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Question 1
ప్రశ్న 1
What is the principal objective of hydraulic design of a water-distribution network?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A distribution network must convey the required quantity of water to the various demand locations. The available hydraulic head must be sufficient to overcome friction and other losses. Adequate residual pressure must remain at critical consumer locations after those losses occur. Pipe sizes should also avoid unnecessary overdesign while satisfying operational requirements. Therefore Option B is correct.
వివరణ:
Question 2
ప్రశ్న 2
Which statement best describes nodal demand in hydraulic modelling of a water-distribution system?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydraulic network models represent consumer withdrawals by assigning demands to appropriate nodes. These nodal demands collectively represent the spatial distribution of water consumption in the service area. A node may represent demand from a group of consumers rather than one individual connection. The assigned demands are used together with pipe characteristics and boundary heads to solve flows and pressures. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
According to current CPHEEO planning guidance, which approach is most appropriate when assigning design demands to nodes of a distribution model?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Current CPHEEO guidance emphasizes realistic assignment of water demand to network nodes. Ward-wise population distribution or density information can be used to distribute projected demand spatially. The appropriate peak factor is then applied for the hydraulic condition being analysed. This gives a more representative network loading than simply dividing demand equally among all junctions. Therefore Option B is correct.
వివరణ:
Question 4
ప్రశ్న 4
A demand node represents 1,200 people, the average demand is 135 L/person/day and the specified peak factor is 2.5. What design nodal demand should be used for this peak condition?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The average daily nodal demand is 1,200 × 135 = 162,000 L/day. Dividing by 86,400 seconds per day gives an average flow of 1.875 L/s. The specified peak factor is 2.5. Hence the design nodal demand is 1.875 × 2.5 = 4.6875 L/s. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
What hydraulic principle must be satisfied at an ordinary junction in a steady-state water-distribution network?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The continuity equation expresses conservation of water mass at a junction. Under steady conditions, water cannot accumulate indefinitely at an ordinary demand node. Total inflow must therefore equal total pipe outflow plus the external water demand taken from the node. Pipe diameters and velocities may differ while this continuity condition is still satisfied. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
At a junction, two pipes bring 22 L/s and 14 L/s into the node. The nodal demand is 8 L/s, and one outgoing pipe carries 11 L/s. What discharge must the other outgoing pipe carry to satisfy continuity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The total inflow is 22 + 14 = 36 L/s. Of this amount, 8 L/s is withdrawn as nodal demand. Another 11 L/s leaves through the first outgoing pipe. The remaining outgoing flow is 36 - 8 - 11 = 17 L/s. Therefore Option C is correct.
వివరణ:
Question 7
ప్రశ్న 7
In a hydraulic distribution model, which is an example of a fixed-head boundary condition?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A reservoir water surface provides a known hydraulic grade or head to the system. It is therefore commonly represented as a fixed-head boundary in a steady hydraulic model. Junction heads and pipe flows are then calculated relative to such boundary conditions. Demand nodes generally have specified withdrawals but unknown hydraulic heads before solution. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
What does the hydraulic grade line at a point in a pressurized distribution pipe represent?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The hydraulic grade line represents piezometric head in a pressurized water system. Piezometric head is the sum of elevation head and pressure head. The energy grade line additionally includes velocity head. As water flows through a resistive pipe, the hydraulic grade generally falls in the direction of flow because of head loss. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
A reservoir water surface is at elevation 142 m. Total head loss from the reservoir to a demand node is 15 m, and the node elevation is 110 m. Neglecting velocity-head differences, what residual pressure head is available at the node?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The hydraulic grade at the node is the source head minus the intervening head loss. Thus the nodal hydraulic grade is 142 - 15 = 127 m. The node elevation is 110 m. The residual pressure head is therefore 127 - 110 = 17 m. Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
For conversion to a 24×7 pressurized water-supply system, which residual nodal pressure criterion at the highest ferrule is stated in current CPHEEO engineering guidance?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Current CPHEEO guidance places greater emphasis on adequate residual pressure in continuously pressurized distribution systems. It specifies approximately 17 to 21 m residual nodal pressure at the highest ferrule for Class I and II cities. For other cities, approximately 12 to 15 m is specified. The older storey-based pressure criteria should not automatically be substituted for this explicitly stated 24×7 conversion criterion. Therefore Option B is correct.
