: +P, -Q

where P denotes real power and Q denotes reactive power.

Thanks in advance.

--Life is the only flaw in an otherwise perfect nonexistence-- Schopenhauer

narkenarke,Jun 8, 2010#11. AdvertisingAds by GoogleHOBUT power metersMultifunction User ProgrammableMulti function Power Meterswww.hobut.co.uk/2. daestromGuestnarke wrote:> Hi,>> In a AC power quadrant diagram, how to understand the physical meanings> of powers that falls in each quadrant?>> I: +P, +Q> II: -P, +Q> II: -P, -Q> IV: +P, -Q>> where P denotes real power and Q denotes reactive power.>> Thanks in advance.>>

What is it you don't understand?

Negative real power simply means that power is flowing in the directionopposite from convention. For a generator, it would mean that power isflowing from the grid/bus into the generator to keep it spinning. Thishappens when the engine/turbine is not generating enough power toovercome friction/windage losses and the electrical bus has to supplypower into the generator to keep it spinning. (it does NOT mean thegenerator shaft has reversed its direction of rotation)

Similarly, negative reactive power means reactive power is flowing inthe direction opposite from convention. Normally a generator suppliesreactive power to a bus to 'feed' the reactive loads on the bus.Convention is that inductive loads consume 'positive reactive power' andcapacitive loads are said to supply 'positive reactive power'. Youcould also argue that capacitive loads supply 'negative' reactive powerwhich cancels out the 'positive' reactive power of inductive loads.

For a generator, reactive power is usually labeled 'positive' when it isover-excited and supplying reactive power to inductive loads. Ifunder-excited, it actually draws reactive power from an infinite bus(its reactive power is 'negative').

daestromdaestrom,Jun 8, 2010#21. AdvertisingAds by GoogleLm339 datasheetDistributors, Inventory and PricingFree Datasheet Downloadswww.datasheets360.com/lm339+datasheet3. narkeGuestOn 2010-06-08, daestrom wrote:> narke wrote:>> Hi,>>>> In a AC power quadrant diagram, how to understand the physical meanings>> of powers that falls in each quadrant?>>>> I: +P, +Q>> II: -P, +Q>> II: -P, -Q>> IV: +P, -Q>>>> where P denotes real power and Q denotes reactive power.>>>> Thanks in advance.>>>>>> What is it you don't understand?>

Thanks for the good explaination, but I still have some difficulties inunderstanding, please see my comments below.

> Negative real power simply means that power is flowing in the direction> opposite from convention. For a generator, it would mean that power is> flowing from the grid/bus into the generator to keep it spinning. This> happens when the engine/turbine is not generating enough power to> overcome friction/windage losses and the electrical bus has to supply> power into the generator to keep it spinning. (it does NOT mean the> generator shaft has reversed its direction of rotation)

Because my domain is three-phase electricity meters that are installedin utilities and high load users. So I like to ask, when the metersobserved negative real power, does it mean that the energy is flowingback to from consumer side to the grid side?

>> Similarly, negative reactive power means reactive power is flowing in> the direction opposite from convention. Normally a generator supplies> reactive power to a bus to 'feed' the reactive loads on the bus.> Convention is that inductive loads consume 'positive reactive power' and> capacitive loads are said to supply 'positive reactive power'. You> could also argue that capacitive loads supply 'negative' reactive power> which cancels out the 'positive' reactive power of inductive loads.

That sounds clear. And, for the follow two AC circuits,

Setup A:------L------+|+----R------Electricity Meter--|------C------+

Setup B:----R------Electricity Meter--

where L is ideal inductor and C is ideal capacitor, and impedance L issame as impedance C in maganitute.

So, can I deduce followings?

1. the apparent energy observed in A is same as that in B;2. real energy in A is same as real energy in B, and that amounts to I^2* R;3. reactive energy in A is same as reactive energy in B, and thatamounts to zero.

