HYDROLOGY FOR ENGINEERS



3

PRECIPITATION

INTRODUCTION

The term precipitation as used in hydrology is meant for all forms of moisture emanating from the clouds and all forms of water like rain, snow, hail and sleet derived from atmospheric vapors, falling to the ground.

Precipitation is one of the most important events of hydrology. Floods and droughts are directly related to the occurrence of precipitation. Water resources management, water supply schemes, irrigation, hydrologic data for design of hydraulic structures and environmental effects of water resources development projects are related to precipitation in one way or the other. So it is important to study various aspects of precipitation.

FORMS OF PRECIPITATION

In the middle latitudes precipitation occurs in many forms, depending on the existing meteorological conditions. These are following.

i. Drizzle

These are the minute particles of water at start of rain. These consist of water drops under 0.5 mm diameter and its intensity is usually less than 1.0 mm/hr. Their speed is very slow and we cannot even feel them. Therefore they cannot flow over the surface but usually evaporate.

ii. Rain

It is form of precipitation in which the size of drops in this case is more than 0.5 mm and less than 6.25 mm in diameter. It can produce flow over the ground and can infiltrate and percolate. Both the duration as well as rate of rainfall are important. If the rainfall per unit time is greater than the rate of infiltration, the rain water can flow over the surface of earth.

iii. Glaze

It is the ice coating formed on drizzle or rain drops as it comes in contact with the cold surfaces on the ground.

iv. Sleet

Sleet is frozen rain drops cooled to the ice stage while falling through air at subfreezing temperatures.

v. Snow

Snow is precipitation in the form of ice crystals resulting from sublimation i.e. change of water vapor directly to ice.

vi. Snowflake

A snowflake is made up of a number of ice crystals fused together.

vii. Hail

Hail is the type of precipitation in the form of balls or lumps of ice over 5 mm diameter formed by alternate freezing and melting as they are carried up and down by highly turbulent air currents. The impact of these is also more. A single hailstone weighing over a pound has been observed.

FACTORS INFLUENCING PRECIPITATION FORMATION

Following four conditions are necessary for the production of precipitation.

a. A lifting mechanism to produce cooling of the air.

b. A mechanism to produce condensation of water vapors and formation of cloud droplets.

c. A mechanism to produce growth of cloud droplets to size capable of falling to the ground against the lifting force of air.

d. A mechanism to produce sufficient accumulation of moisture to account for observed heavy rainfall rates.

a. Mechanism of Cooling

The pressure reduction when air ascends from near the surface to upper levels in the atmosphere is the only mechanism capable of producing the degree and rate of cooling needed to account for heavy rainfall. Cooling lowers the capacity of a given volume of air to hold a certain amount of water vapor. As large degrees of super saturation are not known to occur in the atmosphere, excess moisture over saturation condenses through the cooling process.

b. Condensation of Water Vapor

Condensation of water into cloud droplets takes place on hygroscopic nuclei which are small particles having an affinity for water. The source of these condensation nuclei is the particles of sea salt or such products of combustion of certain sulfurous and nitrous acids. There appears to be always sufficient nuclei present in the atmosphere. Condensation will always occur in air, the lower atmosphere is cooled to saturation, often before the saturation point is reached.

c. Growth of Cloud Droplets

Growth of droplets is required if the liquid water present in the cloud is to reach the ground. The two processes regarded as most effective for droplet growth are:

i. Coalescence of droplets through collision due to difference in speed of motion between larger and smaller droplets.

ii. Co-existence of ice crystals and water droplets.

Co-existence effect generally happens in the temperature range from 100 to 20o F. If in a layer of clouds there is a mixture of water droplets and ice crystals, the saturation vapor pressure over ice is lower than that over water. This leads to the evaporation of water drops and condensation of much of this water on ice crystals causing their growth and ultimate fall through the clouds. This effect is known as Bergeron’s effect. The ice crystals will further grow as they fall and collide with water droplets.

d. Accumulation of Moisture

Heavy rainfall amount over a river basin exceeds by far the amount of water vapor at the atmospheric volume vertically above the basin at the beginning of the rainfall. Clearly there must be a large net horizontal inflow of water vapor into the atmosphere above the basin area. This process is called convergence, which is defined as the net horizontal influx of air per unit area. The moisture added to the atmosphere over a basin may be transported very large distance in the lowest layer of the atmosphere. When this moist current reaches a region of active vertical motion it rises thousands of feets and loses much of its contained water vapor in just a few hours.

Classification of Precipitation Based on the Lifting Mechanism

The precipitation is often classified according to the factor responsible for lifting of air to higher altitudes. Following are the various types of precipitation based on this classification.

i. Convectional Precipitation

In the case of convectional precipitation the main causative element is thermal convection of the moisture laden air. For this to occur, solar radiation is the only source of heat. A major portion of the solar radiation is utilized in heating the earth. As the earth conducts heat slowly, the heat accumulates at the surface of the earth and air which comes in its contact is heated up and the lapse rate near the surface of the earth increases rapidly. With the passage of time as the sun gets higher and higher the lapse rate increases above that of dry adiabatic and air becomes unstable. Vertical currents are then set up which carry heat and the moisture laden air is picked up from the surface to higher levels. Due to convection, the moist air in the lower levels of the atmosphere rises up to the condensation level where clouds develop and with further convection these clouds finally grow into cumunimbus resulting in a thunderstorm. Lightning and thunder, gusty surface winds, showers and sometimes hails accompany a thunderstorm. Each thunderstorm is formed of a cell which updraft or down draft (downward current) turbulence etc. These cells are called under-storm cells. A cell hardly covers an area more than 1-2 square miles.

As the warm moist air from the ground is lifted up, more and more water will condense and the water drops and ice crystals will increase in size till such time that these drops are no longer supported by the existing updrafts. The drops will then begin to fall. Such thunderstorms are called “air mass thunderstorms”. They usually develop by mid-day and reach maximum intensity by afternoon. By late evening such storms start dissipating. Thunderstorms may extend to a height of 20,000 to 25,000 feet in temperate latitude and 30,000 to 40,000 feet in tropics.

ii. Orographic Precipitation

In the orographic precipitation, expansion and condensation occurs because moisture laden air masses are lifted by contact with orographic barriers. This type of precipitation is most pronounced on the windward side of mountain range, generally heaviest precipitation occurs where favorable orographic effects are present. For instance, heaviest precipitation due to south-easterlies in the subcontinent occurs along the Southern slopes of Himalayas and its other ranges. It has been found that the monsoon rainfall (June to October) decreases gradually as the distance from the line of heaviest rainfall increases.

Orographic precipitation also occurs in the inland areas where mountain ranges rise above the surrounding areas in the path of the moisture laden air masses. On these areas, however, orographic precipitation is intermingled with the other types prevalent in the area and, therefore, it becomes difficult to identify quantitatively the amount of precipitation that has occurred due to orographic effects only. On inland areas the orographic precipitation is irregular in occurrence and quantity due to interference by the atmospheric disturbance resulting from cyclonic storm.

Generally, it has been observed that heavy orographic precipitation occurs on the windward side of the orographic barrier, leaving a relatively dry area leeward side. This occurs because the moist air has been forced up the windward side and precipitated its moisture, and upon passing the peak of the barrier, no further orographic lifting occurs so that the rain fall is the residual of previous condensation. If the orographic barrier is sufficiently massive and the low regions on the leeward side are very extensive, the wind may descend on the leeward side, thereby undergoing compression and heating and becoming still more unfavorable for precipitation.

iii. Cyclonic Precipitation

Precipitation in plain regions is generally cyclonic in character. Cyclones are of two general classes, tropical and extra-tropical, so called depending upon whether they occur within or beyond the tropics. In as much as all cyclones occurring in the Indo-Pak Subcontinent are of tropical variety. This kind alone will be discussed here. Moreover, cause of these storms is not of primary concern except if it affects the precipitation. Tropical cyclones are violent storms which are generally formed in the ward maritime air-mass of low latitude where the temperature is high. These are known as typhoons or cyclones when formed in the Indian Ocean.

