A TEXTBOOK OF FLUID MECHANICSAND

HYDRAULIC MACHINES(QUESTIONS FROM VARIOUS UNIVERSITIES OF NIGERIA)

[SUPPLEMENT]

A TEXTBOOK OF

FLUID MECHANICSAND

HYDRAULIC MACHINES(QUESTIONS FROM VARIOUS UNIVERSITIES OF NIGERIA)

[SUPPLEMENT]

By

Dr. M.G. SOBAMOWO Dr. BAYO OGUNMOLA

Mechanical Department Mechanical DepartmentFaculty of Engineering, Lecturer GR II Faculty of Engineering, Lecturer I

University of Lagos University of LagosAkoka-Yaba, Lagos Akoka-Yaba, Lagos

Nigeria Nigeria

Dr. R.K. BANSAL

B. Sc. Engg. (Mech.), M.Tech. Hons. (I.I.T., Delhi)Ph. D., M.I.E. (India)

FormerlyProfessor, Department of Mechanical Engineering,

Dean (U.G. Studies), Delhi College of Engineering, Delhi

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A TEXTBOOK OF FLUID MECHANICS AND HYDRAULIC MACHINES [SUPPLEMENT]

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First Edition : 2015ISBN : 978-93-5138-286-7

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CONTENTSChapters Pages

1. Properties of Fluids ... 32. Pressure and Its Measurement ... 133. Hydrostatic Forces on Surfaces ... 204. Buoyancy and Floatation ... 325. Kinematics of Flow and Ideal Flow ... 426. Dynamics of Fluid Flow ... 537. Orifices and Mouthpieces ... 618. Notches and Weirs ... 669. Viscous Flow ... 70

10. Turbulent Flow ... 8211. Flow Through Pipes ... 9012. Dimensional and Model Analysis ... 10313. Boundary Layer Flow ... 10817. Impact of Jets and Jet Propulsion ... 12418. Hydraulic Machines—Turbines ... 12719. Centrifugal Pumps ... 136

Note: There are no extra questions on Chapters 14, 15, 16, 20 and 21.

SOME MORE IMPORTANT QUESTIONS

(TAKEN FROM EXAMINATION OF VARIOUS NIGERIAN UNIVERSITIES)

1: Properties of Fluids

Problem 1 (a) State Newton’s Law of Viscosity.

(b) An unknown Non-Newtonian fluid flows over a flat surface and developed a velocity profile

u

U

3y

2

1

2

y2

= − FHGIKJδ δ

near the surface. In an attempt to determine the actual shear stress distribution in

the fluid, it was found that the fluid follows a shear stress velocity gradient relationship τ = Adu

dy

nFHGIKJ + B

with the exponential n = 1.3 and developed a shear stress of 205 kN/m2 and 103 kN/m2 at the pointswhere y = 0 and y = δ respectively.

(i) Develop the expressions (in terms of y) for the shear stress distribution over the flat surface.

(ii) If U = 0.25 m/s and δ = 0.05 mm, what is the shear stress of the fluid at the point wherey = δ/2?

(iii) What should be the dynamic viscosity of a Newtonian fluid that will induce the same shearstress value on the surface for similar velocity profile and the same maximum velocity?

Solution. (a) Newton’s law of viscosity states that the shear stress (τ) on a fluid element layer isdirectly proportional to the rate of shear strain.

Mathematically, τ ∝ du

dy

which is τ = µdu

dy

where µ is the constant of proportional which is the coefficient of viscosity.(b) Given that

u

U

y y= FHGIKJ

3

2

1

2

2

δ δ– ...(i)

⇒ u = U3

2

1

2

2y y

δ δ− FHGIKJ

LNMM

OQPP ...(ii)

On differentiating eqn. (ii) with respect to y, we have

du

dyU

y= −LNMOQP

3

2 2δ δ...(iii)

where,du

dy

U

y =

=0

3

2δ...(iv)

anddu

dyU

U

y =

= −LNMOQP =δ δ δ δ

3

2

1

2...(v)

3

4 Fluid Mechanics

(i) Given that, n = 1.3, τ = 205 kN/m2, y = 0n = 1.3, τ = 103 kN/m2, y = δ

τ = Adu

dy

nFHGIKJ + B

then, 205 = A3

2

1 3U

δFHGIKJ

.

+ B ...(vi)

103 = AU

2

1 3

δFHGIKJ

.

