Modern Engineering Thermodynamics

684 CHAPTER 16: Compressible Fluid Flow
Inlet specific kinetic energy,V 2 i /2g c
An ideal diffuser has a negligible exit velocity, so
p es = p ose and
h
h oi
h es
h e
h i
s i
s
s e
p osi
p ose
p e
Maximum (i.e., isentropic)
recoverable portion of the
inlet kinetic energy
p i
Actual diffuser process
FIGURE 16.25
Thermodynamic process path of a diffuser plotted on h–s
coordinates.
T es /T inlet = ðp es /p inlet
Þ ðk−1
Then, the equation for η D becomes
η D =
Now, from Eq. (16.13),
Þ/k = ðp ose /p inlet
ð p ose/p inlet Þ ðk−1Þ/k
− 1
ðk−1ÞM 2 inlet /2
p inlet = p osi 1 + k − 1 − k/ðk−1Þ
2 M2 inlet
Þ ðk−1Þ/k
so that the diffuser efficiency can be written in terms
of the inlet Mach number and the isentropic stagnation
pressure ratio as
Diffuser efficiency
1 + k − 1
η D =
2 M2 inlet
ðp ose /p osi Þ ðk−1Þ/k − 1
ðk − 1ÞM 2 inlet /2 (16.42)
Therefore, for a constant isentropic stagnation pressure ratio, the diffuser efficiency decreases as the inlet Mach
number increases, asymptotically approaching the value (p ose /p osi ) (k–1)/k as the inlet Mach number goes to infinity.
A more direct measure of a diffuser’s ability to convert kinetic energy into pressure is the diffuser pressure recovery
coefficient C p , defined as
C p =
Actual diffuser pressure rise
Isentropic diffuser pressure rise =
Because of flow separation from the diffuser wall, C p values are typically around 0.6.
ð p exitÞ actual
– p inlet
(16.43)
p osi – p inlet
EXAMPLE 16.15
A subsonic diffuser for a spacecraft attitude control thruster has been tested in the laboratory with an inlet Mach number of
0.890, using pure nitrogen as the working gas. The inlet and exit isentropic stagnation pressures are measured and found to
be p osi = 314.7 kPa and p ose = 249.3 kPa. Determine, under these conditions, the diffuser’s (a) efficiency and (b) pressure
recovery coefficient.
Solution
a. The diffuser’s efficiency can be determined directly from the measured conditions using Eq. (16.42) as follows:
η D =
1 + k − 1
2 M2 inlet
ðk−1Þ/k
− 1
pose
p osi
ðk−1ÞM 2 inlet /2
1:40 − 1
1 + 1:40 − 1
0:890 2 249:3 1:40
− 1
=
2
314:7
= 0:529 = 52:9%
1:40 − 1
0:890 2
2
b. The diffuser’s pressure recovery coefficient can be determined from Eq. (16.43), if we can determine the actual inlet and
exit pressures from the given data. Equation (16.13) can be used to relate the actual pressures to their isentropic
stagnation values as
p
ose
= 1 + k − 1 k
p e 2 M2 k−1
exit
and
p
osi
= 1 + k − 1 k
p i 2 M2 k−1
inlet