ES3C9
FLUID MECHANICS FOR MECHANICAL ENGINEERS
Example Class 2
(1) A bellows may be modelled as a deforming wedge-shaped volume as in fig. 1.
The check valve on the left (pleated) end is closed during the stroke. If b is the
bellows width into the paper, derive an expression for outlet mass flow m0 as a
function of stroke   t  .
Figure 1:
For a control volume enclosing the bellows and the outlet flow, using the
continuity equation we obtain:
d
    mout  min  0 ,
dt
where   bhL  bL2 tan   t   is the volume inside the bellows at time t.
There is no inflow in the CV : min  0
Since L is constant, solve for mout  
d
1 d
 bL2 tan      bL2

dt
cos 2  dt
(2) A liquid jet of velocity Vj and section Aj strikes a fixed hollow cone, as in
Fig.2 and deflects back as a conical sheet at the same velocity. Find the
3
cone angle  for which the restraining force F   A jV j2 . The fluid is
2
supposed to be inviscid and incompressible and in a steady state. The
gravitational effects are neglected.
Fig.2
A2
Conical Sheet
y
Jet
θ
A1
F
x
A5
A4
A3
Let the Control Volume enclose the cone, the jet and the sheet.
Then we use the Momentum equation :




 F    Vd     V V .n dA . The flow is steady, so
  Vd    0 .
t  CV
t  CV
 CS

 
Then we distinguish the different surfaces of CV :
 F   V V .n dA   V V .n dA   V V .n dA 
 
CS
 
A1
 
A2
 V V .n dA
A3
Along A1, V and n are in the opposite direction. Along A2 and A3, V and n
are in the same direction. So :  F    V .VdA   V .VdA   V .VdA
A1
A2
A3
Then we apply the relation along x and y direction:
 Fx   V j .V j dA  V j .  V j cos  dA  V j .  V j cos  dA




A1
A2
A3

 Fy   V j . V j sin  dA   V j .  V j sin  dA
A2
A3

 Fx    AjV j .V j    A2  A3 V j .  V j cos  


 Fy    A2  A3  V j . V j sin  
By symmetry, A 2=A3, so Fy=0.
To simplify the expression of F x, we use the continuity equation:




   d      V .n dA  0 . As the flow is steady,
  d   0 .
t  CV
t  CV
 CS

 
So :
  V .n dA    V .n dA    V .n dA    V .n dA  0
CS
A1
A2
A3
Which leads to :  V j Aj  V j A2  V j A3  0
or V j Aj  V j A2  V j A3  m
Replacing in the expression of F x gives : Fx  mV j 1  cos  
Fx is the force applied by the cone on the fluid and is indeed negative.
3
Finally we just have to determine  such as mV j 1  cos     AjV j2
2
3
1
 mV j 1  cos    mV j  cos   , so   60
2
2
Other Method in 1D, steady:  Fx  mout uout  min uin  m  V j cos    mV j =…
(3) The small boat in Fig. 3 is driven at a steady speed Vo by a jet of
compressed air issuing from a 3cm diameter hole at Va  343m / s . Jet exit
conditions are pa = 1atm and Ta  30 C . Air drag is negligible and the hull
drag is kVo2 , where k  19 N .s 2 m2 . Estimate the boat speed V o in m/s.
(Specific gas constant of air R=287 J K-1 kg-1)
Fig.3
Da=3cm
Va
Compressed
air
Vo
Hull drag k(Vo)2
For a control volume enclosing the boat and moving to the right at V o, the air
appears to leave the left side at speed (V o+Va). The density of the air is:
a  Pa RTa  1.165 kg/m3 . The only mass flow across CS is the air moving to the
left. The force balance is in x-direction: Fx  mout uout  min uin  mout uout
So : kV02  a Aa Vo  Va  Vo  Va   kV02   a Aa Vo  Va 
2
 V02  k  a Aa   2a AaVaVo  a AaVa 2  0
 19V02  0.56Vo  95.94  0
Solve for Vo  2.27 m/s . (Pa=101325 Pa, Aa =0.0007 m2)