The second equation expressed the proportionality to surface area in such a way that it is 25 gallons/hour if the area is 1 acre. The water budget for this second pond yields Q1 Qe 2 Q2
Q2 Q1
A2 A1
Qe1 .
The mass balance for the organic material, again in steady state, with no loss of substance by evaporation and with no internal source, is:
0 Q1C 1 Q2 C 2 KV 2 C 2 . Replacing Q2 and V 2 by their respective expressions in terms of A2, we obtain:
A Q1 2 Qe1 KHA2 C 2 Q1C 1 . A1 Since the overall efficiency is to be 95%, the exiting concentration is the remaining 5% of the entering concentration (C 2 = 0.05 C 0 while C 1 = 0.38 C 0 still), we have
A Q1 2 Qe1 KHA2 (0.05) (0.38) Q1 . A1 Grouping the terms with the unknown A2, we have
Q 0.38 KH e1 A2 1 Q1 A 0 . 05 1 Then, plugging in the numerical values of the known quantities and minding the units, we obtain:
(3.34 ft 3 / hr ) 3 4 (4.30 10 / hr )(6 ft ) A2 (7.6 1)(63.50 ft / hr ) 2 ( 43,560 ft ) A2 167,397 ft 2 15,552 m 2 3.84 acres.