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TABLE OF CONTENTS THE PRELIMINARY SURVEY' History Population Prediction Sewage Flow Methods of Treatment THE PLANT DESIGN Outfall'Design Rack Design Grit Chamber Wet Well Pumps Primary Settling Tank Sludge Pumps Aeration Tank Sludge Division Box Final Settling Tank Chlorinator Sludge Digester Gas Chamber Sludge Beds Design of a Wall Design of a Footing Service Building Design of the Heating Unit U'IJI‘Ml-J TABLE OF CONTENTS Determination of Elevations and Tank Elevations Estimate of Costs Total Cost of Plant Capitalized Cost 33 36 M2 East Jordan is a small town located at the southern extremity of Pine Lake, fifteen miles south of Charlevoix. This small town is one of the leaders of the district, espe- cially in the summer when there is a large amount of tourist trade. East Jordan was established in lEZB. Its primary reason for existence was the larggmlugber mills established there, the first mill locating in 1&1]. The city was incorporated in 1903. In 1911 the neighboring village of South Arm was annexed. In 1920 the forest resources had been depleted and the pOpulation started to decline. East Jordan has always been progressive, as is seen by the fact that the city has owned and Operated its own water plant since 1896. The city government is up to date, consist- ing of three commissioners, each being the head of a depart- ment, namely: 1. Public Utilities 2. Streets and Sewers 3. Finances, Health and Sanitation The city is now served by an inadequate sewer system. A separate sewer system was designed for the city in the fall of 19kg. In the thesis covering this design the sources of sewage were given as: 1. Domestic Sewage 2. Trade Wastes...stores, garages, etc. There is a cannery here but it has its own disposal system so does not contribute to the city sewage. f -1.. It is the purpose of this thesis to design an outfall from the end of the proposed sewer line and to design a plant to adequately serve the city‘s needs. The first important consideration in any design is the amount of sewage to be treated. The flow to be expected must be predicted, based on population data. The best source of past population data is the United States Bureau of Census. For the years subsequent to a census it is common to use a city directory as giving the correct population. Another method is to use the ratio between the number of children in school for the year desired to the number of children in school for the census year as the ratio of population of the two years; or do the same thing with the number of telephone services and use the ratio obtained to determine the present population of the city. ' A far more difficult problem is the prediction of a city's population in the future. To do this with any degree of accuracy good Judgment must be used on the part of the engineer. The engineer should know the history of the town, its resources and location as to markets and transportation facilities. After a careful consideration of all the factors effecting future pOpulation a method of prediction must be selected which fits the conditions of the city under consider- ation. There are several of these methods, the most popular being the arithmetical, the uniform percent rate of growth and the curvilinear rate of growth. The method selected for East Jordan was the uniform percent rate of growth with slight modification. An -2- Pop 04 4 7/ o N Caevz/Ze [457' Jon4/gM/cH/64/v — —-— 46 77/41. —"—" [JWMI/ZW § 3.4:- a... D 2?» \ 209 k Ia. T 1.4 \1 ”‘9 [Zoo 3 I... / O Q r ; g s «3 - § § s 3 V5 ,4 . 7650/6750 5r 4/4 [€057- Cour/emso 5r WW Jar /97a ’95? observation of the population data shows a uniform rate of growth except for the period when the lumbering industry was located in East Jordan. If this condition is considered as abnormal and the values for this period ignored, a uniform percent rate of growth will apply to the period from 1882- 1930 and so could be assumed to apply in the future, barring unforseen developments. The usual period of design is 20 years. Therefore, the population will be predicted for 1963. Population Data Year ' 1882 1890 1900 1910 1920 1930 Population 200 731 1205 2516 2123 1523 Computations ngulation 1930 - POpulation 1990 = Increase per Decade 3 1623 — 1205 g 518 = 106 Persons per Decade 3 3 i%%§.x 100% a 70 Persons Increase 19h0 = 1523 * 70 = 1593 106 = 152? x 1593 110 110 + 1593 : 1703 Persons in 1950 1&3; x 1703 = 202 1703 t 202 a 1905 Persons in 1960 1905 x 106 = 3&0 1523 1905 * 340 = 2245 Persons in 1970 Allowing a deviation of approximately 100 persons, the estimated pepulation in 1963 will be 2200, confirming the value obtained by H. L. Frost. ~3~ The best way to predict sewage quantity is to check water consumption. East Jordan is unmetered. Therefore, some other method of prediction must be used. For a city of this type 100 ga1./day/capita is an average value often assumed. A check of similar Michigan cities reveals this to be a good value for East Jordan. The following data was considered: City Population Consumption/gal./day Cass City 1261 57.0 Cassopolis lhu8 117.0 Clare 1h91 167.0 Decatur 1582 63.2 Frankfort 1h6s 102.2 Average 101.3 The infiltration as computed by H. L. Frost was equal to .1823 C.F.S. The quantity of sewage is not as important as the rate. The maximum rate is assumed to be 175% Of the average flow. This gives a value of igggzilg%5;i%gflgi§fi = .5985 