Question

Design a coarse screen and calculate the head loss through the rack, using the following information:

• Peak design wet weather flow 1.98 /s

• Velocity approaching rack at peak wet weather flow 0.80 m/s

• θ = 50°, with a mechanical cleaning device.

• The bars are 12mm thick and opening are 35 mm wide.

• Upstream depth of wastewater 1.52 m (Not depth of screen)

• Use rectangular bars with semicircular upstream face

Calculate:

a)      Screen dimensions: height and width.

b) The number of bars.

c)      Velocity through screen bars.

d)     Width of the channel.

e)      Head losses through the screen.

Answer 1

a) & b) & d) Screen dimensions: height and width & Number of bars.

Clear area through rack opening = peak flow / velocity through rack(vr)

A = 1.98 /0.80

= 2.475 m2

Clear width of rack opening = Clear Area / Depth of flow

w = 2.475 / 1.52

= 1.628 m

No. of opening = w / opening

n = 1.628 /0.035

n = 46.51 47

So. Total No. of bar to be provided = n -1 = 47 -1 = 46 bars

Now Total width of screen chamber= 1.628 + 0.012 x 47

W = 2.192 m

Height of the screen = water depth /sin θ

h   = 1.52 / sin 50°

h = 1.98 m

Providing 0..32 m of freeboard

h = 2.3 m

C)  Velocity through screen bars.

v = Flow / Net Area

Net Area = W x water depth(d)

= 2.192 x 1.52

= 3.33 m

va  = 1.98 / 3.33

va = 0.594 m/s

e)   Head losses through the screen

hL = (vr2 - va2 )/2g x (1/0.7)

= (0.82 - 0.5942)/2x9.81 x (1/0.7)

hL  = 0.021 m

OR

hL = (w /b)4/3 hv sin θ

Where,

= bar shape factor ( 1.67 for rectangular bar with both u/s and d/s face as semicircular).

W = maximum width of bar u/s of flow = 0.012 x 47 = 0.987 m

b = minimum clear spacing between bars = 1.628

hv = velocity head of flow approaching rack = m = v2/2g = 0.82 /2x9.81 = 0.032 m/s

q = angle of inclination of rack with horizontal = 50°

hL = 1.67 x (0.987 /1.628)4/3 x 0.032 x sin 50°

= 0.021 m

Provide higher value of the above two, here it is same

  hL  = 0.021 m