Universidad Nacional Autónoma de México
Dirección General de Cómputo y de Tecnologías de Información y Comunicación
Trabajo realizado con el apoyo del Programa UNAM-DGAPA-PAPIME PE110517

Asignatura de Transporte de Energía
Soluciones de las ecuaciones de transporte

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	Hidrodinámica
	Térmica",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
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	Teorema π",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
tx5=TextCell["Ejemplo, flujo en tuberías
Caso de interés para el IQM: Colada continua",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
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▪Entender la idea de semejanza
▪Saber utlizar el teorema de π
▪Conocer que representan los números adimensionales Re, Nu y Pr",FontColor->White,FontFamily->"Times New Roman",16,TextJustification->1]}],RoundingRadius->5,FrameMargins->Large,Background->RGBColor["#1c3d6d"]]

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Style[" 
\!\(\*FormBox[
RowBox[{
StyleBox[
RowBox[{
RowBox[{
RowBox[{
RowBox[{"ρ", FractionBox[
RowBox[{"∂", "U"}], 
RowBox[{"∂", "t"}]]}], "+", 
RowBox[{SubscriptBox["v", "x"], FractionBox[
RowBox[{" ", 
RowBox[{"∂", "U"}]}], 
RowBox[{"∂", "x"}]]}], "+", 
RowBox[{SubscriptBox["v", "y"], FractionBox[
RowBox[{" ", 
RowBox[{"∂", "U"}]}], 
RowBox[{"∂", "y"}]]}], "+", 
RowBox[{SubscriptBox["v", "z"], FractionBox[
RowBox[{" ", 
RowBox[{"∂", "U"}]}], 
RowBox[{"∂", "z"}]]}]}], "=", 
RowBox[{
RowBox[{"-", 
RowBox[{"∇", "P"}]}], "-", 
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RowBox[{"∇", "▪τ"}], "]"}], "+", "ρg"}]}], "
", "
"}],
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", 
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RowBox[{
RowBox[{"u▪", FractionBox[
RowBox[{" ", 
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RowBox[{"v▪", FractionBox[
RowBox[{" ", 
RowBox[{"∂", "u"}]}], 
RowBox[{"∂", "y"}]]}]}], "=", 
RowBox[{
RowBox[{
RowBox[{"-", FractionBox["1", "ρ"]}], "▪", FractionBox[
RowBox[{" ", 
RowBox[{"∂", "P"}]}], 
RowBox[{"∂", "x"}]]}], "+", 
RowBox[{"γ▪", FractionBox[
RowBox[{" ", 
RowBox[{SuperscriptBox["∂", "2"], "u"}]}], 
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tx1=Style[" 
(∂/(∂t))ρ(Û+(1/2)(u^2))=-(▽▪ρu(Û+(1/2)(u^2)))-(▽▪q)+ρ(u▪g)-(▽▪ρu)-(▽▪[τ▪U])

Estaciona
u▪(( ∂T)/(∂x))+v▪(( ∂T)/(∂y))=α(( (∂^2)T)/(∂(y^2)))+(γ/Cp)▪((( ∂u)/(∂y)))^2
u▪(( ∂u)/(∂x))+v▪(( ∂u)/(∂y))=-(1/ρ)(( ∂P)/(∂y))+γ▪\!\(\*FractionBox[\(\(\\ \)\(\*SuperscriptBox[\(∂\), \(2\)]u\)\), \(∂\*SuperscriptBox[\(y\), \(2\)]\)]\)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx2=Style["Si se desprecian los términos
((∂P)/(∂X))   y (((∂U)/(∂y)))^2",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",18];
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tx4=Style["𝓊▪((∂T)/(∂x))+𝓋▪((∂T)/(∂y))=α▪(((∂^2)P)/(∂(y^2)))",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx5 = Style["α y ϒ se llaman difusividaes, térmica y de cantidad de movimiento, respectivamente 
Su cociente es el número de Prandtl
(γ/α) = Pr",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx6=Grid[{{Style["◼ Región dónde están presentes los gradientes de velocidad o temperatura. Zona ``cercana´´ a un objeto
◼ Hidrodinámica: Gradientes de velocidad
◼ Térmica: Gradientes de temperatura",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16]}}];
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tx7=Style["Resolver la ecuación (vectorial) de transporte de momento (Naver-Stokes permite conocer el campo (vectorial) de velcoidades
Resolver la ecuaciçon de transporte de energía permite conocer el campo de temperaturas
¿Cuáles son las expresiones matemáticas de esos campos? ",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
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tx8=Style["

