Paper II — Q5
(a) What is a flood hydrograph ? The monthly discharge (Q) data of a river at a gauging station are given below : - January: 30…
What is a flood hydrograph ? The monthly discharge (Q) data of a river at a gauging station are given below : - January: 30 m³/s - February: 40 m³/s - March: 50 m³/s - April: 60 m³/s - May: 80 m³/s - June: 90 m³/s - July: 100 m³/s - August: 120 m³/s - September: 110 m³/s - October: 90 m³/s - November: 80 m³/s - December: 70 m³/s
Draw the hydrograph and using the graph find peak flow and minimum flow in the river. 10 marks
Explain the following terms : Evaporation
Transpiration
Evapotranspiration
Infiltration 10 marks
A catchment has five raingauge stations. In a year, the annual rainfall recorded by the raingauges are 72·3 cm, 86·4 cm, 94·2 cm, 103·8 cm and 71·4 cm respectively. For a 5% error in the estimation of mean rainfall, find the additional number of raingauges needed. 10 marks
A high strength wastewater having an ultimate CBOD of 1000 mg/L is discharged to a river at a rate of 2 m³/s. The river has an upstream ultimate CBOD of 10 mg/L and is flowing at a rate of 8 m³/s. Assuming a reaction rate coefficient of 0·1/day, calculate the ultimate CBOD and 5-day CBOD of the river water just after the mixing point of the wastewater (at 0 km) and 20 km downstream from the mixing point. (Assume the velocity of river = 10 km/day) 10 marks
What do you understand by 'per capita demand' of water ? How is it determined ? If average daily water demand is 135 litres per capita per day, determine the following : Maximum daily demand of water;
Maximum weekly demand of water;
Maximum monthly demand of water;
Maximum hourly demand of water. 10 marks
हिंदी में प्रश्न पढ़ें
(क) बाढ़ जलालेख क्या है ? प्रमापी स्टेशन पर नदी के मासिक निस्सरण (Q) के आँकड़े नीचे दिए गए हैं : - जनवरी: 30 घन मीटर प्रति सेकण्ड - फरवरी: 40 घन मीटर प्रति सेकण्ड - मार्च: 50 घन मीटर प्रति सेकण्ड - अप्रैल: 60 घन मीटर प्रति सेकण्ड - मई: 80 घन मीटर प्रति सेकण्ड - जून: 90 घन मीटर प्रति सेकण्ड - जुलाई: 100 घन मीटर प्रति सेकण्ड - अगस्त: 120 घन मीटर प्रति सेकण्ड - सितम्बर: 110 घन मीटर प्रति सेकण्ड - अक्टूबर: 90 घन मीटर प्रति सेकण्ड - नवम्बर: 80 घन मीटर प्रति सेकण्ड - दिसम्बर: 70 घन मीटर प्रति सेकण्ड
जलालेख बनाइए और ग्राफ का उपयोग करके नदी में चरम प्रवाह और न्यूनतम प्रवाह ज्ञात कीजिए। (10 अंक)
(ख) निम्नलिखित पदों की व्याख्या कीजिए : वाष्पन
वाष्पोत्सर्जन (ट्रांसपिरेशन)
वाष्पन-वाष्पोत्सर्जन (इवेपोट्रांसपिरेशन)
अन्तःस्यंदन (10 अंक)
(ग) एक आवाह (जलग्रहण) क्षेत्र में पाँच वर्षामापी स्टेशन हैं । एक वर्ष में, वर्षामापियों द्वारा दर्ज की गई वार्षिक वर्षा क्रमशः 72·3 सेमी, 86·4 सेमी, 94·2 सेमी, 103·8 सेमी और 71·4 सेमी है । औसत वर्षा के आकलन में 5% त्रुटि के लिए, आवश्यक अतिरिक्त वर्षामापियों की संख्या ज्ञात कीजिए । (10 अंक)
(घ) एक उच्च सांद्रण का अपशिष्ट जल, जिसकी चरम CBOD 1000 mg/L है, उसे 2 घन मीटर प्रति सेकण्ड की दर से एक नदी में छोड़ा जाता है । नदी 8 घन मीटर प्रति सेकण्ड की दर से बह रही है और इसका प्रतिप्रवाह चरम CBOD 10 mg/L है । यदि अभिक्रिया दर गुणांक 0·1 प्रतिदिन है, तो अपशिष्ट जल के मिश्रण बिन्दु के ठीक बाद (0 किमी पर) और मिश्रण बिन्दु से अनुप्रवाह के 20 किमी पर नदी के जल की चरम CBOD और 5-दिवसीय CBOD की गणना कीजिए । (नदी का वेग 10 किमी/दिन मान लीजिए) (10 अंक)
(ङ) जल की 'प्रति व्यक्ति माँग' से आप क्या समझते हैं ? इसे कैसे निर्धारित किया जाता है ? यदि जल की औसत दैनिक माँग 135 लीटर प्रति व्यक्ति प्रतिदिन है, तो निम्नलिखित को निर्धारित कीजिए :
जल की अधिकतम दैनिक माँग;
जल की अधिकतम साप्ताहिक माँग;
जल की अधिकतम मासिक माँग;
जल की अधिकतम प्रति घंटा माँग। (10 अंक)
The figure this question refers to, in words
The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.
