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AN ORION“. DESIGN OF A
RAPID SAND FILTER

Thai: fa! tin Dunn of B. S.
MICHIGAN STATE COLLEGE

Alvin L. Funaworth
1947

ya E515

IP-

An Original Design of a
Rapid Send yilter

A Thesis Submitted to
The Faculty of
MICHIGAN SEEIE COLLEGE
of
AGRIGUBTUBE.AND APPLIED SCIENCE

3?

Alvin L. Farnsuorth
an,

Candidate for the Degree of

Bachelor of Seience

December 1947

Tth'lS

 

 

 

A C K N O W L E D G E MCE N T

It is the desire of the author to hereby gratefully
acknowledge the help of Wesley Bintz, Consultant Civil Engi-
neer of Lansing, Huchigan. Mr. Bintz made available informa-
tion which allowed a practical approach to this problem and
consequently made this thesis of more value to the author and

possibly of more value to those who may choose to read it.

Outline of Problem. . . . . .
Design of Roof Slab . . . . .
Design of Walls . . . . . .
Pump Room Wall. . .
Filter Wall . . .
Partition Walls . .
Footings. . . . . . . . . . .
Filter Piping . . . . . . . .

Washwater Trough. .

Filter Cell Outlets, Risers,

and Mixing Chamber Outlet . .

Headers, Pool Inlet and

Mixing Chamber Inlet..

PO01 Outlet . . . .
Pump and motor. . . . . . .
Filter media. . . . . .,. . .
Pump Room.Equipment . . . . .
Uhit Prices for Estimating. .

Estimate of Cost. . . . . . .

Resume of Costs and Conclusion.

References. . . . . . . . . .

l4

15
16
16
18
19
20
22
27
29

OUTLINE Q}; PROBLEM

For the purpose of this thesis we shall assume that
the author has been approached by someone who desires a filter
plant to filter the water in a swimming pool of capacity
68,200 gallons. The filter plant will be Operated 16 hours
each day and it is desired to have a complete turnover of the
water each day. A pump room must be provided to house the
pump and the piping for the filters.

The following is a solution for the above problem.
lore complete details will be worked out as individual pro-
blems but this general solution will tend to give an overall
picture of the design. Since we must filter a pool of water
in 16 hours, the rate per minute will be: 62,200/16 or 64.5
62!. The rate at which the water will pass through the fil-
ters is about 3 GPM per square foot of filter area. However,
the filters must be washed and the washing rate is about 15
cm per square foot of filter area. We will divide the fil-
ter into three cells so that it may be washed part at a time,
thus minimizing the size of pump required. If we now ditide
the total rate per minute by 3 we get the rate per cell per
minute. This will be 21.5 GEM. If we now divide this rate
per cell by the rate of filtering per square foot, we will
have the required number of square feet per cell. This is
7.2 square feet.

Rather than use graded gravel in the cells, we will

use porous underdrain plates to hold the filter sand. An_

arrangement of this kind facilitates cleaning and repair of
filters without the expense of re-grading or replacing the
graded gravel. These plates are about 1 foot square and 1i
inches thick. Since each of the plates gives a filter area
of 1 square foot we will need 8 plates in each cell to meet
the requirements set forth in the proceeding paragraph. Thus,
we will build three filter cells each 2 feet by 4 feet. Also
we will build a similar cell for use see mixing and settling
chamber.

The pool for which the filter is to be used is of
the sunken type and our filters and pump room.will also, theree
fore, be sunken with the floor a continuation of the pool
floor. (The filters being placed at deep end of pool.) With
this condition it will be necessary for pump room door to
open into the pump room, This will require about 3 feet. The
pump and the filter piping will require about 4 feet. The
pump room, then, will be about 8 feet wide if we leave a little
extra space for chemical containers. The length of the room
will be the same as the length of the filters, or about 10
feet. This is assuming the small dimensions of the filter
cells and mixing chamber to run parallel with the length of
the room and the partition walls between filter cells to be
about 8 inches thick. I

Cur problem now becomes that of computing the de-
tails for four 2' by 4’ cells which must be water tight, a

pump room.8' by 10' and the preper system of piping to

circulate the water through the filters and back into the

pool.

