Comprehensive Study Notes on Hydrology, the Hydrological Cycle, and Water Balance Principles

Fundamentals of Hydrology

  • Hydrology is defined as the branch of geophysics that studies water on Earth, encompassing its distribution, circulation, and properties.

  • The field investigates the continuous hydrological cycle and its multi-faceted interactions with the atmosphere, land, and living organisms.

  • Hydrology is closely linked with Earth Sciences, integrating principles from geology, geomorphology, and meteorology to understand water movements and their environmental impacts.

  • Hydrology is fundamental to water resources management, which includes:

    • Provision and management of drinking water supply.

    • Agricultural irrigation design and distribution.

    • Production of hydroelectric energy.

  • The discipline plays a vital role in environmental protection, specifically in flood risk prevention and mitigation, as well as maintaining water quality standards.

Practical Case Study: Gallito Ciego Dam and Hydroelectric Plant

  • Location and Context:

    • Located at the foot of the Gallito Ciego dam in the Cajamarca region of Peru.

    • Primarily utilized for agricultural irrigation activities in the fertile valley of the Jequetepeque River.

  • Ownership and Administration:

    • Statkraft is not the owner of the dam structure.

    • Operation and administrative responsibilities are assigned to relevant competent state entities.

  • Power Generation Capabilities:

    • The dam supplies water to feed the Gallito Ciego Hydroelectric Power Plant.

    • The plant was inaugurated in 19881988 as part of the broader Jequetepeque-Zaña multi-purpose project.

    • Features an installed capacity of 38 MW38\,MW.

    • Generates approximately 134.3 GWh134.3\,GWh of electrical energy annually.

    • Produces sufficient electricity to power over 100,000100,000 households.

  • Operational Mechanics:

    • The power station is controlled and operated remotely from the central dispatch facility in Lima.

    • A compensation reservoir regulates the discharge of water flow into the river bed downstream.

    • This regulated flow ensures the continuation of economic and ecological activities in the lower river basin and enables structured planning for agricultural water demands across the Jequetepeque river valley.

Applications of Hydrology in Civil Engineering

  • Hydrology provides essential data and analytical frameworks required for planning and engineering water-related infrastructure.

  • Structural Infrastructure and Hydraulic Works:

    • Provides calculations of design flows and runoff volumes necessary for designing dams, reservoirs, canals, and drainage networks.

  • Transportation Infrastructure:

    • Determines peak discharge flows and water dynamics during high-intensity rainfall events for the safe structural design of bridges, road culverts, drainage tunnels, and highways.

  • Wastewater Management and Sanitation Systems:

    • Informs the physical design, capacity sizing, and operational management of storm sewer systems, municipal wastewater treatment plants, and urban waste control.

The Hydrological Cycle and Sub-Systems

  • The hydrological cycle describes the uninterrupted movement of water between the atmosphere, terrestrial land, and oceans, continuously driven by solar energy.

  • The Five Core Phases of the Hydrological Cycle:

    • Evaporation: Process where liquid water from Earth's surface, open water bodies, and living organisms transitions into water vapor and ascends into the atmosphere.

    • Condensation: Process where rising water vapor cools in the upper atmosphere, changing phase into liquid water droplets or ice crystals to form clouds.

    • Precipitation: Process where condensed atmospheric moisture falls back to Earth in forms such as rain, snow, sleet, hail, or dew.

    • Runoff: Process where precipitation flows across the land surface under the influence of gravity, feeding streams, rivers, lakes, and oceans.

    • Infiltration: Process where surface water penetrates into the soil layers, replenishing groundwater aquifers and subsurface water bodies.

  • Hydrological Sub-Systems and Water Dynamics:

    • Evapotranspiration: Combined transfer of moisture to the atmosphere from soil evaporation and plant transpiration.

    • Surface and Subsurface Movement: Rainwater either flows over ground surfaces toward channels or infiltrates deep into soil layers to form groundwater reserves.

Distribution of Water on Earth

  • Global Water Volume Allocation:

    • Oceans (Saltwater): 96.5%96.5\% of total planetary water.

