Basics of Fluid Mechanics, 2014a

11.10. INTRODUCTION 479
Example 11.23:
Air enters a pipe with pressure of 3[bar] and temperature of 27 ◦ C at Mach number of
M =0.25. Due to internal combustion heat was released and the exit temperature was
found to be 127 ◦ C. Calculate the exit Mach number, the exit pressure, the total exit
pressure, and heat released and transferred to the air. At what amount of energy the
[
exit temperature will start to decrease? Assume C P =1.004
Solution
kJ
kg ◦ C
The entrance Mach number and the exit temperature are given and from Table (11.7)
or from Potto–GDC the initial ratio can be calculated. From the initial values the ratio
at the exit can be computed as the following.
]
M
T
T ∗ T 0
T 0
∗
P
P ∗ P 0
P 0
∗
ρ ∗ ρ
0.25000 0.30440 0.25684 2.2069 1.2177 0.13793
and
T 2
T ∗ = T 1
T ∗ T 2
T 1
=0.304 × 400
300 =0.4053
M
T
T ∗ T 0
T 0
∗
P
P ∗ P 0
P 0
∗
ρ ∗ ρ
0.29831 0.40530 0.34376 2.1341 1.1992 0.18991
The exit Mach number is known, the exit pressure can be calculated as
P ∗ P 2
P 2 = P 1
P 1 P ∗ =3× 1 × 2.1341 = 2.901[Bar]
2.2069
For the entrance, the stagnation values are
M
T
T 0
ρ
ρ 0
A
A ⋆
P
P 0
A×P
A ∗ ×P 0
F
F ∗
P 02
0.25000 0.98765 0.96942 2.4027 0.95745 2.3005 1.0424
The total exit pressure, P 02 can be calculated as the following:
isentropic
{}}{
P 01
= P 1
P 1
∗
P 0 P 02
∗
P 01 P =3× 1
0 0.95745 × 1 × 1.1992 = 3.08572[Bar]
1.2177
The heat released (heat transferred) can be calculated from obtaining the stagnation
temperature from both sides. The stagnation temperature at the entrance, T 01
T 01
isentropic
{}}{
T 01
= T 1
T 1
= 300/0.98765 = 303.75[K]