IRRIGATION AND DRAINAGE ENGINEERING MARVIN M. CINENSE IF YOU FAIL TO PREPARE, YOU PREPARE TO FAIL.… Please prepare the following before we start: • Calculator • Sheets of paper • Ballpen or pencil • Other things that you think you need during the review proper (etc…) The moisture content of the soil when the gravitational water has been removed. a. Available water b. Field capacity c. Permanent wilting point d. Readily available moisture B – Field capacity • Schematic representations of field capacity and permanent wilting point soil moisture content It is the ratio of the dry weight of soil particles to the weight of an equal volume of water. a. Particle density b. Bulk density c. Real specific gravity d. Apparent specific gravity D - Apparent specific gravity π΅π’ππ ππππ ππ‘π¦ ππ π πππ (π π π π΄π = π·πππ ππ‘π¦ ππ π€ππ‘ππ (1 π π π ππ·π ππ π πππ (π π΅π’ππ ππππ ππ‘π¦ ππ π πππ = π΅π’ππ π£πππ’ππ ππ π πππ (ππ One hundred thousand two hundred and fifty (1,250) cubic meters of water was delivered to a 10 ha farm for the month of June in which consumptive use is estimated at 8 mm/day. The effective rainfall for the period was 150 mm. What is the irrigation efficiency? a. 32% b. 87% c. 72% d. 52% C – 72% Effective Rainfall, ER = 0.15m x 10 ha x 10,000 m2/ha = 15,000 m3 Consumptive Use, CU CU = (0.008 m/day) x 30 days x 10 ha x 10,000 m2/ha = 24,000 m3 Irrigation Efficiency, Ea = (CU – ER)/Wf = (24,000 – 15,000)m3 / 1,250 m3 = 72% These are pipelines built on or near the ground surface to convey water across wide depressions? a. Inverted siphons b. Siphons c. Laterals d. Flumes D – flumes Evapotranspiration in an 8 ha farm is 7 mm/day and percolation losses is 2 mm/day. What is the design discharge of a canal to be able to deliver a 5-day requirement of the farm in 24 hours if irrigation efficiency is 75%? a. 150 m3/hr b. 200 m3/hr c. 175 m3/hr d. 140 m3/hr B - 200 m3/hr Q = Ad/t Q = [(8 ha x 10,000 m2/ha) (7 mm/day + 2 mm/day) (1m/1000mm) (5 days)] / (24 hrs x 0.75) Q = 200 m3/hr Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone c. Parallel drain system d. Double main system B – herringbone How much water should be applied to a 6 ha farm where the rooting depth is 80 cm, if it is in its permanent wilting point? Volumetric moisture contents are 0.15 and 0.32 for permanent wilting point and field capacity, respectively. a. 7,200 m3 b. 6,120 m3 c. 15,360 m3 d. 8,160 m3 D – 8,160 m3 Vol. = (FC – PWP)(D)(A) = (0.32 - 0.15)(0.8m)(60,000 m2) = 8,160 m3 Pv = AsPd Pd = 100 (FW-ODW)/ODW d = PvD/100 = PdAsD/100 What is the depth of water in a trapezoidal channel with a side slope of 2 and carrying a 2.5 m3/s water flow? The channel’s bottom width is 1.5 meters and the flowing water has a velocity of 0.8 m/s. a. 1 m b. 1.2 m c. 0.93 m d. 0.82 m C – 0.93 m Q = AV or A = Q/V A = 2.5 m3/s / 0.8 m/s = 3.125 m2 A = by + zy2 : 3.125 = 1.5y + 2y2 solving for y = 0.93 m How many sprinklers with spacing of 7m x 7m are needed to irrigate a rectangular piece of land 125 m x 190 m if the laterals are set parallel to the longer side of the field? a. 503 b. 504 c. 486 d. 485 C – 486 Number of Laterals, N = 125/7 = 17.86 or 18 Number of Sprinklers/lateral, S = 190/7 = 27.142 or 27 Total number of sprinklers = N x S = 18 x 27 = 486 If the impeller speed of a centrifugal pump is increased from 1800 rpm to 2340 rpm, the resulting power will be how many times the original? a. 1.690 b. 2.197 c. 1.091 d. 1.140 D - 2.197 π1 2340 1800 3 = π2 2.197 π1 = π2 Darcy’s law states that the flow of water through a porous medium is? a. Proportional to the medium’s hydraulic conductivity b. Inversely proportional to the length of flow path c. Both a and b d. Neither a nor b C – both a & b One liter per second is equal to? a. 16.85 gpm b. 15.50 gpm c. 15.85 gpm d. 17.35 gpm C – 15.85 gpm It is the ratio of the volume of voids to the total volume of the soil. a. Void volume b. Bulk density c. Porosity d. Void density n = Vv/VT C – porosity A soil sample was obtained using a cylindrical soil sampler with a 4-inch diameter and 10inch height. After oven-drying, the sample weighed 2,470 grams. What is the soil’s bulk density. a. 12 g/cc b. 1.1 g/cc