Precipitation

A double mass curve (DMC) checks the consistency of rainfall records of a station by comparing it with the average of a group of nearby, reliable stations. It detects changes caused by shifting of the gauge, change in exposure, observer or instrument.

Principle

For a consistent record, the cumulative rainfall at the test station is proportional to the cumulative average rainfall of the base stations, so the plot is a straight line. A break in slope shows that the regime of the test station has changed.

Procedure

1. Select 10 to 25 base stations of the same climate, with long, consistent records.
2. For each year, find the average annual rainfall of the base stations.
3. Compute the cumulative values of the test station ($\Sigma P_x$) and the base average ($\Sigma P_{av}$).
4. Plot $\Sigma P_x$ (ordinate) against $\Sigma P_{av}$ (abscissa).
5. If the points lie on one straight line, the record is consistent. If there is a break in slope, note the year of the break.

 SumPx
   |            . .
   |         .      <- slope Ma (new)
   |      .
   |    .   <- change of regime
   |  .   slope Mc (old)
   | .
   +------------------ Sum Pav

Correction

Records of the earlier (or doubtful) period are adjusted to the conditions of the recent period:

$$P_{cx} = P_x \times \frac{M_c}{M_a}$$

where $P_{cx}$ is the corrected rainfall, $P_x$ the original rainfall, $M_c$ the slope of the line to which the data is to be corrected (usually the recent period) and $M_a$ the slope of the line of the period being corrected.

The same technique is also used for checking consistency of runoff or other hydrologic records.

Rainfall measured at a gauge is point rainfall. To find the average depth over a catchment, three methods are used.

1. Arithmetic mean method

The average of rainfall of all gauges in the area:

$$\bar P = \frac{P_1 + P_2 + \dots + P_n}{n}$$

Suitable for flat areas with uniform gauge distribution and little variation of rainfall. It is simple but ignores the position of gauges and topography.

2. Thiessen polygon method

Gauge stations are joined to form triangles, perpendicular bisectors of the sides are drawn, and the polygon around each station is the area closest to it. Each gauge is weighted by its polygon area:

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

It considers non-uniform gauge density and is more accurate than the arithmetic mean. It does not account for orographic effects, and the polygons must be redrawn if a gauge is added or fails.

3. Isohyetal method

Isohyets (lines of equal rainfall) are drawn using gauge values. The average rainfall between two successive isohyets is taken as their mean, and weighted by the area between them:

$$\bar P = \frac{\sum A_i \dfrac{P_i + P_{i+1}}{2}}{\sum A_i}$$

It is the most accurate, particularly for hilly regions, as it accounts for topography and storm pattern, but requires many gauges and judgement.

   15 cm   10 cm   5 cm   (isohyets)
   ( (  ( o ) ) )    A1, A2, A3 = inter-isohyet areas

IDF curve (Intensity-Duration-Frequency)

An IDF curve shows the relationship between rainfall intensity $i$, duration $D$ and frequency (return period) $T$ of storms at a place. Intensity decreases with duration, and for the same duration a rarer storm (higher $T$) has a higher intensity.

Development: From annual maximum rainfall of various durations (5 min to 24 h), a frequency analysis (Gumbel, Log-Pearson III) gives the depth for each return period. Intensity = depth / duration is plotted against duration on log scales, one curve per return period.

 i (mm/h)
  |\
  | \\   T = 100 yr
  |  \\\  T = 25 yr
  |    \\\ T = 5 yr
  +-------------- D (min)

An empirical form is $i = \dfrac{K T^x}{(D + a)^n}$. IDF curves are used to get design rainfall for storm drains, culverts and small catchments (rational method).

DAD curve (Depth-Area-Duration)

A DAD curve shows the average rainfall depth over an area for storms of different durations. For a given duration, the average depth decreases as the area increases.

Development: From isohyetal maps of a storm for 6, 12, 24 h, etc., the average depth within each isohyet and the area enclosed are computed, and plotted as depth against area for each duration.

 Depth
  | \____  6 h
  |  \____ 12 h
  |   \____ 24 h
  +-------------- Area (km2)

DAD curves are used to estimate the probable maximum precipitation (PMP) and the design storm for large catchments, flood control and spillway design.

Causes of inconsistency in rainfall data

A record is inconsistent when the conditions of observation changed during the record period. Common causes:

• Shifting of the rain gauge to a new site, or change of its elevation.
• Change in the surroundings (new buildings, trees, deforestation) that changes exposure.
• Change of the type of gauge or instrument, or its faults (leaks, damage).
• Change in observation procedure, observer, or time of observation.
• Errors in transcription, instrument calibration, or missing data filled wrongly.
• Changes in environment such as urbanisation or land use.

