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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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       setLayout(new FlowLayout());

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       {

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       Guess game = new Guess();

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3 years ago
An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at
polet [3.4K]

Answer:

q = 1.73 W

Explanation:

given data

small end  = 5 cm

large end = 10 cm

high = 15 cm

small end is held = 600 K

large end at = 300 K

thermal conductivity of asbestos  = 0.173 W/mK

solution

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S = S1 + \frac{S2-S1}{L} x     ...............1

here x is distance from small end and S1 is side of square at small end

and S2 is side of square of large end and L is length

put here value and we get

S = 5 + \frac{10-5}{15} x

S = \frac{0.15 + x}{3}    m

and  

now we get here Area of section at distance x is

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area A = (\frac{0.15 + x}{3})^2    m²

and

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so as per fourier law

heat conduction is express as

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put here value and we get

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it will be express as

q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)  

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q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W

5 0
3 years ago
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