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ES2A7 - Fluid Mechanics Example Classes
Example Questions (Set IV)
Question 1 : Dimensional analysis
a) It is observed that the velocity ‘V’ of a liquid leaving a nozzle depends upon the
pressure drop ‘P’ and its density ρ. Show that the relationship between them is of the
form V =C P ρ
b) It is observed that the frequency of oscillation of a guitar string ‘f’ depends upon
the mass ‘m’, the length ‘l’ and tension ‘F’. Show that the relationship between then is
f =C F ml
c) Find the dimension of the bulk modulus K, knowing its relationship with the speed
of the sound ‘a’ in a liquid and the density ‘ρ’: a =C K ρ
a) Dimension:
[ ] [ ] [ ]
2 3
V = L T , P = M LT , ρ =M L
The relationship should verify:
[ ] [ ] [ ]
a b c
a b 3c b c 2b a
V P 1
L M T 1
ρ
− − + − −
This leads to the following system:
a b 3c 0 2b b 3b 0 a 1
b c 0 b c b 1 2
2b a 0 2b a c 1 2
The relation is thus verified
b) Dimension:
[ ] [ ] [ ] [ ]
2
f = 1 T , m = M, l = L, T =ML T
The relationship should verify:
[ ] [ ] [ ] [ ]
a b c d
c d b d a 2d
f m l T 1
L M T 1
This leads to the following system:
a 1
c d 0 d c
b 1 2
b d 0 d b
c 1 2
a 2d 0 2d a
d 1 2
The relation is thus verified
c) From the relation, we have :
2
K =ρa C
so [ ] [ ][ ]
2
2
3 2 2
M L M
K a
L T LT
= ρ = =
Question 2 : Dimensional analysis
Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number
and find also the velocity required to give the same Reynolds number when the pipe is
transporting air. For the water the kinematic viscosity was 1.31 10-6 m2/s and the
density was 1000 kg/m3. For air those quantities were 15.1 10-6 m2/s and 1.19kg/m3.
Kinematic viscosity is dynamic viscosity over density = ν = μ/ρ.
The Reynolds number is :
Ud Ud
Re
ρ
μ ν
Reynolds number when carrying water:
water 6
Re 24427
−
×
×
To calculate Re air
we know:
air water
Re =Re
air
6
1
air
U 0.
U 18.44m.s
−
−
×
×
Question 3 : Momentum conservation
The figure below shows a smooth curved vane attached to a rigid foundation. The jet
of water, rectangular in section, 75mm wide and 25mm thick, strike the vane with a
velocity of 25m/s. Calculate the vertical and horizontal components of the force
exerted on the vane and indicate in which direction these components act.
From the question:
3 2 1
1 1
a 1.875 10 m , u 25m.s
− −
= × =
Since the jet section is constant a 1
=a 2
, the velocity is also constant u 1
=u 2
(Mass
conservation in the case of an incompressible flow)
The bulk flow is thus
3 1
1 1 2 2
Q a .u a .u 0.0469m .s
−
Calculation of the total force using the momentum equation:
2 1
F = ρQ(V −V )
From the question:
( )
2
1 2
2
2
1 2
1
1 2
P P gh 1000 9.81 30= 294.3 kN/m
a a 0.6 2 0.283m
u u 3m.s
ρ
π
−
= = = × ×
The bulk flow is thus
3 1
1 1 2 2
Q a .u a .u 0.848m .s
−
Calculation of the total force using the momentum equation:
2 1
F = ρQ(V −V )
Projecting this relation along x and y:
x
y
F Qu(cos75 cos0)
1000 0.848 3 (cos75 1 )
1886N
F Qu(sin75 sin0)
1000 0.848 3 (sin75 0)
2457N
ρ
ρ
= × × × −
= × × × −
Calculation of the pressure force:
1 2
P P P
F = F +F
Projecting this relation along x and y:
Px
Py
F aP(cos0 cos75)
0.283 294.3 (1 cos75)
62000N
F aP(sin0 sin75)
0.283 294.3 (0 sin75)
80376N
= × × −
= × × −
There is no body force in the x or y directions.
F is thus equal to sum of the reaction force and the pressure force applied on the fluid.
R B P R P
R P
F F F F F F
so F F F
So force on bend:
x Rx x Px
y Ry y Py
R F F F 1886 62000 63886N
R F F F 2457 80376 82833N
R
x
is directed toward the right and R y
toward the bottom.