In the center of the cyclonic storm there is small low pressure air. The isobars around such a low pressure are very nearly circular in shape. Their intensity is generally greater than the extra-tropical cyclones. On an average the tropical cyclones have a diameter of over 300 to 400 miles. The wind speed around tropical cyclones may be as high as 60 to 90 miles per hour.

The winds around a tropical cyclone are practically parallel to the isobars. As the center or eye of the storm approaches, the high winds suddenly die down as practically dead calm prevails at the center. The diameter of the eye of the storm is about 50 miles. The winds around the center of the storm blow almost in a circular path (especially over the sea or ocean where the frictional force is minimum) and at the same time air is rising. The vertical motion compels the air to be cooled adiabatically. This air being humid, condensation takes place rapidly, resulting copious rainfall and widespread thunderstorm. Over the center of the storm the weather is usually fair.

In the Indo-Pak Subcontinent, the cyclonic storms form in the Bay of Bengal in different months. During April, May and June most of these storms do not reach Pakistan. But some of them affect Bangladesh and give very heavy rainfall there. During the summer monsoon season, the cyclonic storms reach Pakistan and are fed with moisture from the Arabian sea resulting in heavy rainfall over the Northern areas of Pakistan. In September, October and November these storms are very destructive in Bangladesh. Such storms cause considerable loss of life and property over the coastal districts. Cyclonic storms also form in Arabian sea but their number is far less.

MEASUREMENT OF PRECIPITATION

Amount of Precipitation

The amount of precipitation means the vertical depth of water that would accumulate on a level surface, if the precipitation remains where it falls. The amount of precipitation is measured in length units (inches, ft., cm, etc.).

Intensity or Rate of Precipitation

Amount of precipitation per unit time is called the intensity of precipitation or rate of precipitation.

Both the amount and rate of precipitation are important in hydrologic studies.

The precipitation is measured by rain gauges. There are two types of rain gauges:

a. Non-recording rain gauge. (Standard rain gauge)

b. Recording rain gauge

he main difference between these rain gauges is that with the help of recording rain gauges we get the rain recorded automatically with respect to time, so intensity of rain fall is also known whereas an observer has to take readings from non recording rain gauge for rain and he has to record the time also, for calculation of intensity of rain fall.

a. Non-Recording Rain Gauges

The standard gauge of U.S. Weather Bureau has a collector of 200 mm diameter and 600 mm height. Rain passes from a collector into a cylindrical measuring tube inside the overflow can. The measuring tube has a cross sectional area 1/10th of the collector, so that 2.5 mm rain fall will fill the tube to 25 mm depth. A measuring stick is marked in such a way that 1/10 th of a cm depth can be measured. In this way net rainfall can be measured to the nearest 1 mm. The collector and tube are removed when snow is expected. The snow collected in the outer container or overflow can is melted, poured into the measuring tube and then measured. This type of rain gauge is one of the most commonly used rain gauge.

[pic]

Fig. 3.1 Non-Recording Rain Gauge

Sources of Error

Rainfall measured by the rain gauge might have some errors. For example some water is used to wet the surface of instrument; the rain recorded may be less than the actual rainfall due to the direction of the rainfall as affected by wind. Dents in the collector and tube may also cause error. Some water is absorbed by the measuring stick. Losses due to evaporation can also take place. The volume of stick replaces some water which causes some error.

b. Recording Rain Gauges

Recording rain gauges can be divided into the following types:

i. Float type

ii. Weighing type

iii. Tipping bucket type

i. Float Type Rain Gauge

This type of rain gauge also has a receiver and a float chamber along with some recording mechanism or arrangement. In this type the rain is led into a float chamber containing a light, hollow float. The vertical movement of the float as the level of water rises is recorded on a chart with the help of a pen connected to float. The chart is wrapped around a rotating clock driven drum. To provide a continuous record for 24 hours the float chamber has either to be very large, or some automatic means are provided for emptying the float chamber quickly when it becomes full, the pen then returning to the bottom of the chart. This is usually done with some sort of siphoning arrangement. This arrangement activates when the gauge records a certain fixed amount of rain (mostly 10 mm of rainfall.). Snow can not be measured by this type of rain gauge. Fig. 3.2 shows line diagram of float type rain gauge.

[pic]

Fig. 3.2 FLOAT TYPE RAIN GAUGE

ii. Weighing Type Rain Gauge

The weighing type rain gauge consists of a receiver, a bucket, a spring balance and some recording arrangement. The weighing type gauge weighs the rain or snow which falls into a bucket which is set on a lever balance. The weight of the bucket and content is recorded on a chart by a clock driven drum. The record is in the form of a graph, one axis of which is in depth units and the other has time. The records show the accumulation of precipitation. Weighing type gauges operate from 1 to 2 months without stopping. But normally one chart is enough only for 24 hours. This type of rain gauge has advantage of measuring snow also. The receiver is removed when snow is expected. Figure 3.3 shows schematic sketch of a weighing type rain gauge.

[pic]

Fig 3.3 WEIGHING TYPE RAIN GAUGE

iii. Tipping Bucket Type Rain Gauge

This type of gauge used at some Weather Bureau First Order Stations, is equipped with a remote recorder located inside the office which is away from the actual site. The gauge has two compartments pivoted in such a way that one compartment receives rain at one time. A certain amount of rain (usually 0.25 mm fills one compartment and over balances it so that it tips, emptying into a reservoir and bringing the second compartment of the bucket into place beneath the funnel of receiver. As the bucket is tipped by each 0.25 mm of rain it actuates an electrical circuit, causing a pen to mark on a revolving drum. This type of gauge is not suitable for measuring snow without heating the collector. Plotting is similar to that of other recording rain gauges. A Tipping Bucket Type Rain Gauge is shown in Figure 3.4.

[pic]

Fig. 3.4 tipping bucket type rain gauge

SOURCES OF ERROR

All sources of error are similar to those of non-recording rain gauges except the error due to stick. These are mentioned again as below:

a. Dents in the collector.

b. Moistening of inside-surface of the funnel and the tube.

c. Rain drops splashing from the collector.

d. For very intense rain some water is still pouring into the already filled bucket.

e. Inclination of the gauge may result in catching less or more rain than the actual amount.

f. Error in measurement due to wind.

Remedial Measures for Errors in Precipitation Measurement

Removal of error due to dents obviously needs repair of the instrument. For rain recorded with dents a correction should be applied. Errors such as moistening of the inside surfaces of the gauge, splashing of rainwater from the collector and pouring of water into the already filled bucket during an intense rain can only be corrected by some correction factor. Inclined instrument needs to be reinstalled. The correction factor however can be calculated from the angle of inclination. For wind protection certain wind shields are designed and used which are called Splash Guards. Proper setting of gauge above ground level is necessary.

Example 3.1

A rain gauge recorded 125 mm of precipitation. It was found later that the gauge was inclined at an angle of 20 degree with the vertical. Find the actual precipitation.

Solution

P(measured) = 125 mm

Angle of inclination (θ) = 20o with the vertical

P(actual) = P(measured)/cos(θ) = 125/cos20o = 133 mm

Measurement of Precipitation by Radar

This is a modern technique for measurement of rainfall rate. It can also detect local movement of areas of precipitation. The electromagnetic energy released and received back by radar is a measure of rainfall intensity. The measurement is appreciably affected by trees and buildings. However extent of rainfall can be estimated with reasonable accuracy. Use of radar is useful where number of rain gauges installed in an area is not sufficient.

