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Question 1
ప్రశ్న 1
In pipe-flow terminology, what is meant by the major head loss?
Explanation:
Major head loss refers to the distributed mechanical-energy loss associated with wall friction along a pipe length. It is commonly calculated using the Darcy-Weisbach equation. Losses concentrated at entrances, exits, bends, valves and abrupt area changes are conventionally called minor losses. The words major and minor describe the conventional classification and do not guarantee their relative magnitudes in every system. Therefore Option A is correct.
వివరణ:
Question 2
ప్రశ్న 2
Unless otherwise stated, let f denote the Darcy friction factor. Which equation gives the major head loss in a circular pipe?
Explanation:
The Darcy-Weisbach equation using the Darcy friction factor is hf = f(L/D)(V²/2g). The ratio L/D is dimensionless. The velocity-head term V²/(2g) has dimensions of length. Since f is also dimensionless, the resulting value of hf has dimensions of length. Therefore Option B is correct.
వివరణ:
Question 3
ప్రశ్న 3
Which statement about the quantities in the Darcy-Weisbach equation hf = f(L/D)(V²/2g) is correct?
Explanation:
The Darcy friction factor is a dimensionless coefficient. The ratio of pipe length to diameter, L/D, is also dimensionless when consistent units are used. The quantity V²/(2g) has dimensions of length and represents velocity head. Their product therefore has dimensions of length, as required for head loss. Hence Option A is correct.
వివరణ:
Question 4
ప్రశ్న 4
A uniform pipe experiences a frictional head loss of 6.0 m over a length of 300 m. What is the friction slope Sf = hf/L?
Explanation:
The friction slope is defined as the head loss per unit pipe length. Thus Sf = hf/L. Substituting the data gives Sf = 6.0/300. This is equal to 0.020. Therefore Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
Water flows at a mean velocity of 2.0 m/s through a 200 mm diameter pipe 100 m long. If the Darcy friction factor is 0.020, what is the major head loss? Take g = 9.81 m/s².
Explanation:
The Darcy-Weisbach equation is hf = f(L/D)(V²/2g). Here L/D = 100/0.20 = 500. The velocity head is 2²/(2 × 9.81) = approximately 0.2039 m. Therefore hf = 0.020 × 500 × 0.2039 = approximately 2.04 m. Hence Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
For the horizontal constant-diameter water pipe in the previous type of calculation, suppose the frictional head loss is 2.0387 m. If ρ = 1000 kg/m³ and the mean velocities at the two sections are equal, what frictional pressure drop corresponds to this loss?
Explanation:
For a horizontal constant-diameter pipe, equal velocities make the velocity-head terms cancel between the two sections. The frictional pressure drop is then Δp = ρghf. Substitution gives Δp = 1000 × 9.81 × 2.0387 Pa. This is approximately 20000 Pa, or 20.0 kPa. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
A pipe 60 m long and 150 mm in diameter carries fluid at a mean velocity of 2.5 m/s. If the measured major head loss is 3.0 m, what is the Darcy friction factor? Take g = 9.81 m/s².
Explanation:
Rearranging Darcy-Weisbach gives f = hf(D/L)(2g/V²). Substitution gives f = 3.0(0.15/60)[2(9.81)/(2.5²)]. The result is approximately 0.02354. The friction factor is dimensionless. Therefore Option B is correct.
వివరణ:
Question 8
ప్రశ్న 8
A pipe has diameter 120 mm and Darcy friction factor 0.024. Fluid flows through it at 2.2 m/s and experiences a major head loss of 4.0 m. What pipe length produces this loss? Take g = 9.81 m/s².
Explanation:
From Darcy-Weisbach, L = hfD(2g)/(fV²). Substituting hf = 4.0 m, D = 0.12 m, f = 0.024 and V = 2.2 m/s gives the required length. The numerical result is approximately 81.07 m. Rounding appropriately gives 81.1 m. Therefore Option C is correct.
వివరణ:
Question 9
ప్రశ్న 9
A pipe 100 m long carries fluid at 3.0 m/s. If the Darcy friction factor is specified as 0.020 and the permitted major head loss is 5.0 m, what pipe diameter is required when f is treated as constant for this calculation? Take g = 9.81 m/s².
Explanation:
With velocity specified independently, Darcy-Weisbach can be rearranged as D = fLV²/(2ghf). Substitution gives D = 0.020 × 100 × 3²/[2 × 9.81 × 5]. This gives D approximately equal to 0.1835 m. Converting to millimetres gives about 183.5 mm. Therefore Option C is correct under the explicitly stated constant-friction-factor assumption.
వివరణ:
Question 10
ప్రశ్న 10
Fluid flows through a 250 mm diameter pipe 150 m long. If the Darcy friction factor is 0.025 and the major head loss is 6.0 m, what is the mean velocity? Take g = 9.81 m/s².
