when testing transistors, how much minimum voltage difference should be seen across the base-emitter junction? A. 0.1 VDC B. 0.4
1 answer:
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Answer:
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Explanation:
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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