ch12 - Problem 12.1 Problem 12.2 Problem 12.4 (In Excel)...

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Problem 12.4 (In Excel) Given: Data on flow in a passage Find: Pressure at downstream location Solution The given or available data is: R = 286.9 J/kg.K k = 1.4 T 1 = 305 K p 1 = 200 kPa V 1 = 525 m/s M 2 = 2 Equations and Computations: From T 1 and Eq. 11.17 (11.17) c 1 = 350 m/s Then M 1 = 1.50 From M 1 and p 1 , and Eq. 12.7a (using built-in function Isenp ( M , k )) (12.7a) p 01 = 734 kPa For isentropic flow ( p 01 = p 02 ) p 02 = 734 kPa From M 2 and p 02 , and Eq. 12.7a (using built-in function Isenp ( M , k )) p 2 = 93.8 kPa
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Problem 12.8 (In Excel) Given: Data on flow in a passage Find: Flow rate; area and pressure at downstream location; sketch passage shape Solution The given or available data is: R = 286.9 J/kg.K k = 1.4 A 1 = 0.25 m 2 T 1 = 283 K p 1 = 15 kPa V 1 = 590 m/s T 2 = 410 M 2 = 0.75 Equations and Computations: From T 1 and Eq. 11.17 (11.17) c 1 = 337 m/s Then M 1 = 1.75 Because the flow decreases isentropically from supersonic to subsonic the passage shape must be convergent-divergent From p 1 and T 1 and the ideal gas equation ρ 1 = 0.185 kg/m 3
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The mass flow rate is m rate = ρ 1 A 1 V 1 m rate = 27.2 kg/s From M 1 and A 1 , and Eq. 12.7d (using built-in function IsenA ( M , k )) (12.7d) A* = 0.180 m 2 From M 2 and A *, and Eq. 12.7d (using built-in function IsenA ( M , k )) A 2 = 0.192 m 2 From M 1 and p 1 , and Eq. 12.7a (using built-in function Isenp ( M , k )) (12.7a) p 01 = 79.9 kPa For isentropic flow ( p 01 = p 02 ) p 02 = 79.9 kPa From M 2 and p 02 , and Eq. 12.7a (using built-in function Isenp ( M , k )) p 2 = 55.0 kPa
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Problem 12.9 (In Excel) Given: Data on tank conditions; isentropic flow Find: Plot cross-section area and pressure distributions Solution The given or available data is: R = 286.9 J/kg.K k = 1.4 T 0 = 278 K p 0 = 304 kPa p e = 101 kPa m rate = 1 kg/s Equations and Computations: From p 0 , p e and Eq. 12.12d (using built-in function IsenMfromp (M,k)) (12.12d) M e = 1.36 Because the exit flow is supersonic, the passage must be a CD nozzle We need a scale for the area. From p 0 , T 0 , m flow , and Eq. 12.10 (12.10) Then A t = A * = 0.00136 m 2
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For each M , and A *, and Eq. 12.7d (using built-in function IsenA ( M , k ) (12.7d) we can compute each area A . From each M , and p 0 , and Eq. 12.7a (using built-in function Isenp ( M , k ) we can compute each pressure p . L (m) M A (m 2 ) p (kPa) 0.40 0.11 0.00727 301 0.60 0.16 0.00489 298 0.80 0.22 0.00371 294 1.0 0.27 0.00302 289 1.2 0.33 0.00256 282 1.4 0.38 0.00225 275 1.6 0.44 0.00202 267 1.8 0.49 0.00185 258 2.0 0.54 0.00171 249 2.2 0.60 0.00161 239 2.4 0.65 0.00154 228 2.6 0.71 0.00148 218 2.8 0.76 0.00143 207 3.0 0.82 0.00140 196 3.2 0.87 0.00138 186 3.4 0.92 0.00136 175 3.68 1.00 0.00136 161 3.8 1.03 0.00136 154 4.0 1.09 0.00137 144 4.2 1.14 0.00138 135 4.4 1.20 0.00140 126 4.6 1.25 0.00142 117 4.8 1.31 0.00145 109 5.0 1.36 0.00148 101
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Area Variation in Passage 0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 012345 L (m) A (m 2 ) Pressure Variation in Passage 0 50 100 150 200 250 300 350 L (m) p (kPa)
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Problem 12.13 (In Excel) Given: Data on converging nozzle; isentropic flow Find: Pressure and Mach number; throat area; mass flow rate Solution The given or available data is: R = 286.9 J/kg.K k = 1.4 A 1 = 0.05 m 2 T 1 = 276.3 K V 1 = 200 m/s p atm = 101 kPa Equations and Computations: From T 1 and Eq. 11.17 (11.17) c 1 = 333 m/s Then M 1 = 0.60 To find the pressure, we first need the stagnation pressure. If the flow is just choked p e = p atm = p * = 101 kPa From p e = p * and Eq. 11.21a (11.21a) p 0 = 191 kPa
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From M 1 and p 0 , and Eq. 12.7a (using built-in function Isenp ( M , k ) (12.7a) Then p 1 = 150 kPa The mass flow rate is m rate = ρ 1 A 1 V 1 Hence, we need ρ 1 from the ideal gas equation.
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This note was uploaded on 05/02/2008 for the course ME 341 taught by Professor Kennedy during the Spring '08 term at Cal Poly.

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ch12 - Problem 12.1 Problem 12.2 Problem 12.4 (In Excel)...

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