Download MECHANISM CLUSTER First Year B.Eng/M.Eng 2007 Solutions to

MECHANISM CLUSTER
First Year B.Eng/M.Eng
2007 Solutions to Exam Questions
FLUID MECHANICS
Q1. Give answers to the following questions (2 marks each):
1. A gap between two parallel plates 1mm apart is filled with a fluid of viscosity
µ = 0.2Pa s. The area of the upper plate is 10cm2 and it is moving with velocity
1m/s. Find the force resisting the motion.
Solution:
F =τA=µ
U
0.2 P a s × 1 m/s × 10 × 10−4 m2
dU
A=µ A=
= 0.2 N
dy
d
1 × 10−3 m
2. Water depth near a dam wall is 10m. Calculate the total horizontal force and
the overturning moment exerted by the water per unit length of the wall.
Solution:
Pressure near the bottom:
P = ρ g h = 1000 kg/m3 × 9.81 m/s2 × 10 m = 98100 P a
Total force per 1m length for the triangular load:
F =
1
P h = 98100 P a × 10 m/2 = 490500 N/m
2
Force applied at the distance h/3 from the bottom. The moment per 1m
of width:
M=
1
F h = 490500 N/m × 10 m/3 = 1.635 × 106 (N m)/m
3
3. What is the physical meaning of Bernoulli’s equation? What assumptions are
made in deriving it? Where can it be applied?
Solution:
Conservation of energy. Inviscid, incompressible, steady. Along streamlines.
4. Volume of a body is 1m3 and its mass is 2000kg. What is its weight in
(a) water?
(b) oil of relative density 0.9?
Solution:
(a) Density is twice the density of water ⇒ the buoyancy force in water is
half of the gravity force ⇒ weight = G − B is half of the gravity force:
Wwater = m g/2 = 9.81 m/s2 × 2000kg/2 = 981 N
1
(b) Buoyancy in oil is 0.9 B ⇒ Woil = G − 0.9 B = 9.81 m/s2 × 1100 kg =
10791 N
5. A sphere is moving in air (ρ = 1.3kg/m3 ) with velocity 2m/s. What is the
gauge pressure at the front stagnation point?
Solution:
From Bernoulli’s equation:
P0 − P∞ =
2
ρ U∞
2
6. A long cylinder of diameter 6cm is placed perpendicular to a uniform flow of
air (ρ = 1.3kg/m3 , µ = 1.7×10−5 Pa s) with velocity 10m/s. Find the Reynolds
number of this flow. What is the drag of the cylinder per 1m of span if the
drag coefficient CD = 1.2.
Solution:
Reynolds number:
Re =
1.3 kg/m3 × 10 m/s × 0.06 m
ρU d
≈ 4.59 × 104
=
µ
1.7 × 10−5 P a s
Drag per 1m of span:
D = CD
1.3 kg/m3 102 m2 /s2
ρ U2
d = 1.2 ×
× 0.06 m = 4.68 N/m
2
2
7. Water (µ = 10−3 Pa s) flows through a pipe of 1cm in diameter with the flow
rate 1 liter/min. The flow is fully developed. Is this flow laminar or turbulent?
Solution:
Reynolds number:
Re =
ρU d
ρQd
1000 kg/m3 × 10−3 /60 m3 /s × 0.01 m
=
=
≈ 2100
µ
Aµ
(π × 0.012 /4) m2 × 10−3 P a s
The Reynolds number is close to the critical Rec = 2300 and is in transition region.
8. A pipe 100m long and 10cm in diameter connects two large tanks. The fluid
surface in one tank is 2 meter higher then the surface of fluid in another one.
The fluid flows freely between the tanks under the action of gravity with the
mean velocity 1m/s. Neglecting all losses except friction in the pipe calculate
the friction coefficient.
Solution:
The head loss is the pipe is h = 2m. Darcy’s equation:
h=f
2
4 L U2
.
d 2g
Then the friction coefficient:
f=
hdg
2 m × 0.1 m × 9.81 m/s2
=
≈ 0.01
2 L U2
2 × 100 m × 12 m2 /s2
9. A 1/10 model is tested in a flow of water with velocity 3m/s. Find the velocity
when the results of the experiment can be applied to the prototype in the flow
of air. All necessary data can be found in previous questions.
Solution:
Reynolds numbers of a model and a prototype should be the same:
ρwater Um L/10
ρair Up L
=
µwater
µair
Up =
1000 kg/m3 × 1.7 × 10−5 P a s
1 ρwater µair
Um =
×3 m/s = 3.92 m/s
10 µwater ρair
10 × 1.3 kg/m3 × 10−3 P a s
10. A cross-section of a pipe changes gradually from a cylinder of 100mm in diameter to a square with the side 100mm. A fluid enters the pipe from the
cylindrical end with mean velocity 0.5m/s. Find the mean fluid velocity at the
square end.