    • Ice Caps and Glaciers: 2.5%2.5\% of total planetary water.

    • Groundwater: 0.5%0.5\% of total planetary water.

    • Lakes and Rivers: 0.01%0.01\% of total planetary water.

    • Atmosphere: 0.001%0.001\% of total planetary water.

  • Planetary Surface and Fresh Water Breakdown:

    • Total Earth surface covered by water: 70%70\%.

    • Composition of global water: 97.5%97.5\% saltwater and 2.5%2.5\% freshwater.

  • Distribution of the 2.5%2.5\% Freshwater Reserve:

    • Glaciers, permanent snowpack, or ice cover: approximately 70%70\%.

    • Deep or difficult-to-access groundwater: 30%30\%.

    • Readily available freshwater for direct human consumption and ecosystem support: less than 1%1\%.

  • Global Water Extraction by Sector:

    • Agricultural and Livestock Sector: 69%69\%.

    • Industrial Sector: 19%19\%.

    • Municipal and Domestic Sector: 12%12\%.

Hydrological Systems and Modeling

  • Concept of the Hydrological System:

    • Hydrological phenomena are intrinsically complex and impossible to model perfectly in every minute detail.

    • To simplify analysis, hydrology uses the concept of a "system"—a collection of interconnected components operating together as a unified whole.

    • The global hydrological cycle functions as a continuous system with inputs, operators, and outputs.

  • Structural Components of the System:

    • Atmospheric Water Subsystem: Includes interception, evaporation, transpiration, and precipitation.

    • Surface Water Subsystem: Includes overland flow, surface runoff discharging into channels and seas, infiltration, and subsurface flow.

    • Subsurface/Groundwater Subsystem: Includes deep percolation and deep groundwater flow.

  • System Operation Framework:

    • Represented mathematically as an input function I(t)I(t), passing through a system operator function Ω(t)\Omega(t), resulting in an output function Q(t)Q(t).

  • Hydrological Models:

    • Models are analytical tools used to simulate, analyze, and project water behavior within a defined watershed or control volume.

    • Case Study Example: Hydrological modeling of the upper basin of the Tabaconas River utilizing the Soil & Water Assessment Tool (SWAT) model, authored by Renny D. Diaz Aguilar (Undergraduate Environmental Engineering student at Universidad Nacional Agraria La Molina).

    • Model Requirements: Requires historical input datasets such as precipitation rates, river discharge rates, and groundwater levels for precise model calibration and validation.

    • Applications: Utilized for forecasting hydrological events, including flood warnings, drought severity assessment, and regional water supply availability.

Water Balance Equation and Continuity Principle

  • Mass Conservation and Continuity Principle:

    • The water balance equation is grounded in the principle of continuity, stating that the difference between mass inputs and mass outputs over a control volume during a given time interval equals the net change in storage.

  • General Fundamental Equation:

    • I−O=ΔSΔtI - O = \frac{\Delta S}{\Delta t}

    • In verbal terms: Input minus Output equals Change in Storage.

  • Watershed Hydrological Balance Inputs and Outputs:

    • Inputs (II):

    • Direct precipitation (PP).

    • Inter-basin surface water imports.

    • Surface runoff entering from neighboring basins.

    • Groundwater inflow entering from external basins.

    • Outputs (OO):

    • Open water evaporation (EE).

    • Plant transpiration (TT) and evapotranspiration (ETET).

    • Surface runoff exiting to downstream basins or seas (QoutQ_{out}).

    • Inter-basin surface water exports.

    • Subsurface groundwater outflow to adjacent basins (Qg_outQ_{g\_out}).

    • Deep soil infiltration and percolation.

    • Storage Components (ΔS\Delta S):

    • Groundwater storage volume changes.

    • Soil moisture content variations.

    • Surface storage in reservoirs, lakes, or snowpacks.