DESIGN g; ROOF SLAB

The roof is to be built of reinforced concrete and
constructed such that the span will be as shown in Sketch 1.
Since the roof will be used for sun bathing we will design
for a live load of 100# per square foot. Weight of concrete

150? per cubic foot.

Assuming a 6 inch slab we have:

live load 100#
dead load 150/2 75
Total 175? per sq. ft.

Assuming a simply supported slab:
M=tl‘/s = 175 x (10.57f/s =- 2505 ft. lb.
In keeping with the design of the pool to be con-

structed at the same time as the filter plant, we will assume
§320,000#/sq. in., 2f1050#/sq.in., and fg:SOOO#/sq. in.

Then:
k'.344 and J=.885

M‘do =C(d x kd/S)

, game/0:8? “173 1 Rd ”1050/2 x .344d=181d
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M?181d(d - .344d/3)=‘181d(d - .llSd)
=160d‘
dz=M/ldo = 2505/169 = 15.55
d=m = _3_._9_9_ inches
Use effective depth of 4" with 2" cover making a
6' slab.
Steel Required
M=TJd=Agiijd or A5: m/ngd
f,=20,000#/sq.in.
ds4”
j=.885
A;2505/20,0001.885x4 -.0354 sq. in. per inch
i“? c 515" = aces/5.5 =.0356 sq. in. per inch
Steel: Use §-¢’o 5%" c.c. Also use §"¢O 18" c.c.
in apposite direction for temperature
stresses and to tie the slab together.
See detail drawing 2.
Check for Shear:
Using method outlined in references 1 and 3
2,=sooov/7ue
553.4 for i'flf c 5%" c.c.
V=l75 x 1067/z==932.5#'=.93 kips
d=4'
Allowable u = .051" - .05 x 5000 =l50#/Sq. in.
Therefore, no hooks are required on bare.

Allowable v in this case =.02 x f'==60#/sq. in.

vc deve10ped = 932.5/12 x 4 : 19.4#/sq. in.

DESIGN 9; M

In designing walls we have three separate situations:
the pump room walls, the filter wall, and the partition walls.
The maximum stress in the pump room wall will occur after the
backfill is placed and before the filters are filled for op-
eration. thimum.stress in the filter wall will occur when
all filter cells are in operation. In a partition wall the
maximum.strees will occur when the cell on one side is com-
pletely empty and the cell on the Opposite side is in opera-
tion.
Pump Room.Wall

For purposes of computing backfill pressures against
the wall horizontally, we will assume the backfill to be a
liquid weighing 30#/ cubic foot. From the Sketch 2 we will
see that the wall can be designed either as a curtain wall
with the other walls as counterforts or as a canilever retain-
ing wall. (Note that the Joint at the tap of the wall must be
a slip Joint so that the roof slab is simply supported. This

makes the wall a canilever as shown in section AsA.)

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Max” 240 ”7ft?

‘5, developed : 952.5/12 x 4 : 19.4#/sq. in.

DESIGN 9}: M

In designing walls we have three separate situations:
the pump room.walls, the filter wall, and the partition walls.
The maximum stress in the pump room.wall will occur after the
backfill-is placed and before the filters are filled for 0p-
eration. Maximum.stress in the filter wall will occur when
all filter cells are in Operation. In a partition wall the
maximum.strees will occur when the cell on one side is com-
pletely empty and the cell on the Opposite side is in Opera-
tion.

Pump Room Wall

For purposes of computing backfill pressures against
the wall horizontally, we will assume the backfill to be a
liquid weighing 30#/ cubic foot. From the Sketch 2 we will
see that the wall can be designed either as a curtain wall
with the other walls as counterforts or as a canilever retain-
ing wall. (Note that the Joint at the tap of the wall must be
a slip joint so that the roOf slab is simply supported. This

makes the wall a canilever as shown in section AaA.)

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In order to conserve materials and save the neces-
ity of making a thick wall we will use a combination of the
two above mentioned possibilities, designing for 5/8 of the
stresses to be carried each way, thus allowing a safety fac-
tor to take care of the possibility that both sets of rein-
foreement might not come into action simultaneously.

