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Irregular sa Class > II. Irrigation and Drainage Engineering (45%) EASY > Flashcards

II. Irrigation and Drainage Engineering (45%) EASY Flashcards

1
Q
  1. The maximum permissible water velocity for clay loam canal surface based on PAES
    603: 2016.

A. 1.2 m/s
B. 1 m/s
C. 0.9 m/s
D. 0.80 m/s

A

Answer: D

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2
Q
  1. The minimum permissible velocity for water with sediments in lined canals based on PAES 603:2016.

A. 1.2 m/s
B. 1 m/s
C. 0.9 m/s
D. 0.80 m/s

A

Answer: C

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3
Q
  1. Application of water in the soil to supply moisture needed for plant growth.

A. Flooding
B. Sprinkling
C. Irrigation
D. Diverting

A

Answer: C

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4
Q
  1. Loss of water from a channel during transport due to seepage and percolation.

A. Channel loss
B. Seepage loss
C. Percolation loss
D. Conveyance loss

A

Answer: D

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5
Q 
5. Depth of water flow where the energy content is at minimum hence, no other backwater forces are
involved.
A. Minimum depth 
B. Critical depth 
C. Energy depth 
D. Normal depth
A

Answer: B

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6
Q
  1. Ratio of the actual crop evapotranspiration to its potential evapotranspiration.

A. Crop ratio
B. ET ratio
C. Crop coefficient
D. Evaporation ratio

A

Answer: C

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7
Q
  1. Moisture content of the soil when gravitational water has been removed.

A. Soil capacity
B. Gravitational moisture
C. Field capacity
D. Specific capacity

A

Answer: C

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8
Q
  1. Number of days between irrigation applications.

A. Irrigation interval
B. Application interval
C. Dry interval
D. Node interval

A

Answer: A

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9
Q
  1. Removal of excess water.

A. Squeezing
B. Run off
C. Discharging
D. Drainage

A

Answer: D

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10
Q
  1. Elevated section of open channel used for crossing natural depressions.

A. Parshall flume
B. Flume
C. Siphon
D. Elevated channel

A

Answer: B

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11
Q
  1. Surveying instrument used for determining land areas in a topographic maps.

A. Aerometer
B. Erometer
C. Planimeter
D. Lysimeter

A

Answer: C

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12
Q
  1. Elevation of water surface in a stream with reference to a certain datum.

A. Stage
B. Surface elevation
C. Contour
D. Water elevation

A

Answer: A

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13
Q
  1. Facility for determining water consumptive use of crops in an open field.

A. Planimeter
B. Lysimeter
C. Consumeter
D. Crop meter

A

Answer: B

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14
Q
  1. Time required to cover an area with one application of water.

A. Irrigation interval
B. Irrigation period
C. Supply duration
D. Application time

A

Answer: B

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15
Q
  1. At optimal emitter spacing, drip emitter spacing is ___ of the wetted diameter estimated from field tests.

A. 100%
B. 90%
C. 80%
D. 85%

A

Answer: C

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16
Q
  1. Reference crop evapotranspiration is the rate of evapotranspiration from a reference surface which is a hypothetical reference crop with an assumed crop height of 0.2 m and an albedo of _______ (AMTEC, 2020).

A. 0.23
B. 0.25
C. 0.30
D. 0.32

A

Answer: A

17
Q
  1. Manufacturer’s coefficient of variation is the measure of the variability of discharge of a random sample of a given make, model and size of emitter, as provided by the manufacturer and before any field operations or aging has taken place determined through a discharge test of a sample of 50 emitters under a set pressure at ___ ºC.

A. 200
B. 100
C. 50
D. 30

A

do ur part :>

18
Q
  1. Which one is the flattest canal side slope?

A. 1:1
B. 1:4
C. 4:1
D. 2:1

A

Answer: C

19
Q

Moderate Questions
19. Determine the side slope angle Ө with the horizontal plane of an unlined canal with side slope ratio (run: rise) z of 2:1.

A. 16.6º
B. 26.6º
C. 45º
D. 60º

A

Answer: B

Solution:
 tan Ө = rise /run
Ө = arctan (rise /run)
 = arctan (1/2)
 = 26.6º

Notes:
In specifying side slope, run is written first, ex. 4:1 means horizontal run is 4. In computing angles whether bed slope or side slope, rise is the numerator. Some fluid mechanics and hydraulics books use Ө symbol for side angle with the vertical plane, not with the horizontal. In this case subtract Ө from 90° to get angle with the horizontal plane.

20
Q
  1. If the most efficient concrete canal has its side angle Ө with the horizontal plane equal to 60º, what is the z value of the canal sides or the side’s horizontal run in meters per 1 meter rise? This value is commonly used in designing most efficient concrete canals.

A. 0.775
B. 0.757
C. 0.577
D. 1/0.577

A

Answer: C

Solution:
tan Ө = 1 / z
tan 60º = 1 / z
1.732 = 1 / z
z = 0.577
21
Q
  1. What is the top width at water surface level of the most efficient concrete open channel if the design depth is 5 meters? The design discharge is 100 m3/s and velocity is 2 m/s.