  • Mathematical Watershed Balance Equation:

    • P+Qin+Qg_in−E−ET−Qout−Qg_out−ΔS−η=0P + Q_{in} + Q_{g\_in} - E - ET - Q_{out} - Q_{g\_out} - \Delta S - \eta = 0

    • Symbol definitions:

    • PP: Total precipitation.

    • QinQ_{in}: Surface water flow entering the basin.

    • Qg_inQ_{g\_in}: Groundwater flow entering the basin.

    • EE: Evaporation losses.

    • ETET: Evapotranspiration losses.

    • QoutQ_{out}: Surface water outflow leaving the basin.

    • Qg_outQ_{g\_out}: Groundwater outflow leaving the basin.

    • ΔS\Delta S: Net change in internal basin water storage.

    • η\eta: Residual discrepancy error term (accounting for measurement error and unmeasured components).

  • Expression Units:

    • Can be expressed in equivalent depth of water over the basin surface area (mmmm), total accumulated volume (m3m^3), or instantaneous flow rate (m3/sm^3/s).

Practical Evaluation and Knowledge Check

  • Assessment Question 1: What is the process by which water converts into vapor and ascends into the atmosphere?

    • Options: a) Condensation, b) Precipitation, c) Evaporation, d) Infiltration.

    • Correct Answer: c) Evaporation.

  • Assessment Question 2: What is the hydrological cycle?

    • Options: a) A process of circulation of water on Earth, b) A type of animal life cycle, c) An isolated atmospheric phenomenon.

    • Correct Answer: a) A process of circulation of water on Earth.

  • Assessment Question 3: Which term refers to the movement of water over the Earth's surface towards rivers, lakes, or seas?

    • Options: a) Infiltration, b) Condensation, c) Runoff, d) Transpiration.

    • Correct Answer: c) Runoff.

Solved Hydrological Balance Numerical Exercises

  • Exercise 01: Determination of Surface Runoff Depth

    • Problem Parameters:

    • Annual precipitation P=1800 mm/yearP = 1800\,mm/year.

    • Infiltration rate Iinf=35%I_{inf} = 35\% of precipitation.

    • Evapotranspiration rate ET=600 mm/yearET = 600\,mm/year.

    • Step-by-Step Calculation:

    • Calculate annual infiltration volume depth: Iinf=0.35×1800 mm=630 mm/year=0.63 m/yearI_{inf} = 0.35 \times 1800\,mm = 630\,mm/year = 0.63\,m/year.

    • Apply balance equation assuming steady-state storage (ΔS=0\Delta S = 0):

      • ΔS=P−Iinf−ET−Erunoff=0\Delta S = P - I_{inf} - ET - E_{runoff} = 0

      • Erunoff=P−Iinf−ETE_{runoff} = P - I_{inf} - ET

      • Erunoff=1800 mm−630 mm−600 mm=570 mm/yearE_{runoff} = 1800\,mm - 630\,mm - 600\,mm = 570\,mm/year

    • Final Result:

    • Surface runoff E_{runoff} = 570\,mm/year = 0.57\,m/year$.\n\n- Exercise 02: Lake Evaporative Loss Calculation\n - Problem Parameters:\n - Lake surface area A = 1.5\,km^2 = 1.5 \times 10^6\,m^2$.

    • Analysis period t = 1\,month = 30\,days = 2,592,000\,s$.\n - Average monthly inflow Q_{in} = 0.5\,m^3/s$.

    • Average monthly outflow Q_{out} = 0.3\,m^3/s$.\n - Total storage increase \Delta S_{vol} = 0.1\,km^2 \cdot m = 100,000\,m^3$.

    • Total monthly precipitation depth P = 50\,mm$.\n - Soil infiltration loss I_{inf} \approx 0\,mm$.