> For a generator, reactive power is usually labeled 'positive' when it is> over-excited and supplying reactive power to inductive loads. If> under-excited, it actually draws reactive power from an infinite bus> (its reactive power is 'negative').

I don't understant what you mean 'infinite bus'.

>> daestrom


narkenarke,Jun 9, 2010#34. narkeGuestOn 2010-06-09, narke wrote:> On 2010-06-08, daestrom wrote:>> narke wrote:>>> Hi,>>>>>> In a AC power quadrant diagram, how to understand the physical meanings>>> of powers that falls in each quadrant?>>>>>> I: +P, +Q>>> II: -P, +Q>>> II: -P, -Q>>> IV: +P, -Q>>>>>> where P denotes real power and Q denotes reactive power.>>>>>> Thanks in advance.>>>>>>>>>> What is it you don't understand?>>>> Thanks for the good explaination, but I still have some difficulties in> understanding, please see my comments below.>>> Negative real power simply means that power is flowing in the direction>> opposite from convention. For a generator, it would mean that power is>> flowing from the grid/bus into the generator to keep it spinning. This>> happens when the engine/turbine is not generating enough power to>> overcome friction/windage losses and the electrical bus has to supply>> power into the generator to keep it spinning. (it does NOT mean the>> generator shaft has reversed its direction of rotation)>> Because my domain is three-phase electricity meters that are installed> in utilities and high load users. So I like to ask, when the meters> observed negative real power, does it mean that the energy is flowing> back to from consumer side to the grid side?>>>>> Similarly, negative reactive power means reactive power is flowing in>> the direction opposite from convention. Normally a generator supplies>> reactive power to a bus to 'feed' the reactive loads on the bus.>> Convention is that inductive loads consume 'positive reactive power' and>> capacitive loads are said to supply 'positive reactive power'. You>> could also argue that capacitive loads supply 'negative' reactive power>> which cancels out the 'positive' reactive power of inductive loads.>> That sounds clear. And, for the follow two AC circuits,>> Setup A:> ------L------+> |> +----R------Electricity Meter--> |> ------C------+>> Setup B:> ----R------Electricity Meter-->> where L is ideal inductor and C is ideal capacitor, and impedance L is> same as impedance C in maganitute.>> So, can I deduce followings?>> 1. the apparent energy observed in A is same as that in B;> 2. real energy in A is same as real energy in B, and that amounts to I^2> * R;> 3. reactive energy in A is same as reactive energy in B, and that> amounts to zero.>>

Is there an answer? Thanks in advance.

>> For a generator, reactive power is usually labeled 'positive' when it is>> over-excited and supplying reactive power to inductive loads. If>> under-excited, it actually draws reactive power from an infinite bus>> (its reactive power is 'negative').>> I don't understant what you mean 'infinite bus'.>>>>> daestrom>>


narkenarke,Jun 10, 2010#45. GuestGuest"narke" wrote in messagenews:...> On 2010-06-08, daestrom wrote:>> narke wrote:>>> Hi,>>>>>> In a AC power quadrant diagram, how to understand the physical meanings>>> of powers that falls in each quadrant?>>>>>> I: +P, +Q>>> II: -P, +Q>>> II: -P, -Q>>> IV: +P, -Q>>>>>> where P denotes real power and Q denotes reactive power.>>>>>> Thanks in advance.>>>>>>>>>> What is it you don't understand?>>>> Thanks for the good explaination, but I still have some difficulties in> understanding, please see my comments below.>>> Negative real power simply means that power is flowing in the direction>> opposite from convention. For a generator, it would mean that power is>> flowing from the grid/bus into the generator to keep it spinning. This>> happens when the engine/turbine is not generating enough power to>> overcome friction/windage losses and the electrical bus has to supply>> power into the generator to keep it spinning. (it does NOT mean the>> generator shaft has reversed its direction of rotation)>> Because my domain is three-phase electricity meters that are installed> in utilities and high load users. So I like to ask, when the meters> observed negative real power, does it mean that the energy is flowing> back to from consumer side to the grid side?----------------------------Yes------------------>>>>> Similarly, negative reactive power means reactive power is flowing in>> the direction opposite from convention. Normally a generator supplies>> reactive power to a bus to 'feed' the reactive loads on the bus.>> Convention is that inductive loads consume 'positive reactive power' and>> capacitive loads are said to supply 'positive reactive power'. You>> could also argue that capacitive loads supply 'negative' reactive power>> which cancels out the 'positive' reactive power of inductive loads.>> That sounds clear. And, for the follow two AC circuits,>> Setup A:> ------L------+> |> +----R------Electricity Meter--> |> ------C------+>> Setup B:> ----R------Electricity Meter-->> where L is ideal inductor and C is ideal capacitor, and impedance L is> same as impedance C in maganitute.>> So, can I deduce followings?>> 1. the apparent energy observed in A is same as that in B;> 2. real energy in A is same as real energy in B, and that amounts to I^2> * R;> 3. reactive energy in A is same as reactive energy in B, and that> amounts to zero.---------------Your circuit A is hard to follow. It doesn't appear to be either a 3 phaseor a single phase circuit.Source and return paths are omitted.

