TITLE.PM5

COMPRESSIBLE FLOW 891

dharm

\M-therm\Th16-2.pm5

Also,

T

Ts

2

=

p

ps

2

1

F

HG

I

KJ

−γ

γ

=

110

222 2

11

1

.

.4

F .4

HG

I

KJ

−

= 0.818

or, T 2 = 0.818 × 487.45 = 398.7 K
Sonic velocity at outlet section,
C 2 = γRT 2 = 1 4 287 398 7..×× = 400.25 m/s
∴ Velocity at outlet section, V 2 = M 2 × C 2 = 1.05 × 400.25 = 420.26 m/s. Ans.
Now, mass flow at the given section = mass flow at outlet section (exit)
......continuity equation

i.e., ρ 1 A 1 V 1 = ρ 2 A 2 V 2 or
p
RT

1

1

A 1 V 1 =

p

RT

2

2

A 2 V 2

∴ Flow area at the outlet section,

A 2 =

p AVT

TpV

1112

122

200 0 001 170 398 7

473 110 420 26

= ×××

××

..

. = 6.199 × 10

–4 m 2

Hence, A 2 = 6.199 × 10–4 m^2 or 619.9 mm^2. Ans.
(iv) Pressure (pt), temperature (Tt), velocity (Vt), and flow area (At) at throat of
the nozzle :
At throat, critical conditions prevail, i.e. the flow velocity becomes equal to the sonic veloc-
ity and Mach number attains a unit value.

From eqn. (16.22),

T

T

s

t

=^1

1

2

+F −^2

HG

I

KJ

L

N

M

O

Q

P

γ M

t

or,

487 45.

Tt =

1 14 1

2

+F − 12

HG

I

KJ

L ×

N

M

O

Q

P

.

= 1.2 or Tt = 406.2 K

Hence Tt = 406.2 K (or 133.2°C). Ans.

Also,

p

T

t

s

= T

T

t

s

F

HG

I

KJ

−

γ

γ 1

or

pt

222 2. =

406 2

487 45

1

. 11
.

.4

F .4

HG

I

KJ

− = 0.528

or, pt = 222.2 × 0.528 = 117.32 kN/m^2. Ans.

Sonic velocity (corresponding to throat conditions),

Ct = γRTt = 1 4 287 406 2..×× = 404 m/s

∴ Flow velocity, Vt = Mt × Ct = 1 × 404 = 404 m/s

By continuity equation, we have : ρ 1 A 1 V 1 = ρt AtVt

or, p

RT

1

1

A 1 V 1 =

p

RT

t

t

AtVt

∴ Flow area at throat, At =

pAVT

TpV

t

tt

111

1

200 0 001 170 406 2

473 117 32 404

= ×××

××

..

.

= 6.16 × 10–4 m^2

Hence, At = 6.16 × 10–4 m^2 or 616 mm^2 (Ans.)