CHAPTER 2

PRECIPITATION

Learning Objectives

This chapter is designed to assist the students to develop and enhance their ability and knowledge in:

1. defining precipitation, its forms and types 2. illustrating techniques for estimating point and areal precipitation

Learning Outcome

At the end of this chapter, students should be able to:

1. estimate point and areal precipitation amounts from gauge data

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2.1 Introduction Precipitation is a major component of the hydrologic cycle. Precipitation that reaches the surface of the earth can occur in many different forms including rain, storm, snow, hail, drizzle and sleet. Precipitation can be defined as any product of the condensation of atmospheric water vapor (solid or liquid) that is deposited on the earths surface, its form and quantity thus being influenced by the action of other climatic factors such as wind, temperature and atmospheric pressure. 2.2 Formation of Precipitation

Figure 2.1: Formation of precipitation

The formation of precipitation in clouds is illustrated in Figure 2.1. As air rises and cools, water condenses from the vapor to the liquid state. Water

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droplets are formed by nucleation prosess which is condensing of vapor on

tiny solid particles. The diameters of these particles range from 0.001 m to 10

m and the particles are known as aerosols. The tiny droplets grow by condensation and impact with their neighbors as they are carried by turbulent air motion, until they become large enough so that the force of gravity overcomes that of friction and they begin to fall, further increasing in size as they hit other droplets in the fall path. However, as the drop falls, water evaporates from its surface and the drop size diminishes, so the drop may be reduced to the size of an aerosol again and be carried upwards in the cloud through turbulent action. An upward current of only 0.5 cm/s is sufficient to

carry a 10 m droplet. The cycle of condensation, falling, evaporation and rising occurs before the drop reaches a critical size of about 0.1 mm. Up to about 1 mm in diameter, the droplets remain spherical in shape, but beyond this size they begin to flatten out on the bottom until there are no longer stable falling through air and break up into small raindrops and droplets. Normal raindrops falling through the cloud base are 0.1 to 3 mm in diameter.

2.3 Classification of Precipitation

Types of precipitation include hail, sleet, snow, rain, and drizzle. Frost and dew are not classified as precipitation because they form directly on solid surfaces. The general classes of precipitation are as follows: (1) Snow Complex ice crystals. A snowflake consists of

agglomerated ice crystals. The average water content of snow is assumed to be about 10% of an equal volume of water.

(2) Hail Balls of ice that are about 5 to over 125 mm in diameter. Their specific gravity is about 0.7 to 0.9. Thus, hailstones have the potential for agricultural and other property damage. (3) Sleet Results from the freezing of raindrops and is usually a combination of snow and rain. (4) Rain Consists of liquid water drops of a size 0.5 mm to about 7

mm in diameter. (5) Drizzle Very small, numerous and uniformly dispersed water

drops that appear to float while following air currents. Drizzle drops are considered to be less than 0.5 millimeter diameter. The settling velocity is slow, with the intensity rarely exceeding 1 mm / hr (0.04 in./hr). It is also known as warm precipitation.

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2.4 Precipitation Types

Precipitation may be classified according to the conditions that generate vertical air motion. Three major catogories of precipitation are convective, orographic and cyclonic:

Figure 2.2: Precipitation lifting mechanisms

(1) Convective precipitation Convective precipitation is typical of the

tropics such as in South East Asia. It is brought about by heating of the air at the interface with the ground. This heated air expands with a resultant reduction in weight. During this period, increasing quantities of water vapor are taken up, the warm moisture laden air becomes unstable and pronounced vertical currents are developed. Dynamic cooling takes place, causing condensation and precipitation. Convective precipitation maybe in the form of light showers or storms of extremely high intensity (thunderstorms). (2) Orographic precipitation Orographic precipitation results from the

mechanical lifting of moist horizontal air currents over natural barriers such as mountain ranges. This type of precipitation is very common on the West Coast of the United States where moisture laden air from the Pacific Ocean is intercepted by coastal hills and mountains. (3) Cyclonic precipitation - Cyclonic precipitation is associated with the movement of air masses from high pressure regions to low pressure regions. These pressure differences are created by the unequal heating of the earths surface. This precipitation may be classified as frontal or nonfrontal. Any barometric low can produce nonfrontal precipitation as air is lifted through

