1 - Irrigation & Drainage Review Class 2012

1. 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

1. 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

2. 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

1. 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

2. 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

3. One hundred thousand two hundred and fifty 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%

Effective Rainfall, ER = 0.15m x 10 ha x 10,000 m2/ha = 15,000 m3 Consumptive Use, CU = (.008 m/day) x 30 days x 10 x 10,000 m2 = 24,000 m3 Irrigation Efficiency, Ea = (CU – ER)/Wf = (24,000 – 15,000)/100,250 = 72%

3. One hundred thousand two hundred and fifty 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%

4. 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

3. One thousand two hundred and fifty 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%

4. 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

5. 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

5. 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

Q = (Ad)/t = (8x10,000 m2)x(7+2 mm/day)(1m/1000mm (5 days)

24 hrs x .75 = 200 m3/hr

5. 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

Q = (Ad)/t = (8x10,000 m2)x(7+2 mm/day)(1m/1000mm (5 days)

24 hrs x .75 = 200 m3/hr

6. Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone

c. Parallel drain system d. Double main system

6. Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone

c. Parallel drain system d. Double main system

6. Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone

c. Parallel drain system d. Double main system

7. 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

6. Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone

c. Parallel drain system d. Double main system

7. 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

Vol. = (FC – PWP)(D)(A) = (.32 - .15)(.8m)(60,000 m2) = 8,160 m3

6. Subsurface drain system wherein laterals join the submain on both sides alternately. a. Gridiron b. Herringbone

c. Parallel drain system d. Double main system

7. 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

Vol. = (FC – PWP)(D)(A) = (.32 - .15)(.8m)(60,000 m2) = 8,160 m3

8. 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. 1.00 meter b. 0.93 meter

c. 1.20 meters d. 0.82 meter

8. 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. 1.00 meter b. 0.93 meter

c. 1.20 meters d. 0.82 meter

Q = AV or A = Q/V = 2.5/0.8 = 3.125 m2 A = bd + zd2 : 3.125 = 1.5d + 2d2 By trial and error, d = 0.93 m

Check: 3.125 = 1.5(.93) + 2(.93)2

8. 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. 1.00 meter b. 0.93 meter

c. 1.20 meters d. 0.82 meter

9. 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

9. 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

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

8. 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. 1.00 meter b. 0.93 meter

c. 1.20 meters d. 0.82 meter

9. 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

10.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

10.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

11.Darcy’s law states that the flow of water through a porous medium is a. b. c. d.

Proportional to the medium’s hydraulic conductivity Inversely proportional to the length of flow path Both a and b Neither a nor b

10.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

11.Darcy’s law states that the flow of water through a porous medium is a. b. c. d.

Proportional to the medium’s hydraulic conductivity Inversely proportional to the length of flow path Both a and b Neither a nor b

12. One liter per second is equal to

a. 16.85 gpm b. 15.50 gpm

c. 15.85 gpm d. 17.35 gpm

12. One liter per second is equal to

a. 16.85 gpm b. 15.50 gpm 1 li x 60 sec x gal sec

min

3.785 li

c. 15.85 gpm d. 17.35 gpm = 15.85 gpm

12. One liter per second is equal to

a. 16.85 gpm b. 15.50 gpm

c. 15.85 gpm d. 17.35 gpm

13. 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

12. One liter per second is equal to

a. 16.85 gpm b. 15.50 gpm

c. 15.85 gpm d. 17.35 gpm

13. 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

14. A soil sample was obtained using a cylindrical soil sampler with a 4-inch diameter and 10-inch 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

14. A soil sample was obtained using a cylindrical soil sampler with a 4-inch diameter and 10-inch 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

Vb = Ah = (πd2/4)(h) = [π(4x2.54 cm)2/4] x (10x2.54 cm) = 2,059.3 cm3

ρb = OD/Vb = 2,470/2,059.3= 1.2 g/cc = 1200 kg/m3

12. One liter per second is equal to

a. 16.85 gpm b. 15.50 gpm

c. 15.85 gpm d. 17.35 gpm

13. 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

14. A soil sample was obtained using a cylindrical soil sampler with a 4-inch diameter and 10-inch 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

15. 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

15. 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

16.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

16. 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

Specific weight of water, γ γ =(1.3g/cm3)(kg/1000g)(2.2 lbs/kg)[(2.54)3cm3/in3]x(1728 in3/ft3) = 81 lbs/ft3

