fluid_mechanics

11.5 Nonisentropic Flow of an Ideal Gas 619
E XAMPLE 11.13
Effect of Duct Length on Choked Fanno Flow
GIVEN The duct in Example 11.12 is shortened by 50%, but FIND Will shortening the duct cause the mass flowrate through
the duct discharge pressure is maintained at the choked flow value the duct to increase or decrease? Assume that the average friction
for Example 11.12, namely,
factor for the duct remains constant at a value of f 0.02.
p d 45 kPa1abs2
SOLUTION
We guess that the shortened duct will still choke and check our which is read in Fig. D.1 for Ma 1 0.7. Thus,
assumption by comparing p d with p*. If p d 6 p*, the flow is
choked; if not, another assumption has to be made. For choked flow
r 1 10.79211.23 kgm 3 2 0.97 kgm 3 (3)
we can calculate the mass flowrate just as we did for Example 11.12. We get V 1 from
For unchoked flow, we will have to devise another strategy.
For choked flow
(4)
f 1/* / 1 2
D
and from Fig. D.2, we read the values
p* 1.5. With Ma 1 0.70, we use Fig. D.1 and get
p 1
p 0
0.72
Now the duct exit pressure 1 p 2 p*2 can be obtained from
p 2 p* a p*
p 1
b a p 1
p 0,1
b 1 p 0,1 2
10.02211 m2
0.1 m 0.2
a 1 b 10.7223101 kPa1abs24 48.5 kPa1abs2
1.5
and we see that p d 6 p*. Our assumption of choked flow is justified.
The pressure at the exit plane is greater than the surrounding
pressure outside the duct exit. The final drop of pressure
from 48.5 kPa1abs2 to 45 kPa1abs2 involves complicated threedimensional
flow downstream of the exit.
To determine the mass flowrate we use
m # r 1 A 1 V 1
(1)
The density at section 112 is obtained from
Ma 1 0.70 and p 1
V 1
V* 0.73
from Fig. D.2 for Ma 1 0.7. The value of V* is the same as it
was in Example 11.12, namely,
V* 310 ms
Thus, from Eqs. 4 and 5 we obtain
V 1 10.73213102 226 ms
and from Eqs. 1, 3, and 6 we get
m # 10.97 kgm 3 2 c p10.1m22
d 1226 ms2
4
1.73 kgs
(5)
(6)
(Ans)
The mass flowrate associated with a shortened tube is larger than
the mass flowrate for the longer tube, m # 1.65 kgs. This trend is
general for subsonic Fanno flow.
COMMENT For the same upstream stagnation state and
downstream pressure, the mass flowrate for the Fanno flow will
decrease with increase in length of duct for subsonic flow. Equivalently,
if the length of the duct remains the same but the wall friction
is increased, the mass flowrate will decrease.
r 1
r 0,1
0.79
(2)
E XAMPLE 11.14
Unchoked Fanno Flow
GIVEN The same flowrate obtained in Example 11.12 1m
1.65 kgs2 is desired through the shortened duct of Example 11.13
1/ 2 / 1 1 m2. Assume f remains constant at a value of 0.02.
FIND Determine the Mach number at the exit of the duct, M 2 ,
and the back pressure, p 2 , required.
SOLUTION
Since the mass flowrate of Example 11.12 is desired, the Mach
number and other properties at the entrance of the constant area
duct remain at the values determined in Example 11.12. Thus,
from Example 11.12, Ma 1 0.63 and from Fig. D.2
f 1/* / 1 2
0.4
D