TITLE.PM5

876 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th16-1.pm5

Speed of the aircraft, Va :

p 0 V = mRT 0 = m ×

R

M

F 0

HG

I

KJ^ T^0 or ρ^0 =

m

V

pM

RT

=^0

00

where R = characteristic gas constant,

R 0 = universal gas constant = 8314 Nm/mole K.

M = molecular weight for air = 28, and

ρ 0 = density of air.

Substituting the values, we get

ρ 0 = ()35 10^28

8314 235

××^3

×

= 0.5 kg/m^3

Now, using the relation : ps = p 0 +

ρ 0 2

2

Va

...[Eqn. (16.20)]

or, Va =

22654103510

0 05

()(.^33 )

.

pps−

=

×−×

ρ = 348.7 m/s (Ans.)

16.7. AREA-VELOCITY RELATIONSHIP AND EFFECT OF VARIATION OF AREA FOR
SUBSONIC, SONIC AND SUPERSONIC FLOWS
For an incompressible flow the continuity equation may be expressed as :
AV = constant, which when differentiated gives
AdV + VdA = 0 or
dA
A

dV
V
=− ...(16.23)
But in case of compressible flow, the continuity equation is given by,
ρAV = constant, which can be differentiated to give
ρd(AV) + AVdρ = 0 or ρ(AdV + VdA) + AVdρ = 0
or, ρAdV + ρVdA + AVdρ = 0
Dividing both sides by ρAV, we get
dV
V

dA

A

d

++

ρ

ρ = 0 ...(16.24)

or,

dA

A

dV

V

=− –

dρ

ρ ...[16.24 (a)]

The Euler’s equation for compressible fluid is given by,

dp

ρ + VdV + gdz = 0

Neglecting the z terms the above equation reduces to,

dp

ρ + VdV = 0

This equation can also be expressed as :

dp

ρ ×

d

d

ρ

ρ + VdV = 0 or

dp

d

d

ρ

ρ

ρ

× + VdV = 0

But

dp

dρ = C

(^2) ...[Eqn. (16.10)]
∴ C^2 ×
dρ
ρ + VdV = 0 or C
2 dρ
ρ = – VdV or
dρ
ρ = –
VdV
C^2