Depth of lie reactor fi 15

(iii) Surface area requried, A=V/H

(B) Determine the organic loading rate

=(QxSixEA)

=435 (m3/day)x7000 (mg/L)x10-3 (kg/mg.L/m3) x 0.94/870 (m3) =3.29 kg/m3 day <3.52 kg/m3 day, henece OK

(C) Determine the upflow velocity, v=H/6

(D) Determine the SRT

(i) Total COD removed per day=435 (m3/day) x 7000 (mg/L) x 10-3 (kg/mg.L/m3) x 0.94

=2862.3 kg/day

(ii) Biomass produced per day=0.027 (mg/mg) x 2862.3 (kg/day) =77.28 kg/day

(iii) Total biomass in the reactor

=Biomass in the sludge bed + biomass in the sludge blanket

=70 (kg/m3) x n/4 x (14.9)2 (m2) x 1.5 (m) + 4 (kg/m3) x n/4 x (14.9)2 (m2) x

(iv) SRT=Total biomass in the reactor/biomass produced per day

(F) Methane production

(i) Total quantity of methane generated=0.35 (m3/kg CODr) x2862.3 (kg CODr/day)

=1001.8 m3/day

(ii) Methane leaving as dissolved in the effluent=0.028 (m3/m3) x435 (m3/day)

(iii) Usable methane=1001.8-12.18 =989.62 m3/day

(G) Specific gas production

(i) Specific gas production per m3 per m3 of reactor per day

(ii) Specific gas production per m3 per m3 of effluent

(H) Energy equivalent of biogas

=989.62 m3/day x 10,000 (kcal/m3) =989.62 x 104 kcal/day =989.62 x 104 (kcal/day) x 1.1633 x 10-3 (kWh/kcal) =11,512.25 kWh/day

(I) Coal equivalent of biogas =989.62 x 104 (kcal/day)/4000 (kcal/kg) =2474.05 kg/day =2.47 tonnes/day

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