    • Step-by-Step Solution:

    1. Convert volumetric inflow and outflow into equivalent water depth over lake area:

      • Vin=Qin×t=0.5 m3/s×2,592,000 s=1,296,000 m3V_{in} = Q_{in} \times t = 0.5\,m^3/s \times 2,592,000\,s = 1,296,000\,m^3

      • Depthin=1,296,000 m31.5×106 m2=0.864 m=864 mmDepth_{in} = \frac{1,296,000\,m^3}{1.5 \times 10^6\,m^2} = 0.864\,m = 864\,mm

      • Vout=Qout×t=0.3 m3/s×2,592,000 s=777,600 m3V_{out} = Q_{out} \times t = 0.3\,m^3/s \times 2,592,000\,s = 777,600\,m^3

      • Depthout=777,600 m31.5×106 m2=0.5184 m=518.4 mmDepth_{out} = \frac{777,600\,m^3}{1.5 \times 10^6\,m^2} = 0.5184\,m = 518.4\,mm

    2. Convert volumetric storage change into equivalent depth:

      • ΔSdepth=0.1 km2⋅m1.5 km2=0.0666667 m=66.6667 mm\Delta S_{depth} = \frac{0.1\,km^2 \cdot m}{1.5\,km^2} = 0.0666667\,m = 66.6667\,mm

    3. Apply water balance equation to solve for evaporation (EE):

      • ΔSdepth=P+Depthin−Depthout−E\Delta S_{depth} = P + Depth_{in} - Depth_{out} - E

      • E=P+Depthin−Depthout−ΔSdepthE = P + Depth_{in} - Depth_{out} - \Delta S_{depth}

      • E=50 mm+864 mm−518.4 mm−66.6667 mm=328.9333 mmE = 50\,mm + 864\,mm - 518.4\,mm - 66.6667\,mm = 328.9333\,mm

    • Final Result:

    • Total monthly evaporation loss E = 328.9333\,mm$.\n\n- Exercise 03: Reservoir Net Annual Storage Variation\n - Problem Parameters:\n - Annual recharge/precipitation P = 800\,mm/year$.

    • Reservoir evaporation rate E = 400\,mm/year$.\n - Outflow runoff Q_{out} = 300\,mm/year$.

    • Step-by-Step Calculation:

    • ΔS=P−(E+Qout)\Delta S = P - (E + Q_{out})

    • ΔS=800 mm−(400 mm+300 mm)=100 mm/year\Delta S = 800\,mm - (400\,mm + 300\,mm) = 100\,mm/year

    • Final Result:

    • Net annual reservoir storage V_{H2O} = 100\,mm/year$.\n\n- Exercise 04: Reservoir Available Supply Flow Rate\n - Problem Parameters:\n - Drainage basin area A_{basin} = 500\,km^2 = 500 \times 10^6\,m^2$.

    • Historical record duration = 50\,years$.\n - Average annual basin precipitation P = 90\,cm/year = 0.90\,m/year$.

    • Average annual basin runoff depth R = 33\,cm/year = 0.33\,m/year$.\n - Reservoir surface area A_{res} = 1700\,ha = 17\,km^2 = 1.7 \times 10^7\,m^2$.

    • Annual reservoir evaporation rate E_{res} = 130\,cm/year = 1.30\,m/year$.\n - Subsurface infiltration and external basin inflows = 0\,m^3/s$.

    • Step-by-Step Solution:

    1. Perform water balance on the watershed basin:

      • Total basin rain volume: Vrain=P×Abasin=0.90 m/year×500×106 m2=450,000,000 m3/year=14.2599 m3/sV_{rain} = P \times A_{basin} = 0.90\,m/year \times 500 \times 10^6\,m^2 = 450,000,000\,m^3/year = 14.2599\,m^3/s

      • Runoff volume reaching the reservoir: Vrunoff=R×Abasin=0.33 m/year×500×106 m2=165,000,000 m3/year=5.2286 m3/sV_{runoff} = R \times A_{basin} = 0.33\,m/year \times 500 \times 10^6\,m^2 = 165,000,000\,m^3/year = 5.2286\,m^3/s

    2. Perform water balance on the reservoir:

      • Evaporation loss volume: Vevap=Eres×Ares=1.30 m/year×17,000,000 m2=22,100,000 m3/year=0.7003 m3/sV_{evap} = E_{res} \times A_{res} = 1.30\,m/year \times 17,000,000\,m^2 = 22,100,000\,m^3/year = 0.7003\,m^3/s