+ B ...(vii)

If U = 205 m/s, δ = 0.05 mm = 0.00005 m, then equations (vi) and (vii) become205 = 1.09 × 105 A + B ...(viii)103 = 2.641 × 104 A + B ...(ix)

Solving the equation (viii) and (ix) simultaneously, we haveA = 1.2344 × 10–3

B = 70.39

Substituting the values of A and B, and expression du

dy into the given expression, we have

τ = 1.2344 × 10–3 Uy3

2 2

1 3

δ δ−FHGIKJ

LNM

OQP

.

+ 70.39 ...(x)

So, τ = 1.234 × 10–3 × 0.251.3 3

2 0 00005 0 00005 2

1 3

( . ) ( . )

.

−LNM

OQP

y + 70.39

= 2.036 × 10–4 (30000 – 4.0 × 108 y)1.3 + 70.39= 2.036 × 10–4 × (104)1.3 (3 – 40000y)1.3 + 70.39

Thus, the required expression isτ = 32.268425 (3 – 40000y)1.3 + 70.39 Ans.

(ii) For y = δ2

, where δ = 0.00005 m

∴ y = δ2

0 00005

2= .

= 0.000025 m

Now, τ = 32.268425 [3 – 40000 (0.000025)]1.3 + 70.39⇒ τ = 149.844 N/m2 ~− 150 N/m2. Ans.(iii) For the Newtonian fluid

τ = µ du

dy

On the flat surface, y = 0 and at such point the shear stress is given as 205 N/m2.

So, τ µy

y

du

dy==

=0

0

Properties of Fluids 5

i.e., 205 = µ U3

2δFHGIKJ

LNM

OQP

= µ 0 253

2 0 00005.

( . )

FHG

IKJ

LNM

OQP

µ =205 2 0 00005

0 25 3

× ××

.

.

= 2.73 × 10–2 Ns/m2

= 0.0273 Ns/m2 Ans.Problem 2 (a) Differentiate between Newtonian and Non-Newtonian fluids.

(b) (i) A 90 N rectangular solid block slides down a 30° inclined plane. The plane is lubricated by a3 mm thick of oil of relative density 0.90 and kinematic viscosity of 8.9 × 10–4 m2/s. If thecontact area is 0.3 m2, calculate the terminal velocity of the block.

(ii) If the plane is lubricated with a Non-Newtonian fluid, at what value terminal velocity will theblock slide down the inclined plane given that the fluid is dilatant with n = 1.25, µ = 0.96 Ns/m2?Assuming, the fluid develops the same value of the shear stress.

(iii) Calculate the force required to keep the block sliding down the plane and the force requiredto push the block up the plane (under the action of the Non-Newtonian fluid).

Solution. (a) Newtonian fluids are fluids that obey the Newton’s law of viscosity while Non-Newtonian fluids are the fluids that do not obey Newtonian law of viscosity. A fluid, in which theshear stress is directly proportional to shear strain is called Newtonian, otherwise Non-Newtonian.

(b) (i) From the figure,

The shear force at the bottom of the UF = W sin 30°

= 90 sin 30°= 90 × 0.5 = 45 N

The shear force, F = 45 N

τ = F/A = 45

0 3. = 150 N/m2

τ = 150 N/m2

and µ = νρwhere, ρ = 0.9 × 1000 = 900 kg/m3 and ν = 8.9 × 10–4 m2/s

µ = 900 × 8.9 × 10–4 = 0.8 Ns/m2

µ = 0.8 Ns/m2

From the Newton’s law of viscosity

τ = µ du

dy

U

T= µ

τ = µ U

T, where T = 3 mm

30°

3 mm

�W sin 30°

W cos 30°

W sin 30°

W = 90 N

6 Fluid Mechanics

150 = 0.8U

0 003.

U =150 0 003

0 8

× .

. = 0.5625 m/s Ans.

(ii) If the fluid is Non-Newtonian,

τ = µ du

dy

U

T

n nFHGIKJ = FHG

IKJµ

τ = µ U

T

nFHGIKJ

150 = 0.96U

0 003

1 25

.

.FHGIKJ

U = 0.003150

0.961.25 .

U = 0.1707 m/s Ans.(iii) From the figure, the force required to keep the body from sliding down the plane must be

equal to the shear force i.e.,Pkeep = W sin 30° = 90 sin 30° = 90 × 0.5

P = 45 Nwhile the force required to push the block up the plane, must be greater than the shear force in orderto over the shear force that makes the body to slide down the plane.i.e., Pup > 45 N Ans.Problem 3 Water at room temperature flows in between two flat vertical plates which are L m apart

and developed a velocity profile u

U

3x

2–

1

2sin

2

xw

max

3

= FHGIKJδ

πδ

near the surfaces of the plates. If the

water is totally drained out of the plates and a hot air at 59°C is allowed to flow in between the plates

developing a velocity profile u

U

3x

2

1

2

xa

max

3

= − FHGIKJδ δ

.