Para la velocidad se propone un polinomio de tercer grado:
U(y)=a+by+(cy^2)+(dy^3)

Que satisface las condiciones de frontera:
En y=0    U(y)=0
En y=δ   U(δ)=\!\(\*SubscriptBox[\(U\), \(\(∞\)\(\\ \)\)]\) y ((∂u)/(∂y))=0
Con lo cual
a=0, b=(3/2) ((Subscript[U, ∞])/δ),  c=0,  d=-((Subscript[U, ∞])/(2 (δ^3)))
δ=(4.64/(Re^(1/2)))x (x es la distancia desde el borde de ataque)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
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tx9=Style["Para la temperatura se propone un polinomio de tercer grado:
T(y)=e+fy+(gy^2)+(hy^3)

Que satisface las condiciones de frontera:
En y=0  T(0)=(Subscript[T, s])  ((d^(2  T))/(dy^2))=0
En y=(Subscript[δ, 𝓉]) T((Subscript[δ, 𝓉]))=(Subscript[T, ∞]) y ((∂T)/(∂y))=0
°
Con lo cual:
e=(Subscript[T, s]), f=(3/2)((((Subscript[T, ∞]) - (Subscript[T, s])))/(Subscript[δ, 𝓉])), g=0, h=-((((Subscript[T, ∞]) - (Subscript[T, s])))/(2 ((Subscript[δ, 𝓉])^3)))
δ=0.976(Pr^(1/3))δ",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
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tx10=Style["Una placa se coloca en una corriente de aire a presión atomosférica que fluye paralela a ella a una velocidad de 5m/s. La temperatura en la superficie de la placa se mantiene uniforme a 200°C y la correspondiente a la corriente principal de aire es 30°C. Calcule la temperatura y velocidad a una altura de 4mm y una distancia de 30cm del borde de ataque",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];

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Grid[{{img1},{Style["¿Qué es la semjanza?
Dos triángulos semejantes tienen sus «lados adimensionales» iguales.

Son adimensionalmente 'iguales'. Cualquier fenómeno que dependa de la relación de entre los lados, ocurre igual en todos los triángulos semejantes.

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No todas las magnitudes físicas escalan de la misma manera.El peso escala de manera proporcional al volumen
Mg = ((ρL)^3) g
\!\(TraditionalForm\`Una\\ persona\\ 10\\ veces\\ más\\ grande\\ pesa\\ \*SuperscriptBox[\(10\), \(3\)] = 1000\\ veces\\ \(\(mas\)\(.\)\)\)
\!\(TraditionalForm\`La\\ resistencia\\ a\\ la\\ ruptura\\ escala\\ proporcionalmente\\ a\\ la\\ superfice . P = \*FractionBox[\(F\), \(π\\ \*SuperscriptBox[\(r\), \(2\)]\)]\)
\!\(TraditionalForm\`Una\\ persona\\ 10\\ veces\\ más\\ grande\\ resiste\\ \*SuperscriptBox[\(10\), \(2\)] = 100\\ veces\\ \(\(más\)\(.\)\)\)
\!\(TraditionalForm\`Para\\ que\\ la\\ resistencia\\ sea\\ semejante, el\\ cociente\\ peso/resistencia\\ debe\\ mantenerse\\ \(\(constante\)\(.\)\)\)
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\!\(TraditionalForm\`La\\ resistencia\\ a\\ la\\ ruptura\\ escala\\ proporcionalmente\\ a\\ la\\ superfice . P = \*FractionBox[\(F\), \(π\\ \*SuperscriptBox[\(r\), \(2\)]\)]\)
\!\(TraditionalForm\`Una\\ persona\\ 10\\ veces\\ más\\ grande\\ resiste\\ \*SuperscriptBox[\(10\), \(2\)] = 100\\ veces\\ \(\(más\)\(.\)\)\)
\!\(TraditionalForm\`Para\\ que\\ la\\ resistencia\\ sea\\ semejante, el\\ cociente\\ peso/resistencia\\ debe\\ mantenerse\\ \(\(constante\)\(.\)\)\)
\!\(TraditionalForm\`Eso\\ puede\\ ocurrir\\ si\\ las\\ dimensiones\\ horizontales\\ crecen\\ más\\ que\\ las\\ \(\(verticales\)\(.\)\)\)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",18];
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Style["Dos fluidos con el mismo número de Reynolds son hidrodinámicamente semejantes, aunque sus valores de ρ, µ, V y D sean distintos
Lo importante es que la relación entre 'fuerzas' dinámicas ρvD y las de fricción μ tengan una razón constante.
\!\(TraditionalForm\`Re = \*FractionBox[\(ρ\\ V\\ D\), \(μ\)]\)
\!\(TraditionalForm\`Re = \(\*FractionBox[\(ρ\\ \*FractionBox[SuperscriptBox[\(v\), \(2\)], \(l\)]\), \(\(μ\)\(\\ \)\*FractionBox[\(v\), SuperscriptBox[\(l\), \(2\)]]\(\\ \)\)] = \\ \*FractionBox[\(Fuerza\\ de\\ inercia\), \(viscosidad\)]\)\)
Un Reynolds de 2 500 quiere decir que las fuerzas de inercia son 2 500 veces más fuertes que las de fricción. ", FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri"16]}];
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Grid[{{Style["Teorema Pi (π) de Buckingham
Sea: 
m: Variables homogéneas; por ejemplo: diámetro (D), velocidad (v), Temperatura (T), longitud (L), presión (P),… 

