TITLE.PM5

492 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th11-1.pm5

∴ 2(2 × 1) + 2 × 16 → 2(2 × 1 + 16)
i.e., 4 kg H 2 + 32 kg O 2 → 36 kg H 2 O
or 1 kg H 2 + 8 kg O 2 → 9 kg H 2 O
The same are obtained by writing the equation (11.1) as :
H O H O, and this is sometimes done.

proportions

2 +→^1222

L

N

M

O
Q
P
It will be noted from equation (11.1) that the total volume of the reactants is
2 volumes H 2 + 1 volume O 2 = 3 volumes.
The total volume of the product is only 2 volumes. There is therefore a volumetric contrac-
tion on combustion.
Since the oxygen is accompanied by nitrogen if air is supplied for the combustion, then this
nitrogen should be included in the equation. As nitrogen is inert as far as chemical reaction is
concerned, it will appear on both sides of the equation.
With one mole of oxygen there are 79/21 moles of nitrogen, hence equation (11.1) becomes,
2H 2 + O 2 +^7921 N 2 → 2H 2 O +^7921 N 2 ...(11.2)

2. Combustion of carbon
   (i)Complete combustion of carbon to carbon dioxide
   C + O 2 → CO 2 ...(11.3)
   and including the nitrogen,

C + O 2 +^7921 N 2 → CO 2 +^7921 N 2 ...(11.4)

By volume :

0 volume C + 1 volume O 2 +^7921 volumes N 2 → 1 volume CO 2 +^7921 volumes N 2

...[11.4 (a)]

The volume of carbon is written as zero since the volume of solid is negligible in compari-

son with that of a gas.

By mass :

12 kg C + (2 × 16) kg O 2 +^7921 (2 × 14) kg N 2 → (12 + 2 × 16) kg CO 2 +^7921 (2 × 14) N 2

i.e., 12 kg C + 32 kg O 2 + 105.3 kg N 2 → 44 kg CO 2 + 105.3 kg N 2

or 1 kg C +

8

3

kg O 2 +

105 3

12

.

kg N 2 →

11

3

kg CO 2 +

105 3

12

.
kg N 2. ...[11.4 (b)]
(ii) The incomplete combustion of carbon. The incomplete combustion of carbon occurs
when there is an insufficient supply of oxygen to burn the carbon completely to carbon dioxide.
2C + O 2 → 2CO ...(11.5)
and including the nitrogen,

2C + O 2 +

79

21 N^2 → 2CO +

79

21 N^2 ...(11.6)

By mass :

(2 × 12) kg C + (2 × 16) kg O 2 +

79

21 (2 × 14) kg N^2 → 2(12 + 16) kg CO +

79

21 (2 × 14) kg N^2

i.e., 24 kg C + 32 kg O 2 + 105.3 kg N 2 → 56 kg CO + 105.3 kg N 2

or 1 kg C +

4

3 kg O^2 +

105 3

24

.

kg N 2 →

7

3 kg CO +

105 3

24

.

kg N 2 ...[11.6 (a)]