వివరణ:
Question 11
ప్రశ్న 11
A pipeline loses 6 m of head over a length of 1,200 m. What is the average hydraulic gradient due to this head loss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hydraulic gradient is head loss divided by pipe length. The value is therefore 6/1,200 = 0.005 m/m. Multiplying by 1,000 converts this to 5 m of head loss per kilometre. The gradient represents the slope of the hydraulic grade line caused by the stated loss. Therefore Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
For a circular water main flowing full, which relation gives the mean velocity when discharge Q and internal diameter D are known?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Continuity for a full circular pipe gives discharge equal to cross-sectional area multiplied by mean velocity. The cross-sectional area is pi D squared divided by four. Therefore velocity equals Q divided by pi D squared over four. This relation is independent of which friction formula is subsequently used to calculate head loss. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
A 300 mm internal-diameter distribution main carries 60 L/s. What is the approximate mean velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The discharge is 60 L/s, which is 0.060 m3/s. The pipe area is pi × 0.30^2/4, which is approximately 0.07069 m2. Velocity equals discharge divided by area. Thus V = 0.060/0.07069 = approximately 0.85 m/s. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A design discharge of 80 L/s is to be carried at a selected mean velocity of 1.5 m/s. What internal diameter is required from the continuity equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The discharge is 0.080 m3/s. From Q = VA, the required flow area is 0.080/1.5 = 0.05333 m2. For a circular pipe, D equals the square root of 4A divided by pi. This gives D approximately 0.2606 m, or 261 mm. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For a full circular pipe, the current CPHEEO SI Hazen-Williams head-loss equation may be written as hf = 10.68 L Q^1.852/(C^1.852 D^4.87), with Q in m3/s and D and L in metres. Which variable represents the Hazen-Williams roughness coefficient?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In the stated Hazen-Williams equation, C is the empirical coefficient representing the hydraulic smoothness or resistance characteristic of the pipe. A larger C value produces a smaller calculated friction loss when the remaining variables are held constant. L represents pipe length and D represents internal pipe diameter. Q represents discharge through the pipe. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
Using hf = 10.68 L Q^1.852/(C^1.852 D^4.87), what is the approximate friction head loss in a 500 m long pipe carrying 0.040 m3/s when C = 130 and D = 0.20 m?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated quantities are substituted directly into the current CPHEEO form of the SI Hazen-Williams equation. Thus hf = 10.68 × 500 × 0.040^1.852 divided by 130^1.852 × 0.20^4.87. Evaluating this expression gives approximately 4.24 m of friction head loss. Minor rounding of the numerical coefficient does not materially affect the result. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
According to the Hazen-Williams relation hf proportional to Q^1.852, by approximately what factor does friction head loss change if discharge is doubled while L, C and D remain unchanged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
With all other quantities constant, the Hazen-Williams equation gives head loss proportional to discharge raised to the power 1.852. Doubling discharge therefore multiplies head loss by 2^1.852. This evaluates to approximately 3.61. The result demonstrates why relatively modest increases in network flow can cause substantial increases in friction loss. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
Under the Hazen-Williams equation, a pipe diameter is increased from 200 mm to 250 mm while flow, length and C remain unchanged. Approximately what fraction of the original friction head loss remains?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fixed flow, length and C, Hazen-Williams head loss varies approximately as D raised to the power -4.87. The new-to-old head-loss ratio is therefore (0.20/0.25)^4.87. This gives approximately 0.337, or about 0.34 of the original head loss. The strong diameter exponent explains why increasing pipe diameter can greatly reduce friction loss. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
If the Hazen-Williams C value of an existing pipe decreases because of aging, scaling or deterioration, while discharge and diameter remain unchanged, what happens to calculated friction head loss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Hazen-Williams head loss is inversely proportional to C raised to approximately the power 1.852. A reduction in C therefore means greater hydraulic resistance for the same pipe size and discharge. Aging, tuberculation and scaling can consequently reduce carrying capacity or increase head loss. Such deterioration must be considered when assessing an existing distribution network. Therefore Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
Which equation correctly represents Darcy-Weisbach friction head loss in a full pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Darcy-Weisbach equation relates friction head loss to the friction factor, relative pipe length and velocity head. It is written as hf = f(L/D)(V squared/2g). The Darcy friction factor depends on Reynolds number and relative roughness under the relevant flow regime. Unlike the empirical Hazen-Williams coefficient, Darcy-Weisbach has a direct basis in mechanical energy loss. Therefore Option A is correct.
