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11.5 Nonisentropic Flow of an Ideal Gas 611<br />

E XAMPLE 11.11<br />

GIVEN Air 1k 1.42 enters [section 112] an insulated, constant<br />

cross-sectional area duct with the following properties:<br />

T 0 518.67 °R<br />

T 1 514.55 °R<br />

p 1 14.3 psia<br />

Compressible Flow with Friction (Fanno Flow)<br />

FIND For Fanno flow, determine corresponding values of <strong>fluid</strong><br />

temperature and entropy change for various values of downstream<br />

pressures and plot the related Fanno line.<br />

SOLUTION<br />

To plot the Fanno line we can use Eq. 11.75<br />

and Eq. 11.76<br />

to construct a table of values of temperature and entropy change<br />

corresponding to different levels of pressure in the Fanno flow.<br />

We need values of the ideal gas constant and the specific heat<br />

at constant pressure to use in Eqs. 1 and 2. From Table 1.7 we read<br />

for air<br />

or<br />

From Eq. 11.14 we obtain<br />

From Eqs. 11.1 and 11.69 we obtain<br />

and rV is constant for this flow<br />

But<br />

and from Eq. 11.56<br />

Thus, with<br />

T 1rV22 T 2<br />

2c p p 2 R 2 T 0 constant<br />

s s 1 c p ln T T 1<br />

R ln p p 1<br />

R 1716 1ft # lb21slug # °R2 53.3 1ft # lb21lbm # °R2<br />

1112 fts<br />

c p <br />

Rk<br />

k 1<br />

c p 353.3 1ft # lb21lbm # °R2411.42<br />

1.4 1<br />

187 1ft # lb21lbm # °R2<br />

rV p<br />

RT Ma1RTk<br />

rV r 1 V 1 p 1<br />

RT 1<br />

Ma 1 1RT 1 k<br />

T 1 514.55 °F<br />

<br />

T 0 518.67 °R 0.992<br />

Ma 1 A<br />

a 1<br />

0.992 1 b .02 0.2<br />

2RT 1 k 211.42353.3 1ft # lb21lbm # °R241514.55 °R2<br />

196 31ft # lb2lbm4 1 2 3132.2 lbm # fts 2 2lb4 1 2<br />

(1)<br />

(2)<br />

(3)<br />

(4)<br />

Eq. 4 becomes<br />

or<br />

For p 7 psia we have from Eq. 1<br />

or<br />

Thus, since 1 lb 32.2 lbm # fts 2 we obtain<br />

Hence,<br />

where T is in °R.<br />

From Eq. 2, we obtain<br />

or<br />

rV 114.3 psia21144 in.2 ft 2 20.211112 fts2<br />

353.3 1ft # lb21lbm # °R241514.55 °R2<br />

316.7 lbm1ft 2 # s24 2 T 2<br />

T <br />

17 psia2 2 1144 in. 2 ft 2 2 2<br />

23187 1ft # lb21lbm # °R24<br />

353.3 1ft # lb21lbm # °R24 2<br />

518.67 °R<br />

rV 16.7 lbm1ft 2 # s2<br />

2.08 10 3 31lbm # fts 2 21lb # °R24 T 2 T 518.67 °R 0<br />

6.5 10 5 T 2 T 518.67 0<br />

T 502.3 °R<br />

s s 1 3187 1ft # lb21lbm #<br />

502.3 °R<br />

°R24 ln a<br />

514.55 °R b<br />

353.3 1ft # lb21lbm # °R24 ln a 7 psia<br />

14.3 psia b<br />

s s 1 33.6 1ft # lb21lbm # °R2<br />

(Ans)<br />

(Ans)<br />

Proceeding as outlined above, we construct the table of values<br />

shown below and graphed as the Fanno line in Fig. E11.11. The<br />

T, °R<br />

550<br />

500<br />

450<br />

400<br />

350<br />

300<br />

35 40 45 50 55<br />

s – s 1 , ________ (ft • lb)<br />

(lbm•°R)<br />

F I G U R E E11.11

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