(a) A low carbon steel stock of thickness 25 mm is to be rolled in two stages. In the first stage, the reduction is to be from 25…

(a) For flat rolling, use the bite geometry of a roll pair. Let h0 be the incoming thickness, hf the outgoing thickness, R the roll radius, and α the bite angle. The thickness reduction is Δh = h0 - hf. From the entry geometry, the vertical projection of the contact arc equals the reduction, so Δh = R - R cos α = R(1 - cos α). For a specified reduction, a smaller roll radius gives a larger bite angle; therefore the minimum roll radius occurs when α is the maximum permitted bite angle. The bite condition is that friction must be sufficient to draw the stock into the gap, μ ≥ tan α. At the limiting maximum bite angle, the minimum required coefficient of friction is μ = tan α. These relations assume flat rolls, negligible spread, and no slip at the exit.

• Stage 1: h0 = 25 mm, hf = 15 mm, so Δh1 = 10 mm. α1 = 35°. 1 - cos 35° = 0.180848. R1 = 10 mm / 0.180848 = 55.29 mm. D1 = 2R1 = 110.59 mm. μ1 = tan 35° = 0.700.

• Stage 2: h0 = 15 mm, hf = 5 mm, so Δh2 = 10 mm. α2 = 25°. 1 - cos 25° = 0.093692. R2 = 10 mm / 0.093692 = 106.73 mm. D2 = 2R2 = 213.46 mm. μ2 = tan 25° = 0.466.

The exact symbolic forms are D1 = 20/(1 - cos 35°) mm and D2 = 20/(1 - cos 25°) mm. The second stage needs a larger diameter because the same 10 mm reduction is imposed with a smaller bite angle. The numerical values are rounded to three decimal places for diameter and three significant figures for friction; the exact trigonometric expressions should be used if higher precision is required.

Final (a): minimum roll diameters are 110.59 mm and 213.46 mm; minimum coefficients of friction are 0.700 and 0.466.

(b) Use the welding heat balance and the standard efficiency definitions. The electrical power generated at the arc is P arc = V I. The heat transfer efficiency η transfer is the fraction of this arc heat that enters the workpiece, so P work = η transfer P arc. The volume of weld metal melted per second is the joint cross-sectional area multiplied by the travel speed, q v = A v. The heat required to melt that volume per second is P melt = h m q v, where h m is the heat required per unit volume. When heat transfer efficiency is given separately, melting efficiency is the fraction of the heat received by the workpiece that is actually used for melting: η melt = P melt / P work.

• P arc = 15 V × 150 A = 2250 W = 2250 J/s.
• P work = 0.75 × 2250 J/s = 1687.5 J/s.
• q v = 15 mm² × 5 mm/s = 75 mm³/s.
• P melt = 10 J/mm³ × 75 mm³/s = 750 J/s.
• η melt = 750 J/s / 1687.5 J/s = 0.4444 = 4/9.

The volt-ampere product is in watts because 1 V × 1 A = 1 W. The travel speed converts the linear welding rate into a volumetric rate: in one second the torch advances 5 mm, and the 15 mm² cross-section gives a 5 mm long weld segment of volume 75 mm³. The given heat required to melt is treated as the total heat needed to bring the joint metal to the molten state and maintain the observed weld volume; no separate latent heat or specific heat calculation is needed. As a check, the overall efficiency relative to the total arc heat is η transfer × η melt = 0.75 × 4/9 = 1/3 = 33.33 %. The melting efficiency asked for is the value relative to the heat received by the workpiece.

Final (b): melting efficiency is 44.44 %. This assumes the observed joint cross-section is fully melted and no further heat loss is deducted after the 0.75 transfer efficiency.

(c) Use first-order exponential smoothing. The recurrence is: forecast for the next month = α × actual demand of the current month + (1 - α) × forecast for the current month. Here α = 0.25, so the latest actual demand receives 25 % weight and the previous forecast receives 75 % weight. The forecast must be updated month by month because each new forecast uses the immediately preceding forecast. If the April forecast is F-Apr and the April actual is A-Apr, then the May forecast is α A-Apr + (1 - α) F-Apr; the same form is repeated with May and June data. No separate initialization is needed because the April forecast is already given. The result is not rounded in the recurrence; rounding only at the end avoids accumulating small errors.

Given forecast for April = 150 parts.

