DISCUSSION OF PROBLEMS AND ASSIGNED HOME WORK IN CHAPTER 3

PROFESSOR: Dr. Yaw A. Owusu

ASSIGNED HOME WORK: 3.4, 3.7, 3.8, 3.10, 3.15, 3.16, 3.17

DISCUSSED PROBLEMS:

3.1 It is not unusual to find the output of a production process to naturally orient piece parts in an orderly fashion. The classic example is the attachment of injection moldings or die castings to a mold skeleton. To remove the moldings or castings from the mold skeleton is a step which may add cost to the process while it destroys the orderly orientation of the piece parts for the next process. In such cases it can be cheaper to receive piece parts shipped from the vendor in correct orientation, still attached to the mold skeleton.

3.2 Vibration, friction, gravity, centrifugal force, magnetic force, and mechanical force imparted by moving machine parts such as hooks, blades, and wheels. The most versatile and popular parts feeder is the vibratory bowl.

3 . 3 The rotating base hopper uses a stationary hook to gently push the parts to the outer edge of the rotating hopper for gentle feeding of delicate cylindrical parts. The centrifugal hopper uses rapid rotation of the hopper to throw the cylindrical parts against the outer edge of the hopper by centrifugal force for delivery into the discharge chute.

3.4 There would be no working range. Note in Figure 3.15 that there would be no range of slot widths for which the heavy-end-first parts and the light-end-first parts would both fall correctly (light-end-first).

3.5 Reversing the distribution of the input stream increases the number of "passovers" from the light-end-first portion of the input steam. Since the slot-width remains unchanged, all of the pieces that fall do fall correctly. So the effectiveness remains 100%.

a) Effectiveness = 100%
b) Efficiency calculted as follows:

Efficiency = F(0) / F(1)

(c) Output flow rate (i.e., pieces falling in the slot, all of which will be correctly oriented)
i.e = (2000 pcs/hr) x 0.7144 = 1428.8 pcs/hr.
Note:The remaining 2000 - 1428.8 = 571.2 pcs/hr will be light-end-first pieces passing over the slot.

Alternatively, number of passovers as follows:
# of Passovers = F(1) x (% of F(1) entering light-end-first) x (1 - 64.3%)
= 2000 pcs/hr x 80% x 0.357
=571.2 pcs/hr --- must be same as computed above.

3. 6 At a slot width of 0. 65 in., all pieces f rom both the heavy-end-first stream and light-end-first stream eventually fall correctly. True, some of the light-end-first stream will at first bounce back because the slot width is between points G and H. Bounce-backs, however, have nowhere to go except to work their way forward and eventually fall correctly. Efficiency and effectiveness are both 100% and the output flow rate is equal to the full input flow rate of 2000 pcs/hr. Note that we are assuming that the portion of the light-end-first stream that bounces back does not impede the input flow of 2000 pcs per hour.