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Chapter 1 · 2 hours

Introduction

IOE past exam questions

Past questions and answers

18 questions set from this chapter, 4 of them more than once; 3 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 5 of 27 exams
  • Asked 5 times
  • 2081 Bhadra · 1+3 marks
  • 2078 Kartik · 4 marks
  • 2071 Chaitra · 4 marks
  • 2069 Chaitra · 1+3 marks
  • 2066 Magh (old course) · 4 marks

Explain the hydrologic cycle in nature with the help of a neat sketch.

Answer

The hydrologic cycle is the continuous circulation of water between the oceans, atmosphere and land, driven by solar energy and gravity. Water changes state and place, but the total amount on earth stays nearly constant.

   Sun's energy
      |
  Evaporation          Condensation -> Clouds
  + Transpiration  ^        |
      ^            |        v
 [Ocean]---------->|   Precipitation
      ^                     |
      |    Surface runoff   v
      +----- Rivers <--- [Land/Catchment]
      |                     |
      +--- Groundwater <----+  (infiltration,
           flow to sea          percolation)

Main processes

  1. Evaporation and transpiration: Solar heat turns water from oceans, lakes and rivers into vapour. Plants release vapour through leaves (transpiration). Together they are called evapotranspiration.
  2. Condensation: Rising vapour cools, condenses on dust nuclei and forms clouds.
  3. Precipitation: Water returns to earth as rain, snow, hail or sleet. About 78% falls directly on oceans.
  4. Interception and depression storage: Part of the rain is caught by leaves and small surface depressions and later evaporates.
  5. Infiltration and percolation: Water enters the soil, and moves down to the water table to become groundwater.
  6. Surface runoff: Rain beyond infiltration capacity flows over the ground into streams and rivers.
  7. Groundwater flow (base flow): Water moves slowly through aquifers and feeds rivers, springs and the sea.
  8. Snow and glacier melt: In Himalayan basins like Nepal's, snow stored in winter melts in summer and feeds rivers.

Importance

The cycle gives the fresh water used for drinking, irrigation and hydropower, and its study allows engineers to estimate the water available and the floods that may occur.

  • Most repeated · 5 of 27 exams
  • Asked 5 times
  • 2080 Bhadra · 4 marks
  • 2079 Bhadra · 4 marks
  • 2074 Asoj · 2+2 marks
  • 2073 Shrawan · 4 marks
  • 2068 Chaitra · 4 marks

Explain water budget equation. What is the role of water budget equation in hydrology?

Answer

The water budget (water balance) equation is an application of the law of conservation of mass to a hydrologic system: inflow minus outflow equals the change in storage over a time period.

I−O=dSdtorI−O=ΔSI - O = \frac{dS}{dt} \quad\text{or}\quad I - O = \Delta S

For a catchment over a period Δt\Delta t:

P+Qin+Gin−E−T−Qout−Gout=ΔSP + Q_{in} + G_{in} - E - T - Q_{out} - G_{out} = \Delta S

where PP = precipitation, Qin,QoutQ_{in}, Q_{out} = surface inflow and outflow, Gin,GoutG_{in}, G_{out} = groundwater inflow and outflow, EE = evaporation, TT = transpiration, and ΔS\Delta S = change in storage (soil moisture, groundwater, surface water, snow). For a closed natural basin with no inflow, it reduces to

P−R−G−ET=ΔSP - R - G - E_T = \Delta S

with RR = surface runoff, GG = net groundwater outflow and ETE_T = evapotranspiration. Over a long period (many years) ΔS≈0\Delta S \approx 0, so P=R+G+ETP = R + G + E_T.

Role of the water budget equation

  • Estimating an unknown component: If P, R and ET are measured, the groundwater recharge or storage change can be found. Evaporation from lakes and reservoirs can be found in the same way.
  • Water resources assessment: It gives the available yield of a basin for water supply, irrigation and hydropower.
  • Reservoir planning: Inflow, outflow and storage are balanced to find the reservoir capacity and change in level.
  • Checking data: A large non-closure of the budget shows errors in measured data.
  • Impact studies: It helps to study the effect of land-use change, dams or climate change on a basin.
  • Most repeated · 3 of 27 exams
  • Asked 3 times
  • 2078 Bhadra · 1+3 marks
  • 2072 Kartik · 2+2 marks
  • 2071 Shrawan · 4 marks

Explain the hydrologic cycle and the water balance equation.