C.F,S. giving a total maximum flow of .5985 + .1823 = .7808 C.F,S. The minimum rate of flow is important in determining the capacity of the pumps and the wet well capacities, or actually the longest period of detention. The minimum flow will be assumed to be 0% of the avera e or 100 x 2200 _ ' 5 g 2 x 60 x 60 x 24 x 7gh8 ' .1710 C,F.S. The minimum flow will therefore be .1710 4 .1823 = .3533 C.F.S. The values of maximum and minimum flow were used based on a statement by Steel in his book “Water Supply and Sewage“. I -h_ Flow Data in Outfall Sewer From To Length C.F,S. Ground Elev Drain M.H. M.H. Line ' Max. Upper Lower Pipe Grade Fall' 55 Plant 78.0 .7808 107.0 106.6 10" .0077 0.6 Velocity Flowing Full Capacity Flowing Full Upper Lower Invert Invert 3.58'/seC. 1.82 C.F.S. 91.u 90.8 Flow Data in Sewer Above Outfall Population Aver. Flow Min. Flow Max. Flow Infiltration Total 2209 .3h20 0.3533 0.5985 0.1823 0.7807 Values of flow used: Minimum 0.3533 C.F.S. = 157 G.P.M. ' 232 G.P,M. It Average 0.52“} Maximum 0.7808 n 351 G.P.M, The above values being based on the maximum value including infiltration. The best way to determine the physical characteristics is to obtain samples and analyze the samples in the labora— tory. The distance to East Jordan makes this method impossible, A table of average values for typical types of American cities compiled by Metcalf and Eddy gives the following values for a small, residential, agricultural town: 13. 0. D. (5 day) 1113.0 P.P.M. Chlorides h7.0 P.P.M. _Solids - Total 603.0 P.P.M. Volatile 393.0 P.P.M. Fixed 210.0 P.P.M. Suspended Solids - Total 3h2.0 P.P.M. Volatile 260.0 P.P.M. Fixed 82.0 P.P.M. Dissolved Solids — Total 261.0 P.P.M. Fixed 133.0 P.P.M. Volatile 128.0 P.P.M. Fats Variable The first important question to be decided is the method of treatment to be used in the proposed plant. A common primary method of treatment is the plain sedimentation method. The results are 90% reduction in suspended solids and 85% reduction in B.O.D. The cost of treatment averages $h.00/ million gallons treated. This high cost makes it an uneconomical method of treatment, especially since there are no industrial wastes to be treated. The second method of treatment is filtration, either the intermittent sand filter or the trickling filter. The sand filter is now outmoded, so -Q- / it will not be considered. The trickling filter does come in for consideration. The trickling filter gives 85% reduction. in suspended solids. The disadvantages are the high head required and the odors arising during spraying. The third and final method of treatment to be considered is the activated sludge method. This method is a method which is moderately economical. It requires a complicated plant which is difficult to Operate and requires an operator with some degree of skill. The former idea connected with this type of treatment was that it was not applicable to small cities. Recent data has disproven this theory and the expiration of patent rights has done much to further the activated sludge method. The results are reduction of bacteria 90%, reduction of suspended solids 95%, and stability by methylene blue is 72 hours. The important consideration at East Jordan is the purifi- cation of the lake and stream. This leaves but one selection. The activated sludge method is the most applicable due to its, higher efficiency and economical operating costs, so the activated sludge type of plant will be designed. In any project an item of primary importance is the method of financing. The region north of the east west line from Bay City to Lake Michigan is known as "The Sticks“, and any project in this district must have very good security in order to obtain financing from any of the commercial sources. This means that a city must be in very good financial condition to consider a project as large as a sewage treatment plant. East Jordan has its own water plant and no other ~6- recent improvements have been made, so the city is in a good- position to issue bonds. These bonds should be of the revenue type to insure a good market for them. The investigation of the financing of a project is a Subject which requires considerable research. An exact knowledge of the financial condition of a city is something the city regards as purely personal and will not divulge this information unless an engineer were actually hired, so it will suffice Just to mention the generally known facts and theorize on the probable course of action. gutfall Design: Ground elev. at M.H. 55 107.0 Invert elev. at M.H. 55 91.” Depth 15.6 Q = .7808 C.F.S - Maximum Flow Length is 100 ' Use a 10” pipe as that is the largest pipe used in the system before reaching the plant site. Using values obtained from Page 379 “Water and Sewage“ by . Steele, N a .013. Assume lepe‘of .0077. Q = 1.85 C.F.S. at a velocity of 3.5h'/sec. Qizggé a 0.h2h as ratio of capacity when flowing full and when actual maximum flow is flowing. Now, using Figure 1&9, Page 379, Steele, pipe is flowing hex full or .uu x 10' h,h" deep and a velocity of 0.95 giving 0.95(3.5h) 3.4u'/sec. This is a little high, but pipe in system was at approximately this grade, so to be consistent this grade must be maintained. -7- Use a design figure of 20 sq. ft. of rack area per million gallons per day. Use 2” C-0 spacing and 3/8” bars as the first assumption. With a limiting velocity through the rack of l.5'/sec. L%%%§ (1,000,000) a .51(20) 2 10 sq. rt. of rack area, which is high in comparison to areas used in plants already constructed. Use Prof. F. R. Theroux's value of 12 sq. ft./million gal./day and obtain .51(l2) : 6.12 sq. ft. of rack area. The common method is to have the rack on a 1-2 s10pe, so revise the first assumption to give velocity of 1.0'/sec. through the rack and 12 sq. ft./million gal/ daY. First consider the head created, using a formula given by Metcalf and Eddy in their book "Sewage Design” page ”91. h = l§§§_1§ x %_f Where "V” is velocity through rack.and "v“ is velocity above rack in channel, substituting*in the values obtained gives £16%£&£;25L%L£ = .2hu' increase in head. which willonot even fill the 10” pipe, so this assumption of velocity 13 O.K. The manner in which this increase in head may be overcome is to increase the depth of the channel behind the screen by 0.25'. So now the total depth of sewage is .2uu o .h2u . .668'. Qheckinggérea of Screen} Allow 6“ clearance between sewage level and bottom of screen room floor. D = .668 e .5 = 1.168‘ clearance -8... space for racks. 0.5” -g- 3 6120 XI In?" FOR lffldfll 0" 5:25!” 0.6“ m‘_&_ ”’30-‘66. x 13' 1.31 - .56 a 0.75' of screen in sewage per foot length. gf%% = 9* ' which is too long, so use other basis of design of 1'/sec. through screen and change to 1' clear. Area required is .7808 sq. ft. of screen to give the desired velocity. Clear area/ft. length = 9(12-(23 - 1)3/s) = 71 sq, ing/ft. length. Say .57‘length £1808 2 1.53 feet long, which is a little too narrow°gor easy cleaning. So use a 2' rack with 6" clearance above sewage surface. 1 Testing the Bars: ,Assume 1.5' long and 2“ x 3/8" in size. Wt. of bar - 2 x 3/8 x 1.5 x 12 x 17%8 x #90 = 3.35# per bar applied at the center of gravity of the bar. X a gigaf. 1.72# applied perpendicular to the bar. Water Pressure: 632.2” x g- x W = '55# applied .25. from the bottom of tank. Now consider bar as a horizontal beam. £12 055 4 ca: vase ' or! , tag ‘\\\\\\“‘~\\lq ear IMor£'—uifltflarlhnauut Norlnlcumlo I‘m Loan: :41 Java». Shear decrease per foot is 1.15# per foot X : .1.-4.9% r.- .84' from rt. end. 1 . Bending moment a area of shear diagram. Area = .%1.x .SH x 12 = 6 x 0.8M x 0.97 sails W: 8# .1. 1’ a I% x g’x 2 ’ 7 /sg n which is very small for steel. 50 the thickness is merely for rigidity and cleaning with a rake. Sgreeniggs: Use an estimate of 5 C.F./million gallons so provide enough containers to care for the screenings with daily emptying of containers. Ggit Chgmbeg: A grit chamber is not needed or required in this plant due to the fact that only separate sewage is being treated. ~10- Wet Well: ;‘:l970 minimum flow will be a basis of design. Minimum flow a .3533 C.F.S. provide a 10 min. storage. 10 x 60 x .3533 a 21” C.F. = 1600 gallon storage. Using a typical cross section in a 20’ building use a 7' tank. D - 0 So allow 6“ for safety and 6“ free board to bottom of screen room floor. I Pumps: Three pumps must be used for flexibility of operation and safety. When a pump turns on, it must empty the well plus the amount of sewage which will run in during the emptying period. Provide this unit and provide others for economy and safety in case of a breakdown. 1600 gallon tank 250 G.P.M. pump. The time required.to empty the tank in minutes is t - (£26798 ‘ 600 (“52” L601 = 6.11 + .80 - 7.2 min” 250 which is alright. This time of emptying will not -11.. require pumps to be run continuously. A 150 0.2.x. and a 200 G.P.M. pump should be provided to care for flows other than the average. Primary Sgttlinw: To obtain 50% removal of sludge a 1% hour detention period is required, as determined.by diagram Fig. 175, Page 534, Metcalf and Eddy. Assume a sustained flow of 1.5 x 232 = 3M8 G.P.M. V01. - 1.50 x 60 x 3M8 = 3l,u00 gallon capacity Bring; a 11180 OJ". Two identical units should be provided for continuity of service in case of a breakdown. Assume a 6' depth and.ratio of l : 6 for width to length. 6 x W x 6W = #180 w2 - 11.6' Use a 10' tank, giving L = “130 a 70' 60 long. Provide two tanks with an 8' depth, 2' being allowed for free board. Velocity Through Tank: .52h3 x 1.5 - .79 C.F.s. Q : AV V a % ”lg—36 = .013'/sec. =3 .156"/sec., which agrees favorably with.a velocity of .15 - .50"/ sec. given in Metcalf and Eddy, Page 536. Storage Space: Specific gravity 3 1.03 Sludge moisture content of .95 Average flow of 232 G.P.M. 2h hours time between pumping “’12- 50% removal 3H2 P.P.M. suspended solids .50 x 3&2 a 171 P.P.M. removal design on a basis of 200 P.P.M. removal. 13.3%. x 1,000,000 a 515,000 gallons per day 515.0%96%6§65%;£_§QQ. a 860# of dry sludge per day. , 3 Vol. 3 860 ... = 2000 11 r 8.3H_x .05 x 1.03 ga one o sludge storage must be provided. This is %Q%%.- 268 cu.ft. storage. Provide a hopper at inlet end. Try a 10' x 10' x 2' x 2' x h' h0pper. 10 x 102‘ 2 5~§ (h) = 208 cu.ft. of storage, which is not quite enough. Try a 10' x 10' x 2.5' x 2.5' X'5' h0pper. lQ—EQLg—1§§~5~34g~5 (5) = 270 C.F. storage, which is alright. Use a mechanical skimmer of the link belt type and use a 3% bottom slope, as recommended by Metcalf and Eddy. .03 x 70 = 2.1'. depth to bottom of hepper. Invert elev. = 6 + 2 # 2.1 + 5 a 15.l' to bottom of hepper. A scum trough should be provided at the outlet end. The standard type is a lip of concrete projecting down in front of the effluent channel with the lip a little above the water surface. A detail of the tank with fittings is shown on an accompaning sheet. The design of the scum trough necessitates hand -13- ,5 0 removal of scum. The purpose of the lip projecting down in front of the effluent channel is to prevent short circuits. Sludge Pumps: Use a minimum size of 8" as recommended by Metcalf and Eddy, Page 552. A low velocity is recommended, .5 - l'/sec. .785fl(%%h) = .35 sq.ft. cross sectional area of a 8" pipe. ‘ .35 sq.ft. x .75 u .26 cu.ft./sec., so use a 100 gal/min. pump giving 7:%Q%_Ba.