Show that the maximum shear stress developed by the air flow doubled the maximum shearstress developed by the water flow in between the plates and that happened on the left-hand sideplate.

Solution. For water,

u

U

x xw

max

sin= − FHGIKJ

3

2

1

2 2

3

δπ

δ

du

dxU

x xw =FHGIKJFHGIKJ

LNMM

OQPPmax – cos

3

2

1

2 2

3

2

2

3

3

δπ

δπ

δ

30°Wsin

30°W cos 30°

W sin 30°

90 N

D

Properties of Fluids 7

du

dxw = Umax

3

2

3

4 2

2

3

3

δπδ

πδ

− FHGIKJ

LNMM

OQPP

x xcos

From the Newton’s law of viscosity

τ = µ du

dxw

For water, τw = µwdu

dx

τw = µw Umax 3

2

3

4 2

2

3

3

δπδ

πδ

− FHGIKJ

LNMM

OQPP

x xcos

d

dxwτ

= µw Umax 3

4 2

3

2

6

4 2

2

3

2

3

3

3

3πδ

πδ

πδ

πδ

πδ

x x x x xFHGIKJ

FHG

IKJFHGIKJ − F

HGIKJ

LNMM

OQPPsin cos

At the maximum shear stress, d

dxwτ

= 0.

So, 6

4

3

4 2 23

3

3

3 3µ πδ

πδ

πδ

πδ

w U x x x xmax sin cosFHGIKJ − F

HGIKJ

LNMM

OQPP

= 0

⇒ 3

4 2 2

3

3

3 3x x xπδ

πδ

πδ

sin cosFHGIKJ − F

HGIKJ = 0 or

6

4 3

µ πδ

w U xmax = 0

i.e., tan π

δδ

π2

4

3

3 3

3

x

x

FHGIKJ = or x = 0

Since the first expression will definitely lead to an ambiguous result, we say x = 0.

So, τw, max = µw Umax 3

2

3 0

4 20

2

33

δπ

δπ−

LNM

OQP

( )cos ( )

= µw Umax 3

2δFHGIKJ

∴ τw, max =3

2maxµ

δw U

For air, u

U

x xa

max

= − FHGIKJ

3

2

1

2

3

δ δ

du

dx

xa = −FHG

IKJ

3

2

3

2

2

3δ δ Umax

From the Newton’s law of viscosity

τ = µ du

dx

Water

x = 0 x = LVertical plates

8 Fluid Mechanics

So, τa = µa3

2

3

2

2

3δ δ−

FHG

IKJ

xUmax

d

dxaτ

= – µa Umax 6

2 3

x

δFHGIKJ = 0

⇒ x = 0i.e., where the max shear stress occurs.

As x = 0, τa, max = µa 3

2

3 0

2

2

3δ δ−

FHG

IKJ

( ) Umax

τa, max =3

2maxµ

δa U

But µw = µ0 1

1 2+ +FHG

IKJα βT T

= 1.79 × 10–4 1

1 0 03368 25 0 000221 25 2+ +FHG

IKJ. ( ) . ( )

= 1.79 × 10–4 1

1 0842 0138125+ +FHG

IKJ. .

= 1.79 × 10–4 1

1980125.FHG

IKJ

⇒ µw = 9.04 × 10–5 Ns/m2

And µa = µ0 + αT – βT 2

= 1.79 × 10–4 + 0.000000056 (59) – 1.89 × 10–10 (59)2

= 1.79 × 10–4 + 0.000003304 – 0.0000006579= 0.000179 + 0.000003304 – 0.0000006579

⇒ µa = 1.82 × 10–4 Ns/m2

ττ

µδ

µδ

µµ

w

a

w

a

w

a

U

U,max

, max

max

max

=

FHG

IKJ

FHG

IKJ

=

32

32

⇒ ττ

w

a

, max

, max

.

.~ . ~= ×

×− −

−

−9 04 10

182 1005

1

2

5

4

∴ τa, max = 2 τw, max Ans.and this occurs where x = 0 i.e., at the left hand side plate. Ans.Problem 4 A non-Newtonian fluid having a specific gravity 0.95 and a kinetic viscosity of5.75 × 10–4 m2/s flows past a fixed surface. The velocity profile near the surface is given by

u

U

3y

2

1

2sin

2

y 3

= FHIKδ

πδ

– .

A Supplement Of A Textbook Of FluidMechanics And Hydraulic Machines

Publisher : Laxmi Publications ISBN : 9789351382867Author : Dr M. G.Sobamowo, Dr BayoOgunmola, Dr R. K. Bansal

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