MECHANISM CLUSTER
First Year B.Eng/M.Eng
2007 Solutions to Mock Exam Questions
FLUID MECHANICS
Q1. Give answers to the following questions (2 marks each):
1. Write down Bernoulli’s equation and explain physical meaning of its terms.
When this equation can be applied?
Solution:
Bernoulli’s equation
ρ U2
+ ρ g z = const
2
expresses conservation of energy. The terms are: P – pressure work per
unit volume; ρ U 2 /2 – kinetic energy per unit volume; ρ g z – potential
energy due to gravity per unit volume. The equation can be applied along
streamlines of a steady flow of an incompressible inviscid fluid.
P+
2. A cylinder of 1m in diameter is made with material of relative density 0.5 . It
is moored in fresh water by one end and the water level is 1m above the middle
of the cylinder. Find the tension of the mooring cable.
Solution:
When the cylinder is submerged to the middle, the tension is zero (σ =
0.5). Additional buoyancy due to further submerging is compensated by
the cable tension:
T = ρg
π × 12 m2
π d2
l = 1000 kg/m3 × 9.81 m/s2 ×
× 1 m ≈ 7700 N
4
4
3. A Pitot-static tube is place in an air flow (ρ = 1.3kg/m3 ). A connected
manometer shows pressure difference 20mm of water. Whats is velocity of
the flow?
Solution:
Using Bernoulli’s equation for a stagnation point we can write:
p
ρa U 2
P+
= P0 ⇒ U = 2 ∆ P/ρa .
2
The pressure reading is in mm of water, that is ∆ P = ρw g ∆ h. Then
r
ρw
U= 2
g∆h
ρa
and
U=
p
2 × (1000/1.3) × 9.81 m/s2 × 20 × 10−3 m = 17.4 m/s
1
4. A 2mm space between two parallel plates is filled with viscous fluid. One plate
is moving with velocity 1m/s. Find the mean velocity of the flow between the
plates and the flow rate if plates width is 20cm.
Solution:
The velocity profile is linear, and the mean velocity is the average between
velocities on the walls: U = 0.5 m/s. The flow rate is
Q = a b U = 2 × 10−3 m × 0.2 m × 0.5 m/s = 2 × 10−4 m3 /s
5. Water (µ = 10−3 Pa s) flows through a pipe of 5mm in diameter with mean
velocity 0.2m/s. Can the formula f = 16/Re be applied to calculate the
friction factor for this flow? What is the head loss per 1m of pipe length?
Solution:
Reynolds number of the flow:
Re =
ρV d
1000 kg/m3 × 0.2 m/s × 5 × 10−3 m
=
= 1000 .
µ
10−3 P a s
Re < 2000 ⇒ flow is laminar and we can use formula f = 16/Re for
calculating the friction coefficient. The head loss is specified by Darcy’s
equation:
L V2
,
h = 4f
d 2g
and using the laminar formula for f we have
h=
64 × 1 m × 0.22 m2 /s2
64 L V 2
= 0.026 m
=
Re d 2 g
1000 × 5 × 10−3 m × 2 × 9.81 m/s2
6. A mass of a sphere is 3kg and its diameter is 10cm. The sphere falls from a
large height with steady velocity. Assuming that the drag coefficient CD = 0.3
is constant find the velocity of the sphere. Take air density ρ = 1.3kg/m3 .
Solution:
The sphere is moving with a constant speed, which means that the drag
equals to the gravity:
ρ U 2 π d2
m g = CD
2
4
and
s
r
8mg
8 × 3 kg × 9.81 m/s2
U=
=
= 138.6 m/s
π CD ρ d2
π × 0.3 × 1.3 kg/m3 × 0.12 m2
7. A lock gate is 2m high and 3m wide. It can rotate around a vertical hinge
at one side. Calculate the maximal force applied to the gate by water, and
maximal moment about the hinge.
2
Solution:
Maximal possible depth is h = 2 m, the maximal pressure is P = ρ g h, for
the triangular distribution the total load is
F =
PA
1
1
= ρ g h2 b = ×1000 kg/m3 ×9.81 m/s2 ×22 m2 ×3 m = 58860 N .
2
2
2
The load is applied in the middle of the width, therefore the moment about
the vertical axis is:
M = F b/2 = 58860 N × 1.5 m = 88290 N m
8. A cylinder of 90 mm in diameter has length 200mm. It rotates coaxially inside
a tube of 100mm in diameter. The gap between the cylinder and the tube is
filled with an oil with dynamic viscosity µ = 0.2Pa s. Calculate the moment
required to rotate the cylinder with speed 1 revolution per second.
Solution:
Linear velocity of the cylinder surface V = 2 π N R, the gap h = 5mm,
velocity gradient across the gap dU/dy = V /h. Then the shear stress on
the wall is
τ =µ
V
2 × π × 1 rps × 50 × 10−3 m
= 0.2 P a s ×
= 12.6 P a
h
5 × 10−3 m
The shear stress acts on the surface with area A = 2 π R L, and the force
creates the moment
M = τ A R = 2 π τ R2 L = 2×π×12.6 P a×(50×10−3 m)2 ×0.2 m = 0.04 N m
9. A harbour model has scale 1/200. A tidal period of a prototype is 12 hours.
Find the required period of the model neglecting all effects but gravity.
