(a) TIG welding of two sheets of 5 mm thickness is performed using welding current (I) of 200 A and arc voltage (V) of 12 V…

(a) Given: welding current I = 200 A, arc voltage V = 12 V, heat utilisation factor η = 0.70, and heat required for melting steel H = 20 J/mm³.

(a)(i) Power of welding arc Using the arc power relation:

• P_arc = V × I
• P_arc = 12 V × 200 A = 2400 W

Final answer: P_arc = 2400 W

(a)(ii) Rate at which energy is delivered for melting Only 70% of the arc heat is utilised for melting. Therefore:

• Q_melting = η × P_arc
• Q_melting = 0.70 × 2400 W = 1680 W

Final answer: Q_melting = 1680 W

(a)(iii) Volume rate at which weld metal is produced Using the energy balance method, the volume rate is energy delivered for melting divided by heat required per unit volume:

• Volume rate = Q_melting / H
• Volume rate = 1680 J/s ÷ 20 J/mm³ = 84 mm³/s

Final answer: Volume rate = 84 mm³/s

The sheet thickness of 5 mm is not needed for this volume-rate calculation. It would be required only if welding speed or weld cross-sectional area were asked. The result assumes steady arc conditions and that the 70% utilised heat goes entirely into melting the base metal.

(b) Use Taylor’s tool life equation:

• V Tⁿ = C where V is cutting speed, T is tool life, n is the Taylor exponent, and C is a constant for the tool-work pair.

Given:

• V₁ = 60 m/min
• T₁ = 3 hours
• n = 0.27

(b)(i) Life of tool at cutting speed 80 m/min For the same tool-work pair, C is constant. Therefore:

• V₁ T₁ⁿ = V₂ T₂ⁿ
• 60 × 3⁰·²⁷ = 80 × T₂⁰·²⁷
• T₂⁰·²⁷ = (60/80) × 3⁰·²⁷ = 0.75 × 3⁰·²⁷

Now, 3⁰·²⁷ ≈ 1.3453. Hence:

• T₂⁰·²⁷ = 0.75 × 1.3453 = 1.00898
• T₂ = (1.00898) raised to the power 1/0.27
• T₂ ≈ 1.0337 hours
• T₂ ≈ 1.0337 × 60 = 62.02 minutes

Final answer: Tool life ≈ 1.034 hours ≈ 62.0 minutes

(b)(ii) Cutting speed for tool life of 2 hours Again use Taylor’s equation:

• V₂ = V₁ × (T₁/T₂)ⁿ
• V₂ = 60 × (3/2)⁰·²⁷
• V₂ = 60 × 1.5⁰·²⁷

Now, 1.5⁰·²⁷ ≈ 1.1157. Therefore:

• V₂ = 60 × 1.1157 = 66.94 m/min

Final answer: Cutting speed ≈ 66.94 m/min

These results are valid provided the same tool material, workpiece material, feed, depth of cut, and coolant conditions are maintained, and n remains 0.27.

(c) Production industries carry inventories for several operational, economic and strategic reasons. The main reasons are:

• Raw material inventory: It covers the lead time between placing an order and receiving the material. It protects production against supplier delays, transport problems, strikes and material shortages.
• Work-in-process inventory: It decouples successive operations. If one machine breaks down or runs slower, the next stage can continue for some time. This smooths production flow and reduces idle time.
• Finished goods inventory: It allows immediate delivery to customers, improves customer service, and helps meet demand during periods when production is stopped or capacity is insufficient.
• Safety stock: It acts as a buffer against uncertain demand, forecast errors, machine breakdowns, quality rejections and supply delays. It prevents stockouts and loss of sales.
• Cycle stock: Industries often purchase or produce in lots. Larger lots may give quantity discounts, lower ordering costs, lower setup costs, and better transportation economies.
• Speculative inventory: Firms may buy extra material when prices are expected to rise or when seasonal availability is limited. This hedges against inflation and price fluctuations.
• Pipeline or transit inventory: Goods moving between plants, warehouses, suppliers and customers must be held in transit. This is unavoidable in a distributed supply chain.
• Maintenance, repair and operating supplies: Spare parts, lubricants, tools and consumables are kept so that machines and facilities can be maintained without long delays.
• Smoothing production: Inventory allows production to remain level even when demand varies. This reduces frequent hiring, firing, overtime and idle capacity.
• Strategic and competitive reasons: Inventory ensures supply security, supports new product launches, maintains bargaining power with suppliers, and helps in meeting sudden large orders.