Rain Gauge Network

The number of rain gauges and their distribution affect the nature of collected precipitation data. The larger the number of rain gauges the more representative will be the data colleted. But on the other hand we have to observe other factors also, like economy of the project, accessibility of certain areas and topography of the area. So, one has to look for some optimum solution. In this regard the World Meteorological Organization (WMO) has made following recommendations for minimum number of rain gauges in a catchment:

a. In comparatively flat regions of temperate, Mediterranean and Tropical Zones, the ideal is at least one station for 230 – 345 sq. miles. However one station for 345 – 1155 sq. miles is also acceptable

b. In mountainous regions of Temperate, Mediterranean and Tropical Zones, the ideal is at least one station for 35 – 95 sq. miles. However one station for 95 – 385 sq. miles is also acceptable.

c. In arid and polar zones, one station for 575 – 3860 sq. miles is acceptable

ANALYSIS OF PRECIPITATION DATA

Point Data Analysis

Point precipitation data refers to precipitation of a station. This data could be in form of hourly record, daily record, monthly precipitation or annual precipitation. Depending upon the nature of catchment and its area, there could be as many gauging stations as feasible. Before using records from a rain gauge we should check its continuity and consistency. Record may not be continuous and consistent due to many reasons as explained in the following paragraphs. This section deals with estimation of any missing records at a particular gauging station, checking consistency of data and its adjustment accordingly.

1. ESTIMATION OF MISSING PRECIPITATION RECORD

Some precipitation stations may have short breaks in the records because of absence of the observer or because of instrumental failures. It is often necessary to estimate this missing record. In the procedure used by the U.S. Weather Bureau, the missing precipitation of a station is estimated from the observations of precipitation at some other stations as close to and as evenly spaced around the station with the missing record as possible. There are two methods for estimation of missing data - Arithmetic mean method and Normal ratio method.

The station whose data is missing is called interpolation station and gauging stations whose data are used to calculate the missing station data are called index stations.

If the normal annual precipitation of the index stations lies within (10% of normal annual precipitation of interpolation station then we apply arithmetic mean method to determine the missing precipitation record otherwise the normal ratio method is used for this purpose.

Consider that record is missing from a station ‘X’.

Now let,

N = Normal annual precipitation. (Mean of 30 years of annual precipitation data)

P = Storm Precipitation.

Let Px be the missing precipitation for station ‘X’ and Nx , the normal annual precipitation of this station, Na, Nb and Nc are normal annual precipitations of nearby three stations, A, B and C respectively while Pa, Pb and Pc are the storm precipitation of that period for these stations.

Now we have to compare Nx with Na , Nb and Nc separately. If difference of Nx - Na, Nx - Nb, Nx - Nc is within 1/10% of Nx then we use simple arithmetic mean method otherwise the normal ratio method is used.

Simple Arithmetic Mean Method

According to the arithmetic mean method the missing precipitation ‘Px’ is given as:

Px = [pic] where ‘n’ is number of nearby stations, ‘Pi’ is

precipitation at ith station.

In case of three stations 1, 2 and 3,

Px = (P1 + P2 + P3)/3

and naming stations as A, B and C instead of 1, 2 and 3

Px = (Pa + Pb + Pc)/3

Where Pa , Pb and Pc are defined above.

Normal Ratio Method

According to the normal ratio method the missing precipitation is given as:

Px = [pic]

Where Px is the missing precipitation for any storm at the interpolation station ‘x’. Pi is the precipitation for the same period for the same storm at the “ith” station of a group of index stations and Nx and Ni are the normal annual precipitation values for the ‘x’ and ‘ith’ stations. For example, for the symbols defined above for three index stations in a catchment area.

Px= [pic]

Example 3.2

Find out the missing storm precipitation of station ‘C’ given in the following table:

|Station |A |B |C |D |E |

|Storm precipitation (cm) |9.7 |8.3 |---- |11.7 |8.0 |

|Normal Annual precipitation (cm) |100.3 |109.5 |93.5 |125.7 |117.5 |

Solution

In this example the storm precipitation and normal annual precipitations at stations A, B, D and E are given and missing precipitation at station ‘C’ is to be calculated whose normal annual precipitation is known. We will determine first that whether arithmetic mean or normal ratio method is to be applied.

10% of Nc = 93.5x10/100 = 9.35

After the addition of 10% of Nc in Nc, we get 93.5 + 9.35=102.85

And by subtracting 10% we get a value of 84.15

So Na, Nb, Nd or Ne values are to be checked for the range 102.85 to 84.15.

If any value of Na, Nb, Nd or Ne lies beyond this range, then normal ratio method would be used. It is clear from data in table above that Nb, Nd and Ne values are out of this range so the normal ratio method is applicable here, according to which

Px = [pic]

Pc= (1/4 )(93.5 x 9.7/100.3+ 93.5 x 8.3/109.5+ 93.5 x 11.7/125.7+ 93.5 x 8.0/117.5) = 7.8 cm

Example 3.3

Precipitation station “X” was inoperative for part of a month during which a storm occurred. The storm totals at three surrounding stations A, B and C were respectively 10.7, 8.9 and 12.2 cm. The normal annual precipitation amounts at stations X, A, B and C are respectively 97.8, 112, 93.5 and 119.9 cm. Estimate the storm precipitation for station ‘X’.

Solution

Pa = 10.7 cm Na = 112 cm

Pb = 8.90 cm Nb = 93.5 cm

Pc = 12.2 cm Nc = 119.9 cm

Px = ? Nx = 97.8 cm

10% of Nx = 97.8 x 10/100 = 9.78 cm.

( Nx - Na = 97.8 - 112 = -14.2 cm ( More than + 10% of Nx (no need of calculating Nx – Nb and Nx - Nc

So we will use Normal Ratio Method.

( Px = (1/3)( 97.8x 10.7/112+ 97.8x 8.90 /93.5 + 97.8x 12.2 /119.9)

Px = 9.5 cm

2. Consistency of Precipitation Data or Double Mass Analysis

In using precipitation in the solution of hydrologic problems, it is necessary to ascertain that time trends in the data are due to meteorological changes. Quite frequently these trends are the result of the changes in the gauge location, changes in the intermediate surroundings such as construction of buildings or growth of trees, etc. and changes in the observation techniques. Frequently changes in gauge location are not disclosed in the published record. Due to such changes the data might not be consistent. The consistency of the record then is required to be determined and the necessary adjustments be made. This can be achieved by the method called the double mass curve technique.

The double mass curve Fig. 3.5 is obtained by plotting the accumulated precipitation at the station in question along X-axis and the average accumulated precipitation of a number of other nearby stations which are situated under the same meteorological conditions along Y-axis. If the curve has a constant slope, the record of station “X” is consistent. However, if there is any break in the slope of the curve, the record of the station is inconsistent and has to be adjusted by the formula.

Pa = (Sa / So)x Po

Where Pa = Adjusted precipitation.

Po = Observed precipitation.

Sa = Slope prior to the break in the curve

So = Slope after the break in the curve.

All values after break are to be adjusted.

Example 3.4

Check consistency of the data given in table 3.1 below and adjust it if it is found to be inconsistent

Table 3.1 Precipitation Data

|Year |Annual precipitation |Mean of annual |Year |Annual precipitation |Mean of annual |

| |at x (mm) |precipitation of 20 | |at x (mm) |precipitation of 20|

| | |surrounding stations | | |surrounding |

| | |(mm) | | |stations (mm) |

|1972 |188 |264 |1954 |223 |360 |

|1971 |185 |228 |1953 |173 |234 |

|1970 |310 |386 |1952 |282 |333 |

|1969 |295 |297 |1951 |218 |236 |

|1968 |208 |284 |1950 |246 |251 |

|1967 |287 |350 |1949 |284 |284 |

|1966 |183 |236 |1948 |493 |361 |

|1965 |304 |371 |1947 |320 |282 |

|1964 |228 |234 |1946 |274 |252 |

|1963 |216 |290 |1945 |322 |274 |

|1962 |224 |282 |1944 |437 |302 |

|1961 |203 |246 |1943 |389 |350 |

|1960 |284 |264 |1942 |305 |228 |

|1959 |295 |332 |1941 |320 |312 |

|1958 |206 |231 |1940 |328 |284 |

|1957 |269 |234 |1939 |308 |315 |

|1956 |241 |231 |1938 |302 |280 |

|1955 |284 |312 |1937 |414 |343 |

Solution

A double mass curve is plotted by taking cumulative of average precipitation of surrounding stations along x-axis and accumulative precipitation of station ‘X’ along

y-axis for which consistency of data is being investigated.

