<p>ME 362 Compressible Flow with Shock Wave Page 1 of 4</p><p>Solution:</p><p>To show that a normal shock wave exists within the duct, we need to prove that </p><p>PE,exit < Pback < PA,exit</p><p> subsonic P/P A o</p><p>E 1 2 supersonic</p><p> throat exit x</p><p>2   0.05  x   A(x)  0.0021      0.05  </p><p> 2   0.05  0.1 2 At exit (x = 0.1m): Aexit  A(x  0.1)  0.0021     0.004 m   0.05  </p><p> 2    0.05  0.05  2 At throat (x = 0.05m): A  Athroat  A(x  0.05)  0.0021     0.002 m   0.05  </p><p>A 0.004 exit   2 A 0.002</p><p>For curve A:</p><p>From isentropic table (Table B.1):</p><p>A P exit exit  0.9395 at   2.0351 (subsonic) ~ 2  A Po</p><p> PA,exit  0.9395Po  939.5 kPa ME 362 Compressible Flow with Shock Wave Page 2 of 4</p><p>For curve E:</p><p>Before shock (isentropic, supersonic):</p><p>From isentropic table (Table B.1):</p><p>A at exit  2.0050 (supersonic) ~ 2  Ma  2.2 A 1</p><p>Across shock (nonisentropic):</p><p>From shock table (Table B.2):</p><p>P A o2  0.6281 2 at Ma1  2.2  and   1.5920 Po1 A1</p><p> 2 A1  Athroat  0.002 m</p><p>  2 A21.5920 A 1  1.5920  0.002  0.003184 m</p><p>After shock (isentropic, subsonic):</p><p>Aexit 0.004    1.2563 A2 0.003184</p><p>From isentropic table (Table B.1):</p><p>Aexit Pexit at   1.2403 (subsonic)   0.8082 A2 Po2 ~ 1.2563 P P P  exit  exit o2  0.8082 0.6281  0.5076 Po Po2 Po1</p><p> PE,exit  0.5076Po  507.6 kPa</p><p>PE,exit  507.6 kPa < Pback  650 kPa < PA,exit  939.5 kPa</p><p> Normal shock exists within the nozzle [Ans.] ME 362 Compressible Flow with Shock Wave Page 3 of 4</p><p>Iterative procedure for locating normal shock wave:</p><p> sonic subsonic 1 2 supersonic</p><p> throat x exit x shock</p><p>Before shock (isentropic, supersonic): at x = 0.075m:</p><p> 2   0.05  0.075  2 A1  Ashock  A(x  0.075)  0.0021     0.0025 m   0.05  </p><p>A 0.0025 1   1.25 A 0.002</p><p>From isentropic table (Table B.1):</p><p>A at 1  1.2502 (supersonic) ~ 1.25  Ma  1.6 A 1</p><p>Across shock (nonisentropic):</p><p>From shock table (Table B.2):</p><p>P A o2  0.8952 2 at Ma1  1.6  and   1.1171 Po1 A1</p><p>  2 A21.1171 A 1  1.1171  0.002  0.00223 m ME 362 Compressible Flow with Shock Wave Page 4 of 4</p><p>After shock (isentropic, subsonic):</p><p>Aexit 0.004    1.7937 A2 0.00223</p><p>From isentropic table (Table B.1):</p><p>Aexit Pexit at   1.8229 (subsonic) ~ 1.7937   0.9231 A2 Po2</p><p>Pexit Pexit Po2   0.9231 0.8952  0.8264  Pexit  0.8264Po  826.4 kPa [Ans.] Po Po2 Po1</p><p>Determine if correct shock position is upstream or downstream of x = 0.075 m:</p><p>Pexit826.4 kPa  P back  650 kPa  correct shock position > x = 0.075 m</p><p> correct shock position is further downstream [Ans.]</p><p>P /P P > P back o exit back sonic P = P P exit back back shock</p><p> x x x x throat shock exit x = 0.075m</p>