c. 1200 kg/m3 d. 1.3 kg/m3 C – 1200 kg/m3 Vb = Ah = (πd2/4)(h) = [π(4 in x 2.54 cm/in)2/4] x (10 in x 2.54 cm/in) = 2,059.3 cm3 BD = ODW/Vb = 2,470/2,059.3 = 1.2 g/cc = 1200 kg/m3 It is the water retained about individual soil particles by molecular action and can be removed only by heating. a. Permanent wilting point b. Hygroscopic water c. Hydrophobic water d. Microscopic water B – hygroscopic water Compute for the brake horsepower of a pump needed to pump-out a fluid (ρ = 1.3 g/cc) at a rate of 300 gpm with a total head of 6 meters. Assume pump efficiency of 60%. a. 2.5 hp b. 3.0 hp c. 3.5 hp d. 5.0 hp C – 3.5 hp Specific weight of water, γ γ = (1.3g/cm3)(kg/1000g)(2.2 lbs/kg)[(2.54)3 cm3/in3]x(1728 in3/ft3) γ = 81 lbs/ft3 BHP = (300 gal/min x 1 ft3/7.48 gal x 81 lbs/ft3 x 6m x 3.28 ft/m) / (33,000 ft-lbs / min-hp x 0.60) = 3.23 hp A 16-ft thick confined aquifer with hydraulic conductivity of 500 ft/day was tapped by a 4inch diameter shallow tube well. With a radius of influence of 2000 ft, determine the maximum discharge of the STW in lps. Assume an allowable drawdown of 10 ft. a. 16.85 b. 17.55 c. 5.59 d. 6.59 B – 17.55 Q = 2πkt(he – hw) / ln(re/rw) Q = 2π(500 ft/day)(16 ft)(10 ft) / ln[2000 ft/(2/12 ft) Q = 53,515.8 ft3/day Q = 53,515.8 ft3/day x 1m3/ (3.28)3ft3 x 1000 li/m3 x 1 day/86400 sec Q = 17.55 li/sec It refers to the composite parts of the irrigation system that divert water from natural bodies of water such as rivers, streams and lakes. a. Main canal b. Diversion canal c. Irrigation structures d. Headworks D – headworks It is a measure of the amount of water that the soil will retain against a tension of 15 atmospheres. a. Readily available moisture b. Permanent wilting point c. Available moisture d. Field capacity B – PWP Given a shallow tubewell with maximum discharge of 15 lps and a total dynamic head of 7 meters. Determine the power rating of the primemover for the pump if pump and primemover efficiencies are 60% and 55%, respectively. a. 4.0 hp b. 3.5 hp c. 4.5 hp d. 5.0 hp C – 4.5 hp Rated HP = 15 li/sec x 1m3/1000 li x 62.4 lbs/ft3 x 7 m x 3.28ft/m 550 ft-lbs/sec-hp x 0.6 x 0.55 = 4.17 hp , say 4.5 hp What is the discharge in each sprinkler nozzle to irrigate a rectangular piece of land 150m x 180m if the laterals are set parallel to the longer side of the field. Sprinkler spacing is 6m x 6m, irrigation water requirement is 150 mm and irrigation period is 6 hours. a. 0.250 lps b. 0.375 lps c. 0.500 lps d. 0.125 lps A – 0.250 lps Q = 6m x 6m x 0.15m/6hrs x 1hr/3600sec x 1000li/m3 Q = 0.250 lps The International Soil Science Society describes sand as a soil particle with a diameter of a. 0.02 to 2 mm b. 0.2 to 2 mm c. 0.002 to 0.02 mm d. 0.002 to 0.2 mm B – 0.2 to 2 mm Ten m3/hr is equal to a. 2.78 lps b. 44.03 gpm c. Both a and b d. Neither a nor b C – both a & b Determine the irrigation interval for a farm with soil root zone having a field capacity of 200 mm and a wilting point of 105 mm. Assume that the consumptive use for August is 7.5 mm/day with no rainfall and the allowable moisture depletion is 75%. a. 11 days b. 9 days c. 4 days d. 7 days B – 9 days Iint = (FC – WP)(AMD) / CU = ((200 – 105)mm x 0.75) / 7.5 mm/day Iint = 9.5 days or 9 days What is the depth of water in a trapezoidal channel with a side slope of 2 and carrying a 3.2 m3/s water flow? The channel’s bottom width is 1.5 meters and the flowing water has a velocity of 0.85 m/s. a. 1.8 m b. 1.79 m c. 1.05 m d. 1.04 m C – 1.05 m Q = AV A = Q/V = 3.2 m3/s / 0.85 m/s = 3.765 m2 A = by + zy2 3.765 = 1.5y + 2y2 Compute for y = ? The localized lowering of the static piezometric water level due to pumping. a. Groundwater decline b. Drawdown c. Subsidence d. Depression or B – drawdown Any convenient level surface coincident or parallel with mean sea level to which elevations of a particular area are referred a. Datum b. Elevation c. Horizontal surface d. Slope A – datum What is the design discharge of a canal to be able to deliver a 7-day requirement of a 5-ha farm in 12 hours if the irrigation requirement is 8 mm/day? a. 65 m3/s b. 6.5 m3/s c. 0.65 m3/s d. 0.065 m3/s D – 0.065 m3/s Q = (5 ha x 10,000 m2/ha x 8 mm/day