3
General Procedure for
Design Calculations
In this chapter, we shall look into the steps of design calculations of treatment units. The
procedure for design calculations involves the steps for computing the capacity and dimensions
of treatment units based on assumed design criteria or actual given field conditions.
According to shape, the treatment units are mostly of two types, viz. rectangular and
circular. However, functionally, these units can be classified as reactors or settling units.
The units where chemical, biological or biochemical actions take place are generally known
as reactors and the units, which are designed to remove suspended solids or to separate
chemical or biological floes by sedi mentation are called settling basins or sedimentation
tanks or clarifiers. These settling units are required to collect the solids or sludge settled
at the bottom and therefore, they usually have hopper bottom sludge pockets. Figure 3.1
shows schematic diagrams of typical rectangular and circular settling units.
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Sludge hopper
Sludge E
Scraper
Sludge hopper
Plan
(b) Typical circular settling unit
Figure 3.1 Schematic diagrams of typical settling units.
The design of treatment units, in general, means the determination of capacity (volume) and
the dimensions of the treatment tanks. It is also known as sizing of units based on unit
operations and/or processes. In the case of a rectangular unit, the design means computation
of volume or capacity and the leng~h, breadth and depth of the tank or basin. For circular
units, design calculations involve computation of volume or capacity as well as the diameter
and depth of circular basins.
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Sizing treatment units with significant dimensions is called designing the physical
facilities.
In both the cases, for settling units, the dimensions of hopper bottom chambers provided
for collecting settled solids or sludge are determined separately. This is done on the basis
of quantity of solids to be removed in a day as per the need or the disposal method, and
the rate of sludge pumping and process design of that unit.
The detailed design of any treatment plant will also include the sizmg of
influent and effluent conduits (pipes), dimensions of inlet and outlet chambers/
channels, distribution and collection boxes, hopper bottom sludge pockets,
scum removal mechanisms and scum pockets, surface or diffuse aeration system
for biological processes, mixers for chemical processes and mechanical equipment
with pumps, etc. that are necessary for treatment system.
The steps to be followed, in general, to carry out the design calculations for rectangular and
circular units are explained with illustrative examples in the following sections:
Step I: Compute the present and future average daily wastewater flow rate
The most essential data required to start the design calculations for a domestic wastewater
treatment plant will be:
(a) The average daily quantity of wastewater generated by the present population of a
town or community based on the annual demand.
(b) The quantity of wastewater flow rate expected in the future at the end of a design
period, usually 30 years, for the estimated future population.
(c) The various types of fluctuations in the flow rates (as described in Chapter 2).
In most of the cases, such necessary data of flow rates are not available to start the
design calculations. In such cases
o Flow rates are either actually measured or assumed.
o Flow rates are computed on the basis of actual or assumed annual water consumption.
o The various flow rates are used to check the calculations for hydraulic condi-
t;ons, performance of treatment units, conduits, pumping, etc. as mentioned in
Table 2.3.
Following examples illustrate the steps to determine daily average wastewater flow rates
based on different assumptions:
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EXAMPLE 3.1
If it is known or assumed that average daily per capita water supply or consumption of a
town is 250 lpcd (litres per capita per day), calculate the average daily wastewater flow rate
of a town having 50,000 population.
Solution: Assuming the quantity of wastewater generated as 80% of water supply rate,
the average daily flow of wastewater per capita, Qavg = 0.80 x 250
= 200.0 lpcd
Total daily wastewater flow of town, Qavg = 200 x 50000
= 10 X 106lpd (litres per day)
= 10 MLD
= 10 x 103 m3/d
EXAMPLE 3.2
If annual water consumption of a town having 40,000 population is 4000 ML, determine
the per capita per day rate of water supply and the total quantity of wastewater generated.
total quantity of water in a year (L/yr)
number of persons x number of days in a year
4000 X 106 (L/yr)
=-------
40000 x 365 (dlyr)
= 273.97 lpcd
'" 275.0 lpcd
Assuming the flow of wastewater as 80% of water supply,
average daily flow of wastewater, Qavg = 0.80 x 275
= 220.0 lpcd
Total daily wastewater flow of the town = 220 x 40000
= 8.80 X 106 Ipd
= 8.8 MLD
Alternatively, wastewater generated per day = 0.8 x 4000 ML/yr
= 3200 ML/yr
3200 (ML/yr)
=
365 (d/yr)
= 8.75 MLD
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The flow given in terms of MLD (Million Litres per Day), is normally required
to be converted to m3/day or m3/sec for design computations.
