Modern Engineering Thermodynamics

654 CHAPTER 16: Compressible Fluid Flow
h
Isentropic
stagnation
state
V 2
2g c
p os = constant
p o = constant
Actual
stagnation
state
Increase in s due to
irreversibilities in the
real system
p = constant
h o
h
Initial state
FIGURE 16.2
An isentropic and a real deceleration to the stagnation state on a Mollier diagram.
s
and
1
ρ os
ρ = 1 + V2 k−1
2g c c p T
(16.4)
where the os subscript has been added to indicate the isentropic stagnation state condition. These states are shown
schematically in Figure 16.2.
EXAMPLE 16.2
In Example 16.1, it was determined that the stagnation air temperature for air at 20.0°C traveling at 90.0 km/h = 25.0 m/s
was 20.3°C. Now determine the isentropic stagnation pressure and isentropic stagnation density of this air when the atmospheric
pressure is 0.101 MPa.
Solution
First, draw a sketch of the system (Figure 16.3).
Air at 20.0°C
and 90.0 km/h
(25.0 m/s)
Stagnation temperature gauge
reads 20.3°C
Atmospheric pressure = 0.101 MPa
FIGURE 16.3
Example 16.2.
For air, the specific heat ratio is 1.40, and when p = 0.101 MPa, Eq. (16.3) gives the isentropic stagnation pressure as
p os = p 1 + V 2
k
k−1
2g c c p T
0
1
ð25:0 m/sÞ 2 1 kJ/kg
1000 m
= ð0:101 MPaÞ 1 +
2 /s 2
B
C
@ 2ð1Þð1:004 kJ/kg.KÞð20:0 + 273:15 KÞA
= 0:1014 MPa
1:40
1:40−1