COMPRESSIBLE FLOW 889dharm
\M-therm\Th16-2.pm5= 7 84090 1 0 903×−(.) = 238.9 m/s
i.e., V 2 = 238.9 m/s (Ans.)
Example 16.16. A tank fitted with a convergent nozzle contains air at a temperature of
20 °C. The diameter at the outlet of the nozzle is 25 mm. Assuming adiabatic flow, find the mass
rate of flow of air through the nozzle to the atmosphere when the pressure in the tank is :
(i) 140 kN/m^2 (abs.), (ii) 300 kN/m^2
Take for air : R = 287 J/kg K and γ = 1.4, Barometric pressure = 100 kN/m^2.
Sol. Temperature of air in the tank, T 1 = 20 + 273 = 293 K
Diameter at the outlet of the nozzle, D 2 = 25 mm = 0.025 m
Area, A 2 = π/4 × 0.025^2 = 0.0004908 m^2
R = 287 J/kg K, γ = 1.4
(i) Mass rate of flow of air when pressure in the tank is 140 kN/m^2 (abs.) :
ρ 1 =p
RT1
1=
140 10
287 293×^3
×
= 1.665 kg/m^3
p 1 = 140 kN/m^2 (abs.)
Pressure at the nozzle, p 2 = atmospheric pressure = 100 kN/m^2∴ Pressure ratio,p
p2
1=
100
140 = 0.7143
Since the pressure ratio is more than the critical value, flow in the nozzle will be subsonic,
hence mass rate of flow of air is given by eqn. 16.28, asm = A 22
1 112
12
2
11
γ
γρ
γγ
γ
−F
HGI
KJ−
F
HGI
KJL
N
M
M
MO
Q
P
P
P+
p p
pp
p= 0.0004908214
14 1
140 10^3 1 665 0 7143 0 71432
111
× 1
−F
HGI
KJ
××× −L
N
M
MO
Q
P
P+
.
.
.(.)(.).4.4
.4= 0.0004908 1631700 0 7143(. )^1 .4285−(.0 7143)1 7142.
or m = 0.0004908 1631700 0 6184(. −0 5617. ) = 0.1493 kg/s (Ans.)
(ii) Mass rate of flow of air when pressure in the tank is 300 kN/m^2 (abs.) :
p 1 = 300 kN/m^2 (abs.)
p 2 = pressure at the nozzle = atmospheric pressure = 100 kN/m^2∴ Pressure ratio,p
p2
1=
100
300 = 0.33.
The pressure ratio being less than the critical ratio 0.528, the flow in the nozzle will be
sonic, the flow rate is maximum and is given by eqn. (16.32), as
mmax = 0.685 A 2 p 11 ρwhere, ρ 1 =p
RT1
1=300 10
287 293×^3
× = 3.567 kg/m3∴ mmax = 0.685 × 0.0004908 300 10××^3 3 567. = 0.3477 kg/s (Ans.)