Irrigation & Drainage Engineering Flashcards
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
Twelve thousand five hundred (12,500) 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%
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
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
Vol. = (FC – PWP)(D)(A)
= (0.32 - 0.15)(0.8m)(60,000 m2)
= 8,160 m3
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
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
B - 2.197
P1 (2340/1800)^3 = P2
2.197 P1 = P2
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
C – porosity
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
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
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
b. 17.55
c. 5.59
d. 6.59
B – 17.55
Q = 2πkb(h2 – h1) / ln(r2/r1)
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
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
int = (FC – WP)(AMD) / CU
= ((200 – 105)mm x 0.75) / 7.5 mm/day
Iint = 9.5 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 m^3/s / 0.85 m/s = 3.765 m^2
A = by + zy^2
3.765 = 1.5y + 2y^2
Compute for y = 1.05 m
The localized lowering of the static or piezometric water level due to
pumping.
a. Groundwater decline
b. Drawdown
c. Subsidence
d. Depression
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) x7 days) / 12 hrs x 1hr/3600sec
Q = 0.0648 m3/s = 0.065 m3/s