Solution:
Flow rate is constant:
Q = U1 A1 = U2 A2
U2 =
⇒
U2 =
A1
U1
A2
π × 1002 mm2 /4
π
×
0.5
m/s
=
× 0.5 m/s = 0.39 m/s
1002 mm2
4
Q2. A cylinder of diameter d = 30 cm is tested in a wind tunnel with a closed test section
of height 3d. The cylinder is placed across the flow in the middle of the test section.
The reading of a Pitot-static tube in a uniform flow at the inlet of the test section
is 150mm of water. A velocity profile is measured at the outlet of the test section.
It consists of a viscous wake which thickness is half of the working section height,
and an inviscid uniform flow outside the wake. The mean velocity of the flow in the
wake was found to be 10% less then the velocity of the uniform incoming flow.
(a) Assuming constant flow velocity in the wake calculate the drag of the cylinder
per 1m of span and find its drag coefficient. The density of air is 1.3kg/m3 .
[ 30 marks ]
(b) A radio mast has a 10m long cylindrical section of 0.6m in diameter. Find a
wind speed when the results of the experiment described above can be used
for calculating wind load on the mast. Calculate the total load on the section
assuming uniform wind profile.
[ 10 marks ]
3
Solution:
(a) The velocity at the inlet of the working section U1 from the reading of the
Pitot-static tube manometer ∆h using Bernoulli’s equation can be calculated
as:
p
p
U1 = 2 ∆P/ρ = 2 ρw g ∆h/ρ ,
where ∆P = ρw g ∆h is the difference between the total and static pressures,
ρw is the density of water, ρ is the density of air and g is the gravitational
acceleration. Substituting the values we have:
s
2 × 1000 kg/m3 × 9.81 m/s2 × 150 × 10−3 m
= 47.6 m/s .
U1 =
1.3 kg/m3
The mean velocity in the wake is
U3 = 0.9 U1
From the continuity principle we have
Q = U1 d1 = U3 d3 + U2 ( d1 − d3 ) ,
where d1 is the height of the working section, d3 is the thickness of the wake, U2
is the flow velocity outside the wake at the exit of the working section. Then
we can calculate the velocity outside the wake U2 as:
U2 =
d1 − 0.9 d3
d1 − 0.9 d1 /2
U1 =
U1 = 1.1 U1
d1 − d3
d1 − d1 /2
Bernoulli’s equation outside the wake between the inlet and outlet:
P1 +
ρ U12
ρ U22
= P2 +
,
2
2
where P1 and P2 are pressures at the inlet and outlet respectively. Then the
pressure difference between the inlet and outlet is
P1 − P2 =
ρ 2
(U − U12 ) .
2 2
Flow on the outlet is assumed parallel, which means that the pressure does not
change across the flow and P3 = P2 .
By the momentum principle the total force per unit width of the control volume
in horizontal direction including pressure and body reaction (R = −D) is:
(P1 − P2 ) d1 − D = Iout − Iin
The inlet momentum per unit width:
Iin = ρ U12 d1 .
The outlet momentum per unit width is the sum of momenta in and outside
the wake:
Iout = ρ U32 d3 + ρ U22 (d1 − d3 ) .
4
Substitution gives
ρ 2
(U − U12 ) d1 − D = ρ U32 d3 + ρ U22 (d1 − d3 ) − ρ U12 d1
2 2
and after some algebra the expression for the drag per unit width D can be
written as
ρ U12
d1 d1 U22
d3 U22
d3 U32
D=
d
−
+2
−2
,
2
d
d U12
d U12
d U12
where d is the diameter of the cylinder. The value in the brackets is the drag
coefficient:
d1
d3 U22
d3 U32
U22
CD =
1− 2 +2
−2
.
d
U1
d1 U12
d1 U12
Substituting the values we have
CD = 3 × 1 − 1.12 + 2 × 0.5 × 1.12 − 2 × 0.5 × 0.92
= 0.57 ,
and the drag per 1m of cylinder span is
ρ U12
1.3 kg/m3 × 47.62 m2 /s2
d = 0.57 ×
× 0.3 m = 242.4 N/m .
D = CD
2
2
(b) Reynolds numbers of a model and prototype should be the same:
ρ Um dm
ρ Up dp
=
µ
µ
⇒
Up = Um
dm
= 0.5 × 47.6 m/s = 23.8 m/s ,
dp
where subscripts m and p are used for the model and prototype respectively.
At this velocity drag coefficients of the model and prototype are the same.
Then the drag of a mast of the length L = 10 m is
ρ Up2
Dp = CD
dp L
2
Dp = 0.57 ×
1.3 kg/m3 × 23.82 m2 /s2
× 0.6 m × 10 m = 1259 N
2
5