Introduction to Hydrology

Hydrological cycle

The hydrological cycle is the continuous circulation of water between the oceans, atmosphere and land, driven by solar energy and gravity. It has no beginning or end; the total water on earth stays constant.

        Sun
         \   condensation -> clouds
          \   ~~~~~~~~~~~~~~~~~~~~
   evapo- ^        | precipitation
 transpir.|        v (rain, snow)
  _/\_/\__|___  infiltration    evaporation
  vegetation  \_______________   ^
     |   surface runoff ----> rivers --> OCEAN
     v   groundwater flow -----------^

Main processes:

1. Evaporation from oceans, lakes and rivers, and transpiration from plants (together evapotranspiration).
2. Condensation of vapour into clouds as air rises and cools.
3. Precipitation as rain, snow or hail.
4. Interception by vegetation and depression storage.
5. Infiltration into the soil and percolation to groundwater.
6. Surface runoff and groundwater flow (base flow) into rivers, lakes and finally the ocean.

Importance for hydropower: it decides the available river flow, its seasonal variation (monsoon high flows, dry-season low flows in Nepal), snow-fed contribution, and floods for spillway design.

Methods of finding mean precipitation over an area

Rain gauges give point rainfall; the mean (areal) depth over the catchment is found by:

1. Arithmetic mean method

$$\bar{P} = \frac{P_1 + P_2 + \dots + P_N}{N}$$

Simple; good only for flat areas with uniformly spread gauges and small rainfall variation.

2. Thiessen polygon method

• Plot stations on the map and join adjacent stations by straight lines to form triangles.
• Draw perpendicular bisectors of these lines; they form a polygon around each station.
• Measure the polygon area $A_i$ inside the catchment for each station.

$$\bar{P} = \frac{\sum P_i A_i}{\sum A_i}$$

Accounts for uneven gauge spacing; can use stations just outside the catchment. Does not consider orography; polygons must be redrawn if a gauge is added or removed.

3. Isohyetal method

• Draw isohyets (lines of equal rainfall) by interpolating between station values.
• Measure the area $a_i$ between successive isohyets $P_i$ and $P_{i+1}$.

$$\bar{P} = \frac{\sum a_i \left(\frac{P_i + P_{i+1}}{2}\right)}{\sum a_i}$$

Most accurate, especially in hilly areas like Nepal, as the analyst can use knowledge of topography; but needs skill and a dense network.

  Isohyets (mm)          Thiessen polygons
   ___80___               +-----+-----+
  /  _60_  \              |  A  |  B  |
 |  / 40 \  |             |  *  |  *  |
  \ \____/ /              +--+--+--+--+
   \______/                  | C *  |
                             +------+

The average depth of rainfall over a catchment is needed because rain gauges record only point values, while runoff and hydropower flow depend on rainfall over the whole area. Three standard techniques are used.

Methods of finding mean precipitation over an area

Rain gauges give point rainfall; the mean (areal) depth over the catchment is found by:

1. Arithmetic mean method

$$\bar{P} = \frac{P_1 + P_2 + \dots + P_N}{N}$$

Simple; good only for flat areas with uniformly spread gauges and small rainfall variation.

2. Thiessen polygon method

• Plot stations on the map and join adjacent stations by straight lines to form triangles.
• Draw perpendicular bisectors of these lines; they form a polygon around each station.
• Measure the polygon area $A_i$ inside the catchment for each station.

$$\bar{P} = \frac{\sum P_i A_i}{\sum A_i}$$

Accounts for uneven gauge spacing; can use stations just outside the catchment. Does not consider orography; polygons must be redrawn if a gauge is added or removed.

3. Isohyetal method

• Draw isohyets (lines of equal rainfall) by interpolating between station values.
• Measure the area $a_i$ between successive isohyets $P_i$ and $P_{i+1}$.

$$\bar{P} = \frac{\sum a_i \left(\frac{P_i + P_{i+1}}{2}\right)}{\sum a_i}$$

Most accurate, especially in hilly areas like Nepal, as the analyst can use knowledge of topography; but needs skill and a dense network.

