Precipitation Lecture Notes PDF

Summary

These lecture notes cover different types of precipitation, mechanisms of precipitation occurrence, and methods for measuring precipitation. They also discuss the role of various atmospheric factors, such as temperature, moisture content, and pressure.

Full Transcript

Prepared by: ENGR. REYMORE A. INSAS By the end of this lecture, the student must be able to:  define what precipitation is and its different forms;  explain how does precipitation occur;  enumerate the different instruments and methods on how to measure precipitation;  solve problems involv...

Prepared by: ENGR. REYMORE A. INSAS By the end of this lecture, the student must be able to:  define what precipitation is and its different forms;  explain how does precipitation occur;  enumerate the different instruments and methods on how to measure precipitation;  solve problems involving the precipitation process.  All forms of water that reaches the earth surface from the atmosphere.  Moisture is always present in the atmosphere, even on cloudless days.  For precipitation to occur, some mechanism is required to cool the air sufficiently to bring it to near saturation. The large-scale cooling needed for significant amounts of precipitation is achieved by lifting the air. This is accomplished by convective systems resulting from unequal radiative heating or cooling of the earth’s surface and atmosphere or by convergence caused by orographic barriers. the atmosphere must have moisture; there must be sufficient nuclei to aid condensation; weather conditions must be good for condensation of water vapor to take place; and the products of condensation must reach the earth. Rain Drizzle Glaze Sleet Hail Snow 1. Convective Precipitation In meteorology, convection refers primarily to atmospheric motions in the vertical direction. In the atmosphere, convection enables winds to be maintained by an upward and downward transfer of air masses of different temperatures. Convective storms result as warm, humid air rises into cooler overlying air. 1. Convective Precipitation A common form of convective precipitation is the summer thunderstorm.  The earth’s surface is warmed by mid to late afternoon on a hot summer day.  The surface imparts heat to the air mass directly above.  The warmed air rises through the overlying air, and if the air mass has a moisture content equal to the condensation level, moisture will be condensed from the rising, rapidly cooling air.  This may often result in a large volume of rain from a single thunderstorm. 1. Convective Precipitation Rain Clouds Rain Clouds Condensation and subsequent Convective thunderstorm formation Precipitation Air is heated and rises 2. Orographic Precipitation Orography is the study and mapping of variations in the earth’s surface between highlands (mountains) to other land and water features. Orographic precipitation results as warmer air rises over high geographic feature such as a range of mountains and meets cooler air. Precipitation results if the rising air mass has a condensation level of moisture. Consequently, mountain slopes facing prevailing winds get more precipitation than the back or leeward slopes. 2. Orographic Precipitation Condensation Leeward Side Condensation Level Mountain Orographic Precipitation Windward Side 3. Cyclonic Precipitation Cyclonic precipitations are caused by the rising or lifting of air as it converges on an area of low pressure. The movement of air is from high to low pressure areas and the boundary between air masses of different pressure is called a front. Frontal precipitation is formed from the lifting of warm air over cold air. Cold fronts are formed by cold air advancing under warmer air; a warm front is formed by warm air advancing over colder air. The intensity of precipitation associated with a cold front is usually heavy and covers a relatively small area, whereas less intense precipitation is associated with a warm front, but it covers a much larger area. Tornadoes/typhoons and other violent weather phenomena are associated with cold fronts. 3. Cyclonic Precipitation Large regions in the atmosphere that have higher pressure than the surroundings are called high- pressure areas. Regions with lower pressure than the surroundings are called low-pressure areas. Most storms occur in low-pressure areas. Rapidly falling pressure usually means a storm is approaching, whereas rapidly rising pressure usually indicates that skies will clear. 3. Cyclonic Precipitation Cold Front Warm Front Thunderstorm Day-Long Drizzle is is common common Air forced to rise abruptly Cold Air Warm Cold Air Warm Air Air Heavy-short- Light-Long duration duration Precipitation Precipitation  Clouds are created when moist air rises and cools, and water condenses around dust particles to form tiny water droplets or ice crystals.  The ten main types of clouds are classified on the basis of their shape and the height in the atmosphere at which they form.  