Possibly this single phase circuit may help.----1------+---2-----+----3----+source | | |V L C R Impedanceconditions as above| | |-----------+----------+--------- +

Meter connected at 1 (and return) , 2 or 3 real power and energy is measuredAt 1,3 there will be no reactive measured.At 2 there will be reactive measured and it will be negative (capacitiveload).

Does this help?

>>>> For a generator, reactive power is usually labeled 'positive' when it is>> over-excited and supplying reactive power to inductive loads. If>> under-excited, it actually draws reactive power from an infinite bus>> (its reactive power is 'negative').>> I don't understant what you mean 'infinite bus'.----------------This means an ideal voltage source (no internal impedance) and implies, inpractice, that the source is so largethat what you connect to it doesn't measurably affect its voltage.

-------Don Kellycross out to replyGuest,Jun 10, 2010#56. narkeGuestOn 2010-06-10, wrote:>>>> "narke" wrote in message> news:...>> On 2010-06-08, daestrom wrote:>>> narke wrote:>>>> Hi,>>>>>>>> In a AC power quadrant diagram, how to understand the physical meanings>>>> of powers that falls in each quadrant?>>>>>>>> I: +P, +Q>>>> II: -P, +Q>>>> II: -P, -Q>>>> IV: +P, -Q>>>>>>>> where P denotes real power and Q denotes reactive power.>>>>>>>> Thanks in advance.>>>>>>>>>>>>>> What is it you don't understand?>>>>>>> Thanks for the good explaination, but I still have some difficulties in>> understanding, please see my comments below.>>>>> Negative real power simply means that power is flowing in the direction>>> opposite from convention. For a generator, it would mean that power is>>> flowing from the grid/bus into the generator to keep it spinning. This>>> happens when the engine/turbine is not generating enough power to>>> overcome friction/windage losses and the electrical bus has to supply>>> power into the generator to keep it spinning. (it does NOT mean the>>> generator shaft has reversed its direction of rotation)>>>> Because my domain is three-phase electricity meters that are installed>> in utilities and high load users. So I like to ask, when the meters>> observed negative real power, does it mean that the energy is flowing>> back to from consumer side to the grid side?> ----------------------------> Yes> ------------------>>>>>>>> Similarly, negative reactive power means reactive power is flowing in>>> the direction opposite from convention. Normally a generator supplies>>> reactive power to a bus to 'feed' the reactive loads on the bus.>>> Convention is that inductive loads consume 'positive reactive power' and>>> capacitive loads are said to supply 'positive reactive power'. You>>> could also argue that capacitive loads supply 'negative' reactive power>>> which cancels out the 'positive' reactive power of inductive loads.>>>> That sounds clear. And, for the follow two AC circuits,>>>> Setup A:>> ------L------+>> |>> +----R------Electricity Meter-->> |>> ------C------+>>>> Setup B:>> ----R------Electricity Meter-->>>> where L is ideal inductor and C is ideal capacitor, and impedance L is>> same as impedance C in maganitute.>>>> So, can I deduce followings?>>>> 1. the apparent energy observed in A is same as that in B;>> 2. real energy in A is same as real energy in B, and that amounts to I^2>> * R;>> 3. reactive energy in A is same as reactive energy in B, and that>> amounts to zero.> ---------------> Your circuit A is hard to follow. It doesn't appear to be either a 3 phase> or a single phase circuit.> Source and return paths are omitted.>> Possibly this single phase circuit may help.> ----1------+---2-----+----3----+> source | | |> V L C R Impedance> conditions as above> | | |> -----------+----------+--------- +>> Meter connected at 1 (and return) , 2 or 3 real power and energy is measured> At 1,3 there will be no reactive measured.> At 2 there will be reactive measured and it will be negative (capacitive> load).>> Does this help?>

