Welcome to the new and interesting Ivan´s blog.
This project was made to know more about of the threads and its composition.
We´re gonna see the different systems and its respective numbering, I hope you will enjoy this job.
I send you greetings...
Pd. Dedicated for my favorite teacher Judith Gutierrez. (:
WELCOME FUTURES ENGINEERS
Publicadas por UnknownTRANSMISIÓN DE MOVIMIENTO
Publicadas por Unknown
Consiste en dos o más poleas una motriz y otra movida esta es alimentada con cierta velocidad dada en Rpm (Revoluciones por minuto)
ND= nd NZ= nz
ND D1 D2... Dn = nd d1 d2... dn
Ejemplo 1
ND= nd
n= ND= n= (1520 rpm)(18)= 2280rpm
d 12
Ejemplo 2
ND= nd
n= ND= n= (1200 rpm)(15)(10)(15)
d (12)(8)(30)
n= 953.125 rpm
Ejemplo 3
n1= (1440rpm)(8)(10)(40)(50)= 853.33rpm
(10)(30)(60)(15)
n2= (1440rpm)(8)(15)(12)(12)= 3456rpm
(10)(12)(6)(10)
Ejemplo 4
Calcular la velocidad de todos los cilindros
n1= (940 rpm)(8.5)= 532.66 rpm
15
n2= (532.66rpm)(10)(30)= 106.53 rpm
(30)(50)
n3= (532.66rpm)(10)= 177.55rpm
(30)
n4= (532.66rpm)(10)(30)= 133.16rpm
(30)(40)
n5= (532.66rpm)(9)= 798.99rpm
(6)
Ejemplo 1
ND= nd
n= ND= n= (1520 rpm)(18)= 2280rpm
d 12
Ejemplo 2
ND= nd
n= ND= n= (1200 rpm)(15)(10)(15)
d (12)(8)(30)
n= 953.125 rpm
Ejemplo 3
n1= (1440rpm)(8)(10)(40)(50)= 853.33rpm
(10)(30)(60)(15)
n2= (1440rpm)(8)(15)(12)(12)= 3456rpm
(10)(12)(6)(10)
Ejemplo 4
Calcular la velocidad de todos los cilindros
15
n2= (532.66rpm)(10)(30)= 106.53 rpm
(30)(50)
n4= (532.66rpm)(10)(30)= 133.16rpm
(30)(40)
n5= (532.66rpm)(9)= 798.99rpm
(6)
DESARROLLOS
Publicadas por UnknownDesarrollo= π x Ø x n
Ejemplo 1
Desarrolo cil. batidor= π x 11/3” X 532.66rpm
= 2225.62"/min 1yd=36" 1mt= 39.37"
= 61.82yd/min x =2225.62" x= 2225.62"
= 56.53mt/min
Desarrollo cil. regulador= π x 21/8” x 106.53rpm
=711.18"/min 1yd=36" 1mt= 39.37"
= 19.75yd/min x =711.18 x= 711.18
= 18.06mt/min
Desarrollo cil. intermedio= π x 31/7” x 177.55rpm
= 1753.05"/min 1yd=36" 1mt= 39.37"
= 48.22yd/min x =177.55 x= 177.55
= 44.57mt/min
Desarrolo cil. limpiador= π x 23/7 “ x 133.16rpm
= 1015.95"/min 1yd=36" 1mt= 39.37"
= 28.22yd/min x =133.16 x= 133.16
= 25.80mt/min
Ejemplo 2

nC1= (1200rpm)(50)(40)(90) = 450rpm
(160)(50)(60
nC2= (45.45rpm)(19) = 53.97rpm
(160)
nC3= (1200rpm)(50)(40)(22)(20)(30) = 45.45rpm
(160)(90)(80)(11)(55)
nC4= (45.45rpm)(90)(20) = 32.72rpm
(25)(100)
Desarrollo C1= π x 1¾ x 450rpm
= 2474.01"/min 1yd=36" 1mt= 39.37
= 68.72yd/min x =2474.01 x= 2474.01
= 62.83mt/min
Desarrollo C2= π x 1¼ x 53.92rpm
= 211.94"/min 1yd=36" 1mt= 39.37
= 5.88yd/min x =53.92 x= 53.92
= 5.3mt/min
Desarrollo C3= π x 1¾ x 45.45rpm
= 821.01"/min 1yd=36" 1mt= 39.37
= 22.80yd/min x =45.45 x= 45.45
= 20.85mt/min
Desarrollo C4= π x 1¾ x 32.72rpm
= 178.88"/min 1yd=36" 1mt= 39.37
= 4.99yd/min x =32.72 x= 32.72
= 4.5mt/min
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