A. 1.29 m
B. 9.12 m
C. 12.9 m
D. 19.2 m

A

Answer: C

Solution:
If the canal is a trapezoidal concrete and side angle is not given, then it is a Most Efficient
Canal (in which Ө=60º and z=0.577). Specifying it as most efficient canal is often omitted in
the board question.
 Q = AV
 100 = A (2)
 A = 50 m2
 tan Ө = 1 / z
 tan 60 º = 1 / z
 1.732 = 1 / z
 z = 0.577
 A = bd + zd2
 A-zd2 = bd
 b = (A-zd2
) / d
 = [50 – 0.577(5) 2
] / 5
 = 7.12 m
 t = b + (2d / tan Ө)
 = 7.12 m + [2(5) / 1.732]
 = 12.89 m
 Checking 1:
 A = bd + zd2
 50 = (7.12)(5) + 0.577(5)2
 50 = 35.6+14.4
 50 = 50
 Checking 2:
 A = d [(t + b)/2]
 50 = 5 [(12.89+7.12)/2]
 50 = 5[10]
 50 = 50
22
Q
  1. What is the total top width of the most efficient concrete open channel if design depth is 5meters? Design discharge is 100 m3/s and velocity is 2 m/s.
    Use 15% freeboard. 14

A. 12.9 m
B. 13.8 m
C. 18.3 m
D. 8.13 m

A

Answer: B

Solution:
Use the Design Criteria: Most Efficient Canal (Ө=60º and z = 0.577)
 Q = AV
 100 = A (2)
 A = 50 m2
 A = bd + zd2
 A-zd2 = bd
 b = (A-zd2
) / d
 = [50 – 0.577(5) 2
] / 5
 = 7.12 m
 t = b + (2d / tan Ө)
 = 7.12 m + [2(5) / 1.732]
 = 12.89 m
 D= 1.15d
 = 1.15(5)
 = 5.75 m
 T = b + (2D / tan Ө)
 = 7.12 + [2(5.75) / 1.732]
 = 7.12 + 6.64
 = 13.8 m
 Checking 1:
 A = bd + zd2
 50 = (7.12)(5) + 0.577(5)2
 50 = 35.6+14.4
 50 = 50
 Checking 2:
 A = d [(t + b)/2]
 50 = 5 [(12.89+7.12)/2]
 50 = 5[10]
 50 = 50
15
23. What is the base of the most efficient trapezoidal concrete open channel if discharge is 100
m3
/s and velocity is 2 m/s?
A. 6.14 m B. 12.8 m C. 7.21 m D. 14.6 m
 Answer: A

Solution:

 Use the same approach as the previous problem but find the canal base.
Q = AV
100 = A (2)
A = 50 m2
A = 1.732 d2
50/1.732 = d
2
d = 5.4 m
 A = bd + zd2
 b = (A-zd2
) / d
 = [50 – 0.577(5.4) 2
] / 5.4
 = 6.14 m
 Checking 1:
A = bd + zd2
50 = (6.14)(5.4) + 0.577(5.4)2
50 = 33.2+16.8
50 = 50
23
Q
  1. What is the total top width of the most efficient concrete open channel if design depth is 5meters? Design discharge is 100 m3/s and velocity is 2 m/s. Use 15% freeboard 14

A. 12.9 m
B. 13.8 m
C. 18.3 m
D. 8.13 m

A 
Answer: B
Solution:
Use the Design Criteria: Most Efficient Canal (Ө=60º and z = 0.577)
 Q = AV
 100 = A (2)
 A = 50 m2
 A = bd + zd2
 A-zd2 = bd
 b = (A-zd2
) / d
 = [50 – 0.577(5) 2
] / 5
 = 7.12 m
 t = b + (2d / tan Ө)
 = 7.12 m + [2(5) / 1.732]
 = 12.89 m
 D= 1.15d
 = 1.15(5)
 = 5.75 m
 T = b + (2D / tan Ө)
 = 7.12 + [2(5.75) / 1.732]
 = 7.12 + 6.64
 = 13.8 m
 Checking 1:
 A = bd + zd2
 50 = (7.12)(5) + 0.577(5)2
 50 = 35.6+14.4
 50 = 50
 Checking 2:
 A = d [(t + b)/2]
 50 = 5 [(12.89+7.12)/2]
 50 = 5[10]
 50 = 50
24
Q
  1. What is the bottom width for the best hydraulic cross-section (best proportion) of concrete open channel if design depth is 5 meters and side slope is 45º?

A. 3 m
B. 4 m
C. 5 m
D. 6 m

A

Answer: B

Solution:
Since the concrete canal has a side angle other than 60°, then use the design criteria: Best Hydraulic Cross-Section and use the formula for concrete canals (coefficient of d is 2).
 b = 2 d tan (Ө/2)
= 2 (5) tan (45/2)
= 2 (5) tan (22.5)
= 2 (5) (0.41)
= 4.1 m
25
Q
  1. What is the bottom width for best hydraulic cross-section of unlined open channel for minimum seepage if design depth is 5 meters and side slope is 45º?
    A. 3.15 m B. 4.15 m C. 8.15 m D. 6.15 m
A

Answer: C

Solution:
Since the canal is unlined or not concrete, then use the design criteria: Best Hydraulic CrossSection and use the formula for unlined canals (coefficient of d is 4).
 b = 4 d tan (Ө/2)
= 4 (5) tan (22.5)
= 8.2 m
26
Q
  1. What is the bottom width for best hydraulic cross-section of unlined open channel with minimum seepage if design depth is 5 meters and side slope is 2:1?