Applications of mass curves

1. Rainfall mass curve (accumulated rainfall vs time): Slope at any time gives the intensity. From it, the hyetograph and maximum intensity for various durations can be obtained.

Example: Cumulative rain 0, 1.0, 3.0, 4.5 cm at 0, 30, 60, 90 min. Intensity in 30-60 min = (3.0 - 1.0)/0.5 = 4 cm/h, the greatest.

2. Double mass curve: Plot cumulative rainfall of the test station against cumulative average of nearby stations.

• Straight line means consistent.
• A break in slope shows change of regime. The data before the break are corrected by

$$P_{c} = P_{x}\times\frac{M_c}{M_a}$$

Example: If the slope before 1990 is 1.2 and after is 0.9, rainfall before 1990 is multiplied by 0.9/1.2 = 0.75 to make it consistent with the present site.

3. Other uses: The reservoir mass curve (Rippl diagram) of cumulative inflow gives the storage required, and mass curves of runoff are used to find yield and detect inconsistent streamflow records.

Rain gauges give point rainfall. The mean rainfall over a catchment due to a storm is found by three methods.

1. Arithmetic mean method

$$\bar P = \frac{P_1+P_2+\dots+P_n}{n}$$

The simple average of the gauges inside the catchment. Used for flat areas with uniformly spread gauges and rainfall that varies little.

2. Thiessen polygon method

1. Plot the gauges on the catchment map and join adjacent gauges to form triangles.
2. Draw perpendicular bisectors of the sides; they form a polygon around each gauge.
3. Measure the area $A_i$ of each polygon inside the catchment.
4. Compute

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

3. Isohyetal method

1. Draw isohyets (lines of equal rainfall) by interpolating between gauge values.
2. Find the area $A_i$ between consecutive isohyets and the mean of their values $\bar P_i$.
3. Compute $\bar P = \sum A_i \bar P_i / \sum A_i$.

Merits and demerits

For hilly catchments such as those in Nepal, the isohyetal method is preferred if enough gauges exist; otherwise Thiessen is used.

The optimum number of rain gauges for a permissible error $\varepsilon$ in the mean rainfall is

$$N = \left(\frac{C_v}{\varepsilon}\right)^2, \qquad C_v = \frac{100\,\sigma_{n-1}}{\bar P}$$

Data (cm): 130, 142.1, 118.2, 108.5, 165.2, 102.1, 146.9; existing stations $n = 7$.

• Mean: $\bar P = 913/7 = 130.43$ cm
• Standard deviation (sample): $\sigma_{n-1} = 22.545$ cm
• Coefficient of variation: $C_v = \dfrac{100 \times 22.545}{130.43} = 17.29\%$

For $\varepsilon = 5\%$:

$$N = \left(\frac{17.29}{5}\right)^2 = 11.95 \approx 12$$

Additional stations $= 12 - 7 = 5$.

Answer: Total 12 stations are needed, so at least 5 additional rain gauges must be established.

In the isohyetal method the mean depth between two isohyets is taken as the average of their values, and the weighted average is found with the inter-isohyetal areas.

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

For the last band (3-1 cm) the mean is $(3+1)/2 = 2$ cm.

$$\bar P = \frac{4443.5}{600} = 7.41\ \text{cm}$$

Answer: Average depth of precipitation = 7.41 cm (about 74 mm).

A double mass curve plots the cumulative rainfall of station X against the cumulative average rainfall of the 18 base stations. A change in slope shows an inconsistency.

Step 1: Cumulative values

Step 2: Plot and slopes

Plotting cum. X (y-axis) against cum. base average (x-axis), the points follow one straight line up to 1959 and another, flatter line afterwards.

• Slope before the break (to 1959): $M_a = \dfrac{282.4}{223.3} = 1.265$
• Slope after the break (1960-1970): $M_c = \dfrac{272.1}{301.8} = 0.902$

The slopes differ, so the record of station X is not consistent. The change in regime occurred in 1960 (the break appears at the 1959-60 point).

