Please keep at least one language enabled. || కనీసం ఒక భాషను ఎంచుకోండి.
Question 1
ప్రశ్న 1
Which set of assumptions is appropriate for the classical Hagen-Poiseuille equation for flow through a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille solution is derived for steady laminar flow of a Newtonian fluid in a circular pipe. The fluid is treated as incompressible and its viscosity is taken as constant. The flow must be hydrodynamically fully developed so that the axial velocity profile no longer changes with downstream position. The no-slip condition requires the axial velocity to become zero at the stationary pipe wall. Therefore Option A states the appropriate assumptions for the classical solution.
వివరణ:
Question 2
ప్రశ్న 2
Which statement correctly describes the axial velocity distribution for fully developed laminar flow through a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille solution gives a quadratic dependence of axial velocity on radial position. The no-slip boundary condition makes the velocity zero at the pipe wall. Symmetry produces a zero radial velocity gradient at the centreline. The axial velocity therefore reaches its maximum value at the pipe centre and decreases parabolically toward the wall. Hence Option B correctly describes the fully developed laminar velocity profile.
వివరణ:
Question 3
ప్రశ్న 3
For fully developed laminar flow in a circular pipe, what is the relationship between maximum centreline velocity Vmax and cross-sectional mean velocity V?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The velocity profile for fully developed laminar circular-pipe flow is parabolic. Integrating that profile over the circular cross-section gives the mean velocity. The resulting mean velocity is exactly one-half of the centreline maximum velocity. Thus V = Vmax/2, or equivalently Vmax = 2V. Therefore Option C is correct.
వివరణ:
Question 4
ప్రశ్న 4
For fully developed laminar flow in a circular pipe, the velocity profile may be written as u(r) = 2V[1 - (r/R)^2], where V is mean velocity. What is u at r = R/2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Substituting r/R = 1/2 into the parabolic profile gives u = 2V[1 - (1/2)^2]. The squared radius ratio is 1/4. Therefore u = 2V(3/4). This gives u = 1.50V. Hence Option C is correct.
వివరణ:
Question 5
ప్రశ్న 5
In fully developed laminar circular-pipe flow, at what radial position does the local velocity equal the mean velocity V?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The local velocity is u = 2V[1 - (r/R)^2]. Setting u equal to V gives 1 = 2[1 - (r/R)^2]. Therefore (r/R)^2 = 1/2. Taking the positive radial coordinate gives r/R = 1/√2. Hence Option B is correct.
వివరణ:
Question 6
ప్రశ్న 6
How does the magnitude of shear stress vary with radius in fully developed laminar flow through a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fully developed circular-pipe flow, the shear-stress magnitude is proportional to radial distance from the centreline. Symmetry requires the velocity gradient and therefore the shear stress to vanish at r = 0. The velocity gradient increases in magnitude toward the wall. Consequently the shear stress increases linearly from zero at the centreline to its maximum magnitude at the wall. Therefore Option B is correct.
వివరణ:
Question 7
ప్రశ్న 7
For a horizontal circular pipe of length L and diameter D carrying fully developed laminar flow with pressure drop Δp = p1 - p2 > 0, which expression gives the magnitude of wall shear stress?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
A force balance on a cylindrical fluid element relates the axial pressure force to the resisting shear at its surface. For a pipe of radius R, the wall-shear magnitude is τw = ΔpR/(2L). Since D = 2R, this becomes τw = ΔpD/(4L). The same expression is consistent with the radial shear-stress distribution obtained from the velocity gradient. Therefore Option A is correct.
వివరణ:
Question 8
ప్రశ్న 8
What is the shear stress at the centreline of a circular pipe carrying steady fully developed laminar flow?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The laminar velocity profile is symmetric about the pipe centreline. Its radial derivative is therefore zero at r = 0. For a Newtonian fluid, shear stress is proportional to this velocity gradient. Equivalently, the fully developed shear-stress distribution varies directly with radius. Hence the centreline shear stress is zero and Option A is correct.
వివరణ:
Question 9
ప్రశ్న 9
Which expression gives the Hagen-Poiseuille discharge through a horizontal circular pipe in terms of pressure drop Δp, diameter D, length L and dynamic viscosity μ?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille law is obtained by integrating the parabolic laminar velocity profile over the pipe cross-section. In terms of pipe radius, the result is Q = πΔpR⁴/(8μL). Substituting R = D/2 converts R⁴ to D⁴/16. This produces Q = πΔpD⁴/(128μL). Therefore Option B is correct.