The relationship between various units of flow used in the design calculations are given
as follows:
1 MLD = 1 x 106 litres/day
= 10,00,000 litres/day
= 1000 m3/day (1000 litres = 1 m3)
= 1000/(24 x 60 x 60) m3/s (l day = 24 x 60 x 60 s)
= 0.0116 m3/s
Step II: Compute capacity or volume of the treatment unit
Once the daily average quantity of wastewater Qavg is known (either for present or future
conditions), then the capacity or volume of the tank for both rectangular and circular units,
can be computed by assuming the hydraulic detention time from the relation given below:
Volume of treatment unit, V = flow rate x detention time = Q x t
where
V = volume of a tank or reactor, m3
Q = average daily flow rate, m3/d
t = hydraulic detention time, d
EXAMPLE 3.3
If detention time of a tank is 4 hours, compute the volume of a treatment unit for the
conditions given below:
o Average daily flow Qavg= 600 m3/d
o Peaking factor = 2.5
o Minimum flow Qrnin = 400 m3/d
Solution: The volume of the tank for various flow conditions is computed as follows:
CASE I: The volume of the tank for average flow conditions,
Vavg = (Qavg x t)/24
= 600 (m3/d) X 4/24 (d)
3
= 100 m
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Qmax = peaking factor x Qavg
= 2.5 x 600 = 1500 m3/d
V max = (Qmax X t)/24
1500 (m3/d) x 4/24 (d)
= 250 m3
CASE III: The volume of the tank for minimum flow conditions,
Vmin = (Qmin x 4)/24
= 400 (m3/d) x 4/24 (d)
= 66.66 m3
67.0 m3
The dimensions obtained in decimal or fractional values are usually rounded up
to the practically executable dimensions, as explained in above examples.
A tank designed on the basis of average flow conditions should be checked for
other loading criteria at maximum or minimum flow conditions.
Step III: Compute the surface area and cross-sectional area of the treatment unit
o Determination of surface area: If we know the average daily flow rate Qavg and
surface loading rate (SLR), the surface area of rectangular and circular treatment
basins can be determined from the following relationship:
flow rate (m3/d)
Surface area of unit, As = 3 2 as illustrated in Example 3.4.
surface loading rate (m 1m /d)
EXAMPLE 3.4
Determine the surface area of a treatment unit for the given conditions:
Average daily wastewater flow rate, Qavg = 600 m3/d
Surface loading rate (SLR) = 30 m3/m2/d
"I WASTEWATER TREATMENT: CONCEPTS AND DESIGN APPROACH
Solution: Surface area of the treatment unit,
As = flow rate/surface loading rate
= 600 (m3/d)/30 (m3/m2/d)
= 20 m2
o Determination of cross-sectional area: If the average flow and flow through
velocity are known, the cross-sectional area of a rectangular treatment unit can be
determined as shown in Example 3.5.
EXAMPLE 3.5
Determine the cross-sectional area of a Grit Chamber (or any rectangular treatment unit) for
the conditions given below:
Average daily wastewater flow rate, Qavg = 600 m3/d
Flow through velocity, Vh = 1.2 m/min
Qavg
tht:n, the cross-sectional area of the unit, Ax =
600 (m3/d)
1.2 (m/rnin) x 60 (min/h) x 24 (h/d)
"" 0.35 m2
The cross-sectional area of a circular UnIt IS estimated by multiplying the
diameter of tank (D) and the side water depth (d), i.e. "" D x d.
Ax(cir)
Step IV: Compute dimensions of treatment units, assuming suitable design criteria
Once the average daily quantity of wastewater Qavg is known (either for present or future
condition), then dimensions such as the length, breadth (width) and depth for a rectangular
unit and the diameter and depth for a circular unit are computed assuming suitable design
criteria as follows:
o Computation of the net depth of a treatment unit: If we assume average
settling velocity Vs of suspended particles to be removed and the hydraulic retention
time or detention time t, the liquid depth d for the tank is worked out as under:
Assume settling velocity of particles, Vs = 20 mm/s
and hydraulic detention time, t = 90 s
CHAPTER
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3: GENERAL PROCEDURE FOR DESIGN CALCULATIONS 111m
then, the liquid depth in the tank, d = Vs X t
= 20 (mm/s) x 90 (s)
= 1800 mm = 1.8 m
As basins are usually designed and checked for surface loading rate (SLR), the
surface loading criterion is usually assumed in place of settling velocity. In
such cases, the value of settling velocity is first determined from the assumed
SLR before computing the depth of the tank.