Step 3: Adjust to the changed (present) regime

$$P_{cx} = P_x \times \frac{M_c}{M_a} = P_x \times \frac{0.902}{1.265} = 0.713\,P_x$$

Adjusted total for 1952-1959 $= 0.7129 \times 282.4 = 201.3$ cm; the total of 1960-1970 is unchanged $= 272.1$ cm.

$$\bar P_{1952-70} = \frac{201.3 + 272.1}{19} = 24.92\ \text{cm}$$

Answer: Record is inconsistent; change in regime in 1960; mean annual rainfall for 1952-1970 under the changed regime = 24.9 cm (compared with the unadjusted mean of 29.2 cm).

Step 1: Incremental depth and intensity (5-min intervals)

Intensity = increment / (5/60 h) = increment × 12.

Mass curve: plot accumulated rainfall (y) against time (x) from the first two columns, a rising curve that ends at 3.1 cm at 50 min and is steepest between 10 and 20 min.

Hyetograph: a bar chart of the increment (or intensity) against time.

 cm/h
  8.4 |        ##
  7.2 |      # ##
  6.0 |      # ##    #
  3.6 |      # ## #  #
  2.4 |      # ## ####### #
  1.2 | ## ## # ## # ## #
      +-------------------------
       5 10 15 20 25 30 35 40 45 50 min

Maximum depth-duration: for each duration the largest rainfall in any consecutive period is found from the mass curve.

Plot depth (or intensity) against duration. The depth increases with duration while the average intensity falls from 8.4 cm/h (5 min) to 3.72 cm/h (50 min).

Answer: Maximum depths = 0.7, 1.3, 1.6, 1.8, 2.5, 2.9 and 3.1 cm for 5, 10, 15, 20, 30, 40 and 50 min.

The number of gauges needed for a permissible error $\varepsilon$ in the mean is

$$N = \left(\frac{C_v}{\varepsilon}\right)^2, \quad C_v = \frac{100\,\sigma}{\bar P}$$

Given: $\bar P = 143$ mm, $\sigma = 31.6$ mm, existing $n = 7$, $\varepsilon = 7\%$.

$$C_v = \frac{100 \times 31.6}{143} = 22.10\%$$ $$N = \left(\frac{22.10}{7}\right)^2 = 9.97 \approx 10$$

Since $N = 10 > 7$, additional gauges are needed.

Additional gauges $= 10 - 7 = 3$.

Answer: Yes. A minimum of 3 additional rain gauges (10 in total) must be installed.

The maximum average intensity for a duration $D$ is the greatest rainfall depth in any $D$ consecutive hours divided by $D$. Rainfall in each 30-min interval (mm): 6, 6, 5, 8, 5, 9, 13, 6, 4, 3, 2, 0 (total 67 mm).

Checks: 2 h = 8 + 5 + 9 + 13 = 35 mm; 5 h = 6+6+5+8+5+9+13+6+4+3 = 65 mm (the five-hour window starting 06:00; the window 06:30-11:30 gives only 61 mm).

Intensity-duration curve: plot duration (x-axis) against maximum intensity (y-axis):

 mm/h
 26 |*
 22 |   *
 17.5|        *
 13 |                       *
    +------------------------
     0.5   1    2         5   h

The curve falls with duration because the intense part of the storm is spread over a longer time.

Answer: 26.0, 22.0, 17.5 and 13.0 mm/h for 30 min, 1 h, 2 h and 5 h.

The double mass curve plots cumulative X against cumulative base-station average. A single straight line means consistent data.

(i) Cumulative values and consistency

Plotting cum. X against cum. base average shows a straight line up to 1952 and a flatter straight line afterwards.

• Slope up to 1952: $M_a = \dfrac{1654}{1419} = 1.1656$
• Slope after 1952: $M_c = \dfrac{2299}{2718} = 0.8458$

The slope changes, so the data of station X are not consistent.

(ii) Change in regime

The break occurs at the 1952 point, so the regime changed after 1952 (the changed conditions apply from 1953). (The 1954 value of 97 is taken as printed.)

(iii) Mean annual rainfall of X (36 years)

• Without adjustment: $\bar P = \dfrac{3953}{36} = 109.8$ cm
• With adjustment to the present site (recent regime), the 1941-1952 data are corrected by $M_c/M_a = 0.8458/1.1656 = 0.7257$. Corrected total for 1941-52 $= 1200.3$ cm; total for 1953-76 $= 2299$ cm.

$$\bar P_{adj} = \frac{1200.3 + 2299}{36} = 97.2\ \text{cm}$$

(iv) Adjusted annual rainfall for the affected period (1941-1952)

$$P_{cx} = P_x \times 0.7257$$

Answer: Inconsistent; change in regime after 1952; mean = 109.8 cm (unadjusted) and 97.2 cm (adjusted); adjusted values for 1941-52 are listed above (factor 0.726).

