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

(a) Material removal rate and feed rate in electrochemical machining

Given: ρ_e = 5 Ω·cm, V = 15 V, A = 20 mm × 20 mm = 4 cm², gap g = 0.3 mm = 0.03 cm, z = 2, atomic weight 55.85, density 7.86 g/cm³, F = 96 540 C, current efficiency 100%.

Gap resistance: R = ρ_e g/A = 5 × 0.03/4 = 0.0375 Ω. Current: I = V/R = 15/0.0375 = 400 A.

By Faraday's law, mass removal rate = IM/(zF) = 400 × 55.85/(2 × 96 540) = 0.1157 g/s. Volume removal rate = 0.1157/7.86 = 0.01472 cm³/s, so MRR = 14.72 mm³/s (883 mm³/min). Feed rate = MRR/A = 0.01472/4 = 0.00368 cm/s, so f = 0.0368 mm/s (2.21 mm/min).

This assumes a constant gap, no polarisation losses and uniform current density.

(b) Time for a stainless steel surface to reach melting under a laser beam

Absorbed flux: q = 0.1 × 2 × 10⁵ W/mm² = 2 × 10⁴ W/mm² = 2 × 10⁶ W/cm². For a semi-infinite solid under constant surface flux, the surface temperature rise is ΔT = (2q/k)√(αt/π), with α = k/C_v. Hence t = πkC_vΔT²/(4q²).

With k = 0.27 W/cm·°C and C_v = 3.36 J/cm³·°C, and taking the initial temperature as 0°C so ΔT = 1455°C: t = π × 0.27 × 3.36 × 1455²/[4 × (2 × 10⁶)²] = 6.03 × 10⁶/1.6 × 10¹³ = 3.77 × 10⁻⁷ s. If the sheet starts at 25°C, ΔT = 1430°C and t = 3.64 × 10⁻⁷ s, essentially the same.

(c) Trade-off between cost of conformance and cost of non-conformance

Cost of quality has two groups. Cost of conformance is prevention (design review, training, process capability studies, supplier development) plus appraisal (inspection, testing, audits). Cost of non-conformance is internal failure (scrap, rework, downtime) plus external failure (warranty claims, returns, recalls, liability, loss of goodwill).

At a low quality level, failure costs dominate. Each extra rupee spent on prevention and appraisal cuts defects and removes more than a rupee of failure cost, so total quality cost falls. Beyond an optimum, extra conformance spending such as over-inspection or over-specification yields diminishing returns, and total cost rises again. The total cost curve is therefore U-shaped, with its minimum where marginal conformance cost equals marginal non-conformance cost.

Prevention is normally cheaper than failure, and continuous improvement shifts the optimum towards higher quality at lower total cost. Zero defects is not always economic if customer requirements are already met.

(d) Assembly line balancing for 50 units per hour

(i) Network diagram. A precedes both B and C; B leads to D and E; C leads to E and F; F leads to G; G leads to H. E needs both B and C, and no dummy activity is required. D and H are the end activities.

(ii) Cycle time = 3600/50 = 72 s/unit.

(iii) Total task time = 55 + 30 + 22 + 35 + 50 + 15 + 5 + 10 = 222 s. Theoretical minimum stations = ⌈222/72⌉ = ⌈3.08⌉ = 4. Longest-time method with cycle time 72 s: Station 1: A (55 s), idle 17 s, since neither B (30) nor C (22) fits. Station 2: B (30 s) and D (35 s), total 65 s, idle 7 s. Station 3: C (22 s) and E (50 s), total 72 s, idle 0. Station 4: F, G and H (15 + 5 + 10 = 30 s), idle 42 s. Four stations are used, equal to the theoretical minimum, so no extra station is needed.

(iv) Efficiency = 222/(4 × 72) = 222/288 = 77.1%. Balance delay = 1 − 0.771 = 22.9%.

(e) Exponential regression forecast of demand

Take logs of y = ae^(bx): ln y = ln a + bx. With x = 1 to 5 and y = 40, 60, 200, 700, 1000, ln y = 3.689, 4.094, 5.298, 6.551, 6.908. n = 5, Σx = 15, Σx² = 55, Σln y = 26.540, Σx ln y = 88.516. b = (5 × 88.516 − 15 × 26.540)/(5 × 55 − 15²) = 44.472/50 = 0.8894. ln a = (26.540 − 15 × 0.8894)/5 = 2.6397, so a = 14.01. The regression equation is y = 14.01 e^(0.8894x). For year 6: ln y = 2.6397 + 0.8894 × 6 = 7.976, so y ≈ 2911 units (about 2900).