TOPIC 2 : RAIN DATA ANALYSIS

TOPIC 2 :RAINFALL DATA ANALYSISBy : Halina Binti Hamid

TYPES OF PRECIPITATIONPrecipitation can be divided into two categories which are :

Liquid - Rain, DrizzleFrozen Snow, Glaze, Sleet, Hail

Rain mostly occurs in nimbostratus clouds and cumulonimbusclouds. These clouds are capable of producing cloudbursts.

Most rain starts as snow or ice crystals; as the snow falls throughthe cloud it melts.

Drizzle is a fine uniform water droplet with adiameter less than 0.5 mm.

Rain & DrizzleRain is the major form of precipitation in the form of water drops of sizes greater than 0.5 mm. The maximum sizes of rain drop is about 6 mm. based on its intensity, rainfall is classified as light rain (> 2.5 mm/hr), moderate rain (2.5 7.5 mm/hr) and heavy rain ( 10 km2, then the areal reduction factor is required.Therefore, the rainfall intensity I = FA II = 0.9 250 mm/hrI = 225 mm/hr

The rainfall depth P = I tP = 225 0.75P = 168.75 mm

2 hours line 30 km2

0.9

Design Rainfall HyetographA rainfall hyetograph is a graphical representation of the variation of rainfall depth or intensity with time. Rainfall-runoff hydrograph methods require a description of this variation.

The temporal patterns of the west coast and east coast of Peninsular Malaysia are provided in Table 13.B1 and Table 13.B2 of MSMAM.

Example: Using the rainfall depth from previous question, obtain the design hyetograph for 30 minute duration. Assume an area of West Coast Malaysia. Based on Table 13.B1, the temporal depth for 30 minute duration is

Based on the rainfall depths, the design rainfall hyetograph of depth is:

or

Based on the rainfall intensity, the design rainfall hyetograph of intensity is:

AREAL AVERAGE RAINFALL Representing Water Amounts

At a point it is most convenient to represent amounts of water, such as precipitation, runoff, evaporation, or groundwater recharge, as a depth.

Depth, e.g. mm of rainfall, can be thought of as a measure of volume per unit area.

Need for Areal AveragingWhen we want to estimate runoff or compare stream flow amounts to rainfall amounts, it becomes necessary to get an area average based on point measurements.

There are numerous methods to estimate areal average rainfall.

Areal Averaging Methods Arithmetic average Theissen polygon method Isohyetal method

Arithmetic Average MethodSimply take the average of all the station values in the study area. This method works well only if the stations are evenly distributed over the study area.

Example :

P = (0.55 + 0.87 + 2.33 + 5.40 + 1.89 ) / 5

= 2.21

Theissen Polygon MethodThis method weights the contribution of each station based on the area it represents. The areal average is obtained using a area-weighted average of the station values.

Step 1 :

Construct polygons by connecting stations with lines

Example :

Step 2 :

Perpendicular bisectors for each of the sides of the triangle are drawn.These bisectors form a polygon around each station

Step 3 :

Estimate the area of each stations polygon by counting grid squares or other suitable technique for the polygons formed by the bisect lines.In this case the area is 15.0, 33.0, 28.8, 16.4 and 24.3 units for A, B, C, D and E respectively

Step 4 :

Using the table, total the area (A).In this case, A = 117.5 units

Step 5 :

Determine the station weights by dividing the station area by the total are. ( Ai / A)In this case, W1 = 0.128, W2 = 0.281, W3 = 0.245, W4 = 0.140 and W5 = 0.207

StationRainfall, PAreaWeightageWAP1A1A1/ABP2A2A2/ACP3A3A3/ADP4A4A4/AEP5A5A5/ATotalA

Step 6 :

Calculate the average rainfall over the catchment P is given by :

P = P1W1 + P2W2 + + P5W5

In this case, P = 2.03

Problem Based 1 :Calculate the average precipitation over the catchment by the Thiessen Polygons methods

Isohyetal MethodThis method develops areal weights based on the rainfall pattern. The areal average is obtained using a area-weighted average of the isohyetal zones.