The double mass curve is shown in Figure 3.5

[pic]

Figure 3.5 Double Mass Curve

The correction for slope is applied to readings beyond break in slope. The calculations are shown in table 3.2, below.

Slope of Ist line = Sa = 0.854

Slope of deviating line = So = 1.176

Correction to values (multiplying factor) = 0.854/1.176 = 0.70

Now regime changes before 1950. So up to 1950 no correction is required. Before 1950 all readings are multiplied by slopes ratio of 0.7 to get corrected precipitation. Note that data in latter interval (1973-1950) is considered more authentic so kept in initial reach of the graph.

Table 3.2 Adjusted Precipitation

|Year |Cummulative Annual |Cummulative precipitation of 20|Corrected Precipitation |Remarks |

| |precipitation at x (mm) |surrounding stations (mm) | | |

|1972 |188 |264 |188 |No |

| | | | |correction |

|1971 |373 |492 |185 | |

|1970 |683 |878 |310 | |

|1969 |978 |1175 |295 | |

|1968 |1186 |1459 |208 | |

|1967 |1473 |1809 |287 | |

|1966 |1656 |2045 |183 | |

|1965 |1960 |2416 |304 | |

|1964 |2188 |2650 |228 | |

|1963 |2404 |2940 |216 | |

|1962 |2628 |3222 |224 | |

|1961 |2831 |3468 |203 | |

|1960 |3115 |3732 |284 | |

|1959 |3410 |4064 |295 | |

|1958 |3616 |4295 |206 | |

|1957 |3885 |4529 |269 | |

|1956 |4126 |4760 |241 | |

|1955 |4410 |5072 |284 | |

|1954 |4633 |5432 |223 | |

|1953 |4806 |5666 |173 | |

|1952 |5088 |5999 |282 | |

|1951 |5306 |6235 |218 | |

|1950 |5552 |6486 |246 | |

|1949 |5836 |6770 |190.8 |Precipitatio|

| | | | |n of Station|

| | | | |'X' x 0.7 |

|1948 |6329 |7131 |345 | |

|1947 |6649 |7413 |224 | |

|1946 |6923 |7665 |192 | |

|1945 |7245 |7939 |225.4 | |

|1944 |7682 |8241 |306 | |

|1943 |8071 |8591 |272.3 | |

|1942 |8376 |8819 |213.5 | |

|1941 |8696 |9131 |224 | |

|1940 |9024 |9415 |229.6 | |

|1939 |9332 |9730 |215.6 | |

|1938 |9634 |10010 |211.4 | |

|1937 |10048 |10353 |290 | |

Areal Precipitation Data Analysis

To find out runoff from a catchment and most of other hydrologic analyses, it is important to know the average precipitation of a certain part of catchment or for the whole of the catchment area. To find out average precipitation of watershed records of precipitation from different rain gauge stations is used. There are many factors which affect the reliability of average precipitation of watershed determined by using the data from individual stations in the watershed. For example : the total number of rain gauges and their distribution in the catchment (larger the number of rain gauges, the reliable will be the calculated average precipitation), the size and shape of area of catchment, distribution of rainfall over the area and topography of the area. Lastly the method used for determining the average precipitation is also one of these factors. So before using average precipitation in hydrologic analyses, the user should be aware of these factors.

Estimation of Average Precipitation over a Basin

Conversion of point precipitation of various gauging stations into average precipitation of that area a great experience and skill is required. There are three methods to find average precipitation over a basin. Accuracy of estimated average precipitation will depend upon the choice of an appropriate method. These methods are described below:

i. Arithmetic Mean Method

In this method the average precipitation over an area is the arithmetic average of the gauge precipitation values. We take data for only those stations which are within the boundary. This is the simplest method but can be applicable only for flat areas and not for the hilly areas i.e. this method is used when:

a. Basin area is flat and

b. All stations are uniformly distributed (within practical limits) over the area.

c. The rainfall is also nearly uniformly distributed over the area.

According to this method

P (average) = [pic]

or

Pav = [P1+P2+P3+…………+Pn]/n

Where Pi is precipitation at station ‘i’ and there are ‘n’ number of gauges installed in the catchment area from where the data has been collected.

Example 3.5

Six rain gauges were installed in a relatively flat area and storm precipitation from these gauges was recorded as 3.7, 4.9, 6.8, 11.4, 7.6 and 12.7 cm respectively from gauges 1, 2, 3, 4, 5, and 6. Find average precipitation over the catchment.

Solution

As the area is relatively flat so we apply the arithmetic mean method. According to arithmetic mean method

P(average) = (3.7 + 4.9 + 6.8 + 11.4 + 7.6 + 12.7)/6 = 7.85 cm.

ii. Thiessen Polygon Method:

The fundamental principle followed in this method consists of weighing the values at each station by a suitable proportion of the basin area. In this method, a special weighing factor is considered.

The following steps are used to determine average precipitation by Thiessen Polygon Method.

1. Draw the given area according to a certain scale and locate the stations where measuring devices are installed.

2. Join all the stations to get a network of non-intersecting system of triangles.

3. Draw perpendicular bisectors of all the lines joining the stations and get a suitable network of polygons, each enclosing one station. It is assumed that precipitation over the area enclosed by the polygon is uniform.

4. Measure area of the each polygon.

5. Calculate the average precipitation. For the whole basin by the formula

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

Where,

P1 = Precipitation. at station enclosed by polygon of area A1

P2 = Precipitation. at station enclosed by polygon of area A2

and so on

Pn = Precipitation. at station enclosed by polygon of area An

And ‘A’ represents the total area of the catchment.

Example 3.6:

Following is shown map of a catchment having 6 rainfall recording stations.

[pic]

Fig. 3.6 Average Precipitation by Theissen Polygon Method

Find the Average Precipitation over the whole catchment.

The recorded precipitations are shown on the topographic map of the catchment. The Thiessen’s Polygons are constructed by the method explained above. The precipitation and polygon area are given below.

|Station |Precipitation (mm) |Polygon Area (km²) |

|Daggar |48 |5,068.76 |

|Besham |33 |4,349.17 |

|Shinkiari |25 |1,399.25 |

|Phulra |32 |1,693.80 |

|Tarbela |56 |2,196.33 |

|Oghi |30 |2,234.29 |

Solution

The calculations are best done in tabular form as shown in Table 3.3.

Table 3.3 Average Precipitation by Thiessen Polygon Method

|Station |Precipitation P (mm) |Polygon Area A (km²) |P x A (x106 m³) |

|Daggar |48 |5,068.76 |243.30 |

|Besham |33 |4,349.17 |143.52 |

|Shinkiari |25 |1,399.25 |34.98 |

|Phulra |32 |1,693.80 |54.20 |

|Tarbela |56 |2,196.33 |122.99 |

|Oghi |30 |2,234.29 |67.03 |

| |Total |16,941.60 |666.02 |

Mean Precipitation = [pic]

= 666.02x106x10³/16941.60x106

= 39.3 mm

Example 3.7

From the data given in Table 3.4 below, which was obtained from Thiessen Polygon map of a catchment, find out the average precipitation of the catchment.