x (1m/1000mm) x 7 days) / 12 hrs x 1hr/3600sec Q = 0.0648 m3/s = 0.065 m3/s It is a geologic formation which transmits water at a rate insufficient to be economically developed for pumping. a. Aquifer b. Aquiclude c. Aquifuge d. Aquitard B – aquiclude Determine the maximum total head at which a 5-hp centrifugal pump can extract water at a rate of 25 lps if pump efficiency is 65%. a. 25.32 ft b. 32.48 ft c. 33.39 ft d. 35.12 ft B – 32.48 ft BHP = γQH/EP or H = (BHP)(E )/γQ (γ = 62.4 lbs/ft3) H = 5 hp x 0.65 x 550 ft-lbs/sec-hp x 1ft/62.4 lbs 25 li/sec x 1m3/1000 li x (3.28)3 ft3/m3 H = 32.48 ft It is the ratio of the dry weight of the soil to the weight of the water with volume equal to the soil bulk volume. a. Particle density b. Bulk density c. Real specific gravity d. Apparent specific gravity D - ASG It accounts for the losses in an irrigation system from the water source and prior to delivery of water into the field ditches. a. Evaporation b. Application efficiency c. Diversion efficiency d. Conveyance efficiency D – Conveyance efficiency A geologic formation that contains water but do not have the capacity to transmit it. a. Aquifuge b. Aquifer c. Aquitard d. Aquiclude D – Aquiclude Aquifuge - An impermeable body of rock which contains no interconnected openings or interstices and therefore neither absorbs nor transmits water. Aquiclude - Any geological formation that absorbs and holds water but does not transmit it at a sufficient rate to supply springs, wells, etc. An aquifer is an underground layer of water-bearing permeable rock or unconsolidated materials (gravel, sand, or silt) from which groundwater can be extracted using a water well. The study of water flow in aquifers and the characterization of aquifers is called hydrogeology. Related terms include aquitard, which is a bed of low permeability along an aquifer,[1] and aquiclude (or aquifuge), which is a solid, impermeable area underlying or overlying an aquifer. If the impermeable area overlies the aquifer pressure could cause it to become a confined aquifer. Compute the land soaking requirement for a soil (depth of root zone = 60 cm) with residual moisture content of 18% by weight, bulk density of 1,320 kg/m3 and porosity of 50%. Standing water for planting is 20 mm. a. 177.44 mm b. 157.44 mm c. 253.44 mm d. 273.44 mm B – 157.44 mm What is the recommended value for standing water during land preparation. a. 5 mm b. 10 mm c. 15 mm d. 8 mm B – 10 mm Farm water requirement minus the application losses is the. a. Diversion water requirement b. Farm irrigation requirement c. Application efficiency d. Land preparation water requirement B – Farm irrigation requirement What is the root zone depth of a farm with land soaking requirement of 90 mm if the soil porosity is 45%, residual moisture content is 18% (by weight) and bulk density is 1,250 kg/m3? a. 35 cm b. 40 cm c. 45 cm d. 60 cm B – 40 cm This results from overlapping radii of influence of neighboring wells. a. Drawdown b. Groundwater decline c. Well interference d. Drawdown curve C – well interference In furrow irrigation, the rate of water application should be ____ the intake rate of the soil. a. Less than b. Greater than c. Equal to d. Not related to A – less than The International Soil Science Society describes sand as a soil particle with a diameter of a. 0.02 to 2 mm b. 0.2 to 2 mm c. 0.002 to 0.02 mm d. 0.002 to 0.2 mm A – 0.02 – 2.0 mm The localized lowering of the static or piezometric water level due to pumping a. Groundwater decline b. Drawdown c. Subsidence d. Depression B – drawdown For a 10m x 10m sprinkler spacing, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m c. 8.0 m d. 5.0 m B – 7.5 m 5m 2.5m Four liters per second is equivalent to a. 14.4 m3/hr b. 63.41 gpm c. Both a and b d. Neither a nor b C – Both a & b Irrigation method is used for row crops wherein only a part of the surface is wetted a. Basin flooding b. Furrow irrigation c. Border irrigation d. Border-strip flooding B – furrow irrigation It refers to the composite parts of the irrigation system that divert water from natural bodies of water such as rivers, streams and lakes a. Main canal b. Diversion