n: Dimensiones de referencia longitud [L], tiempo [t], masa [M], temperatura [T]… 

Entonces se obtendrán:
(m-n): Números adimensionales

Procedimiento
▪Enumerar las variables que describen el problema
 
▪Seleccionar las dimensiones de referencia que corresponden a las variables

▪Descomponer las variables en sus dimensiones, de manera tabulada

▪Elegrir las variables de referencia según:

	a)Debe ser igual a 'n' variables de referencia
	b)Entre todas deben contener todas las dimensiones
	c)Deben ser sencillos e independientes entre sí

Establecer las m-n ecuaciones adimensionales y obtener los números pi (π)", FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri"16]}}]}];
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tx1=TextCell["Factor de fricción en tuberías",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",18,TextJustification->1];
tx2=Style["
Variables: ΔP, D, L, V, ρ, μ, ϵ.
(ϵ: Rugosidad de la tubería.)
Por tanto m=7

Dimensiones de referencia: [L], [M], [t], 
Por lo tanto n=3

Según el teorema π se obtendrán 7- 3 = 4 números adimensionales.",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",16];
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Referencias = D, v, ρ .

(n=3, variables sencillas e independientes entre si que contienen entre las tres a todas las variables) .

Debe haber 7-3 = 4 números adimensionales:",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
tx5=Style["
(Subscript[π, 1]) = (D^a1)*(V^b1)*(p^c1)*L
(Subscript[π, 2]) = (D^a2)*(V^b2)*(p^c2)*ϵ
(Subscript[π, 3]) = (D^a3)*(V^b2)*(p^c3)*μ
(Subscript[π, 4]) = (D^a4)*(V^b4)*(p^c4)*ΔP
para (Subscript[π, 1])",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",16];
tx6=TextCell["
[L]=a1+b1-3c1+1=0
[t]=0-b1+0+0=0
[M]=0+0+c1+0=0

