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Question 1
ప్రశ్న 1
Which statement best describes water hammer in a pressurized pipeline?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Water hammer is a transient phenomenon associated with rapid changes in fluid velocity in a closed conduit. Rapid valve operation, pump stoppage or other sudden flow changes can generate pressure disturbances. These disturbances propagate through the fluid-pipe system as elastic pressure waves. Water hammer is therefore fundamentally different from the steady distributed pressure loss caused by pipe friction. Hence Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
When a downstream valve in a flowing water pipeline is closed rapidly, what physically propagates upstream from the valve?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Rapid valve closure decelerates the water adjacent to the valve first. The resulting pressure increase compresses the liquid slightly and deforms an elastic pipe wall where applicable. This disturbance is transmitted to neighboring fluid and propagates upstream as a pressure wave. The propagation speed is finite because the fluid and pipe system possess finite elasticity. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
For a sufficiently rapid velocity change in a simple pipeline, which expression gives the magnitude of the Joukowsky pressure change?
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Explanation:
The classical Joukowsky relation connects a rapid change in fluid velocity with the associated pressure-wave magnitude. The pressure change magnitude is Δp = ρa|ΔV|. Here ρ is fluid density, a is pressure-wave celerity and |ΔV| is the magnitude of the velocity change. A rapid reduction in velocity at a downstream valve normally produces a positive pressure rise there. Therefore Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
Which expression gives the corresponding Joukowsky pressure-head change for a rapid velocity change of magnitude |ΔV|?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Pressure head and pressure are related by ΔH = Δp/(ρg). The Joukowsky equation gives Δp = ρa|ΔV|. Dividing by ρg gives ΔH = a|ΔV|/g. Thus the surge head is directly proportional to wave speed and the magnitude of the rapid velocity change. Therefore Option A is correct.
వివరణ:
Question 5
ప్రశ్న 5
Using the sign convention ΔV = Vafter - Vbefore for positive flow toward a downstream valve, what normally happens at the valve when rapid closure causes ΔV < 0?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Rapid downstream valve closure produces a negative change in the signed flow velocity. The corresponding momentum change creates compression of the fluid near the valve. Under the usual water-hammer situation this produces a positive pressure rise at the closing valve. The Joukowsky relation may be written with an appropriate sign or used in magnitude form together with the physical direction of the velocity change. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
If the pipe wall is perfectly rigid, which expression gives the pressure-wave speed in a liquid of bulk modulus K and density ρ?
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Explanation:
For a pressure wave propagating through a liquid in a perfectly rigid conduit, only fluid compressibility contributes to elastic storage. The acoustic wave speed is therefore a = √(K/ρ). A larger bulk modulus means the liquid is less compressible and increases wave speed. A larger density reduces the wave speed when the bulk modulus is unchanged. Therefore Option A is correct.
వివరణ:
Question 7
ప్రశ్న 7
How does allowing an initially rigid pipe model to have an elastic wall generally affect water-hammer wave celerity, all other relevant properties being comparable?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
An elastic pipe wall expands slightly when internal pressure increases. This wall deformation provides additional compliance beyond the compressibility of the fluid itself. Additional compliance causes the pressure disturbance to propagate more slowly than it would in a perfectly rigid pipe. Consequently the elastic-pipe wave speed is generally lower than the rigid-pipe acoustic speed for the same liquid. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
For a thin-walled elastic circular pipe, neglecting Poisson-ratio and axial-restraint corrections for this question, which expression represents the commonly used wave-speed approximation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The stated thin-wall approximation includes both liquid compressibility and circumferential pipe-wall elasticity. The rigid-liquid acoustic term K/ρ appears in the numerator. The additional compliance of the pipe wall produces the denominator 1 + KD/(Ee). Here E is the pipe Young's modulus, D is internal diameter and e is wall thickness under the explicitly stated simplified assumptions. Therefore Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
Water has bulk modulus K = 2.2 × 10^9 Pa and density ρ = 1000 kg/m³. What is its pressure-wave speed in a perfectly rigid pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For a rigid pipe, wave speed is a = √(K/ρ). Substitution gives a = √[(2.2 × 10^9)/1000]. This is √(2.2 × 10^6), which is approximately 1483.2 m/s. The result is consistent with the expected order of acoustic-wave speed in water. Therefore Option C is correct.
