Sewage flow = 150 l/person-day = 9000 m³/day
BOD₅ = 54 g/person-day = 360 mg/l ; BOD_u = 1.47 BOD₅
Total kjeldahl nitrogen (TKN) = 8 g/person-day = 53 mg/l
Phosphorus = 2 g/person-day = 13.3 mg/l
Winter temperature in aeration tank = 18°C
Yield coefficient Y = 0.6 ; Decay constant K_d = 0.07 per day ; Specific substrate utilization rate = (0.038 mg/l)^-1 (h)^-1 at 18°C
Assume 30% raw BOD₅ is removed in primary sedimentation, and BOD₅ going to aeration is, therefore, 252 mg/l (0.7 x 360 mg/l).

Design:
(a) Selection of q_c, t and MLSS concentration:

Considering the operating temperature and the desire to have nitrification and good sludge settling characteristics, adopt q_c = 5d. As there is no special fear of toxic inflows, the HRT, t may be kept between 3-4 h, and MLSS = 4000 mg/l.

(b) Effluent BOD₅:

Substrate concentration, S =  1 (1/q_c + k_d)=         1     (1/5 + 0.07)
                                         qY                (0.038)(0.6)
S = 12 mg/l.
Assume suspended solids (SS) in effluent = 20 mg/l and VSS/SS =0.8.
If degradable fraction of volatile suspended solids (VSS) =0.7 (check later), BOD₅ of VSS in effluent = 0.7(0.8x20) = 11mg/l.

Thus, total effluent BOD₅ = 12 + 11 = 23 mg/l (acceptable).

(c) Aeration Tank:

VX = YQq_c(S_O - S) where X = 0.8(4000) = 3200 mg/l
         1+ k_dq_c
or 3200 V = (0.6)(5)(9000)(252-12)
                       [1 + (0.07)(5)]
      V = 1500 m³

Detention time, t = 1500 x 24 = 4h
                               9000

F/M = (252-12)(9000) = 0.45 kg BOD₅ per kg MLSS per day
          (3200) (1500)

Let the aeration tank be in the form of four square shaped compartments operated in two parallel rows, each with two cells measuring 11m x 11m x 3.1m

(d) Return Sludge Pumping:

If suspended solids concentration of return flow is 1% = 10,000 mg/l

R =        MLSS     = 0.67
      (10000)-MLSS
Q_r = 0.67 x 9000 = 6000 m³/d

(e) Surplus Sludge Production:

Net VSS produced Q_wX_r = VX =  (3200)(1500)(10³/10⁶) = 960 kg/d
                                     q_c                   (5)

or SS produced =960/0.8 = 1200 kg/d

If SS are removed as underflow with solids concentration 1% and assuming specific gravity of sludge as 1.0,

Liquid sludge to be removed = 1200 x 100/1 = 120,000 kg/d
                                                              = 120 m³/d

(f) Oxygen Requirement:

1. For carbonaceous demand,
   oxygen required = (BOD_u removed) - (BOD_u of solids leaving)
                          = 1.47 (2160 kg/d) - 1.42 (960 kg/d)
                          = 72.5 kg/h
2. For nitrification,
   oxygen required = 4.33 (TKN oxidized, kg/d)
   Incoming TKN at 8.0 g/ person-day = 480 kg/day. Assume 30% is removed in primary sedimentation and the balance 336 kg/day is oxidized to nitrates. Thus, oxygen required
                          = 4.33 x 336 = 1455 kg/day = 60.6 kg/h
3. Total oxygen required
           = 72.5 + 60.6 = 133 kg/h = 1.0 kg/kg of BOD_u removed.

Oxygen uptake rate per unit tank volume = 133/1500
                      = 90.6 mg/h/l tank volume

(g) Power Requirement:

Assume oxygenation capacity of aerators at field conditions is only 70% of the capacity at standard conditions and mechanical aerators are capable of giving 2 kg oyxgen per kWh at standard conditions.

Power required =   136   = 97 kW (130 hp)
                        0.7 x 2

                      = (97 x 24 x 365) / 60,000 = 14.2 kWh/year/person