Solución 

b1=0
c1=0
a1=-1

por tanto
",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
tx7=Style["
π1 = (D^(-1)) (V^0) (ρ^0) L =L/D",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",16];
tx8=TextCell["
Mide que tan largo es (adimensionalmente) el tubo.
Analogamente ",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
tx9=Style["
\!\(TraditionalForm\`\*SubscriptBox[\(π\), \(2\)] = \*FractionBox[\(ϵ\), \(D\)]\\ \\ \\ Mide\\ \\ que\\ \\ tan\\ \\ rugoso\\ \\ es\\ \\ \((adimensionalmente)\)\\ \\ el\\ \\ tubo\)
\!\(TraditionalForm\`\*SubscriptBox[\(π\), \(3\)] = \*FractionBox[\(μ\), \(D\\ v\\ ρ\)]\\ \\ \\ Mide\\ \\ que\\ \\ tan\\ \\ viscoso\\ es\\ \\ \((adimensionalmente)\)\\ \\ el\\ tubo . \\ \\ Es\\ \\ el\\ \\ inverso\\ \\ del\\ Re\)
\!\(TraditionalForm\`\*SubscriptBox[\(π\), \(3\)] = \*FractionBox[\(Δ\\ P\), \(\*SuperscriptBox[\(v\), \(2\)]\\ ρ\)]\\ \\ \\ Mide\\ \\ que\\ \\ tan\\ \\ grande\\ \\ es\\ \\ \((adimensionalmente)\)\\ \\ el\\ \\ gradiente\\ \\ de\\ \(\(presion\)\(.\)\)\)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",16];s2=Button[Style["Teorema Pi simulador","Hyperlink"],NotebookLocate[{URL["https://www.dropbox.com/s/scvbfewkrwbo5sq/Teorema%20Pi.nb?dl=0"],None}]];
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tx1=Style["Número de Biot",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",18];
tx2=Style["Relación de la resistencia térmica interna de un cuerpo sólido a su resistencia térmica superficial.
Bi=(hl/(Subscript[k, s]))",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx3= Style["
Número de Nusselt",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",18];
tx4=Style["Coeficiente de transferencia de calor adimensional; relación de transferencia de calor por convección a conducción en una capa de fluído de espesor L
Nu=(((Subscript[h, c]) L)/(Subscript[K, f]))",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx5 = Style["Número de Prandtl",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",18];
tx6=Style["
Relación de difusiviad de la cantidad de movimiento molecular a la difusividad
\!\(\*FormBox[RowBox[{StyleBox["Pr",FontSize->18], StyleBox["=",FontSize->18], RowBox[{StyleBox[FractionBox[RowBox[{SubscriptBox["C", "ρ"], "μ"}], "k"],FontSize->18], StyleBox["=",FontSize->18], FractionBox[StyleBox["v",FontSize->18], "α"]}]}],TraditionalForm]\)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
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tx7=Style["
Re=((((ρv)^2)/l)/(µv/(l^2)))=\!\(\*FractionBox[\(fuerza\\ de\\ inercia\), \(viscosidad\)]\)
Fr=((((ρv)^2)/l)/(pℊ))=\!\(\*FractionBox[\(fuerza\\ de\\ inercia\), \(fuerza\\ de\\ gravedad\)]\)
(Gr/(Re^2))=((ρℊβ ((Subscript[t, 1]) - (Subscript[T, 0])))/(((ρv)^2)/l))=\!\(\*FractionBox[\(Fuerza\\ de\\ inercia\), \(Fuerza\\ de\\ inercia\)]\)
Pe=RePr=((ρCpv ((Subscript[t, 1]) - (Subscript[T, 0]))/l)/(kT ((Subscript[T, 1]) - (Subscript[T, 0)])/(l^2)))=\!\(\*FractionBox[\(Calor\\ transportado\\ por\\ convección\), \(Calor\\ transportado\\ por\\ conducción\)]\)
Br=((µ (((v/l))^2))/(kT ((Subscript[t, 1]) - (Subscript[T, 0]))/(l^2)))=\!\(\*FractionBox[\(Producción\\ de\\ calor\\ por\\ la\\ disicpación\\ de\\ viscosidad\), \(Calor\\ transportado\\ por\\ conducción\)]\)
",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
tx8=Style["El Reynolds
Marca la transición laminar -turbulento
La turbulencia aumenta los valores de los coeficientes de transporte
Importante conocer la longitud de la transición
El punto de transición se conoce como longitud crítica
Puede calcularse a partir de la fórmula del Reynolds

Criterio: Re<5(x10^5)
Re=((ρux)/µ)=5(x10^5)

Laminar /turbulento
En el  caso laminar es posible resolver exactamente la ecuación de energía. La solución puede expresarse como:
Nu =   = 0.332 (Re^0.5) (Pr^(1/3))   (Pr > 0.6)

En el caso turbulento puede construirse la correlación:
Nu =   = 0.0296 (Re^0.8) (Pr^(1/3))   (0.6 <Pr<60)",FontColor->RGBColor["#1C3D6C"],FontFamily->"Times New Roman",16];
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Valor adimensional del cateto opuesto
Porcentaje de ruptura del fémur al aumentar la talla",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];
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Valor adimensional del cateto opuesto
Porcentaje de ruptura del fémur al aumentar la talla",FontColor->RGBColor["#1c3d6d"],FontFamily->"Times New Roman",16,TextJustification->1];

