Solution_week_10 - Solution to week 10 Recitation problem...

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Solution to week 10 Recitation problem Known: A turbojet with afterburner Find: - T-s diagram - Total heat added in the afterburner (in KJ/kg) - Nozzle exit velocity Assumptions - Air as working fluid and is an ideal gas - Constant specific heats with k=1.4 and cp=1.005KJ/kgK - Compressor, turbine, nozzle and diffuser are isentropic - Constant pressure in combustor & afterburner - turbine work output equals the work required to drive the compressor Schematic and given data diffuser compressor T 2  = -33 C P 2  = 35 kPa 1 2 3 Q 34 P 3  = 571.55 kPa T 4  = 1500 K 4 5 Q 56 6 7 afterburner nozzle turbine combustor T 6  = 1500 . . T 1  = -37 C P 1  = 32.84 kPA Analysis a) P 7 = 32.84 kPa V 6 = 0 m/s s (kJ/kgK) T (K) 1 2 3 4 5 6 7 1500 236.15 Without afterburner
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Let’s start by finding p and T for each state Process 2-3 is isentropic, we can find T 3 by applying the isentropic relationship for ideal gas with constant specific heats, (6.45) Process 3-4 is at constant pressure 4 571.55
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This note was uploaded on 03/15/2010 for the course ENGRD 221 taught by Professor Staff during the Fall '08 term at Cornell University (Engineering School).

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Solution_week_10 - Solution to week 10 Recitation problem...

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