  Isohyets (mm)          Thiessen polygons
   ___80___               +-----+-----+
  /  _60_  \              |  A  |  B  |
 |  / 40 \  |             |  *  |  *  |
  \ \____/ /              +--+--+--+--+
   \______/                  | C *  |
                             +------+

Example (Thiessen): stations with 80, 70, 90 mm over areas 100, 150, 80 km² give $\bar{P} = (8000 + 10500 + 7200)/330 = 77.9$ mm, while the arithmetic mean is 80 mm.

a) Thiessen polygon method

The Thiessen polygon method gives the average rainfall by weighting each station's rainfall by the area it represents. It assumes the rainfall at any point equals that at the nearest gauge.

Procedure:

1. Plot all rain gauge stations (inside and near the catchment) on a map of the catchment.
2. Join adjacent stations by straight lines to form a network of triangles.
3. Draw perpendicular bisectors of each side; they meet to form a polygon around each station.
4. Measure (by planimeter or GIS) the area $A_i$ of each polygon inside the catchment boundary.
5. Compute:

$$\bar{P} = \frac{P_1A_1 + P_2A_2 + \dots + P_nA_n}{A_1 + A_2 + \dots + A_n} = \sum P_i\frac{A_i}{A}$$

   +-------+--------+
   |   A   |    B   |   * = rain gauge
   |   *   |    *   |   lines = perpendicular
   +---+---+----+---+   bisectors
       |  C *   |
       +--------+

Merits: allows for uneven gauge spacing. Demerits: ignores orographic effects; polygons must be redrawn when the network changes.

b) Numerical

$$\bar{P} = \frac{\sum P_iA_i}{\sum A_i} = \frac{33500}{450} = 74.44\ \text{mm}$$

Answer: Average rainfall over the catchment ≈ 74.4 mm.

Arithmetic mean method

$$\bar{P} = \frac{30.8 + 34.6 + 32.0 + 24.6}{4} = \frac{122.0}{4} = 30.50\ \text{mm}$$

Thiessen polygon method

Each station's rainfall is weighted by its polygon area.

$$\bar{P} = \frac{\sum P_iA_i}{\sum A_i} = \frac{4664.8}{153} = 30.49\ \text{mm}$$

Answer: Arithmetic mean = 30.50 mm; Thiessen mean = 30.49 mm. The two are nearly equal here because the station areas are fairly similar; the Thiessen value gives slightly less weight to the larger rainfall at station 2.

Factors for selecting a stream gauging site

• Straight reach: about 5–10 times the river width upstream and downstream, with parallel flow lines.
• Stable channel: bed and banks free from scour, deposition and shifting, so the stage–discharge relation stays constant.
• Well-defined control (rock bar, rapids or a weir) downstream, giving a unique stage–discharge relation.
• No backwater effect from confluences, dams or tides; no eddies or dead water.
• Flow contained in one channel at all stages, without overflow at floods.
• Adequate depth and velocity for the current meter in the dry season.
• Easy access in all seasons and safe working (bridge, cableway).
• Away from tributaries, so the measured flow represents the site of interest (e.g. near the intake for a hydropower project).

Measuring velocity with a current meter

A current meter (cup type or propeller type) has a rotor whose speed of rotation $N$ is proportional to water velocity. Its calibration equation is $V = aN + b$.

Procedure:

1. Stretch a tag line across the river and divide the width into 15–25 segments (each carrying ≤ 10% of flow).
2. At each vertical, measure the depth $y$ with a sounding rod or weight.
3. Lower the meter (wading rod for shallow water; cable, weight and cableway or bridge for deep water) to the required depth:
   • Shallow (< 0.6–1 m): one reading at 0.6y from the surface.
   • Deeper: two readings at 0.2y and 0.8y; mean $V = (V_{0.2} + V_{0.8})/2$.
4. Allow the meter to settle, then count the revolutions over a fixed time (40–100 s). $N$ = revolutions/time.
5. Convert to velocity with the rating equation, e.g. $V = 0.51N + 0.03$.
6. Compute discharge by the mid-section or mean-section method: $Q = \sum b_i y_i V_i$.

