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EJEMPLO DE DIFERENCIAS DIVIDIDAS FINITAS En una planta se bombea esencia de trementina, 60 ◦C, desde la base de una columna de fraccionamiento hasta un gran tanque de almacenamiento descubierto. La columna opera a 1,29 atmósferas. En la siguiente tabla se representan los datos relativos los litros por hora que puede bombear la bomba en función de la potencia en watios a la que es necesario que trabaje: Punto 0 Q(l/h) 500 N(w) 365

1 700 361.6

2 900 370.6 4

3 1100 379.6 8

4 1300 384.4 6

5 1500 395.5

6 1700 395.9 5

7 1900 397

PRIMERAS DIFERENCIAS

-

f(X0,

X1) =

-

f(X1,

X2) =

-

f(X2,

X3) =

-

f(X3,

X4) =

-

f(X4,

X5) =

-

f(X5,

X6) =

-

f(X6,

X7) =

f ( x 1) −f ( x 0) x 1−x 0

=

361.6−36 5 700−500

x 2−x 1

=

370.64−361.6 900−700

f ( x 3 )−f ( x 2) x 3−x 2

=

379.68−370.64 1100−900

= 0.045200000

x 4 −x 3

=

384.46−379.68 1300−1100

= 0.023980000

f ( x 5 )−f ( x 4 ) x 5−x 4

=

395.5−384.46 1500−1300

x 6 −x5

=

395.95−395.5 1700−1500 = 0.002250000

f ( x 7 )−f ( x 6) x 7−x 6

=

397−395.95 1900−1700

f ( x 2) −f ( x 1)

f ( x 4 ) −f ( x 3 )

f ( x 6 )−f ( x 5)

SEGUNDAS DIFERENCIAS

= -0.017000000

= 0.045200000

= 0.055200000

= 0.005250000

-

f(X0,

X1, X2) =

f ( x 1 , x 2 )−f (x 0 , x 1)

=

0.0452−(−0.017) 900−500

=

0.0452−0.0452 1100−700

x 4−x 2

=

0.02398−0.0452 1300−900

=

f ( x 4 , x 5 ) −f ( x 3 , x 4) x 5−x 3

=

0.0552−0.02398 1500−1100

=

x 6 −x 4

=

0.00225−0.0552 1700−1300

=

f ( x 6 , x 7 )−f (x 5 , x 6 ) x7 −x 5

=

0.00525−0.00225 1900−1500

x 2−x 0

=

0.000155500 -

f(X1,

X2, X3) =

f ( x 2 , x 3 )−f (x 1 , x 2) x 3−x 1

=

0.0000000000 -

f(X2,

X3, X4) =

f ( x 3 , x 4 ) −f (x 2 , x3 )

-0.000053050 -

f(X3,

X4, X5) =

0.000132535 -

f(X4,

X5, X6) =

f ( x 5 , x 6 )−f ( x 4 , x 5)

-0.000132375 -

f(X5,

X6, X7) =

=

0.000007500 TERCERAS DIFERENCIAS -

f(X0,

X1, X2, X3) =

f ( x 1 , x 2 , x 3 )−f ( x 0 , x 1 , x 2) x 3−x 0

=

0−0.0001555 1100−500

f ( x 2 , x 3 , x 4 ) −f ( x1 , x2 , x3 ) x 4−x 1

=

−0.00005305−0 1300−700

-0.000000259 -

f(X1,

X2, X3, X4) =

= -0.000000088 -

f(X2,

X3, X4, X5) =

f ( x 3 , x 4 , x5 ) −f ( x2 , x3 , x 4 )

0.00007805−(−0.00005305) 1500−900

x 5−x 2

=

=0.000000219

=

-

f(X3,

X4, X5, X6) =

f ( x 4 , x 5 , x6 ) −f ( x 3 , x 4 , x 5 ) x 6−x 3

−0.000132375−0.00007805 1700−1100

-

f(X4,

X5, X6, X7) =

=

=

-0.000000351

f ( x 5 , x 6 , x 7 )−f ( x 4 , x 5 , x 6 ) x 7−x 4

0.0000075−(−0.000132375) 1900−1300

=

= 0.000000233

CUARTAS DIFERENCIAS

-

f(X0,

f ( x 1 , x 2 , x 3 , x 4 ) −f ( x0 , x 1 , x 2 , x 3) x 4−x 0

X1, X2, X3, X4) =

−0.000000088+0.000000259 1300−500

-

f(X1,

X2, X3, X4, X5) =

=0

f ( x 2 , x 3 , x 4 , x5 ) −f ( x1 , x 2 , x 3 , x 4 ) x 5−x 1

0.000000219+ 0.000000088 1500−700

-

f(X2,

X3, X4, X5, X6) =

f(X3,

X4, X5, X6, X7) =

0.0 00000233+ 0.000000351 1900−1100

QUINTA DIFERENCIAS

=

=0

f ( x 3 , x 4 , x5 , x 6 )−f ( x 2 , x 3 , x 4 , x5 ) x 6−x 2

−0. 000000351−0.000000219 1700−900

-

=

=

= -0.0000001

f ( x 4 , x 5 , x6 , x 7 )−f ( x 3 , x 4 , x5 , x 6) x 7−x 3

= 0.000000001

=

f(X0,

-

X1, X2, X3, X4, X5) = 0−0

= 1500−500 f(X1,

-

=

= i

x 5−x 0

=0

X2, X3, X4, X5, X6) =

f ( x 2 , x 3 , x 4 , x5 , x 6 )−f ( x 1 , x 2 , x 3 , x 4 , x 5) x 6−x 1

−0.0000 001−0 =0 1700−700

f(X2,

-

f ( x 1 , x 2 , x 3 , x 4 , x5 ) −f ( x 0 , x 1 , x 2 , x 3 , x 4 )

X3, X4, X5, X6, X7) =

0.000000001+0.00000001 1900−1900

xi

f(xi)

0 500

365

Primeras

f ( x 3 , x 4 , x5 , x 6 , x 7 )−f ( x 2 , x 3 , x 4 , x 5 , x 6) x 7 −x2

=0

Segundas

Tercera s

Cuartas

Quintas

0.0170000 00 1 700

361.6

0.00015550 0 0.0452000 00

2 900

370.6 4

0.00000 0259 0

0.0452000 00 3 110 0

379.6 8

0.00000 0088 0.00005305 0

0.0239800 00 4 130 0

384.4 6

395.9 50

0 0.0000000 01

0.00000 0351 0.00013237 5

0.0022500 00 6 170 0

0

0.00013253 5

395.5

0

0.00000 0219

0.0552000 00 5 150 0

0

0

0.00000 0233 0.0000075

0

0.0525000 00 7 190 0

937