      • Available withdrawal flow rate: ΔV=Vrunoff−Vevap\Delta V = V_{runoff} - V_{evap}

      • ΔV=5.2286 m3/s−0.7003 m3/s=4.5283 m3/s\Delta V = 5.2286\,m^3/s - 0.7003\,m^3/s = 4.5283\,m^3/s

    • Final Result:

    • Average available annual flow rate for community supply Q_{available} = 4.5283\,m^3/s$.\n\n- Exercise 05: Reservoir Constant Outflow Sizing\n - Problem Parameters:\n - Reservoir surface area A = 500\,ha = 5 \times 10^6\,m^2$.

    • Duration t = 30\,days = 2,592,000\,s$.\n - Water level decrease \Delta h = -0.50\,m$.

    • Daily inflow rate Q_{in} = 200,000\,m^3/day$.\n - Seepage/groundwater loss depth I_{sub} = 2\,cm = 0.02\,m$.

    • Total precipitation depth P = 10.5\,cm = 0.105\,m$.\n - Total evaporation depth E = 8.5\,cm = 0.085\,m$.

    • Step-by-Step Solution:

    1. Volumetric storage change (ΔVstorage\Delta V_{storage}):

      • ΔVstorage=Δh×A=−0.50 m×5×106 m2=−2.5×106 m3=−2,500,000 m3\Delta V_{storage} = \Delta h \times A = -0.50\,m \times 5 \times 10^6\,m^2 = -2.5 \times 10^6\,m^3 = -2,500,000\,m^3

    2. Total inflow volume (VinV_{in}):

      • Vin=Qin×30 days=200,000 m3/day×30 days=6×106 m3=6,000,000 m3V_{in} = Q_{in} \times 30\,days = 200,000\,m^3/day \times 30\,days = 6 \times 10^6\,m^3 = 6,000,000\,m^3

    3. Total precipitation volume (VpV_p):

      • Vp=P×A=0.105 m×5×106 m2=5.25×105 m3=525,000 m3V_p = P \times A = 0.105\,m \times 5 \times 10^6\,m^2 = 5.25 \times 10^5\,m^3 = 525,000\,m^3

    4. Total evaporation volume (VevapV_{evap}):

      • Vevap=E×A=0.085 m×5×106 m2=4.25×105 m3=425,000 m3V_{evap} = E \times A = 0.085\,m \times 5 \times 10^6\,m^2 = 4.25 \times 10^5\,m^3 = 425,000\,m^3

    5. Total groundwater loss volume (VsubV_{sub}):

      • Vsub=Isub×A=0.02 m×5×106 m2=1.0×105 m3=100,000 m3V_{sub} = I_{sub} \times A = 0.02\,m \times 5 \times 10^6\,m^2 = 1.0 \times 10^5\,m^3 = 100,000\,m^3

    6. Water balance equation for total output volume (VoutV_{out}):

      • ΔVstorage=Vin+Vp−Vevap−Vsub−Vout\Delta V_{storage} = V_{in} + V_p - V_{evap} - V_{sub} - V_{out}

      • Vout=Vin+Vp−Vevap−Vsub−ΔVstorageV_{out} = V_{in} + V_p - V_{evap} - V_{sub} - \Delta V_{storage}

      • Vout=6,000,000+525,000−425,000−100,000−(−2,500,000)V_{out} = 6,000,000 + 525,000 - 425,000 - 100,000 - (-2,500,000)

      • Vout=8.5×106 m3=8,500,000 m3V_{out} = 8.5 \times 10^6\,m^3 = 8,500,000\,m^3

    7. Constant discharge outflow rate (QoutQ_{out}):

      • Qout=Voutt=8,500,000 m32,592,000 s=3.2793 m3/sQ_{out} = \frac{V_{out}}{t} = \frac{8,500,000\,m^3}{2,592,000\,s} = 3.2793\,m^3/s

    • Final Result:

    • Constant water output rate $$Q_{out} = 3.2793\,m^3/s$.