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horizontal convergence of the inflow into a low pressure area. Frontal precipitation results from the lifting of warm air over cold air at the contact zone between air masses having different characteristics. 2.5 Measurement of Precipitation Precipitation is measured as the vertical depth that would accumulate on a flat level surface if all the precipitation remained where it had fallen. To size water transport and storage systems, quantitative data for rainfall events must be provided. These data can be defined in terms of:

(1) Depth (d), is the sum of rainfall, which is mentioned as depth of water on the flat surface, (2) Intensity (i), or depth of rainfall per unit time, is commonly reported in the units of millimeters per hours. Weather stations utilizing gages that provide continuous records of rainfall can be used to obtain intensity data. These data are typically reported in either tabular form or graphical form. (3) Duration (t) is the duration of a storm is the time from the beginning of rainfall to the point where the mass curve becomes horizontal indicating no further accumulation of precipitation within a certain time after the rain stops.

(4) Frequency, is the frequency of rain, it is usually called as return period (T), for example once in T years, (5) Area (A), is the area of rainfall geographic

Point rainfall can be plotted as accumulated total rainfall or as rainfall intensity at a particular gauge. The first plot is referred to as a cumulative mass curve, which can be analysed for a variety of storms to determine the

frequency and character of rainfall at a given site. A hyetograph is a plot of rainfall intensity (in/hr) versus time. Example 2.1

From the precipitation data given, estimate cumulative rainfall and rainfall intensity.

Time (min)

0 10 20 30 40 50 60 70 80 90

Rainfall (cm)

0 0.18 0.21 0.26 0.32 0.37 0.43 0.64 1.14 3.18

Tme (min)

100 110 120 130 140 150 160 170 170

Rainfall 1.65 0.81 0.52 0.42 0.36 0.28 0.24 0.19 0.17

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(cm)

Solution:

Time (min) Rainfall (cm) Cumulative rainfall (cm)

Rainfall intensity (cm/hour)

0 0 0

10 0.18 0.18 1.08

20 0.21 0.39 1.26

30 0.26 0.65 1.56

40 0.32 0.97 1.92

50 0.37 1.34 2.22

60 0.43 1.77 2.58

70 0.64 2.41 3.84

80 1.14 3.55 6.84

90 3.18 6.73 19.08

100 1.65 8.38 9.90

110 0.81 9.19 4.86

120 0.52 9.71 3.12

130 0.42 10.13 2.52

140 0.36 10.49 2.16

150 0.28 10.77 1.68

160 0.24 11.01 1.44

170 0.19 11.2 1.14

180 0.17 11.37 1.02

Cumulative rainfall = 0.18 + 0.21 = 0.39 cm. Cumulative rainfall is a plot of cumulative rainfall versus time (min) while rainfall intensity (cm/hr) data are typically reported in either tabular form or graphical form (hyetograph).

Time interval, t = 10 minutes = 0.167 hour Rainfall intensity = 0.18 cm / 0.167 hour = 1.08 cm/hour

2.5.1 Types of Rain Gauges Rain gauge is an instrument used to measure how much rain has fallen. There are several different types of rain gauges that are grouped by how they operate: recording rain gauge, non recording rain gauge and rain-intensity gauge.

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(1) Non recording gauges

For non recording gauge, the standard rain gauge shown in Figure 2.3 is a standard 8-inch-diameter rain gauge (203 mm) (U.S. Weather Service). A smaller metal tube may be located in this larger overflow can. An 8-inch-diameter receiver cap may be on top of the overflow can and is used to funnel the rain into the smaller tube until it overflows. The receiver cap has a knife edge to catch rain falling precisely in the surface area of an 8-inch-diameter opening. Measurements are made using a special measuring stick with graduations devised to account for the 8-inch receiver cap opening, funneling water into the smaller tube. When the volume of the smaller tube is exceeded, the volume from the smaller tube is dumped into the larger overflow can.

