TITLE.PM5

(Ann) #1

894 ENGINEERING THERMODYNAMICS


dharm
\M-therm\Th16-2.pm5

Mach number, M 2 :

M 22 =

()
()

γ
γγ

−+
−−

12
21

1
2
1
2

M
M
...[Eqn. (16.49)]

=
()
()

1.4
1.4 1.4

3.6
11.2 0.4

−× +
××− −

=

12 2
22 1

2
2 = 0.333
∴ M 2 = 0.577 (Ans.)
Pressure, p 2 :
p
p

2
1

=

21
1

1
γγ^2
γ

M −−
+

()
() ...[Eqn. (16.46)]

=
2142 141
14 1

11 2 0 4
24

××− −^2
+

.(.)= −
(. )

..

. = 4.5
∴ p 2 = 26.5 × 4.5 = 119.25 kN/m^2 (Ans.)
Density, ρ 2 :
ρ
ρ


2
1
=

()
()

γ
γ

+
−+

1
12

1
2
1
2

M
M ...[Eqn. (16.47)]

= (. )
(. )

.
.

14 1 2
14 1 2 2

96
16 2

2
2

+
−+

=
+

= 2.667

∴ ρ 2 = 0.413 × 2.667 = 1.101 kg/m^3 (Ans.)
Temperature, T 1 :

Since p 1 = ρ 1 RT 1 , ∴ T 1 =

p
R

1
ρ 1
=
26 5 10
0 413 287

.^3
.


×
× = 223.6 K or – 49.4°C (Ans.)
Temperature, T 2 :
T
T

2
1

= [( ) ] [ ( )]
()

[( ) ] [ ( )]
()

γγγ
γ

−+ −−
+

= −+ ××−−
+

122 1
1

12 2 2 2 1
12

12 12
2 12

22
22

MM
M

1.4 1.4 1.4
1.4

= (. )(..
.

16 2 112 0 4)
23 04

+− = 1.6875

∴ T 2 = 223.6 × 1.6875 = 377.3 K or 104.3°C (Ans.)
Velocity, V 1 :
C 1 = γRT 1 = 1 4 287 223 6..×× = 299.7 m/s

Since

V
C

1
1

= M 1 = 2 ∴ V 1 = 299.7 × 2 = 599.4 m/s (Ans.)
Velocity, V 2 :
C 2 = γRT 2 = 1 4 287 377 3..×× = 389.35 m/s

Since

V
C

2
2

= M 2 = 0.577 ∴ V 2 = 389.35 × 0.577 = 224.6 m/s (Ans.)
Free download pdf