• Forecast for May: 0.25 × 250 + 0.75 × 150 = 62.5 + 112.5 = 175 parts.
• Forecast for June: 0.25 × 100 + 0.75 × 175 = 25 + 131.25 = 156.25 parts.
• Forecast for July: 0.25 × 200 + 0.75 × 156.25 = 50 + 117.1875 = 167.1875 parts.

The exact decimal is 167.1875 parts. In practical planning it may be rounded to 167.19 parts or 167.2 parts, depending on the required precision.

Final (c): forecast for July is 167.19 parts.

(d) Use mixed-model line balancing. The cycle time is C = 15 s. A dash in the table means that the task is not required for that product, so its time is taken as 0 s. With no precedence constraints specified, and with tasks treated as indivisible, tasks may be grouped into workstations provided that, for every product, the sum of the task times assigned to any one station is not greater than the cycle time. The required total number of stations is the smallest integer for which such a grouping is feasible.

First compute the total work content for each product:

• Product 1: 5 + 6 + 6 + 7 = 24 s.
• Product 2: 5 + 6 + 7 + 4 = 22 s.
• Product 3: 6 + 6 + 4 = 16 s.
• Product 4: 5 + 6 + 4 = 15 s.
• Product 5: 5 + 6 + 7 = 18 s.

A quick lower bound is N ≥ maximum of Wp/C, rounded up, where Wp is the total work content of product p. Here the largest ratio is 24/15 = 1.6, so the bound gives 2. This bound is necessary but not sufficient; an actual assignment must also be shown. Since Product 1 requires 24 s of work, which is greater than 15 s, one workstation cannot complete Product 1 within the cycle time. Hence at least two workstations are necessary.

Now test whether two workstations can be balanced. Assign:

• Station 1: tasks A and B.
• Station 2: tasks C, D and E.

Check the station load for each product:

• Product 1: Station 1 = 5 + 6 = 11 s; Station 2 = 6 + 7 + 0 = 13 s.
• Product 2: Station 1 = 5 + 6 = 11 s; Station 2 = 0 + 7 + 4 = 11 s.
• Product 3: Station 1 = 0 + 6 = 6 s; Station 2 = 6 + 0 + 4 = 10 s.
• Product 4: Station 1 = 5 + 0 = 5 s; Station 2 = 6 + 0 + 4 = 10 s.
• Product 5: Station 1 = 5 + 6 = 11 s; Station 2 = 0 + 7 + 0 = 7 s.

Every station load is less than or equal to 15 s. Therefore two stations are feasible. The assignment is not unique; any other grouping that keeps every product load at or below 15 s would also prove that two stations are sufficient. Because one station is impossible and two stations are feasible, the minimum feasible total number of workstations is two. If precedence constraints were imposed, the number could increase, but none are given in the question.

Final (d): total number of work stations is 2.

(e) Use the basic hole system. In this system, the basic size is taken as the lower limit of the hole, and the lower deviation of the hole is zero. Here the hole limits are 57.50 mm and 57.52 mm, so the basic size is 57.50 mm. The shaft limits are 57.45 mm and 57.47 mm. The hole tolerance is the difference between the upper and lower hole limits. The shaft tolerance is the difference between the upper and lower shaft limits. Because the smallest hole, 57.50 mm, is larger than the largest shaft, 57.47 mm, the fit is a clearance fit. The allowance for a clearance fit is the minimum clearance, i.e. the smallest hole size minus the largest shaft size.

The deviations from the basic size are: hole upper deviation +0.02 mm, hole lower deviation 0; shaft upper deviation -0.03 mm, shaft lower deviation -0.05 mm. These deviations confirm that the shaft is always smaller than the hole, so no interference is possible. The basic hole system is used here because the hole lower limit is the reference size; the shaft limits are then interpreted as negative deviations from that reference.

• Hole tolerance = 57.52 mm - 57.50 mm = 0.02 mm.
• Shaft tolerance = 57.47 mm - 57.45 mm = 0.02 mm.
• Allowance = 57.50 mm - 57.47 mm = 0.03 mm.

For completeness, the maximum clearance is 57.52 mm - 57.45 mm = 0.07 mm, but the requested allowance is the minimum clearance.

Final (e): allowance = 0.03 mm, hole tolerance = 0.02 mm, shaft tolerance = 0.02 mm.