BHP = 300 gal x ft3 min

7.48 gal

x 81 lbs x 6m x 3.28 ft ft3

33,000 ft-lbs x 0.60

min-hp = 3.23 hp

m

15. 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

16.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

17. A 16-ft thick confined aquifer with hydraulic conductivity of 500 ft/day was tapped by a 4-inch 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 c. 5.59 b. 17.55 d. 6.59

17.A 16-ft thick confined aquifer with hydraulic conductivity of 500 ft/day was tapped by a 4-inch 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 c. 5.59 b. 17.55 d. 6.59 2πkt(he – hw) Q=

2π(500 ft/day)(16 ft)(10 ft) =

=

ln(re/rw)

ln[2000 ft/(2/12 ft)]

ft3 x m3 =

day

(3.28)3

x 1000 li

ft3

m3

x day

86400 sec

=

ft3/day

15. 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

16.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

17. A 16-ft thick confined aquifer with hydraulic conductivity of 500 ft/day was tapped by a 4-inch 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 c. 5.59 b. 17.55 d. 6.59

18.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

18.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

19.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

18.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

19.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

20.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 hp b. 3.5 hp

c. 4.5 hp d. 5 hp

20.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 hp b. 3.5 hp

c. 4.5 hp d. 5 hp 15 li x m3

Rated HP =

x (3.28)3ft3 x 62.4 lbs x 7m x 3.28 ft

sec 1000 li

m3

ft3

550 ft-lbs/sec-hp x 0.6 x 0.55 = 4.17 hp, say 4.5 hp

m

18.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

19.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

20.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 hp b. 3.5 hp

c. 4.5 hp d. 5 hp

21.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

21.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

Q=

c. 0.500 lps d. 0.125 lps

6m x 6m x .15 m x

hr

x 1000 li

6 hrs 3600 sec

m3

= 0.250 lps

21.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

22.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

21.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

22.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

23.Ten m3/hr is equal to a. 2.78 lps b. 44.03 gpm

c. Both a and b d. Neither a nor b

23.Ten m3/hr is equal to

a. 2.78 lps b. 44.03 gpm

c. Both a and b d. Neither a nor b

10 m3 x 1000 li x hr = 2.78 lps 3 hr m 3600 sec 2.78 li x m3 x (3.28)3 ft3 x 7.48 gal x 60 sec = 44.03 gpm 3 3 sec 1000 li m ft min Ans. c. Both a & b

21.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

22.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

23.Ten m3/hr is equal to a. 2.78 lps b. 44.03 gpm

c. Both a and b d. Neither a nor b

24. 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. 10 days b. 9 days

c. 4 days d. 3 days

24. 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. 10 days b. 9 days

c. 4 days d. 3 days

(FC – WP)(AMD) (200 – 105)mm x (.75) Int = = CU 7.5 mm/day = 9.5 days or 9 days

Ans. b

24. 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. 10 days b. 9 days

c. 4 days d. 3 days

25.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. 1.80 m b. 1.79 m

c. 1.05 m d. 1.04 m

25. 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. 1.80 m b. 1.79 m

c. 1.05 m d. 1.04 m

Q = AV or A = Q/V = (3.2 m3/s)/(0.85 m/s) = 3.765 m2 V

d H

A = bd + zd2

b

z = H/V

3.765 = (1.5)d + 2d2 If d = 1.05, then 3.765 =

(1.5)(1.05) + (2)(1.05)2 If d = 1.04, then Ans. c

3.765 = 1.575 + 2.205 = 3.78 (.015) 3.765 = 3.7232 (.0418)

25. 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. 1.80 m b. 1.79 m

c. 1.05 m d. 1.04 m

Q = AV or A = Q/V = (3.2 m3/s)/(0.85 m/s) = 3.765 m2 V

d H

z = H/V

b

A = bd + zd2 3.765 = (1.5)d + 2d2 If d = 1.05, then 3.765 = (1.5)(1.05) + (2)(1.05)2 3.765 = 1.575 + 2.205 = 3.78 (.015) If d = 1.04, then 3.765 = 3.7232 (.0418) Ans. c

26.The localized lowering of the static or piezometric water level due to pumping a. Groundwater decline b. Drawdown

c. Subsidence d. depression

26.The localized lowering of the static or piezometric water level due to pumping a. Groundwater decline b. Drawdown

c. Subsidence d. Depression

27.The aggregation of soil particles into peds.

a. Soil texture b. Soil structure

c. Clay d. Soil porosity

26.The localized lowering of the static or piezometric water level due to pumping a. Groundwater decline b. Drawdown

c. Subsidence d. Depression

27.The aggregation of soil particles into peds.