Figure 2.3: Nonrecording gauge, 8-inch-diameter opening

(2) Recording gauges (Pluviograph) In contrast to the non recording gauge which requires an observer to manually measure the rain at regular intervals (i.e. every 24 hours), recording gauges does not require constant observation. There are at least three types of gauges commonly in use to record depth: i) Weighing gauge ii) Tipping bucket

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iii) Float type i) Weighing gauge

Weighing-type precipitation gauge consists of a storage bin, which is weighed to record the mass. Certain models measure the mass using a pen on a rotating drum, or by using a vibrating wire attached to a data logger. The advantages of this type of gauge to tipping buckets is that it does not underestimate intense rain, and it can measure other forms of precipitation, including rain, hail and snow. However, these gauges are more expensive and require more maintenance than tipping bucket gauges.

The weighing-type recording gauge also contains a device to measure

the quantity of chemicals contained in the locations atmosphere. This is extremely helpful for scientists studying the effects of greenhouse gases released into the atmosphere and their effects on the levels of the acid rain. ii) Tipping bucket

The tipping bucket rain gauge consists of a large copper cylinder set into the ground. At the top of the cylinder is a funnel that collects and channels the precipitation. The precipitation falls onto one of two small buckets or levers which are balanced in same manner as a scale. After an amount of precipitation equal to 0.2 mm (0.007 in) falls the lever tips and an electrical signal is sent to the recorder. The recorder consists of a pen mounted on an arm attached to a geared wheel that moves once with each signal sent from the collector. When the wheel turns the pen arm moves either up or down leaving a trace on the graph and at the same time making a loud click. Each jump of the arm is sometimes referred to as a 'click' in reference to the noise. The chart is measured in 10 minute periods (vertical lines) and 0.4 mm (0.015 in) (horizontal lines) and rotates once every 24 hours and is powered by a clockwork motor that must be manually wound.

The tipping bucket rain gauge is not as accurate as the standard rain gauge because the rainfall may stop before the lever has tipped. When the next period of rain begins it may take no more than one or two drops to tip the lever. This would then indicate that 0.2 mm (0.007 in) has fallen when in fact only a minute amount has. The advantage of the tipping bucket rain gauge is that the character of the rain (light, medium or heavy) may be easily obtained.

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Figure 2.4: Recording tipping bucket gauge

iii) Float type In this type of instrument, the rain passes into a float chamber containing a light float. As the level of the water within the chamber rises, the vertical movement of the float is transmitted, by a suitable mechanism, to the movement of a pen on a chart or a digital transducer. By suitably adjusting the dimensions of the collector orifice, the float, and the float chamber, any desired chart scale can be used. 2.6 Location of Installation Rain Gauge There are a variety of factors that can affect rain gage measurements. The positioning of the gage is very important in order to reduce errors in collecting Buildings, landscaping and trees, and even the wind can impact the amount of precipitation reaching the rain detector. Proper placement is critical to ensure that rain sensor readings are an accurate representation of the actual rain measurement rates and amounts that have fallen. The ideal site for a rain gage is in an open area that is protected from the wind in all directions. Therefore, rain gages should be sited in an open area away from the external factors mentioned above. A good guideline to follow as a minimum distance from these objects is twice their height. 2.7 Missing Data

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Precipitation measuring stations sometimes fail in providing a continous record of precipitation. Instruments do malfunction and back-up systems may not always provide accurate data. A tipping bucket may not function for a short period of time and the back-up volume gage may not provide time related data. For a nonautomatic recording gage, an individual may fail to record the data and miss a visit to the site. Thus, there are generally missing data, the values of which must be estimated.