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Framed[OpenerView[{Style["5. Bibliografía",Darker[Gray],FontFamily->"Calibri",24],Style["Referencias
Kreith F,Manglik R y Bohn Mark \"Principios de transferencia de calor\".Cengage Learning 2012
Levi,E. \"El agua según la ciencia\" Consejo Nacional de Ciencia y Tecnología/Ediciones Castell México 1989.
Bird R B,Stewart W E,Lightfoot E N \"Transport Phenomena\" Wiley 2006
Haber-Schaim,Uri;Dodge,John H.; Gardner, Robert; Shore, Edward A. Walter, Felicie. \"PSSC Physics (7th ed.) \"Dubuque, Iowa: Kendall/Hunt. ISBN 978-0840360250.",FontColor->RGBColor["#1C3D6C"],FontFamily->"Calibri",16]}],RoundingRadius->5,FrameMargins->Large]

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s1=Framed[Manipulate[
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(f3[f2[й2],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й3],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й4],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й5],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й6],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й7],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й8],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й9],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й10],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]]}],Exponentes])==0]]]][{e1,e2,e3,e4,e5,e6,e7,e8,e9,e10}],NumericQ]]]]] ]},{"","",""},{"",
InputField[Dynamic[elemento],FieldSize->5],
Button[Grid[{{"quitar de la lista al elemento numero"},{InputField[Dynamic[elemento],FieldSize->6]}}] ,Dynamic[ListaDeΠ[[elemento]]=Nothing]]}}]}}],TextCell[Text[Style["Teorema Pi",Bold,18]]]],Background->GrayLevel[0.85],RoundingRadius->8],
Initialization:>
{



(*f1 recive las variables directamente y las convierte en su denominador(Segunda entrada distinta de cero para eso) o numerador (segunda entrada igual a cero para esto)*)

ListaDeΠ={};,
O1=Dynamic[MatrixForm[Transpose[Flatten/@{
Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]],
(f3[f2[й1],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й2],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й3],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й4],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й5],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й6],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й7],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й8],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й9],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й10],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]]}]]];,

f1[x_,y_]:=If[y==0,Select[Flatten[List[Characters[ToString[Numerator[x]]]]],LetterQ],Select[Flatten[List[Characters[ToString[Denominator[x]]]]],LetterQ]];,

f2[x_]:=Flatten[Join[Partition[Riffle[Flatten[List[f1[x,0]]],Flatten[List[(Exponent[x,ToExpression[#]])&/@f1[x,0]]]],2],Partition[Riffle[Flatten[List[f1[x,1]]],Flatten[List[(Exponent[x,ToExpression[#]])&/@f1[x,1]]]],2]]];,
f3[y_,x_]:=If[ContainsAny[y,{Extract[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]],{x}]}],Extract[y,Position[y,Extract[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]],{x}]]+1],{0}];,

ro={e1->0,e2->0,e3->0,e4->0,e5->0,e6->0,e7->0,e8->0,e9->0,e10->0};

MultiplicaLista[x_]:=Times[Take[x,{1}],Take[x,{2}],Take[x,{3}],Take[x,{4}],Take[x,{5}],Take[x,{6}],Take[x,{7}],Take[x,{8}],Take[x,{9}],Take[x,{10}]];,

ValoresConstruccionMatriz=Dynamic[Most[Drop[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]];,

Exponentes=List[e1,e2,e3,e4,e5,e6,e7,e8,e9,e10];,
f4[x_]:=Flatten[Take[Most[Drop[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]],{x}]^(Exponentes.Flatten[Take[Transpose[Flatten/@{

(f3[f2[й1],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й2],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й3],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й4],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й5],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й6],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],
(f3[f2[й7],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й8],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й9],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]],(f3[f2[й10],#])&/@ Range[1,Length[Most[Drop[AlphabeticSort[Map[ToString,DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]]]],Length [Select[DeleteDuplicates[Flatten[List[f2[й1],f2[й2],f2[й3],f2[й4],f2[й5],f2[й6],f2[й7],f2[й8],f2[й9],f2[й10]]]],NumberQ]]]]]]}],{x}]])];
,n=2}],18];