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Question 1
ప్రశ్న 1
For a gravity sewer carrying sewage below the crown with a free surface, which hydraulic description is correct?
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Explanation:
A gravity sewer flowing partially full has a free surface exposed to atmospheric pressure. Its behaviour can therefore be analysed as open-channel flow. Manning and Chezy equations describe the mean velocity in steady uniform turbulent flow. Under steady uniform conditions, the energy slope equals the sewer bed slope. This equality should not be assumed without qualification for nonuniform or surcharged flow.
వివరణ:
Question 2
ప్రశ్న 2
Which expression gives Manning mean velocity V in SI units for steady uniform gravity flow, where R is hydraulic radius, S is energy slope and n is Manning roughness coefficient?
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Explanation:
Manning velocity is V = (1/n)R^(2/3)S^(1/2). Here R is the flow area divided by the wetted perimeter. The coefficient n represents the resistance associated with the sewer boundary. The energy slope S is dimensionless and is equal to the bed slope under steady uniform flow. Discharge is then obtained by multiplying the velocity by the flowing area, Q = AV.
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Question 3
ప్రశ్న 3
Which pair of equations correctly represents Chezy mean velocity and discharge for uniform gravity sewer flow?
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Explanation:
Chezy mean velocity is expressed as V = C sqrt(RS). Multiplication by the flowing area yields Q = AC sqrt(RS). The hydraulic radius R equals the flow area divided by the wetted perimeter. The coefficient C represents a resistance parameter with dimensions consistent with the units used. The relation applies to uniform flow and should not be treated as a purely geometric discharge formula.
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Question 4
ప్రశ్న 4
What is the hydraulic radius of a circular sewer of internal diameter 0.80 m when its full circular section is used as the hydraulic reference?
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Explanation:
The hydraulic radius is defined as R = A/P, with P denoting wetted perimeter. For a full circular section, area A = pi D^2/4. Its wetted perimeter is P = pi D. The ratio therefore becomes R = D/4. Substituting D = 0.80 m gives R = 0.20 m.
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Question 5
ప్రశ్న 5
For a 0.60 m diameter circular sewer flowing full, what are its flow area and wetted perimeter, approximately?
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Explanation:
The full circular flow area is A = pi D^2/4. With D = 0.60 m, the area is approximately 0.2827 m^2. The corresponding wetted perimeter is P = pi D. This gives a perimeter of approximately 1.8850 m. Their ratio also confirms the full-section hydraulic radius of 0.15 m.
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Question 6
ప్రశ్న 6
A 0.60 m diameter circular sewer is evaluated at full-flow reference conditions with Manning n = 0.013 and uniform energy slope S = 0.001. What is the mean velocity?
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Explanation:
For a full circular section, R = D/4 = 0.15 m. Manning velocity is V = R^(2/3)sqrt(S)/n. Substituting R = 0.15 m, S = 0.001 and n = 0.013 gives approximately 0.6867 m/s. The flow area is not needed to calculate the mean velocity. Thus the nearest option is 0.687 m/s.
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Question 7
ప్రశ్న 7
For the same 0.60 m diameter sewer, Manning n = 0.013 and uniform slope S = 0.001, what is its full-flow reference discharge?
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Explanation:
The full-section area of the 0.60 m diameter pipe is 0.28274 m^2. Its hydraulic radius is 0.15 m. Manning calculation gives mean velocity of approximately 0.68673 m/s for the specified values. Discharge is the product Q = AV = 0.28274 times 0.68673. The result is approximately 0.1942 m^3/s, or 194.2 L/s.
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Question 8
ప్రశ్న 8
What uniform energy slope is needed to obtain a mean full-flow reference velocity of 1.00 m/s in a 0.60 m diameter circular sewer with Manning n = 0.013?
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Explanation:
The full-section hydraulic radius is R = 0.60/4 = 0.15 m. Rearranging Manning equation gives S = [Vn/R^(2/3)]^2. Substitute V = 1.00 m/s and n = 0.013 into this expression. The calculated slope is approximately 0.0021205, dimensionless. This corresponds to a fall of about 1 m in 472 m under the uniform-flow assumption.
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Question 9
ప్రశ్న 9
Two sewers have identical geometry, flow depth and energy slope, but Manning n increases from 0.012 to 0.015. Under Manning flow, what is the ratio of new velocity to original velocity?
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Explanation:
At fixed hydraulic radius and slope, Manning velocity is inversely proportional to n. Therefore Vnew/Vold = nold/nnew. Substituting the coefficients gives 0.012/0.015 = 0.80. The mean velocity decreases by 20 percent when the roughness coefficient increases in this way. Discharge also decreases by the same factor if the flowing area remains unchanged.
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Question 10
ప్రశ్న 10
For a circular sewer of fixed diameter and fixed fractional depth, Manning n is constant. If uniform energy slope changes from 1 in 1600 to 1 in 400, how does velocity change?