The procedure for estimating daily totals relies on the data from adjacent stations. The locations of the adjacent stations are such that they are close to and approximately evenly spaced around the site with the missing data. 2.7.1 Point Precipitation Precipitation events are recorded by gauges at specific locations. Precipitation measured at a rain gauge is called point rainfall.

2.7.1.1 Arithmatic Mean Method If the average annual precipitation at each of the stations differs from the average at the missing data station by + 10%, the following formula is used to estimate the missing daily data:

PX = 1/M (P i)

or; PX = 1/M [ P1+P2+P3+..........+PM] where

PX = estimated daily precipitation volume at the missing data site, X (depth)

P1, P2, P3, ..PM = estimated daily precipitation volume at the adjacent stations, 1, 2, 3 M (depth)

Example 2.2: Rain gauge X was out of operation for a month during which there was a storm. The rainfall amounts at three adjacent stations A, B, and C were 37, 42, dan 49 mm, respectively. The average annual precipitation amounts for the gauges are X = 694, A = 726, B = 752 and C = 760 mm. Using the arithmatic method, estimate the amount of rainfall for gauge X.

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Stations Amounts of precipitation

(mm)

Normal annual precipitation

(mm)

A 37 726

B 42 752

C 49 760

X ? 694

Solution: If NX = 694 10% from NX = 69.4 Precipitation allowed = 624.6 mm ~ 763.4 mm Since all annual precipitations (726, 752 and 760) are within the ranges, arithmetic method can be applied: PX = 1/3{37+ 42+ 49} = 42.7 mm

2.7.1.2 Normal Ratio Method If the difference between the average annual precipitation at any of the adjacent stations and the missing data station is greater than 10%, a normal ratio is used.

PX = NX/M (Pi/Ni ) or PX = NX/M [ P1/N1 +P2/N2 + P3/N3 +..........+ PM/NM ] where NX = average annual precipitation at the missing data site, X (cm) Ni = average annual precipitation at the adjacent sites

(cm)

Example 2.3: The average annual precipitation amounts for the gauges A, B, C and D are 1120, 935, 1200 and 978 mm. In year 1975, station D was out of operation.

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Stations A, B and C recorded rainfall amounts of 107, 89 and 122 mm, respectively. Estimate the amount of precipitation for station D in year 1975.

Station Normal annual

precipitation

Amounts of precipitation year 1975

A 1120 107

B 935 89

C 1200 122

D 978 X

Solution If NX = 978 10% from NX = 97.8 Maximum precipitation allowed = 880.2 mm ~ 1075.8 mm Since the average annual precipitation amounts for the gauges A and C exceeded 1075.8 mm, normal ratio method is used; PX = 978 107 + 89 + 122 3 1120 935 1200 PX = 95.3 mm 2.7.1.3 Quadrant Method

This is station weighting technique. A grid of point estimates is made based on a distance weighting scheme. Each observed point value is given a unique weight for each grid point based on the distance from the grid point in question. The grid point precipitation value is calculated based on the sum of the individual station weight multiplied by observed station value. Once the grid points have all been estimated they are summed and the sum is divided by the number of grid points to obtain the areal average precipitation.

The U.S.A. National Weather Service has developed a procedure for this method. a) Consider that rainfall is to be calculated for point X. Establish a set of axes running through X and determine the absolute coordinates of the nearest surrounding points P, Q, R, S, T and U.

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b) The estimated precipitation at X is determined as a weighted average of the

other six points. The weights are reciprocals of the sums of the squares of X

and Y; that is,

Figure 2.5: Four quadrants sorrounding precipitation station X

L2 = X

2 + Y

2 and W = 1/ L

2

d) The estimated rainfall at the point of interest is given by;

PX = (P x W)

W Example 2.4:

Stations A, B, C, D and E are the gauge stations. Rain gauge at station A was out of operation. Precipitation amounts for other stations were 40, 45, 37.5, 50 and 42.5 mm. Calculate the rainfall depth at station A with coordinates (0,0) using the quadrant method.

