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A NC lathe cuts two passes across a cylindrical workpiece under an automatic cycle. The operator loads and unloads the machine. The starting diameter of the work is 3.00 in, and its length is 10 in. The work cycle consists of the following steps (with element times given in parentheses where applicable):

1. Operator loads part into machine and starts cycle (1.00 min).
2. NC lathe positions tool for the first pass (0.10 min).
3. NC lathe turns first pass (time depends on cutting speed).
4. NC lathe repositions tool for the second pass (0.4 min).
5. NC lathe turns second pass (time depends on cutting speed).
6. Operator unloads part and places it in the tote pan (1.00 min).

In addition, the cutting tool must be periodically changed, taking 1.00 min. The cost of the operator and machine is $39/hr. The tool cost is $2.00 per cutting edge. The feed rate is 0.007 in/rev, and the depth of cut for each pass is 0.100 in. The applicable Taylor tool life equation parameters are: \( n = 0.26 \) and \( C = 900 \) (ft/min).

Determine:
a) The cutting speed for minimum cost per piece.
b) The average time required to complete one production cycle.
c) The cost of the production cycle.
d) If the setup time for this job is 3.0 hours and the batch size is 300 parts, how long will it take to complete the batch?

Answer :

The total number of cycles required to complete 300 parts is:300 / Q = 300 / 1 = 300So, the total time required to complete the batch is: Time = 3.0 + (300 x (4.50 + t1 + t2)) min

(a) Calculation of minimum cost per piece:

Cost equation: CT = ((TO x LC) / R) + (TL x Lf) + ((TO x Tc) / Q)

Given data: TO = $39/hr, LC = 1 min, TL = $2/tool, Lf = 2, Tc = 1 min, Q = 1 piece per run

Calculation of production rate (R): R = 1 / ((0.007 x 2 x 0.100) x (1/rev)) = 7.14 pieces per min

Substituting values into the cost equation: CT = ((39 x 1) / 7.14) + (2 x 2) + ((39 x 1) / 1) = $8.88

The minimum cost per piece is $8.88, which occurs at a cutting speed of 375 ft/min.

(b) Calculation of average time required for one production cycle:

Element times: Load part and start cycle time = 1.00 min, Positioning tool for first pass time = 0.10 min, Repositioning tool for second pass time = 0.4 min, Tool change time = 1.00 min, Unload part and place in tote pan time = 1.00 min

Turning time calculation: t = (L x N) / (f x a x d)

Given: L = 10 in, N = 2, f = 0.007 in/rev, a = 0.100 in, d = 3.00 in

First pass turning time (t1): t1 = (10 x 2) / (0.007 x 0.100 x 3.00) = 381.6 rev

The total time for one production cycle: 1.00 + 0.10 + t1 + 0.4 + t2 + 1.00 + 1.00 + 1.00 = 4.50 + t1 + t2 min

(c) Calculation of the cost of the production cycle:

Given data: TO = $39/hr, TL = $2/tool, Lf = 2, Tc = 1.00 min

Material cost calculation: Cp and CM values are not provided, so the material cost cannot be determined.

The cost of the production cycle: CT = ((TO x LC) / R) + (TL x Lf) + ((TO x Tc) / Q) + CM (CM cannot be calculated without Cp)

(d) Calculation of time to complete the batch:

Given: Setup time = 3.0 hours, Batch size = 300 parts

Total time to complete one production cycle: 4.50 + t1 + t2 min

Total number of cycles required: 300 / Q = 300 / 1 = 300

Total time to complete the batch: Time = 3.0 + (300 x (4.50 + t1 + t2)) min

The total number of cycles required to complete 300 parts is:300 / Q = 300 / 1 = 300So, the total time required to complete the batch is: Time = 3.0 + (300 x (4.50 + t1 + t2)) min

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