powerpoint presentation€¦ · powerpoint presentation author: pasquale volpintesta created date:...
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Problem
a Físico
Modelación
Modelo Simple:Haga las simplificaciones
necesarias para obtener una
ecuación que modele el
problema.
Modelo
Complejo:Deje las complejidades que
parezcan importantes, aún si
dificultan la solución del
problema
SoluciónSolución
Exacta:Solución con
“fórmula”.
Solución
Aproximada:Debido a la complejidad
del modelo, sólo se
puede obtener una
aproximación.
0
0
FrMM
FR
T
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Potencia
130 Kv – 400 Kv
10 MVA – 200 MVA
100 KUSD a 4 MDD
Distribución Industriales
6 KV – 26 KV
15 KVA – 1 MVA
1 KUSD a 10 KUSD
130 KV – 260 KV
1 MVA – 15 MVA
20 KUSD a 200 KUSD
Comerciales
Secos
440 V – 13 KV
15 KVA – 1 MVA
1 KUSD a 10 KUSD
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Modelo de Transferencia de Calor vía
Conducción FEM
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°
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DESIGN-EXPERT Plot
TempX = A: GapY = B: Heigh
Actual FactorC: q'' = 750.00
24.75
49.1825
73.615
98.0475
122.48
Tem
p
0.38 0.41
0.44 0.47
0.50
6.00
9.00
12.00
15.00
18.00
A: Gap
B: Heigh
Final Equation in Terms of Actual
Factors:
Temp =
+79.45992
-432.59626 * Gap
+6.26123 * Heigh
+0.10049 * q''
+547.93270 * Gap ^2
-0.064757 * Heigh ^2
-1.00488E-005 * q’’^2
-8.96000 * Gap * Heigh
-0.16944 * Gap * q''
+3.85667E-003 * Heigh * q''
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