TITLE.PM5

516 ENGINEERING THERMODYNAMICS

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\M-therm\Th11-2.pm5

Air applied on the basis of conditions at exit

= 80 6 80

33 7 9 0

.

(. )

×

+

= 24.73 kg

∴ Air leakage = 24.73 – 23.45 = 1.28 kg of air per kg of fuel. (Ans.)
For each kg of fuel burnt, the ash collected is 15% i.e., 0.15 kg.
∴ Weight of fuel passing up the chimney = 1 – 0.15 = 0.85 kg
∴ Total weight of products
= Weight of air supplied per kg of fuel

• Weight of fuel passing through chimney per kg of fuel
  = 23.45 + 0.85 = 24.3 kg
  Heat in flue gases per kg of coal
  = Weight of flue gases × Specific heat × Temperature rise above 0°C
  = 24.3 × 1.05 × (410 – 0) = 10461 kJ
  Heat in leakage air = Weight of leakage air × Specific heat
  × Temperature rise of air above 0°C
  = 1.28 × 1.005 × (20 – 0) = 25.73 kJ.
  We can still consider, in the mixture, the gas and the air as separate and having their own
  specific heats, but sharing a common temperature t.
  For heat balance :
  (1.005 × 1.28 + 24.3 × 1.05) t = 10461 + 25.73
  26.8 t = 10486.73
  ∴ t = 391.3°C
  ∴ Fall in temperature as a result of the air leakage into the economiser
  = 410 – 391.3 = 18.7°C. (Ans.)

AIR-FUEL RATIO AND ANALYSIS OF PRODUCTS OF COMBUSTION
Example 11.12. The chemical formula for alcohol is C 2 H 6 O. Calculate the stoichiometric
air/fuel ratio by mass and the percentage composition of the products of combustion per kg of
C 2 H 6 O. (GATE 1998)
Solution. Chemical equation for complete combustion of given fuel can be written as follows :
C 2 H 6 O + 3O 2 → 2CO 2 + 3H 2 O
(1 × 46) + (3 × 32) = (2 × 44) + (3 × 18)
For complete combustion of 1 kg of C 2 H 6 O, oxygen required

=^332

46

× = 2.087 kg of oxygen

= 2.087 ×

100

23

= 9.074 kg of air

A : F ratio for complete combustion = 9.074 : 1. Ans.

Also 46 kg of fuel produces products of combustion = 88 + 54 = 142 kg

L∴=

NM

O

QP

1kgof fuelproduces^142 3.087 kgofproducts of combustion (. ., CO and H O) 22

46

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