  tag line  0   1   2   3   4   5
  ~~~~~~~~~~|~~~|~~~|~~~|~~~|~~~|~~~~
   \        o   o   o   o   o      /  o = meter at 0.6y
    \_______o___o___o___o_________/      or 0.2y & 0.8y

The mid-section method is used: each vertical represents a segment extending halfway to the neighbouring verticals, so $\Delta Q_i = w_i\,y_i\,V_i$ with $w_i = (x_{i+1} - x_{i-1})/2$. Velocity is from the rating equation $V = 0.51N + 0.03$, with $N$ = revolutions/second at 0.6 depth (gives mean velocity).

Computation

(At the edges, x = 0 and 12 m, depth and velocity are zero, so they add nothing.)

Sample: at x = 5 m, $N = 112/150 = 0.7467$ rev/s, $V = 0.51 \times 0.7467 + 0.03 = 0.4108$ m/s, $\Delta Q = 2.0 \times 2.5 \times 0.4108 = 2.054$ m³/s.

$$Q = \sum \Delta Q_i = 6.24\ \text{m}^3/\text{s}$$

Answer: Discharge ≈ 6.24 m³/s.

(Subramanya's textbook version gives the end verticals a width $\bar{W}_1 = (W_1 + W_2/2)^2/(2W_1) = 2.0$ m instead of 1.5 m, which gives Q ≈ 6.45 m³/s. Either method is accepted if stated.)

The mid-section method is used: each vertical represents the strip extending halfway to the adjacent verticals, $w_i = (x_{i+1} - x_{i-1})/2$, and $\Delta Q_i = w_i\,y_i\,V_i$. Velocity at 0.6 depth (taken as mean velocity) is from $V = 0.55N + 0.04$, with $N$ = revolutions per second.

Computation

Sample: at x = 10 m, $V = 0.55 \times 0.8333 + 0.04 = 0.4983$ m/s; $\Delta Q = 2.0 \times 2.4 \times 0.4983 = 2.392$ m³/s.

$$Q = 0.780 + 2.775 + 5.040 + 2.392 = 10.99\ \text{m}^3/\text{s}$$

Answer: Discharge ≈ 10.99 m³/s.

(If Subramanya's end-segment rule $\bar{W} = (W_1 + W_2/2)^2/(2W_1)$ is used for the first and last verticals, their widths become 3.125 m and Q ≈ 12.77 m³/s.)

The data table was not printed in the paper, so the full method is shown; the student puts in the given values.

Principle

Discharge is found by the velocity–area method: the cross-section is divided into vertical segments, the velocity in each is measured by a current meter, and

$$Q = \sum \Delta Q_i = \sum w_i\,y_i\,\bar{V}_i$$

Steps

1. Revolutions per second at each vertical: $N_i = \dfrac{\text{revolutions}}{\text{time (s)}}$.
2. Velocity at 0.6 depth from the calibration equation:

$$V_i = 0.55N_i + 0.04\ \text{m/s}$$

The velocity at 0.6y below the surface is taken as the mean velocity of that vertical. 3. Width of each segment (mid-section method):

$$w_i = \frac{x_{i+1} - x_{i-1}}{2}$$

where $x$ is the distance of the vertical from the bank. 4. Segment discharge: $\Delta Q_i = w_i\,y_i\,V_i$. 5. Total discharge: $Q = \sum \Delta Q_i$. The edge verticals (depth 0) give zero flow.

Table layout

Illustration

For a vertical at 4 m (neighbours at 1 m and 7 m) with depth 2.5 m and 60 revolutions in 100 s: $N = 0.6$ rev/s, $V = 0.55 \times 0.6 + 0.04 = 0.37$ m/s, $w = (7 - 1)/2 = 3$ m, $\Delta Q = 3 \times 2.5 \times 0.37 = 2.775$ m³/s.

Repeating for every vertical and adding gives the stream discharge. (With the 2072 Asoj data of the same form, this method gives Q ≈ 10.99 m³/s.)

Mean velocity in each vertical is the average of the velocities at 0.2d and 0.8d: $\bar{V} = (V_{0.2} + V_{0.8})/2$. The mid-section method is used; the verticals are equally spaced at 1.5 m, so each interior vertical represents a 1.5 m wide strip. Depth is zero at both banks.

Computation

$$Q = \sum w_i\,y_i\,\bar{V}_i = 6.28\ \text{m}^3/\text{s}$$

Answer: Discharge ≈ 6.28 m³/s.

The two-point method is used: mean velocity in each vertical $\bar{V} = (V_{0.2d} + V_{0.8d})/2$. Discharge is by the mid-section method; verticals are 2 m apart, so each interior vertical represents a 2 m wide strip. Depth and velocity are zero at the banks (0 and 12 m).

Computation

 0    2    4    6    8    10   12  (m)
 ~~~~~|~~~~|~~~~|~~~~|~~~~|~~~~~
  \   |    |    |    |    |   /
   \__|    |    |    |____|__/
      |____|____|____|

$$Q = \sum w_i\,d_i\,\bar{V}_i = 8.56\ \text{m}^3/\text{s}$$

Answer: Discharge of the stream ≈ 8.56 m³/s.