(Nate: For wall design, reference 1 will be used
instead of the long method used in the roof design.)

Canilever Steel

At base of wall:

P=240l2 x 8 =950# (See Sketch 2)
M=960 18/5= 2840 ft.lb. or 2.56 kip ft.
2.56 x 5/8 =l.60 kip ft.

From table 2 or reference 1: Req'd. d=5fi"

wall thickness 3%” plus 3' cover::6§'
(See "Negative Steel")

(maximum.moment for the counterfort part of the de-
sign will occur at the bottom foot of the wall where the aver-
age pressure per running foot will be 7.5/8 1 240 225# Then:

M=el‘/lO(end span) =225 x (8.5)/lO =1525 ft. lb.

From this we see that the moment developed in the
canilewer part of the design is greater and the wall thickness
established in that part of the design will govern.)

L At point 2'.above base of wall:
11% I 30/2 I 6 x 2 .1030 ft.lb.
1.08 kips ft.X5/8 *.675 kip ft.

1t point 4' above base of wall:
11:4 1 30/2 I 4 x 4/5 =- 520 ft. lb.
.32 kip ft. x 5/8 = .20 kip ft.
At point 6' above base of wall:
11:2 1 50/2 at 2 x 2/3 -- 40 ft. lb.
.04 kip ft. x 5/8 =- .025 kip ft.

Height above Moment As: M/ad Steel
footing kip ft. Table 3 Ref. 1
O 1.60 .318 i“? O 7%” 0.0.
2 .68 .154 two 15* c.c.

Since we don't want a spacing greater than 183 we
will go no higher than the 15" spacing shown for 2’ above
footing. See Sketch 3.

15“ 75p of" Backfi//

 

 

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Horizontal53teel

For the following computation "1' will be taken as
8'-6' which is the longest span. Not as much steel would be
required in the shorter span but we will keep the wall uni-

form regardless.

Height of horizontal

1-foot strip above Mewl/lo A3=h/ad Steel
footing x 5/8 Table 3 Ref. 1
O to 1 1.015 .21 east e 11%” c.c.
2 to 5 ' .745 .15 two c l6” c.c.
4 to 5 .467 .09“ i'é 0 18' c.c.

All steel above 4' point to be 18' c.c.

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Negative Steel .

The canilever steel already required that the wall
be o§~ thick in order to give 3" cover of the canilever steel
on the backfill side. By increasing the wall thickness to 7”
we can place our steel in the center of the wall and have it
serve both as negative and positive steel.

m 292.12 19.11 1so be 1: islet.

Bond

A é”! fully stressed at 20,000#/sq. in. will carry
20,000 x .1965“3920#. The bond between the concrete and the
bar will carry: 5.1416 x p x 150'*256#. Thus, 3920/236==16.6
inches will be required for prOper bond. make bars extend 18"

beyond corners or points of lapping.

Filter Wall

Since the partition walls provide us supports at
2'-6* centers it will no doubt be most economical to design
the filter wall as a curtain wall.

Pressure at bottom of wall is 8 x 62.4==499#/ft.
The average pressure on the bottom l-foot strip of wall is:
7.5/8 1 499 = 467f/ft.

”12/10 =46? x (2151710 =292 ft.lb. or .292 kip ft.

From table 2 of Reference 1, the depth required is
less than 2". Since the wall will be cpen into the pump room
2" cover will be enough on the pump room side. As in the case
of the pump room.walls we will make the same steel act as posi-
tive and negative steel. We will want 3' cover on the side
toward the filter cells and the gall thickness then becomes:

2' plus 3' or §:.

F'Mt’er Cc//S J, "
W’T'fl’ - Al’ I .
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6K: raw 6 SK! 717/ 6‘

10

Anchorage in the partition walls will have to be
checked after the partition walls are designed. At corners
continue steel for 18" after bending around the corner.
Steel Required

M/ad=As =.292/1.44 x z =.1oz sq. in./ft.

From table 3 of reference 1, use §"¢Hs 18" c.c.
horizontally. Use same amount of steel also vertically for
temperature stresses. (Theoretically, less steel could be
'used higher up on the wall but we don't want spacing greater
than 18" c.c.)

