Transcript Control volume analysis - My FIT (my.fit.edu)

MAE 5360: Hypersonic Airbreathing Engines
Integral Forms of Mass and Momentum Equations
Mechanical and Aerospace Engineering Department
Florida Institute of Technology
D. R. Kirk
Kinematic Properties: Two ‘Views’ of Motion
1. Lagrangian Description
– Follow individual particle trajectories
– Choice in solid mechanics
– Control mass analyses
– Mass, momentum, and energy usually formulated for particles or systems of fixed
identity
• ex., F=d/dt(mV) is Lagrangian in nature
2. Eulerian Description
– Study field as a function of position and time; not follow any specific particle paths
– Usually choice in fluid mechanics
– Control volume analyses
– Eulerian velocity vector field:
V r , t   V x, y, z, t   u x, y, z, t iˆ  vx, y, z, t  ˆj  wx, y, z, t kˆ
–
Knowing scalars u, v, w as f(x,y,z,t) is a solution
CONSERVATION OF MASS

d
dV    U  nˆ dS  0

dt CV
S
Relative to CS
 
d
dV    U  U CS  nˆdS  0

dt CV
S
Inertial
•
•
•
•
This is a single scalar equation
– Velocity doted with normal unit vector results in a scalar
1st Term: Rate of change of mass inside CV
– If steady d/dt( ) = 0
– Velocity, density, etc. at any point in space do not change with time, but may vary from
point to point
2nd Term: Rate of convection of mass into and out of CV through bounding surface, S
3rd Term (=0): Production or source terms
Integral vs. Differential Form
 

• Integral form of mass conservation
d
dV    V  nˆ dS  0

dt V
S
   AdV   A  nˆdS
V
• Apply Divergence (Gauss’) Theorem
S
 


 V  nˆdS     V dV
S
• Transform both terms to volume integrals
V
 dV  0

 
   V


t
V 
 


   V  0
t

• Results in continuity equation in the form of
a partial differential equation
• Applies to a fixed point in the flow
• Only assumption is that fluid is a continuum
– Steady vs. unsteady
– Viscous vs. inviscid
– Compressible vs. incompressible
Summary: Incompressible vs. Constant Density
 D
  V 
0
Dt


   V  0
t
• Two equivalent statements of conservation
of mass in differential form
 
• In an incompressible flow
 u v w 
D

  



Dt

x

y

z



• Says particles are constant volume, but not
 V  0
necessarily constant shape
D
• Density of a fluid particle does not change
0
as it moves through the flow field
Dt
• Incompressible: Density may change within the flow field but may not change
along a particle path
• Constant Density: Density is the same everywhere in the flow field
MOMENTUM EQUATION: NEWTONS 2nd LAW

 

d
UdV   U U  nˆ dS   F

dt CV
Inertial
S
Relative to CS

  

d
UdV   U U  U CS  nˆdS   F

dt CV
S

•
•
•
•

This is a vector equation in 3 directions
1st Term: Rate of change of momentum inside CV or Total (vector sum) of the momentum of all parts of the CV at any
one instant of time
–
If steady d/dt( ) = 0
–
Velocity, density, etc. at any point in space do not change with time, but may vary from point to point
2nd Term: Rate of convection of momentum into and out of CV through bounding surface, S or Net rate of flow of
momentum out of the control surface (outflow minus inflow)
3rd Term:



 F    pnˆ dS  t dS   gdV   F
ext
S
•
S
CV
–
Notice that sign on pressure, pressure always acts inward
–
Shear stress tensor, t, drag
–
Body forces, gravity, are volumetric phenomena
–
External forces, for example reaction force on an engine test stand
Application of a set of forces to a control volume has two possible consequences
1. Changing the total momentum instantaneously contained within the control volume, and/or
2. Changing the net flow rate of momentum leaving the control volume
Application to Rocket Engines
Chemical
Energy
F
Thermal
Energy
Rocket Propulsion (class of jet propulsion) that
produces thrust by ejecting stored matter
•
Propellants are combined in a combustion chamber
where chemically react to form high T&P gases
•
Gases accelerated and ejected at high velocity
through nozzle, imparting momentum to engine
•
Thrust force of rocket motor is reaction experienced
by structure due to ejection of high velocity matter
•
Same phenomenon which pushes a garden hose
backward as water flows from nozzle, gun recoil
Kinetic
Energy
F  m eVe  Pe  Pa Ae
F  m eVe
Application to Airbreathing Engines
Chemical
Energy
Thermal
Energy
F  m eVe  m oVo  Pe  Pa Ae
F  m Ve  Vo 
• Flow through engine is conventionally called THRUST
– Composed of net change in momentum of inlet and exit air
• Fluid that passes around engine is conventionally called DRAG
Kinetic
Energy