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Mechanical Engineering 2025 Paper I 50 marks Calculate

Paper I — Q6

(a) (i) If the power source characteristic in a Metal Inert Arc welding process is V_P = 38 - I/60, and the arc characteristic is…

(a)
(i)

If the power source characteristic in a Metal Inert Arc welding process is V_P = 38 - I/60, and the arc characteristic is V_a = 3L_a + 27, where V_P and V_a is voltage, I is current and L_a is arc length in mm. Calculate the change in power of the arc if the arc length is changed from 1 mm to 3 mm.

If the maximum current capacity of the power source is 300 Amps, then determine the maximum arc length that can be sustained. 10 marks

(ii)

An electro-discharge machining process is used for cutting a 6 mm deep cavity in a high carbon steel workpiece using the following:

(I) Copper-tungsten electrode, and

(II) Copper electrode.

Assuming the wear ratio for copper-tungsten electrode as 9 : 1 and for copper electrode as 3 : 1, determine the required spindle movement for cutting this cavity. 10 marks

(b)
(i)

A long hole having 15 mm diameter and 125 mm depth is required to be drilled in high carbon steel using electro-chemical machining process. Calculate the time required to drill this hole if the supplied current magnitude is 45 Amp and electrolyte used is 15% NaCl. Consider that the valency of iron is 2, atomic weight of iron is 56 and density of steel is 7·8 gm/cm³. 10 marks

(ii)

An electro-chemical machining process is used for machining of Nimonic 75 alloy. The composition (% by weight) of Nimonic 75 alloy is given here:

NiCrFeSiMnCuTi
72·519·55·01·01·00·60·4

Consider the following data:

MetalGram atomic weightValency of dissolutionDensity gm/cm³
Nickel58·712/38·90
Chromium51·992/3/67·19
Iron55·852/37·86
Silicon28·0942·33
Manganese54·942/4/6/77·43
Copper63·571/28·96
Titanium47·93/44·51

Using the lowest valency of dissolution for each element, determine the material removal rate, when a current of 1050 Amp is applied. 10 marks

(c)

With the help of a line diagram, explain the different types of flow patterns used in plant layouts. What are the conditions to be satisfied by an ideal flow pattern ? 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

यदि एक धातु अक्रिय आर्क वेल्डिंग प्रक्रिया में शक्ति स्रोत अभिलक्षण V_P = 38 - I/60 तथा आर्क अभिलक्षण (characteristic) V_a = 3L_a + 27 है, जहाँ V_P तथा V_a वोल्टेज है, I धारा है और L_a आर्क लंबाई mm में है। यदि आर्क की लंबाई 1 mm से 3 mm में बदलती है, तो आर्क की शक्ति में परिवर्तन की गणना कीजिए।

यदि शक्ति स्रोत की अधिकतम धारा क्षमता 300 Amps है, तो सहनीय अधिकतम आर्क लंबाई निर्धारित कीजिए। 10

(ii)

एक इलेक्ट्रो-डिस्चार्ज मशीनिंग प्रक्रिया का उपयोग उच्च कार्बन इस्पात (स्टील) कार्यखंड में 6 mm गहरे कोटर (cavity) में कटाई के लिए निम्नलिखित का उपयोग करते हुए किया जाता है :

(I) कॉपर-टंगस्टन इलेक्ट्रोड, और

(II) कॉपर इलेक्ट्रोड।

कॉपर-टंगस्टन इलेक्ट्रोड का निष्कर्षण अनुपात (wear ratio) 9 : 1 तथा कॉपर इलेक्ट्रोड के लिए 3 : 1 मानते हुए इस कोटर को काटने के लिए आवश्यक स्पिंडल गति निर्धारित कीजिए। 10

(b)
(i)

उच्च कार्बन इस्पात (स्टील) में विद्युत-रासायनिक मशीनिंग प्रक्रिया का उपयोग करके 15 mm व्यास तथा 125 mm गहराई का एक लंबा छिद्र ड्रिल करने की आवश्यकता है। यदि प्रदान (आपूर्ति) की गई धारा का परिमाण 45 Amp तथा उपयोग किया गया विद्युत-अपघट्य (Electrolyte) 15% NaCl है, तो इस छिद्र को ड्रिल करने के लिए आवश्यक समय की गणना कीजिए।

लोह की संयोजकता 2, लोह का परमाणु भार 56 तथा इस्पात का घनत्व 7·8 gm/cm³ मानिए। 10

(ii)

निमोनिक 75 मिश्रधातु के मशीनिंग के लिए एक विद्युत-रासायनिक मशीनिंग प्रक्रिया का उपयोग किया जाता है। निमोनिक 75 मिश्रधातु की संरचना (% भार द्वारा) यहाँ दी गई है:

निम्नलिखित आँकड़े मानिए :