Station Precipitation, P (mm)

X (cm)

Y (cm)

L2 W x 10

3 P x (W x 10

3)

A PA 0 0 0 0 0

B 40 4 2 20 50 2000

C 45 1 6 37 27 1215

D 37.5 3 2 13 76.9 2883.75

E 50 3 3 18 55.6 2780.0

= 209.5 8,878.75

PA = 8878.75 209.5

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PA = 42.4 mm

2.8 Gage Consistency

Estimating missing data is one problem that hydrologists need to address. A second problem occurs when the catch at rain gauges is inconsistent over a period of time and adjustment of the measured data is necessary to provide a consistent record.

Double Mass Curve analysis is a technique commonly employed to

detect changes in data-collection procedures or conditions at a given location. The changes may result from changes in instrumentation, changes in observation procedures or changes in gauge location or surrounding conditions. A double mass curve is a plot of the accumulation of the observed

element over time for one location (test station) versus the accumulation over

time for a reference location (base station). The mass curve is approximately a straight line if the variations at both test and base stations are quite consistent. Any break point in the curve suggests a possible change at the test station in relation to the base station. If a change in slope is evident, then the record needs to be adjusted, with either the early or later period of record adjusted.

The first step is to form the double mass curve and compute the slopes are as follows:

i) Add the annual precipitation of base stations. ii) Cummulate the sums of Step 1. iii) Cummulate the annual precipitation for station X. iv) Plot graph accumulated annual precipitation Station X versus

accumulated precipitation of Base stations and compute the

slope Mo and Ma.

Mo = Po

P and

Ma = Pa

P

v) Adjust the measured precipitation of gauge X using the general equation:

Pa = Po Ma

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Mo where Pa = adjusted precipitation value at station X Po = original precipitation value at Station X Ma = adjusted slope Mo = original slope

EXAMPLE 2.5 Measured annual precipitation gauge for five stations (A, B, C, D and E) from 1926 until 1942 are given below. After 5 years, gauge A was relocated at a new location due to changes in land use that make it impractical to maintain the gauge at the old location. You are required to adjust the record for the period from 1926 to 1930 using the records at gauges B, C, D and E.

YEAR Annual precipitation (mm) Total Cummulative

precipitation

(mm)

A B C D E B+C+D+E A B+C+D+E

P

Pox

Pax Ma

Mo

Accum

ula

ted

tota

l p

recip

itati

on a

t S

tati

on

X

Accumulated precipitation of base stations

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1926 32.9 39.8 45.7 30.7 37.4 153.6 33 154

1927 28.1 39.6 38.5 41.0 30.9 150 61 304

1928 33.5 42.0 48.3 40.4 42.0 172.7 95 476

1929 29.6 41.4 34.6 32.5 39.9 148.4 124 625

1930 23.8 31.6 45.1 36.7 36.3 149.7 148 774

1931 58.4 56.5 53.3 62.4 36.6 208.8 206 983

1932 46.3 48.1 40.1 47.9 38.6 174.7 253 1158

1933 30.8 39.9 29.6 32.7 26.9 129.1 283 1287

1934 46.8 45.4 41.7 36.1 32.4 155.6 330 1443

1935 38.1 44.9 48.1 30.7 41.6 165.3 368 1608

1936 40.8 32.6 39.5 35.4 31.3 138.8 411 1747

1937 37.9 45.9 44.1 39.2 44.1 173.3 449 1920

1938 50.7 46.1 38.9 43.3 50.6 178.9 499 2099

1939 46.9 49.8 41.6 49.9 41.1 182.4 546 2281

1940 50.5 47.3 49.7 47.9 39.0 183.9 597 2465

1941 34.4 37.1 31.9 32.2 34.5 135.7 631 2601

1942 47.6 45.9 38.2 52.4 47.3 183.8 679 2785

P nb = P x Ma Mo i) P1926 = 32.9 0.19 0.26 P1926 = 32.9 (0.73) P1926 = 24 ii) P1927 = 28.1 (0.73) P1927 = 20.5

