jan 2007 3 cm 0.25 m/s y v plate fixed surface 1.figure q1 shows a plate with an area of 9 cm 2 that...
TRANSCRIPT
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
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