Partition Walls

For the partition walls we have the same maximum .
pressures as for the filter wall but "1" will be 4'-6'.

M =uf/1o =46? x (4.5)2/10 945 ft. lb.

Required 'd” is 2%". However, since the walls come
in contact with water on both sides we will want 3' cover of
the steel On each side and the wall will, therefore, be 6*
thick. The steel will be placed in the center of the wall
and will act both as negative and positive as required.
Steel Required
JHeight of horizontal

l-foot strip above u=fwlz/ 10 A,= M/ad Steel
footing
o to 1 .945 .219 t's‘ e 11' c.c.
z to 3 .695 .160 iv e 15' c.c.
4 to 5 .441 .102 i”? e 18" c.c.

All steel above 4' to be 18' c.c.

11

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Check for Anchorage of Filter Wall

The steel in the partition walls will also serve as
tie steel to tie the filter wall to the partition walls. The
b0tt0m.l-foot strip of the filter wall has an average pressure
of zasfi/eq. ft. The shear at the partition wall connection
would be 225 x 2.5 =563#. At least there will be one fi'wbat
this point which will carry 20,000 x .1963=39303‘* Thus, the
steel in the partition walls will be sufficient to tie the

filter wall to the partition walls.

FOOTINGS
The footing under the swimming pool walls will be

continued to serve the filter plant also. See drawings of
filter walls for footing details.

The following is an investigation to determine if
the footings as used for the pool wall are sufficient for our
purposes.

In Sketch 8 in" equals weight of roof slab, plus

weight of wall, plus weight of footing. (Considering a 1-
foot length of structure.) "E” is the weight of backfill

12

over footing. 1?; equals the side pressure of the backfill.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

El :
‘ :30 = 'flH;Z%.* **’”’/‘My’
1.93.3 F5“;
;'1; ;f**¢;jfijflh
: 11/ A; x \
<____.__é::;:T-——_—T%%2: ‘\fl
5&3” ,_
Roof slab & r 671 150 = 450
Wall 7/12 x 9 x 150 '= 78?
Footing 8/12 x l x 1505*; £99
_ Total 2, : p 15373
34/12 x a x 100 = 266';
MR equals resisting moment of floor
section.
Hanged As’.2'0
d=2*
«a = 1.44

H =2 x 1.44 x .2 =.576 kip ft.= 5'76
The moment produced at the critical
system shown_in sketch 8 (at left of critical
It: (1125 x 27)-(1337 x lbw-(286 1
From this we see that the resisting

at critical

ft. lb.

section by the
section):
17):.844 ft.lb.

moment of 576

ft. lb. produced by the slab at the critical section isn't

enough to resist the 885 ft. lb. moment set up by the eiternal

13

forces. Therefore, we will widen the footing. Making the
footing l'-6' wide by adding 3" on the inside of the wall the
externally caused moment now becomes:
M 2(1125 x 27)-(l337 x l4.5)-(266 x 20)4=477 ft. lb.
Our footing will thus become 8" x l'-6". See de-

tail drawings.

PILTER PIPING

In reference 4 we find the following recommenda-

tions:
Hydraulics:
Piping:
Inlet Max. Velocity 4 fps
Outlet w '. 4 fps
Waste " " ' 4 fps
Backwash" ' lO fpsyo
20 GPM] sq. ft. filter area
Controllers etc. 6 fps

Washwater Trough
While water is being filtered it runs down the trough

from.the mixing chamber and overflows the sides of the trough
into the filter cells. No particular size of trough would be
required for this part of the Operation. However, when back-
washing we wash one cell at a time and we want the water to
come up through the filter and run out the trough into the

mixing chamber without overflowing into the other two clean

14

cells. 'Since the water can only escape through one end of
the trough this one end acts as a wier. It will be a 90°
V-notch wier as shown in sketch 9. It has been found that
the V-notch trough works nicely in this type of set-up be-
cause it allows the filter to be evenly washed since the

washwater that comes up directly under the trough can easily

get into the trough. L.- 27! :4
g

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\/ fl
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When specifying the pump we will assume a back-
wash rate of 15 GEM] sq. ft. of filter area. However, in
designing the piping we will assume a rate of wash of 20 GPM/
sq. ft. of filter area.

