(a) In an electrochemical machining process of an iron surface using sodium chloride solution as electrolyte, following…

(a) Given: electrolyte specific resistance ρ = 5 Ω cm, voltage V = 15 V, workpiece area A = 20 mm × 20 mm = 400 mm² = 4 cm², gap g = 0.3 mm = 0.03 cm, valency z = 2, atomic weight M = 55.85 g/mol, density ρm = 7860 kg/m³ = 7.86 g/cm³, Faraday constant F = 96540 C, current efficiency η = 1.

Using Ohm’s law for the electrolyte gap, R = ρg/A = 5 × 0.03 / 4 = 0.0375 Ω. Current, I = V/R = 15 / 0.0375 = 400 A.

By Faraday’s law of electrolysis, mass removal rate is ṁ = η I M / (z F) = 400 × 55.85 / (2 × 96540) = 0.1157 g/s = 1.157 × 10⁻⁴ kg/s.

Volume removal rate, Q = ṁ / ρm = 0.1157 / 7.86 = 0.01472 cm³/s = 14.72 mm³/s.

Thus material removal rate = 14.72 mm³/s = 883.2 mm³/min.

Electrode feed rate is f = Q/A = 0.01472 / 4 = 0.00368 cm/s = 0.0368 mm/s = 2.208 mm/min.

This assumes 100% current efficiency, constant gap, no polarization, and uniform current density.

(b) Given incident intensity I₀ = 2 × 10⁵ W/mm². Only 10% is absorbed, so absorbed heat flux is q = 0.1 × 2 × 10⁵ = 2 × 10⁴ W/mm² = 2 × 10⁶ W/cm².

For a semi-infinite solid heated by a constant surface heat flux, the surface temperature rise is ΔT = (2q/k) √(αt/π), where α = k/(ρc) = k/Cv.

Here k = 0.27 W/cm-°C, Cv = 3.36 J/cm³-°C. Taking initial temperature as 0°C, as no ambient value is given, ΔT = 1455°C.

Solving for time, t = π k Cv ΔT² / (4q²) = π × 0.27 × 3.36 × (1455)² / [4 × (2 × 10⁶)²] = 6.0336 × 10⁶ / 1.6 × 10¹³ = 3.77 × 10⁻⁷ s.

If initial temperature is taken as 25°C, ΔT = 1430°C and t ≈ 3.64 × 10⁻⁷ s.

(c) Cost and quality trade-off is analysed through cost of quality, divided into cost of conformance and cost of non-conformance.

Cost of conformance includes prevention costs and appraisal costs. Prevention costs are incurred to stop defects before they occur: design review, training, supplier development, process capability studies and quality planning. Appraisal costs are incurred to detect defects: inspection, testing, audits and gauging.

Cost of non-conformance includes internal failure and external failure. Internal failure costs arise before delivery: scrap, rework, re-inspection, downtime and yield loss. External failure costs arise after delivery: warranty claims, returns, recalls, liability, complaint handling and loss of goodwill.

At low quality levels, non-conformance costs dominate because defects, rework and customer complaints are high. Spending more on prevention and appraisal reduces defects and hence reduces failure costs. Initially the reduction in failure costs exceeds the extra conformance cost, so total quality cost falls.

Beyond an optimum point, additional conformance spending gives diminishing returns. Over-inspection, excessive testing, over-specification and slowed throughput raise cost without proportional reduction in failure cost. The total quality cost curve is therefore roughly U-shaped, with a minimum where marginal conformance cost equals marginal non-conformance cost. Modern quality practice emphasises prevention because it is usually cheaper than failure, and continuous improvement can shift the optimum toward higher quality and lower total cost. However, zero defects may not always be economically optimal if customer requirements are already met.

(d)(i) The network diagram is:

A → B A → C B → D B → E C → E C → F F → G G → H

Thus E depends on both B and C, D depends on B, F depends on C, and H is reached through G. No dummy activity is needed.

(d)(ii) Desired output = 50 units/hour. Cycle time, C = available time per hour / output = 3600 / 50 = 72 seconds/unit.

(d)(iii) Total task time = 55 + 30 + 22 + 35 + 50 + 15 + 5 + 10 = 222 seconds.

Theoretical minimum number of workstations: Nmin = ceil(222/72) = ceil(3.083) = 4.

Using the longest-time method, assigning tasks without exceeding 72 seconds and respecting precedence:

• Station 1: A (55 s). Idle = 17 s.
• Station 2: B (30 s), D (35 s). Total = 65 s. Idle = 7 s.
• Station 3: C (22 s), E (50 s). Total = 72 s. Idle = 0 s.
• Station 4: F (15 s), G (5 s), H (10 s). Total = 30 s. Idle = 42 s.

Number of workstations used = 4. This is equal to the theoretical minimum, so no, more workstations than the theoretical minimum were not used.

(d)(iv) Efficiency = total task time / (actual stations × cycle time) = 222 / (4 × 72) = 222 / 288 = 0.7708 = 77.08%.

Balance delay = 1 − efficiency = 1 − 0.7708 = 0.2292 = 22.92%.

(e) For exponential demand, y = a exp(bx). Taking natural logarithms, ln y = ln a + b x.

Data: x = 1, 2, 3, 4, 5 y = 40, 60, 200, 700, 1000 ln y = 3.6889, 4.0943, 5.2983, 6.5511, 6.9078.

Sums: n = 5 Σx = 15 Σx² = 55 Σln y = 26.54038 Σx ln y = 88.51562.

Normal equations: 26.54038 = 5 ln a + 15b 88.51562 = 15 ln a + 55b.

Solving, b = (5 × 88.51562 − 15 × 26.54038) / (5 × 55 − 15²) = 44.47244 / 50 = 0.88945.

ln a = (26.54038 − 15 × 0.88945) / 5 = 2.63973.

So a = exp(2.63973) = 14.009.

Thus the regression equation is y = 14.009 exp(0.88945x).

For the 6th year, x = 6: ln y₆ = 2.63973 + 0.88945 × 6 = 7.97642.

y₆ = exp(7.97642) = 2911.5 units.

Forecast demand for the 6th year = approximately 2912 units.