The ultimate analysis of a mixture of sewage shudge and solid waste subjected to
ID: 707259 • Letter: T
Question
The ultimate analysis of a mixture of sewage shudge and solid waste subjected to composting shows the following resultsx 70, y 98, z 28 and a 1. The plant composts 100 tonnes day of the mixture. Assuming that 20% excess air is used, calculate the cubic meters per minute o pressure (STP) of 0°C and 1 atm pressure for complete biological combustion. f air a blower must force into the pile measured at a standard temperature and X a CO y+4x-3a-2z-2b y+4x-3a-2z-2by y-5a-2b 20+4x-3a-2z-2b) Molar masses : C 12, H 1; ? 16, N 14 Assum e air is 21% O2 and 79% N2 1 mole of a gas at STP - 22.4 liters of a gasExplanation / Answer
In the reaction given above, lets balance the nitrogen as it appears in only one product. On reactant side we have
= (1 / y + 4x -3a - 2z - 2b) Na
= (1/98 + 280 - 3*1 - 2*28-2b) N70
= ( 1/319 - 2b)N70
Total N in reactants = 70* ( 1/319 - 2b) = 70 / (319 - 2b)
Similarly on the product side,
= (1 / y + 4x -3a - 2z - 2yb) N
= (1/98 + 280 - 3*1 - 2*28-2*98*b) N
= ( 1/319 - 196b)N
Total N in products = 1* ( 1/319 - 196b) = 1 / (319 - 196b)
N in reactants = N in products
70 / (319 - 2b) = 1 / (319 - 196b)
b = 1.6
The reactants coefficient , (1 / y + 4x -3a - 2z - 2b) = 1/319 - 2*1.6 = 0.00316
For 0.00316 kmol of sewage, 0.25 kmol of O2 is required.
Molar mass of the sewage mix = 12*70 + 98*1 + 16*56 + 14*1 + 32*1.6 = 1899.2 Kg/kmol
Given 100 tonnes of sewage mix is composted per day .
Moles of sewage mix composted per day = Mass / Molar mass = 100*1000/1899.2 = 52.65 Kmol/day
From stoichiometry, For 0.00316 kmol of sewage, 0.25 kmol of O2 is required. Then for 52.65 Kmol/day.
0.25*52.65/0.00316 = 4165.35 Kmol/day of O2 is required.
Air comprising 4165.35 kmol of O2 = Molar rate of O2/ Mole fraction of O2 in air = 4165.35/0.21 = 19835 kmol/day
20%excess air is supplied , thus Total air to be supplied = Theoretical required air * 1.2 = 19835*1.2 = 23802 Kmol/day
Converting the units mol/min :
Ait to be supplied = 23802 kmol/day * 1000 mol/kmol * day/1440 min = 165291.67 mol/min
At STP 1 mol = 22.4 L
then 165291.67 mol/min = 165291.67 mol/min * 22.4 L/mol = 3702533.408 L/min
1000 L = 1 m3
3702533.408 L/min = 3702.53 m3/min
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