where The thermal efficiency of the Otto cycle assuming c, is constant (cold air
ID: 1766717 • Letter: W
Question
where The thermal efficiency of the Otto cycle assuming c, is constant (cold air standard) becomes: Processes 1-2 and 3-4 are isentropic, and V2 = V3 and v-V1. Thus T (vi T, Substituting these equations into the thermal efficiency relation and simplifying gives where r =-=-arr =- and k = V, Y2 EXPERIMENTAL PROCEDURE Fill the fuel line of the petrol generator. Start the engine and measure the drop in fuel consumption for each electric load condition. The electric load can be varied by turning the light bulbs on or off. With each load, measure the air intake velocity using the anemometer, the inlet air temperature (ambient lab temperature) and the outlet air temperature. Also measure the air inlet diameter of the engine. Useful information: Petrol lower heating value (LHV) 43.7 M/kg Specific heat air 1.0 kJ/kg K Density of petrol-737.22 kg/m . Density of atmospheric air 1.2 kg/m3 k = G/G+ 1.2 to 1.4 depending on exhaust gas composition 0.73722 g/ml . ·Explanation / Answer
This can best be comprehend by the 2nd law of thermodynamics which states that "There can be no machine which can produce work while exchanging heat with a single reservoir such a machine is called PMM-2 (Perpatual motion machine of 2nd kind) and hence PMM-2 is impossible. Efficiency of PMM-2 is 100% and hence from 2nd law , 100% efficiency of the system is impossible."
The petrol engine comprises of four basic processes as it works on the principal of otto cycle. The processes are as follows :-
Process 1-2 :- Adeabatic compression
Process 2-3 :- Constant Volume heat addition
Process 3-4 :- Adeabatic Expansion
Process 4-1 :- Constant volume heat rejection.
As we know no process in the universe (system+surrounding) is 100% reversible so there is loss of energy due to the irreversibilities. These irreversibilities mostly in the moving system comes in the form of heat and changes the entropy of the system because of which we need to use cooling systems to reject heat generated to the surrounding.
Loss of heat will be equivalent to = Heat supplied- Work Done = Qs- W.
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