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Explanation:
The initial uniform slope is 1/1600 and the new slope is 1/400. Their ratio is four because the second slope is four times as steep. Under fixed geometry and roughness, Manning velocity is proportional to the square root of slope. The velocity ratio is therefore sqrt(4) = 2. Hence the new mean velocity is twice the original value.
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Question 11
ప్రశ్న 11
Using Chezy equation, find the mean velocity for C = 60 m^(1/2)/s, hydraulic radius R = 0.16 m and uniform energy slope S = 0.0004.
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Explanation:
Chezy equation gives V = C sqrt(RS). The product RS is 0.16 times 0.0004, equal to 0.000064. Its square root is 0.008. Multiplying by C = 60 gives V = 0.48 m/s. The result uses the supplied Chezy coefficient in consistent SI units.
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Question 12
ప్రశ్న 12
Under Chezy uniform-flow conditions, what slope is required for V = 0.90 m/s if C = 50 m^(1/2)/s and R = 0.09 m?
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Explanation:
Chezy equation states that V = C sqrt(RS). Squaring and rearranging gives S = V^2/(C^2 R). The numerator is 0.90^2 = 0.81. The denominator is 50^2 times 0.09 = 225. Thus the required dimensionless slope is 0.81/225 = 0.00360.
వివరణ:
Question 13
ప్రశ్న 13
Under consistent SI units and uniform-flow conditions, Manning and Chezy expressions are equated for R = 0.125 m and n = 0.0125. What Chezy coefficient C follows?
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Explanation:
Equating Chezy and Manning mean velocities gives C = R^(1/6)/n. The hydraulic radius is specified as 0.125 m. Its one-sixth power is approximately 0.7071. Dividing by the specified n of 0.0125 gives approximately 56.6. The result is the Chezy coefficient corresponding to Manning representation at this hydraulic radius.
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Question 14
ప్రశ్న 14
For the same circular sewer, bed slope and constant Manning n, how do mean velocity and discharge at exactly half-full depth compare with their full-section reference values?
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Explanation:
At exactly half-full depth, the flow area is half the full circular area. The wetted perimeter is also half the full circular circumference. Their ratio is therefore still R = D/4. Under the same slope and Manning coefficient, mean velocity is unchanged from the full-section reference. Since Q = AV and the area is halved, the discharge is one-half of full-flow discharge.
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Question 15
ప్రశ్న 15
A circular sewer of diameter 0.80 m runs exactly half full under steady uniform flow with Manning n = 0.013 and S = 0.001. What is its discharge, approximately?
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Explanation:
For a half-full circular sewer, A = pi D^2/8 = 0.25133 m^2. Its hydraulic radius is R = D/4 = 0.20 m. Manning velocity is (0.20)^(2/3)sqrt(0.001)/0.013 = approximately 0.8319 m/s. The discharge is 0.25133 multiplied by 0.8319. This gives approximately 0.209 m^3/s, which is one-half the full-section discharge.
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Question 16
ప్రశ్న 16
A circular sewer carries flow to a depth y = D/4. What is the central angle subtended by the wetted arc, in degrees?
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Explanation:
For a circular section, the wetted central angle in radians is theta = 2 arccos(1 - 2y/D). At y/D = 1/4, the cosine argument becomes 1 - 1/2 = 1/2. The arccosine of 1/2 is 60 degrees. Multiplying this half-angle by two gives 120 degrees. The wetted perimeter is correspondingly one-third of the full circumference.
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Question 17
ప్రశ్న 17
For a circular sewer flowing steadily at depth y = 0.75D, what is its hydraulic radius approximately, expressed in terms of internal diameter D?
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Explanation:
At depth 0.75D, the wetted central angle is 240 degrees, or 4pi/3 radians. The flow area is A = (D^2/8)(theta - sin theta). The wetted perimeter is P = D theta/2. Their ratio evaluates to approximately 0.3017D. Therefore R is closest to 0.302D rather than the full-section value of 0.250D.
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Question 18
ప్రశ్న 18
Under constant Manning n and slope, a circular sewer flows at 0.75 of its diameter. Which approximate ratios of partial-flow velocity and discharge to full-flow reference values are correct?
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Explanation:
For y/D = 0.75, the wetted area is approximately 0.8045 of the full-section area. The hydraulic radius is about 1.2067 times the full-section hydraulic radius. Manning velocity ratio is therefore (1.2067)^(2/3), approximately 1.133. Multiplying that by the area ratio gives a discharge ratio of approximately 0.912. Thus velocity exceeds full-section reference velocity while discharge remains below full-section discharge.
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Question 19
ప్రశ్న 19
For a circular gravity sewer at y/D = 0.80, with the same constant Manning n and slope as the full-flow reference case, which approximate comparison is correct?
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Explanation:
At 80 percent depth, the wetted perimeter and area must be calculated from circular segment geometry. With constant n and slope, the hydraulic radius makes mean velocity about 1.14 times the full-flow reference velocity. The flow area is less than the full circular area. Consequently, discharge is approximately 0.97 to 0.98 times the full-section reference discharge. This comparison illustrates why sewer velocity and discharge cannot be inferred directly from depth fraction.