Mass curve ordinates (mm) at 15-min intervals from 07:00: 0, 9.5, 17.0, 27.0, 40.5, 49.0, 63.0, 84.0, 95.0, 102.0, 110.0, 112.0, 112.0.

Incremental rainfall (mm) per 15 min: 9.5, 7.5, 10.0, 13.5, 8.5, 14.0, 21.0, 11.0, 7.0, 8.0, 2.0, 0.

For each duration $D$, the maximum depth is the largest difference of mass curve ordinates $D$ minutes apart (found by sliding the window along the curve), and $i_{max} = \text{depth}/D$.

Checks: 30 min = 14.0 + 21.0 = 35 mm; 1 h = 84.0 - 27.0 = 57 mm; 2 h = 95.0 - 0 = 95 mm.

Intensity-duration graph: plot duration on the x-axis and maximum intensity on the y-axis:

 mm/h
 84 |*
 70 |   *
 61 |     *
 57 |       *
 52 |            *
 47.5|                *
 37 |                          *
    +---------------------------
     15 30 45 60  90 120   180 min

The intensity falls steadily as the duration increases.

Answer: Maximum intensities = 84, 70, 61.3, 57, 52, 47.5 and 37.3 mm/h for 15, 30, 45, 60, 90, 120 and 180 min.

Setup: Semicircle of radius $r$ in the upper half-plane, area $A = \pi r^2/2$. Gauges: $O(0,0)$, $P_1(r/2, r/2)$, $P_2(-r/2, r/2)$.

        y
        |   P2 .  | . P1
        |   (-r/2,r/2) (r/2,r/2)
   -----+----O-----+---- x
       -r          r

Step 1: Perpendicular bisectors

• Between $P_1$ and $P_2$: they are at the same height, so the bisector is $x = 0$.
• Between $O$ and $P_1$: points equidistant from both satisfy $x^2+y^2 = (x-r/2)^2 + (y-r/2)^2$, which gives $x + y = r/2$.
• Between $O$ and $P_2$: similarly $-x + y = r/2$.

Step 2: Polygon of the gauge at O

The area closer to $O$ than to $P_1$ or $P_2$ is bounded by the x-axis, $y = r/2 - x$ and $y = r/2 + x$. This is a triangle with vertices $(-r/2, 0)$, $(r/2, 0)$ and $(0, r/2)$ (all inside the semicircle):

$$A_O = \tfrac{1}{2} \times r \times \frac{r}{2} = \frac{r^2}{4}$$ $$W_O = \frac{A_O}{A} = \frac{r^2/4}{\pi r^2/2} = \frac{0.5}{\pi}$$

Step 3: Remaining two gauges

The rest of the semicircle is divided by $x = 0$ into two equal halves (by symmetry):

$$A_{P_1} = A_{P_2} = \frac{1}{2}\left(\frac{\pi r^2}{2} - \frac{r^2}{4}\right)$$ $$W_{P_1} = W_{P_2} = \frac{1}{2}\left(1 - \frac{0.5}{\pi}\right) = 0.5 - \frac{0.25}{\pi}$$

Numerically: $W_O = 0.159$, $W_{P_1} = W_{P_2} = 0.4204$; the sum $= 1.000$. Hence proved.

Method: Draw perpendicular bisectors between neighbouring stations P(50,25), Q(100,25), R(100,-25), S(50,-25). The catchment is divided by the lines $x = 75$ (between P-Q and S-R) and $y = 0$ (between P-S and Q-R). Each polygon is clipped by the catchment boundary A(0,0), B(50,75), C(100,70), D(150,0), E(75,-50).

      B(50,75)  C(100,70)
   A(0,0)  P    |   Q          D(150,0)
   --------+----+---+-----------
           S    |   R
              E(75,-50)
          x=75 divides left/right, y=0 upper/lower

Polygon areas (by coordinates, shoelace formula):

Total area check: pentagon area $= 11000$ km².

$$\bar P = \frac{\sum A_i P_i}{\sum A_i} = \frac{1\,144\,218.75}{11\,000} = 104.02\ \text{mm}$$

Answer: Mean rainfall by Thiessen polygon method = 104.0 mm.

Precipitation forms when moist air is lifted, cools and condenses. Based on the lifting mechanism there are three types.