Alternatively, as settling velocity is numerically equal to SLR, we can compute the
liquid depth in a tank, if SLR and t are known as follows:
Let SLR = 30 m3/m2_d
and t = 90 min
Then, settling velocity, Vs = SLR = 30 m3/m2_d (mid)
= 30/24 x 60 x 60 (rnIs)
= 0.000347 x 1000 (mm/s)
"" 0.35 mmls
.. d = Vs X t = 0.35 (mm/s) x 90 x 60 (s) = 1890 mm = 1.89 m "" 1.9 m
o Computation of the length and breadth (or sizing) of a rectangular tank: If
we assume the flow through velocity Vi> and hydraulic detention time t, we can find
the length of the tank L and assuming a suitable L : B ratio the breadth (or width),
B of the tank is determined as illustrated below:
Let flow through velocity, Vi> = 0.30 mls
and detention time, t = 1.0 min (i.e. 60 s)
then, length of tank, L = Vi> X t
= 0.30 (rnIs) x 60 (s)
= 18.0 m
Assuming L: B = 2.5 : 1, B = L/2.5
= 18.012.5
= 7.20 m
Alternatively, the dimensions of a rectangular tank can also be calculated by assuming
a suitable SLR and other design criteria as illustrated below:
Let us assume SLR = 30 m3/m2_d = 0.35 mmls and t = 1 h
Then, the surface area, As = QISLR
= 600 (m3/d)/30 (m3/m2_d) (assuming Q = 600 m3/d)
= 20 m2
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So, providing L = 7.0 ill and B = 2.9 m,
surface area = 7.0 m x 2.9 m = 20.3 m2 (acceptable as greater than 20 m2)
Assuming Vs = 0.35 mm/s,
depth of the tank, D = v, X t
= 0.35 (mm/s) x 1 x 60 x 60 (s)
= 1.26 m "" 1.3 m
Thus, L = 7.0 m, B = 2.9 m, and D = 1.3 m.
Design computation will provide the net or effective dimensions of the unit. To
calculate the total or overall dimensions of the unit for all practical purposes, some
percentage values of dimensions for inlet and outlet chambers and channels, sludge
collective zones and hopper bottom (wherever applicable), freeboard, etc. are
assumed and added to the computed net values as illustrated in Example 3.6.
EXAMPL 3.6
Assuming suitable design criteria, compute the net and overall dimensions of rectangular
and circular settling basins for a wastewater flow of 1000 m3/d.
Solution: CASE I: COMPUTATION OF NET AND OVERALL DIMENSIONS OF THE RECTANGULAR
BASIN
For the given daily flow rate, assume SLR = 40 m3/m2_d and HRT = 2 hours,
(a) Computation of net dimensions
Surface area, As = QISLR
= 1000.0 (m3/d)/40 (m3/m2_d)
= 25.0 m2
Assuming L B = 2.5 : 1,
= 2.5 B x B = 2.5 B2 = 25 m2
L x B
,'. B = 3.16 m = 3.2 m, and L = 3.16 x 2.5 = 7.9 "" 8.0 m
So, the provided surface area = 8.0 m x 3.2 m = 25.6 m2 (acceptable as greater than 25 ro2)
.. Depth of tank, D = Vs X t (.,' V, = SLR)
= 40 (mid) x 2/24(d)
= 3.3 m
Thus, L = 8.0 m, B = 3.2 m, and D = 3.3 m
The volume of the tank V required = Q x t = 1000 (m3/d) x 2/24 (d)
= 83.5 m3
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However, the tank volume provided = 8.0 (m) x 3.2 (m) x 3.3 (m) = 84.5 m3 (acceptable
as greater than 83.5 m3)
Summary of net dimensions of the basin
o Volume, V =
84.5 m3
o Length, L = 8.0 m
o Breadth, B 3.2 m=
o Depth, D = 3.3 m
(b) Computation of overall or total dimensions
Assuming free board as 0.3 m and sludge zone depth as 0.7 m,
D(lola!) = net depth + free board + sludge zone depth
= 3.3 + 0.3 + 0.7 = 4.3 m
Similarly, assuming 10% of net length for each in:et and outlet zones,
L(IOlal) = net length + length of inlet zone + length of outlet zone
= L + 0.1 L + 0.1 L
= 8.0 + 0.8 + 0.8
= 9.6 m
However, provide 10.0 m length to include wall thickness, etc.