Explanation:
Darcy-Weisbach gives V = sqrt[2ghfD/(fL)]. Substituting the stated values gives V = sqrt[2(9.81)(6)(0.25)/(0.025 × 150)]. The value under the square root is approximately 7.848. This gives V approximately equal to 2.80 m/s. Therefore Option C is correct.
వివరణ:
Question 11
ప్రశ్న 11
Water flows through an 80 m long, 200 mm diameter pipe. If the Darcy friction factor is 0.020 and the allowed major head loss is 4.0 m, what discharge can the pipe carry? Take g = 9.81 m/s².
Explanation:
First solve Darcy-Weisbach for mean velocity. The result is V = sqrt[2ghfD/(fL)] = approximately 3.132 m/s. The pipe area is A = π(0.20)²/4 = approximately 0.031416 m². Therefore Q = AV = 0.031416 × 3.132 = approximately 0.0984 m³/s. Hence Option B is correct.
వివరణ:
Question 12
ప్రశ్న 12
A pipe must carry Q = 0.050 m³/s over a length of 200 m with a maximum major head loss of 10 m. If the Darcy friction factor may be treated as constant at 0.020, what theoretical pipe diameter is required? Take g = 9.81 m/s².
Explanation:
For a circular pipe, V = 4Q/(πD²). Substitution into Darcy-Weisbach gives hf = 8fLQ²/(π²gD⁵) when f is treated as constant. Rearranging gives D = [8fLQ²/(π²ghf)]^(1/5). Substituting the numerical data gives D approximately equal to 0.1526 m. Therefore the theoretical required diameter is about 152.6 mm and Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
If Darcy friction factor, pipe length and mean velocity are held constant, how does major head loss vary with pipe diameter?
Explanation:
Darcy-Weisbach gives hf = f(L/D)(V²/2g). If f, L and V are all held constant, only the factor 1/D varies. Therefore head loss is inversely proportional to pipe diameter under these particular conditions. Doubling D would halve the calculated head loss. Hence Option C is correct.
వివరణ:
Question 14
ప్రశ్న 14
For a circular pipe carrying a fixed discharge Q, assume the Darcy friction factor and pipe length remain constant. How does the major head loss vary with diameter?
Explanation:
At fixed discharge, the mean velocity is V = 4Q/(πD²). Thus V² is proportional to Q²/D⁴. Darcy-Weisbach contains an additional factor L/D. With f and L treated as constant, hf is therefore proportional to Q²/D⁵. Hence Option D is correct.
వివరణ:
Question 15
ప్రశ్న 15
A uniform pipe is replaced by another pipe of twice the length but with the same diameter, Darcy friction factor and mean velocity. How does the major head loss change?
Explanation:
The Darcy-Weisbach equation contains pipe length linearly through L/D. With diameter, friction factor and mean velocity unchanged, head loss is directly proportional to length. Doubling the length therefore doubles the major head loss. This represents the accumulation of distributed frictional resistance over a greater distance. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
For a given pipe with constant Darcy friction factor, what happens to the major head loss if the mean velocity is doubled?
Explanation:
Darcy-Weisbach contains the square of the mean velocity. Thus hf is proportional to V² when f, L and D remain fixed. Doubling V multiplies V² by four. The head loss consequently becomes four times its original value under the stated constant-f assumption. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
For a fixed circular pipe, assume the Darcy friction factor remains unchanged. If the discharge is doubled, what happens to the major head loss?
Explanation:
For a fixed pipe area, mean velocity is directly proportional to discharge. Doubling discharge therefore doubles the mean velocity. Darcy-Weisbach head loss is proportional to V² when the friction factor is treated as constant. The head loss consequently increases by a factor of four. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
Water flows at 0.030 m³/s through a pipe in which the major frictional head loss is 8.0 m. What hydraulic power is dissipated by pipe friction? Take ρ = 1000 kg/m³ and g = 9.81 m/s².
Explanation:
Hydraulic power associated with a head change is ρgQH. For frictional dissipation, H is the friction head loss hf. Thus Ploss = 1000 × 9.81 × 0.030 × 8.0 W. This gives 2354.4 W, or approximately 2.35 kW. Therefore Option B is correct.
వివరణ:
Question 19
ప్రశ్న 19
A liquid flow of 0.025 m³/s experiences a frictional pressure drop of 120 kPa through a pipe section. What rate of mechanical-energy loss is associated with this pressure drop when elevation and mean velocity are unchanged between the ends?
Explanation:
For incompressible flow with unchanged elevation and mean velocity, the frictional hydraulic power loss may be written as ΔpQ. Substituting Δp = 120000 Pa and Q = 0.025 m³/s gives 3000 W. This is equal to 3.00 kW. The same result would be obtained from ρgQhf using hf = Δp/(ρg). Therefore Option B is correct.