dam c. Irrigation structures d. Headworks D - headworks It is a measure of the amount of water that the soil will retain against a tension of 15 atmospheres a. Readily available moisture b. Permanent wilting point c. Available moisture d. Field capacity B - PWP The infiltration equation based on the exhaustion process is the a. Lewis-Kostiakov’s b. Horton’s c. Philip’s d. Darcy’s B – Horton’s Determine the irrigation interval for a farm with soil root zone having a field capacity of 200 mm and a wilting point of 140 mm. Assume that the consumptive use is 6 mm/day with no rainfall and the allowable moisture depletion is 75% a. 8 days b. 9 days c. 7 days d. 10 days C – 7 days Int = (FC-WP)(AMD)/CU = ((200-140)mm x 0.75)/6 mm/day = 7.5 or 7 days What is the design discharge of a canal to be able to deliver a 6-day requirement of a 6-ha farm in 9 hours if the irrigation requirement is 8 mm/day? a. 88.9 m3/s b. 8.89 m3/s c. 0.889 m3/s d. 0.0889 m3/s D – 0.0889 m3/s Q = Ad/t Q = (6 ha x 10,000 m2/ha x 0.008 m/day x 6 days x 1hr/3600 sec) / 9 hrs Q = 0.0889 m3/sec The amount of drainage water to be removed per unit time per unit area is the a. Drainage requirement b. Drainage coefficient c. Drain spacing d. Drainage volume B – drainage coefficient In Hooghoudt’s drain spacing formula, it is assumed that a. The water table is in equilibrium with the rainfall or irrigation water b. The drains are evenly spaced c. Darcy’s law is valid for flow through soils d. All of the above D – all of the above A mathematical expression for the macroscopic flow of water through a porous system a. Steady state groundwater flow equation b. Darcy’s Law c. Laplace’s equation d. Scobey;s equation B – Darcy’s Law It is the soil moisture constant describing the amount of moisture retained by the soil against a suction pressure of 1/3 atmosphere a. Field capacity b. Hygroscopic water c. Permanent wilting point d. Saturation point A – field capacity Run-off is the difference between the gross depth of irrigation water and the a. Net depth requirement b. Crop evapotranspiration c. Depth that infiltrated d. Water use rate A – net depth requirement It is the type of sprinkler irrigation system where the number of laterals installed is equal to the total number of lateral positions a. Hand move system b. Periodic move c. Special type d. Set system D – set system In a drip system, the allowable pressure variation in a lateral with minimum and average head of 1.51 and 1.38 meters, respectively is a. 0.185 psi b. 0.462 psi c. 0.370 psi d. 0.130 psi B – 0.462 psi A soil with root zone depth of 1.2 meters has 24% initial volumetric moisture content, volumetric field capacity and permanent wilting point of 30% and 15%, respectively and 50% allowable moisture depletion. The initial depth of water in the soil is a. 43.2 mm b. 270 mm c. 288 mm d. 360 mm C – 288 mm Pv = d/D Initial depth, d = Pv x D d = (0.24) (1.2 m x 1000 mm/m) = 288 mm In previous Problem, when the soil reaches _____, irrigation should be done a. 43.2 mm b. 270 mm c. 288 mm d. 360 mm B – 270 mm FC = .3 x 1,200 = 360 mm AMD = .5(FC-WP) = (0.5)(360-180) = 90 mm WP = .15 x 1,200 = 180 mm WP + AMD = 180 + 90 = 270 mm } 90 mm In previous Problem, natural drainage occurs when the soil water reaches or exceeds a depth of _______. a. 43.2 mm b. 270 mm c. 288 mm d. 360 mm D – 360 mm FC = .3 x 1,200 = 360 mm AMD = .5(FC-WP) = (0.5)(360-180) = 90 mm WP = .15 x 1,200 = 180 mm } 90 mm Since FC is 360 mm, then natural drainage will occur when the soil water reaches FC or 360 mm It is a surface irrigation system where the area is subdivided by dikes and water flows over these dikes from one subdivision to another. a. Border irrigation b. Furrow irrigation c. Basin irrigation d. Corrugation irrigation C – basin irrigation Distribution control structures placed across an irrigation ditch to block the flow temporarily and to raise the upstream water level. a. Turnouts b. Checks c. Culverts d. Weirs B – checks Using Scobey’s equation for friction loss, a lateral with 10 sprinklers has a reduction coefficient of a. 0.3766 b. 0.100 c. 0.200 d. 0.3964 D – 0.3964 Scobey’s Formula: Hf = [ksLQ1.9 /(d4.9)]1.45 x 10-8 d = ft, L = ft, Q = gpm, H = ft Which is not a component of the impact arm of an impact sprinkler? a. Nozzle b. Counterweight c. Vane d. Spoon A - nozzle A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4inch diameter shallow tube well. With a radius of influence of 2,500 ft, determine the maximum discharge of the STW in liters per second. Assume an allowable drawdown of 12 ft? a. 22.17 b. 20.57 c. 62.71 d. 25.63 B – 20.57 2πkt(he – hw) 2π(400 ft/day)(20 ft)(12 ft) Q= = ln(re/rw) ln(2,500/.17) = 67.52 ft3 /day 67.52 ft3 x day m3 x day (3.28)3 ft3 86,400 sec x 1000 li m3 = 20.57 lps In surface irrigation, the ratio between the gross amount of irrigation water and the net requirement of the crop is the a. Application efficiency b. Deep percolation c. Seepage d. Runoff A – APPLICATION EFFICIENCY It is an orderly sequence of planting crop in an area for a 365-day period a. Cropping pattern b. Crop combination c. Crop sequence d. Cropping schedule A – Cropping pattern A 21.6 mm/day water requirement is equivalent to a. 23.8 gpm/ha b. 0.9 lps/ha c. 2.5 lps/ha d. 14.3 gpm/ha C – 2.5 lps/ha 21.6 mm x day m x 10,000 m2 x day 1000 mm = 2.5 lps/ha ha x 1000 li 86,400 sec m3 In a 5 ha area, it was determined that the soil volumetric field capacity and permanent wilting point are 25% and 15%, respectively. Crop consumptive use is 5 mm/day, application efficiency is 80% and irrigation application rate is 32 m3/hr. The allowable soil moisture depletion is 60%, apparent specific gravity is 1.2 and the depth of root zone is 0.8 m. The net depth of irrigation water to be applied is a. 80 mm b. 48 mm c. 24 mm d. 36 mm B – 48 mm dnet = (AMD)(FC-WP)(D) = (.6)(.25-.15)(800mm) = 48 mm In previous problem, the gross depth of irrigation water to be applied is a. 100 mm b. 60 mm c. 64 mm d. 38 mm B – 60 mm dgross = dnet /Ea = 48/.8 = 60 mm In previous problem, the irrigation interval, in days is a. 10 b. 5 c. 9 d. 4 C – 9 days Int = dnet /CU = (48 mm)/(5 mm/day) = 9.6 or 9 days In previous problem, the irrigation period, in hours is a. 93 b. 47 c. 230 d. 94 D - 94 Irrigation period = Adgross /Q 5 ha x 10,000 m2 x .06 m ha 32 m3/hr = 93.75 hrs The head in an emitter discharging 4 liters of water per hour and with discharge coefficient of 0.798 and exponent of 0.5 is a. 6.4 m b. 5.0 m c. 10 m d. 1.8 m 25 m In furrow irrigation, it is the difference between the depth of water that infiltrated and the net depth requirement a. Runoff b. Application losses c. Deep percolation d. Seepage C – deep percolation A rectangular piece of land 180m x 240m is laid out with one-sided sprinkler irrigation system. Laterals are set parallel to the longer side of the field. Sprinkler spacing is 6m x 6m, irrigation water requirement is 150 mm and irrigation period is 6 hours. Laterals are set on only one side of the mainline. The sprinkler discharge is a. 0.50 lps b. 0.375 lps c. 0.250 lps d. 0.125 lps C – 0.250 lps (6m) x (6m) x (.15 m) x hr Q= 6 hrs 3600 sec x 1000 li m3 = 0.25 lps In previous problem, determine the number of lateral positions a. 30 b. 20 c. 40 d. 60 A - 30 240 m 180 m No. of lateral positions = 180/6 = 30 In previous problem, determine the number of sprinklers/lateral a. 30 b. 20 c. 40 d. 60 c - 40 240 m 180 m No. of sprinkler/lateral = 240/6 = 40 The field in previous problem is installed with hand-moved system and 2 sets of laterals can be installed per day. Calculate the minimum number of laterals that can be installed per set if there are 5 operating days per irrigation interval a. 2 b. 4 c. 3 d. 1 C–3 No. of laterals/set = total no. of laterals/5days = 30/5 = 6 lat/set Since there are 2 sets of laterals, then the minimum no. of laterals per set = 6/2 = 3 laterals In-line canal structure designed to convey canal water from a higher level to a lower level, duly dissipating the excess energy resulting from the drop in elevation a. Drop b. Flume c. Weir d. None of the above A – drop Longitudinal Section of a Vertical Drop (Source: PAES 606:2016 ) Longitudinal Section of a Rectangular Inclined Drop (Source: PAES 606:2016 ) Longitudinal Section of a Baffled Apron Drop (Source: PAES 606:2016 ) Amount of rainwater that falls directly on the field and is used by the crop for growth and development excluding deep percolation, surface runoff and interception a. Average rainfall b. Effective rainfall c. Rainfall depth d. None of the above B – effective rainfall Applicator used in drip, subsurface, or bubbler irrigation designed to dissipate pressure and to discharge a small uniform flow or trickle of water at a constant rate that does not vary significantly because of minor differences in pressure a. Drippers b. Emitters c. Nozzle d. None of the above B – emitters Closed conduit designed to convey canal water in full and under pressure running condition, to convey canal water by gravity under roadways, railways, drainage channels and local depressions a. Siphon b. Inverted siphon c. Elevated flumes d. None of the above B – inverted siphon • Inverted siphons – closed conduit designed to convey canal water in full and under pressure running condition, to convey canal water by gravity under roadways, railways, drainage channels and local depressions. – for crossing wide deep hollows, depressions, or canyons through them under pressure. Cross-section of a proposed inverted siphon at a river showing the outline of the conduit (Source: PAES 606:2016 ) Amount of water required in lowland rice production which includes water losses through evaporation, seepage and percolation and land soaking a. Land soaking water requirements b. Land preparation water requirements c. Irrigation water requirements d. None of the above B – land preparation requirements Portion of the pipe network between the mainline and the laterals a. Connector b. Valve c. Manifold d. None of the above C – manifold Constant flow depth along a longitudinal section of a channel under a uniform flow condition a. Normal depth b. Critical depth c. Uniform depth d. None of the above A – normal depth Water flow that is conveyed in such a manner that top surface is exposed to the atmosphere such as flow in canals, ditches, drainage channels, culverts, and pipes under partially full flow conditions a. Open channel flow b. Canal flow c. Pipe flow d. None of the above A – open channel flow Tube or shaft vertically set into the ground at a depth that is usually less than 15 m for the purpose of bringing groundwater into the soil surface whose pumps are set above the water level a. Shallow tubewell b. Deep well c. Pipe d. None of the above A – shallow tubewell Ratio of the horizontal to vertical dimension of the channel wall a. Slope b. Channel gradient c. Side slope d. All of the above C – side slope Slope of the water surface profile plus the velocity head in open channels a. Energy grade line slope b. Water surface slope c. Channel bottom slope d. Hydraulic grade line slope A – energy grade line slope Slope of the free water surface in open channel a. Energy grade line slope b. Water surface slope c. Channel bottom slope d. Hydraulic grade line slope D – Hydraulic grade line slope Occurs when flow has a constant water area, depth, discharge, and average velocity through a reach of channel a. Normal flow b. Critical flow c. Uniform flow d. Varied flow C – uniform flow Accounting of water inflows, such as irrigation and rainfall; and outflows, such as evaporation, seepage and percolation. a. Water cycle b. Water balance c. Water flow d. All of the above B – water balance Area which contributes runoff or drains water into the reservoir. a. Watershed b. River network c. Streams d. All of the above A - watershed Diameter of the circular area wetted by the sprinkler when operating at a given pressure and no wind. a. Wetted diameter b. Wetted perimeter c. Diameter of throw d. All of the above A – wetted diameter Portion of the perimeter of the canal that is in contact with the flowing water. a. Wetted diameter b. Wetted perimeter c. Diameter of throw d. All of the above A – wetted perimter Moisture left in the soil before the initial irrigation water delivery which describes the extent of water depletion from the soil when the water supply has been cut-off. a. Current soil moisture content b. Residual moisture content c. Allowable moisture depletion