Primary effluent wastewater at 60^oF (v = 1.217 x 10^-5) is pumped to the top of a trickling filter (TF) a total of 75 vertical feet through schedule-40 welded steel pipe. The flow rate is 8.64 mgd (6,000 gpm). The suction piping is 24 in diameter straight piping (velocity of 4.8 fps), has a length of 10 ft and an energy loss of 0.03 ft. The discharge piping is 18 inch diameter (velocity of 8.6 fps), has a length of 177 ft and an energy loss of 3.6 ft. The total length of the suction and discharge piping is 187 ft. Minor losses exist in the discharge piping only and total an equivalent length of 137 ft. The wet well and TF distribution arms are open to the atmosphere. The pump efficiency is 85%. 1.  Determine the head added by the pump h_A:

1. Determine the head added by the pump h_A:

h_A = (P_D/γ - P_S/ γ) + ( v2 _D/2g - v2 _S/2g) + Z_D - Z_S + hf Z_D - Z_S = 75 ft

v2 _dis/2g = (8.6) 2 /64.4 = 1.15 ft v2 _suc/2g = (4.8) 2 /64.4 = 0.35 ft

total h_f = 0.03 ft + 3.6 ft = 3.62 ft

h_A = 0 + (1.15 ft - 0.35 ft) + 75 ft + 3.62 ft = 79.4 ft

2.  Determine the hydraulic horsepower required:   

whp = h_AQ(sp gr)/3956=((79.4 ft)(6000 gpm)(1))/3956 = 120 HP

3.  Determine the brake horsepower required:

bhp = whp/pump efficiency = 120 HP / 0.85 = 142 HP

4.  The friction or heat horsepower is

bhp-whp = 142 HP - 120 Hp = 22 HP