Jan 2007

3403

1yyV

3 cm

0.25 m/s

y

V

Plate

Fixed surface

1.FIGURE Q1 shows a plate with an area of 9 cm2 that moves at a constant velocity of 0.25

m/s. A Newtonian fluid with a specific gravity of 0.92 and a kinematic viscosity of 4 10-4

m2/s, is filled between the plate and a fixed surface at a constant thickness of 3 cm. Due to

the ‘no-slip’ condition, the velocity of the fluid, V, at any thickness of the fluid, y, can be

determined through the velocity profile, which is given as:

where:V = Velocity at any thickness of the fluidy = Thickness of the fluid measured from the fixed

surface

FIGURE Q1: Velocity profile of a Newtonian fluid

Determine:

a.The dynamic viscosity of the fluid.[3 marks]

b. The magnitude and direction of shear stress developed on the surface of the plate.[4 marks]

c. The shear stress at the fixed surface.[4 marks]

d. The magnitude of force required to move the plate. [4 marks]

e. If all values remain constant and the velocity of the fluid follows a linear profile, determine:

i.The new velocity gradient.[2 marks]

•The magnitude of force required to move the plate.[3 marks]

a.The dynamic viscosity of the Newtonian fluid.

Dynamic viscosity, = = (4 10-4)(1000)(0.92)= 0.368 kg/m.s

b. The magnitude and direction of shear stress developed on the surface.[4 marks]

For Newtonian fluid, shear stress is proportional to velocity gradient.

Pa 0.0123

0.051203

10.368

120y3

10.368

dy

dVμτ

120y3

1

dy

dV

40yy3

1V

2

2

2

3

Direction : to the left

c. The shear stress at the fixed surface.[4 marks]

At the fixed surface, y = 0

Pa 0.123

3

10.368

120y3

10.368

dy

dVμτ 2

d. Magnitude of force required to move the plate. [4 marks]

At the moving plate, = 0.0123 PaF = A = (0.0123)(9 10-4) = 1.107 10-5 N

e. If all values remain constant and the velocity of the fluid follows a linear profile, determine:

i. The new velocity gradient.[2 marks]

Since it follows a linear profile,

sy

V

dy

dV 133.8

03.0

25.0

ii. The magnitude of force required to move the plate.[3 marks]

Pa 3.067

33.80.368dy

dVμτ

33.8dy

dV

F = A = (3.067)(9 10-4) = 2.76 10-3 N

2. a. FIGURE Q2a shows a pipe carries water and tapers uniformly from a diameter of 0.5 m at section A to 0.8 m at section B. The pipe with a length of 2 m slopes upwards from section A to section B at an angle of 30o. Pressure gages are installed at sections A, B and also at C, the midpoint of AB. The gage pressures recorded at A and B are 2.0 bar and 2.3 bar, respectively.

A

C

= 30o

B

FIGURE Q2a: Tapering pipe carries water

Assuming the pipe is frictionless, determine:

i. The velocity at section A.[4 marks]

ii. The volumetric flow rate through the pipe.[3 marks]

iii. The pressure recorded at C.[4 marks]

b. FIGURE Q2b shows a piston-cylinder device connected to a

pressurized tank. The diameter of the piston is 0.5 m and located in

a cylinder containing oil having a specific gravity of 0.92. The

cylinder is connected to a pressurized tank containing water and

oil. The gage pressure of the air above the oil in the tank is 20 kPa.

A force of magnitude F is applied to the piston.

FIGURE Q2b: Piston-cylinder connected to pressurized tank.

0.25 m

0.25 m

Water

Oil

Air pressure = 20 kPag

0.5 m

0.5 m

F

Tank

Piston

Cylinder

i. If the system is under equilibrium, calculate the required

force, F, acting on the piston.

[3 marks]

ii. Determine the pressure head, expressed in meter of water,

acting on the bottom of the tank.

[3 marks]

iii. Greater force is exerted to the piston resulting in a decrease

of the oil level in the cylinder by an amount of h1 and an

increase of oil level in the tank by an amount of h2 . Find the

relationship between the air pressure in the tank and the force

exerted if h1 = h2.

[3 marks]

i. The velocity at section A.

