Thermodynamics filled in class notes_Part_23

# Thermodynamics filled in class notes_Part_23 - 2.5 BRAYTON...

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Unformatted text preview: 2.5. BRAYTON 53 Expansion through the nozzle completes the cycle. Assume an isentropic nozzle, therefore T 5 = T 4 parenleftbigg P 5 P 4 parenrightbigg ( k − 1) /k , (2.318) = T 4 parenleftbigg P 1 P 4 parenrightbigg ( k − 1) /k , (2.319) = = (2013 ◦ R ) parenleftbigg 5 psia 39 . 7 psia parenrightbigg (1 . 4 − 1) / 1 . 4 , (2.320) = 1114 ◦ R. (2.321) Now consider the energy balance in the nozzle: h 5 + v 5 · v 5 2 = h 4 + v 4 · v 4 2 bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright small , (2.322) = h 5 − h 4 + v 2 5 2 , (2.323) = c P ( T 5 − T 4 ) + v 5 · v 5 2 , (2.324) v 5 = radicalbig 2 c P ( T 4 − T 5 ) , (2.325) = radicaltp radicalvertex radicalvertex radicalbt 2 parenleftbigg . 240 Btu lbm ◦ R parenrightbigg (2013 ◦ R − 1114 ◦ R ) parenleftBigg 25037 ft 2 s 2 1 Btu lbm parenrightBigg , (2.326) = 3288 ft s . (2.327) Now find the thrust force magnitude | F | = ˙ m ( v 5 − v 1 ) = parenleftbigg 100 lbm s parenrightbiggparenleftbigg 3288...
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Thermodynamics filled in class notes_Part_23 - 2.5 BRAYTON...

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