In the mid-section method, each vertical represents a strip extending halfway to its neighbours; here all verticals are 1.2 m apart, so each strip is 1.2 m wide. Mean velocity in a vertical:

• Shallow verticals (one reading): $\bar{V} = V_{0.6y}$
• Deeper verticals (two readings): $\bar{V} = (V_{0.2y} + V_{0.8y})/2$

Computation

(The bank points at 0 and 7.2 m have zero depth and give no flow.)

$$Q = \sum w_i\,y_i\,\bar{V}_i = 4.80\ \text{m}^3/\text{s}$$

Answer: River discharge ≈ 4.80 m³/s.

Mean velocity in each vertical is taken as the average of the velocities at 0.2d and 0.8d. The mid-section method is used with equal spacing of 0.6 m, so each interior vertical represents a strip 0.6 m wide; the bank points (0 and 6.6 m) have zero depth.

Computation

Sample: at 2.4 m, $\bar{V} = (0.87 + 0.60)/2 = 0.735$ m/s, $\Delta Q = 0.6 \times 2.55 \times 0.735 = 1.1246$ m³/s.

$$Q = \sum 0.6\,d_i\,\bar{V}_i = 4.83\ \text{m}^3/\text{s}$$

Answer: Stream flow ≈ 4.83 m³/s.

Mean velocities are given directly, so the mid-section method is applied: each vertical represents a strip from halfway to the previous vertical to halfway to the next, $w_i = (x_{i+1} - x_{i-1})/2$, and $\Delta Q_i = w_i d_i V_i$.

Computation

$$Q = \sum w_i d_i V_i = 232.1\ \text{m}^3/\text{s}$$

Answer: Discharge ≈ 232 m³/s.

(By the mean-section method, averaging depth and velocity of adjacent verticals for each 10 m panel, $Q = 225.9$ m³/s; the mid-section result is normally reported.)

The slope-area method is an indirect method of finding discharge, mainly used for flood peaks when direct current-meter measurement is not possible. Discharge is computed from the water surface slope (from high flood marks) and the channel cross-sections, using Manning's equation.

Field work

1. Select a fairly straight, uniform reach (length $L$, about 75 m or more, fall ≥ 0.15 m).
2. After the flood, survey high water marks (silt lines, debris) at two (or more) sections 1 and 2.
3. Survey the cross-sections to get area $A$, wetted perimeter $P$ and hydraulic radius $R$ at each section.
4. Estimate Manning's $n$ from bed and bank conditions.

  Sec 1                   Sec 2
  ___ HFL ___  Dz (fall)
  |          ------____
  |  y1        L        ----___ HFL
  |____________________________y2_|
  =========== bed ============>

Computation

Energy equation between sections 1 and 2:

$$h_f = (Z_1 - Z_2) + \left(\frac{V_1^2}{2g} - \frac{V_2^2}{2g}\right) - h_e$$

where $Z$ = water surface elevation, $h_e$ = eddy loss $= K_e\left|\frac{V_1^2}{2g} - \frac{V_2^2}{2g}\right|$ ($K_e \approx 0$–0.1 for contracting, 0.3 for expanding reach).

Conveyance $K = \dfrac{1}{n}AR^{2/3}$ at each section; average $K = \sqrt{K_1K_2}$.

$$Q = K\sqrt{S_f}, \qquad S_f = \frac{h_f}{L}$$

Since $V$ depends on $Q$, solve by trial:

1. First take $h_f = Z_1 - Z_2$ (fall), find $S_f$ and $Q$.
2. Compute $V_1 = Q/A_1$, $V_2 = Q/A_2$; correct $h_f$ with the velocity head and eddy loss terms.
3. Repeat until $Q$ stops changing (usually 2–3 trials).

Limitations: accuracy depends on the choice of $n$ and on reliable flood marks; the reach must be free of backwater and sudden changes.

The slope-area method is used: discharge from Manning's equation with friction slope found by the energy equation, solved by trial.

Data

$b = 10$ m, $y_1 = 3.0$ m, $y_2 = 2.9$ m, $L = 200$ m, fall $\Delta Z = 0.12$ m, $n = 0.025$.