[4 marks]

02

VVzzg

ρ

PP 2A

2B

ABCB

Taking zA = 0 (datum level)

zB = L sin = (2) sin 30o = 1 m

Through continuity equation

VAAA = VBAB VB = 2B

2A

AB

AA

D

DV

A

AV

s

m 9.693

0181.91000

000,200000,302

8.0

0.51

2

zzgρ

PP

D

D1

2V

zzgρ

PP

2

D

DVV

2

VVzzg

ρ

PP

4

4

ABCB

4B

4A

A

ABCB

4B

4A2

A2A

2B

2A

ABCB

ii. The volumetric flow rate through the pipe.[3

marks]

Solution:Q = QA = AAVA

= 693.95.04

2

= 1.903 m/s

m 0.652

0.80.52

DD BA

sm

5.7355

65.05.0

9.693

D

DV

AA

VV

2

2C

2A

AC

AAC

iii. The pressure recorded at C.[4 marks]

Solution:Taking point A and point C as reference.DC =

From Bernoulli’s equation

bar 2.26 kPa 225.62P

25624200,000P

25.624

1sin3009.812

5.73559.693

2

VVzzg

ρ

PP

C

C

O22

2C

2A

CAAC

b(i). If the system is under equilibrium, calculate the required force, F, acting on the piston.

[3 marks]Answer:

A = (/4)D2 = (/4)(0.6)2 = 0.09 m2

Pair + oilgh1 - oilgh2 = Ppiston

30000 + (0.9)(1000)(9.81)[0.5 – 1.5] = Ppistion

Ppiston = 21171 Pa

Fpiston =Ppiston A = (0.09)(21171)

= 6.0 kN

b(ii). Determine the pressure head, expressed in meter of water, acting on the bottom of the tank.

[3 marks]

Answer:

Pair + oilgh1 + watergh2 = Pbottom

30000 + (0.9)(1000)(9.81)(1) + (1000)(9.81)(1) = 48639 Pa

P = watergh

h = (48639) / (1000)(9.81) = 4.96 m H2O

2.b(iii). Greater force is exerted to the piston resulting in a decrease of the oil level in the cylinder by an amount of h1 and an increase of oil level in the tank to a height h2. Find the relationship between the air pressure in the tank and the force exerted if h1 = h2.

[3 marks]

Answer Pair + oilgha – oilghb = Ppiston Pair + (0.9)(1000)(9.81)(0.5 + h1) – (0.9)(1000)(9.81)(1.5 – h2) = F (/4)(0.6)2 Pair + 4414.5 + 8829h1 – 13243.5 – 8829 h2 = 0.09F Pair + 8829(h1 – h2 – 1) = 0.09F If h1 = h2, then Pair – 8829 = 0.09?F

3. a. FIGURE Q3a shows a closed tank of fixed volume used for the

continuous mixing of two liquids. Liquid A enters through inlet 1

while Liquid B enters through inlet 2. Both liquids are completely

mixed as Liquid C and discharged through outlet 3. The data for

the flow characteristics is given in TABLE Q3.

FIGURE Q3a: Liquid mixing in a tank

TABLE Q3: Flow characteristics

Inlet/

Outlet Liquid

Pipe Size

(Schedule 40)

Velocity, V

(m/s)

Specific

Gravity

(SG)

1 A 3 inch 0.15 0.93

2 B 2 inch 0.20 0.87

3 C 4 inch X Y

1

2

3

i. Determine the mass flow rate at outlet 3.

[3 marks]

Solution:

s

kg1.042

2.00525.00.8715.007793.00.93 10004

VAρVAρ

mmm

22

BBBAAA

BAC

i. Calculate the specific gravity and specific weight of the

mixture discharged.

[4 marks]

Solution:

33

322

95.890081.93.907

9073.01000

3.907

3.9072.00525.015.007793.0

4

042.1

m

N

m

kgg

SG

m

kg

QQ

mm

C

Water

C

CA

BA

C

ii

i. Determine the velocity of Liquid C.

