1. Top End

2. COMPONENTS

3. HEAD

4. CYLINDER

5. THE FOUR STROKE ENGINE (THE OTTO CYCLE) http://www.engr.colostate.edu/~dga/mech324/handouts/

6. ANIMATION

7. RESTORATION PROCESS

8. BORING THE CYLINDER

9. USING THE LATHE

10. A PERFECT FIT

11. SANDBLASTING

12. AND SOME MORE…

13. AFTER THE SAND BLASTING

14. MASKING TAPE AND SILVER SPRAY PAINT

15. LOOKING GOOD

16. VALVE GUIDES

17. CUTTING THE SEATS

18. ANGLED BLADES

19. BLUING THE VALVES

20. REASSEMBLY OF THE HEAD

21. EXPLODED VIEW - VALVES

22. RE-INSERTING THE SPRINGS

23. SUCCESS!

24. ROCKER HEAD - ALMOST

25. FINISHED

27. TAKING MEASUREMENTS FOR THE SPRING CONSTANT CALCULATIONS

28. WHICH SPRING SET TO USE: CALCULATIONS OF SPRING CONSTANT Force applied (lbs)Force (N) Spring Length (in) Δ X (in) Δ X (m) Spring Constant (k) 001.7100- 1148.9281.6250.0850.00215922662.34368 25111.21.4960.2140.005435620457.72316 42186.8161.3730.3370.008559821824.80899 59262.4321.2440.4660.011836422171.60623 Average k=21779.12051 k~22000 N/m OLD SPRINGS (SINGLE)

29. Force applied (lbs)Force (N) Spring Length (in) Δ X (in) Δ X (m) Spring Constant (k) 001.7100 - 1253.3761.6250.0850.00215924722.55674 63280.2241.4960.2140.005435651553.46236 87386.9761.3730.3370.008559845208.53291 Average k=40494.85067 k~40000 N/m WHICH SPRING SET TO USE: CALCULATIONS CONTINUED NEW SPRINGS (DOUBLE)

30. Assuming: the intake valve allows 200cc air into the engine during each first downward stroke the air is at average atmospheric concentrations (C O2 =21%) the gas is at standard temperature (25°C, 298K) and pressure (101.3 kPa) the mixture of Gasoline with Oxygen is perfectly stoichiometric the gasoline mixture is 87% Octane (by mass) (the remainder can be disregarded as alcohol additives and impurities which do not combust as readily as the gasoline) This is of course heavily simplified. Full calculations would have to take into consideration the rate of flow of air into the cylinder (as determined by the valve timings) and therefore the actual volume of air taken in per stroke, the temperature of the gases present and the possible incomplete combustion of the fuel among other factors. ENERGY PER EXPLOSION

31. The moles of oxygen present for each explosion: All gas: PV=nRT n= PV/RT n= (101.325*200*10 -6 )/(8.31*298) n= 8.18*10 -3 moles n O2 =n*0.21 =1.72x10 -3 moles C 8 H 18 + 12.5O 2  8CO 2 + 9H 2 O Therefore according to the reaction equation the moles of octane taken in per intake ought to be: N c8h18 = n O2 /12.5 N c8h18 = (1.72x10 -3 )/12.5 N c8h18 =1.38* 10 -4 moles Density( ρ )= 730 kgm -3 = 730 000g/m 3 Concentration= (% purity* ρ )/(Molar mass) = (0.87*730000)/((8*12)+(18*1)) = 5.57 x10 3 mol/m 3 =5.57*10 -3 mol/cc Volume gasoline necessary per stroke = N/C = (1.38* 10 -4 )/ ( 5.57x10 -3 ) = 2.48*10 -5 << volume air Thus may be considered negligible The Δ H c of C 8 H 18 = −5.53–−5.33 MJ mol −1 (http://en.wikipedia.org/wiki/Octane)http://en.wikipedia.org/wiki/Octane Thus the energy released in each explosion to these specifications = Δ H c *N ~ (5.40x10 6 )( 1.38* 10 -4 ) ~745.2 J per explosion CALCULATIONS CONTINUED…

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