I fell it can be very helpful ... just now the format of the graph notgood, it's hard to get the information. Would you please redraw thecircuit to make it clear? (yes, single phase is assumed)

Many thanks!

>>>>>>> For a generator, reactive power is usually labeled 'positive' when it is>>> over-excited and supplying reactive power to inductive loads. If>>> under-excited, it actually draws reactive power from an infinite bus>>> (its reactive power is 'negative').>>>> I don't understant what you mean 'infinite bus'.> ----------------> This means an ideal voltage source (no internal impedance) and implies, in> practice, that the source is so large> that what you connect to it doesn't measurably affect its voltage.>


narkenarke,Jun 11, 2010#67. Andrew GabrielGuestIn article ,narke writes:> Hi,>> In a AC power quadrant diagram, how to understand the physical meanings> of powers that falls in each quadrant?>> I: +P, +Q> II: -P, +Q> II: -P, -Q> IV: +P, -Q>> where P denotes real power and Q denotes reactive power.>> Thanks in advance.

I wrote a java applet which allows you to play with phase shift andsee the effect on the reactive power draw. You can see how the systemeffectively borrows energy from the source and returns it at differentpoints in the cycle. This excess energy you repeatedly borrow and returnis the reactive power, whereas the energy you take, use, and don'treturn is the real power.

http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained

--Andrew Gabriel[email address is not usable -- followup in the newsgroup]Andrew Gabriel,Jun 11, 2010#78. GuestGuestply

"narke" wrote in messagenews:...> On 2010-06-10, wrote:>>>>>>>> "narke" wrote in message>> news:...>>> On 2010-06-08, daestrom wrote:>>>> narke wrote:

> I fell it can be very helpful ... just now the format of the graph not> good, it's hard to get the information. Would you please redraw the> circuit to make it clear? (yes, single phase is assumed)>> Many thanks!----1------+---2-----+----3----+source | | |V L C R| | |-----------+----------+--------- +

L, C and R in parallel:Source at 1 and L between 1 and 2, C between 2 and 3 and R after 3meter at 1, 2 or 3 with its voltage leads to bottom line

KWH reading the same at all locations and the same as if L and C didn'texistReactive (KVARH) same as without L and C for meter at 1 or 3 (zero)Reactive negative with meter at 3Meter reads KWH and KVARH "downstream" of its location.-------Don Kellycross out to reGuest,Jun 12, 2010#89. narkeGuestOn 2010-06-11, Andrew Gabriel wrote:> In article ,> narke writes:>> Hi,>>>> In a AC power quadrant diagram, how to understand the physical meanings>> of powers that falls in each quadrant?>>>> I: +P, +Q>> II: -P, +Q>> II: -P, -Q>> IV: +P, -Q>>>> where P denotes real power and Q denotes reactive power.>>>> Thanks in advance.>> I wrote a java applet which allows you to play with phase shift and> see the effect on the reactive power draw. You can see how the system> effectively borrows energy from the source and returns it at different> points in the cycle. This excess energy you repeatedly borrow and return> is the reactive power, whereas the energy you take, use, and don't> return is the real power.>>http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained>

That's so much wonderful!!! Actualy, in the way of learning theseconcepts these days, I also came with an idea that is to do exactylywhat you did after I really understand the knowledge. But I plan toimplement the demo application in other language. So I am thinking,could you share your source code to me, so I can see the algorithm. Ifyou don't want to share, it is okay. You've already help anyway.