Solution:
A non-dimensional
group proportional to time and including length and
p
gravity is T g/L. This parameter must be the same for both model and
prototype:
r
r
g
g
Tp
12 h
Tm
= Tp
⇒ Tm = √
=√
= 0.85 h
L/200
L
200
200
10. Calculate the minimal head of the pump required to pump 1m3 of water per
second to the height 10m trough a pipe with cross section area 0.1m2 . Assume
constant friction coefficient f = 0.01 and neglect all losses in fittings.
Solution:
The pump head is the gravity head plus losses, which can be found by
Darcy’s equation and are minimal when the pipe is vertical:
H = Z + h = Z + 4f
3
L Q2
d 2 g A2
H = 10 m + 4 × 0.01 ×
10 m × (1 m3 /s)2
= 12 m
2 × 9.81 m/s2 (0.1 m2 )2
Q2. A 1/5 scale model of a wind turbine rotor has diameter 1m. It is tested in a wind
tunnel with a closed cylindrical test section of 3m diameter. The flow at the inlet of
the test section is uniform and reading of a manometer connected to a Pitot-static
tube in this flow is 26.5mm of water. A longitudinal velocity profile is measured at
the outlet of the test section. It consists of a viscous wake of 1.5m diameter and an
inviscid uniform flow outside the wake. The mean velocity of the flow in the wake
was found to be 30% less then the velosity of the uniform incoming flow.
(a) Assuming constant longitudinal flow velocity in the wake, calculate the drag
of the model and specify its drag coefficient. The density of air is 1.3kg/m3 .
[ 30 marks ]
(b) Specify wind velocity when experimental results can be applied to a prototype.
What is the drag of the prototype at this velocity? What is the speed of
rotation of the prototype if the model rotates at 10 revolutions per second?
[ 10 marks ]
Solution:
(a) Velocity at the inlet of the working section U1 can be calculated from the
reading of the Pitot-static tube manometer ∆h using Bernoulli’s equation:
p
p
U1 = 2 ∆P/ρ = 2 ρw g ∆h/ρ ,
where ∆P = ρw g ∆h is the difference between total and static pressures, ρw is
the density of water, ρ is the density of air, g is the gravitational acceleration.
Substituting values we have:
s
2 × 1000 kg/m3 × 9.81 m/s2 × 26.5 × 10−3 m
= 20 m/s .
U1 =
1.3 kg/m3
The mean velocity in the wake is
U3 = 0.7 U1 = 0.7 × 20 m/s = 14 m/s .
From the continuity principle we have
Q = U1
π d21
π d2
π
= U3 3 + U2 ( d21 − d23 ) ,
4
4
4
where d1 is the diameter of the working section, d3 is the diameter 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 =
d21 − 0.7 d23
32 m2 − 0.7 × 1.52 m2
U
=
× 20 m/s = 22 m/s
1
d21 − d23
32 m2 − 1.52 m2
4
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
(U2 − U12 ) .
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 on the control volume in horizontal
direction including pressure and body reaction (R = −D) is:
(P1 − P2 )
π d21
− D = Iout − Iin
4
The inlet momentum:
π d21
.
4
The outlet momentum is the sum of momenta in and outside the wake:
Iin = ρ U12
Iout = ρ U32
π d23
π
+ ρ U22 (d21 − d23 ) .
4
4
Substituting we have
2
2
ρ 2
2 π
2
2
2
2 π d3
2 π d1
(U − U1 ) − F = ρ U3
+ ρ U2 (d1 − d3 ) − ρ U1
2 2
4
4
4
and after some algebra for the drag D we have
π d21 1 2 1 d23 2 d23 2
D=ρ
U −
−
U2 − 2 U3 .
4
2 1
2 d21
d1
Substituting the values we obtain:
1.3 kg/m3 × π × 32 m2
1 2 1 2 1 2
D=
×
20 − 22 − 14
m2 /s2 ≈ 276 N
4
2
4
4
Drag coefficient:
CD =
D
8 × 276 N
=
= 1.35 .
(π d2 /4) (ρ U12 /2)
π × 12 m2 × 1.3 kg/m3 × 202 m2 /s2
(b) Reynolds numbers of a model and prototype should be the same (dynamic
similarity):
ρ Um dp /5
ρ Up dp
=
µ
µ
5
⇒
Up = Um /5 = 4 m/s ,
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
(CD = CD (Re)):
Dm
Dp
CD =
=
;
2 /2
Am ρ Um
Ap ρ Up2 /2
Dp = Dm
Ap Up2
d2p (Um /5)2
=
D
= Dm = 276 N .
m
2
2
A m Um
(dp /5)2 Um
The ratio of the flow velocity to the blade end velocity should be the same for
the model and prototype (kinematic similarity):
dp Np
dm Nm
=
;
Up
Um
Np = Nm
Nm
dp /5 Um /5
10 rps
dm Up
= Nm
=
=
= 0.4 rps .
Um dp
Um dp
25
25
6