వివరణ:
Question 10
ప్రశ్న 10
Two geometrically similar circular pipes carry the same Newtonian fluid under the same pressure gradient in fully developed laminar flow. If the diameter of the second pipe is twice that of the first, what is Q2/Q1?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fixed pressure gradient and viscosity, Hagen-Poiseuille flow gives Q proportional to D⁴. Doubling the pipe diameter therefore multiplies the discharge by 2⁴. This produces a factor of 16. The strong fourth-power dependence explains why small changes in tube diameter can strongly affect laminar discharge. Therefore Option D is correct.
వివరణ:
Question 11
ప్రశ్న 11
For fully developed laminar flow through a given circular pipe under a fixed pressure drop, what happens to the discharge if the dynamic viscosity is doubled while all other quantities remain unchanged?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille equation shows that discharge is inversely proportional to dynamic viscosity. Greater viscosity produces greater resistance to shear deformation in the pipe. If μ is doubled while Δp, D and L remain fixed, Q is divided by two. This conclusion applies under the same fully developed laminar assumptions. Therefore Option C is correct.
వివరణ:
Question 12
ప్రశ్న 12
A Newtonian liquid of dynamic viscosity 0.050 Pa·s flows steadily and fully developed through a horizontal circular pipe of diameter 40 mm and length 10 m at a mean velocity of 0.20 m/s. What pressure drop is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fully developed laminar circular-pipe flow, Δp = 32μVL/D². Substituting μ = 0.050 Pa·s, V = 0.20 m/s, L = 10 m and D = 0.040 m gives Δp = 32(0.050)(0.20)(10)/(0.040²). The numerator is 3.2 and the denominator is 0.0016. Thus Δp = 2000 Pa, or 2.00 kPa. Therefore Option C is correct.
వివరణ:
Question 13
ప్రశ్న 13
A liquid of dynamic viscosity 0.10 Pa·s flows steadily and fully developed through a 20 mm diameter horizontal pipe 5 m long under a pressure drop of 4.0 kPa. What discharge is predicted by the Hagen-Poiseuille equation?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille equation is Q = πΔpD⁴/(128μL). Here D⁴ = (0.020)⁴ = 1.6 × 10^-7 m⁴. Substitution gives Q = π(4000)(1.6 × 10^-7)/[128(0.10)(5)]. This evaluates to approximately 3.14 × 10^-5 m³/s. Therefore Option B is correct.
వివరణ:
Question 14
ప్రశ్న 14
A viscous liquid of dynamic viscosity 0.080 Pa·s flows at 5.0 × 10^-4 m³/s through a horizontal circular pipe of diameter 40 mm and length 12 m. Assuming fully developed laminar flow, what pressure drop is required?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Rearranging the Hagen-Poiseuille equation gives Δp = 128μLQ/(πD⁴). Substituting μ = 0.080 Pa·s, L = 12 m, Q = 5.0 × 10^-4 m³/s and D = 0.040 m gives the required pressure difference. The numerical result is approximately 7639 Pa. This is about 7.64 kPa. Therefore Option C is correct.
వివరణ:
Question 15
ప్రశ్న 15
An oil of density 850 kg/m³ and dynamic viscosity 0.10 Pa·s flows steadily and fully developed through a 50 mm diameter horizontal pipe 20 m long at a mean velocity of 0.20 m/s. What is the frictional head loss? Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pressure drop for laminar pipe flow is Δp = 32μVL/D². Substitution gives Δp = 32(0.10)(0.20)(20)/(0.050²) = 5120 Pa. The corresponding head loss is hf = Δp/(ρg). Thus hf = 5120/(850 × 9.81) = approximately 0.614 m. Therefore Option C is correct.
వివరణ:
Question 16
ప్రశ్న 16
For the oil flow in a 50 mm diameter pipe with μ = 0.10 Pa·s and mean velocity V = 0.20 m/s, what is the wall shear-stress magnitude under fully developed laminar conditions?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fully developed laminar circular-pipe flow, the wall-shear magnitude can be written as τw = 8μV/D. Substituting μ = 0.10 Pa·s, V = 0.20 m/s and D = 0.050 m gives τw = 8(0.10)(0.20)/0.050. This evaluates to 3.20 Pa. The same value would result from τw = ΔpD/(4L) if the corresponding pressure drop were used. Therefore Option C is correct.
