infrome 2.docx
TRANSCRIPT
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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