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

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

Answer: D

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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

Answer: C

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

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

Answer: C

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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

Answer: D

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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

Answer: B

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Ratio of the actual crop evapotranspiration to its potential evapotranspiration.

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

Answer: C

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

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

Answer: C

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Number of days between irrigation applications.

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

Answer: A

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Removal of excess water.

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

Answer: D

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Elevated section of open channel used for crossing natural depressions.

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

Answer: B

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

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

Answer: C

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

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

Answer: A

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

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

Answer: B

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

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

Answer: B

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

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

Answer: C

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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

Answer: A

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

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Which one is the flattest canal side slope?

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

Answer: C

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

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.

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

Answer: C

Solution:
tan Ө = 1 / z
tan 60º = 1 / z
1.732 = 1 / z
z = 0.577

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

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

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

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

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

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

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

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. 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

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. 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

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. 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

``` 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. 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

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. 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

Answer: C ``` Solution: A = d2 50 = d2 d = 7.1 m Ө = 45º for efficient triangular canals. ```

30. 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

``` Answer: C Solution: A= Q/V = 100/5 = 20 m2 A = 1.732 d2 20 = 1.732 d2 d = 3.4 m ```

31. 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

``` 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. 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

``` 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 ```