TITLE.PM5

890 ENGINEERING THERMODYNAMICS

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\M-therm\Th16-2.pm5

Example 16.17. At some section in the convergent-divergent nozzle, in which air is flow-
ing, pressure, velocity, temperature and cross-sectional area are 200 kN/m^2 , 170 m/s, 200°C and
1000 mm^2 respectively. If the flow conditions are isentropic, determine :
(i)Stagnation temperature and stagnation pressure,
(ii)Sonic velocity and Mach number at this section,
(iii)Velocity, Mach number and flow area at outlet section where pressure is 110 kN/m^2 ,
(iv)Pressure, temperature, velocity and flow area at throat of the nozzle.
Take for air : R = 287 J/kg K, cp = 1.0 kJ/kg K and γ = 1.4.
Sol. Let subscripts 1, 2 and t refers to the conditions at given section, outlet section and
throat section of the nozzle respectively.
Pressure in the nozzle, p 1 = 200 kN/m^2
Velocity of air, V 1 = 170 m/s
Temperature, T 1 = 200 + 273 = 473 K
Cross-sectional area, A 1 = 1000 mm^2 = 1000 × 10–6 = 0.001 m^2
For air : R = 287 J/kg K, cp = 1.0 kJ/kg K, γ = 1.4
(i) Stagnation temperature (Ts) and stagnation pressure (ps) :

Stagnation temperature, Ts = T 1 +

V

cp

1

2

2 ×

= 473 +

170

2 1 0 1000

2

××(. )

= 487.45 K (or 214.45°C) (Ans.)

Also,

p

p

s

1

=

T

T

s

1

1

1

487 45 11

473

F

HG

I

KJ

=FHG IKJ

− −

γ

γ.

.4

.4

= 1.111

∴ Stagnation pressure, ps = 200 × 1.111 = 222.2 kN/m^2 (Ans.)

(ii) Sonic velocity and Mach number at this section :

Sonic velocity, C 1 = γRT 1 = 1 4 287 473.×× = 435.9 m/s (Ans.)

Mach number, M 1 =

V

C

1

1

170

435 9

=

. = 0.39 (Ans.)
(iii) Velocity, Mach number and flow area at outlet section where pressure is
110 kN/m^2 :
Pressure at outlet section, p 2 = 110 kN/m^2 ...(Given)

From eqn (16.17),

p

p

s

1

=^1

1

2 2

+F −^21

HG

I

KJ

L

N

M

O

Q

P

γ −

γ

γ

M

222 2

110

. = 1 14 1
2 0

2

1

11

+F −

HG

I

KJ

L

N

M

O

Q

P

. −

.4

.4

M = (1 + 0.2 M 02 )3.5

or, (1 + 0.2 M 22 ) =

222 2

110

1

F.^35.

HG

I

KJ = 1.222

or, M 2 =
1 222 1
02

. 1/ 2
.

F −

HG

I

KJ = 1.05 (Ans.)