Our trough must handle the water from one cell or
8 sq. ft. of filter area or 8 x 20==160 GPM.

From page 87 of reference 5, we get the formula
0 2.523” . From table 37 on page 110 of reference 5, we
find that for q = .356cfe, H =.455 ft. or 5.45 inches.

Use H==7' to allow some freeboard.

Filter Cell Outlets, Risers, and Mixing Chamber Outlet

The sizes of these pipes will all depend on the

washwater flow since the maximum requirement will be while

washing. Since we wash one filter at a time, the maximun flow

through these pipes will be:

14

cells. 'Since the water can only escape through one end of
the trough this one end acts as a wier. It will be a 90°
V-notch wier as shown in sketch 9. It has been found that
the V-notch trough works nicely in this type of set-up be-
cause it allows the filter to be evenly washed since the

washwater that comes up directly under the trough can easily

get into the trough. L‘J 34/ :4

l

, 1‘
\/ i”
Y .w

§@£1§H9

 

When specifying the pump we will assume a back-
‘wash rate of 15 GPM/ sq. ft. of filter area. However, in
designing the piping we will assume a rate of wash of 20 GPM/
sq. ft. of filter area.

Our trough must handle the water from one cell or
8 sq. ft. of filter area or 8 x 20==160 GPM.

From page 87 of reference 5, we get the formula
Q 2.52H147 . From table 37 on page 110 of reference 5, we
find that for Q = .355c~°s , H =.455 ft. or 5.45 inches.

Use H==7' to allow some freeboard.
Filter Cell Outlets, Risers, and Mixing Chamber Outlet

The sizes of these pipes will all depend on the
washwater flow since the maximum requirement will be while

washing. Since we wash one filter at a time, the maximun flow

through these pipes will be:

15

8 sq. ft. x 20 GPM/sq. ft. filter area: 160 GPM.
Converting this to cubic feet per second we have:

160/60 x 7.48 z .357 cfs

For computing pipe sizes we will use the simple
formula Q, =AV where:

Q::quantity in cubic feet per second

A::area (inside cross-section) of pipe in sq. ft.

V=Lvelocity of water in feet per second

Recommended V during backwash is 10 fps.

A:Q,/V= .357/10 :.0557 sq. ft. or 5.13 sq. in.

Use 5" pipe.

Since the flow out of the mixing chamber is by grav-
ity, we must check to see if the flow will equal the 10 fps
rate upon which we have based the design of the pipe size.

w = C fig—h where:

‘v=-velocity of flow in feet per second

C =coefficient of discharge

g==gravity constant

h =—head in feet

 

v = .ezV?x 32.2 x 4.75: 14.4 fps

From this we see that the water would flow even
meter than the 10 fps rate assumed and the design is on the
conservative side.
Headers, Pool Inlet, and maxing Chamber Inlet:

These pipes handle the filtered water from all three

cells at once. At a filter rate of 3 GPM.per square foot of

16

filter area this will be: 3 x 8 x 3::72 GPM.‘ Velocity will
be assumed as 4 fps as recommended in reference 4.
A =72 x 144/60 1 7.48 x 4:=5.77 sq. in.
Use 3'4plpe.

In order to better distribute the clean water, the
pool inlet will be divided into two 1%" pipes as shown on
”Isometric" of filter piping.

Pool Outlet:

The size of the pool outlet is now based on the fil-
ter plant Operation. It will be 6" in the case of this size
pool, the size being determined by the pool designer. This
6" pipe is larger than would be required to operate the fil-
ters, but such a size is desirable in order to facilitate the
quick drainage of the entire pool. See "Isometric" of filter

piping.

mall-am
Our pump has two Jobs to perform. It must wash the
filters and also return Clean water to the pool when filters
are filtering. (There Operations cannot, obviously, be per-
formed at the same time.) We must find the head loss (total)
in each case.
Losses During Washing
We must supply water for washing at the rate of 160
GPM or .356 cubic feet per second. Velocity in the 6" pipe

will be:

l7

V::l44 x .556/28.89l =l.77 fps.

