Modern Engineering Thermodynamics

16.3 Isentropic Stagnation Properties 653
For an ideal gas or a low-pressure vapor with constant specific heats, this equation can be written as
or
V 2 /ð2g c Þ = h o − h = c p ðT o − TÞ
T o
T = 1 + V2
2g c c p T
(16.2)
where T o is the stagnation temperature (the temperature at zero velocity).
EXAMPLE 16.1
While driving in your new sports car at 90.0 km/h in still air at 20.0°C, you put your hand out the window with your palm
toward the front of the car. What is the air temperature on the center of your palm?
Solution
First, draw a sketch of the system (Figure 16.1).
When your hand is placed perpendicular to the air flow, you should feel the stagnation pressure and temperature of the air
flow. The stagnation temperature is given by Eq. (16.2) as
T o = T 1 + V2
2g c c p T
= ð20:0 + 273:15Þ 1 + ð90:0km=hÞð1000 m=kmÞð1h=3600 sÞð1kJ=kg=1000 m2 /s 2
Þ
2ð1Þð1:004 kJ/kg.KÞð20:0 + 273:15 KÞ
= 294 K = 20:3°C
So the stagnation temperature rise is not very much at this speed.
Exercises
1. How fast would the sports car in Example 16.1 have to travel
to produce a 1.00°C stagnation temperature rise at the center
of your hand? Answer: V = 161 km/h.
2. Suppose it is winter and the temperature of the air in
Example 16.1 is 0.00°C. What would be the stagnation
temperature rise at the center of your hand if all the other
variables remain the same? Answer: T o – T = 0.310°C
(independent of the value of T).
3. Now, you are in an aircraft traveling at 800. km/h in air at
20.0°C. If you put your hand out the window now, what
is the air temperature at the center of your hand? Answer:
T o = 318 K = 44.6°C.
20.0°C air
at 90.0 km/h
FIGURE 16.1
Example 16.1.
16.3 ISENTROPIC STAGNATION PROPERTIES
If, in addition, we decelerate the flow reversibly (i.e., without friction or other losses) and aergonically, then the
entire process becomes isentropic and Eq. (7.38) of Chapter 7 can be combined with Eq. (16.2) to provide
equations for the isentropic stagnation pressure p os and the isentropic stagnation density ρ os based on the isentropic
stagnation temperature T o = T os as
Then, Eq. (16.2) becomes
T os
T = p ðk−1Þ/k
os
= v 1 − k k−1
os ρ = os
(7.38)
p
v ρ
k
p os
p = 1 + V2 k−1
2g c c p T
(16.3)