= .2h cu.ft. per sec. Inlet Channel: Tank is 10' wide, use 3 inlets...one at each end and one in the center. End outlets to be 1' from outside wall. ® @ @ g Velocity through openings to be 2'/sec. Assume sustained flow of 3U8 G.P.M. = .7808 C.F.S. Each inlet will be 31§Q%.= .13 sq.ft. in cross section. 2 1.1%.2 = .13 d a 5" Area = .136 sq.ft. Channel Design: ‘ Velocity of 3'/sec. in all sections of tank. Area -1h. at (1) a1%5 = .26 sq.ft., use a 6“ x 6" section. Width at (2) iii—4...; 25 a .17. use a 5" x 5" section and continue on to end as a 5” channel, due to the fact that if a smaller size is used a st0ppage is apt to occur. Grade of Channel Bottom: n = .012 . .. A z: 4 2 = 2 i_ a diagram. 3 a .oou or .016' fall in the h ft. of channel. R(2 - 3) = .138 Slope - .010 or .ou' fall. .0” + .16 = LéQ.a .025 average slope. 50 use %" fall/ft. of channel. Use a 1.5' submergence as recommended by Prof. F. R. Theroux. A baffle in front of each inlet should be provided; or since there is a lip of the scum trough extending down, use it for a baffle. Outlet: The outlet should be of the Wier type with a baffle in front of it extending to a depth of 2' below the surface. Velocity must be relatively low and QT%15'= .0775 if 1' deep, but maximum depth is about 2“. Therefore, velocity is .0775 x lg a 0.45'/sec. At this velocity particles of ordinary size will not be carried over. Therefore, this Wier is alright. Make adjustable to 3 .1' from elevation to be determined later. -15- AgggtépngTagkz Detention period of 6 - 10 hours. Use an updraft type of aerator. Use 1.0 cu.ft. of air per cu.ft. of sewage treated, as recommended by Steele, Page 567. Circulation to be provided by aerator is 2 - 3 times per hour. Flow of 351 G.P.M. or .7808 C.F.S. Also returned activated sludge must be accounted for. Assume 25% return. 1.25 x .7808 = .98 C.F.S. 8 x 60 x 60 x .98 = 28,000 C.F. of tank must be provided. Assume 121 deep tanks 20' wide. L - %§Q%st . 117.5' long. So if the area which can be served by an aerator is about 25' in diameter and maintains a rectagular shape, five 25‘ tanks, 20‘ wide and 12' deep will fulfill the requirements. Round the bottom edges up to aid circulation. Use an aerator of the updraft type which the manufactur— er guarantees to furnish 3600 cu.ft. of air and able to provide at least 18000 cu.ft. of circulation per hour. Inlet: The inlet can.be a channel from the pipe flowing into the tank from the primary tank. The influent channel should be well baffled with a perforated plate containing 1" circular holes. The slope of the channel will be determined by the following method: Q = .98 C.F.S. Assume four -16- Openings in the tank wall, the end openings to be 1.0' in from the ends. Assume the sewage enters from one end. 0 e e @ N . .013 Diagram Page 130, Metcalf & Eddy Assume 2‘/sec. velocity. Area :3 9, - .98 = l sq.ft. in cross section, so use a 12" square cfiannel. H.R. 1 - 2 a A a %-= .33 Se .0013 W.P. Area 2 - 3 a .75 Use 12' deep and 9" wide. H.R. 2 - 3 .33 a 512.. .33 + .30 - .315. s a .001u .1 Area 3 - h a .5 12" deep gives 6“ width. Keep this same width the remainder of the length for ease in cleaning. Round the end of the channel in order to obtain smooth flow into the last port. Depth (3) to (h) is 4%.: 12" deep. W.P. 9 2(1) + .5 : 2.5 Area = .5 H.R.°= gég - .20 s = .031 Depth (h) to (5) is ‘%5 =’.5 is 6" deep. so w.P. a 2(.5) a .5.- 1.5 I H.R. = {2%.= 0.016 s a .ooul Average slope 3 .qul T '0031 a '0013 T “OOIBZ= .0022 0.0022 x 20 = .Ohh’ total. +17“ Outlet: The outlet can be a simple Wier.adjustable to £0.1' and emptying into a channel. Design of channel Assume an even flow over the Wier. Design in U parts and use average slope. Use a minimum width of h” to obtain ease of cleaning. Velocity in channel to be l%'/sec. at a Vol. 2 .25 C.F.S. o s. Av a :31: £2213. . .167 sq.ft. .1673"? = .111' ‘3 or 5”. 'Av H.R. (0) to (A) 3 %g'* f; = 1,} + lgh-z —-—'o—.- 2 1.35. s a .00015 '*2 ‘ V01. At B: .5 cu.ft. A =3 *5 - «33 .332 a 6" wide 6 H.R. (a) to (B) s l'ug’ T8' = 1.70 S = .00012 v01. at c - .75 cu.ft. A = fig = .50 .505 = .707 Use 9” depth 2.0 e 81 H.R. (B) to (C) a _____"§1,= 2.5 S a 0.001 2 Vol. at D s 1.0 cu.ft.. Depth 2 (%4%J% a .807 Use 10" 11.11, (C) to (D) = 3'0 7‘ L173 = 3.17 s I! 0.00005 2 Average lepB = .00005 f .OOOlfii .0001 + .QQOIE = 0.00011 .00011 x 20 a i" fall from end to end of channel. Outlet: The outlet can be a simple Wiervadjustable to £0.1‘ and emptying into a channel. Design of channel Assume an even flow over the Wier. Design in h parts and use average sIOpe. Use a minimum width of h” to obtain ease of cleaning. Velocity in channel to be l%'/sec. at a Vol. 2 .25 C.F.S. Q = Av a =13 : s§5-= .167 sq.ft. .167% = .hl' ° “5 16.25 or5”. AvH.R.(0)to(A)--12 15 2113111.: 1.35. s a .00015 “‘2 ' ‘ Vol. At B'= .5 cu.ft. A a .5 - .33 .332 a 6” wide 1.