Velocity in the 3" pipe will be:

V: 144 x .556/7.593=6.95 fps.
Losses (References 4, 5, and 6)

Entrance 'KV/2g=:.04 x (l.77)/2 x 32.2 Negligible

50' of 6' pipe i x .38'= .19 ft.

Tee in 6” pipe .3 x.38:. .ll

Contraction to 3": .28

2 Tees & 1 E11 in 5" pipe 10.5 x 50/100=~ 5.25

12’ of 3" pipe 10.5 x 12/100: 1.26

Porous plates & filter media; gégg
Total 9.09 ft.

Water after washing is at a level in the trough
2'-6" below pool water level. Therefore, we subtract this
from the 9.09 ft. leaving 6.59 feet.

Losses During Filtering

The filters will be Operated at 3 GPM per square
foot of filter area. Each filter produces 3 x 8-=24 0PM;
Velocity in the cell outlets will be:

v: 144 x 24/7,.48 x 60 x 7.1595 =1.05 fps.

Since the pool inlet is the same size as the cell
outlets but carries the water from all three cells at the same
time the velocity in the pool inlet will be::

V¢3 x 1.05:5.15 fps.

Velocity in the 1%" pipes will be:

V: 144 x 36/7.48 x 60 x 2.04 =5.67 fps.

18

Suction Lift (References 4,5,& 6)
Height from floor to pump ‘ 2.00 ft.
Entrance Negligible

12' of 5*, 1 Ell, & 5 Tees .58 x 87/100: .55
03 1.05 fps)

9' of 5”, 1 Tee & 1 E11 2.57 x 43/100: 1.11
(0 3.15 fps)

Total ' 3.44 ft.
Head Losses-Discharge Side (References 4, 5, & 6)
100' of 3" pipe 0'5.15 fps 2 2.57 ft.
5 Ells & 1 Tee 2.57 x 95/100 = 2.44
20' of 1%", 4 Elle, o 5.67 fps

161 52/100:. 8.50
Height from pump to water level:: 6.00
Total 19.51 ft.

From.this we can see that the demand on the pump
is much greater during filtering than during washing.

From page 7 of reference 6, we see that a 3" Type-B
Gardner-Denver centrifugal pump with a 1% hp motor will be
required or its equal. The efficiency will be 70% and it will

Operate at 1150 RPM;

FILTERLQEQLA

Sand only will be required since the porous plates
take the place of the usual graded gravel. We will want 24"
of sand over the porous plates. The recommended effective

size is .55mm. and this with a uniformity coefficient of 1.75.

19

See reference 4, #12-3-1660.

PULP; M EQUIPMENT (When required by health authorities or
by Owner.)
Disinfector ‘

This shall be a chlorinator similar to that manu-
factured by the Everson Filter 00., 627 W. Lake Street, Chicago,
Illinois. Use size recommended by manufacturer.

Pressure Gauge

Attach to influent side of pump a gauge that will
show pressure or vacuum. Relative readings will indicate when
washing of filters is necessary.

Flow Indicator

Attach rate of flow indicator to discharge side of
pump to indicate filtering and washing rates.

Coagulating Device

An automatic device for feeding of lump crystal Pot—
ash Alum or Ammonia Alum to the water will be connected to
the supply piping of the filters.
Alkalinity Device

Connect this device to venturi tube of the coagula-
ting device.
Emir Strainer (or Catcher)

Connect this item into riser pipe below entrance

into mixing chamber.

