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Question 1
ప్రశ్న 1
What is meant by a hydraulically most economical open-channel section when flow area, channel slope and resistance characteristics are prescribed?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydraulic radius is R = A/P, where A is flow area and P is wetted perimeter. For a prescribed area, reducing the wetted perimeter increases the hydraulic radius. Both Chezy and Manning resistance equations predict greater conveyance when hydraulic radius increases, if slope and resistance coefficient are unchanged. Thus minimizing wetted perimeter for a given area is equivalent to maximizing hydraulic efficiency. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
For two channel shapes having the same flow area, slope and Manning roughness coefficient, which geometric change maximizes the uniform-flow discharge?
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Explanation:
Manning discharge is Q = (A/n)R^(2/3)S^(1/2). When A, n and S are fixed, discharge increases as hydraulic radius R increases. Since R = A/P, maximizing R for a fixed area requires minimizing wetted perimeter P. This is the geometric basis of the most economical channel section. Therefore Option D is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a rectangular channel of width b and flow depth y, what is the condition for the most economical rectangular section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a rectangular channel, A = by and P = b + 2y. For a prescribed area A, b may be written as A/y, so P = A/y + 2y. Differentiating P with respect to y and setting the derivative to zero gives A = 2y². Since A = by, this gives b = 2y. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
What is the hydraulic radius of a most economical rectangular channel of flow depth y?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the most economical rectangular section, b = 2y. The area is therefore A = 2y². The wetted perimeter is P = b + 2y = 4y. Hence R = A/P = 2y²/(4y) = y/2. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
A most economical rectangular channel has a flow area of 18 m². What are its depth, bed width and hydraulic radius?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the economical rectangular section, b = 2y. Therefore A = by = 2y². Setting 2y² = 18 gives y² = 9 and hence y = 3 m. The width is b = 6 m, and R = y/2 = 1.5 m. Therefore Option D is correct.
వివరణ:
Question 6
ప్రశ్న 6
The wetted perimeter of a most economical rectangular channel is 16 m. What are its flow depth and bed width?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the most economical rectangle, b = 2y. Its wetted perimeter is therefore P = b + 2y = 4y. With P = 16 m, the depth is y = 16/4 = 4 m. The corresponding width is b = 2y = 8 m. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
Which relationship among width b, depth y and wetted perimeter P is valid for a most economical rectangular channel?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The economical rectangular condition is b = 2y. Its wetted perimeter is P = b + 2y. Substituting b = 2y gives P = 4y. Since 2b is also equal to 4y, the relation becomes P = 2b = 4y. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
A most economical rectangular channel is to carry approximately 21.33 m³/s. Using Manning's equation with n = 0.015 and bed slope S = 0.0016, what dimensions are required approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For an economical rectangle, A = 2y² and R = y/2. Manning's equation gives Q = (1/n)(2y²)(y/2)^(2/3)S^(1/2). Substituting n = 0.015, S = 0.0016 and Q = 21.33 m³/s gives y approximately 2.00 m. The economical width is then b = 2y = approximately 4.00 m. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A most economical rectangular channel carries 16 m³/s under uniform flow. If Chezy coefficient C = 50 m^(1/2)/s and slope S = 0.0016, what are the required dimensions approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the economical rectangle, A = 2y² and R = y/2. Chezy velocity is V = C√(RS), so Q = 2y²C√[(y/2)S]. Substituting C = 50 and S = 0.0016 shows that y = 2 m gives V = 2 m/s. The corresponding area is 8 m² and Q = 8 × 2 = 16 m³/s, while b = 4 m. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
For a trapezoidal channel with a prescribed side slope, which geometrical condition applies to the most economical section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a trapezoid with fixed side slope, the economical proportions are obtained by minimizing wetted perimeter for a prescribed area. The resulting geometric condition is that half of the top width equals the length of one sloping side. If T is top width and l is one sloping side, the condition is T/2 = l. This relationship leads directly to the standard formula for economical bottom width. Therefore Option D is correct.