Thanks again.


narkenarke,Jun 12, 2010#910. narkeGuestOn 2010-06-12, wrote:> ply>> "narke" wrote in message> news:...>> On 2010-06-10, wrote:>>>>>>>>>>>> "narke" wrote in message>>> news:...>>>> On 2010-06-08, daestrom wrote:>>>>> narke wrote:>>> I fell it can be very helpful ... just now the format of the graph not>> good, it's hard to get the information. Would you please redraw the>> circuit to make it clear? (yes, single phase is assumed)>>>> Many thanks!> ----1------+---2-----+----3----+> source | | |> V L C R> | | |> -----------+----------+--------- +>> L, C and R in parallel:> Source at 1 and L between 1 and 2, C between 2 and 3 and R after 3> meter at 1, 2 or 3 with its voltage leads to bottom line>> KWH reading the same at all locations and the same as if L and C didn't> exist

Cleared for this, many thanks!

> Reactive (KVARH) same as without L and C for meter at 1 or 3 (zero)

Also cleared! Thanks.

> Reactive negative with meter at 3

I assume you meant to say 'Reactive negative with meter at 2', right?

> Meter reads KWH and KVARH "downstream" of its location.

Can you explain what mean "downstream" ?

For the serial connection of L,C,R in the following circuit:

--- AC Source V -------L-------R-------C---------1 2 3

Please also give me an answer for meter readings of kWh and kVAh inpoint 1, 2, 3 respectively.

Thanks.


narkenarke,Jun 12, 2010#1011. narkeGuestOn 2010-06-12, narke wrote:> On 2010-06-11, Andrew Gabriel wrote:>> In article ,>> narke writes:>>> Hi,>>>>>> In a AC power quadrant diagram, how to understand the physical meanings>>> of powers that falls in each quadrant?>>>>>> I: +P, +Q>>> II: -P, +Q>>> II: -P, -Q>>> IV: +P, -Q>>>>>> where P denotes real power and Q denotes reactive power.>>>>>> Thanks in advance.>>>> I wrote a java applet which allows you to play with phase shift and>> see the effect on the reactive power draw. You can see how the system>> effectively borrows energy from the source and returns it at different>> points in the cycle. This excess energy you repeatedly borrow and return>> is the reactive power, whereas the energy you take, use, and don't>> return is the real power.>>>>http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained>>>> That's so much wonderful!!! Actualy, in the way of learning these> concepts these days, I also came with an idea that is to do exactyly> what you did after I really understand the knowledge. But I plan to> implement the demo application in other language. So I am thinking,> could you share your source code to me, so I can see the algorithm. If> you don't want to share, it is okay. You've already help anyway.>> Thanks again.>>

Andrew,

I have another qeustion after read your web page. When you demostratethe concepts, your were using the areas of the curve. The areas (red,grean), i think, are energes. And, I feel it can be decuded from yourtratement, thatVAH = kWh + kvarh (1)where,* kWh is (red area) - (green area)* kvarh = (green area)* VAH = (red area)

Am I right?

However, we knowS^2 = P^2 + Q^2 (2)where, S is apparent power, P is real power or true power, Q is reactivepower.

Now I get problem, since (2) can not be deduced from (1). Did you seeit? Please help.


narkenarke,Jun 12, 2010#1112. Andrew GabrielGuestIn article ,narke writes:>> Andrew,>> I have another qeustion after read your web page. When you demostrate> the concepts, your were using the areas of the curve. The areas (red,> grean), i think, are energes. And, I feel it can be decuded from your> tratement, that> VAH = kWh + kvarh (1)> where,> * kWh is (red area) - (green area)> * kvarh = (green area)> * VAH = (red area)>> Am I right?