Table 3.4 Precipitation Data

|Sr No |Gauge |Area of Thiessen Polygon |Sr No |Gauge precipitation |Area of Thiessen Polygon |

| |precipitation |enclosing the station | |(cm) |enclosing the station |

| |(cm) |(sq. km) | | |(sq. km) |

|1 |10.2 |416 |4 |9.4 |520 |

|2 |8.1 |260 |5 |15.2 |390 |

|3 |12.7 |650 |6 |7.6 |325 |

Solution

According to Thiessen Polygon Method

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

The calculations are shown in tabular form in Table 3.5

Table 3.5 Average Precipitation by Thiessen Method

|Gauge precipitation (cm) |Area of Thiessen Polygon enclosing |Volume = Pi x Ai (x104 m³) |

| |the station (sq. km) | |

|(1) = Pi |(2) = Ai |(3) = (1)x(2) |

| | | |

|10.2 |416 |4243.20 |

|8.1 |260 |2106.00 |

|12.7 |650 |8255.00 |

|9.4 |520 |4888.00 |

|15.2 |390 |5928.00 |

|7.6 |325 |2470.00 |

|Total |2561 |27890.20 |

So P (average) = 27890.20 ÷ 2561=10.9 cm

Example 3.8:

There are 10 observation stations, 7 being inside and 3 in neighborhood of a catchment. Thiessen Polygons were drawn for a storm data from these observation stations and the data given in Table 3.6 below was obtained. Find out the average precipitation of the catchment.

Table 3.6 Average Precipitation by Thiessen Polygon Method

|Sr No |Gauge precipitation (cm) |Area of Thiessen Polygon enclosing |

| | |the station (sq. km) |

|1 |5 |100 |

|2 |3 |160 |

|3 |4 |200 |

|4 |3.5 |215 |

|5 |4.7 |250 |

|6 |6 |175 |

|7 |4 |100 |

| | | |

Solution

According to Thiessen Polygon Method

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

|Gauge precipitation (cm) |Area of Thiessen Polygon enclosing the station |Volume = PixAi (x104 m³) |

| |(sq. km) | |

|(1) = Pi |(2) = Ai |(3) = (1)x(2) |

|5 |100 |500 |

|3 |160 |480 |

|4 |200 |800 |

|3.5 |215 |752.50 |

|4.7 |250 |1175 |

|6 |175 |1050 |

|4 |100 |400 |

|Total |1200 |5157.50 |

So P (average) = 5157.5 ÷ 1200 = 4.3 cm

iii. Isohyetal Method

The most accurate method of averaging precipitation over an area is the isohyetal method.

For estimation of average precipitation of the catchment by isohyetal method the following steps are used:

1. Draw the map of the area according to a certain scale.

2. Locate the points on map where precipitation measuring gauges are installed.

3. Write the amount of precipitation for stations.

4. Draw isohyets (Lines joining points of equal precipitation).

5. Measure area enclosed between every two isohyets or the area enclosed by an isohyet and boundary of the catchment.

6. Find average precipitation by the formula.

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

Where,

P1= Mean precipitation of two isohyets 1 and 2

A1= Area between these two isohyets.

P2 = Mean precipitation of two isohyets 2 and 3

A2 = the area b/w these two isohyets.

and, so on

Pn = Mean precipitation of isohyets n-1 and n

An = the area between these two isohyets.

It may be noted that the last and first areas mentioned above may be between an isohyet and boundary of the catchment. In this case the precipitation at the boundary line is required which may be extrapolated or interpolated.

Example 3.9

From the data given in table 3.7 below, which was obtained from isohyetal map of a catchment, find out the average precipitation of the catchment.

Table E3.7 Data from Isohyetal Map.

|Isohyet No. |Isohyetal precipitation (cm) |Area enclosed between two isohyets. (sq km) |

|1 |2.5 |390 |

|2 |5.0 |520 |

|3 |7.5 |650 |

|4 |10.0 |390 |

|5 |10.0 |390 |

|6 |7.5 |442 |

|7 |5.0 |546 |

|8 |2.5 | |

Note that the isohyet No. 1 and 8 were out of the boundary of the catchment. The area between isohyet No. 2 and the boundary was estimated to be 312 sq. km and that of between isohyet No. 8 and boundary was 494 sq. km. Precipitation on these boundaries was interpolated as 3.0 and 3.1 cm, respectively.

Solution

In isohyetal method we have to calculate the average precipitation of every two consecutive isohyets. This is given in Table E3.8 below:

Table E3.8 Average Precipitation by Isohyetal Method

|Isohyet No. |Isohyetal |Average of precipitation of|Area enclosed between two |Volume |

| |precipitation (cm) |two consecutive isohyets |isohyets |(x104 m³) |

| | |(cm) |(sq km) | |

|(1) |(2) |(3) |(4) |(5) = (3) x (4) |

|1 |2.5 |3 (for isohyet and |312 (for isohyet and |936.00 |

| | |boundary) |boundary) | |

|2 |5.0 |6.25 |520 |3250.00 |

|3 |7.5 |8.75 |650 |5687.50 |

|4 |10.0 |10.0 |390 |3900.00 |

|5 |10.0 |8.75 |390 |3412.50 |

|6 |7.5 |6.25 |442 |2762.50 |

|7 |5.0 |3.1 (for isohyet and |546 (for isohyet and |1692.60 |

| | |boundary) |boundary) | |

|8 |2.5 | | | |

| | |∑ |3250 |21641.1 |

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

= 21641.1/3250=6.66 cm

Example 3.10

In a catchment of area 1,000 sq km, there are 8 rain gauges, 5 inside the area and 3 outside, in its surroundings. Isohyets were drawn from the data of these rain gauges for a storm. From the isohyetal map the following information was obtained: areas between 1 and 2 cm isohyets, 2 and 3 cm, 3 and 4 cm and 4 and 5 cm isohyets was 105, 230, 150 and 220 sq. km, respectively. The area between one end boundary which has 0.75 cm rainfall and 1 cm isohyet was 120 sq. km and the other end boundary which has precipitation of 5.5 cm and isohyet of 5 cm was 175 sq. km. Find average precipitation.

Solution

In the isohyetal method we have to calculate the average precipitation of every two consecutive isohyets. This is given in Table 3.9 below.

Table E3.9 Average Precipitation by Isohyetal Method

|Isohyet No |Isohyetal |Average of precipitation of two |Area enclosed between two |Volume |

| |precipitation (cm) |consecutive isohyets (cm) |isohyets |(x104 m³) |

| | | |(sq km) | |

|Boundary |0.75 |0.875 (for isohyet and boundary)|120 (for isohyet and |105.00 |

| | | |boundary) | |

|1 |1 |1.5 |105 |157.50 |

|2 |2 |2.5 |230 |575.00 |

|3 |3 |3.5 |150 |525.00 |

|4 |4 |4.5 |220 |990.00 |

|5 |5 |5.25 (for isohyet and boundary) |175 |918.75 |

|Boundary |5.5 | | | |

| | |∑ |1000.00 |3271.25 |

P (average) = (P1 A1 + P2 A2 + ...........+ Pn An)/A

= 3271.25/1000

=3.27 cm

Example 3.11:

From the isohyetal map shown in Fig. 3.3 below find out average precipitation.

[pic]

Fig. 3.7 Isohyetal Map

Solution:

The isohyets are drawn on the topographic map by interpolating rainfall depths at given stations. Once isohyets are drawn, the area enclosed between consecutive isohyets is determined either by planimeter or other suitable more precise method.

The calculations for average precipitation are given in table 3.10 below.

Table 3.10 Average Precipitation by Isohyetal Method

|Isohyte value (mm) |Av. Isohyte Value (mm) |Area Between Consecutive |Volume (x106 m³) |

| | |Isohytes (km²) | |

|Boundary and 25 |25.0 |310.53 |7.76 |

|25 and 30 |27.5 |2220.71 |61.07 |

|30 and 35 |32.5 |2968.38 |96.47 |

|35 and 40 |37.5 |2231.86 |83.69 |

|40 and 45 |42.5 |2303.52 |97.90 |

|45 and 50 |47.5 |2731.90 |129.77 |

|50 and 55 |52.5 |2689.70 |141.21 |

|55 and Boundary |55 |1484.99 |81.67 |

| |Total |16,941.60 |699.54 |

Mean Precipitation Depth = Volume/Area

= 699.54x106x10³/16941.60x106

= 41.29 mm

INTENSITY OF PRECIPITATION

The rate of occurrence of precipitation is called intensity of precipitation or precipitation occurring in unit time is known as intensity of precipitation.

