TITLE.PM5

874 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-1.pm5

(ii) Stagnation temperature, Ts :

The stagnation temperature is given by,

Ts = T 0 1

1

2 0

+F −^2

HG

I

KJ

L

N

M

O

Q

P

γ M

...[Eqn. (16.22)]

= 265^1

14 1

2

L + − ×0 851 2

NM

O

QP

..
= 303.4 K or 30.4°C (Ans.)
(iii) Stagnation density, ρs :
The stagnation density (ρs) is given by,
ps
ρs
= RTs or ρs = p
RT

s

s

or, ρs =

126 1 10

287 303 4

.^3
.

×

× = 1.448 kg/m

(^3) (Ans.)
Example 16.9. Air has a velocity of 1000 km/h at a pressure of 9.81 kN/m^2 in vacuum
and a temperature of 47°C. Compute its stagnation properties and the local Mach number. Take
atmospheric pressure = 98.1 kN/m^2 , R = 287 J/kg K and γ = 1.4.
What would be the compressibility correction factor for a pitot-static tube to measure the
velocity at a Mach number of 0.8.
Sol. Velocity of air, V 0 = 1000 km/h =
1000 1000
60 60
×
× = 277.78 m/s
Temperature of air, T 0 = 47 + 273 = 320 K
Atmospheric pressure, patm = 98.1 kN/m^2
Pressure of air (static), p 0 = 98.1 – 9.81 = 88.29 kN/m^2
R = 287 J/kg K, γ = 1.4
Sonic velocity, C 0 = γRT 0 = 1 4 287 320.×× = 358.6 m/s
∴ Mach number, M 0 =
V
C
0
0
277 78
358 6
=.

. = 0.7746
Stagnation pressure, ps :
The stagnation pressure is given by,

ps = p 0 1 1

2 0

+F −^21

HG

I

KJ

L

NM

O

QP

γ −

γ

γ

M ...[Eqn. (16.17)]

or, ps = 88.29 1 1
2

L + − ×0 7746 2 1

NM

O

QP

1.4 −

1.4

. 1.4

= 88.29 (1.12)3.5 = 131.27 kN/m^2 (Ans.)

Stagnation temperature, Ts :

Ts = T 0 1 1

2 0

+F −^2

HG

I

KJ

L

N

M

O

Q

P

γ M ...[Eqn. (16.22)]

or, Ts = 320^1

14 1

2

L + − ×0 7746 2

NM

O

QP

..
= 358.4 K or 85.4°C (Ans.)