धातुग्राम परमाणु भारविलयन की संयोजकताघनत्व gm/cm³
निकल58·712/38·90
क्रोमियम51·992/3/67·19
लोहा55·852/37·86
सिलिकॉन28·0942·33
मैंगनीज54·942/4/6/77·43
ताँबा63·571/28·96
टाइटेनियम47·93/44·51

प्रत्येक तत्व के लिए विलयन की न्यूनतम संयोजकता का उपयोग करते हुए, जब 1050 Amp की धारा प्रयुक्त की जाती है, तो पदार्थ पृथक्करण दर का निर्धारण कीजिए। 10

(c)

रेखाचित्र की सहायता से संयंत्र के अभिन्यास में उपयोग होने वाले विभिन्न प्रकार के प्रवाह प्रतिरूपों (Flow Patterns) को समझाइए । एक आदर्श प्रवाह प्रतिरूप के लिए कौन-सी शर्तें पूरी की जानी चाहिए ? 10 अंक

Q6 of the 2025 UPSC Mains Mechanical Engineering Paper I, as printed
The question as printed in the 2025 Mechanical Engineering paper
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Model answer

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(a)(i)

For stable arc, V_P = V_a. V_P = 38 − I/60, V_a = 3L_a + 27. Thus 38 − I/60 = 3L_a + 27 ⇒ I = 60(11 − 3L_a).

At L_a = 1 mm: V_a = 30 V, I = 60(11 − 3) = 480 A. P_1 = V_a I = 30 × 480 = 14,400 W = 14.4 kW.

At L_a = 3 mm: V_a = 36 V, I = 60(11 − 9) = 120 A. P_3 = 36 × 120 = 4,320 W = 4.32 kW.

Change in arc power: ΔP = P_3 − P_1 = 4,320 − 14,400 = −10,080 W. ΔP = −10.08 kW (decrease by 10.08 kW).

If maximum current capacity is 300 A: I ≤ 300 ⇒ 60(11 − 3L_a) ≤ 300 ⇒ 11 − 3L_a ≤ 5 ⇒ L_a ≥ 2 mm. So 300 A sets the minimum sustainable arc length at 2 mm. The theoretical maximum sustainable arc length is set by open-circuit voltage at I → 0: 38 = 3L_a + 27 ⇒ L_a = 11/3 mm ≈ 3.67 mm. Maximum arc length = 11/3 mm ≈ 3.67 mm; minimum due to 300 A limit = 2 mm. Note: L_a = 1 mm would require 480 A, so it is not sustainable if the source is limited to 300 A.

(a)(ii)

Let wear ratio = workpiece volume removed : electrode volume worn. For constant cross-section, this equals cavity depth : electrode wear length. Electrode wear w = cavity depth / wear ratio. Spindle movement = cavity depth + w = 6 + w.

Copper-tungsten electrode, wear ratio 9 : 1: w = 6/9 = 2/3 mm. Spindle movement = 6 + 2/3 = 20/3 = 6.667 mm.

Copper electrode, wear ratio 3 : 1: w = 6/3 = 2 mm. Spindle movement = 6 + 2 = 8 mm.

Assumption: uniform end wear and same cross-section of electrode and cavity.

(b)(i)

Volume of hole: V = (π/4)d²h = (π/4)(1.5 cm)²(12.5 cm) = 7.03125π cm³ ≈ 22.089 cm³. Mass removed: m = ρV = 7.8 × 22.089 ≈ 172.294 g.

By Faraday’s law for ECM: dm/dt = A I/(z F) = 56 × 45 / (2 × 96500) = 0.013057 g/s.

Time: t = m / (dm/dt) = 172.294 / 0.013057 ≈ 13,195.8 s = 3.665 h ≈ 3 h 40 min. t = (940875π/224) s ≈ 1.32 × 10⁴ s ≈ 3.67 h. Assumption: 100% current efficiency and uniform dissolution.

(b)(ii)

Using lowest valency: Ni = 2, Cr = 2, Fe = 2, Si = 4, Mn = 2, Cu = 1, Ti = 3. For alloy, S = Σ(w_i z_i / A_i).

Ni: 0.725 × 2/58.71 = 0.02470 Cr: 0.195 × 2/51.99 = 0.00750 Fe: 0.05 × 2/55.85 = 0.00179 Si: 0.01 × 4/28.09 = 0.00142 Mn: 0.01 × 2/54.94 = 0.000364 Cu: 0.006 × 1/63.57 = 0.000094 Ti: 0.004 × 3/47.9 = 0.000251

S ≈ 0.03612.

Mass removal rate: dm/dt = I/(F S) = 1050/(96500 × 0.03612) = 0.3012 g/s = 18.07 g/min.

Alloy density by inverse rule: 1/ρ = Σ(w_i/ρ_i) ≈ 0.12214 cm³/g ρ ≈ 8.19 g/cm³.

