A screw extruder is 50 mm in diameter, 1 m long, has a 50mm lead, a channel 5 mm deep and a flight 3 mm wide. The circular die through which the extruded material forms the shape of a rod is of diameter 4 centimeters and length 5 cms. The viscosity of the thermoplastic fiber suspension that goes through the die to form the rod is 100 Pa.s.
If you want to manufacture 3600 solid rods of diameter 4 centimeters and length 25 cms each day in a shift of 10 hours what should be the RPM of the screw? Also find the power requirements for this extruder. What will be the pressure build up within the extruder?

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batolisis batolisis · Answer 1 · 2021-04-24T19:32:56+02:00

Answer:

A) 105.7 rpm

B) 11.32 kw

C) 20.85 NPa

Explanation:

Number of solid rods to be manufactured = 3600

a) Determine the RPM of the screw

we will apply the relation below

discharge rate ( Qd ) = 0.5 π^2 * D^2 * N di * sinA * cos A ------- ( 1 )

where : D = 50 mm , di = 5 mm , N = ?

Tan A = p / πD = 50 / π*50 ∴ A = 17.65°

Insert values into equation ( 1 )

Qd = 17.83 * 10^-6 * N

required discharge rate ( Q ) = [tex]\frac{\frac{\pi D^2}{4}*L*N }{Time}[/tex] ------ ( 2 )

where : D = 0.01 , L = 25 * 10^-2 , N = 3600 , time = 10 * 3600

input value into 2

Q = 31.415 * 10^-6 m^3/s

Hence the RPM of the screw ( N )

= Q / Qd = 31.415 * 10^-6 / 17.83 * 10^-6 = 1.76 rev/s = 105.7 rpm

b) Determine the power requirements of the extruder

max power requirement = Pm * A * πDN / 60

= ( 20.85 * π * ( 50 )^2 / 4 ) * π * 150 *1.76

max power requirement = 11.32 kw

c) What is the pressure buildup within the extruder

Pressure buildup within the extruder = ( 6π*D*N*L* η * cot A ) / di^2

= ( 6π * 0.05 * 1.76 * 1 * 100 * cot17.65 ) / ( 5 * 10^-3 )^2

therefore ; Pm = 20.85 NPa