Therefore, the provided total volume of the tank = lOx 3.2 x 4.3 = 137.6 m3.
Summary of overall or total dimensions of the basin
o Volume, V =
137.6 m3
o Length, L = 10.0 m
o Breadth, B 3.2 m=
o Depth, D =
4.3 m
CASE II: COMPUTATION OF NET AND OVERALL DIMENSIONS OF THE CIRCULAR BASIN
For the given daily flow rate, assume SLR = 40 m3/m2_d and HRT = 2 hours,
(a) Computation of Net Dimensiom
Surface area, As = Q/SLR
= 1000.0 (m3/d)/40 (m3/m2_d)
= 25.0 m2
4
The diameter of a circular basin, d __~ :,. As __~ 4 :I. 25
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The depth of tank, D = volume of tank/surface area of tank
= 83.5 (m )/25 (m2)
3
= 3.34 m :::::3.4 m
Summary of net dimensions of the basin
o Volume, V 83.5 m3=
o Diameter, d 5.70 m =
o Depth, D = 3.40 m
(b) Computation of Overall or Total Dimensions
Assuming free board as 0.3 m and sludge zone depth as 0.50 m,
D(lolal) = net depth + free board + sludge zone depth
= 3.4 + 0.3 + 0.50 = 4.2 m
Similarly, assuming 10% of net diameter for the width of an outlet channel,
total diameter, d(lolal) = net diameter + width of the outlet channel
= d + 0.1 d
= 5.70 + 0.57
= 6.27 m "'" 6.5 m
Therefore, provided total volume of tank = -7r x (6.5)2 x 4.2 = 139.4 m3
4
Summary of overall or total dimensions of the basin
o Volume, (capacity) V = 139.4 m3
o Diameter, d = 6.50 m
o Depth, D = 4.20 m
The overall depth in the above examples does not include the depth of hopper
bottom of the tank provided for sludge coli xtion. Computation for the same
is shown in ex·ample 6.1 in Chapter 6.
Figure 3.2 shows the details of net and overall or total dimensions for rectangular
treatment units.
In Figure 3.2,
I =
Inlet zone
II = Outlet zone
III =
Settling zone
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IV = Sludge zone
Vs = Settling velocity of particles
Vh = Flow through velocity or horizontal velocity of particle.
AB = Resultant settling path of particle. The average settling path adopted by particles
during detention period t of flow in the tank.
Dnet = Side Water Depth (SWD) = net depth of tank.
Lnet = Net or effective length of the tank
FE = Free Board
Total depth, D(lOtal) = net depth + free board + sludge zone depth
=D + 0.3 + D1
= D + 0.3 + 10% of D
Sizing of treatment units based on process parameters has been illustrated in the
examples of Chapters 7 and 8.
SUMMARY
In this chapter you have learnt the following:
D Treatment units may be rectangular or circular in shape.
D Design of treatment units, in general, involves the computation of their capacity and
dimensions based on design criteria or process parameters.
D Computation of capacity and dimensions of basins or reactors is known as sizing
of the treatment units.
D Wastewater treatment plants are, in general, designed for average daily flows and
then the designed units are checked for maximum and minimum flow conditions.
D Computations based on design criteria provide net or effective dimensions.
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o Certain percent of net dimensions are added to net dimensions to obtain the overall
dimensions of the units.
o Total or overall dimensions are used to estimate the cost of treatment units.
EXERCISES
1. What is sizing of a treatment unit? Discuss the general procedure to compute the
size of a sedimentation tank.
2. An industrial township has 250 staff quarters, all occupied by the staff. How would
you compute the flow rate of domestic wastewater generated in order to design a
treatment plant for the township?
3. A town having 75,000 population is annually consuming 5000 ML of water; determine
the per capita per day rate of water supply and the total quantity of wastewater
generated.
4. What do you understand by the net and overall dimensions of a unit? Illustrate the
same with a neat sketch for a rectangular treatment unit.
5. What is the significance of settling velocity and surface overflow rate in sizing the
treatment units? Explain with suitable examples how they can be used to determine
the size of treatment units.
