fluid_mechanics
618 Chapter 11 ■ Compressible Flow For the maximum flowrate condition, the constant area duct must be choked, and f 1/* / 1 2 D With k 1.4 for air and the above calculated value of D 0.4, we could use Eq. 11.98 to determine a value of Mach number at the entrance of the duct [section 112]. With k 1.4 and Ma 1 known, we could then rely on Eqs. 11.101, 11.103, 11.107, and 11.109 to obtain values of T 1T*, V 1V*, p 1p*, and p 0,1p* 0 . Alternatively, for air 1k 1.42, we can use Fig. D.2 with f 1/* / 1 2 D 0.4 and read off values of Ma 1 , T 1T*, V 1V*, p 1p*, and p 0,1p* 0 . The pipe entrance Mach number, Ma 1 , also represents the Mach number at the throat 1and exit2 of the isentropic, converging nozzle. Thus, the isentropic flow equations of Section 11.4 or Fig. D.1 can be used with Ma 1 . We use Fig. D.1 in this example. With Ma 1 known, we can enter Fig. D.1 and get values of T 1T 0 , p 1p 0 , and r 1r 0 . Through the isentropic nozzle, the values of T 0 , p 0 , and r 0 are each constant, and thus T 1 , p 1 , and r 1 can be readily obtained. Since T 0 also remains constant through the constant area duct 1see Eq. 11.752, we can use Eq. 11.63 to get T*. Thus, T* T 0 2 k 1 2 1.4 1 0.8333 Since T 0 288 K, we get from Eq. 2, T* 10.833321288 K2 240 K T 2 With T* known, we can calculate V* from Eq. 11.36 as V* 1RT*k f 1/ 2 / 1 2 D 231286.9 J21kg # K241240 K211.42 310 1Jkg2 1 2 (1) (2) (3) (Ans) 1ms2 2 , V A , r , Thus, since 1 Jkg 1 N # mkg 1 1kg # ms 2 2 # mkg Now can be obtained from V* and V 1V*. Having and V* 310 ms V 2 (4) (Ans) we obtain 1 V 1 we can get the mass flowrate from 1 1 m # r 1 A 1 V 1 Values of the other variables asked for can be obtained from the ratios mentioned. Entering Fig. D.2 with f 1/* /2D 0.4 we read Ma 1 0.63 T 1 T* 1.1 V 1 V* 0.66 p 1 p* 1.7 p 0,1 p* 0 1.16 10.02212 m2 0.4 10.1 m2 f 1/* / 1 2 (5) (7) (8) (9) (10) (11) Entering Fig. D.1 with Ma 1 0.63 we read T 1 0.93 T 0 p 1 0.76 p 0,1 r 1 0.83 r 0,1 Thus, from Eqs. 4 and 9 we obtain V 1 10.6621310 ms2 205 ms From Eq. 14 we get r 1 0.83r 0,1 10.83211.23 kgm 3 2 1.02 kgm 3 and from Eq. 5 we conclude that From Eq. 12, it follows that Equation 13 yields m # p10.1 11.02 kgm 3 m22 2 c d 1206 ms2 4 1.65 kgs T 1 10.9321288 K2 268 K p 1 10.7623101 kPa1abs24 77 kPa1abs2 (12) (13) (14) (Ans) (Ans) (Ans) (Ans) The stagnation temperature, T 0 , remains constant through this adiabatic flow at a value of T 0,1 T 0,2 288 K (Ans) The stagnation pressure, p 0 , at the entrance of the constant area duct is the same as the constant value of stagnation pressure through the isentropic nozzle. Thus p 0,1 101 kPa1abs2 (Ans) To obtain the duct exit pressure 1 p 2 p*2 we can use Eqs. 10 and 13. Thus, p 2 a p* b a p 1 b 1 p p 1 p 0,1 2 a 1 b 10.7623101 kPa1abs24 0,1 1.7 45 kPa1abs2 (Ans) For the duct exit stagnation pressure 1 p 0,2 p* 0 2 we can use Eq. 11 as p 0,2 a p* 0 b 1 p p 0,1 2 a 1 b 3101 kPa1abs24 0,1 1.16 87.1 kPa1abs2 (Ans) The stagnation pressure, p 0 , decreases in a Fanno flow because of friction. COMMENT Use of graphs such as Figs. D.1 and D.2 illustrates the solution of a problem involving Fanno flow. The T–sdiagram for this flow is shown in Fig. E.11.12b, where the entropy difference, s 2 s 1 , is obtained from Eq. 11.22.