d. None of the above B – residual moisture content Pressure required to overcome the elevation difference between the water source and the sprinkler nozzle, to counteract friction losses and to provide adequate pressure at the nozzle for good water distribution. a. Average pressure b. Design pressure c. Pressure requirement d. None of the above B – design pressure An overflow structure built perpendicular to an open channel axis to measure the rate of flow of water. a. Weir b. Flume c. Orifice d. None of the above A – weir Types of Weir, Description and Discharge Evaluation Types of Weir, Description and Discharge Evaluation In-line structure with a geometrically specified constriction built in an open channel such that the center line coincides with the center line of the channel in which the flow is to be measured. a. Weir b. Flume c. Orifice d. None of the above B – flume Types of Flumes, Description and Discharge Evaluation Types of Flumes, Description and Discharge Evaluation Measuring device with a well-defined, sharpedged opening in a wall through which flow occurs such that the upstream water level is always well above the top of this opening. a. Weir b. Flume c. Orifice d. None of the above C – orifice Types of Orifices, Description and Discharge Equation Determine the seepage and percolation losses within the canal if the difference in flow rates at 200-meter distance is 2 m3/s. a. 1 liter/s-m b. 0.1 m3/s-m c. 0.01 m3/s-m d. None of the above C – 0.01 m3/s-m A device with individual scales on the rods to provide data to plot furrow depth as a function of the lateral distance where data can then be numerically integrated to develop geometric relationships such as area verses depth, wetted perimeter versus depth and top-width verses depth. a. Infiltrometer b. Profilometer c. Penetrometer d. None of the above B – profilometer Profilometer Application of water by gravity flow to the surface of the field. a. Surface irrigation b. Furrow irrigation c. Basin irrigation d. None of the above A – surface irrigation Method of irrigation where water runs through small parallel channels as it moves down the slope of the field. a. Surface irrigation b. Furrow irrigation c. Basin irrigation d. None of the above B – furrow irrigation Recommended slope for furrow irrigation method. a. 0.05 % to 3.0 % b. 2.0% to 5.0% c. ≤ 0.1% d. None of the above A – 0.05% to 3.0% a. 0.05 % to 3.0 % ο¬ Furrow ο¬ Border b. 2.0% to 5.0% ο¬ Basin c. ≤ 0.1% What should be the depth of water to be applied if the needed water depth is 6 mm and the application efficiency is 75%. a. 4.5 mm b. 6.0 mm c. 8.0 mm d. None of the above C -8.00 mm What should be the net depth of water to be applied if the depth of soil to be irrigated is 600 mm. The soil has a FC of 28%, PWP of 14% and the allowable moisture depletion is 50%. a. 36 mm b. 42 mm c. 50 mm d. None of the above B – 42 mm What should be the irrigation frequency if the depth of water needed is 42 mm. The potential ET is 7.6 mm/day and the crop coefficient is 0.95. a. 4 days b. 6 days c. 8 days d. None of the above B – 6 days ππππ‘ πΌπ = πππππΈππ πΈππ = 0.95 π₯ 7.6 ππ/πππ¦ πΈππ = 7.22 ππ/πππ¦ 42 ππ πΌπ = ππ 7.22 πππ¦ π°π = 5.82 πππ¦π = π π πππ What is the maximum sprinkler spacing, in square pattern, if the average wind speed is 5 km/hr and below? a. 55% b. 50% c. 45% d. None of the above A – 55% What should be the sprinkler precipitation rate if the sprinkler spacing is 6m by 6m. The sprinkler discharge is 210 lph at 3 bars pressure with 14 meter wetted diameter. a. 4.2 mm/hr b. 5.8 mm/hr c. 6.3 mm/hr d. None of the above B – 5.8 mm/hr A trapezoidal channel has a base width of 6 meters and a side slope of 1H:1V. The channel bottom slope is 0.0002 and the Manning roughness coefficient is 0.014. What is the depth of flow if Q = 12.1 m3/s? a. 1.5 m b. 1.3 m c. 1.7 m d. 1.1 m A – 1.5 m π΄ = π¦ (π + π§π¦ π = π + 2π¦ 1 + π§ 2 1 π = π΄π 2 3 π 1 2 π 1 π¦ (6 + 1π¦ 12.1 = π¦ (6 + 1π¦ 0.014 6 + 2π¦ 1 + 12 By trial and error: π¦ = 1.5 π 2 3 0.00021 2 A trapezoidal channel has a base width of 6 meters and a side slope of 1H:1V. The channel bottom slope is 0.0002 and the Manning roughness