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# Poiseuille’s Law

1 - Question

For flow of water in tube, the velocity at surface is ___________

a) zero

b) equal to velocity at 1/3 of diameter

c) equal to velocity at 1/2 of diameter

d) equal to velocity at centre

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Answer: aExplanation: The surface of the tube is wet forming a very thin layer of water that is immobile due to molecular attraction. So the velocity at the surface is zero.

2 - Question

In Poiseuille’s law of flow, the variation of the velocity is given by _______

a) v = hγw4ηL(r2−R2)

b) v = hγw14ηL(R2−r2)

c) v = hγw4ηL(R2−r2)

d) v = hγw8ηL(R2−r2)

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Answer: cExplanation: Consider a capillary tube of length L and radius R.

In equilibrium the sum of the forces is zero, πr2h1γw– πr2h1γw-τ(2πrL)=0 on simplification, dv=−hγw2ηLrdr on integration, v=−hγw4ηLr2+CwhereC=hγw4ηLR2 ∴ v=hγw4ηL(R2−r2).

3 - Question

The velocity of flow of water at the centre of tube is _______ when the head is 0.1m, η=8.9*10-3 dynes-s/cm2, γw=1 g/cm3, the length is 10m and the diameter is 20cm.

a) 20 cm/s

b) 22 cm/s

c) 28 cm/s

d) 30 cm/s

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Answer: cExplanation: Given, h=0.1m η=8.9*10-3 dynes-s/cm2 γw=1 g/cm3 L= 10m d= 20cm. Since v=hγw4ηL(R2−r2) v=10∗14∗8.9∗10−3∗1000(102−02) v=28 cm/s.

4 - Question

The quantity of water flowing in thin cylindrical sheet of thickness dr is ____________

a) dq=hγw4ηL(R2−r2)2πrdr

b) dq=hγw4ηL(R2−r2)4πrdr

c) dq=hγw4ηL(R2−r2)8πrdr

d) dq=hγw4ηL(R2−r2)12πrdr

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Answer: aExplanation: Consider a capillary tube of length L and radius R. dq=(2πrdr)v substituting for velocity, dq=(2πrdr)∗hγw4ηL(R2−r2) ∴dq=hγw4ηL(R2−r2)2πrdr.

5 - Question

The total quantity of water flowing through the capillary tube is given by _________

a) q=hγw4ηLR2

b) q=hγw4ηLR4

c) q=hγw8ηLR2

d) q=hγw8ηLπR4

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Answer: dExplanation: The total quantity of flow is obtained by integrating with the limits 0 to R in the differential discharge. dq=hγw4ηL(R2−r2)2πrdr q=hγw4ηL∫R0(R2−r2)2πrdr ∴q=hγw8ηLπR4

6 - Question

The total quantity of water with respect to the hydraulic gradient flowing through the capillary tube is given by _________

a) q=hiγw4ηLR2

b) q=hγw4ηLR4/i

c) q=iγw8ηπR2

d) q=iγw8ηπR4

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Answer: dExplanation: The hydraulic gradient i is the head loss divided by the length travelled, i=h/L ∴q=hγw8ηLπR4=iγw8ηπR4.

7 - Question

The average velocity of the tube of area a is given by _________

a) vav=iγw8ηπa

b) vav=iγw8ηπa

c) vav=iγw2ηπa

d) vav=iγw4ηπa

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Answer: aExplanation: The average velocity is given by, Vav=q/a Vav=q/πr2 Substituting for q, ∴ vav=iγw8ηπa.

8 - Question

The hydraulic radius RH is the ratio of the area of the tube to its volume.

a) True

b) False

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Answer: aExplanation: The hydraulic radius RH is defined as the ratio of the area of the tube to the wetted perimeter. It is used to generally designate the velocity of flow in engineering hydraulics.

9 - Question

The hydraulic radius RH of circular tube is given by ________

a) RH=R/4

b) RH=R/6

c) RH=R/2

d) RH=R/12

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Answer: cExplanation: The hydraulic radius RH is defined as the ratio of the area of the tube to the wetted perimeter. ∴ RH=A/P=πR2/2πR ∴ RH=R/2.

10 - Question

The hydraulic radius RH is related with the voids ratio as ________

a) RH=Vs/eAs

b) RH=eVsAs

c) RH=e/AsVs

d) RH=eVs/As

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Answer: dExplanation: The hydraulic radius RH is given by, RH=A/P=AL/PL AL=eVs PL= As ∴ RH=eVs/As.

11 - Question

The quantity of flow for circular tube with respect to the hydraulic radius is given by ________

a) q=12γwR2Hη

b) q=12γwR2Hηia

c) q=12γwR2Hηi

d) q=12γwR2Hηa

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Answer: bExplanation: The quantity of flow for the capillary tube of any geometrical shape is given by, q=CSγwR2Hηia, where Cs is the shape constant. For circular tube Cs=1/2 ∴q=12γwR2Hηia.

12 - Question

The relationship between the quantity of flow with respect to the hydraulic radius and the voids ratio is given by ________

a) q=γwR2Hηe1+eiA

b) =CSγwR2Hηe1+e

c) =CSγwR2HηiA

d) =CSγwR2Hηe1+eiA

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Answer: dExplanation: Let the total cross-section of the soil be A and the porosity be n, ∴ a=nA ∴ q=12γwR2Hηia=12γwR2HηinA Since n=e/(1+e), ∴ q=CSγwR2Hηe1+eiA.