Answer

Hydrologic cycle

The hydrologic cycle is the endless movement of water from the oceans to the atmosphere, to the land and back to the oceans by evaporation, condensation, precipitation, runoff, infiltration and groundwater flow. The sun supplies the energy and gravity moves the water.

   Sun's energy
      |
  Evaporation          Condensation -> Clouds
  + Transpiration  ^        |
      ^            |        v
 [Ocean]---------->|   Precipitation
      ^                     |
      |    Surface runoff   v
      +----- Rivers <--- [Land/Catchment]
      |                     |
      +--- Groundwater <----+  (infiltration,
           flow to sea          percolation)
  • Evaporation/transpiration puts vapour in the air.
  • Condensation forms clouds, and precipitation brings water to earth.
  • Part of the precipitation is intercepted, part infiltrates, and the rest becomes surface runoff that reaches streams and the sea.
  • Infiltrated water feeds groundwater, which flows slowly to rivers and oceans.

Water balance equation

By the law of conservation of mass, for any basin and time interval:

Inflow−Outflow=ΔS\text{Inflow} - \text{Outflow} = \Delta S P−R−G−E−T=ΔSP - R - G - E - T = \Delta S

Here PP is precipitation, RR surface runoff, GG net groundwater outflow, EE evaporation, TT transpiration and ΔS\Delta S the change in storage. Over a long term ΔS≈0\Delta S \approx 0, so P=R+G+ETP = R + G + ET.

For the global balance, the land receives more precipitation than it evaporates, and the extra flows to the oceans as runoff. The equation is used to find unknown components, to size reservoirs and to assess the water yield of a basin.

  • Asked 2 times
  • 2079 Baisakh · 4 marks
  • 2076 Asoj · 3 marks

Discuss the significance of hydrology in civil engineering applications and engineering design with appropriate examples.

Answer

Hydrology is the science that deals with the occurrence, circulation, distribution and properties of water on earth. Civil engineering structures are exposed to or use water, so hydrologic data gives the design basis.

ApplicationHydrologic inputExample
HydropowerMean flows, flow duration curve, design floodFixing the installed capacity and spillway of a project like Kaligandaki
IrrigationRainfall, ET, dependable flowCanal capacity of the Sunsari-Morang scheme
Water supplyDependable yield of sourceMelamchi water supply sizing
Flood controlDesign flood, flood routing, frequencyEmbankments on Narayani or Koshi
Bridges and culvertsPeak discharge for a return periodWaterway opening, afflux and foundation depth
DrainageRainfall intensity, runoffStorm sewers in Kathmandu
ReservoirMass curve, sedimentation, storageLive storage of a dam

Significance in design

  1. Design floods: Spillway, bridge opening and embankment height depend on a flood of selected return period (for example 100 years).
  2. Water availability: The dependable flow (75% or 90%) decides whether the demand can be met.
  3. Economic design: An overestimated flood makes the structure costly, and an underestimate risks failure and loss of life.
  4. Safety and operation: Flood forecasting and reservoir operation reduce damage.
  5. Environmental flows: The minimum river flow for aquatic life is fixed from hydrologic records.

Failures such as bridge washouts in the monsoon are often due to poor hydrologic estimates, so hydrology is a first step in water resources and infrastructure design.

  • 2075 Asoj · 6 marks

The catchment area of a reservoir is 1600 ha. A uniform precipitation of 8 mm/hr for 2 hour was observed on a particular day. 55% run off reached the reservoir. A canal carrying a flow of 1 m3^3/s is taken from the reservoir. The rate of evaporation was 0.8 mm/h/m2^2. Assuming seepage loss is 40% of evaporation loss, find the change in the reservoir level for 6 hours, if the water spread of the reservoir was 45 ha.

Similar questions: Reservoir level change, water balance, 15.5 km2 (2076 Chaitra)

Answer

Given: catchment area A=1600A = 1600 ha =16×106= 16 \times 10^6 m²; rainfall 8 mm/h for 2 h; runoff coefficient 0.55; canal flow 1 m³/s; evaporation 0.8 mm/h/m² of water spread; seepage = 40% of evaporation; water spread =45= 45 ha =4.5×105= 4.5 \times 10^5 m²; period T=6T = 6 h.