(a) Table with two columns: Month | Q (m3/s) January | 30 February | 40 March | 50 April | 60 May | 80 June | 90 July | 100 August | 120 September | 110 October | 90 November | 80 December | 70
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) A flood hydrograph is a plot of discharge (Q) against time at a gauging site for a flood caused by a storm. It has a rising limb (runoff building up), a crest (peak discharge) and a recession or falling limb (depletion of storage). The flow below a base-flow separation line is base flow (groundwater), and the area above it is direct runoff. The time from the start of rise to the crest is the time to peak, and the duration of direct runoff is the time base.
The given data are mean monthly discharges, so their plot is an annual (flow) hydrograph, not the hydrograph of a single storm. The method of plotting and reading it is the same.
Plot month on the x-axis (each value at mid-month) and Q in m³/s on the y-axis, then join the points:
- January 30, February 40, March 50, April 60
- May 80, June 90, July 100, August 120
- September 110, October 90, November 80, December 70
Reading from the graph:
- Peak flow = 120 m³/s, in August (highest ordinate, after the monsoon build-up from May to July).
- Minimum flow = 30 m³/s, in January (lean season, sustained only by base flow).
(b)
(i) Evaporation is the change of water from liquid to vapour at a free water surface or a wet soil surface. Its rate follows Dalton's law, E = C(e_w − e_a), where e_w is the saturation vapour pressure at the water temperature and e_a the actual vapour pressure of the air. It increases with temperature, solar radiation, wind speed and vapour-pressure deficit, and decreases with humidity and with salinity of water. It is measured with a Class A pan, and lake evaporation ≈ 0.7 × pan evaporation.
(ii) Transpiration is the loss of water as vapour from living plants, mainly through the stomata of leaves. Water taken up by the roots moves up the plant and escapes to the air. It depends on the plant type and stage of growth, soil moisture, sunlight (stomata open by day), temperature, wind and humidity. It is measured with a phytometer.
(iii) Evapotranspiration (ET) is the total water lost from a vegetated area by evaporation from soil and water surfaces plus transpiration by plants; in practice the two cannot be separated. Potential ET (PET) is the rate when water supply to the soil and plants is unlimited; actual ET (AET) is the rate under the actual soil moisture, so AET ≤ PET. ET is the consumptive use of crops and is measured with lysimeters or estimated by the Penman or Blaney–Criddle methods.
(iv) Infiltration is the entry of water into the soil through the ground surface. The maximum rate at which a soil can absorb water at a given time is its infiltration capacity f_p; the actual rate equals rainfall intensity when intensity is below f_p, and equals f_p otherwise, the excess becoming surface runoff. Capacity falls with time as the soil wets, as in Horton's equation f = f_c + (f_0 − f_c)e^(−kt). It depends on soil texture, initial moisture, vegetation cover, compaction and rainfall intensity, and is measured with a ring infiltrometer. Downward movement of water already in the soil is percolation, not infiltration.
(c) Given P = 72.3, 86.4, 94.2, 103.8 and 71.4 cm, n = 5 and ε = 5%.