“ + 1Q H.R. (A) to (B) a 18 = 1.70 S = .00012 2 Vol. at c - .75 cu.ft. A = .15 = .50 .502 = .707 Use 9" depth 2.0 e 81 H.R. (B) to (C) a 27 = 2.5 s a 0.001 2 Vol. at D a 1.0 cu.ft. Depth 2 (14902 a .807 Use 10" 1’5 3 0 130 .+' .. ell-1's- -3-17 S H.R. (C) to (D) 3 2 Average 81028 = .00005 t g0001 a .0001 + .00015 = ll 0.00005 0.00011 .00011 x 20 s i" fall from end to end of channel. Sludge Division Box: The sludge division box must be so designed so as to distribute sludge from the final tank to the wet well aerator on primary tank in any desired pr0portion. The details of the division box are shown on the tank details plate. Einal SettnggTank: Use a 2% hr. detention period. Use a circular tank. ‘ Use as first assumption 1600 gal/sq.ft. of tank surface. .7808 C.F.3. = 506,000 G.P.D. Surface area “.5%g§%9 = 316 sq.ft. of area. Volume of tank.6 2.5 x 60 x 60 x .7808 = 7020 cu.ft. 10%% = 20+ ft. which is very mush too deep, so try a valae of 850 gal/sq.ft. of surface area based on average flow as given in Metcalf and Eddy. J33??? x. 6116300 :- 336,000 gal/day. Surface area a 396 sq.ft. Volume required is *ngi4&j%?%Q—l—3L5 = #630 C.F. H§%%.- l2+ ft. deep, which is very close to the desirable depth. Assume a 10' depth for convenience. Vol. - 11,300 based on maximum flow. LQ_5EQ§}llflu= 11,300 x 8 3h‘ in diameter the tank should be to give the desired volume. Final SettlinggTank Storage Storage must be provided for fluctuations in flow. The fluctuations in flow will be the maximum and minimum values. Assume a maximum occurs immediately after a -19- minimum and exists in length at least one flowing through period. 150 - 50 = 100% fluctuation of flow flowing through period. Primary 1% hr. Ae rat 0r fir . Total 9% hr. An inspection of flow diagrams for various cities shows that the maximum flow is approximately 1h hours duration. 80 assuming”this condition the volume of sewage flowing through the tank is .7808 x 9.5 x 3600 = 26,676 cu.ft. of sewage, or assuming 95% moisture and 150 P.P.M. removal the volume of sludge is 26L§76_x 7.h8 x 8.39 x 150 x 1.02.2 252 cu.ft., which 1,000,000 may'be conveniently cared for in the bottom of the tank if placed on the Conventional one on twelve lepe. r = 17 %% a 1.42' (Consider as 1' deep) V01. = A2 3.1h x 1/3 = 17 é=lg x 3-lflie 300 cu.ft. of storage Which is o.k. islet: The inlet will be a pipe in the center discharging verticallywand surrounded by a perforated baffle extend- ing two feet below the water surface. Provide a 12” free board in this tank to provide protection from wind currents. - 91:11:21.; The outlet will be an adjustable Wier around the circumference of the tank. Collect the effluent in a -20- channel of uniform size. The effluent to have a velocity of l'/sec. or .7808 = d2 d a 10" wide and 10” deep, but to obtain a better depth make 6” wide and 12“ deep, giving a velocity of 1.5'/sec. Chlorinator: Use a straight through tank 15 min. detention maximum flow of 351 G.P.M. 3 .7808 C.F.S. 351 x 15 3 5250 gal. or 700 C.F. capacity. Use a 5' wide tank 6' deep. L a Z%%.= 23.3' long, which is a long tank, so use two l2’ltanks placed side by side. The flow going around the end acts as a baffle. Baffles should also be placed alternately rising to within 2' of the surface and going down to within 2' of the bottom. Use four baffles. An investigation of the head losses will now be made. 2 At end baffle the head loss is g; ‘ 2 At vertical baffles the head loss is %E 2 Total head loss due to baffles is 5%8 Besides this loss there is also the loss in the channel. Use N a .013 - A - 0 H.R..- W??? - %E'3 1.87 From chart page 130, Metcalf and Eddy, the problem is unsolvable, so solve by the Manning formula. V = AZ§%§.. .026'/sec. _ 2 1 1 86 = - ‘ 026 2- v KRFSB K ' 0.01 11'” S ' (11.511.52) ‘ .00002'. So total loss in this tank is 2 Egg'* 2u(.00002) = 026 . at (.00002) = .076 So use a 1” fall and a cascading Wier to the river as -21- an outlet. Sludge_21gester: BMO P.P.M. suspended solids 50% removal in primary tank. S.G. = 1.02 92% moisture h0% removal in final tank. S.G. a 1.005 98% moisture 75% volatile 30% digestible Primary Tank: 3&0 x .5 = 170 P.P.M. Removal. 17o — 170 (.75)(.3) = 132 P.P.M. per day. 132 x 1,000,000 x 8.3% a llOO#/million gallons 1,000,000 .505 x 1100 = 550#/day V01. a .08x Eége x 62:E = 108 cu.ft. of storage per day needed. Assuming 120 day detention, Vol. = 108.0 cu.ft. x 120 = 12,800 anlem: 40 x 3&0 : 136 P.P.M. removed 136 - 136 (.75)<.3) = 105 P.P.M. - #05 x 1.000.000 x 8.33 x .505 x 50 V°1 ’ 1,000,000 x 1.005 x 62.u ‘ 3 352 C°F°/ day. Vol = 352 x 120 = h2,500 cu.ft. of storage, which is obviously a mistake in reasoning, so assume 0 90% removal 90% moisture content 30% digestible 75% volatile 3' for supernatant -22... 3UO x 9 a 296.0 P.P.M. removal 296.0 - 296 (.3)(.75) = 228 P.P.M. new each day. lho x 120 a 17,h00 cu.ft./day l%%%%.= 7.9 cu.ft. storage per capita, which is well above the minimum of 50 cu.ft/capita standard set up by the Michigan State Health Department. Dimensions of tank: Assume a 25' tank 3.1u r3 d = 17,u00 d a 37% 30%57 = 35' deep, which is unreasonable. So assume a 15' tank with 3' of space allowed for supernatant. .2 15 (31.1%..1.) = 17100 d2 = %£%1% 11180 d a 38.5' Use a 39’ tank 15' deep with a 3' provision for supernatent. ' Use valves of the ring type as used in East Lansing with provisions for drawing the sludge from three different levels. Use a floating bottom with.a safety valve which will go off under a 2' head of water. 0 . Qgsign of_Valgg: .. . 