వివరణ:
Question 20
ప్రశ్న 20
In a fixed pipe, assume the Darcy friction factor remains unchanged. If the discharge increases by 20%, by what factor does the major head loss change?
Explanation:
For a fixed pipe, velocity is directly proportional to discharge. Under the stated constant-friction-factor assumption, Darcy-Weisbach gives hf proportional to V² and therefore to Q². Increasing discharge by 20% multiplies Q by 1.20. The new head loss is consequently multiplied by 1.20² = 1.44. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
Water flows at 0.120 m³/s through a 300 mm diameter pipe 250 m long. If the Darcy friction factor is 0.018, what is the major head loss? Take g = 9.81 m/s².
Explanation:
The pipe area is π(0.30)²/4, giving a mean velocity of approximately 1.698 m/s. Darcy-Weisbach then gives hf = 0.018(250/0.30)[1.698²/(2 × 9.81)]. Evaluating the expression gives approximately 2.203 m. Rounding to the listed precision gives 2.20 m. Therefore Option B is correct.
వివరణ:
Question 22
ప్రశ్న 22
For a circular pipe with fixed Darcy friction factor, the major-loss relation may be written hf = RQ². For f = 0.020, L = 100 m and D = 0.20 m, what is R approximately when g = 9.81 m/s²?
Explanation:
For a circular pipe, substitution of V = 4Q/(πD²) into Darcy-Weisbach gives hf = [8fL/(π²gD⁵)]Q². Therefore R = 8fL/(π²gD⁵). Substituting f = 0.020, L = 100 m and D = 0.20 m gives approximately 516.4 s²/m⁵. The unusual units are required so that RQ² has dimensions of length. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A calculation uses the Darcy-Weisbach form hf = f(L/D)(V²/2g). A source instead supplies a Fanning friction factor fF = 0.005. What value should be used for f in the stated Darcy-Weisbach equation?
Explanation:
The Darcy friction factor is four times the Fanning friction factor for the same flow. Thus fD = 4fF. With fF = 0.005, the required Darcy factor is 4 × 0.005 = 0.020. Using the Fanning value directly in the stated Darcy equation would underestimate the head loss by a factor of four. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
For steady flow through a uniform constant-diameter pipe with no pump or turbine and with only distributed pipe friction considered, which statement about the friction slope and energy grade line is correct?
Explanation:
Friction dissipates mechanical energy as fluid moves along a pipe. If hf is the frictional head loss over a uniform length L, the average friction slope is Sf = hf/L. The energy grade line therefore falls in the direction of flow by hf over that length. For a constant-diameter pipe with constant mean velocity, the hydraulic and energy grade lines remain separated by the constant velocity head. Hence Option A is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements about major pipe losses using the Darcy friction factor: I. For specified f, D and V, Darcy-Weisbach head loss is directly proportional to pipe length. II. For specified f, L and Q in a circular pipe, head loss is proportional to D^-5. III. Frictional hydraulic-power loss is ρgQhf. IV. A so-called minor loss can exceed the distributed major loss in a sufficiently short system. Which option is correct?
Explanation:
Statement I follows directly from the L/D factor in the Darcy-Weisbach equation. Statement II follows after substituting V = 4Q/(πD²) while treating f as specified and constant. Statement III is the hydraulic power associated with the frictional head dissipated at discharge Q. Statement IV is also correct because the labels major and minor do not guarantee relative magnitude, especially in short systems containing restrictive fittings or abrupt geometry changes. 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. Distributed loss caused primarily by wall friction over the length of a pipe సమాధానం: A.
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Question 2 ప్రశ్న 2Answer: B. hf = f(L/D)(V²/2g) సమాధానం: B.
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Question 3 ప్రశ్న 3Answer: A. Both f and L/D are dimensionless సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. 0.020 సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. 2.04 m సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: B. 20.0 kPa సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. 0.0235 సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: C. 81.1 m సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: C. 183.5 mm సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: C. 2.80 m/s సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 0.0984 m³/s సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. 152.6 mm సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: C. hf is proportional to 1/D సమాధానం: C.
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Question 14 ప్రశ్న 14Answer: D. hf is proportional to D^-5 సమాధానం: D.
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Question 15 ప్రశ్న 15Answer: C. It doubles సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: C. It becomes four times as large సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. It becomes four times as large సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: B. 2.35 kW సమాధానం: B.
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Question 19 ప్రశ్న 19Answer: B. 3.00 kW సమాధానం: B.
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Question 20 ప్రశ్న 20Answer: C. 1.44 సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. 2.20 m సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: C. 516.4 s²/m⁵ సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: C. 0.020 సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: A. The magnitude of the energy-grade-line slope is hf/L సమాధానం: A.
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Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.
