Problem 13.25

# Problem 13.25 - Isenp M k(13.7a Then p 1 = 150 kPa The mass...

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Problem 13.25 [Difficulty: 2] 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. 12.18 (12.18) 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. 12.22a (12.22a) p 0 = 191 kPa From M 1 and p 0 , and Eq. 13.7a (using built-in function

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Unformatted text preview: Isenp ( M , k ) (13.7a) Then p 1 = 150 kPa The mass flow rate is m rate = 1 A 1 V 1 Hence, we need 1 from the ideal gas equation. 1 = 1.89 kg/m 3 The mass flow rate m rate is then m rate = 18.9 kg/s The throat area A t = A * because the flow is choked. From M 1 and A 1 , and Eq. 13.7d (using built-in function IsenA ( M , k ) (13.7d) A* = 0.0421 m 2 Hence A t = 0.0421 m 2...
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## This note was uploaded on 01/20/2012 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 13.25 - Isenp M k(13.7a Then p 1 = 150 kPa The mass...

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