Qavg=1,260,000 gallons each day. This drops the water level in a tank from El 14
ID: 1844782 • Letter: Q
Question
Qavg=1,260,000 gallons each day. This drops the water level in a tank from El 140 to El 110. Water is pumped up from a constant El 35. The tank is refilled 11 PM to 5AM due to low power cost. In pumping Qavg= 3500 gpm, average static Dz =125 – 35 = 90’. However, at start of pumping, Dz =110 – 35 = 75’, so Qstart~ 4500 gpm. As the tank approaches being filled, the pump must push harder (Dz =140 -35 = 105’) so Qend ~ 2500 gpm. The two are normally set for alternating flow, but can operate in parallel if there is more drawdown in a day. In any case, water is pushed 1000’ in an 18” dia main, f = 0.018. S Kl =15
a For the average (half-filled) 3500 gpm, static Dz = 90’, find vavg and Shl from friction and local hl.
b. At that normal day average, a point on the service or demand curve, find TDH & impeller (Fig 9.25).
c. The Q over a typical fill interval varies with water level Dz, so the system or demand curve is concave down. This contrasts with the Q as per Fig 9.26, concave up. Anyway, calculate the TDH’s from 2500 to 4500 and plot on Fig 9.25. Does this follow that impeller’s curve over a good efficiency range?
d When a high demand day drains the tank more than 1,260,000 gallons, say to El 90, it’s still desired to refill in a 6 hour “window”. In parallel pumping, both units feed the same 1000’ of 18” pipe. Thus, the service curve for higher Q is concave up, as per Fig 9.26. Add two pump curves for parallel flow (Fig 9.26 shows pumps in parallel). At the higher Qavg-2 and its TDH, will two parallel pumps work?
Explanation / Answer
a. average velocity,(vavg) = total distance travel / total time taken
= 1000/3
= 333.3m/s
average velocity =333.3m/s
Distance = 1000' as given , time taken to fill is 11am to 5pm i.e 6hrs for full tank filling so for half tank it become 3 hrs
suction head, Hs= hd-H (H = system head, Hd = discharge head)
suction head = 180 -105
suction head Shl = 75'
b. to find TDH
TDH = static hieght+static lift+friction loss
friction loss = [C(Q/100)2 L] / 100 (C =friction loss cofficient,Q=flow rate,L= length)
= [15(3500/100)2x1000] /100 (cofficient of friction loss is given K)
= 15x1225x1000 /100
18375x1000/100
18375000/100 = 183750m
friction loss= 183750m
TDH = 35+90+183750
TDH = 183875psi
impeller = Q1 /Q2 = D1/ D2
4500/2500 = 1000/D
1.8 = 1000D
Impeller diameter =1.8x10-03
c. Total dynamic head (TDH)
=P2-P1 / pg => 4500-2500 / 105
= 200/105
total head=1.90'
TDH = total head+static lift +friction loss
= 1.90+90+183750
TDH = 183842psi
yes, this follow impeller curve over good efficiency of range
d. Qavg = 1226000gpm
average static El = 90'
time taken = 6hrs
diameter = 1000' and 18'
pump 1
diameter of 1000' and Dz = 75' and the velocity is of 4500gpm and El = 110
same as pump 1
=> pump 3 has a diametr of 18' and Dz = 105' and the velocity is of 2500gpm and El = 140
so the pump 2 can work parallel to both the pumps 1and 3