7-12

11-75

9-07


8-45


4-85


3-01


13-20
15-30
U-75


100


67-20

73-0

74-3


99-98

100-05

99-98



It is thus necessary to be able to calculate backwards from the
exhaust analysis results to obtain the mixture proportions. A
concrete example will make the method clear.
The composition of the exhaust gases in an actual case was : —



CO 2
CO

0.
H,
CH,



= 12-5 per cent.
= 2-6

= 0-2

= 0-9

= 0-3

= 83-5



100-0 per cent.
volimies C

. c
, c



Now 12-5 volumes of CO 2 contain 6-25
2-6 „ „ CO „ 1-3
0-3 „ „ CH4 „ 0-15
Hence total volumes of C =7-7.
Again, 12-5 volumes of CO 2 contain 12-5 volumes of O2
2-6 „ „ CO „ 2-5

0-2 „ „ 0, „ 0-2



0.
Total 15-2 volumes of O2

* Some interesting curves of exhaust gas analysis results for alcohol and
benzol mixtures are given in a Paper entitled " Benzol, etc., as Fuels for
Internal Combustion Engines " {W. Watson), Proc. Inst. Auto Engra. 1914-15.
EXPLOSION AND COMBUSTION IN THE ENGINE 73

Since the composition of air by volume is 79-1 per cent. N and

20-9
20-9 per cent. 0, it follows that 1 volume of N is mixed with —--

or -266 volumes of in atmospheric air.
Hence the 83-5 volumes of N in the exhaust are mixed with
83-5 X -266 or 22-2 volumes of oxygen.
As we have found 15-2 volumes of a in the exhaust gases, we
can argue that the remainder, or 22-2 — 15-2, has gone to form water,
i.e., 7-0 volumes of Oj, which is equivalent to 14-0 volumes of H
in water.
