Question

3) As shown in the figure below, water (o 1.94 slugs/ ft') is pumped from a large tank (1) at a rate of 3 ft'/s through a main supply line to the top of an aeration column (2), which is a large free surface open to the atmosphere. The head loss through the supply line is 4 ft. In the acration column, the water is entrained with air as it moves downward via gravity through the column and exits to the atmosphere (3). [30 pts] a) Calculate the power of the pump (Note 1 horsepower-550 ft-lb/s). [15 pts) Aerator column 10 t 3 ft Pump b) Calculate the head loss through the aeration column if the average velocity at the exit (3) is 6 t/s [15 pts) r 2)

Answer 1

Ans) Given ,

Density of water = 1.94 slugs/ft³ or 62.4 lb/ft³

Discharge rate = 3ft³/sec

Head loss = 4ft

Apply Bernoulli equation between point 1 and 2,

P1/ $\rho$ g + V1² /2g + H_p + Z1 = P2/ $\rho$ g + V2²/2g + Z2 + H_L

where, Hp = head added by pump

Z1 = height of point 1 from datum

Z2 = height of point 2 from datum

H_L = Head loss

Since, point 1 and 2 are open to atmosphere, the pressure is only atmospheric pressure, hence gauge pressure P1 = P2 = 0

Also, Velocity at point 1 and 2 = 0 .

Let center of pump be datum .Hence, above equation can be written as ,

Hp + 5 = 13 + 4

Hp = 12ft

We know, power of pump (P),

P = $\rho$ g Q Hp

P = 62.4 x 32.2 x 3 x 12

P = 72334.08 lb ft²/s³

P = 2264.5 lb-ft/s

P = 2264.5/550 HP

P = 4.08 HP

Ans b) To determine head loss in aeration column apply Bernoulli equation between point 2 and 3

     P2/ $\rho$ g + V2² /2g + Z2 = P3/ $\rho$ g + V3²/2g + Z3 + H_L

Since, point 2 and 3 are open to atmosphere, the pressure is only atmospheric pressure, hence gauge pressure P2 = P3 = 0

Also, velocity at point is zero, therefore,

Z2 = V₃²/2g + Z3 + H_{L ,} where, V3 is velocity at exit

13 = 6²/ (2x32.2) + 3 + H_L

H_L = 13 - 3.56

H_L = 9.44 ft