Question

You have been asked to provide a design for a rapid mix tank at a water treatme nt plant which has an average water temperature of 50F. The tank is to have a detention time (td= HRT) of 1.5 seconds, a mixing intensity (G) of 2,000/second, and treat a flow (Q) of 10 mgd.

a.

Identify the volume of reactor required (in ft^3 and gallons)

b.

Find the reactor dimensions (i.e., Depth and Diameter, in feet) if the rapid mix tank is cylindrical with a depth 2x the basin diameter.

c.

Calculate the water power required (in HP) to achieve the desired mixing.

d.

If motors are available with an efficiency of 68%, what motor power is required (in HP)?

Answer 1

Given-: td = 1.5 sec G = 2,000/second   Q =10 mgd. = 15.472 ft3/sec =  115.740 gallon / second

Note-: 1 mgd = 1.5472 ft3/sec

1 mgd = 133680.556 cubic foot /day

1 mgd = 3.78 x 103 m3/d

a) Reactor Volume

volume of reactor required = Q x td

= 15.472 x 1.5

   V (in ft3) = 23.208 ft3

Volume (in gallons) = 115.740 x 1.5

Volume (in gallons)   = 173.61 gallons

b) Reactor dimensions

Volume = (π/4) x d2 x depth

23.208 = (π/4) x d2 x 2d ( Given-: depth = 2x the basin diameter )

Diameter of tank (d) = 2.453 feet

Depth of tank = 2 x 2.453

Depth = 4.907 m

c)  power Required

P = V G2

- Dynamic viscosity of water

V - Volume of mixing tank

G- mixing intensity or velosity gradient

At T = 50 ºF = 10 ºC ,    = 1.3076 x 10-3 N-s/m2 = 2.731 x 10-5 lbf-s/ft2

P = (2.731 x 10-5) x 23.208 x 2000

P = 1.2676 ft⋅lbf/s

P = 0.002304 HP (1ft⋅lbf/s = 1.818×10−3 horsepower)

d) Motor Power Required

efficiency = 68%

Preq. = 0.002304 /0.68

Preq. = 0.00338 HP