1. APPLIED THERMODYNAMICS UNIT- 2 Gas power cycle 1 Department of Mechanical Engineering,A.I.E.T.,Mijar 3)Air Standard Diesel Cycle/ Constant Pressure cycle:

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3. 3 An important application of thermodynamics is the analysis of power cycles through which the energy absorbed as heat can be continuously converted into mechanical work. A thermodynamic analysis of the heat engine cycles provides valuable information regarding the design of new cycles or for improving the existing cycles.

4.  Definition of a cycle!!!  Gas power cycles in which the working fluid does not undergo any phase change.  Vapour power cycles in which the working fluid undergoes a phase change during the cyclic process  In the thermodynamic analysis of power cycles, our chief interest lies in estimating the energy conversion efficiency or the thermal efficiency.  The thermal efficiency of a heat engine is defined as the ratio of the network delivered to the energy absorbed as heat.  Air Standard Efficiency!!!  Relative Efficiency!!! 4 Department of Mechanical Engineering,A.I.E.T.,Mijar

5. 5  Assumptions: The working substance consists of a fixed mass of air and behaves as a perfect gas. The closed system is considered which under goes a cycle process. Therefore, there are no intake or exhaust process. The combustion process is replaced by an equivalent heat addition process form an external source. Thus there is no change in the chemical equilibrium of the working fluid and also composition. There is no exhaust process; this is replaced by an equivalent heat rejection process. Compression and expansion processes in the cycle are considered as reversible adiabatic process. The specific heats Cp and Cv of air do not vary with temperature.

6. Department of Mechanical Engineering,A.I.E.T.,Mijar 6 1)Carnot Cycle:

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9. 9 1.A Carnot engine operates between two reservoirs at temperatures T 1 and T 2.The work output of the engine is 0.6 times the heat rejected. The difference in temperatures between the source and the sink is 200 0 C.Calculate the thermal efficiency,source temperature and the sink 2. A Carnot engine rejects heat to the sink at 32 0 C and has a thermal efficiency of 52.3%.The work output from the engine is 120 kJ. Determine: i)The maximum working temperature of the engine ii)The heat added in kJ iii)The change in entropy during the heat rejection. 3.0.5kg of air(ideal gas)executes a Carnot cycle having a thermal efficiency of 50%.The heat transfer to the air during the isothermal expansion 40kJ.At the beginning of the isotheral expansion the pressure is 7bar and the volume is 0.12m 3. Determine i)The maximum and minimum temperature for the cycle in K ii)The volume at the end of isothermal expansion in m 3 iii)The heat transfer for each of the four processes in kJ for air C v =0.721 kJ/kgK and C p =1.008 kJ/kgK

10. Department of Mechanical Engineering,A.I.E.T.,Mijar 10 4.In a Carnot cycle,the maximum pressure and temperature are limited to 18bar and 410 0 C.The ratio of isentropic compression is 6 and isothermal expansion is 1.5.Assuming the volume of the air at the beginning of isothermal expansion to be 0.18m 3, Determine i)The temperature and pressure at main points in the cycle ii)Change in entropy during the isothermal expansion iii)Mean thermal efficiency of the cycle iv)Mean effective pressure of the cycle v)The theoretical power if there are 210 working cycles per minute. 5.An inventor claims that a new heat cycle will develop 0.4 kW for a heat addition of 32.5 kJ/min. The temperature of heat source is 1990K and that of sink is 850K.Is his claim possible. 6.A Carnot engine working between 400 0 C and 40 0 C produces 130kJ of work. Determine i)The engine thermal efficiency ii)The heat added iii)The entropy changes during the heat rejection.

11. Department of Mechanical Engineering,A.I.E.T.,Mijar 11 7.A Carnot engine works with isentropic compression ratio of 5 and isothermal expansion ratio of 2.The volume of air at the beginning of the isothermal expansion is 0.3m 3.If the maximum temperature and pressure is limited to 550K and 21 bar, determine : i)Minimum temperature of the cycle ii)Thermal efficiency of the cycle iii)pressure at all the salient points, iv)change of entropy during the isothermal expansion, v)work done per cycle,and vi)Mean effective pressure. Take ratio of specific heats as 1.4.

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15. Department of Mechanical Engineering,A.I.E.T.,Mijar 15 2)Otto Cycle/Constant Volume Cycle/Petrol Engine Cycle Process 1-2: Reversible adiabatic compression of air. Process 2-3: Heat addition at constant volume. Process 3-4: Reversible adiabatic expansion of air. Process 4-1: Heat rejection at constant volume.

16. Department of Mechanical Engineering,A.I.E.T.,Mijar 16 3)Air Standard Diesel Cycle/ Constant Pressure cycle: Process 1-2: Reversible adiabatic Compression. Process 2-3: Constant pressure heat addition. Process 3-5: Reversible adiabatic Compression. Process 4-1: Constant volume heat rejection.

17. Department of Mechanical Engineering,A.I.E.T.,Mijar 17  Dual Combustion Cycle/Limited Pressure Cycle/Mixed Cycle/Semi- Diesel cycle Process 1-2: Reversible adiabatic compression. Process 2-3: Constant volume heat addition. Process 3-4: Constant pressure heat addition. Process 4-5: Reversible adiabatic expansion. Process 5-1: Constant volume heat rejection.

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19. Department of Mechanical Engineering,A.I.E.T.,Mijar 19  COMPARISON OF OTTO, DIESEL AND DUAL COMBUSTION CYCLES  Following are the important variable factors which are used as a basis for comparison of the cycles : Compression ratio. Maximum pressure Heat supplied Heat rejected Net work 1)Efficiency Versus Compression Ratio For a given compression ratio Otto cycle is the most efficient while the Diesel cycle is the least efficient. (η otto > η dual > η diesel ).

20. Department of Mechanical Engineering,A.I.E.T.,Mijar 20 2)For the Same Compression Ratio and the Same Heat Input Since all the cycles reject their heat at the same specific volume, process line from state 4 to 1, the quantity of heat rejected from each cycle is represented by the appropriate area under the line 4 to 1 on the T-s diagram. η = 1 – (Heat rejected /Heat supplied) (η otto > η dual > η diesel )

21. Department of Mechanical Engineering,A.I.E.T.,Mijar 21 3)For Constant Maximum Pressure and Heat Supplied the maximum pressure the points 3 and 3′ must lie on a constant pressure line. — On T-s diagram the heat rejected from the Diesel cycle is represented by the area under the line 4 to 1 and this area is less than the Otto cycle area under the curve 4′ to 1 ; Hence the Diesel cycle is more efficient than the Otto cycle for the condition of maximum pressure and heat supplied.

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