FUELS AND COMBUSTION 537

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\M-therm\Th11-2.pm5

+Example 11.37. The higher heating value of kerosene at constant volume whose ulti-
mate analysis is 88% and 12% hydrogen, was found to be 45670 kJ/kg. Calculate the other three
heating values.
Solution. Combustion of 1 kg of fuel produces the following products :

CO 2 =

44

12

× 0.88 = 3.23 kg

H 2 O =

18

2

× 0.12 = 1.08 kg

At 25°C : (ug – uf) i.e., ufg = 2304 kJ/kg

hfg = 2442 kJ/kg

(i)(LHV)v :

(LHV)v = (HHV)v – m(ug – uf)

= 45670 – 1.08 × 2304 = 43182 kJ/kg

Hence (LHV)v = 43182 kJ/kg. (Ans.)

(ii)(HHV)p, (LHV)p :

The combustion equation is written as follows :

1 mole fuel + x

32

O 2 → 323

44

. CO
2 +

108

18

. H
2 O

i.e.,
x
32

323

44

108

18 2

..+

×

or x = 3.31 kg

i.e., 1 kg fuel + 3.31 kg O 2 = 3.23CO 2 + 1.08H 2 O

Also, ∆H = ∆U + ∆nR 0 T

i.e., – (HHV)p = – (HHV)v + ∆nR 0 T

or (HHV)p = (HHV)v – ∆nR 0 T

where ∆n = np – nR

=^323

44

331

32

F ..−

HG

I

KJ

n

n

p

R

=

=

L

N

M

M

O

Q

P

P

number of moles of gaseous products

number of moles of gaseous reactants

Since in case of higher heating value, H 2 O will appear in liquid phase

(HHV)p = 45670 –^323

44

331

32

F ..−

HG

I

KJ

× 8.3143 × (25 + 273)

= 45744 kJ/kg. (Ans.)

(LHV)p = (HHV)p – 1.08 × 2442 = 45774 – 1.08 × 2442

= 43107 kJ/kg. (Ans.)

Highlights

1. A chemical reaction may be defined as the rearrangement of atoms due to redistribution of electrons.
   ‘Reactants’ comprise of initial constituents which start the reaction while ‘products’ comprise of final
   constituents which are formed by the chemical reaction.


2. A chemical fuel is a substance which releases heat energy on combustion.


3. The total number of atoms of each element concerned in the combustion remains constant, but the atoms
   are rearranged into groups having different chemical properties.