a. Soil texture b. Soil structure

c. Clay d. Soil porosity

28.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

28.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

5 ha x 10,000 m2 x 8 mm x (1m/1000mm) x 7 days x hr ha day 3600 sec Q= 12 hrs = .0648 m3/s or .065 m3/s

26.The localized lowering of the static or piezometric water level due to pumping a. Groundwater decline b. Drawdown

c. Subsidence d. Depression

27.The aggregation of soil particles into peds.

a. Soil texture b. Soil structure

c. Clay d. Soil porosity

28.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

29. 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

29. 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

30.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

30. 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

BHP = γQH/EP or H = (BHP)(E )/γQ

H=

5 hp x .65 x 550 ft-lbs x ft sec-hp 62.4 lbs m3

25 li x x sec 1000 li

(3.28)3ft3

m3

(γ = 62.4 lbs/ft3)

= 32.48 ft

29. 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

30.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

31.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

29. 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

30.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

31.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

32.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 efficency

c. Diversion efficiency d. Conveyance efficiency

32.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

33.A geologic formation that contains water but do not have the capacity to transmit it. a. Aquifuge b. Aquifer

c. Aquitard d. Aquiclude

32.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

33.A geologic formation that contains water but do not have the capacity to transmit it. a. Aquifuge b. Aquifer

c. Aquitard d. Aquiclude

34.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

32.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

33.A geologic formation that contains water but do not have the capacity to transmit it. a. Aquifuge b. Aquifer

c. Aquitard d. Aquiclude

34.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

LSR = [P – (Res. M.C. X As)]D + Standing Water

35.Farm water requirement minus the application losses is the

a. Diversion water requirement c. Application efficiency b. Farm irrigation requirement d. Land preparation water requirement

35.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

36.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. 45 cm

c. 40 cm d. 60 cm

35.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

36.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. 45 cm

c. 40 cm d. 60 cm

37.This results from overlapping radii of influence of neighboring wells. a. Drawdown b. Groundwater decline

c. Well interference d. Drawdown curve

35.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

36.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. 45 cm

c. 40 cm d. 60 cm

37.This results from overlapping radii of influence of neighboring wells. a. Drawdown b. Groundwater decline

c. Well interference d. Drawdown curve

38.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

38.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

39.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

38.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

39.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

40.The localized lowering of the static or piezometric water level due to pumping

a. Groundwater decline b. Drawdown

c. Subsidence d. depression

38.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

39.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

40.The localized lowering of the static or piezometric water level due to pumping

a. Groundwater decline b. Drawdown

c. Subsidence d. depression

41.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 m d. 5 m

41.For a 10m x 10m sprinkler spacing, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m

5m

c. 8 m d. 5 m

2.5m

Ans. b. 7.5 m

41.For a 10m x 10m sprinkler spacing, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m

5m

c. 8 m d. 5 m

2.5m

Ans. b. 7.5 m

41.For a 10m x 10m sprinkler spacingm, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m

c. 8 m d. 5 m

42.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

42.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

4 li x m3 x 3600 sec = 14.4 m3/hr sec 1000 li hr 14.4 m3 x (3.28)3 ft3 x 7.48 gal x hr = 63.35 gpm 3 3 hr m ft 60 min

Ans. C. Both a and b

41.For a 10m x 10m sprinkler spacingm, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m

c. 8 m d. 5 m

42.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

43.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

41.For a 10m x 10m sprinkler spacingm, what is the design sprinkler throw for a 50% overlap? a. 15 m b. 7.5 m

c. 8 m d. 5 m

42.Four liters per second is equivalent to

a. 14.4 m /hr b. 63.41 gpm

c. Both a and b d. Neither a nor b

43.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

44.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

44.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

45.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

44.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

45.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

46.The infiltration equation based on the exhaustion process is the a. Lewis-Kostiakov’s b. Horton’s

c. Philip’s d. Darcy’s

44.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

45.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

46.The infiltration equation based on the exhaustion process is the a. Lewis-Kostiakov’s b. Horton’s

c. Philip’s d. Darcy’s

47.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 for June 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

47.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 for June 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

(FC – WP)(AMD)/100 (200-140)mm x (.75) Int = = CU 6 mm/day

Int = 7.5 or 7 days Ans. c. 7 days

47.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 for June 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

48.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

48.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

Q = Ad/t Q = discharge, A = area to be irrigated, d = depth of water to be applied, t = irrigation time 6 ha x 10,000 m2 x .008 m x 6 days x hr ha day 3600 sec Q= 9 hrs = 0.0889 m3/sec Ans. d. 0.0880 m3/sec

47.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 for June 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

48.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

49.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

47.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 for June 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

48.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

49.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

50.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

50.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 51.A mathematical expression for the macroscopic flow of water through a porous system. a. b. c. d.

