Problem 5. Facultative lagoon: Consider a facultative lagoon located inMonth kWh
ID: 1732633 • Letter: P
Question
Problem 5. Facultative lagoon: Consider a facultative lagoon located inMonth kWh/m2/day Pittsburgh, PA that has an average flow rate of 18,000 m3/day, CBOID influent of 320 mg/L, and algae conversion efficiency of 4% with 23,000 kJ of sunlight per kg of algae biomass. Dec Nov Oct Sep Aug Jul Jun May Apr Mar Feb Jan 1.88 2.63 4.14 4.64 5.41 5.26 5.4 5.17 4.91 4.24 3.44 2.64 a) What solar radiation quantity should be used for the design and what is the actual insolation? (Use table to the right) b) What is the theoretical maximum surface loading rate? c) What is the required lagoon area?Explanation / Answer
Given;
Avg flow Q=18000m^3/d=18000*1000 litre/d
CBOD= 320mg/l=320/(10^6) kg/litre
Efficiency =4% =0.04
Sunlight per kg biomass=23000KJ / kg =23000*0.0002778=6.3894KWh / kg
Now, Algae produced per day is computed as;
=18000*1000l/d * 320/(10^6) kg/l * 0.04 = 230.4 kg/d
? average solar quantity (per day ) requires is = 230.4 * 6.3894 = 1472.117 KWh/d = 1472.117/24 = 61.3382 KW.
Now solar radiation is expressed in watt/m^2.
So we have to find out area required in different months which is computed below.;
KWh/m^2/d / (KWh/d) = m^2
Like in first 1.88/1472.117 = 783.04 m^2.
Area requires of the lagoon is maximum of above i.e =783.04 m^2 (Ans).
Now therotical maximum surface loading rate is compute as = flow/ min area = 18000/272.11 = 66.149 m/d (Ans).
Solar radiation quantity (KWh/m^2)should be used in design is (power required/area provided) = 61.33/783.04 = 0.078 KW/m^2 = 78.32 watt/m^2 (Ans)
Actual insolation for every month is provided in table above.
Month KWh/m^2/d Area (m^2) Actual Insolation (Watt/m^2) Dec 1.88 783.04 78.32 Nov 2.63 559.74 109.58 Oct 4.14 355.58 172.5 Sept 4.64 317.26 193.33 Aug 5.41 272.11 225.41 Jul 5.26 279.87 219.16 Jun 5.4 272.61 225 Ma 5.17 284.74 215.41 Apr 4.91 299.82 204.58 Mar 4.24 347.19 176.66 Feb 3.44 427.94 143.33 Jan 2.64 557.62 109.99Related Questions
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