GATE Civil Engineering (CE) 2020 Shift 2 Solved Paper

Question : 44 of 65

Marks: +1, -0

A waste to energy plant burns dry solid waste of composition: Carbon

=35 \%

, Oxygen

=

26 \%

, Hydrogen

=10 \%

, Sulphur

=6 \%

, Nitrogen

=3 \%

and Inerts

=20 \%

. Burning rate is 1000 tonnes/d. Oxygen in air by weight is

23 \%

. Assume complete conversion of Carbon to

{CO}_{2}

, Hydrogen to

{H}_{2} {O}

, Sulphur to

{SO}_{2}

and Nitrogen to

{NO}_{2} .

Given Atomic weights:

{H}=1, {C}=12, {N}=14, {O}=16, {S}=32

The stoichiometric (theoretical) amount of air (in tones/d, round off to the nearest integer) required for complete burning of this waste, is

Solution:

\;{ {C}}↙{(12)}+ {O}_{(2)}↙{(32)} ⟶ {CO}_{2}↙{(44)}

Oxygen required for

1 {t} /

day

=\;{32}{/}{12}=2.6667

moles
Amount of air required for combustion of carbon

=2.6667 × 350=933.33

tonnes/day Similarly,
for hydrogen

\;{32}/{4} × 100=800

for Sulphur

\;{32}/{32} × 60=60

for nitrogen

{32}/{14} × 30=68.57

∴

total oxygen required

=933.34+800+60+68.57 ≈ 1862 {t} {/} {day}

Available amount of oxygen

=260 {t} {/} {day}

∴

required amount of oxygen

=1862-260=1602 {t}

/day

∴

Amount of air required

=1602{/}.23=6965.2

tonnes/day

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