Question

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?

Answer 1

A) 105.7 rpm

B) 11.32 kw

C) 20.85 NPa

Step-by-step 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 ) =
`((\pi D^2)/(4)*L*N )/(Time)` ------ ( 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