Hydrology 1

Precipitation and Hydrologic Cycle

Precipitation Interpretation:
- Average precipitation in an area.
- Simpler method: add all precipitation.
- Consider the area; closer to the form, combination of flow of water.
- Groundwater flow is part of the hydrologic cycle.
- Surface runoff: flow of water.
- Water amount decreases due to:
  - Voids.
  - Volume of water and air affecting infiltration.
Infiltration Processes:
- Difference in pressure.
- Difference in temperature.
- Occurs in mountains due to difference in elevation.

Hydrologic Cycle

Sun:
- Main driver for the hydrologic cycle, which has no start and end.
Hydrologic Cycle Processes:
- Solid to water.
- Vapor evaporation in plants.
- Sublimation.
- Deposition.
- Free evaporation.
- Condensation.
States of Matter:
- Plasma (4th state of matter).
- Bose-Einstein Condensate (5th state).
  - Reached at absolute zero temperature (approximately -273^\circ C).
Hydrologic Budget:
- Measure or balance the hydrologic budget.
- Ecosystem and water under the ground.
Hydrology vs. Hydraulics:
- Hydrology: study of water.
- Hydraulics: foundation for water supply engineering.

System Inflow and Outflow

Example:
- Projecting the water level of Angat Dam.
System Dynamics:
- Q{in} = Q{out} \pm \Delta S_s = 0
  - Q_{in}: Inflow.
  - Q_{out}: Outflow.
  - \Delta S_s: Change in storage.
- Combined inflow (Q{in}) and outflow (Q{out}) equation.
- (Q{in} - Q{out}) = 0

Water Distribution

Distribution Types:
- Atmospheric.
- Surface (rivers, lakes).
- Subsurface.
- Biological.
Volume of Water:
- World water quantity.
Flowrate:
- Measured in km^3/year.
Global Annual Water Balance:
- Ocean.
- Land.

Water Movement

Atmosphere to Ground:
- Water travels from the atmosphere to the ground.
Phase Changes:
- Water vapor to liquid.
- Water vapor to solid (ice).
Particle Behavior:
- Tendency of small particles to become large particles.
Frozen Precipitation (Snow):
- Sponge texture (large voids).
- Symmetrical shape.
- Fast velocity (big or solid, bumabagsak).
Rainfall Droplets:
- Movement of rainfall from atmosphere to earth's surface.
- For pressure 101.3 kPa and density 1000 kg/m^3.
- As diameter increases, terminal velocity increases.
- Velocity is due to the diameter.
Exact Computation:
- Diameter calculation using volume of sphere relative to density of water and air.
Rainfall Measurement:
- Measured in heights.

Rainfall Intensity

Measurement Intervals:
- Every 5 minutes.
- Example: 6:00, 6:05, 6:10, 6:15, 6:20, 6:25, 6:30.
Rainfall Intensity Calculation:
- Convert 5-minute rainfall to 30-minute rainfall interval.
- I_{max} = 5.28 \text{ in/hr}
- Maximum depth.

Rainfall Gauges

Interpretation:
- Interpret data from one rainfall gauge.
- If multiple gauges, simply get the average.
- Consider the area.
Precipitation:
- Total area.

Areal Rainfall Methods

Thiessen Method:
1. Connect the rain gauges (no intersection to each other).
2. Get the midway of each connecting line and draw a perpendicular line.
3. Get the area corresponding to a particular gauge.
4. Compute areal rainfall (weighted average).
  - Formula: \text{Average Rainfall} = \frac{\sum Ai Pi}{\sum A_i}
  - Example Data:
    - P1 = 10.0 \text{ mm or in}, A1 = 0.22 \text{ km}^2 \text{or mi}^2, A1P1 = 2.2
    - P2 = 20.0 \text{ mm or in}, A2 = 2.02 \text{ km}^2 \text{or mi}^2, A2P2 = 40.5
    - P3 = 30.0 \text{ mm or in}, A3 = 1.35 \text{ km}^2 \text{or mi}^2, A3P3 = 40.5
    - P4 = 40.0 \text{ mm or in}, A4 = 1.60 \text{ km}^2 \text{or mi}^2, A4P4 = 64.0
    - P5 = 50.0 \text{ mm or in}, A5 = 1.95 \text{ km}^2 \text{or mi}^2, A5P5 = 97.5
    - \sum Ai = 9.14 \text{ km}^2 \text{or mi}^2, \sum AiP_i = 284.6
  - Average rainfall = 284.6 / 9.14 \approx 31.1 \text{ mm or in}
Isohyetal Method:
- Overcomes difficulties by constructing isohyets.
- Uses observed depths at rain gauges and interpolation between adjacent gauges.
- With a dense network of rain gauges, isohyetal maps can be constructed using computer programs for automated contouring.
  - Example Data:
    - Average Rainfall (mm or in), enclosed Area (km^2 or mi^2) and Rainfall Volume (mm or in).
    - Between isohyets 0-10, Area = 0.88, Average Rainfall = 5, Rainfall Volume = 4.4
    - Between isohyets 10-20, Area = 1.59, Average Rainfall = 15, Rainfall Volume = 23.9
    - Between isohyets 20-30, Area = 2.24, Average Rainfall = 25, Rainfall Volume = 56.0
    - Between isohyets 30-40, Area = 3.01, Average Rainfall = 35, Rainfall Volume = 105.4
    - Between isohyets 40-50, Area = 1.22, Average Rainfall = 45, Rainfall Volume = 54.9
    - Between isohyets 50-60, Area = 0.20, Average Rainfall = 53, Rainfall Volume = 10.6
    - Estimated Total Area: 9.14, Total Rainfall Volume: 255.2
  - Average rainfall = 255.2 / 9.14 \approx 27.9 \text{ mm or in}
Reciprocal Squared Distance Method