వివరణ:
Question 10
ప్రశ్న 10
For the simplified thin-wall relation a = √[(K/ρ)/(1 + KD/(Ee))], take K = 2.0 × 10^9 Pa, ρ = 1000 kg/m³, D = 0.50 m, E = 200 × 10^9 Pa and e = 0.010 m. What is a approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
First evaluate the wall-compliance term KD/(Ee). It is (2.0 × 10^9 × 0.50)/(200 × 10^9 × 0.010) = 0.50. The denominator is therefore 1.50. Hence a = √[(2.0 × 10^6)/1.50] = approximately 1154.7 m/s. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
Water of density 1000 kg/m³ flows initially at 2.0 m/s in a pipe having water-hammer wave speed 1200 m/s. If a rapid valve closure reduces the velocity essentially to zero, what is the Joukowsky pressure rise?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The magnitude of the rapid velocity change is 2.0 m/s. Joukowsky's relation gives Δp = ρa|ΔV|. Substitution gives Δp = 1000 × 1200 × 2.0 = 2.4 × 10^6 Pa. This is equal to 2.40 MPa. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
For the rapid complete closure in the preceding type of situation, with a = 1200 m/s and initial velocity 2.0 m/s, what is the water-hammer head rise? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Joukowsky head rise is ΔH = a|ΔV|/g. The velocity change magnitude is 2.0 m/s. Therefore ΔH = 1200 × 2.0/9.81. This gives approximately 244.65 m of water head. Hence Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A pipeline initially has a gauge pressure of 300 kPa at a downstream valve. A rapid complete closure produces a calculated positive Joukowsky pressure rise of 2.40 MPa at that instant. Neglecting other transient effects for this calculation, what is the resulting gauge pressure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The initial gauge pressure is 300 kPa, which is 0.30 MPa. The calculated water-hammer increment is a positive 2.40 MPa. Under the explicitly stated simplified superposition, the resulting gauge pressure is 0.30 + 2.40 MPa. This equals 2.70 MPa. Therefore Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
Water of density 1000 kg/m³ flows at 3.0 m/s. A rapid partial valve closure reduces the velocity to 1.0 m/s, and the wave speed is 1000 m/s. What is the magnitude of the initial Joukowsky pressure rise?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The velocity decreases from 3.0 m/s to 1.0 m/s. The magnitude of the change is therefore 2.0 m/s. Joukowsky's relation gives Δp = ρa|ΔV| = 1000 × 1000 × 2.0 Pa. This equals 2.0 × 10^6 Pa, or 2.00 MPa. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
For a simple reservoir-pipe-valve system of pipe length L and water-hammer wave speed a, which time is conventionally used to distinguish rapid from slower valve closure?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A pressure wave requires a time L/a to travel from the valve to the upstream reservoir. It requires another L/a for the reflected disturbance to return to the valve. The round-trip or characteristic pipe period is therefore 2L/a. Valve closure completed within approximately this period is conventionally treated as rapid or sudden for the simple system. Hence Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A valve is at the downstream end of a 600 m long pipeline in which the pressure-wave speed is 1200 m/s. What is the characteristic round-trip wave time 2L/a?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The characteristic round-trip time is 2L/a. Substitution gives 2 × 600/1200 seconds. This simplifies to 1.00 second. It represents the characteristic period used in the simple reservoir-pipe-valve water-hammer classification. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
For the 600 m pipeline with wave speed 1200 m/s, a valve closes completely in 0.60 s. Using the conventional criterion that closure is rapid when tc ≤ 2L/a, how should the closure be classified?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For this pipe, 2L/a equals 1.00 s. The stated closure time is 0.60 s. Since 0.60 s is less than the characteristic 1.00 s round-trip time, the closure satisfies the adopted rapid-closure criterion. Under the classical simple-pipe approximation, the rapid-closure Joukowsky surge relation is therefore applicable. Hence Option A is correct.
వివరణ:
Question 18
ప్రశ్న 18
Compared with sufficiently rapid complete closure of the same initial flow, what is generally expected when the valve is closed over a time substantially longer than 2L/a?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
When closure takes longer than the wave round-trip time, the flow does not experience the entire velocity reduction as one rapid wave event. Reflected disturbances can return while the valve is still moving. This generally reduces the maximum surge compared with sufficiently rapid complete closure of the same initial flow. The exact transient also depends on the valve closure law, friction and the overall hydraulic system. Therefore Option A is correct.
వివరణ:
Question 19
ప్రశ్న 19
For this question, use the simplified Michaud estimate for gradual linear valve closure, ΔH = 2LV0/(gtc). A 500 m pipe has wave speed a = 1000 m/s, carries water initially at 2.0 m/s, and its valve closes in 4.0 s. What surge head does this approximation give? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The characteristic time is 2L/a = 2 × 500/1000 = 1.0 s. Since the stated closure time is 4.0 s, it is gradual under the adopted simple-pipe criterion. The supplied Michaud estimate gives ΔH = 2 × 500 × 2.0/(9.81 × 4.0). This evaluates to approximately 50.97 m. Therefore Option B is correct under the explicitly stated simplified gradual-closure approximation.