Volume removal rate: MRR = 0.3012 / 8.19 = 0.0368 cm³/s = 2.21 cm³/min = 2207 mm³/min.

MRR ≈ 0.301 g/s = 18.07 g/min = 2.21 cm³/min (≈2207 mm³/min).

(c)

Common flow patterns in plant layouts (line diagram, figure below):

Diagram: Plant layout flow patterns
  • Straight-line (I) flow: a simple unidirectional flow for sequential operations in a long, narrow building.
  • U-flow: the line turns back so that receiving and dispatch are at the same end, saving length and sharing docks.
  • S (serpentine or zig-zag) flow: a long line folded several times to fit a compact area.
  • L-flow: a 90° turn to suit the building shape or the location of services.
  • Circular flow: material returns to the starting point, used where containers, fixtures or pallets recirculate.
  • Comb (odd-angle) flow: branches or sub-assembly lines feed a main aisle or assembly line.

Conditions to be satisfied by an ideal flow pattern:

  • Operations must follow the required sequence.
  • No backtracking, cross-movement or congestion.
  • Shortest possible travel distance and time.
  • Unidirectional, continuous and smooth flow.
  • Minimum material handling and work-in-process.
  • Flexibility for product/volume changes.
  • Safe working, good space utilisation and easy supervision.

How this answer will be evaluated

What earns full marks: Correct application of governing equations (Faraday, Power balance) with clear steps and units.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a)(i) Change in arc power for length change and maximum sustainable arc length. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Stable arc: V_P = V_a ⇒ I = 60(11 − 3L_a)
    • L_a = 1 mm: V = 30 V, I = 480 A, P = 14.4 kW; L_a = 3 mm: V = 36 V, I = 120 A, P = 4.32 kW
    • ΔP = −10.08 kW
    • 300 A limit ⇒ minimum L_a = 2 mm; maximum sustainable L_a = 11/3 ≈ 3.67 mm (OCV)

    Loses marks

    • Fails to equate Vp and Va to find current
    • Calculates power without finding current first

    Earns more

    • Explicitly states Vp = Va at operating point
    • Shows unit consistency in calculations

    Extra mark

    • Sketches the intersection of the two characteristic curves
  2. (a)(ii) Required spindle movement for 6 mm cavity using two electrode types. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Wear ratio = workpiece : electrode volume = depth : electrode wear
    • Cu-W (9:1): wear 2/3 mm ⇒ spindle movement 6.667 mm
    • Cu (3:1): wear 2 mm ⇒ spindle movement 8 mm
    • Assumptions stated (uniform end wear, equal cross-sections)

    Loses marks

    • Confuses wear ratio direction (Electrode:Workpiece)
    • Ignores electrode wear in spindle movement calculation

    Earns more

    • Clearly distinguishes between the two electrode cases

    Extra mark

    • Notes that spindle movement compensates for electrode loss
  3. (b)(i) Time required to drill a 15 mm x 125 mm hole via ECM. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Volume 22.089 cm³; mass 172.294 g
    • Faraday: dm/dt = 56×45/(2×96500) = 0.013057 g/s
    • t = 172.294/0.013057 = 13,196 s
    • = 3.67 h (3 h 40 min); exact 940875π/224 s

    Loses marks

    • Uses wrong valency or atomic weight
    • Forgets to convert volume to mass using density

    Earns more

    • States Faraday's constant value used
    • Checks dimensional consistency of units

    Extra mark

    • Mentions efficiency factor if applicable (though not given)
  4. (b)(ii) Material removal rate for Nimonic 75 alloy at 1050 A. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Lowest valencies: Ni 2, Cr 2, Fe 2, Si 4, Mn 2, Cu 1, Ti 3
    • S = Σwz/A ≈ 0.03612
    • dm/dt = 1050/(96500×0.03612) = 0.3012 g/s = 18.07 g/min
    • ρ ≈ 8.19 g/cm³ ⇒ MRR = 2.21 cm³/min ≈ 2207 mm³/min

    Loses marks

    • Uses highest valency instead of lowest
    • Ignores the weight percentage of alloy components

    Earns more

    • Presents data in a clear table format
    • Explicitly lists the chosen lowest valencies

    Extra mark

    • Notes that MRR is proportional to current
  5. (c) Types of flow patterns in plant layouts and conditions for ideal flow. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Six flow patterns shown in the line diagram (straight, U, S, L, circular, comb)
    • Each pattern's geometry/use explained
    • Conditions for ideal flow: sequence, no backtracking/crossing, shortest movement, unidirectional/smooth
    • Flexibility, safety, space use and supervision listed

    Loses marks

    • Describes patterns without diagrams
    • Fails to distinguish between different flow types

    Earns more

    • Labels diagrams clearly (e.g., U-shape, I-shape)
    • Mentions 'no backtracking' as an ideal condition

    Extra mark

    • Provides a real-world example of a specific layout
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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