వివరణ:
Question 21
ప్రశ్న 21
Using Darcy-Weisbach, calculate the approximate friction head loss for f = 0.020, L = 1000 m, D = 0.30 m and V = 1.20 m/s. Take g = 9.81 m/s2.
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Darcy-Weisbach gives hf = f(L/D)(V squared/2g). Substituting gives hf = 0.020 × (1000/0.30) × (1.20^2/(2 × 9.81)). The velocity-head term is approximately 0.0734 m. The complete calculation gives approximately 4.89 m of friction head loss. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
Two pipe reaches are connected in series without any intermediate withdrawal. If their friction head losses are 3.5 m and 4.2 m at the common discharge, what is the total friction head loss between the ends?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pipes in series carry the same discharge when there is no intermediate withdrawal or inflow. The energy losses through successive reaches are additive. Therefore the total friction loss is 3.5 + 4.2 = 7.7 m. This total loss must be deducted from the upstream hydraulic head when determining the downstream hydraulic grade. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A fitting has a minor-loss coefficient K = 4.0 and the pipe velocity is 1.5 m/s. Using hL = K V^2/(2g) with g = 9.81 m/s2, what is the approximate minor head loss?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The velocity head is V squared divided by 2g. For 1.5 m/s, this is 2.25/19.62 = approximately 0.1147 m. Multiplying by K = 4 gives approximately 0.459 m. Hence the minor head loss is approximately 0.46 m. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
A demand node is at elevation 118 m and must retain 15 m of residual pressure head. If the total head loss from the source boundary to that node is 7 m, what minimum source hydraulic grade is required, neglecting velocity-head differences?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The required hydraulic grade at the demand node is its elevation plus the required pressure head. This gives 118 + 15 = 133 m. The source must additionally supply the 7 m of head lost between the source and the node. The minimum source hydraulic grade is therefore 133 + 7 = 140 m. Therefore Option C is correct.
వివరణ:
Question 25
ప్రశ్న 25
A pipe resistance relation is written in the form hf = RQ^n. What does R represent in such a network formulation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Many network equations combine geometric and roughness properties of a pipe into a resistance coefficient. For a selected friction law, R incorporates quantities such as pipe length, internal diameter and the applicable roughness parameter. The exponent n depends on the head-loss relationship being used. Expressing head loss in this compact form is particularly useful in network calculations. Therefore Option A is correct.
వివరణ:
Question 26
ప్రశ్న 26
Consider the following statements about hydraulic design of distribution networks: I. Continuity must be satisfied at demand junctions. II. Residual pressure depends on hydraulic grade and node elevation. III. Increasing pipe diameter generally reduces friction loss for the same discharge. IV. Reservoir heads and nodal demands are examples of important boundary or loading data for a hydraulic model. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because conservation of flow must hold at every ordinary junction. Statement II is correct because pressure head is obtained from the difference between hydraulic grade and node elevation. Statement III is correct because increasing diameter substantially lowers hydraulic resistance for the same discharge. Statement IV is correct because fixed source heads and assigned nodal demands provide essential boundary and loading information for solving a network. 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: B. To satisfy required nodal demands while maintaining acceptable residual pressures and hydraulic performance సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. It is the water demand assigned to or withdrawn at a network junction or demand node సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: B. Assign demands using realistic spatial population or demand distribution and apply an appropriate peak factor సమాధానం: B.
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Question 4 ప్రశ్న 4Answer: C. 4.688 L/s సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. The algebraic sum of inflows equals the algebraic sum of outflows including nodal demand సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 17 L/s సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: A. A reservoir with a specified water-surface elevation సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. Elevation head plus pressure head సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: B. 17 m సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: B. 17–21 m for Class I and II cities and 12–15 m for other cities సమాధానం: B.
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Question 11 ప్రశ్న 11Answer: B. 0.005, or 5 m/km సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: A. V = Q/(pi D^2/4) సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: B. 0.85 m/s సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: C. Approximately 261 mm సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. C సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: C. 4.24 m సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: C. 3.61 సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. 0.34 సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. It increases సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: A. hf = f(L/D)(V^2/2g) సమాధానం: A.
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Question 21 ప్రశ్న 21Answer: B. 4.89 m సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 7.7 m సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 0.46 m సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: C. 140 m సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. A coefficient containing the effect of pipe length, diameter and roughness for the selected head-loss equation సమాధానం: A.
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Question 26 ప్రశ్న 26Answer: D. I, II, III and IV సమాధానం: D.