1. Convective precipitation

Heated ground warms the air above it; the warm, light, moist air rises, cools and condenses into cumulus clouds. It gives short, intense, local showers and thunderstorms, common in the tropics and in Nepal's pre-monsoon season.

2. Orographic precipitation

Moist air is forced to rise over a mountain barrier, cools and condenses. The windward side gets heavy rain and the leeward side lies in the rain shadow. The southern slopes of the Himalaya (for example Lumle-Pokhara, with heavy rainfall) are a good example.

3. Cyclonic precipitation

Caused by the movement of air masses towards a low-pressure area. Two kinds:

• Frontal: a warm air mass rides over a cold one (warm front, gentle, long rain) or a cold front pushes under warm air (steep, short, heavy rain).
• Non-frontal (tropical cyclone / monsoon depression): air converges into a low-pressure centre and rises. The monsoon is of this kind.

 Convective       Orographic         Cyclonic (front)
    ^ ^ ^          clouds /|           warm ///  cold
   (cloud)        wind-> / | rain      -------\---
  hot ground     ______/   |_____      ground

Ways of presenting precipitation data

Drawing an IDF curve

1. Collect long-term rainfall records from an autographic (recording) gauge.
2. For each year, find the maximum rainfall depth for durations such as 5, 10, 15, 30, 60, 120, 360 and 1440 min (annual maximum series).
3. For each duration, fit a probability distribution (Gumbel, log-Pearson III) and get the depth for chosen return periods (2, 5, 10, 25, 50, 100 years).
4. Convert depth to intensity: $i = P/D$.
5. Plot intensity (y, usually log scale) against duration (x, log scale) with one curve for each return period.
6. If needed, fit $i = \dfrac{K T^x}{(D+a)^n}$.

 i
  |\
  | \\   100-yr
  |   \\\  25-yr
  |      \\\ 5-yr
  +---------------- D

The curve is read for a design duration (usually the time of concentration) and chosen return period to get the design intensity.

Causes of inconsistency

• Shifting of the gauge to a new location or elevation.
• Changes in the surroundings: new buildings or trees, or clearing of trees, changing exposure.
• Change in the type or condition of the instrument, or its errors (leaks, damaged funnel).
• Change in observer, observing time or procedure.
• Changes in the environment such as urbanisation, or errors in recording and transcription.

Correction for future use

1. Test the record with a double mass curve: plot cumulative annual rainfall of the test station against the cumulative mean of 10-25 nearby consistent stations.
2. If the points form a single straight line, the record is consistent. A break in slope shows inconsistency, and the year of break is noted.
3. Correct the earlier (or doubtful) data to the conditions of the recent period:

$$P_{cx} = P_x \times \frac{M_c}{M_a}$$

where $M_c$ is the slope of the segment to which the data are to be corrected (usually the recent one) and $M_a$ is the slope of the segment being corrected.

4. Use the adjusted series for frequency analysis, design and further studies.

The values are the rainfall (cm) recorded in each 30-minute interval ending at the stated time. Total rainfall $= 7+5+8+9+13+10+8+6+5+3+1 = 75$ cm.

(i) Hyetographs

30-min interval: the bar chart of the 30-min depths (intensity = depth × 2 cm/h).

2-hour interval (sum of four 30-min depths):

 cm/h (30-min hyetograph)
 26 |             ##
 20 |             ## ##
 16 |       ##    ## ## ##
 12 |    ## ## ## ## ## ## ##
  6 | ## ## ## ## ## ## ## ## ## ##
    +--------------------------------
     6:00        9:00       12:00

(ii) Maximum average intensities

(For 2 h the window 7:00-9:00 holds 8 + 9 + 13 + 10 = 40 cm.)

Intensity-duration curve: plot duration (x) against maximum intensity (y): points (0.5 h, 26), (1 h, 23), (2 h, 20) joined by a smooth falling curve.

Answer: Maximum intensities = 26, 23 and 20 cm/h for 30 min, 1 h and 2 h.

The optimum density of rain gauges balances accuracy and cost. The number needed to estimate the mean areal rainfall within a permissible error is found statistically.