Sample Problem: Thiessen Method

Problem: Compute the mean aerial rainfall using Theissen's Method for a small urban watershed with four rainfall gauges.
Given:
- Rainfall recorded at each gauge during a storm event.
- Gage A: 81.50 mm
- Gage B: 73.00 mm
- Gage C: 75.25 mm
- Gage D: 76.25 mm
Solution:
- Areas:
  - A_1 = (800 \times 300) / 2 = 120,000 m^2
  - A_2 = ((800 + 200) / 2) \times 400 = 200,000 m^2
  - A3 = A2 = 200,000 m^2
  - A_4 = (700 \times 800) - (400 \times 300) / 2 - (400 \times 400) / 2 = 280,000 m^2
- Mean Aerial Rainfall:
  - P = \frac{81.5A1 + 73A2 + 75.25A3 + 76.25A4}{A1 + A2 + A3 + A4} = 75.98 \text{ mm}

Return Period

Determine the number of years of data, n.
Set rainfall duration for analysis (5 minutely, hourly, daily, etc.).
Find the maximum depth for the duration in each year.
Rank the depths from highest to lowest for all years (greatest amount at top of list, rank = m = 1).
Compute return period:
- T = \frac{n+1}{m}
  - where n is the number of years of data, m is the rank of data from highest (m=1) to lowest (m=n).
- Corresponding probability = 1/T
  - Example: For T = 100 year event, the probability = 0.01.

Partial Duration Series:
- Algorithm swaps out maximum for during n years with n maximum in n years (e.g., more than 1 value per year allowed).

Additional Notes on Return Period

n = 20 years.
Ocanerank.
Reciprocal of return period.
Chance na marit.
Probability is inversely proportional to area.
n = 26 years.
1/20.
Dependent on historical data.
Change of state from water to water vapor.

Comparing Water Vapor to Moist Air

Specific Humidity:
- Mass of water vapor per unit mass of moist air.
Vapor Pressure:
- Dalton's law of partial pressures states that the pressure exerted by a gas (its vapor pressure) is independent of the presence of other gases.
- Based on ideal gas law:
  - If the total pressure exerted by the moist air is p, then p - e is the partial pressure due to the dry air:
    - p - e = \rhoa Ra T
    - e = \rhov Rv T
    - T: Absolute Temperature in K
    - e: Vapor pressure
    - R_v: Gas constant of water vapor
    - \rho_v: Density of water vapor
Specific Humidity Approximation:
- q_v = 0.622 \frac{e}{p}
- R_a: Gas constant for dry air (287 J/kg\cdot K)
- \rho_a: Density of dry air
- \rho = \rhoa + \rhov
- R_v: Gas constant for water vapor
- p = [\rhoa + (0.622)\rhov] R_a T
Gas Constants Relationship:
- Ra = Ra (1 + 0.608qv) \approx 287 (1 + 0.608qv)
Saturated Vapor Pressure:
- For a given air temperature, there is a maximum moisture content the air can hold, and the corresponding vapor pressure is called saturation vapor pressure.
- e_s = 611 e^{\frac{17.27T}{237.3 + T}}
  - where e_s is Saturated vapor pressure (Pa = N/m^2) and T is Temperature (\degree C)
  - Example Values:
    - T = -20, e_s = 125
    - T = -10, e_s = 286
    - T = 0, e_s = 611
    - T = 5, e_s = 872
    - T = 10, e_s = 1227
    - T = 15, e_s = 1704
    - T = 20, e_s = 2337
    - T = 25, e_s = 3167
    - T = 30, e_s = 4243
    - T = 35, e_s = 5624
    - T = 40, e_s = 7378
Relative Humidity:
- Ratio of the actual vapor pressure to its saturation value at a given air temperature:
  - Rh = \frac{e}{es}
Dew Point Temperature:
- The temperature at which air would just become saturated at a given specific humidity.
Sample Problem:
- Given:
  - Air pressure = 100 kPa
  - Air temperature = 20 ^\circ C
  - Dew-point temperature = 16 ^\circ C
- Calculate:
  - Specific humidity
  - Relative Humidity: \frac{e}{e_s}
    - e_s = 611 e^{\frac{17.27T}{237.3 + T}} \approx 2339 Pa
    - e = 611 e^{\frac{17.27T}{237.3 + T}} \approx 1819 Pa
  - R_h = \frac{1819}{2339} \approx 78\%%
  - q_v = 0.622 \frac{e}{p} \approx 0.0113 \frac{\text{kg of water}}{\text{kg of moist air}}

