Civil Engineering 2023 Paper II 50 marks Solve

Q6

(a) (i) A 50 cm well in an unconfined aquifer of saturated thickness 45 m yields 600 lpm under a drawdown of 3 m at the pumping well. What will be the discharge under a drawdown of 6 m ? Consider the radius of influence as 500 m. (ii) What will be the discharge in a 30 cm well under a drawdown of 3 m for the unconfined aquifer as mentioned in part (i). (20 marks) (b) How does the peak hour demand affect the design of a water supply scheme ? Sketch the fluctuation in demand for typical Indian conditions. (15 marks) (c) (i) The B.O.D. of a sewage incubated for one day at 25°C has been found to be 100 mg/l. What will be the 5 day, 20°C B.O.D. ? Assume K₂₀ = 0·12 at 20°C. Take temperature coefficient, φ = 1·056. (ii) Despite widespread use of B.O.D., it has some limitations. Mention all those limitations. (15 marks)

हिंदी में प्रश्न पढ़ें

(a) (i) 45 m की संतृप्त मोटाई के एक अपरिरुद्ध जलभृत में 50 cm का एक कुआँ, एक पम्पिंग कुएँ में 3 m अपकर्ष (ड्रॉडाउन) पर 600 lpm का उत्सर्जन देता है । 6 m अपकर्ष पर इसका निस्सरण क्या होगा ? प्रभाव त्रिज्या को 500 m मान लीजिए । (ii) भाग (i) में उल्लिखित अपरिरुद्ध जलभृत के लिए 3 m अपकर्ष पर 30 cm के कुएँ का निस्सरण क्या होगा ? (20) (b) चरम घंटा माँग एक जलप्रदाय परियोजना को कैसे प्रभावित करती है ? विशिष्ट भारतीय परिस्थितियों के लिए माँग के उच्चावचन को आरेखित कीजिए । (15) (c) (i) 25°C पर एक दिन के लिए उद्भवन किए गए एक अपशिष्ट का बी.ओ.डी. 100 mg/l पाया गया । अपशिष्ट के 20°C पर पाँच दिन के बी.ओ.डी. का मान क्या होगा ? 20°C पर K₂₀ = 0·12 मान लीजिए । तापमान गुणांक, φ = 1·056 लीजिए । (ii) बी.ओ.डी. के व्यापक उपयोग के बावजूद इसकी कुछ सीमाएँ हैं । इन सभी सीमाओं का उल्लेख कीजिए । (15)

Directive word: Solve

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How this answer will be evaluated

Approach

Solve the numerical problems in parts (a) and (c) using appropriate groundwater and BOD formulae, while explaining and sketching for part (b). Allocate approximately 40% time to part (a) given its 20 marks, 30% to part (b) for the demand curve sketch and explanation, and 30% to part (c) for temperature correction calculations and limitations discussion. Present derivations clearly with stated assumptions before substituting values.

Key points expected

  • Part (a)(i): Apply Dupuit-Thiem equation for unconfined aquifer to find discharge at 6m drawdown, recognizing that transmissibility remains constant and discharge is proportional to (2H-s)s
  • Part (a)(ii): Calculate discharge for 30cm well using same aquifer properties, applying well radius scaling in the logarithmic term of Dupuit equation
  • Part (b): Explain peak hour factor (typically 1.5-2.5 for Indian cities) and its impact on distribution main sizing, service reservoir capacity, and pumping station design
  • Part (b): Sketch typical Indian diurnal demand curve showing morning and evening peaks, with minimum night flow and seasonal variation for summer/winter
  • Part (c)(i): Apply temperature correction formula K_T = K_20 × φ^(T-20) to find K_25, then use BOD_t = L_0(1-e^(-Kt)) relationship to find ultimate BOD and 5-day 20°C BOD
  • Part (c)(ii): Enumerate BOD limitations: 5-day delay in results, inhibition by toxic substances, nitrification interference, dilution requirements, and non-representative of actual stream conditions

Evaluation rubric

DimensionWeightMax marksExcellentAveragePoor
Concept correctness20%10Correctly identifies Dupuit-Thiem equation for unconfined aquifer in (a), recognizes proportionality Q ∝ (2H-s)s for same well; explains peak hour factor impact on infrastructure sizing in (b); applies correct temperature correction and first-order BOD kinetics in (c)Uses correct general equations but makes minor errors in identifying aquifer type or temperature correction direction; basic explanation of peak demand without specific Indian contextConfuses confined/unconfined aquifer equations, omits temperature correction entirely, or misunderstands BOD as linear rather than first-order reaction
Numerical accuracy25%12.5Accurate calculations: (a)(i) Q_6 ≈ 1150-1200 lpm using exact Dupuit; (a)(ii) Q_30cm ≈ 580-620 lpm; (c)(i) correct K_25 = 0.158, L_0 ≈ 730 mg/l, BOD_5,20 ≈ 325-335 mg/l with proper unit handlingCorrect method but arithmetic errors or unit conversion mistakes (e.g., cm/m, lpm to m³/s); approximate values within 10-15% of correct answerMajor calculation errors, wrong formula substitution, or order-of-magnitude mistakes; missing critical steps leading to nonsensical results
Diagram quality15%7.5Clear hand-drawn sketch for (b) showing: x-axis (24 hours), y-axis (flow as % of average), two distinct peaks (morning 6-9 AM, evening 6-9 PM), minimum night flow (~20-30%), labeled Indian conditions; may include cone of depression for (a)Basic curve showing general trend but missing peak timings, unlabeled axes, or poor proportion between peaks and average flowNo diagram for (b), or completely incorrect sketch showing random fluctuations without typical diurnal pattern; missing labels and scale
Step-by-step derivation25%12.5Explicit statement of all formulae (Dupuit-Thiem, temperature correction, BOD rate equation), clear substitution of values with units, logical flow from given data to final answer; shows intermediate steps for K_25 and L_0 before final BODSome formulae stated but skips intermediate steps; jumps from given data to answer without showing working; minor gaps in logical progressionNo derivations shown, only final answers stated; or incorrect formulae used throughout with no explanation of method
Practical interpretation15%7.5Interprets (a) results for well design efficiency and specific capacity; relates (b) peak factor to Indian urban water supply challenges (e.g., Bangalore, Delhi intermittent supply); connects (c) BOD limitations to need for COD/TOC alternatives in pollution controlBrief mention of practical relevance without specific examples; generic statements about water demand or pollution without Indian contextPurely mathematical treatment with no physical interpretation; fails to explain what results mean for engineering design or environmental management

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