Water balance: ΔS=Inflow−Canal outflow−Evaporation−Seepage\Delta S = \text{Inflow} - \text{Canal outflow} - \text{Evaporation} - \text{Seepage}

Step 1: Inflow

Rainfall depth =8×2=16= 8 \times 2 = 16 mm =0.016= 0.016 m

Vin=0.55×0.016×16×106=140 800 m3V_{in} = 0.55 \times 0.016 \times 16\times10^6 = 140\,800\ \text{m}^3

Step 2: Outflows over 6 h

  • Canal: 1×6×3600=21 6001 \times 6 \times 3600 = 21\,600 m³
  • Evaporation: 0.0008×4.5×105×6=2 1600.0008 \times 4.5\times10^5 \times 6 = 2\,160 m³
  • Seepage: 0.4×2160=8640.4 \times 2160 = 864 m³

Step 3: Change in storage

ΔS=140 800−21 600−2 160−864=116 176 m3\Delta S = 140\,800 - 21\,600 - 2\,160 - 864 = 116\,176\ \text{m}^3

Step 4: Change in level

Δh=116 1764.5×105=0.258 m\Delta h = \frac{116\,176}{4.5\times10^5} = 0.258\ \text{m}

Answer: The reservoir level rises by about 0.258 m (25.8 cm) in 6 hours.

  • 2076 Chaitra · 6 marks

The catchment area of a reservoir is 15.5 km2^2. A uniform precipitation of 0.5 cm/h for 2 hr was observed on a particular day. 65% of precipitation reached into the reservoir. A canal carrying a flow of 1.1 m3^3/s is taken from the reservoir. The rate of evaporation was 0.6 mm/hr/m2^2. Assuming seepage loss to be 45% of evaporation loss, find the change in reservoir level for 10 hrs, if the water spread of the reservoir was 0.6 km2^2.

Similar questions: Reservoir level change, water balance, 1600 ha (2075 Asoj)

Answer

Given: catchment area A=15.5A = 15.5 km² =15.5×106= 15.5\times10^6 m²; rainfall 0.5 cm/h for 2 h =1= 1 cm =0.01= 0.01 m; runoff reaching reservoir 65%; canal 1.1 m³/s; evaporation 0.6 mm/h per m²; seepage 45% of evaporation; water spread =0.6= 0.6 km² =6×105= 6\times10^5 m²; period T=10T = 10 h.

ΔS=Vin−Vcanal−Vevap−Vseep\Delta S = V_{in} - V_{canal} - V_{evap} - V_{seep}

Step 1: Inflow

Vin=0.65×0.01×15.5×106=100 750 m3V_{in} = 0.65 \times 0.01 \times 15.5\times10^6 = 100\,750\ \text{m}^3

Step 2: Outflows in 10 h

  • Canal: 1.1×10×3600=39 6001.1 \times 10 \times 3600 = 39\,600 m³
  • Evaporation: 0.0006×6×105×10=3 6000.0006 \times 6\times10^5 \times 10 = 3\,600 m³
  • Seepage: 0.45×3600=1 6200.45 \times 3600 = 1\,620 m³

Step 3: Storage change

ΔS=100 750−39 600−3 600−1 620=55 930 m3\Delta S = 100\,750 - 39\,600 - 3\,600 - 1\,620 = 55\,930\ \text{m}^3

Step 4: Level change

Δh=55 9306×105=0.0932 m\Delta h = \frac{55\,930}{6\times10^5} = 0.0932\ \text{m}

Answer: The reservoir level rises by about 0.093 m (9.3 cm) in 10 hours.

  • 2076 Asoj · 4 marks

In a certain catchment, inflow rate into the catchment due to rainfall is given by equation I=2tI = 2t m3^3/s. If loss in the catchment is neglected, determine the change in storage in catchment within 3 hr duration.

Answer

The water balance gives I−O=ΔSI - O = \Delta S. Losses are neglected, so outflow O=0O = 0 and the change in storage equals the total inflow volume.

Given: I=2tI = 2t m³/s, with tt in seconds; duration T=3T = 3 h =3×3600=10800= 3 \times 3600 = 10800 s.