To find out intensity of a certain interval the points on graph of accumulative precipitation vs time are chosen in such a way that we get the maximum difference for the given interval. It is explained in the following example.

Example 3.12

Find out intensity of precipitation of 5, 10 & 15 minutes for rainfall data given in

Table 3.11.

Table 3.11 Rainfall Data

|Time (Minutes) |(P (cm) |

|0 |00 |

|05 |12.5 |

|10 |20 |

|15 |42.5 |

|20 |62.5 |

Solution

For 5-minutes interval the maximum difference is 22.5

so, intensity for 5-minutes interval = 22.5 / 5 = 4.5 cm/min.

For 10-minutes intensity = 42.5 / 10 = 4.25 cm/min.

For 15-minutes intensity = 50 / 15 = 3.33 cm/min.

Depth - Area Relationships

The distribution of rainfall is usually not uniform over the area. The precipitation is maximum at the centre of storm and decreases as we move away from the centre of storm. For rainfall of a given duration, the average depth decreases with the area in an exponential manner.

Mass Curve

Precipitation recorded by a weighing type and float type recording rain gauge is in form of a graph. It is a plot of cumulative precipitation against time in chronological order. This is called mass curve of rainfall data. Intensity of rainfall for a certain duration is determined from this graph. This graph gives a complete history of a storm regarding the duration and magnitude of precipitation. In the case of non-recording rain gauges the mass curve is to be plotted from the data in which both duration and magnitude of precipitation have been observed for different time intervals during storm.

Depth-Area-Duration Curves

Analysis of both the time and areal distribution of a storm is required in many hydrologic studies. Depth area duration curves provide requisite information for such studies. It is necessary to have information on the maximum amount of precipitation of various durations occurring over various sizes of areas. The development of a relationship between maximum depth-area-duration for a region is called depth area duration analysis (DAD analysis). In this analysis first the isohyetal maps and mass curves of worst storms which have occurred in past in the region are prepared. For a storm with a single major centre the isohyets are taken as the boundaries of individual area. The average storm precipitation within each isohyet is computed. The storm total is distributed through successive increments of time (say 3 hours) in accordance with the distribution record at nearby stations. This gives data showing the time distribution of average rainfall over areas of various sizes. From this data the maximum rainfall for various durations (3, 6, 9, 12 hours) can be selected for each size of area. The maximum values for every duration plotted versus area gives what are called depth area duration curves.

Typical depth-area-duration curves are shown in figure 3.8.

[pic]

Fig. 3.8 Typical Depth Area Curves

EXAMPLE 3.13:

The isohyetal map shown in Fig 3.9 is for 10 hour storm over a catchment area. Area enclosed between two consecutive isohyets is shown on the map and isohyetal interval is 5 cm with storm center having precipitation of 30 cm. Find:

a. The average precipitation of the catchment by isohyetal method

b. The Equivalent Uniform Depth of rain for depth area duration curve.

[pic]

Fig. 3.9 Isohyet Map

Solution

The calculations are performed in Table 3.13. The average precipitation is found by summing up area enclosed by consecutive isohyets multiplied by average isohyte value and whole sum divided by total area.

The EUD is found by dividing cumulative volume by cumulative area. The Fig. 3.10 shows variation of EUD with area.

Table 3.13 Finding EUD

|Isohyte (cm) |Area enclosed |Cumm. Area |Mean |Volume |Cumm. Volume|EUD (cm) |

| |(km²) |enclosed (km²) |Isohyte |(x106) |(x106) | |

| | | |(cm) | | | |

|1 |2 |3 |4 |5 |6 |7 = 6÷3 |

|30 |60 |60 |30.00 |18.00 |18.00 |30.00 |

|30 & 25 |100 |160 |27.50 |27.50 |45.50 |28.44 |

|25 & 20 |90 |250 |22.50 |20.25 |65.75 |26.30 |

|20 & 15 |130 |380 |17.50 |22.75 |88.50 |23.29 |

|15 & 10 |200 |580 |12.50 |25.00 |113.50 |19.57 |

|10 & 5 |400 |980 |7.50 |30.00 |143.50 |14.64 |

|Mean Precipitation = 14.64 cm | |  |  |  |

[pic]

Fig. 3.10 Depth-Area Curve

EXAMPLE 3.14:

Plot depth area duration curves from the rainfall data for six stations given in Table 3.14

