TITLE.PM5

872 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-1.pm5

Substituting the value of M 02 in eqn. (16.18), we get

ps = p 0 +

pV

p

0020 M M

0

02 04

2

1

4

2

24

γρ

γ

×++F −γ +

HG

I

KJ

or, ps = p 0 +

ργ 002 02

2 1 4 04

2

24

VMF ++− M +

HG

I

KJ

... ...(16.19)

Also, ps = p 0 +

ρ 002

2

V
(when compressibility effects are neglected) ...(16.20)
The comparison of eqns. (16.19) and (16.20) shows that the effects of compressibility are
isolated in the bracketed quantity and that these effects depend only upon the Mach number. The

bracketed quantity ie. ., 1 M M ...
4

2

24

0

2

0

F ++− (^4) +
HG
I
KJ
L
N
M
M
O
Q
P
P
γ may thus be considered as a compressibility
correction factor. It is worth noting that :
l For M < 0.2, the compressibility affects the pressure difference (ps – p 0 ) by less than 1 per cent
and the simple formula for flow at constant density is then sufficiently accurate.
l For larger value of M, as the terms of binomial expansion become significant, the
compressibility effect must be taken into account.
l When the Mach number exceeds a value of about 0.3 the Pitot-static tube used for
measuring aircraft speed needs calibration to take into account the compressibility
effects.

16.6.2. Expression for stagnation density (ρs)

From eqn. (ii), we have

ρ

ρ

00 γ

1

ss

p

p

=

F

HG

I

KJ

or

ρ

ρ

ssp γ

00 p

1

=

F

HG

I

KJ

or ρs = ρ 0

p

p

s

0

1

F

HG

I

KJ

γ

Substituting the value of

p

p

s

0

F

HG

I

KJ from eqn. (iv), we get

ρs = ρ 0 1 1

2 0

2 1

1

+F −

HG

I

KJ

R

S

T

U

V

W

L

N

M

M

M

O

Q

P

P

P

γ −

γ

γ

γ

M

or, ρs = ρ 0 1

1

2 0

2

1

1

+

F −

HG

I

KJ

L

N

M

O

Q

P

γ γ−

M ...(16.21)

16.6.3. Expression for stagnation temperature (Ts)

The equation of state is given by :

p

ρ = RT

For stagnation point, the equation of state may be written as :

ps

ρs

= RTs or Ts =

1

R

ps

ρs