2 2
x y
1 y
x
R R R 104kN
R
tan 52
R
θ
−
-4 -2 0 2 4
0
50
100
R
h
Question 5 : Mass conservation + Bernoulli
A water clock is an axisymmetric vessel with a small exit pipe in the bottom. Find the
shape for which the water level falls equal heights in equal intervals of time.
Continuity equation:
out in
d
m m
dt
There is no inflow and the fluid is incompressible.
Let ‘a’ be the cross-section of the exit pipe:
out
d t
m aU t
dt
0
t
υ t = − a U t dt
Bernoulli equation between the upper free surface and the exit section:
2
U t
gh t 0 0
ρ
ρ + = +
U t = 2gh t
So using the first relation:
0
0
t
t
t a gh t dt
t a g h t dt
From the question, we know that the variation of h is linear in time:
0
h t = h −xt
So
0
0
3 2
0
3 2
t
t a g h xtdt
a
t g h xt
x
a
t g h t
x
The volume of fluid can always be written as
( ) ( )
( ) 2
2
0 0 0 0
R z h t h t
t rdrd dz R z dz
π
So
( )
3 2 2 1 2 2
0
h t
a a
R z dz g h t R h t g h t
x π x π
2 2
1
P 2
2
Q A
F 1
2 A
ρ
- Calculation of the body force
The only body force is the weight due to gravity in the y-direction - but we need not
consider this as the only forces we are considering are in the x-direction.
- Calculation of the resultant force
R P
2 2
2 1
R P 2
2 1 2
2 2
1
R 2
2 2 1
F F F
1 1 Q A
F F F Q 1
A A 2 A
Q A 2 2
F 1
2 A A A
ρ
ρ
ρ
So the fireman must be able to resist the force of
2 2
1
R 2
2 2 1
Q A 2 2
R F 1
2 A A A
ρ
This force increases when A 1
increase or A 2
decrease, which correspond to the
expected behaviour.
Question 7 : Momentum conservation
Consider a rocket of mass m r
traveling at a speed u r
as measured from the ground.
Exhaust gases leave the engine nozzle (area A e
) at a speed U e
relative to the nozzle
of the rocket, and with a pressure that is higher than local atmospheric pressure by
an amount p e
. The aerodynamic drag force on the rocket is D. Derive an equation
for the acceleration of the rocket.
We selected a control volume enclosing the rocket and moving with the rocket.
Because a reference frame attached to the rocket would be noninertial, we chose a
stationary reference frame fixed to the launch pad. We selected the y-direction
momentum equation. The force diagram shows three forces: weight, aerodynamic
drag force, and the pressure force acting at the nozzle exit. From the force diagram
y e e
F = P A − D −W
The momentum diagram shows an accumulation term and an outflow term. The
momentum accumulation term is not zero because the momentum of the rocket is
changing with time as the rocket accelerates. The momentum of the rocket is the
mass of the rocket times the velocity of the rocket, and the accumulation term is
given by (d/dt)(m r
u
r
). This can be formally developed by using integration:
y r r r
CV CV
d d d
u dV u dV = (m u )
dt dt dt
ρ ρ
In this development, we assumed that all parts of the rocket were traveling at the
same speed u r
. The outward momentum flow is
o
mu
, where u o
is relative to the
stationary reference frame:
o e r
u = U −u
So : ( )
o e r
mu = m U −u
From the momentum diagram
y y
y o o i i r r e r
cs cs
CV
d d
u dV m u mu = (m u )+ m U u
dt dt
ρ
Substituting force and momentum terms into the momentum equation gives:
e e r r e r
d
P A D W= (m u )+ m U u
dt
The accumulation term can be expanded using the product rule for differentiation.
This term may be evaluated by applying continuity to the control volume:
r
dm
m
dt
Combining the two previous equations gives:
r r r r r
d d
(m u )=m (u ) mu
dt dt
Substituting this result into the momentum equation yields:
r r e e e
d
m (u ) mU P A D W
dt
The mass flow rate exiting the rocket nozzle is:
e e e
m =ρA U
Combining the above two equations results in:
2
r r e e e e e
d
m (u )= A U P A D W
dt
ρ + − −
The acceleration of the rocket (du r
/dt) depends on the instantaneous mass of the
rocket and the 4 terms on the right side of the right side of this equation. The term
2
e e e e e
T = ρA U + P A is known as the thrust of the rocket motor. The exit pressure
of the exhaust jet (P e
) is often expressed using absolute pressure, so the thrust is
given by
2
e e e e e e 0
T = ρA U + P A P − P , where P 0
is the local atmospheric pressure,
which will change with altitude as a rocket ascends.