వివరణ:
Question 20
ప్రశ్న 20
For a circular conduit analysed using Manning formula with constant n and fixed slope, which depth ranges approximately correspond to maximum mean velocity and maximum discharge in partially full uniform flow?
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Explanation:
Under constant Manning roughness and slope, mean velocity depends on hydraulic radius. Its maximum occurs when the wetted-section hydraulic radius is greatest, around 0.81D depth. Discharge also depends on the flow area and consequently peaks at a different depth. For a circular section, maximum Manning discharge occurs near 0.94D. These results describe mathematical uniform-flow comparisons and do not imply that a practical gravity sewer should be designed nearly full.
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Question 21
ప్రశ్న 21
If the diameter of a full circular sewer is doubled while the uniform energy slope and Manning n remain unchanged, by what factor does its Manning reference discharge increase?
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Explanation:
For a full circle, flow area A is proportional to D^2. The hydraulic radius R = D/4 is proportional to diameter. Manning velocity is therefore proportional to D^(2/3) at fixed n and slope. Discharge Q = AV is consequently proportional to D^(8/3). Doubling diameter gives a discharge factor of 2^(8/3), approximately 6.35.
వివరణ:
Question 22
ప్రశ్న 22
A 0.60 m diameter sewer is laid with a bed fall of 0.40 m over 200 m. Assuming steady uniform flow at the full-section reference, with n = 0.013, what is the Manning mean velocity?
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Explanation:
The sewer bed slope is 0.40/200 = 0.002. In steady uniform flow this equals the energy slope used in Manning equation. The full-section hydraulic radius is D/4 = 0.15 m. Substituting n = 0.013 and S = 0.002 yields a mean velocity of approximately 0.9712 m/s. The result is sqrt(2) times the velocity for an otherwise identical sewer at slope 0.001.
వివరణ:
Question 23
ప్రశ్న 23
When computing hydraulic properties for a circular gravity sewer flowing partially full, which statement about wetted perimeter and hydraulic depth is correct?
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Explanation:
Wetted perimeter refers to the length of the solid boundary in contact with the flowing water. The free surface is not part of the wetted perimeter used in hydraulic radius. Thus hydraulic radius is the wetted flow area divided by its wetted boundary length. Hydraulic depth is a different quantity defined as flow area divided by top width at the free surface. Confusing hydraulic depth with hydraulic radius gives incorrect Manning and Chezy calculations.
వివరణ:
Question 24
ప్రశ్న 24
Which approach best distinguishes calculation of a free-surface gravity sewer from a fully surcharged sewer or sewage pumping main?
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Explanation:
Partially full gravity sewers behave as open-channel conduits with an atmospheric free surface. Their capacity is commonly assessed with Manning equation using the actual flowing area and wetted perimeter. A surcharged conduit or pumping main operates under pressure and its hydraulic grade must be evaluated accordingly. Darcy-Weisbach and, in suitable water or sewage applications, Hazen-Williams are examples of pressure-flow loss formulations. Sewer bed slope alone is not sufficient to determine energy slope in general pressure-flow conditions.
వివరణ:
Question 25
ప్రశ్న 25
Under otherwise identical full-section Manning conditions, the roughness coefficient n becomes 1.20 times its original value. By what factor must uniform energy slope increase to maintain the original mean velocity?
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Explanation:
Manning equation is V = R^(2/3)sqrt(S)/n. With the same diameter and flow geometry, hydraulic radius remains unchanged. To maintain V, sqrt(S) must increase in proportion to n. Increasing n by a factor of 1.20 therefore requires S to increase by 1.20 squared. The required slope factor is 1.44, representing a 44 percent increase.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. It behaves as open-channel flow, and the energy slope equals the bed slope for steady uniform flow. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. V = (1/n)R^(2/3)sqrt(S) సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. V = C sqrt(RS) and Q = AC sqrt(RS) సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 0.20 m సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 0.2827 m^2 and 1.8850 m సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. Approximately 0.687 m/s సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. Approximately 0.1942 m^3/s సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. Approximately 0.00212 సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: A. 0.80 సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: B. It becomes twice the original velocity. సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: C. 0.48 m/s సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: D. 0.00360 సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: A. Approximately 56.6 m^(1/2)/s సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. Velocity is equal and discharge is half. సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: C. 0.209 m^3/s సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: D. 120 degrees సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: A. 0.302D సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: B. V/Vfull = 1.13 and Q/Qfull = 0.91 సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: C. Velocity is about 1.14 of full value and discharge about 0.97–0.98 of full value. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. Maximum velocity is near y/D = 0.81 and maximum discharge is near y/D = 0.94. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: B. Approximately 6.35 సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. Approximately 0.971 m/s సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. Wetted perimeter includes only the boundary in contact with water, while hydraulic depth is area divided by top width. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. Use open-channel formulations such as Manning for gravity free-surface flow; analyse pressure-flow friction and hydraulic grade using suitable pressure-pipe relations. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. 1.44 సమాధానం: A.