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Question 1
ప్రశ్న 1
Which statement best describes turbulent flow through a pipe?
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Explanation:
Turbulent pipe flow contains velocity fluctuations about a time-averaged or mean velocity field. These fluctuations produce strong mixing and transport of momentum across the pipe section. Molecular viscosity still acts in turbulent flow and remains especially important near the wall. A geometrically smooth pipe can also carry turbulent flow when the Reynolds number is sufficiently high. Therefore Option B correctly describes the essential characteristics of turbulent pipe flow.
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Question 2
ప్రశ్న 2
In the Reynolds decomposition commonly used to describe turbulent flow, an instantaneous streamwise velocity u is represented most appropriately as:
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Explanation:
A turbulent velocity signal is commonly separated into a mean component and a fluctuating component. The instantaneous velocity may therefore be written as u = U + u'. The time average of the fluctuating component is zero when the averaging interval is appropriate for a statistically steady turbulent flow. The fluctuations contribute to turbulent momentum transport even though their individual time averages vanish. Therefore Option A is correct.
వివరణ:
Question 3
ప్రశ్న 3
Unless a question explicitly states otherwise, let f denote the Darcy friction factor in this test. Which Darcy-Weisbach head-loss equation is then correct for a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Darcy-Weisbach equation written with the Darcy friction factor is hf = f(L/D)(V²/2g). Here L is the pipe length, D is the pipe diameter and V is the cross-sectional mean velocity. The Darcy friction factor is dimensionless. A factor of four appears only when the same loss relation is written using the Fanning friction factor. Therefore Option A is correct under the convention explicitly adopted in the question.
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Question 4
ప్రశ్న 4
How are the Darcy friction factor fD and Fanning friction factor fF related for the same pipe flow?
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Explanation:
The Darcy and Fanning friction factors use different normalizations of wall shear stress. For the same physical pipe flow, the Darcy friction factor is four times the Fanning friction factor. Thus fD = 4fF. Confusing the two conventions produces a factor-of-four error in head-loss calculations. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
Which expression relates the Darcy friction factor f to wall shear-stress magnitude τw for fully developed flow in a circular pipe?
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Explanation:
The Darcy friction factor can be related to the wall shear stress through the pipe force balance. The resulting relation is τw = fρV²/8. By comparison, the Fanning friction factor is defined directly as τw divided by ρV²/2. These two definitions are consistent with fD = 4fF. Therefore Option A is correct.
వివరణ:
Question 6
ప్రశ్న 6
For turbulent flow through a circular pipe, what is meant by the relative roughness?
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Explanation:
Absolute roughness ε represents a characteristic height of wall-surface irregularities. The hydraulic influence of those irregularities also depends on the pipe size. Relative roughness is therefore formed as the dimensionless ratio ε/D. Turbulent friction-factor correlations and the Moody chart use this ratio rather than absolute roughness alone. Hence Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
Which statement correctly describes a hydraulically smooth pipe in turbulent flow?
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Explanation:
Hydraulically smooth does not mean that the physical wall has absolutely no microscopic roughness. It means that the roughness elements do not significantly protrude through the viscous near-wall region under the flow conditions considered. The turbulent friction factor then follows the smooth-pipe behaviour and depends primarily on Reynolds number. Even a hydraulically smooth pipe has nonzero wall shear and nonzero head loss. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
Which statement correctly describes the transitionally rough turbulent regime in a pipe?
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Explanation:
The transitionally rough regime lies between hydraulically smooth turbulent flow and fully rough turbulent flow. Wall roughness has become dynamically important, but viscous effects near the wall still influence the resistance. The friction factor therefore depends on both Reynolds number and relative roughness. The fully rough limit, in contrast, becomes essentially independent of Reynolds number. Hence Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
In the fully rough turbulent regime of a circular pipe, which statement is most accurate for a fixed relative roughness?
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Explanation:
At sufficiently high Reynolds number, the roughness elements dominate the near-wall resistance. In this fully rough regime, the Darcy friction factor approaches a value controlled mainly by relative roughness. Further increases in Reynolds number then have little effect on the friction factor. On a Moody chart, the constant-relative-roughness curves approach nearly horizontal asymptotes in this region. Therefore Option A is correct.
వివరణ:
Question 10
ప్రశ్న 10
Which two quantities are principally required to determine the Darcy friction factor from the turbulent region of a Moody chart for a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The turbulent portion of the Moody chart expresses friction factor as a function of Reynolds number and relative roughness. Reynolds number represents the balance between inertial and viscous effects. Relative roughness describes the wall irregularity compared with the pipe diameter. Pipe length is required later when the friction factor is used to calculate head loss but is not a direct Moody-chart input. Therefore Option A is correct.