Example :

Draw lines of equal precipitation.Estimate precipitation in each grid area within basinTotal the values in each grid areaDivide areas to obtain a basin areal estimate of precipitationAreal estimate is 1.90 in this case

IsohytesAverage value of PAreaFraction of total area(col. 3 / Total area, A)Weighted P(col. 2 x col. 4)12345TOTAL

Problem Based 2 :Calculate the average precipitation over the catchment by the Isohyetal methods

Estimation of Missing DataIncomplete records of rainfall are sometimes possible due to operator error or equipment malfunction.In the case, it is often necessary to estimate the missing record.There are two methods of correction:Arithmetic MethodNormal Ratio MethodQuadrant Method

Normal Ratio MethodNormal ratio method - Get observed rainfall at surrounding stationsThese are weighted by the ratio of the normal annual rainfall at station X and normal annual rainfall at that station.

where Px = estimation of ppt at station x, Pi = ppt at Nx = normal ppt at station X and Ni is ppt at the ith surrounding station .

Example :The normal annual rainfall at stations A,B,C, and D in a basin are 80.97, 67.59, 76.28 and 92.01 cm respectively. In the year 1975, the station D was inoperative and the station A,B and C recorded annual precipitations of 91.11, 72.23 and 79.89 cm respectively. Estimate the rainfall at station D in that year.

Solution :As the normal rainfall values vary more than 10% the normal ratio method is adopted. Using the equation :

= 99.48 cm

Quadrant MethodsFour quadrants are delineated by north-south and east-west lines passing through raingauge station where the missing rainfall is to be estimated.One raingauge station in each quadrant, which is the nearest to the raingauge station under question in that quadrat , is selected.The weight applicable to each of these four station is computed as the reciprocal of the square of the distance between this station and the origin of the quadrants.Then the rainfalls recorded at the four stations in the four quadrants are multiplied by their respective weights and added.The resulting sum is divided by the sum of the weights to yield the missing rainfall.

Where Px, is the missing rainfall at station X. P1, P2, P3 and P4 are the rainfalls recorded at the raingauge stations in the four quadrants which are nearest to station X in the respective quadrants, and which are at distances of r1, r2, r3 and r4 from station X respectively.

Example :Station X failed to report the rainfall recorded during a storm. With respect to east-west and north-south axes set up at station X, the coordinates of 4 surrounding gauges, which are the nearest to station X in the respective quadrants, are (10,15), (-8,5), (-12,-9) and (5,-15) km respectively. Determine the missing rainfall at X, if the storm rainfalls at the four surrounding gauges are 73, 89, 68 and 57 mm respectively

Solution :

Test For Consistency of RecordSome of the common causes for inconsistency of record include :

Shifting of raingauge station to a new locationThe neighborhood of the station undergoing a marked change.

Double Mass CurveThis technique is based on the principle that when each recorded data comes from the same parent population, they are consistent.

A group of n (usually 5 to 10) base stations in the neighborhood of the problem station X is selected.Annual (or monthly mean) rainfall data of station X and also the average rainfall of the group of base stations covering a long period is arranged in the reverse chronological order (i.e. the latest record as the first entry and the oldest record as the first entry and the oldest record as the last entry in the list).

Example :The annual rainfall at station X, and the average of annual at 25 surrounding base stations in cm are given below for a period of 36 years starting from 1941.

Check whether the data of station X is consistent.In which year a charge in regime indicated ?Compute the mean annual rainfall for station X at its present site for given 36 year period first without adjustment and secondly with the data adjusted for the change in regime.Compute the adjusted annual rainfalls at station X for the affected period.

Solution :

As the double mass curve is not having a uniform slope, it can be concluded that the record at station X is not consistent.Since there is a break in the slope of the double mass curve in the year 1953, it may be said that the data at station X prior to 1953 is inconsistent and needs adjustment.From Col. (3), the total of 36 year rainfall at station X without adjustment = 3953 cm so, The man annual rainfall at X without adjustment = 3953 = 109.8 cm 36

From the graph of double mass curve, the cumulative rainfall at station X for 36 years after correction = 3480 cm.So, The mean annual rainfall at X after adjustment = 3480 = 96.7 cm 36iv) The slope of the adjusted mass curve for the affected period of 1941 to 1953, that is the slope of the line CB = 0.85. The slope of the unadjusted mass curve, that is the slope of the line CB = 1.17 Therefore the correction factor for adjustment = 0.85 = 0.7265 1.17

The adjusted annual rainfalls at station X for the period 1941 to 1952 are given below.

Problem Based 3 :The annual precipitation at station Z and the average annual precipitation at 10 neighboring stations are as follows:

Estimate the missing precipitation data at station Y.

YearPrecipitation at Z (mm)10 station Average (mm)19723528197337291974393119753527197630251977252119782017197924211980302619813131198235361983383919844044198528321986253019872123

The End