వివరణ:
Question 17
ప్రశ్న 17
In fully developed laminar pipe flow, the wall shear-stress magnitude is 4.0 Pa. What is the shear-stress magnitude at r = R/2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The shear-stress magnitude in fully developed circular-pipe flow varies linearly with radius. Thus |τ(r)|/τw = r/R. At r = R/2, the local magnitude is one-half of the wall value. One-half of 4.0 Pa is 2.0 Pa. Therefore Option B is correct.
వివరణ:
Question 18
ప్రశ్న 18
A fully developed laminar flow in a circular pipe has a mean velocity of 0.25 m/s. What is the centreline velocity?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For fully developed laminar flow in a circular pipe, the centreline velocity is the maximum velocity. The parabolic profile gives Vmax = 2V. With a mean velocity of 0.25 m/s, the centreline velocity is 2 × 0.25 m/s. This equals 0.50 m/s. Therefore Option D is correct.
వివరణ:
Question 19
ప్రశ్న 19
A 60 mm diameter circular pipe carries fully developed laminar flow with mean velocity 0.30 m/s. What is the local axial velocity at a radial distance of 20 mm from the centreline?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The pipe radius is R = 30 mm and the specified radial coordinate is r = 20 mm. The velocity distribution is u = 2V[1 - (r/R)²]. Therefore u = 0.60[1 - (20/30)²]. This gives u = 0.60(5/9) = approximately 0.333 m/s. Hence Option C is correct.
వివరణ:
Question 20
ప్రశ్న 20
For fully developed laminar circular-pipe flow with velocity profile u = Vmax[1 - (r/R)^2], what fraction of the total discharge passes through the central region 0 ≤ r ≤ R/2?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The discharge through a circular annulus is obtained from dQ = u(2πr dr). Integrating the parabolic profile from zero to a radius aR and dividing by the total discharge gives Q(a)/Q = 2a² - a⁴. For a = 1/2, the fraction becomes 2(1/4) - 1/16. This equals 7/16, or 43.75%. Therefore Option C is correct.
వివరణ:
Question 21
ప్రశ్న 21
For fully developed laminar flow of the same fluid through the same circular pipe at the same mean velocity, what happens to the pressure drop if the pipe length is doubled?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For Hagen-Poiseuille flow, Δp = 32μVL/D². With μ, V and D held constant, pressure drop is directly proportional to pipe length. Doubling L therefore doubles Δp. Equivalently, the pressure gradient remains constant while the pressure loss accumulates over twice the distance. Therefore Option C is correct.
వివరణ:
Question 22
ప్రశ్న 22
Using the Darcy-Weisbach equation with the Darcy friction factor f, which relation is consistent with fully developed laminar flow in a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille pressure-loss expression can be written in the Darcy-Weisbach form. Equating the two relations for fully developed laminar circular-pipe flow gives the Darcy friction factor f = 64/Re. This is the Darcy factor, not the Fanning friction factor. The corresponding Fanning friction factor would be 16/Re. Therefore Option C is correct.
వివరణ:
Question 23
ప్రశ్న 23
A fully developed laminar pipe flow has Reynolds number 800. Using the Darcy friction factor and the Darcy-Weisbach equation, determine the head loss in a 40 mm diameter pipe of length 10 m carrying fluid at mean velocity 0.15 m/s. Take g = 9.81 m/s².
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
For laminar circular-pipe flow, the Darcy friction factor is f = 64/Re = 64/800 = 0.08. The length-to-diameter ratio is 10/0.040 = 250. The velocity head is V²/(2g) = 0.15²/(2 × 9.81) = approximately 0.001147 m. Darcy-Weisbach gives hf = 0.08 × 250 × 0.001147 = approximately 0.0229 m. Therefore Option C is correct.