|DATE |TIME |STATION & PRECIPITATION |

| |(HOURS) | |

| | |BESHAM |PHULRA |DAGGAR |OGHI |SHINKIARI |TARBELA |

|16-02-2003 |0 |- |- |- |- |- |- |

| |1 |- |- |- |- |- |- |

| |2 |- |- |- |- |- |- |

| |3 |- |- |- |- |- |- |

| |4 |- |- |- |- |- |- |

| |5 |- |- |- |- |- |- |

| |6 |- |- |- |- |1 |- |

| |7 |1 |- |- |- |- |- |

| |8 |- |- |- |- |- |- |

| |9 |1 |- |- |- |- |- |

| |10 |2 |1 |2 |3 |- |- |

| |11 |2 |2 |2 |2 |1 |- |

| |12 |3 |3 |1 |2 |3 |4 |

| |13 |3 |3 |1 |1 |2 |6 |

| |14 |2 |2 |1 |2 |2 | |

| |15 |2 |1 |2 |3 |3 |2 |

| |16 |1 |2 |1 |2 |1 |- |

| |17 |2 |1 |1 |3 |1 |- |

| |18 |1 |1 |1 |3 |- |- |

| |19 |1 |- |1 |3 |- |- |

| |20 |1 |- |2 |2 |- |- |

| |21 |2 |- |2 |3 |- |- |

| |22 |3 |2 |2 |4 |- |3 |

| |23 |5 |3 |1 |3 |3 |4 |

| |24 |- |3 |1 |3 |4 |2 |

|17-02-2003 |1 |4 |3 |3 |3 |5 |- |

| |2 |5 |4 |3 |2 |3 |1 |

| |3 |5 |3 |4 |1 |2 |- |

| |4 |8 |3 |5 |1 |- |4 |

| |5 |3 |1 |6 |4 |- |- |

| |6 |1 |1 |3 |3 |- |1 |

| |7 |3 |1 |2 |3 |- |1 |

| |8 |3 |6 |2 |2 |3 |8 |

| |9 |2 |4 |- |9 |2 |5 |

| |10 |2 |4 |- |9 |3 |5 |

| |11 |7 |5 |6 |6 |1 |4 |

| |12 |4 |3 |11 |5 |- |8 |

| |13 |6 |4 |6 |7 |3 |4 |

| |14 |4 |4 |9 |8 |3 |6 |

| |15 |5 |5 |3 |6 |4 |8 |

| |16 |5 |6 |6 |6 |3 |9 |

| |17 |5 |3 |8 |4 |5 |8 |

| |18 |5 |5 |9 |5 |1 |9 |

| |19 |4 |3 |10 |5 |3 |10 |

| |20 |5 |4 |9 |4 |4 |15 |

| |21 |5 |4 |9 |5 |4 |9 |

|DATE |TIME |STATION & PRECIPITATION |

| |(HOURS) | |

| | |BESHAM |PHULRA |DAGGAR |OGHI |SHINKIARI |TARBELA |

| |22 |6 |5 |8 |7 |6 |11 |

| |23 |6 |8 |7 |5 |4 |6 |

| |24 |7 |9 |6 |4 |5 |4 |

|18-02-2003 |1 |5 |6 |3 |5 |4 |4 |

| |2 |5 |8 |3 |4 |6 |1 |

| |3 |3 |5 |2 |6 |8 |- |

| |4 |2 |4 |2 |5 |9 |1 |

| |5 |2 |5 |1 |7 |1 |- |

| |6 |1 |- |1 |10 |5 |- |

| |7 |3 |- |1 |6 |6 |- |

| |8 |5 |5 |1 |9 |3 |- |

| |9 |3 |7 |- |10 |6 |1 |

| |10 |1 |2 |- |9 |7 |6 |

| |11 |1 |5 |1 |5 |4 |5 |

| |12 |3 |3 |1 |4 |6 |6 |

| |13 |4 |4 |3 |1 |3 |9 |

| |14 |5 |2 |5 |1 |1 |6 |

| |15 |5 |2 |2 |5 |2 |1 |

| |16 |10 |5 |1 |3 |4 |3 |

| |17 |4 |4 |2 |4 |5 |4 |

| |18 |1 |2 |- |2 |1 |2 |

| |19 |1 |4 |- |1 |5 |3 |

| |20 |2 |6 |- |3 |4 |5 |

| |21 |3 |5 |- |2 |3 |4 |

| |22 |2 |3 |1 |4 |4 |- |

| |23 |1 |1 |4 |2 |5 |- |

| |24 |- |1 |3 |2 |3 |- |

|19-02-2003 |1 |3 |1 |3 |2 |1 |3 |

| |2 |2 |2 |2 |5 |- |- |

| |3 |1 |1 |2 |1 |3 |- |

| |4 |2 |2 |1 |1 |- |1 |

| |5 |1 |1 |1 |2 |- |- |

| |6 |1 |1 |1 |1 |1 |- |

| |7 |1 |1 |1 |1 |1 |1 |

| |8 |2 |2 |- |1 |1 |4 |

| |9 |1 |2 |1 |3 |3 |4 |

| |10 |2 |2 |2 |4 |1 |5 |

| |11 |3 |1 |2 |2 |- |3 |

| |12 |1 |- |2 |1 |- |1 |

| |13 |2 |- |1 |1 |- |1 |

| |14 |- |- |1 |- |- |2 |

| |15 |- |- |- |- |- |- |

| |16 |- |- |- |- |- |- |

| |17 |- |- |- |- |- |- |

| |18 |- |- |- |- |- |- |

| |19 |- |- |- |- |- |- |

| |20 |- |- |- |- |- |- |

| |21 |- |- |- |- |- |1 |

|DATE |TIME |STATION & PRECIPITATION |

| |(HOURS) | |

| | |BESHAM |PHULRA |DAGGAR |OGHI |SHINKIARI |TARBELA |

| |22 |- |- |- |1 |- |1 |

| |23 |- |- |- |1 |- |1 |

| |24 |3 |- |- |1 |- |- |

|20-02-2003 |1 |- |- |- |1 |- |- |

| |2 |- |- |- |2 |- |- |

| |3 |1 |1 |0 |- |- |2 |

| |4 |2 |1 |0 |- |5 |- |

| |5 |1 |1 |1 |1 |- |- |

| |6 |1 |1 |0 |1 |- |- |

| |7 |- |1 |0 |1 |- |- |

| |8 |3 |- |2 |1 |1 |- |

| |9 |1 |- |- |1 |- |- |

| |10 |1 |- |- |- |- |- |

| |11 |- |- |1 |- |- |- |

| |12 |- |- |- |- |- |- |

| |13 |- |- |- |2 |- |- |

| |14 |- |1 |- |1 |- |- |

| |15 |- |- |- |- |- |- |

| |16 |- |- |- |- |- |- |

| |17 |- |- |- |- |- |- |

| |18 |- |- |- |- |- |- |

| |19 |- |- |1 |- |- |- |

| |20 |- |- |- |- |- |- |

| |21 |- |- |- |- |- |- |

| |22 |- |- |- |- |- |- |

| |23 |- |- |- |- |- |- |

| |24 |- |- |- |- |- |- |

Solution

Step 1:

First we plot average un-weighted mass curve for total rainfall over the whole catchment. The calculations are shown in table 3.15 and figure 3.11. Total precipitation is found by summing up precipitation recorded at individual stations for each hour of record. Similarly average for precipitation is found by averaging precipitation recorded at six stations. Now the average un-weighted mass curve is plotted by taking time along x-axis and cumulative average precipitation along y-axis. The purpose of this plot is to find maximum precipitation for 6 hours, 12 hours and 24 hours rain over the whole catchment.

The maximum precipitation during selected intervals of 6, 12 and 24 hours can also be found by calculations. For example for 6-hour maximum rain total precipitation is added for every 6 consecutive hours. Once it has been done for all data maximum value is noted along with the period in which it occurred. In this case, this value comes out to be 224 mm. Similarly total precipitation is summed for consecutive 12 and 24 hours. These values come out to be 418 mm and 697 mm. Note that these are rains for all six given sections for selected time interval.

Table 3.15 Depth Area Duration Analysis

|DATE |TIME |TIME |CUMM. TIME |6 STATION PRECIP. (mm) |6-HR |12-HR |24-HR |

| |(HOURS) |INTERVAL | | | | | |

| | | | |TOTAL |AV. |CUMM. AV. |PRECIP. |

| | | | |TOTAL |AV. |CUMM. AV. |PRECIP. |

| | | | |TOTAL |AV. |CUMM. AV. |

| |(mm) |(km²) | |(x106 m³) |(x106 m³) |(mm) |

|Boundary and 25 |25 |310.53 |310.53 |7.76 |7.76 |24.99 |

|25 and 30 |27.5 |2220.71 |2531.24 |61.07 |68.83 |27.2 |

|30 and 35 |32.5 |2968.38 |5499.62 |96.47 |165.3 |30.06 |

|35 and 40 |37.5 |2231.86 |7731.48 |83.69 |248.99 |32.2 |

|40 and 45 |42.5 |2303.52 |10035 |97.9 |346.89 |34.57 |

|45 and 50 |47.5 |2731.9 |12766.9 |129.77 |476.66 |37.34 |

|50 and 55 |52.5 |2689.7 |15456.6 |141.21 |617.87 |39.97 |

|55 and Boundary |55 |1484.99 |16941.6 |81.67 |699.54 |41.3 |

| |Total |16,941.60 | |699.54 | | |

|Average Precipitation over the area = 699.54x1000 ÷ 16941.60 = | |41.29 mm |

Table 3.17 : 12 - Hour Rainfall

|Isohyet Value (mm) |Av. Isohyet |Area Between |Cumulative Area |Volume |Cumulative Volume |Equivalent |

| |Value |Consecutive |(km²) | | |Uniform Depth |

| | |Isohyets | | | | |

| |(mm) |(km²) | |(x106 m³) |(x106 m³) |(mm) |

|Boundary and 45 |25 |675.27 |675.27 |16.88 |16.88 |25.00 |

|45 and 50 |47.5 |698.88 |1,374.15 |33.20 |50.08 |36.44 |

|50 and 55 |52.5 |805.13 |2,179.29 |42.27 |92.35 |42.38 |

|55 and 60 |57.5 |1,360.77 |3,540.06 |78.24 |170.59 |48.19 |

|60 and 65 |62.5 |960.18 |4,500.24 |60.01 |230.60 |51.24 |

|65 and 70 |67.5 |1,396.98 |5,897.21 |94.30 |324.90 |55.09 |

|70 and 75 |72.5 |1,777.90 |7,675.11 |128.90 |453.80 |59.13 |

|75 and 80 |77.5 |1,871.56 |9,546.67 |145.05 |598.84 |62.73 |

|80 and 85 |82.5 |1,863.69 |11,410.36 |153.75 |752.60 |65.96 |

|85and 90 |87.5 |1,677.95 |13,088.30 |146.82 |899.42 |68.72 |

|90 and 95 |92.5 |1,875.49 |14,963.79 |173.48 |1,072.90 |71.70 |

|95 and Boundary |95 |1,977.81 |16,941.60 |187.89 |1,260.79 |74.42 |

| |Total |16,941.60 | |1,260.79 | | |

|Average Precipitation over the area = 1260.79x1000 ÷ 16941.60 = |74.42 mm |

Table 3.18 : 24 - Hour Rainfall

|Isohyet Value |Av. Isohyet|Area Between |Cumulative Area |Volume |Cumulative Volume |Equivalent |