వివరణ:
Question 20
ప్రశ్న 20
For a 500 m pipeline with wave speed a = 1000 m/s, initial velocity V0 = 2.0 m/s and valve-closure time tc = 4.0 s, use ΔHrapid = aV0/g and the stated simplified gradual-closure estimate ΔHgradual = 2LV0/(gtc). What is ΔHgradual/ΔHrapid?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The ratio of the two stated expressions is [2LV0/(gtc)]/[aV0/g]. The common factors V0 and g cancel. The ratio becomes 2L/(atc). Substitution gives 1000/(1000 × 4) = 0.25. Therefore Option B is correct.
వివరణ:
Question 21
ప్రశ్న 21
Using the simplified thin-wall wave-speed model a = √[(K/ρ)/(1 + KD/(Ee))], what happens to wave celerity if the pipe Young's modulus E decreases while K, ρ, D and e remain unchanged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Young's modulus measures the stiffness of the pipe-wall material. Reducing E makes the wall more compliant. In the stated formula, reducing E increases the term KD/(Ee) in the denominator. A larger denominator produces a smaller wave speed. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
Using the same simplified thin-wall wave-speed model, what is the effect of increasing pipe-wall thickness e while K, ρ, D and E remain unchanged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Increasing wall thickness makes the pipe circumferentially less compliant under otherwise identical conditions. In the simplified expression, a larger e reduces KD/(Ee). The denominator therefore moves closer to unity. As a result, wave speed increases toward the rigid-pipe value √(K/ρ). Hence Option A is correct.
వివరణ:
Question 23
ప్రశ్న 23
For the same liquid and the same rapid velocity reduction, an elastic pipe has wave speed 1000 m/s while an ideal rigid-pipe model gives 1500 m/s. How do the corresponding Joukowsky pressure-rise magnitudes compare?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Joukowsky pressure rise is proportional to wave speed when density and velocity change are the same. Therefore the ratio of elastic-pipe to rigid-pipe pressure rise is 1000/1500. This ratio equals 2/3. Pipe-wall compliance can consequently reduce the immediate Joukowsky surge by reducing wave celerity. Therefore Option A is correct.
వివరణ:
Question 24
ప్రశ్న 24
Which measure is most directly intended to mitigate excessive pressure transients in a pipeline system?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Water hammer is intensified by rapid changes in flow velocity. Controlled valve operation can spread the velocity change over a longer time and reduce peak transients. Air vessels or similar accumulators can provide a compressible cushion, while surge tanks can accommodate transient exchange of water in appropriate systems. Pressure-relief devices may also limit excessive surge pressure when properly designed. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about water hammer: I. A rapid decrease in pipe-flow velocity can generate a positive pressure surge. II. For rapid change, Joukowsky pressure magnitude is proportional to wave celerity and velocity change. III. Pipe-wall elasticity generally reduces wave speed relative to an ideal rigid-pipe model. IV. In a simple reservoir-pipe-valve system, 2L/a is an important time scale for distinguishing rapid and slower valve closure. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because rapid deceleration at a downstream valve commonly produces compression and a positive surge. Statement II is the magnitude form of the classical Joukowsky relationship. Statement III is correct because elastic wall deformation adds compliance and normally lowers wave celerity. Statement IV correctly identifies the characteristic wave round-trip time of the simple pipeline. 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 transient pressure phenomenon produced by a rapid change in flow velocity సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: A. An elastic pressure wave associated with changes in fluid velocity and compression సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. Δp = ρa|ΔV| సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: A. ΔH = a|ΔV|/g సమాధానం: A.
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Question 5 ప్రశ్న 5Answer: A. A positive pressure surge is generated సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: A. a = √(K/ρ) సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: A. It reduces wave celerity because some elastic deformation occurs in the pipe wall సమాధానం: A.
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Question 8 ప్రశ్న 8Answer: A. a = √[(K/ρ)/(1 + KD/(Ee))] సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: C. 1483 m/s సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: C. 1155 m/s సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: C. 2.40 MPa సమాధానం: C.
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Question 12 ప్రశ్న 12Answer: C. 244.65 m సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. 2.70 MPa సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: C. 2.00 MPa సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: B. 2L/a సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. 1.00 s సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: A. Rapid or sudden closure సమాధానం: A.
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Question 18 ప్రశ్న 18Answer: A. The peak pressure rise is generally reduced because reflected waves can influence the continuing valve motion సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: B. 50.97 m సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. 0.25 సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. Wave celerity decreases సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: A. Wave celerity generally increases toward the rigid-pipe value సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: A. The elastic-pipe pressure rise is two-thirds of the rigid-pipe value సమాధానం: A.
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Question 24 ప్రశ్న 24Answer: A. Using controlled valve closure and, where appropriate, surge-control devices such as air vessels, surge tanks or pressure-relief devices సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.