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ME 200 L36 Ground Transportation (Continued) (Air Standard Otto Cycle) 9.1 and 9.2 Material not picked up this week may be recycled! ME 200 L36 Ground Transportation (Continued) (Air Standard Otto Cycle) 9.1 and 9.2 Kim Sees Office ME Gatewood Wing Room 2172 Examination and Quiz grades are available Blackboard Examinations and Quizzes can be picked up all of this week from Gatewood Room 2172 Material not picked up this week may be recycled! https://engineering.purdue.edu/ME200/ ThermoMentor © Program Spring 2014 MWF AM J. P. Gore Gatewood Wing 3166, Office Hours: MWF TAs: Robert Kapaku Dong Han

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Stroke Compression ratio, r : volume at bottom dead center divided by volume at top dead center Displacement volume: volume swept by piston when it moves from top dead center to bottom dead center Top dead center Bottom dead center Introducing Engine Terminology

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The Otto cycle consists of four internally reversible processes in series: Process 1-2 : isentropic compression. Process 2-3 : constant-volume heat addition to the air from an external source. Process 3-4 : isentropic expansion. Process 4-1 : constant-volume heat transfer from the air. The Otto cycle compression ratio is: Air-Standard Otto Cycle

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4 Air-Standard Otto Cycle: r is important! (1) Define Compression Ratio (2) Ideal Gas Law and eq. (1) (3) Isentropic compression (4) Combine (2) & (3) Compression Ratio is Very IMPORTANT!

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Ignoring kinetic and potential energy effects, closed system energy balances for the four processes of the Otto cycle reduce to give The thermal efficiency is the ratio of the net work to the heat added and for air standard cycle with constant specific heats, it reduces to solely a function of compression ratio:: Thermal Efficiency of Air-Standard Otto Cycle Compression Ratio is Very IMPORTANT!

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6 Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv 1.2 = constant instead of the isentropic process. Find: (i) heat added, (ii) heat rejected, (iii) work done in the compression and expansion strokes, (iv) thermal efficiency, and (v) break mean effective pressure. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K

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7 Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv 1.2 = constant instead of the isentropic process. Find: (i) heat added, (ii) heat rejected, (iii) work done in the compression and expansion strokes, (iv) thermal efficiency, and (v) break mean effective pressure. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K

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8 Now consider an Otto cycle with a compression ratio of 8.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv 1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K

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9 Now consider an Otto cycle with a compression ratio of 8.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv 1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K

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Since the air-standard Otto cycle is composed of internally reversible processes, areas on the T-s and p- v diagrams can be interpreted as heat and work, respectively: On the T-s diagram, heat transfer per unit of mass isTds. Thus, Area 2-3-a-b-2 represents heat added per unit of mass. Area 1-4-a-b-1 is the heat rejected per unit of mass. The enclosed area is the net heat added, which equals the net work output. T-s Diagram for Air Standard Otto Cycle

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On the p- v diagram, work per unit of mass ispd v. Thus, Area 1-2-a-b-1 represents work input per unit of mass during the compression process. Area 3-4-b-a-3 is the work done per unit of mass in the expansion process. The enclosed area is the net work output, which equals the net heat added. P-v Diagram for Air-Standard Otto Cycle

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The compression ratio, r = V 2 /V 1, is an important operating parameter for reciprocating internal combustion engines as brought out by the following discussion centering on the T-s diagram: An increase in the compression ratio changes the cycle from to Since the average temperature of heat addition is greater in cycle , and both cycles have the same heat rejection process, cycle has the greater thermal efficiency. Accordingly, the Otto cycle thermal efficiency increases as the compression ratio increases. Effect of Compression Ratio Air-Standard Otto Cycle

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13 Now consider an Otto cycle with a compression ratio of 12.5 with (a) constant specific heats, (b) specific heats as a function of temperature, and (c) compression and expansion with pv 1.2 = constant instead of the isentropic process. Find: (i) heat added (kJ/kg), (ii) heat rejected (kJ/kg), (iii) work done in the compression (kJ/kg) and expansion strokes (kJ/kg), (iv) thermal efficiency, and (v) break mean effective pressure. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K Efficiency increases Significantly as compression Ratio is increased.

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14 internal irreversibilities onlyConsider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv 1.2 = constant instead of the isentropic processes. Considering internal irreversibilities only, find: (I) Entropy change and (II) Entropy generation during each of the processes: 1-2, 2-3, 3-4, and 4-1. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K Back to lower compression ratio and efficiency to study entropy production For the cycle, entropy change is zero and so is entropy production because the Internal irreversibilities are assumed to be zero!

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15 Consider an Otto cycle with a compression ratio of 8.5 with constant specific heats, specific heats as a function of temperature, and compression and expansion with pv 1.2 = constant instead of the isentropic processes. Considering internal and external irreversibilities, find: (I) Entropy change and (II) Entropy generation during each of the processes: 1-2, 2-3, 3-4, and 4-1. Data: T 1 = 300 K, T 3 = 1020 K; k=1.4, c p =1.005 kJ/kg-K, c v = kJ/kg-K, T LT = 270 K,T HT = 1400 K For the air, entropy change is zero and so is internal entropy production However, for the universe (and as a result of the cycle), entropy change is positive as a result of the entropy production at the boundaries of the cycle!

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