ΔS=∫0TI dt=∫0108002t dt=[t2]010800\Delta S = \int_0^{T} I\,dt = \int_0^{10800} 2t\,dt = \left[t^2\right]_0^{10800} ΔS=(10800)2=1.1664×108 m3\Delta S = (10800)^2 = 1.1664 \times 10^{8}\ \text{m}^3

Answer: Change in storage = 1.1664 × 10⁸ m³ (116.64 million m³) increase in 3 hours.

(If tt were taken in hours, the answer would be different; tt in seconds is consistent with the unit m³/s.)

  • 2075 Chaitra · 3+2 marks

Explain hydrologic cycle with neat sketches and justify its need in Engineering Hydrology.

Answer

Hydrologic cycle

The hydrologic cycle is the continuous movement of water between the ocean, atmosphere and land through evaporation, condensation, precipitation, runoff and infiltration. It is driven by solar energy and gravity.

   Sun's energy
      |
  Evaporation          Condensation -> Clouds
  + Transpiration  ^        |
      ^            |        v
 [Ocean]---------->|   Precipitation
      ^                     |
      |    Surface runoff   v
      +----- Rivers <--- [Land/Catchment]
      |                     |
      +--- Groundwater <----+  (infiltration,
           flow to sea          percolation)

Land-phase sketch of a catchment

        Precipitation (P)
   \ \ \ \ \ \ \ \
    Interception   Evapotranspiration
   -------------------------------------
    Depression storage | Surface runoff -> River
    Infiltration  \    |
   ~~~~~~~~~ Water table ~~~~~~~~~~~~~~~~
      Groundwater flow -> River / sea

Processes: evaporation, transpiration, condensation, precipitation, interception, infiltration, percolation, surface runoff, groundwater flow and snowmelt.

Need of the hydrologic cycle in engineering hydrology

  • It is the framework of hydrology: every topic (precipitation, losses, runoff, streamflow) is a part of the cycle.
  • The water balance of a basin (P−R−ET−G=ΔSP - R - ET - G = \Delta S) comes from the cycle and is used to find water availability.
  • Engineers estimate design floods, yield and droughts by quantifying the cycle components.
  • It shows how human activities (dams, deforestation, urbanisation, pumping) change evaporation, runoff and recharge.
  • It helps in planning water use for supply, irrigation and hydropower in a sustainable way.
  • 2072 Chaitra · 4 marks

Define the following terms: hydrological cycle, runoff, water balance and catchment.

Answer

  • Hydrological cycle: The continuous circulation of water from oceans to atmosphere, to land and back to the oceans through evaporation, condensation, precipitation, infiltration and runoff, driven by solar energy and gravity.
  • Runoff: The part of precipitation that flows over the land surface and in channels to reach a stream, lake or sea. It includes surface runoff, interflow and base flow, and is usually expressed as discharge (m³/s) or depth (mm) over the catchment.
  • Water balance: The accounting of inflows, outflows and storage change of a region over a given time, based on conservation of mass: P−R−G−E−T=ΔSP - R - G - E - T = \Delta S.
  • Catchment (watershed or drainage basin): The area of land that drains to a given point on a stream, bounded by a ridge line called the water divide. All rain falling inside it (apart from losses) flows to that outlet.
  • 2070 Asar · 2+2 marks

Why is the study of hydrology important for engineers for planning and designing of water resources projects in Nepal? Explain the significant features of global water balance with necessary equation.

Answer

Importance of hydrology in Nepal

Nepal has more than 6,000 rivers and a large hydropower potential, but rainfall is highly seasonal (about 80% in June to September). Hydrology is therefore important to:

  • Hydropower: estimate mean flow, flow duration and design flood for projects (Arun, Kali Gandaki, Upper Tamakoshi).
  • Irrigation: find dry-season dependable flows for canals such as the Mahakali and Sunsari-Morang schemes.
  • Water supply: assess the yield of springs and rivers for towns (Melamchi).
  • Flood and landslide control: design embankments, bridges and culverts; handle monsoon floods and GLOF risk.
  • Planning and climate change: assess snowmelt, glacier retreat and drought for sustainable use.

Global water balance

