Chapter 1 Gas Power Cycle

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(b) w net = q H –q L , similar to q 23 ; -q L =q 41 = u 1 –u 4 3-4 Isentropic proc. v r4/v r3 = v 4/v 3 , v 4/v 3 =r v = 8, � v r4 = v r3r v = 6.108*8.0 = 48.864 Table A-17 : at v r4 = 48.864 � T 2 = 795.6 K and u 4 =588.74 kJ/kg, 4-1 Constant volume heat rejected, 1 st law q 41 = w 41 + u 1 –u 4 ; w 41 = 0 q 41 = u 1 –u 4 = 206.91 - 588.74 = -381.83 kJ/kg q L = -q 41 = 381.83 kJ/kg w net = q H –q L = 800 – 381.83 = 418.17 kJ/kg answer (c) η th = w net /q H = 418.83/800 = 0.523 or 52.3 % answer (d) MEP = w net /(v 1-v 2 ) ; P 1v 1 = RT 1 � v 1 = 0.832 m 3 /kg , v 2 =v 1/8, MEP = 574.4 kPa answer Spark-ignition Engine Diesel cycles: the ideal cycle for compression-ignition engine Ideal diesel cycle consists of four internally reversible processes: Process 1-2 Isentropic Compression Process 2-3 P = constant, heat added Process 3-4 Isentropic expansion Process 4-1 v = constant, heat rejection P 2 q in v 2 =v 3 3 Pv k = c Pv k = c 4 1 v 1 =v 4 q out v T 1 2 P = const. v = const. q in q out s 1 =s 2 s3 =s 4 4 3 s
1st law : closed system Ideal gas: and Analysis of Ideal Diesel <strong>Cycle</strong> Constant pressure heat transfer 2 P = const. ⇒ w = P ( v − v ) and 2 2 4 1 3 3 3 2 = − q = C ( T −T ) Isentropic Process of Ideal gases 3 1 4 3 2 3 1 k 1v1 4 3 Pv = RT, dh = C dT q = q = C (T −T ) in 2 2 q = u − u = −C ( T −T ) q q = u − u + w out k Pv = P k p 3 v = P 2 2 4 2 k 2v2 P ⎛ 2 v ⎞ ⎛ 1 V ⎞ 1 = ⎜ ⎟ = ⎜ ⎟ P1 ⎝ v2 ⎠ ⎝V2 ⎠ ( k −1) / k T ⎛ ⎞ ⎛ ⎞ 2 P2 v1 = ⎜ ⎟ = ⎜ ⎟ T1 ⎝ P1 ⎠ ⎝ v2 ⎠ 3 q = u − u + P ( v − v ) = h − h v 2 3 4 1 = constant k p k −1 2 1 3 2 P T 2 q in v 2 =v 3 2 1 s 1 =s 2 3 Pv k = c P = const. Pv k = c q in v = const. Analysis of Ideal Diesel <strong>Cycle</strong> Cut off ratio (r c ) is the ratio of the cylinder volume after and before the combustion process Thermal w η th = q from at net H r c efficiency q = 1 − q V3 v3 = = V v T1 ( T4 / T1 − 1) = 1 − kT ( T / T − 1) 2 definition processes 3 out in of 2 2 1 − 2 and 3 − 4 2 T4 − T1 = 1 − k ( T − T ) r c and k 1 ⎡ r ⎤ c − 1 η th = 1 − k −1 ⎢ ⎥ r ⎣ k ( rc − 1) ⎦ 3 2 isentropic P T 2 q in v 2 =v 3 2 3 Pv k = c 1 s 1 =s 2 Pv k = c P = const. q in v = const. q out 4 1 v 1 =v 4 q out 4 1 v 1 =v 4 4 3 s3 =s 4 q out v 4 3 s3 =s 4 q out v s s
Page 1 and 2: Chapter 1 Gas Power Cycle Content:
Page 3 and 4: Example 8.1 Show that the thermal e
Page 5 and 6: P Mean Effective Pressure, MEP Conc
Page 7 and 8: P Air Standard Otto Cycle Ideal Ott
Page 9: Example 8.2 An ideal Otto cycle has
Page 13 and 14: Ideal Brayton Cycle Brayton Cycle c

Chapter 1 Gas Power Cycle

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