M E 320
Professor John M. Cimbala
Lecture 41
Today, we will:
•
•
Discuss isentropic, compressible, adiabatic flow in ducts
Discuss converging-diverging ducts, and introduce choking and shock waves
2. One-Dimensional Isentropic Adiabatic Flow in Ducts
a. Setup and equations
Pressurized
reservoir
Duct with non-constant cross-sectional area
Stagnation
conditions:
V≈0
T = T0
P = P0
h = h0
Goal: Predict P, T, ρ, V, Ma,
etc. at some downstream
location in the duct
For simplicity, we approximate the flow as isentropic (negligible friction and other
irreversibilities) and adiabatic (no heat transfer from the air to the surroundings – insulated
duct walls).
Equations for isentropic, compressible, adiabatic flow of an ideal gas:
1
k
T
k −1
ρ ⎛ k − 1 2 ⎞ k −1 P0 ⎛ k − 1 2 ⎞ k −1
Ma 2 0 = ⎜1 +
Ma ⎟
= ⎜1 +
Ma ⎟
• For any ideal gas: 0 = 1 +
ρ ⎝
2
P ⎝
2
T
2
⎠
⎠
2.5 P
3.5
T
ρ
0
= (1 + 0.2Ma 2 )
• For air (k = 1.4): 0 = 1 + 0.2Ma 2 0 = (1 + 0.2Ma 2 )
T
P
ρ
ρ ⎛T ⎞
• Or, for air in terms of temperature, 0 = ⎜ 0 ⎟
ρ ⎝T ⎠
2.5
P0 ⎛ T0 ⎞
=⎜ ⎟
P ⎝T ⎠
3.5
1/ 2
c ⎛T ⎞
also 0 = ⎜ 0 ⎟
c ⎝T ⎠
Subsonic water nozzle
Supersonic rocket nozzle
A
1 (1 + 0.2Ma
For air: * =
A Ma
1.728
)
2 3
and
T0 k + 1
=
= 1.2 from which we get
T*
2
ρ0 ⎛ T0 ⎞
P ⎛T ⎞
c ⎛T ⎞
= ⎜ * ⎟ = 1.22.5 = 1.577 0* = ⎜ 0* ⎟ = 1.23.5 = 1.893 0* = ⎜ 0* ⎟
*
ρ ⎝T ⎠
P ⎝T ⎠
c ⎝T ⎠
2.5
3.5
1/ 2
= 1.21/ 2 = 1.095
But these are usually listed in terms of their inverses, i.e.,
T*
2
ρ*
P*
c*
=
= 0.8333
= 0.6339
= 0.5283
= 0.9129 = sonic or critical values
T0 k + 1
ρ0
P0
c0
We also define a critical Mach number, Ma*, as
V V c Ma ⋅ c Ma ⋅ kRT
T
Ma * = * =
=
=
= Ma *
*
*
c
cc
c
T
kRT *
We plot all these as functions of Mach number in Appendix A-13 (for air):
Should be ρ/ρ0