II. Irrigation and Drainage Engineering (45%) EASY Flashcards
1
Q
- 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
2
Q
- 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
3
Q
- Application of water in the soil to supply moisture needed for plant growth.
A. Flooding
B. Sprinkling
C. Irrigation
D. Diverting
A
Answer: C
4
Q
- 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
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
6
Q
- 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
7
Q
- 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
8
Q
- Number of days between irrigation applications.
A. Irrigation interval
B. Application interval
C. Dry interval
D. Node interval
A
Answer: A
9
Q
- Removal of excess water.
A. Squeezing
B. Run off
C. Discharging
D. Drainage
A
Answer: D
10
Q
- Elevated section of open channel used for crossing natural depressions.
A. Parshall flume
B. Flume
C. Siphon
D. Elevated channel
A
Answer: B
11
Q
- Surveying instrument used for determining land areas in a topographic maps.
A. Aerometer
B. Erometer
C. Planimeter
D. Lysimeter
A
Answer: C
12
Q
- 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
13
Q
- Facility for determining water consumptive use of crops in an open field.
A. Planimeter
B. Lysimeter
C. Consumeter
D. Crop meter
A
Answer: B
14
Q
- 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
15
Q
- 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
16
Q
- 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
- 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
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18
Q
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
