# 10th-mfo77 - 10 th International Physics Olympiad 1977...

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Unformatted text preview: 10 th International Physics Olympiad 1977, Hradec Kr alov e, Czechoslovakia Problem 1. The compression ratio of a four-stroke internal combustion engine is = 9 . 5. The engine draws in air and gaseous fuel at a temperature 27 o C at a pressure 1 atm = 100 kPa. Compression follows an adiabatic process from point 1 to point 2, see Fig. 1. The pressure in the cylinder is doubled during the mixture ignition (23). The hot exhaust gas expands adiabatically to the volume V 2 pushing the piston downwards (34). Then the exhaust valve opens and the pressure gets back to the initial value of 1 atm. All processes in the cylinder are supposed to be ideal. The Poisson constant (i.e. the ratio of specific heats C p /C V ) for the mixture and exhaust gas is = 1 . 40. (The compression ratio is the ratio of the volume of the cylinder when the piston is at the bottom to the volume when the piston is at the top.) p p = p p p p V 1 1 2 2 3 4 1 2 3 4 V V Figure 1: 1 a) Which processes run between the points 01, 23, 41, 10? b) Determine the pressure and the temperature in the states 1, 2, 3 and 4. c) Find the thermal efficiency of the cycle. d) Discuss obtained results. Are they realistic? Solution: a) The description of the processes between particular points is the following: 01 : intake stroke isobaric and isothermal process 12 : compression of the mixture adiabatic process 23 : mixture ignition isochoric process 34 : expansion of the exhaust gas adiabatic process 41 : exhaust isochoric process 10 : exhaust isobaric process Let us denote the initial volume of the cylinder before induction at the point 0 by V 1 , after induction at the point 1 by V 2 and the temperatures at the particular points by T , T 1 , T 2 , T 3 and T 4 . b) The equations for particular processes are as follows. 01 : The fuel-air mixture is drawn into the cylinder at the temperature of T = T 1 = 300 K and a pressure of p = p 1 = 0 . 10 MPa. 12 : Since the compression is very fast, one can suppose the process to be adiabatic. Hence: p 1 V 2 = p 2 V 1 and p 1 V 2 T 1 = p 2 V 1 T 2 ....
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10th-mfo77 - 10 th International Physics Olympiad 1977...

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