Section properties and conveyance

$$\begin{aligned} A_1 &= 30\ \text{m}^2,\; P_1 = 16\ \text{m},\; R_1 = 1.875\ \text{m} \\ A_2 &= 29\ \text{m}^2,\; P_2 = 15.8\ \text{m},\; R_2 = 1.835\ \text{m} \\ K_1 &= \frac{1}{n}A_1R_1^{2/3} = \frac{30 \times 1.875^{2/3}}{0.025} = 1824.7 \\ K_2 &= \frac{29 \times 1.835^{2/3}}{0.025} = 1738.9 \\ K &= \sqrt{K_1K_2} = 1781.3 \end{aligned}$$

Trial solution

Area decreases downstream (contracting reach), so eddy loss is taken as zero. Then

$$h_f = \Delta Z + \frac{V_1^2 - V_2^2}{2g}, \qquad Q = K\sqrt{h_f/L}$$

Sample (trial 1): $Q = 1781.3\sqrt{0.12/200} = 1781.3 \times 0.02449 = 43.63$ m³/s.

Answer: Flood discharge ≈ 42.3 m³/s (taking $K_e = 0$ for the contracting reach).

A stage–discharge curve (rating curve) is a graph or equation relating the stage (water surface level, G) at a gauging site to the discharge (Q) of the river at that site. Once it is established from a number of current-meter measurements over a range of flows, discharge can be read directly from a simple daily water level reading.

Development

1. Measure Q (by current meter) and the stage G at the same time, many times over low, medium and high flows.
2. Plot G against Q; fit a smooth curve.
3. Usual equation:

$$Q = C_r(G - a)^{\beta}$$

where $a$ = gauge reading for zero flow, and $C_r$, $\beta$ are found by plotting $\log Q$ against $\log(G - a)$ (least squares). 4. Extend to high flood levels by log-log extrapolation, or by Stevens or conveyance methods.

 Stage G (m)
   |                  .  *
   |             *  .
   |         * .
   |      *.
   |   *.
 a |*_____________________ Q (m3/s)

Applications in a hydropower project

• Converts long records of daily gauge readings into a discharge series, from which hydrographs, flow duration curves and firm flow are drawn for fixing the design discharge and installed capacity.
• Gives the tailwater level for each discharge, needed for net head, turbine setting and draft tube design.
• Gives water levels at the headworks/intake for different flows: intake sill level, weir crest, freeboard.
• Gives flood levels for design of the dam/weir, spillway, river training works and powerhouse floor level.
• Used in real-time operation: estimating inflow, managing pondage and spills.

Limitations: a shifting river bed (common in Nepal's sediment-laden rivers) changes the curve, so it must be checked regularly.

Peak flow (design flood) is the maximum discharge expected at a site for a chosen return period. It is needed to design the spillway, diversion works, cofferdams and river training works of a hydropower project.

1. Empirical methods

They relate peak flow mainly to catchment area $A$ (km²) with a regional constant. Simple but only valid in the region where they were developed.

In Nepal, regional methods such as WECS/DHM (1990) and modified Dicken's are commonly used, e.g. $Q_{100} = 14.63\,(A_{3000} + 1)^{0.7342}$, where $A_{3000}$ is the catchment area (km²) below 3000 m elevation.

Example: $A = 500$ km², $C_D = 11$: $Q_p = 11 \times 500^{0.75} = 11 \times 105.7 = 1163$ m³/s.

2. Gumbel's extreme value distribution

A statistical method using the annual maximum flood series (one peak per year for N years).

Steps:

1. Find mean $\bar{x}$ and standard deviation $\sigma_{n-1}$ of the annual peaks.
2. Reduced variate for return period T:

$$y_T = -\ln\left[\ln\left(\frac{T}{T-1}\right)\right]$$

3. Frequency factor:

$$K = \frac{y_T - \bar{y}_n}{S_n}$$

where $\bar{y}_n$ and $S_n$ are the reduced mean and reduced standard deviation from Gumbel's table for sample size N (for very large N, $\bar{y}_n = 0.577$, $S_n = 1.2825$). 4. Flood of return period T:

$$x_T = \bar{x} + K\sigma_{n-1}$$

5. Optionally plot on Gumbel probability paper and give confidence limits.

Example: $\bar{x} = 800$ m³/s, $\sigma = 250$ m³/s, large N, $T = 100$ yr: $y_T = 4.600$, $K = (4.600 - 0.577)/1.2825 = 3.137$, $x_{100} = 800 + 3.137 \times 250 = 1584$ m³/s.