|(mm) |Value |Consecutive |(km²) |(x106 m³) |(x106 m³) |Uniform Depth |

| |(mm) |Isohyets | | | |(mm) |

| | |(km²) | | | | |

|Area enclosed by|145 |138.50 |138.50 |20.08 |20.08 |145.00 |

|145 | | | | | | |

|145 and 140 |142.5 |83.10 |221.59 |11.84 |31.92 |144.06 |

|140 and 135 |137.5 |592.76 |814.36 |81.50 |113.43 |139.29 |

|135 and 130 |132.5 |893.50 |1,707.85 |118.39 |231.82 |135.74 |

|130 and 125 |127.5 |1,164.95 |2,872.80 |148.53 |380.35 |132.40 |

|125 and 120 |122.5 |1,582.02 |4,454.82 |193.80 |574.14 |128.88 |

|120 and 115 |117.5 |3,303.32 |7,758.14 |388.14 |962.28 |124.04 |

|115 and 110 |112.5 |1,205.31 |8,963.45 |135.60 |1,097.88 |122.48 |

|110 and 105 |107.5 |1,179.19 |10,142.64 |126.76 |1,224.65 |120.74 |

|105 and 100 |102.5 |1,620.80 |11,763.44 |166.13 |1,390.78 |118.23 |

|100 and Boundary|100 |5,178.16 |16,941.60 |517.82 |1,908.59 |112.66 |

| |Total |16,941.60 | |1,908.59 | | |

|Average Precipitation over the area = 1908.59x1000 ÷ 16941.60 = |112.66 mm |

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Figure 3.15 Depth-Area-Duration Curves for 6, 12 & 24 hours rainfall

QUESTIONS

1. What is meant by Precipitation?

2. Define the terms ‘Amount of Rainfall’ and ‘Intensity of Rainfall’?

3. What are the types of precipitation based on the mechanisms that produce it?

4. What are different forms of precipitation on the basis of lifting mechanism?

5. How precipitation is measured?

6. Explain how a network of rain-gauge stations is designed?

7. Categorize different instruments used for measurement of precipitation with respect to their merits and limitations?

8. Explain various methods used for analysis of precipitation data. What is meant by reliable data?

9. Describe the ‘Double Mass Curve Technique’ for checking the consistency of precipitation data?

10. How missing precipitation data is interpolated?

11. How precipitation data is checked for consistency and homogeneity?

12. Explain the utility of “Mass Curve” and “Double Mass Curve”?

13. What are Depth-Area-Duration Curves? What important information is extracted from these curves?

14. Explain procedure for Depth-Area-Duration Analysis?

15. Define Probable Maximum Precipitation (PMP) of a catchment?

EXERCISE

Problem 3.1

The names and locations of the rain gauge stations with their mean annual precipitation are as given in the table below. Calculate average precipitation of the catchment area by Isohyetal method.

|Name of Station |Latitude |Longitude |Mean Annual |

| | | |Precipitation (in) |

| |Degrees |Minutes |Degrees |Minutes | |

|A |16 |03 |103 |04 |39.85 |

|B |16 |35 |104 |44 |56.58 |

|C |14 |58 |102 |07 |45.73 |

|D |17 |25 |104 |50 |85.12 |

|E |17 |27 |101 |44 |47.51 |

|F |17 |10 |104 |09 |70.59 |

|G |14 |53 |103 |29 |52.67 |

|H |16 |25 |101 |08 |50.83 |

|I |15 |15 |104 |53 |60.57 |

|J |17 |26 |102 |46 |56.22 |

|K |15 |45 |102 |02 |41.50 |

|L |16 |26 |102 |51 |46.81 |

Latitude and Longitude of the Catchment Area are 140 30’ to 180 30’ and 1010 00’ to 1040 00’, respectively.

Problem 3.2

Assuming the rain falling vertically, express the catch of a gauge inclined 15° from the vertical as a percentage of the catch of a gauge installed vertically.

Problem 3.3

A 3-hour storm occurred at a place and the precipitation in the neighboring rain-gauge stations P, Q and R were measured as 3.8, 4.1 and 4.5 cm respectively. The precipitation in the neighboring station S could not be measured since the rain gauge was damaged. The normal annual precipitation in the four stations P, Q, R and S were 45, 48, 53 and 50 cm, respectively. Estimate the storm precipitation at station ‘S’.

Problem 3.4

In a catchment one precipitation station ‘A’, was inoperative for part of a month during which storm occurred. The respective storm totals at three surrounding stations 1, 2 and 3 are 35, 40 and 30 mm, respectively. The normal annual precipitations at A, 1, 2 & 3 are respectively 985, 1125, 940 and 1210 mm. Estimate the storm precipitation for Station ‘A’.

Problem 3.5

The average annual precipitation for the four sub-basins constituting a large river basin is 58, 67, 85 and 80 cm. The sub-basin areas are 900, 690, 1050 and 1650 km², respectively. What is the average annual precipitation for the basin as a whole?

Problem 3.6

The annual precipitation at station ‘A’ and the average precipitation at

15 surrounding stations are given in the Table 3.19 given below:

a. Determine the consistency of the record at station ‘A’.

b. In which year, there is a regime change indicated?

Table 3.19 Data Table

|Year |Annual Precipitation |Year |Annual Precipitation |

| |Station A (cm) |15 Station Average (cm)| |Station A (cm) |15 Station Average |

| | | | | |(cm) |

|1971 |50.5 |71.5 |1981 |36.0 |27.5 |

|1972 |90.0 |57.0 |1982 |42.0 |60.5 |

|1973 |16.0 |27.5 |1983 |18.0 |55.0 |

|1974 |21.5 |25.0 |1984 |30.0 |38.5 |

|1975 |50.5 |60.0 |1985 |54.0 |38.5 |

|1976 |62.5 |22.0 |1986 |48.0 |47.5 |

|1977 |69.5 |55.0 |1987 |12.0 |49.5 |

|1978 |36.0 |57.0 |1988 |36.0 |24.0 |

|1979 |42.0 |36.5 |1989 |42.0 |44.0 |

|1980 |42.0 |19.5 |1990 |36.0 |60.5 |

Problem 3.7

The Table 3.20 given below gives the annual rainfall at a station ‘A’ and the average annual rainfall of 10 stations in the vicinity for a period of 17 years. It is suspected that there has been a change in the location of the rain gauge at Station ‘A’ during the period of this record. Determine the year when this change occurred and the corrected rain gauge readings prior to this year.

Table 3.20 Rainfall Data

|Year |Rainfall |Year |Rainfall |

| |Station A (mm) |Average Annual | |Station A (mm) |Average Annual |

| | |Rainfall of 10 | | |Rainfall of 10 |

| | |Stations (mm) | | |Stations (mm) |

|1931 |505 |615 |1941 |420 |605 |

|1932 |900 |570 |1942 |180 |550 |

|1933 |60 |275 |1943 |300 |385 |

|1934 |215 |250 |1944 |540 |385 |

|1935 |505 |605 |1945 |480 |475 |

|1936 |625 |220 |1946 |120 |495 |

|1937 |695 |550 |1947 |360 |240 |

|1938 |360 |570 | | | |

|1939 |420 |365 | | | |

|1940 |360 |275 | | | |

Problem 3.8

Point rainfalls due to a storm at several rain-gauge stations in a basin are shown in Fig. 3.16. Determine the mean areal depth of rainfall over the basin.

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Figure 3.16 Isohyetal Map

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