However, excess inventory increases holding cost, obsolescence risk, insurance, storage space and tied-up working capital. Therefore, industries try to balance inventory levels against service level and cost.

(d) Given sales for the first five months:

• S₁ = 5 units
• S₂ = 17 units
• S₃ = 29 units
• S₄ = 41 units
• S₅ = 39 units

(d)(i) Sales forecast for the next month

• Last value method: Forecast for month 6 = last observed value = S₅ = 39 units. Final answer: 39 units

• Averaging method: Forecast = (S₁ + S₂ + S₃ + S₄ + S₅) / 5 = (5 + 17 + 29 + 41 + 39) / 5 = 131 / 5 = 26.2 units. Final answer: 26.2 units

• Moving average method with the 3 most recent months: Forecast = (S₃ + S₄ + S₅) / 3 = (29 + 41 + 39) / 3 = 109 / 3 = 36⅓ units ≈ 36.33 units. Final answer: 36.33 units

(d)(ii) Suitability of the methods The sales pattern shows a strong upward trend for the first four months: 5 → 17 → 29 → 41, an increase of 12 units each month. In the fifth month, sales fall slightly to 39. This suggests either a turning point or a random fluctuation after a strong rise.

• The averaging method is clearly inappropriate here. It gives equal weight to old low values and recent high values. Because of the strong upward trend, it heavily underestimates the next month’s sales. Its forecast of 26.2 units is far below the latest values of 41 and 39 units.

• The last value method may also be inappropriate if the fall from 41 to 39 is only random. It ignores the earlier trend and reacts only to the latest observation. It would be reasonable only if the process has just shifted to a new stable level near 39 units.

• The 3-month moving average is better than simple averaging, but it still lags behind a trend. It includes the relatively low value of 29, so its forecast, 36.33 units, is below the recent 39–41 units. It cannot quickly capture a sudden turning point.

Therefore, the simple averaging method is unsuitable. The last value and moving average methods should be used cautiously. A trend-adjusted forecasting method, such as regression or exponential smoothing with trend, would be more appropriate for this sales pattern.

(e) The hole is to be machined to diameter 30⁺⁰·⁰⁶ mm. Therefore:

• Lower limit of hole = 30.000 mm
• Upper limit of hole = 30.060 mm
• Work tolerance = 30.060 − 30.000 = 0.060 mm

Given:

• Gauge tolerance = 10% of work tolerance = 0.10 × 0.060 = 0.006 mm
• Wear allowance = 10% of work tolerance = 0.006 mm

For checking a hole, a plug gauge is used. In the unilateral system of tolerances, the GO gauge checks the lower limit and the NO-GO gauge checks the upper limit.

GO plug gauge: The wear allowance is added to the lower limit of the hole:

• Basic GO size = 30.000 + 0.006 = 30.006 mm
• Gauge tolerance is placed on the plus side, so:
• GO gauge limits = 30.006 mm to 30.006 + 0.006 = 30.012 mm

GO gauge dimensions: 30.006 mm to 30.012 mm

NO-GO plug gauge: The basic size is the upper limit of the hole:

• Basic NO-GO size = 30.060 mm
• Gauge tolerance is placed on the minus side, so:
• NO-GO gauge limits = 30.060 − 0.006 = 30.054 mm to 30.060 mm

NO-GO gauge dimensions: 30.054 mm to 30.060 mm

Wear allowance is normally not added to the NO-GO gauge because wear makes it smaller, which only causes additional rejection of good holes, not acceptance of oversize holes.

Final answer: GO plug gauge = 30.006 mm to 30.012 mm; NO-GO plug gauge = 30.054 mm to 30.060 mm.