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Thermodynamics filled in class notes_Part_29

Thermodynamics filled in class notes_Part_29 - 65 2.9...

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2.9. STIRLING 65 Now one can calculate the cycle efficiency: η = 1 u 4 u 1 h 3 h 2 , (2.443) = 1 parenleftBig 1011 . 98 kJ kg parenrightBig parenleftBig 205 . 756 kJ kg parenrightBig parenleftBig 2733 . 15 kJ kg parenrightBig parenleftBig 933 . 15 kJ kg parenrightBig , (2.444) = 0 . 552098 . (2.445) Now one has q L = u 1 u 4 = parenleftbigg 205 . 756 kJ kg parenrightbigg parenleftbigg 1011 . 98 kJ kg parenrightbigg = 806 . 224 kJ kg . (2.446) So, w net = q H + q L = parenleftbigg 1800 kJ kg parenrightbigg parenleftbigg 806 . 224 kJ kg parenrightbigg = 993 . 776 kJ kg . (2.447) So P meff = W net V max V min , (2.448) = w net v max v min , (2.449) = w net v 1 v 2 , (2.450) = w net RT 1 P 1 RT 2 P 2 , (2.451) = 993 . 776 kJ kg parenleftBig 0 . 287 kJ kg K parenrightBig ( 288 K 100 kPa 900 K 6250 kPa ) , (2.452) = 1265 . 58 kPa. (2.453) 2.9 Stirling Another often-studied air-standard engine is given by the Stirling cycle. This is similar to the Otto cycle except the adiabatic processes are replaced by isothermal ones. The efficiency can be shown to be equal to that of a Carnot engine. Stirling engines are difficult to build.
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