UNIT PRICES LOB ESTIMATING
Concrete (l:1%:5) Per Cubic Yard
Materials
Cement 1.85 Bbls o 5.60 s 6.66
Sand .42 Cqu o 1.75 .74

Gravel 1.84 Cqu o 2.75 2.51

Lime .15 Bbls 0 2.60 .54

Total Materials $10.05
labor (Mixing in small batch mixer and

placing with buggies or wheelborrows)

Common

labor 3.75 Hrs. @' 1.00 $ 3.75

Mixer .58 Hrs. 0 2.00 .76
Foreman .50 Hrs. 0 1.50 .34
Total Labor 5 4.96

Total per Cu.Yd. concrete in place
Formwork (100 sq.ft. contact area)
Materials

Lumber 225 BdFt 5100.00 $22.50

Nails .5 lbs. 0 .07 .04
Wire 2.5 lbs. 0 .08 .20
Total materials $22.74

Labor (Erection and Wrecking)

Carpenter

5.85 Hrs o 2.00 511.70

515.01

20

21

Formwork (continued)
Helper 5.10 Hrs. 3 1.50 7.65
Foreman .60 Hrs. 8 2.25 1.55
Total Labor $20.70
Total Cost Formwork per 100 sq. ft.
contact area $45.44
Cost Formwork per sq. ft.
contact area . .45
Excavation (Per Cubic Yard)
Hand Excavation ’
Common
labor 1.50 Hrs. 0 1.00 $ 1.50
Total Cost Hand Excavation per
Cubic Yard 5 1.50
Power Equipment Excavation
Operator .06 Hrs.-3 2.25 5 .14
Equipment.06 Hrs. o 9.00 .54
Total Cost per Cu.Yd. for power equipment

Excavation (no haul) 5 .68

ESTIMATE pg COST

Earthwork (Assuming level ground at water level in pool)

 

Excavation
Filter rOOm 80.0 Cqu
Access stairs 15.0 "

95.0 Cqu 5 51.20 5114.00

Gravel Sub-drainage

Under floor .2'54 Cqu
Around Walls 4.45 ”
Under Stairs 2.65 "

9.40 curd e 52.75 5 25.85
Backfill
Around filter room 21.0 Cqu 0 $1.00 5 21.00
Total Earthwork. . . . . .,. . . . . . . . . . . .5160.85
Concrete

Floor, Footings, Pump Pad,

and Pedestal 5.76 Cqu G $20.80 3 75.20
Con. F1. & Scum Cutter .96 s c 55.00 55.60
Roof Slab 2.84 " 0 45.00 127.80
Walls 7.86 7 5 70.00 550.20
Stairs _ 1.54 n a 25.00 55.50
Trough .27 " 5 110.00 29.70

Total Concrete. . . . . . . . . . . . . . . . . . $850.00

 

gpinforcing Steel 5" Lin. Ft.
Roof 400

Floor and Footings 418

25

ESTIMATE 9; COST

Reinforcing Steel (Continued) Lin. Ft.

 

 

Partition Walls 160
Filter Wall 155
"Pump Room Wall 685
Stairs 170
Trough 115
Miscellaneous 1 - 50
Total . 2,109 Lin.Ft.
Weight 2,109 x .668/2,OOO ' '.705 Tons
5" .705 Tone 55120.00 584.50
Labor ' w - .705 Tons e 40.00 28.17

Total Reinforcing Steel. . . .'. . . . . . . . . .5112.67

Woodwork
Door & Frame with Hardware 1 o $50.00 $50.00
Window Frames & Lights 2‘ o 5.00 10.00

Total Woodwork. . . . . . . . . . . . . . . . ... .$40.00
Underground

Drain Tile 4O Lin.Ft. o 5 .64 $25.60

Tbtal Underground. . . . . . . . . . . . . . . . . $25.60

Painting
Woodwork 6O Sq. Ft. 0 5 .10 $ 6.00
Pumproomn& Stairway 550 Sq. Ft. 0 .08 28.00

Total Painting. . . . . . . . . . . . . . . . . . .354.00

ESTIMATE‘QF cost

-Plastering

 

Filter Cell walls ‘ 25 Squ o 5 .84

Total Plastering. O O O O O 0 O O O O O O O O O

 

EleCtrical
Fixtures 2 a $50.00
Wall Plug l c 10.00
Switch 1 0 12.00
Connection to Pump 1 e 40.00

Total Materials
Overhead & Incidentals 50%

Total E-lCCtricale 0 O O C O O O O O O O O O O O 0

Plumbing
' Water Service to Pool .
Pipe 2" 50 L.F. c S .26
Valve 2" b 1 o 7.74
Tee 2" 1 «s 1.45
Plug 2" l o .50
E113 2" 5 0 .74

