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Thermodynamics filled in class notes_Part_79

# Thermodynamics filled in class notes_Part_79 - 165 5.3...

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5.3. FIRST LAW ANALYSIS OF REACTING SYSTEMS 165 The combustion is stoichiometric. Assume that no small concentration species are generated. The global reaction is given by CH 4 + 2 O 2 CO 2 + 2 H 2 O. (5.142) The first law analysis for the closed system is slightly different: U 2 U 1 = 1 Q 2 1 W 2 . (5.143) Since the process is isochoric, 1 W 2 = 0. So 1 Q 2 = U 2 U 1 , (5.144) = n CO 2 u CO 2 + n H 2 O u H 2 O n CH 4 u CH 4 n O 2 u O 2 , (5.145) = n CO 2 ( h CO 2 RT 2 ) + n H 2 O ( h H 2 O RT 2 ) n CH 4 ( h CH 4 RT 1 ) n O 2 ( h O 2 RT 1 ) , (5.146) = h CO 2 + 2 h H 2 O h CH 4 2 h O 2 3 R ( T 2 T 1 ) , (5.147) = ( h o CO 2 ,f + Δ h CO 2 ) + 2( h o H 2 O,f + Δ h H 2 O ) ( h o CH 4 ,f + Δ h CH 4 ) 2( h o O 2 ,f + Δ h O 2 ) 3 R ( T 2 T 1 ) , (5.148) = ( 393522 + 28030) + 2( 241826 + 21937) ( 74873 + 0) 2(0 + 0) 3(8 . 314)(900 298) , (5.149) = 745412 kJ. (5.150) For the pressures, one has P 1 V 1 = ( n CH 4 + n O 2 ) RT 1 , (5.151) V 1 = ( n CH 4 + n O 2 ) RT 1 P 1 , (5.152) = (1 kmole + 2 kmole ) parenleftBig 8 . 314 kJ kg K parenrightBig (298 K ) 101 . 325 kPa , (5.153) = 73 . 36 m 3 . (5.154) Now V 2 = V 1 , so P 2 = ( n CO 2 + n H 2 O ) RT 2 V 2 , (5.155) = (1 kmole + 2 kmole )

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Thermodynamics filled in class notes_Part_79 - 165 5.3...

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