TITLE.PM5

536 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th11-2.pm5

Given that the enthalpies of combustion at 25°C are :

C 6 H 6 = – 3169500 kJ/mole

C 8 H 18 = – 5116200 kJ/mole

Both the above figures are for the case where the water in the products is in the vapour

phase.

Solution. (i) Air and benzene vapour :

For benzene, the combustion equation is as follows :

C 6 H 6 (g) + 7.5O 2 (g) + 7.5

79

21

F

HG

I

KJ N^2 (g) = 6CO^2 (g) + 3H^2 O(g) + 7.5

79

21

F

HG

I
KJ N^2 (g)
Since the water in the products is in vapour phase, therefore, the given value of enthalpy of
combustion corresponds to the lower heating value at constant pressure.
i.e., (LHV)p = 3169500 kJ/mole

(LHV)v per kg of mixture =

3169500

12 6 6 1 7 5 32 7 5^79

21

()(.).()×+× + × + FHG IKJ 28

=

3169500

78 240 790++ = 2861 kJ/kg. (Ans.)

Now, (HHV)p = (LHV)p + mhfg

where (HHV)p = Higher heating value at constant pressure,

(LHV)p = Lower heating value at constant pressure,

m = Mass of water formed by combustion.

= 3 × 18 = 54 kg/kg mole of fuel, and

hfg = Latent heat of vapourisation at given temperature per unit mass of water

= 2442 kJ/kg at 25°C.

∴ (HHV)p = 3169500 + 54 × 2442 = 3301368 kJ/mole

Thus, (HHV)p per kg of mixture =

3301368

78 240 790++ = 2980 kJ/kg. (Ans.)

(ii)Air and octane vapour :

(LHV)p = 5116200 kJ/mole of C 8 H 18

For octane, the combustion equation is written as follows :

C 8 H 18 (g) + 12.5O 2 (g) → 8CO 2 (g) + 9H 2 O(g) + 12.5

79

21

F

HG

I

KJ N^2 (g)

(LHV)p per kg of mixture =^5116200

12 8 18 1 12 5 32 12 5^79

21

()..×+ × + × + × × 28

=

5116200

114 400 1317++ = 2794 kJ/kg. (Ans.)

(HHV)p = (LHV)p + mhfg

m = 9 × 18 = 162 kJ/kg mole of fuel

∴ (HHV)p = 5116200 + 162 × 2442 = 5511804

Hence, (HHV)p per kg of mixture =

5511804

114 400 1317++ = 3010 kJ/kg. (Ans.)