8 95 8 73 Air at 96 kPa 27 \u00baC enters an eight cylinder four stroke Otto cycle

# 8 95 8 73 air at 96 kpa 27 ºc enters an eight

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8- 95 8-73 Air at 96 kPa, 27 ºC enters an eight-cylinder, four-stroke Otto cycle that operates at 3,000 RPM. Each cylinder has a bore of 9 cm and a stroke of 8.5 cm. At top dead center, the volume is 15% of the cylinder volume at bottom dead center. The maximum temperature in the cycle is 2200 K. Using an air-standard analysis, determine: a. the net work per cycle per cylinder (in kJ) b. the cycle thermal efficiency c. the power developed (in kW and hp). Approach: For a cycle net work is equal to net heat transfer. Apply conservation of energy to the heat addition and heat rejection processes to determine the heat transfer. Evaluate properties using the ideal gas tables. Assumptions: 1. Neglect potential and kinetic energy effects. 2. All processes are ideal. 3. Air is the working fluid and is an ideal gas. Solution: a) The net work per cycle 23 41 net net W = Q Q Q = Applying the closed system energy equation to these processes, neglecting potential and kinetic energy effects, and noting that no work occurs during processes 2-3 and 4-1: ( ) 23 3 2 Q m u u = and ( ) 41 4 1 Q m u u = Evaluating the properties from the air tables and using the ideal gas equation to determine the mass: 1 1 1 PV M m RT = Volume 1 needs to be determined. From the given geometry ( )( ) ( ) 2 2 4 1 2 4 4 0.09m 0.085m 5.41 10 m V V B S π π = = = × 3 1 V We are also given . Therefore, combining these two equations we obtain 2 0.15 V = and 4 3 1 6.36 10 m V = × 4 3 2 0.954 10 m V = × The compression ratio is 1 2 6.67 V r V V = = ( )( ) ( ) ( )( ) 2 4 3 4 1 1 1 96kN m 6.36 10 m 28.97kg kmol 7.09 10 kg 8.314kJ kmol K 300K PV M m RT × = = = × From the air table, Appendix A-9: ( ) 1 300K 214.02kJ kg u = 1 621.2 r v = ( ) ( ) 2 1 2 1 621.2 1 6.67 93.18 r r v v V V = = = by interpolation 2 457.1kJ kg u = ( ) 3 2200 K 1872.4kJ kg u = 3 2.012 r v = ( ) ( ) 4 3 4 3 4 2.012 6.67 13.41 959.9kJ kg r r v v V V u = = = = Therefore, ( ) ( ) 4 23 7.09 10 kg 1872.4 457.1 kJ kg 1.003kJ Q = × = Answer ( ) ( ) 4 41 7.09 10 959.9 214.07 0.529kJ Q = × = Answer 1.003 0.529 0.475kJ net W = = Answer Note that these results are per cylinder b) Cycle thermal efficiency 23 0.475kJ 0.473 1.003kJ net net cycle in W W = = Q Q η = = Answer c) Power developed can be calculated following the procedure given in Example 8-15. ( ) kJ rev 1min 1kWs 0.475 8cylinders 3000 cylinder-powerstroke min 60s 1 kJ rev 2 powerstroke net net W n N W X ⎞⎛ ⎟⎜ ⎠⎝ = = ± 95kW 127hp = = Answer 8- 96 8-74 An eight-cylinder, four-stroke Otto cycle runs at 5000 rpm. The compression ratio is 12.2, and the engine displacement is 396 in 3 . Air enters the engine at 14.4 psia, 70 ºF. The maximum temperature in the engine is 4000 ºF. Using an air-standard analysis, determine: a. the net work per cycle per cylinder (in Btu) b. the cycle thermal efficiency c. the net power output (in kW and hp) d. the airflow through the engine (in ft 3 /min). Approach: The approach used in Example 8-15 can be followed. Assumptions: 1. Neglect potential and kinetic energy effects.  #### You've reached the end of your free preview.

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• Spring '08
• BORCA-TASCIUC
• Energy, kg, kinetic energy effects, WNET QIN
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