COMPRESSIBLE FLOW 873

dharm

\M-therm\Th16-1.pm5

Substituting the values of ps and ρs from eqns. (16.17) and (16.18), we get

Ts =

1

1 1

2

1 1

2

002

1

00

2

1

R 1

pM

M

+F −

HG

I

KJ

L

NM

O

QP

+F −

HG

I

KJ

L

N

M

O

Q

P

−

−

γ

ρ γ

γ

γ

γ

=

1

R^

p (^0) M
0
02
1
1
1
1 1
ρ 2
γ
γ
γγ
L +FHG − IKJ
N
M
O
Q
P
− − −
F
HG
I
KJ

1
R
p
(^0) M
0
0
2
1
1
1
1
ρ 2
γ
γ
γ

• F −
  HG
  I
  KJ
  L
  N
  M
  O
  Q
  P
  −
  −
  F
  HG
  I
  KJ
  or, Ts = T 0 1 1
  2 0
  +F −^2
  HG
  I
  KJ
  L
  NM
  O
  QP
  γ M ...(16.22) Q p (^0) RT
  0
  ρ =^0
  F
  HG
  I
  KJ
  Example 16.8. An aeroplane is flying at 1000 km/h through still air having a pressure of
  78.5 kN/m^2 (abs.) and temperature – 8 °C. Calculate on the stagnation point on the nose of the
  plane :
  (i)Stagnation pressure, (ii)Stagnation temperature, and
  (iii)Stagnation density.
  Take for air : R = 287 J/kg K and γ = 1.4.
  Sol. Speed of aeroplane, V = 1000 km/h =
  1000 1000
  60 60
  ×
  × = 277.77 m/s
  Pressure of air, p 0 = 78.5 kN/m^2
  Temperature of air, T 0 = – 8 + 273 = 265 K
  For air : R = 287 J/kg K, γ = 1.4
  The sonic velocity for adiabatic flow is given by,
  C 0 = γRT 0 = 1 4 287 265.×× = 326.31 m/s
  ∴ Mach number, M 0 =
  V
  C
  0
  0
  277 77
  326 31
  =.
  .
  = 0.851
  (i) Stagnation pressure, ps :
  The stagnation pressure (ps) is given by the relation,
  ps = p 0 1 1
  2 0
  +F −^21
  HG
  I
  KJ
  L
  N
  M
  O
  Q
  P
  γ −
  γ
  γ
  M ...[Eqn. (16.17)]
  or, ps = 78.5^1
  14 1
  2
  0 851^2
  1
  11
  +FHG − IKJ×
  L
  NM
  O
  QP

. −
.

.4

.4

= 78.5 (1.145)3.5 = 126.1 kN/m^2 (Ans.)