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2.9 Mean Areal Precipitation The representative precipitation over a defined area is required in engineering application, whereas the gaged observation pertains to the point precipitation. The areal precipitation is computed from the record of a group of rain gages within the area by the following methods:

2.9.1 Arithmatic - mean Method

The arithmetic average method uses only those gaging stations within the topographic basin and is calculated using: P = P1 + P2 + P3 + ....... + Pn n

P = Pi n di mana, P = average precipitation depth (mm)

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Pi = precipitation depth at gage (i) within the topographic basin (mm) n = total number of gaging stations within the topographic basin

2.9.2 Thiessen Polygon Method Another method for calculating average precipitation is the Thiessen method. This technique has the advantage of being quick to apply for multiple storms because it uses fixed sub-areas. It is based on the hypothesis that, for every point in the area, the best estimate of rainfall is the measurement physically closest to that point. This concept is implemented by drawing perpendicular bisectors to straight lines connecting each two raingages. This procedure is not suitable for mountainous areas because of orographic influences. The procedure involves: i) Connecting each precipitation station with straight lines; ii) Constructing perpendicular bisectors of the connecting lines and forming polygons with these bisectors; iii) The area of the polygon is determined.

Average precipitation = Polygon area for each station x precipitation

Total polygon area

=++++

++++=

=

n

i

ii

n

nn

A

pA

AAAA

pApApApAP

1321

332211

......

...... (17)

If Ai/A = wi, then wi is the percentage of area at station 1 in which the sum of total area is 100%.

==1i

ii pwP (18)

Where: A = total area p = average precipitation depth

p1, p2, pn = depth of precipitation at rainfall station A1, A2, An = sub area at station 1,2,3, .n

St2

St3

Catchment boundary

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St1

St4

St5 St6

St1

A1

A2

A3

A4

A5 A6

St2

St3

St4

St5 St6

Catchment boundary

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Example 2.6 Using data given below, estimate the average precipitation using Thiessen method.

Station Area (km2) Precipitation

(mm) Area x precipitation

(km2.mm)

A 72 90 6480

B 34 110 3740

C 76 105 7980

D 40 150 6000

E 76 160 12160

F 92 140 12880

G 46 130 5980

H 40 135 5400

I 86 95 8170

J 6 70 420

Total 568 1185 69210

Average precipitation = Area x precipitation

Area Average precipitation = 69210 568 Average precipitation = 121.8 mm 2.9.3 Isohyetal Method

The isohyetal method is based on interpolation between gauges. It

closely resembles the calculation of contours in surveying and mapping. The first step in developing an isohyetal map is to plot the rain gauge locations on a suitable map and to record the rainfall amounts. Next, an interpolation between gauges is performed and rainfall amounts at selected increments are plotted. Identical depths from each interpolation are then connected to form isohyets (lines of equal rainall depth). The areal average is the weighted average of depths between isohyets, that is, the mean value between the

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isohyets. The isohyetal method is the most accurate approach for determining average precipitation over an area.

A

pp

A

App

AAA

App

App

App

P

n

i

ii

n

ii

i

n

i

ii

n

nnn

+

=

+

=+++

+++

++

+

= =

=

=

1

1

1

1

1

21

12

211

1022

........2

........22

where: P = mean areal precipitation A = Area

p1, p2, pn = precipitation depth for each station A1, A2, An = area for each site

Example 2.7

Use the isohyetal method to determine the average precipitation depth within the basin for the storm.