1#31??? 0\ El“ 4:: Assume an 8“ relieve hole, 1' of water - sQ.1n. 2 x .h33 = .866#/sq.in. Total P = .866 x 14” ( 81H X_§E) = 43.7# acting up. 3.11 r2 td a h3.7# Use 0. I. d . h90#/cu.rt. t a h 3 3" thiCK. Use a 3" plate with . x x 28 guides which are easy running and prevent binding, ~23- Gas,Chapbgg: Assume a gas production of l cu.ft./day/capita. 2200 x l = 2200 cu.ft. of gas storage must be pro- vided. General Operating head is 8" - 9“ of water. 8" water a .3u6M#/sq.in. 1“ x 2200 a 2100 cu.ft. of storage must be pro~ 15.0 vided to care for the gas. This means a chamber iallnm_323_é = 2200 d = 2 Provide a waste vent for burning the excess gas. Sludge Beds: Use 1 sq.ft. of sludge bed per capita. 1 x 2200 a 2200 sq.ft. of bed required. Use beds 20' wide. Egggu= 110' long, so use 2 - 55' beds 20‘ wide. The beds are to be of t0p 6" graded sand and 15” of graded gravel. Design of a lflll‘ ' Proving the assumption and empirical rule that mini- mum wall thickness be 12" correct. Design of final settling tank wall. Design when full and empty. lO' sewage and 2‘ free board, total h = 12' P = Egg - éggB-x 10 x 1g = 3120# applied 3.33' from bottom of tank 2 2 Ma 2 3.33 x 3120 = 10,h00 #' M = bKd2 ~2h- ll’ 4—— 23F...- [— s0 = 750 F8 = 18000# N = 15 K = 127 d = (%d)% = (19:02 §2%2)% = 8.99 effective depth + 3" protective covering 8 12.00" A = M - 19§QQ_§_1§______ = .88 sq.in. of steel/ft. s fst - 18000 x g x 8.99 check when tank is empty. 2 . P = E§§Egl Use 1' footing 9‘2] X 10g x 11 X ll 8 .1630# applied.at 3.66' from bottom of tank, which gives a smaller moment than the water. Steel must be placed in the exterior face due to this cause, so 2-16micL66x12 a .. d ' “”'12 x 127 d 6'8 using a 9" effective depth so AS - fs%a . lgogoxx Z6: 9 = 8 .Ohl sq.in. So use %" steel rods spaced 12" on center. Use a 7“, but we are Check bond. u a 2%?5 3 Z x 16%3 x 9 = 262#/sq.in., which is too high. 8 fi So, 20 = 16 O = 1.66 3.11L = 1.66 d = .53 X X ‘ -25- Use 7/8“ circular rode on front face spaced 12" on centers to give corrent bond. Check shear. - v , _g_1630, 3-..... Jd g x l2li'9 Now to return to the steel resisting the water pressure. = 18#/sq.in., which is very safe. AS = .88 sq.in. Try 1' sq.rds. 11“ on centers, giving u a .3120 : # 2 x‘h x Z x 9 92 /sq.in., which is o.k. 8 51-1 v = 3120 = # s .in. which is safe. 12 x,1 x 9 33 / q ’ 8 Design of the footing: Tank is 10' wide so use g 8 10' for extension of footing. Front to have a 6“ extension. CD @ ® X W1: 12 x 1 x 150 = 1500 we: 1 x 11.5 x 150 1730 W3: 62.h x 10 x 10 = 62$; Total 9770# ZMo = O lsgg (1.0) c 1;;0O(5.75} c 62u0 (6.5) = 5,3u' resultant acts from toe. Vertically, reaction cuts base at -26- = 3;20 x hi}? = 1.375 9770 I 5.3h _ 1.35 : 3.96' from toe. e = 5.75 _ 3,95 . 1.76 1%15 = 1.92 allowable. U . 6 ‘3 - a E E ‘1‘ I. + or ( I 7 ) a A j; v 5 ft. which is none too good and should be corrected. N Maximum bending moment occurs at section A - A, .H. 2 Area of shear diagram. A 1 r I B I507£1M D'- a. awb‘ B n 2 (1650 -128) 10 x 10 {_lg . 駓§;lo x 10 _ ' ' 11.5 x 2 x 3 2 ‘1 o2u0 x 5 - 150 (10)(l%) a 12,300#' 1 D = (12100)2 = 9.8" + 3" cover = 13' deep. So increase 127 depth to 13”. The footing could be redesigned, but there is only one flexible dimension and that is the thickness of the footing, so the new thickness would have to do. The footing is designed with a consideration that l. e is too large 2. Footing thickness is not quite thick enough 3, Variation of pressure is too great Design of Tank When Empty: Consider a 5' head of water operating on the tank. The uniform load up is 5 x 62.u - 150 (l) = 162# acting up per lineal foot. 2 Consider as a continuous beam B.M, = WL ' 10 ' 15213_EQQ.: 6280#, which is less than B.M. tank was designed, and so for speed and convenience put the same amount of steel in top and bottom of footing. s a i20881§_:2i_I5.= 0.96 sq.in. per lineal ft. Try 1' sq. rds. 12" on centers. 0 = 2 # a Q 11 a “fizz—m 50 /-q.in., which is too great. Use 2" rods 3“ on centers. u = ETx g7EOZX 10 = 125#/sq.in., which is o.k. s = .1.. 9Z7O :.-. # v de a 12 x g x 10 93 /sq.in., which is too high and should be corrected to a new depth of %g X 10 = 18.5“ to insure safety. If this increase is made the area of the steel may be cut down to 12300 11. : As s g x 18.5 x 12 .53 sq.in. per lineal foot. So try 5" sq. bars — 6" on centers, giving .50 sq.in. -28.. of steel per lineal foot. u a E“x 7 g—Tg a 156#/ sq.in., which is too high. 8 l ‘ . So use g3: sq. bars placed 3" on centers giving :: 9170 122# 1 .a u E_+ 20 x l x 18 8 /sq. n., which i- o.k. 1’6 3 Place temperature steel in wall to .3 of 1% of wall area, 66% in front on exposed wall (tank interior) and 33% in rear wall. Service Building: The service building should be of good pleasing architectural appearance and yet be very functional. The building should have two basements containing: 1. Sub basement — wet well, sewage pumps, slump to provide drainage and a ventilator. 