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Taya2010 [7]
3 years ago
14

An extruder barrel has a diameter of 4.22 inches and a length of 75 inches. The screw rotates at 65 revolutions per minute. The

screw channel depth = 0.23 in, and the flight angle = 21.4 degrees. The head pressure at the die end of the barrel is 705 lb/in2. The viscosity of the polymer melt is given as 145 x 10-4 lb·sec/in2. Calculate the volume flow rate in in3/sec of the plastic through the barrel.
Engineering
1 answer:
Vlada [557]3 years ago
7 0

Answer:

volume flow rate Q  = 53.23 in³/s

Explanation:

given data

diameter = 4.22 inches

length = 75 inches

screw rotates = 65 revolutions per minute

depth = 0.23 in

flight angle = 21.4 degrees

head pressure = 705 lb/in²

viscosity = 145 x 10^{-4} lb·sec/in²

solution

we get here volume flow rate of platstic in barrel that is express as

volume flow rate Q = volume flow rate of die - volume flow of extruder barrel   ................1

here

volume flow rate extruder barrel is

flow rate   = \frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }    

flow rate = 60.10  

and

volume flow rate of die is express as

volume flow rate = 0.5 × π² × D² × Ndc ×  sinA × cosA   .............2

put here value and w eget

volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4

volume flow rate = 6.866

so put value in equation 1 we get

volume flow rate Q  = 60.10 - 6.866  

volume flow rate Q  = 53.23

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A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
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Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

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old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

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back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

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hence ; 28.65 * s = 678 * 0.2956

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S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

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back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

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Answer:

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Explanation:

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back to equation ( 1 )

( M / I ) * y

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4d^2 = 34560000 / 940300

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