10mm

20mm

36mm

45mm

57mm

42mm

51mm

p0=10mm

p1=20mm p2=30mm

p3=40mm

p4=50mm p5=60mm

A1 A2

A3

A4 A5

A6

P6=70mm

70mm

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Isohyetal interval

Average precipitation

(cm)

Area (km 2) Area x

Average precipitation

(km2.cm)

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Station Precipitation (mm)

Polygon Thiessen area (cm

2)

A 20 112.25

B 13 53.5

C 18.3 120.0

D 12.5 62.5

E 10 119.0

F 5.8 144.0

G 6.7 72.0

H 14.8 130.0

I 13.9 62.5

J 11 85.0

K 5.5 110.0

L 3.7 40.0

Answer: 11.6 mm

Q6. Lines delineating a particular measurement are drawn within the watershed. Precipitation data and the area in between each isohyet are shown in the table below:

Isohyetal line (cm)

30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90-100

Area (cm2) 32 162 155 92 228 120 65

Estimate the average precipitation within this watershed. Answer: 66.03 cm Q7. Rain gauge X was out of operation for a month during which there was a storm. The rainfall amounts at three adjacent stations A, B, and C were 10.3 cm, 9.5 cm dan 13.0 cm, respectively. The average annual precipitation amounts for the gauges are X = 96.3, A = 111.3, B = 93.0 and C = 106.4 cm. Estimate the amount of rainfall for gauge X. Answer: 10.2 cm Q8. Estimate the amount of precipitation for gauge X in Problem 4 if three adjacent stations recorded precipitation amounts were A = 42.4, B = 35.3 and C = 38.4 cm. Answer: 36 cm Q9. Estimate the precipitation depth at station X with coordinates (0,0) using

four quadrant method.

Quadrant Rain gauge

Precipitation depth (cm)

Coordinat X Y

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I A B C

10 3.5 9.5

3 18 6

9 26 4

II D E

4.0 0.5

11 14

-8 -26

III F G

2.3 7.6

-4 -10

-22 -5

IV H 2.3 -21 19

Answer: 7.9 cm

Q10. A watershed has a system of three rainfall gauges as shown on the map below. The total storm rainfall depths is: A = 74 mm, B = 67 mm and C = 82 mm. Determine the spatial average rainfall for the network of rainfall gauges using Arithmetic Average and Thiessen Polygon methods.

Answer: 74 ~ 77 cm

1.0 km A

C

B

1.0 km

Figure 1.0

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Q11. Gauge X was installed in January 1975 and removed from its original location in January 1962. Adjust the record as in Table 5 for the period from 1958 to 1962 using the records at gauges P, Q and R.

Answer: Mo = 0.3, Ma = 0.37

Table 5: Annual rainfall for each station

Year

Annual rainfall (cm)

P Q R X

1958 54 50 56 50 1959 60 60 66 58 1960 64 58 70 60 1961 68 66 74 62 1962 58 58 60 52 1963 56 52 54 60 1964 64 68 68 72 1965 70 68 72 76 1966 62 58 68 72 1967 56 54 58 62 1968 50 44 50 56 1969 56 50 58 60 1970 66 60 68 74

Q12. Gauge X was temporarily moved in January 1975 and will be returned to its original location in January 1979. Adjust the record for the period from 1975 to 1978 using the records at gauges D, E, F, F and G.

Answer: Mo = 0.27, Ma = 0.22

YEAR

Annual rainfall (cm)

D E F G X

1969 21 19 20 23 17 1970 27 24 27 29 24 1971 29 28 27 29 25 1972 25 23 23 26 22 1973 19 22 17 23 16 1974 21 20 18 22 20 1975 23 20 22 25 24

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1976 17 16 18 20 20 1977 18 16 18 20 22 1978 22 19 20 25 25

Summary

Precipitation input is the main driver of the hydrologic cycle, as it relates to river flow, water supply and urban drainage. Too much or too little can mean the difference between prosperity and disaster. In between these extremes are the normal precipitation event that are experienced with a frequency and intensity related mainly to geographic position and topographic features.

At the end of this chapter you should be able to estimate point and areal precipitation amounts from gauge data and conceptualize simple hydrologic process models. References

Bedient, P. B., Huber, W. C. and Vieux, B. E. (2003) Hydrology and

Floodplain Analysis, 4th ed., Prentice Hall.

Viessman, W., and G. L. Lewis. (2003). Introduction to Hydrology, 5th

ed., Prentice Hall.