No. It's simply red area is energy drawn from supply (quadrants I andIII in your original post), and green is energy returned to supply(quadrants II and IV in your original post). So red includesboth real and reactive energy draw, but green can only be reactive.Since the reactive energy returned to the supply (green) will besame as reactive energy drawn from supply, you can assume that apart of the red area equal in size to the green area is also reactiveenergy, and the remainder of the red is real power. However, itdoesn't make any sense to try and identify that part because you can'tsensibly say which particular bit of the energy drawn from the supply(red) is reactive and will be returned, verses another bit of the redwhich is not reactive and will be burned in the load. i.e. at a phaseangle of 60 (PF=0.5), 1/3rd of the red area is reactive and 2/3rdsis real, but there's no way to say which of the red pixels is reactiveor which is real.

So:Wh is fixed in this example. It's same as red area at phase angle = 0.VAh will be red + green.VArh will be 2*green.

> However, we know> S^2 = P^2 + Q^2 (2)> where, S is apparent power, P is real power or true power, Q is reactive> power.

That comes from the vector diagram equalatral triangle.It should be same result, but at 01:05 in the middle of the nighthere, I can't immediately think how you could prove it.

--Andrew Gabriel[email address is not usable -- followup in the newsgroup]Andrew Gabriel,Jun 13, 2010#1213. narkeGuestHi, Andrew, I just came back for a travel. Sorry for so late followingthe thread.

On 2010-06-13, Andrew Gabriel wrote:> In article ,> narke writes:>>>> Andrew,>>>> I have another qeustion after read your web page. When you demostrate>> the concepts, your were using the areas of the curve. The areas (red,>> grean), i think, are energes. And, I feel it can be decuded from your>> tratement, that>> VAH = kWh + kvarh (1)>> where,>> * kWh is (red area) - (green area)>> * kvarh = (green area)>> * VAH = (red area)>>>> Am I right?>> No. It's simply red area is energy drawn from supply (quadrants I and> III in your original post), and green is energy returned to supply> (quadrants II and IV in your original post). So red includes> both real and reactive energy draw, but green can only be reactive.> Since the reactive energy returned to the supply (green) will be> same as reactive energy drawn from supply, you can assume that a> part of the red area equal in size to the green area is also reactive> energy, and the remainder of the red is real power. However, it> doesn't make any sense to try and identify that part because you can't> sensibly say which particular bit of the energy drawn from the supply> (red) is reactive and will be returned, verses another bit of the red> which is not reactive and will be burned in the load. i.e. at a phase> angle of 60 (PF=0.5), 1/3rd of the red area is reactive and 2/3rds> is real, but there's no way to say which of the red pixels is reactive> or which is real.>> So:> Wh is fixed in this example. It's same as red area at phase angle = 0.> VAh will be red + green.> VArh will be 2*green.>

Okay, you fixed my error in understanding what is apparent power. So,let me rearrange the result as:

Wh = red - greenVAh = (red - green) + (green + green) = Wh + WArhVArh = 2*green

So, it still shows that VAh = Wh + WArh. To my mind, it still conflictswith S^2 = P^2 + Q^2, becaseu I think S is WAh/h, P is Wh/h, Q isWArh/h.

>> However, we know>> S^2 = P^2 + Q^2 (2)>> where, S is apparent power, P is real power or true power, Q is reactive>> power.>> That comes from the vector diagram equalatral triangle.> It should be same result, but at 01:05 in the middle of the night> here, I can't immediately think how you could prove it.

As stated above, yes I don't see a sign that I can prove it. So pleasehelp

-narke

http://www.sma.de/en/solutions/medium-power-solutions/knowledgebase/sma-shifts-the-phase.html

como hacer un match entre dos columnas en excel

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