coefficient is 0.014. What will be the state of flow if the depth of flow is 1.5 m? a. Critical b. Sub-critical c. Supercritical d. Laminar B – sub-critical πΉ= π ππ· π· = π΄/π π = π + 2π¦ π΄ = π¦ π + π§π¦ = 1.5 (6 + 1 1.5 = 11.25 π2 π = 6 + 2 1.5 = 9π 11.25 π·= = 1.25 π 9 π 12.1 π= = = 1.076 π/π π΄ 11.25 πΉ= 1.076 (9.81 (1.25 = 0.307 πππππ 0.307 < 1, π‘βπππππππ, π‘βπ ππππ€ ππ πΊππππππππππ What is the required power if the needed system capacity is 25 m3/hr and the total dynamic head is 40 meters. The pump efficiency is 60%. a. 3 hp b. 5 hp c. 7 hp d. None of the above C – 7 hp Determine the nominal pipe diameter for the 20 m3/hr flow if the allowable velocity in pipe is 2 m/s. a. 50 mm b. 63 mm c. 75 mm d. None of the above B – 63 mm Q = AV Q = (πD2/4) V D = (4Q/ πV)1/2 ; allowable V = 2 m/s = 7200 m/hr Dm= (4 x 20/ π x 7200)1/2 Dm = 59 mm Any barrier constructed to store water. a. Reservoir b. Dam c. Tank d. None of the above B – dam Volume of water stored in reservoir between the minimum water level and normal water level. a. Active storage b. Dead storage c. Storage capacity d. None of the above A – active storage Maximum elevation the water surface which can be attained by the dam or reservoir without flow in the spillway. a. Maximum storage elevation b. Dam crest elevation c. Normal storage elevation d. None of the above C – normal storage elevation Part of water impounding system that stores the runoff. a. Watershed b. Reservoir c. Dam d. None of the above C – dam What is the critical depth for a grassy triangular channel with 2H:1V side slopes and a 0.5% slope when the flow is 3.00 m3/s? a. 1.23 m b. 1.36 m c. 0.86 m d. 0.78 m C – 0.86 m 3 2 π΄ π = π π Where: A = flow area (m2, ft2) T = top width of flow (m, ft) Q = channel flow rate (m3/s, ft3/s) g = gravitation acceleration (m/s2, ft/s2) What should be the state of flow on previous Problem if the channel is actually flowing at a depth of 1.2 m? a. Critical b. Sub-critical c. Supercritical d. Laminar The critical depth is 0.86 m. The actual flow depth of 1.2 m is greater than the critical depth, so the flow is subcritical B – sub-critical A trapezoidal channel carrying 400 cfs is built with nonerodible bed having a slope of 0.0016 and n = 0.025. Determine the depth of water, y if b = 20 ft and z = 2 a. 3.36 ft b. 2.87 ft c. 5.72 ft d. 4.13 ft A – 3.36 ft π΄π 2 3 = 0.025 π₯ 400 1.486 0.0016 = 167.7 Where: π΄ = π + π§π¦ π¦ π = π + π§π¦ π¦ π + 2π¦ 1 + π§ 2 Substituting A and R in the above equation: π + π§π¦ π¦ 5/3 π + 2π¦ 1 + π§ 2 20 + 2π¦ π¦ 2/3 5/3 20 + 2π¦ 1 + 22 π = π. ππ ππ = 167.7 2/3 = 167.7 On previous problem, determine the crosssectional area, A of the channel. a. 72.36 ft2 b. 89.78 ft2 c. 63.56 ft2 d. 83.92 ft2 π΄ = 20 + 2 π₯ 3.36 π₯ 3.36 = 89.78ππ‘ 2 B – 89.78 On previous problem, determine the best hydraulic section? a. y = 3.36 ft; b = 3.74 ft b. y = 4.52 ft; b = 6.26 ft c. y = 6.60 ft; b = 7.62 ft d. y = 7.81 ft; b = 9.76 ft C – y=6.6 ft; b=7.62 ft For Best Hydraulic Section (Trapezoidal section): π΄ = 3π¦ 2 1 π = π¦ 2 Since section factor, π΄π 2 3 = 167.7 Simplify and compute for y; Then π = π. π ππ The water area, π΄ = 3(6.6 2 = 75.45ππ‘ 2 Since the best hydraulic trapezoidal section is the half hexagon, the side slopes are 1 on 3/3. Computing for bottom width, b: π΄ 74.45 π = − π§π¦ = − π¦ 6.6 3 6.6 = π. ππ ππ 3 CHANNEL GEOMETRY Flow Area (A) Wetted perimeter (P) Hydraulic π Radius (R=π Top Width (T) Hydraulic π¨ Depth (D=π») by b + 2y ππ² π + ππ² b y b + 2zy π + π³π² π² π + ππ³π² 2zy π π π ππ¨ ππ π π π (b + zy )y b + 2y π + ππ zy2 2y π + ππ π π»π π π ππ π»+ ππ» π + π³π² π² π + ππ² π + ππ πππ ππ π + ππ ππ»π π ππ»π + πππ π£12 π£22 π§1 + π¦1 + 2π = π§2 + π¦2 + 2π + βπ ; hf =0 π£12 π£22 π§1 + π¦1 + = π§2 + π¦2 + ; constant 2π 2π Specific energy D1cos θ π£2 E= y + 2π π£2 π E= y + 2π ; v=π΄ ππΈ π 2 ππ΄ =1− 2 ππ¦ ππ΄ ππ¦ π£2 π2 E= y + 2π ; v=π΄2 π£ 2 ππ΄ ππ΄ π΄ 1− ; = π ;π· = ππ΄ ππ¦ ππ¦ π ππΈ π£ 2π π£2 π£ =1− =1− ; = 1; πΉ = 1 ππ¦ ππ΄ ππ· ππ· π£ 2 = ππ·; π2 π΄ = g π΄2 π π΄3 π = ππ; π΄3 π 2 = π π Specific energy π£2 E= y + 2π BEST HYDRAULIC SECTION πΉ= π΄3 π 2 ; = π π ππ· π£
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