Mean: P̄ = 428.1/5 = 85.62 cm
Deviations (P − P̄) and their squares:
- 72.3: −13.32, 177.42
- 86.4: +0.78, 0.61
- 94.2: +8.58, 73.62
- 103.8: +18.18, 330.51
- 71.4: −14.22, 202.21
Σ(P − P̄)² = 784.37
Standard deviation (n − 1 in the denominator): σ = √(784.37/4) = √196.09 = 14.00 cm
Coefficient of variation: Cv = 100σ/P̄ = 100 × 14.00/85.62 = 16.36%
Optimum number of gauges: N = (Cv/ε)² = (16.36/5)² = 10.7, so N = 11
Additional raingauges needed = 11 − 5 = 6
(d) Given Q_w = 2 m³/s, L_w = 1000 mg/L; Q_r = 8 m³/s, L_r = 10 mg/L; k = 0.1/day (base e); u = 10 km/day.
At 0 km (just after mixing), by mass balance: L₀ = (Q_r L_r + Q_w L_w)/(Q_r + Q_w) = (8 × 10 + 2 × 1000)/10 = 2080/10 = 208 mg/L (ultimate CBOD)
BOD₅ = L₀(1 − e^(−5k)) = 208(1 − e^(−0.5)) = 208 × 0.3935 = 81.8 mg/L
At 20 km downstream, travel time t = 20/10 = 2 days: L₂₀ = L₀e^(−kt) = 208e^(−0.2) = 208 × 0.8187 = 170.3 mg/L (ultimate CBOD)
BOD₅ = 170.3 × 0.3935 = 67.0 mg/L
So in 2 days of travel about 38 mg/L of CBOD has been exerted (208 − 170.3), and this oxygen demand is what creates the dissolved-oxygen sag downstream. The results assume complete mixing at the outfall, plug flow, no other inflows, and k to base e. If k were to base 10, BOD₅ at 0 km would be 142.2 mg/L, and at 20 km L = 131.2 mg/L and BOD₅ = 89.7 mg/L.
(e) Per capita demand is the annual average quantity of water needed per person per day, in litres per capita per day (Lpcd). It includes domestic, commercial and institutional, industrial and public uses, fire demand, and losses in the system.
It is determined as: q = total yearly quantity of water required by the town (litres) ÷ (365 × design population)
In practice the total is built up from the separate demands, or taken from records of metered supply, and checked against CPHEEO norms. The CPHEEO norms are 135 Lpcd for towns with sewerage, 150 Lpcd for metropolitan cities and 70 Lpcd for towns without sewerage. It varies with climate, size of the city, standard of living, industries, metering and pricing, supply pressure and system losses.
Consumption is not uniform, so peak factors are applied to the average q = 135 Lpcd. Goodrich's formula p = 180 t^(−0.10) (p in % of average, t in days) gives the factors for daily, weekly and monthly peaks.
(i) Maximum daily demand (t = 1 day, p = 180%): = 1.8 × 135 = 243 Lpcd
(ii) Maximum weekly demand (t = 7 days, p = 180 × 7^(−0.1) = 148%): = 1.48 × 135 = 200 Lpcd
(iii) Maximum monthly demand (t = 30 days, p = 180 × 30^(−0.1) = 128%): = 1.28 × 135 = 173 Lpcd
(iv) Maximum hourly demand = 1.5 × average hourly demand of the maximum day = 1.5 × 1.8 × average = 2.7 × average: = 2.7 × 135 = 364.5 Lpcd (as a rate), i.e. 364.5/24 ≈ 15.2 litres per person per hour
Maximum daily demand is used to size the source, intake and treatment plant. Maximum hourly demand (or maximum daily plus fire demand, whichever is larger) is used to size the distribution pipes.