Steady state groundwater flow equation Darcy’s Law Laplace’s equation Scobey;s equation

50.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 51.A mathematical expression for the macroscopic flow of water through a porous system. a. b. c. d.

Steady state groundwater flow equation Darcy’s Law Laplace’s equation Scobey;s equation

52.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

52.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

53. 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

52.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

53.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

54.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

52.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

53.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

54.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

55. 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

54. 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

55.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 m

c. 10 m d. 1.8 m

54. 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

55.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 m

c. 10 m d. 1.8 m

56.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

54. 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

55.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 m

c. 10 m d. 1.8 m

56.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

54. 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

Initial depth = Pv x D = (.24)(1.2 m x 1000 mm/m) = 288 mm Ans. c.

57.When the soil in Problem 56 reaches ____, irrigation should be done. a. 43.2 mm b. 270 mm

c. 288 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

WP + AMD = 180 + 90 = 270 mm Ans. b

}

90 mm

58. Natural drainage occurs in Problems 56 & 57 when the soil water reaches or exceeds a depth of a. 43.2 mm c. 288 mm b. 270 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. Ans. d.

57.When the soil in Problem 56 reaches ____, irrigation should be done. a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

58.Natural drainage occurs in Problems 56 & 57 when the soil water reaches or exceeds a depth of a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

57.When the soil in Problem 56 reaches ____, irrigation should be done. a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

58.Natural drainage occurs in Problems 56 & 57 when the soil water reaches or exceeds a depth of a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

59.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

57.When the soil in Problem 56 reaches ____, irrigation should be done. a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

58.Natural drainage occurs in Problems 56 & 57 when the soil water reaches or exceeds a depth of a. 43.2 mm b. 270 mm

c. 288 mm d. 360 mm

59.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

60.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

60.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

61.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

Scobey’s Formula: Hf = [ksLQ1.9 /(d4.9)]1.45 x 10-8 d = ft, L = ft, Q = gpm, H = ft

60.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

61.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

62.Which is not a component of the impact arm of an impact sprinkler? a. Nozzle b. Counterweight

c. Vane d. spoon

60.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

61.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

62.Which is not a component of the impact arm of an impact sprinkler? a. Nozzle b. Counterweight

c. Vane d. spoon

63.A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4-inch 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

63.A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4-inch 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

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

63.A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4-inch 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

64.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

63.A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4-inch 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

64.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

65.It is an orderly sequence of planting crop in an area for a 365day period. a. Cropping pattern b. Crop combination

c. Crop sequence d. Cropping schedule

63.A 20-ft thick confined aquifer with hydraulic conductivity of 400 ft/day was tapped by a 4-inch 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

64.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

65.It is an orderly sequence of planting crop in an area for a 365day period. a. Cropping pattern b. Crop combination

c. Crop sequence d. Cropping schedule

66.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

21.6 mm x m x 10,000 m2 x day x 1000 li day 1000 mm ha 86,400 sec m3 = 2.5 lps/ha Ans. c.

66.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

21.6 mm x m x 10,000 m2 x day x 1000 li day 1000 mm ha 86,400 sec m3 = 2.5 lps/ha Ans. c.

67.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

dnet = (AMD)(FC-WP)(D) = (.6)(.25-.15)(800mm) = 48 mm Ans. b.

67.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

dnet = (AMD)(FC-WP)(D) = (.6)(.25-.15)(800mm) = 48 mm Ans. b.

68.The gross depth of irrigation water to be applied in Prob. 67.

a. 100 mm b. 60 mm

c. 64 mm d. 38 mm

dgross = dnet /Ea = 48/.8 = 60 mm

68.The gross depth of irrigation water to be applied in Prob. 67.

a. 100 mm b. 60 mm

c. 64 mm d. 38 mm

dgross = dnet /Ea = 48/.8 = 60 mm Ans. b

69.The irrigation interval, in days in Prob. 67 is

a. 10 b. 5

c. 9 d. 4

Int = dnet /CU = (48 mm)/(5 mm/day) = 9.6 or 9 days

69.The irrigation interval, in days in Prob. 67 is

a. 10 b. 5

c. 9 d. 4

Int = dnet /CU = (48 mm)/(5 mm/day) = 9.6 or 9 days Ans. c

70.The irrigation period, in hours in Prob. 67 is

a. 93 b. 47

c. 230 d. 94

Irrigation period = Adgross /Q 5 ha x 10,000 m2 x .06 m ha 32 m3/hr Ans. d

= 93.75 hrs

69.The irrigation interval, in days in Prob. 67 is

a. 10 b. 5

c. 9 d. 4

70.The irrigation period, in hours in Prob. 67 is a. 93 b. 47

c. 230 d. 94

71.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

69.The irrigation interval, in days in Prob. 67 is

a. 10 b. 5

c. 9 d. 4

70.The irrigation period, in hours in Prob. 67 is a. 93 b. 47

c. 230 d. 94

71.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

72.A rectangular piece of land 180m x 240m is laid out with onesided 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. 0.500 lps b. 0.375 lps