2. Basement - screen room, heating plant, sludge pumps. The main floor should have a good floor plan and con- tain a laboratory, a chlorination room, an office, a lavatory and storage space for mOps, brooms, shovels, etc. The walls should be brick exterior and 2“ glazed wall tile interior. There should be an electrical control panel to cover all the Operations of the plant and many safety features for the chlorination room. The light switches should be on the outside of the room. The doors should swing out from the room and for this reason all of the doors are made to open outward instead of inward. The outside door to the chlorinating room should be of the ventilating type with a mechanical ventilator provided as an -29- additional safety measure. A gas mask should also be pro- vided. The laboratory should be equipped to run chemical analysis and B,O.D. tests, but not too extensively equipped, as the Operator will probably be the type of man who knows very little or nothing.about the technical side of sewage plant operation. Design_of the Heating Unit: Factors to be used. K0 is conductivity of material in B.T.U./sq.ft./hr./°dff. F. fl is the transmissibility coefficient Of inside surface, which will be omitted in the case of a liquid in contact with a surface. fo is the transmissibility coefficient of the internal surface. K is the transmissibility coefficient. Xah 1 _l l + i e 1 f1 K f0 Allowance should be made for wind direction and exposure. Sludge Tank: Use a 12' wall and 6“ bottom and tOp for design purposes. 6" concrete, K = 0.77 12” concrete,K = 0.48 Assume 2' of cinder cover and complete submergence of tank in ground and 1' of air and gas space in the top of the tank. K a _ 1 = .0319 Top loss 3 .0319 x 1160 = 370 B.T.U. per hr. Bottom, K = 35 from lecture notes by F,R, Theroux 80,500 B.T.U./hr. 35 x 1160 Side Walls: K = 20.8 from notes on heat, lecture by Prof. F. R, Theroux. Area 3 18 x 3,1h d = 18 x 3.1” x 39 = 2200 sq.ft. 2200 x 20.8 a #6000 B.T.U. Sludge itself: Flow of 2080 gal./hr. 2080 x 1:921x 8.3h x 3H2 x .95 g # ' 1,000,000 ‘“ 53 /hP- of dry sludge solids at 90% moisture. 10 x 58 . 580# of sludge per hr. Assume #00 F. for temperature of incoming sludge and 85° F. temperature of sludge tank. (85—40)'x 1 x 580 = 25,800 B.T.U./hr. Total B.T.U. lost in tank: Top 370 Bottom 40,500 Sides M6,000 Sludge 25,800 Total 112,670 B.T.U./hr must be provided. The Service Building: 1 The top floor only will be heated. Area - 8.5 x 20 x 20 = 3h00 cu.ft. Exposure factors: North 1.30 East 1.15 South 1.0 West 1.15 -31- Item Area in Coef. Exp.Factor Temp Diff. B.T.U./' sq.ft. 0 8. hr. Windows N 28 1.13 1.30 70 2600 E 31 1.13 1.15 70 2720 s #0 1.13 1.00 70 3160 w 21 1.13 1.15 70 1920 Doors 5 21 .75 1.00 70 1100 E 21 .75 1.15 70 1260 v 21 .75 1.30 70 1&25 Walls N 150 .M6 1.30 70 6300 E 1M8 .h6 1.15 70 5500 s 139 .ue 1.00 70 N800 w 179 .M6 1.15 70 7150 Floor Top too .56 1.00 35 7800 Bottom uoo .62 1.00 20 .3529. Total 50335 Typical computation - a north window 28 x 1.13 x 1.30 x 70 a 2600 B.T.U./hr. Leakage loss = two complete air changes per hour. 2 x 3&00 = 6800 cu.ft./hr. .2h x .075 x l x 70 x 6800 = 8600 B.T.U./hr. Buildhg total = 65.000 B.T.U./hr. Total 8.1.0. required a 65000 + 112670 a 177,670 Piping in Tank: Use a 2.5'/sec. velocity of flow. -32- 1 sq. ft. of pipe is equivalent to 15 B,T.U,/hp/0F Assume a 30° F. drop in temperature. - 112 670 Q - _ -L ___= llahébo = 250 8 .ft. 15 (125-957 15 x 35 ‘1 Assume a 2" pipe. 3W: l""I'.02 .ft. 1,1 3.1“ d 12 3%r_. 5 sq /l nea- ft. of pipe. 259. - 500' of pipe so -_5%Qa.- 500__, = h. 0502 , 3.1 d 3.1—u. x 39‘ 5 times around tank, so go around tank 5 times on a 32' diameter. Heating of Service Building: Total B.T.U./hr required a 50,000 B.T.U./hr. Radiation a E?2T%i"§2%%%£gg’ $2 : 133 T1 - T2 = 65 K - 1.9 - 1.h, use 1.7 1%%£3%5 e 1160 sq.ft. of radiation are required, provide 500 sq.ft. Gas Pipe Size: Maximum production assuming 1 C.F,/cap/day a 2200 C.F./day. 2200 C.F. of gas can be carried in a 3" pipe on a 0.2” drOp in pressure. Determination of Pumping_Head & Tank Elevations: Invert elevation of outfall = 90.63 Ground elevation at primary tank a 105.8 - 2.0 a 103.8 as elevation of inlet wier. Head pumped against a 103.8 - 90.6 = 13.6 static head. Head Losses Pump to Inlet Wier: Loss in a 5" check valve 1.0 E; = 0.25 Loss in a 5“ gate valve 0.5 32% a 0.12 Loss in a 90" elbow .25 g; - 0.06 ~33— Loss in 20' or 5" pipe (250 GPM) = 0.80 Loss in 60' of 8" pipe a 0.15 _Loss in 2. 8”, 90° elbows 0.25 g; - 0.02 Loss in l - 8" tee 1.50 g; a 0.06 Loss in 1 - 8" gate valve 0.5 g; = 0.02 Loss in channel computed - = 0.20 Loss thru inlet wier laggli 3 0‘0“ Total 1.3h A11 losses are computed according to Schoder & Dawson, Page 179. Pipe losses determined by diagram cover page, Babbit and Dolan. Total head pumped against a 13.2 $ 1.33 = 13.5%. Most manufacturers recommend a pump which will deliver against a 1h' head at the desired capacity. Elevation of inlet a 103.8 - 0.2 a 103.6 Loss through Primary Tank: R ._., 3% = 2.79 V = 0.013'/sec. S u 0.0000001 from diagram, Page 307, Schoder and Dawson, which is too small to be considered. 80 this loss will not be computed for the other tanks. Loss Of head over Wier in primary tank using Francis formula ha = __91__ h = 10' Q - .7808 C.F.S. 3 3.33h 2 hg - O 808. 0.02 h (0.02 )2 h - .00 ' 3 3T331x 10 3 35 u 35 3 head is required for this discharge, allowing 0.5' depth of channel and a 0.20' loss through the channel, giving an elevation of 103.6 - 0.5 - 0.2 = 10? Q as the elevation of the outlet pipe to the primary tank. -21. / Loss in pipe between primary & aerator = 0.15 Loss in 2 - 90° - 8" elbows = 0.02 Loss in 1 - 8" check valve = 0.02 Loss in channel (computed) = 0.0hh Loss in entry port : Q‘g§_ Total 0.2995 Elevation Of inlet port invent = 102.9 - 0.30 = 102.6 The loss through the aerator will be considered neglible. Loss over Outlet Wier: L a 20 Q =~0.98 C.F.S. h = (37%3393' a (jjgg‘gggan =_Q.002' head loss over the wier. Allowing an 8" depth Of outlet channel and 0.1' loss through the outlet pipe gives 102.6 - 0.6u . 