వివరణ:
Question 11
ప్రశ్న 11
A trapezoidal channel has side slope z horizontal to 1 vertical, depth y and bed width b. For the most economical section when z is prescribed, which expression for b is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The top width of the trapezoid is T = b + 2zy. The sloping-side length is l = y√(1 + z²). For the economical section, T/2 = l. Thus b/2 + zy = y√(1 + z²), which gives b = 2y[√(1 + z²) - z]. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the most economical trapezoidal section with a prescribed side slope, what is the hydraulic radius in terms of flow depth y?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a trapezoid with prescribed side slope z, optimization gives b = 2y[√(1 + z²) - z]. Substitution of this condition into the expressions for area and wetted perimeter simplifies their ratio. The resulting hydraulic radius is R = y/2. This important result is independent of the particular prescribed value of z once the section is economically proportioned. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A most economical trapezoidal channel has side slope 1 horizontal to 1 vertical and flow depth 2.0 m. What are its bed width and hydraulic radius approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Here z = 1 and y = 2 m. The economical bed width is b = 2y(√2 - 1) = 4(0.4142), which is approximately 1.657 m. The hydraulic radius of an economical trapezoid with fixed side slope is R = y/2. Therefore R = 1.0 m and Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A most economical trapezoidal channel has prescribed side slope z = 0.5 horizontal to 1 vertical and bed width b = 2.0 m. What is its flow depth approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the economical trapezoid, b = 2y[√(1 + z²) - z]. With z = 0.5, the bracket becomes √1.25 - 0.5 = approximately 0.6180. Hence 2.0 = 2y(0.6180). Solving gives y approximately 1.618 m. Therefore Option D is correct.
వివరణ:
Question 15
ప్రశ్న 15
A most economical trapezoidal channel has side slope 1 horizontal to 1 vertical and a required flow area of approximately 7.314 m². What are its depth and bed width approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For z = 1, the economical width is b = 2y(√2 - 1). The area becomes A = y(b + y) = y²(2√2 - 1). Setting this equal to approximately 7.314 m² gives y approximately 2.0 m. The corresponding bottom width is 2 × 2(√2 - 1) = approximately 1.657 m. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
If both the bed width and side slope of a trapezoidal channel are allowed to vary to obtain the hydraulically most efficient trapezoidal section, what is the optimum side slope?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
When the side slope itself is also optimized, the wetted perimeter must be minimized with respect to both channel depth proportions and side inclination. The resulting side slope is z = 1/√3 horizontal to 1 vertical. A line having horizontal projection 1/√3 for a unit vertical rise makes an angle of 60 degrees with the horizontal. This produces the familiar half-hexagonal form of the most efficient trapezoid. Therefore Option A is correct.
వివరణ:
Question 17
ప్రశ్న 17
A fully optimized trapezoidal channel has flow depth y = 3.0 m, with both bed width and side slope chosen for maximum hydraulic efficiency. What are the bed width, top width and hydraulic radius approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the fully optimized trapezoid, z = 1/√3 and b = 2y/√3. With y = 3 m, b = 6/√3 = approximately 3.464 m. The top width is T = b + 2zy = 4y/√3 = approximately 6.928 m. The hydraulic radius remains R = y/2 = 1.5 m. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Which geometric relationship is true for the fully optimized trapezoidal channel in which the side slope is also allowed to vary?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the fully optimized trapezoid, z = 1/√3. The bed width is b = 2y/√3. The length of one sloping side is y√(1 + 1/3) = 2y/√3, which equals b. The top width is b + 2zy = 4y/√3 = 2b. Therefore Option D is correct.
వివరణ:
Question 19
ప్రశ్న 19
For a fully optimized trapezoidal channel of depth y, which expression gives its flow area?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the fully optimized trapezoid, z = 1/√3 and b = 2y/√3. The trapezoidal area is A = y(b + zy). Substitution gives A = y[(2y/√3) + (y/√3)]. This simplifies to A = √3 y². Therefore Option A is correct.