వివరణ:
Question 11
ప్రశ్న 11
For a hydraulically smooth turbulent pipe within an applicable smooth-pipe correlation range, what is the general trend of Darcy friction factor as Reynolds number increases?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In the hydraulically smooth turbulent regime, friction factor depends primarily on Reynolds number. Common smooth-pipe correlations show a gradual decrease in Darcy friction factor as Reynolds number increases. This decrease is much weaker than the inverse Reynolds-number dependence of laminar flow. The trend changes as wall roughness becomes increasingly influential. Therefore Option A is correct.
వివరణ:
Question 12
ప్రశ్న 12
At the same Reynolds number in the turbulent regime, what is the usual effect of increasing relative roughness ε/D?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Increasing relative roughness makes the wall irregularities more significant compared with the pipe diameter. In turbulent flow this generally increases form drag and near-wall resistance. The corresponding Darcy friction factor therefore increases at a given Reynolds number once roughness effects are relevant. This trend is represented by the separated roughness curves on the Moody chart. Therefore Option A is correct.
వివరణ:
Question 13
ప్రశ్న 13
For comparison with turbulent behaviour, which statement is correct for fully developed laminar flow in a circular pipe when the Darcy friction factor is used?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The exact fully developed laminar solution in a circular pipe gives the Darcy friction factor f = 64/Re. In that classical regime, friction factor is governed by viscous shear and does not depend on the usual roughness parameter. This differs fundamentally from turbulent pipe flow, where roughness can strongly affect resistance. The laminar relation is included here only as a contrast with the turbulent Moody-chart behaviour. Therefore Option A is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a hydraulically smooth turbulent circular pipe, use the stated Blasius correlation f = 0.3164 Re^-0.25. What is f at Re = 1.0 × 10^5?
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Explanation:
The problem explicitly supplies the Blasius correlation using the Darcy friction factor. Substituting Re = 1.0 × 10^5 gives f = 0.3164/(10^5)^0.25. Evaluating the fourth-root dependence gives f approximately equal to 0.01779. Rounding appropriately gives 0.0178. Therefore Option B is correct.
వివరణ:
Question 15
ప్రశ్న 15
Water flows through a 200 mm diameter pipe at a mean velocity of 3.0 m/s over a length of 100 m. If the Darcy friction factor is 0.025, what is the frictional head loss? Take g = 9.81 m/s².
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Explanation:
Using the Darcy convention, the head loss is hf = f(L/D)(V²/2g). Here L/D = 100/0.20 = 500. The velocity head is 3²/(2 × 9.81) = approximately 0.4587 m. Therefore hf = 0.025 × 500 × 0.4587 = approximately 5.73 m. Hence Option B is correct.
వివరణ:
Question 16
ప్రశ్న 16
A circular pipe has length 80 m, diameter 0.10 m and mean velocity 2.5 m/s. If the measured frictional head loss is 4.0 m, what is the Darcy friction factor? Take g = 9.81 m/s².
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Explanation:
Rearranging Darcy-Weisbach gives f = hf(D/L)(2g/V²). Substitution gives f = 4.0(0.10/80)[2(9.81)/(2.5²)]. The numerical value is approximately 0.01570. The result is dimensionless, as required for a friction factor. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
Water of density 1000 kg/m³ flows at 4.0 m/s in a pipe whose Darcy friction factor is 0.028. What is the wall shear-stress magnitude?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the Darcy convention, wall shear is τw = fρV²/8. Substitution gives τw = 0.028 × 1000 × 4²/8. Since 4² = 16, the calculation becomes 0.028 × 2000. This gives 56 Pa. Therefore Option C is correct.
వివరణ:
Question 18
ప్రశ్న 18
A Fanning friction factor of 0.006 is reported for a turbulent pipe flow. What Darcy friction factor represents the same physical resistance?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Darcy friction factor is four times the Fanning friction factor. Thus fD = 4 × 0.006. This gives fD = 0.024. Using 0.006 directly in the Darcy-Weisbach equation would therefore produce an incorrect loss. Hence Option C is correct.
వివరణ:
Question 19
ప్రశ్న 19
A commercial pipe has an absolute roughness ε = 0.15 mm and internal diameter D = 150 mm. What is its relative roughness?
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Explanation:
Relative roughness is ε/D provided both dimensions are expressed in the same units. Here ε/D = 0.15/150. The result is 0.001. This dimensionless quantity is the roughness parameter used with Reynolds number in turbulent friction-factor calculations. Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
Two pipes have the same absolute roughness ε, but Pipe 2 has twice the diameter of Pipe 1. What is the relative roughness of Pipe 2 compared with Pipe 1?
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Explanation:
Relative roughness is defined as ε/D. The absolute roughness is unchanged between the two pipes. Doubling the diameter therefore divides ε/D by two. Pipe 2 consequently has one-half the relative roughness of Pipe 1. In a roughness-influenced turbulent regime, this tends to correspond to a lower Darcy friction factor when the other relevant conditions are comparable.
