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Universidad internacional del Ecuador

Rodrigo CaleroIng. Automotriz04/11/2015

Bom as !iesel

Q 1 = 29,16 x 10 -6 [m3 / seg ]

Q 2 = 63 x 10 7 [m3 / seg ]

= 4,8 x 10 -7 [m2 / seg ] 20

= 5,6 x 10 -7 [m2 / seg ] 40

"ramo# Salida de la trampa de agua= 0.01035 m

A= .d2

4 = (0,01035 )2

4 = $%41 & 10'5 m 2

V =Q 1 A =

29,16 x 10 6

8,41 x 10 5 = 0%(4) [m / seg ]

20 = V . d =

0,347 (0,01035 )

4,8 x 10

7 = )4$2%1* +,. "ransici-n

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40 = V . d

=0,347 (0,01035 )

5,6 x 10 7 = 41(%(0 +,.

"ransici-n

"ramo# Ingreso a la trampa de agua= 0.01415 m

A= .d2

4 = (0,01415 )2

4 = 1%5) & 10'4 m 2

V =Q 1 A =

29,16 x 10 6

1,57 x 10 4 = 0%1$5 [m / seg ]

20 = V . d =

0,185 (0,01415 )4,8 x 10

7 =

545(% 4 +,. "ransici-n.

40 = V . d

=0,185 (0,01415 )

5,6 x10 7 =

4 )4%55 +,. "ransici-n.

"ramo# Ingreso al filtro de combustible= 0.0131 m

A= .d2

4 = (0,0131 )2

4 = 1%(4) & 10'4 m 2

V =Q 1

A =

29,16 x 10 6

1,347 x 10 4 = 0%21 [m / seg ]

20 = V . d =

0,216 (0,0131 )4,8 x 10

7 = 5$*5 +,. "ransici-n

40 = V . d

=0,216 (0,0131 )

5,6 x10 7 = 5052%$5 +,. "ransici-n

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"ramo# Salida del filtro de combustible= 0.01975 m

A= .d2

4 = (0,01975 )2

4 = (%0 & 10'4 m 2

V =Q 1 A =

29,16 x 10 6

3,06 x 10 4 = 0%0*5 [m / seg ]

20 = V . d =

0,095 (0,01975 )4,8 x 10

7 = (*0$%$5 +,.

"ransici-n

40

= V . d

=

0,095 (0,01975 )5,6 x10 7 = ((50%45 +,. "ransici-n

"ramo# Ingreso a la bomba= 0.0044 m

A= . d2

4 = (0,0044 )2

4 = 1%52 & 10'5 m 2

V =

Q 1 A =

29,16 x 10 6

1,52 x 10 5 = 1%*1$ [m / seg

]

20 = V . d =

0,192 (0,0044 )4,8 x 10

7 = 1) *1% +,. "ur ulento

40 = V . d

=0,185 (0,0044 )

5,6 x 10 7 = 150$5%)1 +,. "ur ulento

"ramo# Salida de la bomba= 0.0122 m

A= .d2

4 = (0,0122 )2

4 = 1%1) & 10'4 m 2

V =Q 2 A =

63 x 10 7

1,17 x 10 4 = 0%05( [m / seg ]

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20 = V . d =

0,053 (0,0122 )4,8 x 10

7 = 1(4)%0$ +,. aminar

40 = V . d

=0,053 (0,0122 )

5,6 x10 7 = 1154% 4 +,. aminar

"ramo# Ingreso a los inyectores= 0.0044 m

A= .d2

4 = (0,0044 )2

4 = 1%52 & 10'5 m 2

V = Q 2 A =

63 x10 7

1,52 x10 5 = 0%414 [m / seg ]

20 = V . d =

0,414 (0,0044 )4,8 x 10

7 = ()*5 +,.

aminar

40 = V . d

=0,414 (0,0044 )

5,6 x 10 7 = (252.$5 +,.

aminar

"ramo# Inyectores

= 0.002 m

A= . d2

4 = (0,002 )2

4 = (%141 & 10'

m 2

V =Q 2 A =

63 x 10 7

13,1416 x 10 6 = 0%4)*4 [m / seg ]

20 = V . d =

0,4794 (0,002 )4,8 x 10

7 = 1**).5 +,. aminar

40 = V . d

=o , 4794 (0,002 )

5,6 x 10 7 = 1712,14 +,. aminar

"ramo# Retorno de los inyectores

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= 0.0066 m

A= .d2

4 = (0,0066 )2

4 = (%42 & 10'5 m 2

V =Q 2 A =

63 x 10 7

3,42 x 10 5 = 0.1$4 [m / seg ]

20 = V . d =

0,184 (0,0066 )4,8 x10

7 = 25(2%$* +,. aminar

40 = V . d

=0,184 (0,0066 )

5,6 x 10 7 = 21 $%5) +,. aminar

"ramo# Retorno al filtro= 0.01380 m

A= .d2

4 = (0,01380 )2

4 = 1%4*5 & 10'5 m 2

V =Q 1 A =

29,16 x 10 6

1,495 x 10 5 = 0%1*5 [m / seg ]

20 = V . d =

0,195 (0,01380 )4,8 x 10

7 = 5 0 %25 +,. "ransici-n

40 = V . d

=0,195 (0,01380 )

5,6 x10 7 = 4$05%( +,. "ransici-n

"ramo# Conducto de retorno= 0.0064 m

A= . d2

4 = (0,01415 )2

4 = (%22 & 10'5 m 2

V =Q 1 A =

29,16 x 10 6

3,22 x 10 5 = 0%*0 [m / seg ]

20 = V . d =

0,906 (0,0064 )4,8 x10

7 = 120$0 +,. "ur ulento

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40 = V . d

=0,906 (0,0064 )

5,6 x 10 7 = 10(54%2$ +,.

"ur ulento