Outlet Piping from Pool

Tee 6' x 5* x 6" 1 <3 $14.16
Pipe 2" ' 5O L.F. o .55
Valve 5" 5 c 15.54
E11 5" 2 e. 2.27

Tee 5” 2 o 2.45

24

£21.00

.521.OO

$60.00
10.00
12.00
40.00
$122.00

56.60

$158.60

5 7.80
7.74
1.45

.50
2.22

514.16
15.90
40.62

4.54

4.90

25

ESTIMATE 9; COST
Plumbing (continued)
Filter Outlets and Backwash Line

Pipe 5" 55 L.F. o S .55 $18.55
Tee 5" 7 0 2.45 17.15
Valves 5" 4 c 15.54 64.16
Ell 5” 2 o 2.27 4.45
Plugs 5" 5 o .45 1.55
Mixing Chamber Outlet
Pipe 5" 5 L.F. o 5 .55 S 1.59
Valve 5" . 1 e 15.54 15.54
E11 5" 1 0 2.27 2.27

Pump Discharge Inlets

Pipe 5” 9O L.F. o 3 .54 $47.70
Pipe 14" 20 7 c .19 '5.80
Valve 5" 2 c 15.54 27.08
E11 5' '7 o 2.27 15.89
Tee 5" 5 o 2.45 7.55
Bushings 5" to 1%" 4 0 .58 2.52
Filter media
Porous Plates 24 SqFt
Sand-special 1.8 Cqu
$90.00
Miscellaneous
Caulking lead 10 Lbs. a.$ .20 S 2.00

Caulking Oakum. 5 ” o .15 .75

26

ESTIMATE OF COST
Plumbing (continued)
Special Equipment

Pump and MOtor 1 5 $250.00 $250.00
Alkalinity Pot l 5 These
Coagulent Pot l 8
Items
Chlorinator 1 0
Flow Indicator 1 0 OptiOnal
Total materials, $669.65
Labor and Incidentals 100% 669.65

 

TOtal Pleing o 0 e e e e 'e e e e e e e e 0 $1 ’ 559 0 30
Hflscellaneous

WOOd baffles, mixing<flamber 67 BdFt ‘0 S .15 5 10.05

Fine Grading ' 500 SqFt 0' .20 60.00
manhole Rings and Covers 2 Each 5 15.00 50.00

Total Miscellaneous. . . . . . . . . . . . . . . .$100.05

27

RESUME or; COSTS

 

Earthwork $ 160.85
Concrete _ 850.00
Reinforcing Steel 112.67
Woodwork & Sash 40.00
Underground 25.60
Painting 1 54.00
Plastering . 21.00
Electrical 158.60
Plumbing 1,539.50
Miscellaneous 100.05

Total $2,842.07

Misc. and Incidentals

 

(Overhead) 5% 142.10
. Total ‘ $2,984.17
Contractors Profit 10% 298.42

 

Grand Total. . . . . . . . .'. . .35,2sé.59
Cost per square foot of Filter. . . . . . . . .$156.77

It might be well to note in conclusion that the prices
used here are those for central Michigan in August 1947 as
near as could be determined.

A swimming pool of the size for which this filter plant
is designed would cost approximately $3,000.00 or, in other
words, only as much as the filter plant itself. When we get

into larger sizes of pools, we find that the filtering equipment

28

becomes a lesser percentage of the total cost. However, even
in large jobs of this kind (swimming pool with recirculation
system) the filtering arrangement remains one of the largest

items and demands much attention.

2.
3.

4.

5.
6.

29

BEEEBEEQEE
”Reinforced Concrete Design Handbook" of the American
Concrete Institute.
"Reinforced Concrete Design" by Sutherland & Reese.
5Report of Joint Committee" of the American Concrete
Institute.
”Camp Porous Bottoms for Rapid Sand Filters”, Bulletin
12 S 12 & 12 S 1660 of Walker Process Equipment, Inc.
"Handbook of Hydraulics" by King.
'GardneréDenver'Centrifugal Pumps Types B & C A-lOl.

 

 

 

 

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