How this answer will be evaluated
What earns full marks: All parts answered with correct calculations, clear definitions, and proper units.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Define a flood hydrograph, plot the given monthly discharges and read peak and minimum flow from the graph. 10 marks
describe— define → structure or process in order → labelled diagram → significance
Must cover
- Define flood hydrograph: discharge vs time at a gauging site for a flood event, with rising limb, crest and recession limb
- Plot all 12 monthly discharges with labelled axes (month; Q in m³/s)
- Peak flow read from the graph = 120 m³/s (August)
- Minimum flow read from the graph = 30 m³/s (January)
Loses marks
- No definition of flood hydrograph
- Values stated without a plotted graph, or axes/units missing
Earns more
- Note that monthly data give an annual flow hydrograph, not a single-storm flood hydrograph
- Show base flow, direct runoff or time to peak on a sketch of a flood hydrograph
Extra mark
- Labelled sketch of storm-hydrograph components with base-flow separation
- (b) Explain evaporation, transpiration, evapotranspiration and infiltration with their controlling factors. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Evaporation: liquid to vapour from water/soil surfaces; controlled by vapour-pressure difference, temperature, wind, radiation
- Transpiration: water lost as vapour from plants through stomata; depends on plant, soil moisture, climate
- Evapotranspiration = evaporation + transpiration from a vegetated area; PET vs AET distinguished
- Infiltration: entry of water into soil through the surface; infiltration capacity and its decline with time
Loses marks
- Confusing infiltration with percolation, or evaporation with transpiration
- One-line definitions with no factors or units
Earns more
- Measurement: evaporation pan (pan coefficient ≈ 0.7), phytometer, lysimeter, infiltrometer
- Factors affecting infiltration: soil texture, initial moisture, cover, compaction, rainfall intensity
Extra mark
- Dalton's law E = C(e_w − e_a) or Horton's equation f = f_c + (f_0 − f_c)e^(−kt)
- (c) Find the additional raingauges needed for 5% error in mean rainfall using the coefficient of variation. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Mean rainfall = 428.1/5 = 85.62 cm
- Standard deviation with (n − 1): σ = 14.00 cm; Cv = 16.36%
- N = (Cv/ε)² = (16.36/5)² = 10.7, rounded up to 11
- Additional raingauges = 11 − 5 = 6
Loses marks
- Dividing by n instead of (n − 1) without comment (gives Cv 14.6%, N = 9, additional 4)
- Giving 11 as the additional number without subtracting the existing 5
Earns more
- Tabulate deviations and their squares
- State the formula N = (Cv/ε)² with symbols defined
Extra mark
- Mention recommended raingauge densities (WMO/IMD) as a check
- (d) Calculate ultimate and 5-day CBOD just after mixing and 20 km downstream using mass balance and first-order decay. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Mixing: L₀ = (8 × 10 + 2 × 1000)/(8 + 2) = 208 mg/L
- BOD₅ at 0 km = 208(1 − e^(−0.5)) = 81.8 mg/L
- Travel time 20/10 = 2 d; L = 208e^(−0.2) = 170.3 mg/L
- BOD₅ at 20 km = 170.3(1 − e^(−0.5)) = 67.0 mg/L
Loses marks
- Averaging concentrations instead of flow-weighted mixing
- Using distance instead of travel time in the decay term
Earns more
- State L_t = L₀e^(−kt) and BOD_t = L₀(1 − e^(−kt))
- State assumptions: complete mixing, plug flow, no other load
Extra mark
- Note that a base-10 k would give BOD₅ 142.2 mg/L at 0 km and L 131.2, BOD₅ 89.7 mg/L at 20 km
- (e) Define per capita demand, explain how it is determined, and compute maximum daily, weekly, monthly and hourly demands for 135 Lpcd. 10 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Per capita demand = total annual water requirement ÷ (population × 365), in Lpcd; includes domestic, commercial, industrial, public uses and losses
- Maximum daily demand = 1.8 × 135 = 243 Lpcd
- Maximum weekly ≈ 1.48 × 135 = 200 Lpcd; maximum monthly ≈ 1.28 × 135 = 173 Lpcd (Goodrich p = 180t^(−0.10))
- Maximum hourly = 1.5 × maximum daily = 2.7 × 135 = 364.5 Lpcd (≈ 15.2 L per person per hour)
Loses marks
- Using peak factors without stating them
- Hourly value given without clear units (Lpcd rate vs litres per hour)
Earns more
- Factors: climate, city size, living standard, industry, metering, pressure, losses
- CPHEEO norms: 135 Lpcd with sewerage, 150 for metros, 70 without sewerage
Extra mark
- Link each peak factor to the component it sizes (source, treatment, pipes)
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
Practice this exact question
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