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
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618 Chapter 11 ■ Compressible Flow<br />
For the maximum flowrate condition, the constant area duct<br />
must be choked, and<br />
f 1/* / 1 2<br />
D<br />
With k 1.4 for air and the above calculated value of<br />
D 0.4, we could use Eq. 11.98 to determine a value of Mach<br />
number at the entrance of the duct [section 112]. With k 1.4 and<br />
Ma 1 known, we could then rely on Eqs. 11.101, 11.103, 11.107,<br />
and 11.109 to obtain values of T 1T*, V 1V*, p 1p*, and p 0,1p* 0 .<br />
Alternatively, for air 1k 1.42, we can use Fig. D.2 with f 1/* / 1 2<br />
D 0.4 and read off values of Ma 1 , T 1T*, V 1V*, p 1p*, and<br />
p 0,1p* 0 .<br />
The pipe entrance Mach number, Ma 1 , also represents the Mach<br />
number at the throat 1and exit2 of the isentropic, converging nozzle.<br />
Thus, the isentropic flow equations of Section 11.4 or Fig. D.1 can<br />
be used with Ma 1 . We use Fig. D.1 in this example.<br />
With Ma 1 known, we can enter Fig. D.1 and get values of<br />
T 1T 0 , p 1p 0 , and r 1r 0 . Through the isentropic nozzle, the values<br />
of T 0 , p 0 , and r 0 are each constant, and thus T 1 , p 1 , and r 1 can be<br />
readily obtained.<br />
Since T 0 also remains constant through the constant area duct<br />
1see Eq. 11.752, we can use Eq. 11.63 to get T*. Thus,<br />
T*<br />
T 0<br />
2<br />
k 1 2<br />
1.4 1 0.8333<br />
Since T 0 288 K, we get from Eq. 2,<br />
T* 10.833321288 K2 240 K T 2<br />
With T* known, we can calculate V* from Eq. 11.36 as<br />
V* 1RT*k<br />
f 1/ 2 / 1 2<br />
D<br />
231286.9 J21kg # K241240 K211.42<br />
310 1Jkg2 1 2<br />
(1)<br />
(2)<br />
(3) (Ans)<br />
1ms2 2 ,<br />
V A , r ,<br />
Thus, since 1 Jkg 1 N # mkg 1 1kg # ms 2 2 # mkg <br />
Now can be obtained from V* and V 1V*. Having and<br />
V* 310 ms V 2 (4) (Ans)<br />
we obtain<br />
1<br />
V 1 we can get the mass flowrate from<br />
1 1<br />
m # r 1 A 1 V 1<br />
Values of the other variables asked for can be obtained from the<br />
ratios mentioned.<br />
Entering Fig. D.2 with f 1/* /2D 0.4 we read<br />
Ma 1 0.63<br />
T 1<br />
T* 1.1<br />
V 1<br />
V* 0.66<br />
p 1<br />
p* 1.7<br />
p 0,1<br />
p* 0<br />
1.16<br />
10.02212 m2<br />
0.4<br />
10.1 m2<br />
f 1/* / 1 2<br />
(5)<br />
(7)<br />
(8)<br />
(9)<br />
(10)<br />
(11)<br />
Entering Fig. D.1 with Ma 1 0.63 we read<br />
T 1<br />
0.93<br />
T 0<br />
p 1<br />
0.76<br />
p 0,1<br />
r 1<br />
0.83<br />
r 0,1<br />
Thus, from Eqs. 4 and 9 we obtain<br />
V 1 10.6621310 ms2 205 ms<br />
From Eq. 14 we get<br />
r 1 0.83r 0,1 10.83211.23 kgm 3 2 1.02 kgm 3<br />
and from Eq. 5 we conclude that<br />
From Eq. 12, it follows that<br />
Equation 13 yields<br />
m # p10.1<br />
11.02 kgm 3 m22<br />
2 c d 1206 ms2<br />
4<br />
1.65 kgs<br />
T 1 10.9321288 K2 268 K<br />
p 1 10.7623101 kPa1abs24 77 kPa1abs2<br />
(12)<br />
(13)<br />
(14)<br />
(Ans)<br />
(Ans)<br />
(Ans)<br />
(Ans)<br />
The stagnation temperature, T 0 , remains constant through this<br />
adiabatic flow at a value of<br />
T 0,1 T 0,2 288 K<br />
(Ans)<br />
The stagnation pressure, p 0 , at the entrance of the constant area<br />
duct is the same as the constant value of stagnation pressure<br />
through the isentropic nozzle. Thus<br />
p 0,1 101 kPa1abs2<br />
(Ans)<br />
To obtain the duct exit pressure 1 p 2 p*2 we can use Eqs. 10 and<br />
13. Thus,<br />
p 2 a p* b a p 1<br />
b 1 p<br />
p 1 p 0,1 2 a 1 b 10.7623101 kPa1abs24<br />
0,1 1.7<br />
45 kPa1abs2<br />
(Ans)<br />
For the duct exit stagnation pressure 1 p 0,2 p* 0<br />
2 we can use Eq.<br />
11 as<br />
p 0,2 a p* 0<br />
b 1 p<br />
p 0,1 2 a 1 b 3101 kPa1abs24<br />
0,1 1.16<br />
87.1 kPa1abs2<br />
(Ans)<br />
The stagnation pressure, p 0 , decreases in a Fanno flow because of<br />
friction.<br />
COMMENT Use of graphs such as Figs. D.1 and D.2 illustrates<br />
the solution of a problem involving Fanno flow. The T–sdiagram<br />
for this flow is shown in Fig. E.11.12b, where the entropy<br />
difference, s 2 s 1 , is obtained from Eq. 11.22.