c. 0.250 lps d. 0.125 lps

72.A rectangular piece of land 180m x 240m is laid out with onesided 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. 0.500 lps b. 0.375 lps

c. 0.250 lps d. 0.125 lps

72.A rectangular piece of land 180m x 240m is laid out with onesided 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. 0.500 lps b. 0.375 lps

c. 0.250 lps d. 0.125 lps

(6m) x (6m) x (.15 m) x hr Q=

Ans. c

6 hrs

3600 sec

x 1000 li

m3

= 0.25 lps

72.A rectangular piece of land 180m x 240m is laid out with onesided 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. 0.500 lps b. 0.375 lps

c. 0.250 lps d. 0.125 lps

73.Determine the number of lateral positions in Prob. 72. a. 30 b. 20

c. 40 d. 60

73.Determine the number of lateral positions in Prob. 72.

a. 30 b. 20

c. 40 d. 60 240 m

180 m

No. of lateral positions = 180/6 = 30

73.Determine the number of lateral positions in Prob. 72.

a. 30 b. 20

c. 40 d. 60 240 m

180 m

No. of lateral positions = 180/6 = 30 Ans. a

73.Determine the number of sprinklers/lateral in Prob. 72.

a. 30 b. 20

c. 40 d. 60 240 m

180 m

No. of sprinkler/lateral = 240/6 = 40

73.Determine the number of sprinklers/lateral in Prob. 72.

a. 30 b. 20

c. 40 d. 60 240 m

180 m

No. of sprinkler/lateral = 240/6 = 40 Ans. c

72.A rectangular piece of land 180m x 240m is laid out with onesided 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. 0.500 lps b. 0.375 lps

c. 0.250 lps d. 0.125 lps

73.Determine the number of lateral positions in Prob. 72. a. 30 b. 20

c. 40 d. 60

74.Determine the number of sprinklers per lateral in Prob. 72. a. 30 b. 20

c. 40 d. 60

75.The field in Prob. 72 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

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

75.The field in Prob. 72 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

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 Ans. c

76.An unconfined aquifer is situated above a horizontal impervious base and is composed of san with a hydraulic conductivity of 8 m/day. In this aquifer, two fully penetrating ditches from a strip of land with a constant width of 1200 m. The water level in the left and right ditches rises to 6 and 10 m, respectively above the impervious base. The recharge by rainfall is 7 mm/day. Evaporation losses amount to 2 mm/day. Determine the location of the stagnation point of the groundwater flow. a. 557 from left drain b. 643 from left drain

c. 557 from right drain d. 587 from right drain

76.An unconfined aquifer is situated above a horizontal impervious base and is composed of san with a hydraulic conductivity of 8 m/day. In this aquifer, two fully penetrating ditches from a strip of land with a constant width of 1200 m. The water level in the left and right ditches rises to 6 and 10 m, respectively above the impervious base. The recharge by rainfall is 7 mm/day. Evaporation losses amount to 2 mm/day. Determine the location of the stagnation point of the groundwater flow. a. 557 from left drain c. 557 from right drain b. 643 from left drain d. 587 from right drain

77.Compute the flow into the right drain of in Prob. 76.

a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

77.Compute the flow into the right drain in Prob. 76.

a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

78.Compute the flow into the left drain in Prob. 76. a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

77.Compute the flow into the right drain in Prob. 76.

a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

78.Compute the flow into the left drain in Prob. 76. a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

79.Compute the maximum and minimum elevation of water table inside the strip of land in Prob. 76.

a. 11.259 m b. 12.302 m

c. 11.921 m d. 12.426 m

77.Compute the flow into the right drain in Prob. 76.

a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

78.Compute the flow into the left drain in Prob. 76. a. 2.787 m3/day b. 3.213 m3/day

c. 0.213 m3/day d. 4.287 m3/day

79.Compute the maximum and minimum elevation of water table inside the strip of land in Prob. 76.

a. 11.259 m b. 12.302 m

c. 11.921 m d. 12.426 m