The types of clouds provide clues to atmospheric conditions.  Classification identifies clouds by height of cloud base. For example, cloud names containing the prefix “cirr-”, as in cirrus clouds, are located at high levels while cloud names with the prefix “alto-”, as in altostratus, are found at middle levels. 1. High-Level Clouds Form above 6 km and since the temperatures are so cold at such high elevations, these clouds are primarily composed of ice crystals. High-level clouds are typically thin and white in appearance, but can appear in a magnificent array of colors when the sun is low on the horizon. 2. Mid-Level Clouds The bases of mid-level clouds typically appear between 2 to 6 km. Because of their lower altitudes, they are composed primarily of water droplets, however, they can also be composed of ice crystals when temperatures are cold enough. 3. Low-Level Clouds Are mostly composed of water droplets since their bases generally lie below 2 km. However, when temperatures are cold enough, these clouds may also contain ice particles and snow. Cumuloni Cumulus Stratus mbus Stratocumulus Altocumulus Altostratus Nimbostratus Cirrus Cirrocumulus Cirrostratus  Precipitation is expressed in terms of the depth to which rainfall water would stand on an area if all the rain were collected on it.  Apparatus/Device for measuring rain 1. Rain Gages – used to measure rain that falls in a certain place during a specific period of time. Rain gage is shaped like a cylinder and has a removable cover. a. The ground must be level and in the open and the instrument must present a horizontal catch surface. b. The rain gage must be set as near the ground as possible to reduce wind effects but it must be sufficiently high to prevent splashing and flooding. c. The instrument must be surrounded by an open fenced area of at least 5.5 m by 5.5 m. No object should be nearer to the instrument than 30 m or twice the height of the obstruction. a. Non-recording gauges I. Standard gauge  The parts includes the support, overflow can, measuring tube, and the collector or receiver.  Inside the cylinder is a long narrow tube, where the rain is measured.  The top of the tube is connected with a funnel. The rain falls into the funnel and flows into the tube.  The collector or receiver has a diameter of 8 inches (203 mm).  The mouth of the funnel has an area 10 times that of the tube.  The rain in the tube is measured by a “ruler”. With this ruler, a depth of 10 inches (25 cm) gives a reading of 1 inch (25 mm) of rainfall. a. Non-recording gauges II. Symon’s gauge  Consists of a circular collecting area of 12.7 cm (5 inches) diameter connected to a funnel.  The rim of the collector is set in a horizontal plane at a height of 30.5 cm above the ground level. b. Recording rain Gages I. Tipping bucket type  This is a 30.5 cm size rain gage.  The catch from the funnel falls onto one pair of small buckets.  The buckets are so balanced that when 0.25 mm, 0.1 mm of rainfall collects in one bucket, it tips and brings the other one in position.  The water from the tipped bucket is collected in a storage can.  The tipping actuates an electrically driven pen to trace a record on clock-work-driven chart. b. Recording rain Gages II. Weighing bucket type  The rain gage catch from the funnel empties into a bucket mounted on a weighing scale.  The weight of the bucket and its contents are recorded on a clock – work – driven chart.  The clock work mechanism has the capacity to run for as long as one week.  This instrument gives a plot of the accumulated rainfall against the elapsed time. b. Recording rain Gages III. Natural-Siphon type (Float type gage)  The rainfall collected by a funnel-shaped collector is led into a float chamber (special reservoir of mercury or oil) causing a float to rise.  As the float rises, a pen attached to the float through a lever system records the elevation of the float on a rotating drum driven by a clock- work mechanism.  A siphon arrangement empties the float chamber when the float has reached a pre-set maximum level.  Radar is an electronic instrument that sends out radio waves that are reflected by raindrops.  The reflected waves, called “echoes”, appear on a screen as spots of light.  The brightness of the echoes depends chiefly on the sized and number of raindrops.  The echoes indicate the amount and intensity of rainfall.  Radar measures scattered showers missed by rain gauges, which are too far apart to measure precipitation in all places.  Satellite observations can provide information on the areal distribution of precipitation working on the principle that the atmosphere selectively transmit radiation at various wavelengths, more particularly in the visible and thermal infra-red wavelengths.  The visible wavelengths are the order of 0.77 to 0.91 μm and give information on the distribution of clouds, and therefore possible areal location of rainfall.  