Download Class02a Forces

Turbomachinery
Lecture 2a
- Conservation of Mass, Momentum
- Bernoulli Equation
- Performance Parameters
1
Fluid Flow
• Microscopic: atomic- molecular motion
• Macroscopic: continuum motion approach
2
Eulerian Control Volume Approach
• Lagrangian Approach:
– Follow fluid particle or droplet on path [called path line].
– Fluid particle is an aggregate of discrete particles of
fixed identity. Focus on particles as they move through
flow.
– Each particle is labeled by its original position.
– Examples are particle tracking for sprays and coatings,
continuum mechanics, oceanographics [flow meters
drift along prevailing currents]
t0+t;
xi=xi(t0+t)
t0;
xi0=xi(t0)
3
Eulerian Control Volume Approach
• Eulerian or Control Volume Approach:
– Watch a fixed point in space, not one particle, as time
proceeds.
– CV is designated in space and the bondary known.
The amount and identity of the matter in CV may
change with time, but shape is fixed.
– Property field, e.g. V=velocity=V(x,y,z,t) [streamline] is
defined.
– Path lines and streamlines are identical in a steady
flow.
V
4
Eulerian Control Volume Approach
• Eulerian – Lagrangian
– Extensive property: dependent on mass in volume
under consideration: N
– Intensive property: independent of mass in volume
under consideration: n
N   n  dV   ndm
For example
M    n  1  dV   dm
5
Eulerian Control Volume Approach
• Elemental Fixed Volume Cube
dV = dx*dy*dz
uu
u
dy
dz
dx
6
Brief Math Review
•
•
•
•
•
•
Cartesian, cylindrical coordinate systems
Unit vectors [i, j, k, n]
Dot products
Grad, curl operators
Green’s integral theorem
Ordinary, partial, and material derivatives
• Be careful, many variables have same symbol
– V=volume, V=velocity, etc.
7
Eulerian Control Volume Approach
• Scalar convection
– Substantial derivative [for a coordinate system] of an extensive
property N
– Consider some arbitrary extensive property N of fluid associated
with CV
N 
  ndV
– n = N per unit mass and dV is the elemental mass
• Differentiate along particle path  a Lagrangian process
 dN 
 DN 




 
 dt  system
 Dt 
N
sources
where total or Stokes derivative reflects the change in N over time
and space
8
Scalar Conservation
Time=t
Time=t+t
V
III
dN N  N

  N source
dt
dt

1 
   ndV   ndV   ndV   ndV 
t  II /

II
I
III /

 
   ndV    nV  d A
t  II
 I  III  II
I II
/
Time rate of change
convection
9
Conservation Laws
DN 
  n dV   n V  n dA  N source
Dt t


Mass [ particle identity ]
N  M    dV
n 1
Ns 0
n  u, v
Ns 
n  e0
N s  Qadded  Won system
Momentum [ Newton]
N  M  ax , a y 
Energy [1st law of thermo]
V2
N  Eo  E 
2
F ,F
x
y
10
Conservation Laws
n
2D Steady Flow
p
 V  n dA  0



V
V

n
   dA    pndA  F  ...





e
V

n
dA


p
V

n
  
   dA 
viscous
0
Qconduction  Qdissipation  ...
No. equations = 5
No. unknowns = , u, v, w, p, h0
Therefore need additional relations
to close system
p   RT
dh  c p dT
11
Conservation Laws
2D Steady Flow of Energy





e
V

n
dA


p
V

n
0
  
   dA 
Qconduction  Qdissipation  ...



e
V

n
0
   dA   
p
V  n  dA 


Qconduction  Qdissipation  ...



h
V

n
0
   dA  Qconduction  Qdissipation  ...
12
Conservation Laws
2D Steady Flow
 V  n dA  0
  r 2 
u2 (r )  uCL 1    
  R  
 What is uCL ?
R
1u1 i   i    R  2   2u2 (r )i  i rdr  0
2
0
u1
13
Conservation Laws
n
2D Steady Flow
 V  n dA  0
p



V
V

n
   dA    pndA  Fviscous  ...
x  component , no friction



u
V

n
   dA    pn  idA
14
Mass to Continuity

 dV    V  n dA  0  F .V .

t
Apply Green ' s / Divergence Theorem
Integral
 B  n dA     B dV

 dV     V dV  0

t
or integro  differential equation
 





V
  t
 dV  0
or partial differential equation

   V  0
t
or
d
    V   0
dt
 F .E.
 F .D.
15
Bernoulli’s Equation
Assume Steady , Inviscid Flow [no gravity , no shear stresses ]
Integral form
  uV  ndA     pn  i  dA
Differential form
 V   V  p
Vector Identity
V 2 
V   V     V 
 2 


where
   V
16
Bernoulli’s Equation
Differential form
V 2 
     V    p
 2 
Option 1: Steady, Irrotational [  0]
Option 2 : Along a streamline [dr ]
V 2 
    0