వివరణ:
Question 24
ప్రశ్న 24
Why should the Hagen-Poiseuille velocity distribution not be applied directly within the hydrodynamic entrance region of a circular pipe?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
The Hagen-Poiseuille profile assumes that the axial velocity distribution no longer changes with downstream coordinate. In the entrance region, boundary layers grow from the pipe wall and progressively reshape the velocity profile. The axial profile therefore remains a function of downstream position before becoming fully developed. The entrance flow can still be laminar, but it does not yet satisfy the fully developed assumption used in the classical derivation. Therefore Option B is correct.
వివరణ:
Question 25
ప్రశ్న 25
Consider the following statements for steady fully developed laminar flow of a Newtonian fluid through a circular pipe: I. The axial velocity profile is parabolic. II. The maximum velocity is twice the mean velocity. III. The shear-stress magnitude is zero at the centreline and maximum at the wall. IV. For fixed pressure gradient and viscosity, discharge is proportional to the fourth power of pipe diameter. Which option is correct?
Select an option first. || ముందుగా ఒక సమాధానాన్ని ఎంచుకోండి.
Explanation:
Statement I is correct because the Hagen-Poiseuille solution produces a quadratic radial velocity distribution. Statement II is correct because cross-sectional integration of that profile gives Vmax = 2V. Statement III is correct because the shear stress varies linearly with radial distance and vanishes by symmetry at the centreline. Statement IV is correct because the Hagen-Poiseuille discharge contains D⁴ when pressure gradient and viscosity are fixed. Therefore all four statements are correct and Option D is the valid answer.
వివరణ:
Answer Key సమాధానాల పట్టిక
-
Question 1 ప్రశ్న 1Answer: A. Steady, incompressible, Newtonian, fully developed laminar flow in a straight circular pipe with constant viscosity and no slip at the wall సమాధానం: A.
-
Question 2 ప్రశ్న 2Answer: B. It is parabolic, with maximum velocity at the centreline and zero velocity at the wall సమాధానం: B.
-
Question 3 ప్రశ్న 3Answer: C. Vmax = 2V సమాధానం: C.
-
Question 4 ప్రశ్న 4Answer: C. 1.50V సమాధానం: C.
-
Question 5 ప్రశ్న 5Answer: B. r/R = 1/√2 సమాధానం: B.
-
Question 6 ప్రశ్న 6Answer: B. It varies linearly from zero at the centreline to a maximum at the wall సమాధానం: B.
-
Question 7 ప్రశ్న 7Answer: A. τw = ΔpD/(4L) సమాధానం: A.
-
Question 8 ప్రశ్న 8Answer: A. Zero సమాధానం: A.
-
Question 9 ప్రశ్న 9Answer: B. Q = πΔpD⁴/(128μL) సమాధానం: B.
-
Question 10 ప్రశ్న 10Answer: D. 16 సమాధానం: D.
-
Question 11 ప్రశ్న 11Answer: C. The discharge is halved సమాధానం: C.
-
Question 12 ప్రశ్న 12Answer: C. 2.00 kPa సమాధానం: C.
-
Question 13 ప్రశ్న 13Answer: B. 3.14 × 10^-5 m³/s సమాధానం: B.
-
Question 14 ప్రశ్న 14Answer: C. 7.64 kPa సమాధానం: C.
-
Question 15 ప్రశ్న 15Answer: C. 0.614 m సమాధానం: C.
-
Question 16 ప్రశ్న 16Answer: C. 3.20 Pa సమాధానం: C.
-
Question 17 ప్రశ్న 17Answer: B. 2.0 Pa సమాధానం: B.
-
Question 18 ప్రశ్న 18Answer: D. 0.500 m/s సమాధానం: D.
-
Question 19 ప్రశ్న 19Answer: C. 0.333 m/s సమాధానం: C.
-
Question 20 ప్రశ్న 20Answer: C. 43.75% సమాధానం: C.
-
Question 21 ప్రశ్న 21Answer: C. It doubles సమాధానం: C.
-
Question 22 ప్రశ్న 22Answer: C. f = 64/Re సమాధానం: C.
-
Question 23 ప్రశ్న 23Answer: C. 0.0229 m సమాధానం: C.
-
Question 24 ప్రశ్న 24Answer: B. Because the velocity profile is still developing with downstream position and is not yet fully developed సమాధానం: B.
-
Question 25 ప్రశ్న 25Answer: D. I, II, III and IV సమాధానం: D.