A. 4.72 m
B. 7. 42 m
C. 2.47 m
D. 7.24 m

A

Answer: A

Solution:
Compute for the side angle then use the design criteria: Best Hydraulic Cross-Section and
then use the formula for unlined canals (coefficient of d is 4).
 Ө = arctan (rise /run)
= arctan (1/2)
= 26.6º
 b = 4 d tan (Ө/2)
= 4 (5) tan (26.6/2)
= 4 (5) tan (13.3)
= 4 (5) (0.236)
= 4.72 m
27
Q
  1. Estimate the width and depth of concrete-lined rectangular open channel for water velocity of 2 m/s and discharge of 10 m3/s.

A. 6.1 m, 2.3 m
B. 3.2 m, 1.6 m
C. 2.5m, 5.0 m
D. 13.6 m, 3.1 m

A 
Answer: B
Solution:
17
It is specified that the concrete canal is rectangular. Hence, it cannot qualify for Most
Efficient criterion. Use the design criteria: Efficient Canal and then use the formula for
concrete-lined rectangular open channels.
Expressing b in terms of d:
 b = 2d
Determining A:
 A = Q/V
= 10/2
= 5 m2
Solving for b and d:
 A= bd
 5 = (2d)d
 d
2 = 5/2
 d = 1.58 m
 b = 2 (1.58)
= 3.16 m
28
Q
  1. What should be the base and depth of concrete-lined rectangular open channel for a cross-sectional area of 50 m2
    ? Design for efficiency over proportion.

A. 10 m, 5 m
B. 12 m, 6 m
C. 2.5 m, 5 m
D. 3 m, 6 m

A

Answer: A

Solution:
From this item until Item 32 in Irrigation and Drainage Engineering, determine what criteria
to apply and what formulas to use as part of your practice in board problem analysis.
A = 2 d2
50 =2 d2
d = 5 m
b = 2d
b = 2(5)
b = 10 m
 Checking:
 A = bd
 50 = (10)(5)
 50 = 50
29
Q
  1. What should be the depth and side angle with the horizontal of concrete-lined triangular open\channel for a cross-sectional area of 50 m2?

A. 5 m, 16.6º
B. 6 m, 26.6º
C. 7 m, 45º
D. 8 m, 60º

A

Answer: C

Solution:
A = d2
50 = d2
d = 7.1 m
Ө = 45º for efficient triangular canals.
30
Q
  1. What design depth of open channel would you recommend to carry 100 cumecs or cubic meters/sec with a velocity of 5 mps? Use the most efficient of all trapezoidal cross-sections.

A. 1.4 m
B. 2.4 m
C. 3.4 m
D. 1.3 m

A 
Answer: C
Solution:
 A= Q/V
= 100/5
= 20 m2
 A = 1.732 d2
 20 = 1.732 d2
 d = 3.4 m
31
Q
  1. If the most efficient of all trapezoidal cross-sections can be used, what actual depth of open channel would you recommend to carry 100 cumecs with a velocity of 5 mps? Use 15%
    freeboard.

A. 3.9 m
B. 3.5 m
C. 3.6 m
D. 1.3 m

A 
Answer: A
Solution:
 A= Q/V
= 100/5
= 20 m2
 A = 1.732 d2
 20 = 1.732 d2
 d = 3.4 m
 D = 1.15 d
 = 1.15 (3.4)
 = 3.91 m
32
Q
  1. If an unlined trapezoidal canal with best hydraulic cross-section can be used, what actual depth of open channel would you recommend to carry 10 cumecs with a velocity of 1 mps?
    Use 2:1 side slope and 15% freeboard.

A. 1.12 m
B. 2.12 m
C. 21.2 m
D. 2.21 m

A 
Answer: B
Solution:
A = Q/V
 = 10/1
 = 10 m2
Ө = arctan (rise /run)
 = arctan (1/2)
 = 26.565º
b = 4 d tan (Ө/2)
 = 0.944 d
A = bd + zd2,
10 = 0.944d + zd2
but z = 2
10 = 0.944d + 2d2
but z = 2
10 = 2.944d2
d = 1.84 m
D = 1.15 d
 = 1.15 (1.84 m)
 = 2.12m
Checking:
tan (26.565º) = 0.5
b = 0.944 d
 = 0.944 (1.84m)
 = 1.74 m
t = b + (2d / tan Ө)
 = 1.74 + [2(1.8 4) / 0.5]
 = 9.11 m
A = d [(t + b)/2]
10 = 1.84 [(9.11+1.74)/2]
10 = 10

Irregular sa Class (6 decks)

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