About 97.5% of the earth's water is saline (oceans) and only 2.5% is fresh, most of it locked in ice and groundwater. Features of the global balance:

  • Evaporation from the ocean is greater than precipitation on it, and the reverse holds over land.
  • The surplus vapour carried to land (about 10% of ocean evaporation) returns to the sea as river runoff.
  • Oceans: Eo>PoE_o > P_o; land: Pl>ElP_l > E_l; and Pl−El=RP_l - E_l = R.
  • Over the earth as a whole, total evaporation equals total precipitation.
Pglobal=Eglobal,Pl−El=Rl (runoff and groundwater to oceans)P_{global} = E_{global}, \qquad P_l - E_l = R_l \ (\text{runoff and groundwater to oceans})

Approximate values (per year): ocean evaporation about 85% and ocean precipitation 77% of the global cycle; land precipitation about 23% and land evaporation about 15%.

  • 2070 Chaitra · 4 marks

Explain different prospects of Hydrological study.

Answer

Hydrology can be studied from different viewpoints, each serving a practical need:

  1. Scientific hydrology: the study of the occurrence, movement and properties of water for knowledge, such as the physics of precipitation, evaporation and infiltration.
  2. Engineering hydrology: use of hydrologic data for planning, design and operation of water projects, such as design floods for spillways and bridges, and reservoir yield.
  3. Surface water hydrology: rainfall, runoff, streamflow, floods, droughts and sediment.
  4. Groundwater hydrology (hydrogeology): aquifers, recharge, well yield and groundwater development.
  5. Hydrometeorology: the link between atmosphere and water, such as storms, probable maximum precipitation and forecasting.
  6. Statistical and stochastic hydrology: frequency analysis and time-series models for floods and droughts.
  7. Water quality and environmental hydrology: pollution, sediment, ecology and climate change effects.
  8. Snow and glacier hydrology: important for the Himalayan rivers of Nepal.

Each aspect supports water supply, irrigation, hydropower, flood control, drainage and navigation.

  • 2067 Shrawan (old course) · 6 marks

Briefly describe the role of ground water in irrigation development.

Answer

Groundwater is water stored in the pores and fractures of soil and rock below the water table. It is a major, reliable source for irrigation, especially where surface water is limited or seasonal.

Role in irrigation development

  1. Dependable dry-season source: Rivers and canals often carry little water in winter and spring. Wells give water when it is needed.
  2. Supplementary irrigation: Where canal water is short, tubewells give extra water for the second and third crops.
  3. Terai development: In Nepal's Terai the shallow aquifer allows shallow tubewells (STW) and deep tubewells (DTW). The Bhabar zone recharges the aquifer.
  4. Individual control: Farmers can irrigate at the time and amount they need, which raises yields.
  5. Quick and cheaper development: Small wells need less capital and time than large canal systems, and have no conveyance loss or large land acquisition.
  6. Conjunctive use: Using surface water and groundwater together lowers waterlogging and salinity by keeping the water table low.
  7. Drought protection: Stored groundwater reduces the risk of crop loss in dry years.

Limits

Over-pumping lowers the water table, energy cost is high, and water quality (salts, arsenic in some Terai districts) must be checked. Pumping should not exceed the recharge.

  • 2067 Shrawan (old course) · 10 marks

A 30 cm well fully penetrates an unconfined aquifer of saturated depth 25 m. When a discharge of 2100 lpm was being pumped for a long time, observation wells at radial distances of 30 and 90 m indicated drawdowns of 5 and 4 m respectively. Estimate the co-efficient of permeability and transmissibility of the aquifer. What is the drawdown at the pumping well?

Answer

Given: unconfined aquifer, saturated depth H=25H = 25 m; well diameter 30 cm (rw=0.15r_w = 0.15 m); Q=2100Q = 2100 lpm =0.035= 0.035 m³/s; observation wells r1=30r_1 = 30 m, s1=5s_1 = 5 m and r2=90r_2 = 90 m, s2=4s_2 = 4 m.

Water-table depths: h1=25−5=20h_1 = 25 - 5 = 20 m, h2=25−4=21h_2 = 25 - 4 = 21 m.

(a) Coefficient of permeability

For a steady unconfined flow (Dupuit):