Procedure

1. Use the existing gauges ($n$ stations) and their mean annual (or storm) rainfall values $P_1, \dots, P_n$.
2. Compute the mean $\bar P$ and the standard deviation

$$\sigma_{n-1} = \sqrt{\frac{\sum (P_i - \bar P)^2}{n-1}}$$

3. Compute the coefficient of variation $C_v = \dfrac{100\,\sigma_{n-1}}{\bar P}$.
4. Choose the permissible error $\varepsilon$ in the mean rainfall (usually 10%, or 5% for precise work).
5. The optimum number of gauges is

$$N = \left(\frac{C_v}{\varepsilon}\right)^2$$

6. If $N > n$, additional gauges $= N - n$ are installed (placed to cover under-represented areas and elevation zones).

Also considered

• WMO recommended minimum densities: flat regions 1 gauge per 600-900 km², hilly regions 1 per 100-250 km², mountainous regions 1 per 25 km² (with some relaxation).
• Cost, accessibility of sites and purpose of data (design vs research).
• Because $N$ depends on $C_v$, the process may be repeated as new data come in.

Intensity-Duration Curve

It shows how the maximum average rainfall intensity of a storm changes with duration. The intensity is high for short durations and falls for long durations.

• Prepared from a recorded storm: for each duration $D$ the greatest rainfall in any period of length $D$ is found from the mass curve, and $i_{max} = \text{depth}/D$.
• Plotted as intensity (y) against duration (x). With frequency (return period) added it becomes the IDF curve.

 i
  |*
  |  *
  |     *
  |         *
  +--------------- D

Uses: design storm for drainage, culverts, small dams and rational method calculations.

Depth-Area Curve

It shows how the average depth of rainfall of a storm decreases as the area increases. Rainfall is greatest at the storm centre and falls off outward.

• Prepared from an isohyetal map: the average depth inside each isohyet and the area enclosed are computed, and plotted as depth (y) against area (x). Curves for different durations form the DAD curve.

 Depth
  |*
  |  *
  |     *
  |         *
  +--------------- Area

Uses: to convert point rainfall to areal rainfall (areal reduction), design of large catchments, estimating probable maximum precipitation (PMP).

Precipitation is any form of water that falls from the atmosphere to the earth. The main forms are:

Of these, rain and snow contribute most to the water resources of Nepal.

Rain gauges measure the depth of rainfall (mm) at a point. They are of two kinds: non-recording and recording.

A. Non-recording (Symons') gauge

A funnel of 12.7 cm diameter sits on a cylindrical vessel and collects rain into a bottle; the depth is read by a graduated measuring jar once a day (8:45 am in Nepal).

   |<-- 12.7 cm -->|
    \             /   Funnel
     \___________/
      |  bottle  |   <- collecting can
      |__________|
   ========ground=======

It is cheap and simple, but gives only daily totals and needs an observer.

B. Recording gauges

1. Tipping-bucket gauge: Rain falls through a funnel onto a two-compartment bucket balanced on a pivot. When 0.25 or 0.5 mm of rain fills one compartment it tips, empties, and brings the other under the funnel. Each tip closes a switch and is recorded electronically. It gives the intensity but may under-record in very heavy rain.

     funnel
       \ /
     [ \/ ]   two-compartment bucket
      /\  tips at fixed volume
   pivot   -> pulse to recorder

2. Weighing-bucket gauge: The collecting bucket sits on a spring balance; the increasing weight moves a pen over a clock-driven drum chart, giving a continuous mass curve. It records snow and hail also.

3. Float-type (natural siphon) gauge: A float in a chamber rises with the rain collected, moving a pen on a rotating drum. When the chamber is full, a siphon empties it automatically and the pen returns to zero. It gives a continuous mass curve (Dines or Hellman type).

Comparison

Siting

The gauge is placed on open level ground, with its funnel rim horizontal 30 cm above the ground, away from obstructions at a distance of at least twice their height.

Four gauges at the corners of a rectangle are equally spaced from the centre, so the perpendicular bisectors are the two lines through the centre parallel to the sides. The rectangle divides into four equal rectangles, one for each gauge.

  G1 +--------+--------+ G2
     | A/4    | A/4    |
     +--------+--------+
     | A/4    | A/4    |
  G4 +--------+--------+ G3

Total area $= 100 \times 150 = 15\,000$ km²; each gauge area $= 3750$ km² (weight 0.25).

$$\bar P = \frac{\sum A_i P_i}{\sum A_i} = \frac{3750(1200+1300+1500+1100)}{15\,000} = \frac{5100}{4} = 1275\ \text{mm}$$

Answer: Average rainfall by Thiessen polygon method = 1275 mm (equal to the arithmetic mean).