Terminologies

Evaporation:
- Process by which liquid water passes directly to the vapor phase.
Transpiration:
- Process by which liquid water passes from liquid to vapor through plant metabolism.
Sublimation:
- Process by which water passes directly from the solid phase to the vapor phase.
Vapor Pressure:
- Water vapor normally behaves as an ideal gas.
- Partial pressure of water (vapor pressure) adds to partial pressures of the other gaseous constituents.
- Water vapor is about 1-2% of the total pressure.
Humidity:
- Quantity of water vapor present in air (absolute, specific, or a relative value).
Specific Humidity:
- Ratio of the mass of water vapor in moist air to the mass of air.
Dew Point Temperature:
- Temperature at which air becomes saturated at a given specific humidity.

Factors Influencing Evaporation

Energy supply for vaporization (latent heat):
- Solar radiation.
Transport of vapor away from evaporative surface:
- Wind velocity over surface.
- Specific humidity gradient above surface.
- Vegetated surfaces.
Supply of moisture to the surface:
- Evapotranspiration (ET).
  - Potential Evapotranspiration (PET): moisture supply is not limited.

Evaporation from Pan:
- National Weather Service Class A type.
- Installed on a wooden platform in a grassy location.
- Filled with water to within 2.5 inches of the top.
- Evaporation rate is measured by manual readings or with an analog output evaporation gauge.

Methods for Estimating Evaporation

Energy Balance Method:
- Parameter is energy from on.
- Rn - Hs - G = Lv \rhow E
  - R_n = net radiation, W/m^2
  - L_v = latent heat of vaporization, kJ/kg
  - \rhow = density of water, kg/m^3 HS = Sensible heat to air
    G = Heat conducted to the ground

Example:
- For a particular location, the average net radiation is 185 W/m^2, air temperature is 28.5°C, relative humidity is 55%, and wind speed is 2.7 m/s at a height of 2m. Determine the open water evaporation rate in mm/d using the energy method.
  - L_v = 2.501 \times 10^6 - 2370T = 2501 - 2.37 \times 28.5 = 2433 kJ/kg
  - E = \frac{Rn}{\rho Lv} = \frac{185}{2433 \times 10^3 \times 996.3} = 7.63 \times 10^{-8} m/s = 6.6 mm/d

Aerodynamic Method:
- Ea = B(es - e)
  - B is the vapor transfer coefficient with units of mm/day
  - B = \frac{0.1022}{[ln(z/z_0)]}
    - where uz is the wind velocity (m/s) measured at height z (cm) and z0 is the roughness height (0.01-0.06 cm) of the water surface.
Combined Method:
- E = \frac{\Delta}{\Delta + \gamma} Er + \frac{\gamma}{\Delta + \gamma} Ea
  - Er is the vapor transport term and Ea is the aerodynamic term.
  - \gamma is the psychrometric constant (approximately 66.8 Pa/°C).
  - \Delta is the gradient of the saturated vapor pressure curve.
    - \Delta = \frac{4098e_s}{(237.3 + T)^2}
Priestly Taylor Method:
- Similar to the combined method, with constant.
- E = 1.3 \frac{\Delta}{\Delta + \gamma} E_r

Vapor Flow Rate

mv = \rhow A E
V = A \frac{dh}{dt}
\rho_w \approx 1000 kg/m^3