Q=πK(h22−h12)ln⁡(r2/r1)Q = \frac{\pi K (h_2^2 - h_1^2)}{\ln(r_2/r_1)} K=Qln⁡(r2/r1)π(h22−h12)=0.035×ln⁡(3)π(441−400)=0.035×1.0986128.8K = \frac{Q\ln(r_2/r_1)}{\pi (h_2^2-h_1^2)} = \frac{0.035 \times \ln(3)}{\pi (441 - 400)} = \frac{0.035\times 1.0986}{128.8} K=2.985×10−4 m/s  (=25.8 m/day)K = 2.985\times10^{-4}\ \text{m/s} \;(= 25.8\ \text{m/day})

(b) Transmissibility

T=KH=2.985×10−4×25=7.46×10−3 m2/s  (≈645 m2/day)T = K H = 2.985\times10^{-4} \times 25 = 7.46\times10^{-3}\ \text{m}^2/\text{s} \;(\approx 645\ \text{m}^2/\text{day})

(c) Drawdown at the pumping well

Apply the same equation between rwr_w and r1r_1:

h12−hw2=Qln⁡(r1/rw)πK=0.035×ln⁡(200)π×2.985×10−4=197.7 m2h_1^2 - h_w^2 = \frac{Q \ln(r_1/r_w)}{\pi K} = \frac{0.035 \times \ln(200)}{\pi \times 2.985\times10^{-4}} = 197.7\ \text{m}^2 hw2=400−197.7=202.3⇒hw=14.22 mh_w^2 = 400 - 197.7 = 202.3 \Rightarrow h_w = 14.22\ \text{m} sw=25−14.22=10.78 ms_w = 25 - 14.22 = 10.78\ \text{m}

Answer: K = 2.985 × 10⁻⁴ m/s (25.8 m/day); T = 7.46 × 10⁻³ m²/s (about 645 m²/day); drawdown at the well = 10.78 m.

  • 2066 Magh (old course) · 6 marks

In a recuperation test, the static water level in an open well was depressed using pumps by 3 m and it recuperated 1.5 m in 1 hour. If the diameter of the well is 3.0 m and the safe working depression head is 2.4 m, find out the average specific yield of the soil and specific capacity of the soil.

Answer

A recuperation test measures how fast water rises in a depressed open well. The rate of recovery is proportional to the depression head hh at that time:

dhdt=−QA=−Kih  ⇒  Ki=1tln⁡h1h2\frac{dh}{dt} = -\frac{Q}{A} = -K_i h \;\Rightarrow\; K_i = \frac{1}{t}\ln\frac{h_1}{h_2}

where KiK_i is the specific capacity (per unit area, per unit depression) of the soil, also called the constant of the well/soil, and AA is the well area.

Given: initial depression h1=3h_1 = 3 m; after t=1t = 1 h the depression h2=3−1.5=1.5h_2 = 3 - 1.5 = 1.5 m; well diameter 3.0 m; safe depression head H=2.4H = 2.4 m.

Ki=11ln⁡31.5=0.693 h−1K_i = \frac{1}{1}\ln\frac{3}{1.5} = 0.693\ \text{h}^{-1}

Well area: A=π4(3.0)2=7.069A = \frac{\pi}{4}(3.0)^2 = 7.069 m².

Specific capacity

Specific capacity of the well (discharge per unit depression):

QH=KiA=0.693×7.069=4.90 m3/h per m of depression\frac{Q}{H} = K_i A = 0.693 \times 7.069 = 4.90\ \text{m}^3/\text{h per m of depression}

The specific capacity of the soil per unit area is Ki=0.693K_i = 0.693 m³/h per m² per m depression (0.693 h−10.693\ \text{h}^{-1}).

Specific (safe) yield at the working depression

Q=KiAH=0.693×7.069×2.4=11.76 m3/hQ = K_i A H = 0.693 \times 7.069 \times 2.4 = 11.76\ \text{m}^3/\text{h}

Answer: Specific capacity = 0.693 m³/h/m² of well area per metre depression (4.90 m³/h per metre depression for this well); safe yield (average yield) = 11.76 m³/h (about 282 m³/day).

  • 2082 Bhadra · 3 marks

Explain briefly humankind's interference in various parts of hydrological cycle.

Answer

Human activity changes many parts of the hydrologic cycle:

  • Precipitation: Cloud seeding for rain, and climate change from greenhouse gases and urban heat change rainfall patterns.
  • Evaporation and transpiration: Reservoirs and irrigation raise evaporation; deforestation lowers transpiration; planting forests raises it.
  • Interception and infiltration: Roads, roofs and pavements reduce infiltration and raise runoff and flood peaks; agriculture and terracing may increase infiltration.
  • Surface runoff: Dams and barrages store and regulate flow; embankments and river training change flood patterns; urbanisation shortens time of concentration.
  • Groundwater: Heavy pumping lowers the water table, while artificial recharge and irrigation seepage raise it.
  • Water quality: Sewage, industry and fertilisers pollute surface and groundwater.
  • Sediment: Land clearing and mining increase erosion and siltation of reservoirs.
  • 2082 Baisakh · 1+3 marks

Define hydrology. Why is hydrology considered important in case of water resources engineering?

Answer

Hydrology is the science that deals with the occurrence, circulation, distribution and movement of water on and below the earth's surface and in the atmosphere, and its properties and relation to the environment and living things.

