Transcript Gas Dynamics ESA 341 Chapter 1

Gas Dynamics
ESA 341
Chapter 1
Dr Kamarul Arifin B. Ahmad
PPK Aeroangkasa
Chapter 1 : Introduction

Review of prerequisite elements
 Perfect
gas
 Thermodynamics laws
 Isentropic flow
 Conservation laws

Speed of sound
 Analogous
concept
 Derivation of speed of sound

Mach number
Review of prerequisite elements

Perfect gas:
Equation of state
P  RT
For calorically perfect gas
u  u (T )
h  u  RT
dh  c p dT
du  cv dT
c p  cv  R
cp

cv
Entropy
ds 
q
T
ds 
du  Pdv dh  vdP

T
T
Entropy changes?
T 
 
s2  s1  cv ln  2   R ln  1 
 T1 
 2 
T 
P 
s2  s1  c p ln  2   R ln  2 
 T1 
 P1 
T2 
s  s   
 exp 2 1  2 
T1 
cv  1 
T2 
s2  s1  P2 
 exp
 
T1 
c p  P1 
R cv
R cp
Review of prerequisite elements
Forms of the 1st law
q  w  e
Tds  de  pd
Tds  dh  dp
The second law
ds 
q
T
Cont.
Review of prerequisite elements
If ds=o
For an isentropic flow
R
cv
 
T2   2 
     2 
T1  1 
 1 
T2  P2 
  
T1  P1 
Cont.
R
cp
P 
  2 
 P1 
 1
 1

T2  P2 
  
T1  P1 
 1

P2   2 
  
P1  1 
P


 2 
  
 1 

 constant
 1
Review of prerequisite elements
Cont.
1  m
2
m
Conservation of mass (steady flow):
1V1 A1   2V2 A2
Rate of mass
Rate of mass
enters control = leaves control
volume
volume
VA  (   d )(V  dV )( A  dA)
VAd  AdV  VdA  0
d

V
A
1
flow
dx
  d
2
V  dV
A  dA


dV dA

0
V
A
If  is constant (incompressible):
dV
dA

V
A
Review of prerequisite elements
Cont.
Conservation of momentum (steady flow):
Net force on
Rate momentum
gas in control = leaves control
volume
volume
Rate momentum
enters control
volume
F  Fp  m V 2  m V 1
p

V
A
1
flow
dx
2
p  dp
  d
V  dV
A  dA
Euler equation (frictionless flow):
V2
dp

 constant
2

Review of prerequisite elements
Cont.
Conservation of energy for a CV (energy balance):
Basic principle:
• Change of energy in a CV is related to
energy transfer by heat, work, and energy in
the mass flow.
2
2




dECV
V
V
i
e
 Q  W  m i  ui 
 gzi   m e  ue 
 gze 
dt
2
2




heat transfer
work transfer
energy transfer due to mass flow
Review of prerequisite elements
Cont.
Analyzing more about Rate of Work Transfer:
• work can be separated into 2 types:
• work associated with fluid pressure as mass entering or leaving the CV.
• other works such as expansion/compression, electrical, shaft, etc.
Work due to fluid pressure:
• fluid pressure acting on the CV boundary creates force.
Fp  pA  W p  FpV
W  WCV  W p
m  AV  m v  AV
W p  m pv
W p  W e  Wi
2
2




dECV
V
V
 Q  WCV  m e pe ve  m i pi vi   m i  ui  i  gzi   m e  u e  e  gze 
dt
2
2




2
2




dECV
V
V
i
e







 Q  WCV  mi  ui  pi vi 
 gzi   me  u e  pe ve 
 gze 
dt
2
2




h  u  pv
2
2




dECV
V
V
i
e
 Q  WCV  m i  hi 
 gzi   m e  he 
 gze  Most important form
dt
2
2



 of energy balance.
Review of prerequisite elements
Cont.
2
2


V
V
e
i
  m i  hi 

Q  W  m e  he 


2 
2 



Ve2  Vi2 
dq  dw  he  hi  
2
For calorically perfect gas (dcp=dcv=0):
T
h
V
1
flow
dx
2
T  dT
h  dh
V  dV
h  c pT
For adiabatic flow (no heat transfer)
and no work:
c p dT  VdV  0
Review of prerequisite elements
Cont.
Conservation laws
Conservation of mass
(compressible flow):
Conservation of momentum
(frictionless flow):
Conservation of energy
(adiabatic):
m 1  m 2 
d


dV dA

0
V
A
F  Fp  m V 2  m V 1 
V
dq  dw  h  h  
2
e
e
i
dP


 VdV  0
 Vi 2
 c p dT  VdV  0
2
Group Exercises 1
1. Given that standard atmospheric conditions for air at 150C are a
pressure of 1.013 bar and a density of 1.225kg, calculate the gas
constant for air. Ans: R=287.13J/kgK
2. The value of Cv for air is 717J/kgK. The value of R=287 J/kgK.
Calculate the specific enthalpy of air at 200C. Derive a relation
connecting Cp, Cv, R. Use this relation to calculate Cp for air using
the information above. Ans: h=294.2kJ/kgK,Cp=1.004kJ/kgK
3. Air is stored in a cylinder at a pressure of 10 bar, and at a room
temperature of 250C. How much volume will 1kg of air occupy
inside the cylinder? The cylinder is rated for a maximum pressure of
15 bar. At what temperature would this pressure be reached? Ans:
V=0.086m2, T=1740C.
Speed of sound
Sounds are the small pressure disturbances in the gas around us,
analogous to the surface ripples produced when still water is disturbed
Sound wave
Sound wave
P  dP
  d
T  dT
V  dV
P

T
V 0
Sound wave moving
through stationary gas
P  dP
  d
T  dT
V  a  dV
P

T
V a
Gas moving through
stationary sound wave
Speed of sound
cont.
Combination of mass and momentum
Derivation of speed of sound
a
dp
d
Conservation of mass
m  aA    d a  dV A
d
dV


a

P2   2 
For
  
P1  1 
isentropic flow
P
 constant

Conservation of momentum
PA  P  dP  A  m a  dV   m a
dP  adV
Finally
dP P

d 
a
P
 RT

Mach Number
M=V/a
M<1
Subsonic
M=1
Sonic
M>1
Supersonic
M>5
Hypersonic
Distance traveled =
speed x time = 4at
Distance traveled = at
Zone of
silence
If M=0
Source of
disturbance
Region of
influence
Mach Number
cont.
Original location
of source of
disturbance
Source of
disturbance
If M=0.5
Mach Number
ut
ut
ut
ut
cont.
Original location
of source of
disturbance
Direction
of motion
Source of
disturbance
Mach wave:
If M=2
at 1
sin   
ut M