The infra-red wavelengths, 8 to 9.2 μm and 17 to 22 μm, can be used to locate high clouds and their associate convective precipitation cells.  Satellite is commonly used in areas of low inhabitation, where rain gauges or radar are not available, and particularly remote island locations.  If there are already some rain gage stations in a catchment, the optimal number of stations that should exist to have an assigned percentage of error in the estimation of the mean rainfall is obtained by the statistical analysis as; 𝟐 𝐂𝐯 𝟏𝟎𝟎𝐒 𝐍= 𝐂𝐯 = 𝛆 ഥ 𝐏 where: N = optimal number of stations e = allowable degree of error in percent Cv = Coefficient of variation in percent S = Standard deviation 𝟏 σ𝐌 ഥ 𝐢=𝟏 𝐏𝐢 − 𝐏 𝟐 𝟐 𝐒= 𝐌−𝟏 σ𝐌 𝐢=𝟏 𝐏𝐢 ഥ= 𝐏 = 𝐦𝐞𝐚𝐧 𝐩𝐫𝐞𝐜𝐢𝐩𝐢𝐭𝐚𝐭𝐢𝐨𝐧 𝐌  A catchment has six rain gage stations. In a month, the monthly rainfalls recorded by the rain gauges are as follows (given table below). For a 10% error in the estimation of the mean rainfall, calculate the optimum number of stations in the catchment. Given: M = 6 rain gauge ε = 10% Required: N = optimum number of stations in the catchment Solution: a. Solve for the mean precipitation, 82.6 + 102.9 + 180.3 + 110.3 + 98.8 + 136.7 ഥ P= = 𝟏𝟏𝟖. 𝟔𝐦𝐦 6 Solution (continued): b. Determine the standard deviation; Rainfall 1 Station (Pi – Pmean)2 (mm) σM i=1Pi − ഥ P 2 2 S= A 82.0 1339.56 M−1 B 102.9 246.49 C 180.3 3806.89 6181.48 S= = 𝟑𝟓. 𝟏𝟔𝟏 𝐦𝐦 D 110.3 68.89 5 E 98.8 392.04 F 136.7 327.61 Sum = 711.0 6181.48 Solution (continued): c. Solve for the coeff. of variation & optimum no. of stations = 29.647= 100S 100 𝑥 35.161 Cv = ഥ = P 118.6 2 2 Cv 29.647 N= = = 8.79 𝐬𝐚𝐲 𝟗 𝐬𝐭𝐚𝐭𝐢𝐨𝐧𝐬 ε 10  If the normal annual precipitation at various stations within about 10% of the normal annual precipitation station X; Arithmetic mean method 𝟏 𝐏𝐱 = 𝐏𝟏 + 𝐏𝟐 + ⋯ + 𝐏𝐌 𝐌 If the normal annual precipitations vary considerably, use the normal ratio method 𝐍𝐱 𝐏𝟏 𝐏𝟐 𝐏𝐌 𝐏𝐱 = + + ⋯+ 𝐌 𝐍𝟏 𝐍𝟐 𝐍𝐌 where: Px = estimated precipitation volume at the missing data station X, depth P1, P2, PM = estimated precipitation volume of the stations 1, 2, and M, depth Nx = average annual precipitation at the missing data station X, depth N1, N2, NM = average annual precipitation at the adjacent stations, depth The normal annual rainfall at stations A, B, C, and D in a basin are 81, 68, 76, and 92 cm respectively. In the year 1980, the station D was inoperative and stations A, B, C recorded annual precipitations of 91, 72, and 80 cm respectively. Estimate the rainfall at station D in that year. Given: Required: PD = precipitation at station D in the year 1980 Solution: 𝑁𝐷 𝑃𝐴 𝑃𝐵 𝑃𝐶 𝑃𝐷 = + + 𝑀 𝑁𝐴 𝑁𝐵 𝑁𝐶 92 91 72 80 PD = + + 3 81 68 76 𝐏𝐃 = 𝟗𝟗. 𝟐𝟎 𝐜𝐦  Rain gauges rainfall represent only point sampling of the areal distribution of a storm  The important rainfall for hydrological analysis is a rainfall over an area, such as over the catchment  To convert the point rainfall values at various stations to in to average value over a catchment, the following methods are used: Arithmetic – Mean Method Thiessen Method Isohyetal Method Reciprocal – distance – squared method 1. Arithmetic Mean Method ✓ the simplest method of determining areal average rainfall ✓ Satisfactory if the gages are uniformly distributed over the area; and ✓ the individual gage measurements do not vary greatly about the mean N 1 ഥ P = ෍ Pi N i=1 where: Pi = rainfall at the ith station N = no. of rain gauge stations within the catchment Thiessen Method ✓ Rain gages are considered more representative of the area in question than others; relative weights may be assigned to the gages in computing the areal average ✓ Assumption: At any point in the watershed, the rainfall is the same as that of the nearest gage so the depth recorded at a given gage is applied out to a distance halfway to the next station in any direction. P1 A1 A2 P3 A3 P2 A5 P4 P5 A4 A7 P7 A6 P6 Thiessen Method N 𝑁 1 𝐴𝑖 ഥ P = ෍ 𝐴𝑖 Pi = ෍ 𝑃𝑖 A 𝐴 i=1 𝑖=1 where: Pi = rainfall at the ith station Ai = corresponding area of the Thiessen polygon at the ith station. N = no. of rain gauge stations within the catchment Ai/A = weightage factor for each station Isohyetal Method An isohyet is a line joining points of equal rainfall magnitude. 𝑷𝟏 + 𝑷𝟐 𝑷 𝟐 + 𝑷𝟑 𝑷𝒏−𝟏 + 𝑷𝒏 𝑨𝟏 + 𝑨𝟐 + ⋯ + 𝑨𝒏−𝟏 ഥ= 𝟐 𝟐 𝟐 𝐏 𝑨 where: P1,P2 ,...,Pn = values of the isohyets A1,A2,...,An = inter-isohyets area respectively A = total catchment area. 12.0 10.0 8.0 14.2 6.0 6.8 A D A5 10.8 12.0 C A4 A1 A3 E 10.0 B 9.6 8.1 5.2 G A2 8.0 7.4 F 6.0 Bedient, et al. (2013). Hydrology and Floodplain Analysis. Pearson Education Limited. England. Chow, et al.(1988). Applied Hydrology. McGraw-Hill Book Co. Singapore. Mays, L.W. (2011). Water Resources Engineering. John Wiley & Sons, Inc. USA. Subramanya, K. (2008). Engineering Hydrology. Tata McGraw-Hill Publishing Company Limited. India.

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