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 t
1 answer:
Answer:
A) 105.7 rpm
B) 11.32 kw
C) 20.85 NPa
Explanation:
Number of solid rods to be manufactured = 3600
<u>a) Determine the RPM of the screw</u>
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 ) = ------ ( 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
<em>Hence the RPM of the screw ( N )</em>
= Q / Qd = 31.415 * 10^-6 / 17.83 * 10^-6 = 1.76 rev/s = 105.7 rpm
<u>b) Determine the power requirements of the extruder</u>
max power requirement = Pm * A * πDN / 60
= ( 20.85 * π * ( 50 )^2 / 4 ) * π * 150 *1.76
max power requirement = 11.32 kw
<u>c) What is the pressure buildup within the extruder</u>
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
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Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