 2 
V 2 
dp
    d  2   0
p
or
What is meant by incompressible?
Bernoulli
- Steady
- Inviscid
- Irrotational   0
or on streamline
-Incompressible
17
Pitot – Kiel Head Pressure Probe
ps
V
1
1 V 2
2
p0  ps  V  ps 
2
2 g
lbf
ft
lbm
p 2
V
 3
ft
sec
ft
lbm
ft
g

lbf sec 2
lbf
lbf  lbm   ft 2   lbf
ft 



ft 2
ft 2  ft 3   sec 2   lbm sec2 
p0
Inviscid Momentum Equation
• Adding additional force terms (gravity, magnetism, etc.)
on flow, gives more general form of inviscid, integral
momentum equation:
  PdA  F 

Vd   V V dA

t 


• Equation is basis of C.V. approach to many problems
• Note - for steady flow calculate force on immersed
object from flow variables on surface of C.V.!!!
• To solve unsteady and/or viscid flows must integrate
throughout the volume - orders of magnitude more
19
difficult!
Steady Inviscid Momentum Equation
• Integral form of Inviscid
Momentum Equation
n

 
  
  PdA  F   V V  dA

• n is outward normal from surface area. In 2D:
n  i cos   j sin 
n  n  i  i cos 2   j  j sin 2   1
V  ui  vj  V i cos   j sin    V n
n V  u cos   v sin 
20
Steady Inviscid Momentum Equation
For cylindrical surface, use cylindrical coordinates
n  i r  i cos   j sin 
V  ui  vj  ur ir  u i
n V  u cos   v sin   ur
y
r

x
21
Steady Inviscid Momentum Equation
• Substituting into vector equation:
  P  i cos   j sin  dA  F 
   ui  vj   u cos  v sin   dA
• Writing this as two scalar equations:
  P cos dA  Fx   u u cos  v sin  dA
  P sin  dA  Fy   vu cos  v sin  dA
22
Steady Inviscid Momentum Equation
• The math works - but don't lose common sense. Still:
F  ma  mV
• Pressure force is positive to right, acceleration is
positive to right, so:
 
  
  PdA  F   V V  dA


P1  P2 A  F  m V2  V1 
23
Steady Inviscid Momentum Equation
• Examples:
– Circular cylinder in flight: Compressible flow homework
– Circular cylinder in duct: Compressible flow homework
– Jet Engine In Flight
e
d
f
c
g
Vo, Ao
Vj, Aj
h
i
j
b
a
24
Inviscid Momentum Equation
• Example: Application to Jet Engine in Flight
Fx
 P cos dA
AB
BC
CD
DE
EF
FG
GH
HI
IJ
JA
0
0
0
0
0
0
0
0
0
0
 uu cos  v sin  dA
cancels IJ
 jV j 2 A j
cancels EF
1
 0V0 2 A j  A0 
2
cancels CD

1
 0V0 2 A j  A0 
2
  0V0 A0
2

1
 0V0 2 A j  A0 
2
cancels AB
1
 V A  A 
2
2
0
0
j
0
25
Inviscid Momentum Equation
• Summing terms:
Fx   jV j A j   0V0 A0
2
2
Fx  mV j  V0 

• Jet engine control volume chosen to eliminate  PdA
• Same result for any control volume fully enclosing the
engine
• Generally cannot eliminate
 PdA
for internal flows
26
Effect of Ambient Pressure [patm]
Uninstalled (Ideal) Thrust
• So From Force Considerations (Control
Volume Analysis), the Uninstalled (Ideal)
Thrust for an Engine is:
F 
(m air  m fuel ) Vexit  m air Vinlet
gc
 ( Pexit  Pinlet ) A
• Why now pressure term?
• What is effect of Pambient term?
28
Specific Fuel Consumption
• The Rate of Fuel Used by the Propulsion System per
Unit of Thrust:
m fuel
– uninstalled:
S 
– installed:
m fuel
• So
TSFC 
T
F

mass flow rate of fuel
installed engine thrust
S  TSFC (1 inlet   nozzle )
29
3D Steady Flow Energy Equation





e
V

n
dA


p
V

n
0
  
   dA



e
V

n
0
   dA 
 h
0
V  n  dA 


p


V

n
    dA  0
  h dm  Q  0
0
Energy based
on 1st law
If flow velocity brought to zero adiabatically, apply
Gibbs equation at stagnation properties h01  h02  T01  T02
 T02    1  P02 
s2  s1
  1  P02 
 ln   
ln 
ln 


cp

 T01  
 P01 
 P01 
P02
 s 
 exp    1
P01
R
Was
derived
earlier
30
Other Important Equations
p   RT
Eq. of state
V2
h0  e0  pv  e0   h 

2
cp
cP  cv  R
=
cv  1 / 
p
T2  P2 
 
T1  P1 
 P2 
 
T1 or P1 
 1 / T2
T  CP 
 1 /
Enthalpy def .
Kinetic theory
or
T  CP 1 / 
Adiabatic eq. of state
if s  0
31