వివరణ:
Question 20
ప్రశ్న 20
For a symmetric triangular channel whose side slope is free to vary, what geometry gives the most economical triangular section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a symmetric triangular channel, A = zy² and P = 2y√(1 + z²). At fixed area, substitution for y reduces the wetted-perimeter minimization to minimizing z + 1/z. Differentiation gives the optimum at z = 1. Thus each side has slope 1 horizontal to 1 vertical and the included vertex angle is 90 degrees. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
A most economical symmetric triangular channel carries water to a depth of 2.0 m. What are its flow area and hydraulic radius?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The most economical symmetric triangular channel has z = 1. Its area is A = zy² = 1 × 2² = 4.0 m². The wetted perimeter is P = 2y√(1 + z²) = 4√2 = approximately 5.657 m. Therefore R = A/P = 4/5.657 = approximately 0.707 m. Hence Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
Two channel sections have the same flow area of 12 m², the same Manning n and the same slope. Their wetted perimeters are 8 m and 10 m respectively. What is the ratio of the discharge through the first section to that through the second section?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The hydraulic radii are R1 = 12/8 = 1.5 m and R2 = 12/10 = 1.2 m. Since the areas, roughness coefficients and slopes are identical, Manning's equation gives Q1/Q2 = (R1/R2)^(2/3). Thus Q1/Q2 = (1.5/1.2)^(2/3) = 1.25^(2/3). This evaluates to approximately 1.160. Therefore Option D is correct.
వివరణ:
Question 23
ప్రశ్న 23
A trapezoidal channel with side slope 1 horizontal to 1 vertical is to be proportioned as the most economical section for that prescribed side slope. It must carry approximately 15.42 m³/s with Manning n = 0.015 and slope S = 0.001. What are the required depth and bed width approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For z = 1, the economical bed width is b = 2y(√2 - 1). The area is A = (2√2 - 1)y² and the hydraulic radius is R = y/2. Substituting these expressions into Manning's equation with n = 0.015 and S = 0.001 gives y approximately 2.00 m for Q = 15.42 m³/s. The corresponding bed width is approximately 1.657 m. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which statement is most appropriate when applying the concept of a hydraulically most economical section to an actual canal design?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The most economical section is a hydraulic-geometric optimum based primarily on minimizing wetted perimeter for a required area or maximizing conveyance. Real channels are also subject to geotechnical, structural and operational requirements. Side slopes may be limited by soil stability, while permissible velocity may depend on erosion, lining and sediment considerations. Freeboard and construction practicality can also require dimensions different from the mathematical optimum. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about economical open-channel sections: I. A most economical rectangular channel satisfies b = 2y. II. For a most economical trapezoid with prescribed side slope, R = y/2. III. For that trapezoid, half the top width equals one sloping-side length. IV. If the side slope is also optimized, the trapezoidal side makes 60 degrees with the horizontal. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is the standard condition obtained by minimizing the wetted perimeter of a rectangular section for fixed area. Statement II is the resulting hydraulic-radius property of the optimized trapezoid with a prescribed side slope. Statement III is the equivalent geometric condition T/2 = l for that trapezoidal optimization. Statement IV follows when the side inclination itself is additionally optimized, giving a 60-degree side angle to the horizontal. Therefore all four statements are correct and Option D is the valid answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: A. A section having the minimum wetted perimeter and hence the maximum hydraulic radius for the prescribed area సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: D. Minimizing wetted perimeter సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: B. b = 2y సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. R = y/2 సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: D. y = 3 m, b = 6 m, R = 1.5 m సమాధానం: D.
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Question 6 ప్రశ్న 6Answer: B. y = 4 m and b = 8 m సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: A. P = 2b = 4y సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: C. y = 2.0 m and b = 4.0 m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. y = 2.0 m and b = 4.0 m సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: D. Half the top width equals the length of one sloping side సమాధానం: D.
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Question 11 ప్రశ్న 11Answer: A. b = 2y[√(1 + z²) - z] సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: C. R = y/2 సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. b = 1.657 m and R = 1.0 m సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: D. 1.618 m సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: C. y = 2.0 m and b = 1.657 m సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: A. z = 1/√3 horizontal to 1 vertical, so the side makes 60 degrees with the horizontal సమాధానం: A.
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Question 17 ప్రశ్న 17Answer: B. b = 3.464 m, T = 6.928 m and R = 1.5 m సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: D. Each sloping side equals the bed width and the top width equals twice the bed width సమాధానం: D.
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Question 19 ప్రశ్న 19Answer: A. A = √3 y² సమాధానం: A.
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Question 20 ప్రశ్న 20Answer: C. Side slope 1 horizontal to 1 vertical, giving a 90-degree vertex angle సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. A = 4.0 m² and R = 0.707 m సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: D. 1.160 సమాధానం: D.
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Question 23 ప్రశ్న 23Answer: A. y = 2.0 m and b = 1.657 m సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: C. The hydraulic optimum is an important starting point, but practical design may also be governed by side-slope stability, lining, sediment, freeboard and construction requirements సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.