Importance in water resources engineering

  1. Quantity of water available: Rainfall, runoff and groundwater data decide how much water a river or aquifer can supply for water supply, irrigation or hydropower.
  2. Design floods: Spillways, dams, bridges, culverts and embankments are designed for flood discharge of a chosen return period.
  3. Storage planning: Reservoir capacity is found from the mass curve of inflows and demands.
  4. Hydropower: The flow duration curve gives firm power and installed capacity.
  5. Drought and low flow: It gives the minimum flow for water supply and environmental needs.
  6. Drainage and flood control: Runoff estimates size drains, sewers and flood protection works.
  7. Operation and forecasting: Flood warning and reservoir operation use hydrologic models.
  8. Safety and economy: Reliable hydrologic design avoids both failure and over-design.
  • 2081 Baisakh · 2+2 marks

Describe Hydrological cycle. Discuss about the uses of Engineering Hydrology.

Answer

Hydrological cycle

The hydrological cycle is the continuous exchange of water between the oceans, atmosphere and land. Solar energy evaporates water, vapour condenses to clouds and falls as precipitation; on land it is intercepted, infiltrates or runs off to streams, and finally returns to the sea or atmosphere.

   Sun's energy
      |
  Evaporation          Condensation -> Clouds
  + Transpiration  ^        |
      ^            |        v
 [Ocean]---------->|   Precipitation
      ^                     |
      |    Surface runoff   v
      +----- Rivers <--- [Land/Catchment]
      |                     |
      +--- Groundwater <----+  (infiltration,
           flow to sea          percolation)

Main components: evaporation, transpiration, condensation, precipitation, interception, infiltration, percolation, runoff and groundwater flow.

Uses of engineering hydrology

  • Water supply: yield of surface and groundwater sources.
  • Irrigation: crop water demand (evapotranspiration), dependable flow and canal design.
  • Hydropower: flow duration, design flood and storage.
  • Flood control and drainage: design floods, flood routing, storm drains.
  • Hydraulic structures: waterway of bridges and culverts, spillways, weirs, dams.
  • Reservoir operation and sedimentation: capacity and life of reservoirs.
  • Environmental and navigation needs: minimum flows and water-level records.
  • 2080 Baisakh · 4 marks

As a civil engineer how do you justify the need and importance of Engineering Hydrology for the design of Bridge?

Answer

A bridge crosses a river, so its safety depends on how much water the river carries and how high it rises. Engineering hydrology supplies this information.

Need and importance

  1. Design discharge: The peak flood for a selected return period (for example 100 years for major bridges) is found from rainfall-runoff relations, rational method, unit hydrograph or flood frequency analysis of gauged data.
  2. Design high flood level (HFL): From the discharge and the cross-section (rating curve), the HFL is found. The bridge deck level is fixed above HFL with a freeboard.
  3. Waterway (linear opening): The span and number of piers are chosen so that the flood passes without excess afflux, using the discharge and regime width (Lacey).
  4. Scour depth: Scour depends on discharge per unit width and silt factor, and decides the foundation depth of piers and abutments.
  5. Afflux and backwater: Narrowing the channel raises the upstream level; this needs hydrologic and hydraulic computation to avoid flooding upstream.
  6. River training and protection: Guide bunds, spurs and pitching are based on flow and flood data.
  7. Economy and safety: An underestimate causes overtopping, scour and failure (common in Nepal's monsoon), and an overestimate wastes money.

A civil engineer therefore cannot fix the span, deck level or foundation depth without hydrologic data.

Questions from Old Question Collection (CE 606) (IOE exam papers 2066 Magh to 2079 Baisakh) and Old Question Collection (CE 606) (IOE exam papers 2076 Chaitra to 2082 Bhadra, plus older repeats). Answers are written for this site; check them against your class notes.

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