Basics of Fluid Mechanics, 2014a

390 CHAPTER 11. COMPRESSIBLE FLOW ONE DIMENSIONAL
and for perfect gas
dT
T = k − 1
k
dP
P
(11.43)
Thus, the temperature varies in the same way that pressure does.
The relationship between the Mach number and the temperature can be obtained
by utilizing the fact that the process is assumed to be adiabatic dT 0 =0. Differentiation
of equation (11.25), the relationship between the temperature and the stagnation
temperature becomes
(
dT 0 =0=dT
and simplifying equation (11.44) yields
dT
T
1+ k − 1 )
M 2 + T (k − 1)MdM (11.44)
2
− 1) MdM
= −(k
1+ k − 1
(11.45)
M
2
2
11.4.3.2 Relationship Between the Mach Number and Cross Section Area
The equations used in the solution are energy (11.45), second law (11.43), state (11.35),
mass (11.32) 2 . Note, equation (11.39) isn’t the solution but demonstration of certain
properties of the pressure profile.
The relationship between temperature and the cross section area can be obtained
by utilizing the relationship between the pressure and temperature (11.43) and the
relationship of pressure with cross section area (11.39). First stage equation (11.45) is
combined with equation (11.43) and becomes
(k − 1)
k
Combining equation (11.46) with equation (11.39) yields
ρU 2
1 A
k
dP
P
dA
1 − M 2
P
− 1) MdM
= −(k
1+ k − 1
(11.46)
M
2
2
= −
MdM
The following identify, ρU 2 = kMP can be proved as
kM 2 P = k
M 2
{}}{
U 2
c 2
P
{}}{
ρRT = k U 2
kRT
2 The momentum equation is not used normally in isentropic process, why?
1+ k − 1
(11.47)
M
2
2
P
{ }} {
ρRT = ρU 2 (11.48)