వివరణ:
Question 21
ప్రశ్న 21
Which equation is the standard Colebrook-White form for the Darcy friction factor f in turbulent circular-pipe flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Colebrook-White relation combines Reynolds-number and relative-roughness effects for turbulent pipe flow. In Darcy-factor form it is implicit because f occurs both outside and inside the logarithm. The first logarithmic term represents wall roughness through ε/D, while the second contains the Reynolds-number contribution. The equation bridges hydraulically smooth, transitionally rough and increasingly rough turbulent behaviour over its applicable range. Therefore Option A is correct.
వివరణ:
Question 22
ప్రశ్న 22
A turbulent circular-pipe flow has Re = 1.0 × 10^5 and relative roughness ε/D = 0.001. Using the Darcy-form Colebrook equation 1/√f = -2 log10[ε/(3.7D) + 2.51/(Re√f)], what friction factor is obtained approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Colebrook equation is implicit in the Darcy friction factor and therefore requires iteration or an equivalent numerical solution. Substituting Re = 100000 and ε/D = 0.001 and iterating gives f approximately equal to 0.02217. Rounding to four significant figures gives 0.0222. This value is higher than the corresponding ideal smooth-pipe value because the specified wall roughness contributes additional resistance. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
In the fully rough limit, the Colebrook equation reduces to 1/√f = -2 log10[ε/(3.7D)]. For ε/D = 0.001, what is the Darcy friction factor approximately?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
In the fully rough limit, the Reynolds-number term in the Colebrook equation becomes negligible. Substituting ε/D = 0.001 gives 1/√f = -2 log10(0.001/3.7). Evaluating this expression and solving for f gives approximately 0.01964. Rounding gives f approximately equal to 0.0196. Therefore Option B is correct.
వివరణ:
Question 24
ప్రశ్న 24
If a Fanning friction factor is mistakenly inserted directly into the Darcy-Weisbach equation hf = f(L/D)(V²/2g) without converting conventions, how will the calculated head loss compare with the correct value?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For the same flow, the Darcy friction factor is four times the Fanning friction factor. The Darcy-Weisbach equation in the form stated requires the Darcy factor. If the smaller Fanning value is inserted without conversion, the factor used in the equation is only one-quarter of the required Darcy value. Because head loss is directly proportional to friction factor in that equation, the predicted loss will also be one-quarter of the correct result. Therefore Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements for circular-pipe friction: I. Darcy friction factor is four times the Fanning friction factor. II. In hydraulically smooth turbulent flow, friction factor depends mainly on Reynolds number. III. In the fully rough turbulent limit, friction factor for fixed ε/D becomes essentially independent of Reynolds number. IV. In the transitionally rough turbulent regime, friction factor depends on both Reynolds number and relative roughness. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because fD = 4fF for the same pipe flow. Statement II is correct because roughness elements are hydraulically submerged in the smooth turbulent regime and the friction-factor behaviour is governed primarily by Reynolds number. Statement III is correct because the fully rough asymptote is controlled principally by relative roughness. Statement IV is correct because both viscous and roughness effects remain important in the transitionally rough zone. 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. The motion contains irregular velocity fluctuations superimposed on a mean flow and generally produces enhanced momentum mixing సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: A. u = U + u', where U is the mean velocity and u' is the fluctuating component సమాధానం: A.
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Question 3 ప్రశ్న 3Answer: A. hf = f(L/D)(V²/2g) సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. fD = 4fF సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. τw = fρV²/8 సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: B. ε/D సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. Its roughness elements are sufficiently small relative to the near-wall viscous region that friction factor behaves essentially like that of a smooth pipe సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: A. The Darcy friction factor depends on both Reynolds number and relative roughness సమాధానం: A.
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Question 9 ప్రశ్న 9Answer: A. The Darcy friction factor is approximately independent of Reynolds number and depends mainly on relative roughness సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: A. Reynolds number and relative roughness సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: A. It generally decreases సమాధానం: A.
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Question 12 ప్రశ్న 12Answer: A. The Darcy friction factor generally increases సమాధానం: A.
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Question 13 ప్రశ్న 13Answer: A. f = 64/Re and it is independent of wall roughness in the classical laminar solution సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: B. 0.0178 సమాధానం: B.
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Question 15 ప్రశ్న 15Answer: B. 5.73 m సమాధానం: B.
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Question 16 ప్రశ్న 16Answer: C. 0.0157 సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: C. 56 Pa సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: C. 0.024 సమాధానం: C.
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Question 19 ప్రశ్న 19Answer: B. 0.0010 సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: B. One-half as large సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: A. 1/√f = -2 log10[ε/(3.7D) + 2.51/(Re√f)] సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 0.0222 సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. 0.0196 సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: A. It will be one-quarter of the correct value సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.