12-Practice Problems - Chapter 12 Compressible Flow Problem...

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Unformatted text preview: Chapter 12 Compressible Flow Problem 12.1 Methane at 25 o C ( g85 = 518 J/kg/K, g110 = 1 g61 31 ) is g32 owing in a pipe at 400 m/s. Is the g32 ow subsonic, sonic, supersonic, or hypersonic? Solution The speed of sound in methane is g102 = g115 g110g85g87 = g115 1 g61 31 × 518 × 298 = 450 m/s Because the velocity is less than the speed of sound, the g32 ow is subsonic . Problem 12.2 Air ( g85 = 1716 ft-lbf/slug/ o R, g110 = 1 g61 40) with a velocity of 1500 ft/s, a pressure of 10 psia, and a temperature of 40 o F passes through a normal shock wave. Find the velocity, pressure, and temperature downstream of the shock wave. Solution First g31 nd the upstream Mach number and then use relationships for normal shock waves. The speed of sound is g102 = g115 g110g85g87 = p 1 g61 4 × 1716 × (460 + 40) = 1096 ft/s 109 110 CHAPTER 12. COMPRESSIBLE FLOW The upstream Mach number is g80 1 = g89 g102 = 1500 1096 = 1 g61 37 The Mach number behind the shock wave is g80 2 2 = ( g110 g3 1) g80 2 1 + 2 2 g110g80 2 1 g3 ( g110 g3 1) = g61 4 × 1 g61 37 2 + 2 2 × 1 g61 4 × 1 g61 37 2 g3 g61 4 = 0 g61 567 g80 2 = 0 g61 753 The temperature ratio across the wave is g87 2 g87 1 = 1 + g110 g3 1 2 g80 2 1 1 + g110 g3 1 2 g80 2 2 = 1 + 0 g61 2 × 1 g61 37 2 1 + 0 g61 2 × g61 753 2 = 1 g61 24 Thus the temperature is g87 2 = 1 g61 24 g87 1 = 1 g61 24 × 500 = 620 o R = 160 o g73 The pressure ratio is g115 2 g115 1 = 1 + g110g80 2 1 1 + g110g80 2 2 = 1 + 1 g61 4 × 1 g61 37 2 1 + 1 g61 4 × g61 753 2 = 2 g61 022 Thus the pressure is g115 2 = 2 g61 022 g115 1 = 2 g61 022 × 10 = 20 g61 22 psia The speed of sound behind the shock wave is g102 2 = g115 1 g61 4 × 1716 × 620 = 1220 ft/s The velocity behind the shock wave is g89 2 = g80 2 g102 2 = 0 g61 753 × 1220 = 919...
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12-Practice Problems - Chapter 12 Compressible Flow Problem...

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