The average depth between two isohyets is the mean of their values. Weighted by the area, the catchment average is

$$\bar P = \frac{\sum A_i \bar P_i}{\sum A_i}$$ $$\bar P = \frac{4194}{442} = 9.49\ \text{cm}$$

Answer: Average rainfall of the Bagmati basin = 9.49 cm (about 95 mm).

Rainfall data can be presented in the following ways:

1. Tables of hourly, daily, monthly and annual rainfall.
2. Hyetograph (bar chart of depth or intensity vs time).
3. Mass curve (cumulative rainfall vs time).
4. Intensity-duration and IDF curves.
5. Depth-area-duration (DAD) curves.
6. Isohyetal maps (contours of equal rainfall).
7. Frequency curves and annual/monthly bar graphs.

Two methods explained

1. Hyetograph A hyetograph is a bar chart of rainfall depth (or intensity) in successive equal time intervals, plotted against time. It is obtained from the mass curve by taking the difference between successive ordinates.

 mm/h
  |      ##
  |   ## ## ##
  | ## ## ## ## ##
  +------------------ time

It shows the storm pattern, the time of peak intensity and the total depth (area under the bars). It is the input for rainfall-runoff analysis such as the unit hydrograph.

2. Mass curve of rainfall A mass curve plots the accumulated rainfall (y-axis) against time (x-axis) from the start of the storm. It is always rising or flat.

 cm
  |        _____
  |      /
  |    /      <- steep = high intensity
  |__/
  +------------------ time

• The slope at any point gives the rainfall intensity.
• A horizontal portion means no rain.
• It is used to find the maximum depth for a duration, to derive the hyetograph, and to prepare intensity-duration curves.

The tipping bucket gives cumulative rainfall at 10-min intervals. The incremental depth in each interval is the difference of successive values, and intensity = depth / (10/60 h) = depth × 6.

(i) Hyetograph

 mm/h
 85 |                  ##
 50 |       ##         ##
 43 |       ## ##      ## ##
 30 |    ## ## ## ##   ## ##
 10 | ## ## ## ## ## ## ## ## ##
    +--------------------------------
     10 20 30 40 50 60 70 80 90 min

(Bar heights are the intensities in the table; the peak of 85.2 mm/h occurs in the 50-60 min interval.)

(ii) Maximum intensity-duration curve

For each duration the maximum depth is the greatest difference of cumulative ordinates that far apart.

Example: 30 min: 46.8 (at 70 min) - 20.2 (at 40 min) = 26.6 mm, intensity 53.2 mm/h.

Plot duration (x) against maximum intensity (y). The curve falls from 85.2 mm/h at 10 min to 33.9 mm/h at 90 min.

Answer: Peak 10-min intensity = 85.2 mm/h; maximum intensities are 85.2, 67.8, 53.2, 49.5, 48.2, 45.2, 41.0, 37.1 and 33.9 mm/h for 10 to 90 min.

The Thiessen polygon method gives the mean rainfall by weighting each gauge by the area nearest to it.

Procedure

1. Draw the catchment boundary to scale and mark the positions of all rain gauges (inside, and a few just outside).
2. Join adjacent stations by straight lines to form a network of triangles (avoid very acute triangles).
3. Draw the perpendicular bisector of each side. The bisectors meet at the circumcentres of the triangles and form a polygon around each station.
4. Each polygon is the area closer to its station than to any other. Trim the polygons at the catchment boundary.
5. Measure the area $A_i$ of each polygon inside the catchment (planimeter, grid or coordinates).
6. Find the weight $W_i = A_i/A$, where $A = \sum A_i$ is the catchment area.
7. Mean rainfall:

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

        *1
       /|\
      / | \     perpendicular bisectors
   *2---+---*3   form polygons around
      \ | /     each station
        *4

Remarks

It is better than the arithmetic mean for uneven gauge spread, but it does not include orographic effects. The polygons must be redrawn when a gauge is added or fails.

The data are first arranged in chronological order (2013 to 2022) and the cumulative values are found.

Consistency check

Plot cumulative X (y) against cumulative average of the 15 stations (x). The points follow one straight line up to 2016 and a flatter line from 2017.

• Slope up to 2016: $M_a = \dfrac{1402}{1089} = 1.287$
• Slope after 2016: $M_c = \dfrac{1687}{1699} = 0.993$

The slopes differ, so the record of station X is not consistent. The change in regime occurred after 2016 (from 2017).

Adjustment

The earlier data (2013-2016) are adjusted to the present regime (slope $M_c$):

$$P_{cx} = P_x\times\frac{M_c}{M_a} = P_x\times\frac{0.993}{1.287} = 0.771\,P_x$$

Data for 2017-2022 are unchanged.