[3 marks]

Solution:

cQc = cAcVc

s

m

A

mV

CC

C

C 140.0

10226.04

3.907

042.1

2

iii.

b. FIGURE Q3b shows a siphon used to empty water out of Tank A into

Tank B. The siphon is made of a plastic pipe with an inner diameter

of 0.025 m. The water level in Tank A is 1.5 m above the discharge

of the siphon, and 1 m below the centerline of the crest. Both tanks

are opened to the atmosphere and very large compared to the

siphon.

1 m

1.5 m

FIGURE Q3b: Siphon–tanks system.

Tank A

Tank B

Assuming the pipe is frictionless, determine:

i. The volumetric flow rate of water flowing out from the siphon.

State ONE (1) assumption.

[4 marks]

Solution:

Taking Point 1 = Inlet

Point 2 = Outlet

Point 3 = crest

Assumption:

No work interaction.

The pipe is frictionless

Tank is very large that the velocity of gas at point 1 is relatively small,

V1 = 0

No velocity change at point 2 and 3 V3 = V2

Pressure at point 1 and 2 = Patm = 0 kPag

Steady flow

s

m102.665.420.025

4

πAVQ

s

m 5.42

1.5-09.812

zz2gV

02

VVzzg

332

212

21

22

12

i. The minimum time to withdraw 20 L of water from Tank A to

Tank B.

[3 marks]

s 7.51 102.66

1002

AV

Vt

3

3

ii.

i. The pressure at the centerline of the crest.

[3 marks]

Solution:

kPa 5342

2.509.811000

zzρgP

0zzgρ

PP

313

1313

.

iii.

4. a. Discuss the following terms:

i. Laminar and turbulent flow.

[3 marks]

Solution:

A laminar flow is one in which the fluid particles move in layers or laminae

with one layer sliding over the other and hence no transfer of lateral

momentum to the adjacent layers. Examples are

o Flow of oil in measuring instruments

o Flow of oil in the narrow tubes of governor used for prime movers.

o Flow through filters and sand beds in groundwater.

o Rise of water in plants from their roots

o Movement of blood in a human body

Turbulent Flow:

When the flow attains a certain velocity, it no longer remaining steady

and eddy currents appear. The velocities in such a flow varies from point

to point in magnitude and directions as well as from instant to instant. All

the fluid particles are disturbed and they mix with each other. Thus there

is a continuous transfer of momentum to adjacent layer. Such diffuse flow

is called turbulent flow. When there is lateral mixing between two layers

of fluid flows

i. The Hagen-Poiseuille equation.

[3 marks]

Solution:

In general the variation of head loss hf due to uniform laminar flow in a

length L of a pipe of diameter D is given by:

2

32

D

LVh f

This is known as Hagen-Poiseulle Equation. Since the mean velocity

as 2

4D

QV

Where Q = discharge

Therefore 4

128

D

LQh f

ii.

i. Velocity distribution of a laminar flow of an incompressible

fluid in a circular pipe.

[3 marks]

This is the equation of parabolic. Therefore the velocity distribution over

the pipe section is parabolic. The maximum value of the local velocity is

denoted by umax and is located at the centre of the pipe. The value of

umax is found by substituting 0 for r, giving

2

wmax

)rr

(1u

u

)rr(μr2

τu 22

ww

w

iii.

b. Oil of an absolute viscosity, , of 1.5 poise and specific gravity (SG)

of 0.85, flows through a 12 inch, Schedule 40 pipeline. If the head

loss in the 3000 m length of pipe is 20 m, determine:

i. The shear stress developed at the pipe wall.

[4 marks]

Solution:

Wall shear stress

For 12 in, Schedule 40 pipe – din = 11.938 in = 0.30323 m

LD

4

w

fh

Pa 8.43

3000

2081.9100085.0

2

30323.0

)L

h(

2

R f

w

i. Shear stress at a radial distance of 10 cm from the pipe axis.

[4 marks]

Solution:

Pa 5.56

15162.0

10.08.43

r

r

W

O

ii

i. The Fanning friction factor, f, by assuming the flow to be

laminar.

[3 marks]

m/s 1.065

300015.032

30323.081.910000.8520

L32

DhV

D

LV32h

2

2f

2f

Check the Reynolds number,

Re =

15.0

30323.0065.1100085.0

μ

ρVD = 1829.8 < 2100

Therefore our assumption is correct.

f = 310744.88.1829

16

Re

16