Answer: Station X is inconsistent; change in regime from 2017; the 2013-2016 values are multiplied by 0.771 to give 337.0, 262.2, 235.2 and 246.8 mm.

The maximum depth for a duration $D$ is the largest sum of $D$ consecutive hourly values; the maximum intensity is that depth divided by $D$. Hourly rainfall (mm) in time order from 7 am: 10, 8, 6, 12, 5, 0, 2, 5, 8, 12, 7, 6, 0, 0 (total 81 mm).

1-hour duration

Largest single hour $= 12$ mm, so $i_{max} = 12/1 = 12$ mm/h.

3-hour duration

Sums of 3 consecutive hours: 24, 26, 23, 17, 7, 7, 15, 25, 27, 25, 13, 6. Maximum $= 8 + 12 + 7 = 27$ mm, so $i_{max} = 27/3 = 9.0$ mm/h.

5-hour duration

Sums of 5 consecutive hours: 41, 31, 25, 24, 20, 27, 34, 38, 33, 25. Maximum $= 10+8+6+12+5 = 41$ mm (first five values), so $i_{max} = 41/5 = 8.2$ mm/h.

Answer: Maximum intensities = 12.0 mm/h (1 h), 9.0 mm/h (3 h) and 8.2 mm/h (5 h).

The optimum number of gauges for a permissible error $\varepsilon$ is

$$N = \left(\frac{C_v}{\varepsilon}\right)^2, \qquad C_v = \frac{100\,\sigma_{n-1}}{\bar P}$$

Data (cm): 240, 252.2, 228.3, 218.6, 275.3, 212.2; $n = 6$.

• Mean: $\bar P = 1426.6/6 = 237.77$ cm
• Standard deviation: $\sigma_{n-1} = 23.38$ cm
• $C_v = \dfrac{100\times23.38}{237.77} = 9.83\%$

For $\varepsilon = 10\%$:

$$N = \left(\frac{9.83}{10}\right)^2 = 0.97 \approx 1$$

Answer: The optimum number is about 1 gauge. The existing 6 stations are more than enough for a 10% error, so no additional gauge is needed.

The normal annual rainfall of the base stations A and C differs from that of B by more than 10% (A: 22%; C: 15%; D: 4%), so the normal ratio method is used:

$$P_B = \frac{1}{n}\left[\frac{N_B}{N_A}P_A + \frac{N_B}{N_C}P_C + \frac{N_B}{N_D}P_D\right]$$

with $n = 3$ stations, $N_A = 1150$, $N_B = 940$, $N_C = 1080$, $N_D = 980$ and $P_A = 1000$, $P_C = 970$, $P_D = 890$ mm.

$$\begin{aligned} \frac{N_B}{N_A}P_A &= \frac{940}{1150}\times1000 = 817.39\\ \frac{N_B}{N_C}P_C &= \frac{940}{1080}\times970 = 844.26\\ \frac{N_B}{N_D}P_D &= \frac{940}{980}\times890 = 853.67 \end{aligned}$$ $$P_B = \frac{817.39 + 844.26 + 853.67}{3} = 838.4\ \text{mm}$$

Answer: Missing precipitation at station B = 838.4 mm (about 838 mm).

Procedure for developing a DAD curve

1. Select a major storm and collect rainfall data of all gauges for the storm period.
2. Prepare isohyetal maps of the storm for different durations (for example 6, 12, 24, 48 h) using the maximum depth for each duration at every station.
3. For each duration, measure the area enclosed by each isohyet, starting from the storm centre outward, and compute the cumulative area.
4. Compute the average depth over each cumulative area = (sum of area × mean depth) / cumulative area.
5. Plot average depth (y) against area (x) for each duration, giving one curve per duration.
6. Do the same for several storms and draw the envelope curves of the maximum values. These are used for design.

Calculation for the 12-hour isohyetal data

Assumptions: the mean depth of the top class (> 70 cm) is taken as 75 cm and that of the lowest class (< 40 cm) as 35 cm. Other classes use the mid value.

Plot the average depth against the cumulative area (125, 350, 565, 915, 1065 km²) for the 12-h duration; the depth falls from 75 cm to 53.4 cm as the area increases.

Answer: Average 12-h depth over 125, 350, 565, 915 and 1065 km² = 75.0, 68.6, 63.4, 56.4 and 53.4 cm respectively.