0.1 = 102.6 - 0.7% = 101.86 as the elevation of the outlet to the aeration tank. Loss in 100' - 8“ pipe = 0.25 Loss in 2 - 900 4 8" elbows : ngg Total friction loss = 0.27 Loss in inlet channel - 0.20 Loss in inlet port %2 = 9‘03 Total loss = 0.51 Elevation of inlet to final tank a 101.86 - 0.51 a 101.35. Loss over outlet Wier has been shown in the previous tanks to be so small that the i 0.1' adjust— ment provided in the wier will easily take care of any variation which arises. Using a 10" channel and;9-9§ loss in the channel the elevation of the outlet to the final tank which is also the inlet to the chlorinator is: 101.35 - 0.81 - 0.05 = 100.h9 The head loss through the chlorination chamber has been computed to be 0.08', so elevation of outlet to Chlorinator is _ 100.19 - 0.08 = 100.80. Total loss through plant is 103.8 - 100.h = 3.h' which is consistent with.a good economy Of head. Estimate of Costs: The estimate or cost will be in two parts. The first will be an estimate Of concrete work and service build- ing. The second part will be an estimate Of the mechanical equipment, such as pumps, piping, skimmers, etc. The cost of concrete and forms is based on a value Obtained from Arthur Sargent, estimator for the Michigan State Parks Division. The cost of mechanical equipment is based on values obtained from Professor F. R. Theroux. Excavation: Ground Elev.~corners V01 . Aver— Base cu. Item N,E, S.W. S.E. N.W. age Elev. Diff.Size yd. Cost Service . Bldg. 106.2 106.5 106.h 106.3 106.u 89.h 17.0 20x20 252 215.00 Primary Tank 105.1 105.8 105.7 105.2 105.5 97.0 8.5 10x1u0 52 uu.30 Aeration Tank 10u.3 10u.7 10u.6 10h.5 10u.5 90.5 1u.0 20x120 89 76.00 Final Tank 10u.2 10u.h 10u.3 10u.3 101.3 91.8 12.5 3u' #22 358.00 Chlor- inator 10h.3 10h.3'10n.3 10u.3 10h.3 97,3 7,0 12x12 33 32.50 Dist. Box 1o’1.8 101;.7 1014.7 101+.8 1011.7 101.7 3.0 11x6 1 0.85 Sludge Dig. 10n.9 105.0 105.1 10u.9 105.0 86.5 19.5 20' 230 196,00 Sludge Drying ' Beds . 103.2 103.8 103.9 103.3 103.0 103.0 0.6 20x110 5 h.25 Ditching 2200 lineal ft. at an average depth of 6'5 #90 u15.00 Total $1,3Mo.90 Based on a value of $0.85/cu.yd. as advised by A,L. Sargent. ' Service building figured on a basis of $1.00/cu.ft. 8 x 20 x 20 = $3#00 for building and laboratory equipment such as shelves, racks, special sinks, ovens, etc. Concrete: Concrete based on $6.50/cu.yd. Item Length' Width' Depth' Vol.1n cu.yd. Cost Service Bldg. 1. 3 floors 20 20 0.5 22.? $lu5.00 2. Walls 20(5) 18.5 1.0 69.0 uu5,oo .1 / 7.. Item Length' Width' Depth' 3. Footings 80 3 h. Wet Well 60 10 Primary Tank 1. Walls 3(70) 8.5 2. End wa11e 2(23) 8.5 3. Bottom 10 1&0 h. Footings 3 256 5. Hoppers 5 5 Aerator 1. Walls 60(2) 1h 2. walls #0 1h 3. Walls “0(3) 1h h. Wall 20 1h 5, Bottom 20 120 6. Footings 260 3 Final Tank 1. Wall 107 12.5 2. Bottom (17 x 3.142) 3, Footings 107 3.0 Chlorinators 1. wa11e 12(3) 7.0 2. wa11e 10(3) 5.0 3. Cascade 10 5 h. Footings #6 3 Sludge Division Box 1. Walls 6(2) 3 2. Walls h(2) 3 3. Bottom h 6 1.0 0.5 1.0 1.0 0.5 1.0 0.5 1.0 1.0 1.0 1.0 a .J 1.0 O 1.0 0.5 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 Vol.1n cu.yd. 8.8 11.1 66.0 1h.5 25.80 28.5 2.0 62 21 63 11 #6 WNNKO Cost 3 56.10 71.80 “29.00 9h.50 167.50 186.00 13.00 hoh.00 136.50 M10.0o 71.50 300.00 188.50 325.00 110.50 78.00 58.50 13.00 13.00 32.50 6.50 6.50 6.50 Item Length' Width' Depth' Vol.in cu.yd. Cost . h. Footings 20 3 11.0 3 $ 19.50 Sludge Digester 1. Wall 63 19.5 1.0 05 293.00 2. Cover (102)3.1u 0.5 6 39.00 3, Bottom (102)3.1u 0.5 6 39.00 u, Footing 63 3.0 1.0 7 1*5.50 Sludge Drying Beds 1, walls llo<3> 1.0 0.5 6 '39.00 2. Footings 330 1.0 0.5 6 39.00 Total for concrete u,180.80 Steel: Steel to be computed at 35¢ per pound in place. The weight of steel to be 2#/sq.ft. of area as recommended by Arthur Sargent. Area of concrete = 13312.0 sq.ft. 0 area of tank. Bottom = 6800 Total area = 20,120 sq.ft. 20,120 x 2 = 00,200# of steel to be used. 00,200 x 3.5 = $1,408.H0, cost of steel in place. Form Work at 10¢ per sq.ft.: Item Length Depth Area sq.ft.(2 sides) Cost Service Bldg. 1, Walls 20(5) 18.5 3700.0 $370.00 2. Wet Well ' (3 sides 2 wells) 20 10 '600.0 60.00 3, Footings' 80 1.0 160.00 16.00 Primary Tank 1. 3 walls 70 8.5 3580 358.00 2. 3. 4. Item 2 Walls Footings Inlet & Outlet Aerator QWPWMH Walls Wall Wall Wall Footings Outlet & Inlet Length 20 250 00 2(60) 20 260 80 Final Settling Tank 1. 2. Wall Footings 3. Motor Mount h. Outlet & Inlet Chlorinator l. U'IPWN Walls Walls Cascade Footings Outlet & Inlet 107 210 u 107 12(3) 10(3) 70 M6 10 Sludge Division Box 1. 2. 3. Walls Walls Footing 6 u no Depth 8.5 1.0 1.0 11) 11) 14 12.5 1.0 10 1.0 1.0 1.0 1.0 Area sq.ft.(2 sides) 680 500 80 Cost 3 68.00 50.00 8.00 336.00 112.00 536.00 56.00 52.00 16.00 268.00 21.140 16.00 21.00 50.h0 30.00 7.00 9.20 2.00 7.20 h.80 h.00 Item Length Depth Area sq.ft.(2 sides) Cost Sludge Digester 1. Wall 63.0 19.5 2520 $252.00 2. Cover 31% 3M” 3l.U0 3, Footings 126 1.0 126 12.60 0. Control Box 20 1.0 “0 ”.00 Sludge Beds 1. 3 Walls 330 1.0 660 66.00 2. 3 Footings 330 1.0 660 66.00 Total 32,662.u0 Mechanical Equipment Required: Service Bldg. 250 G.P.M. pump to deliver against a 14' head 3 350.00 200 0,2.m. pump to deliver against a in. head u00.00 150 G,P.M. pump to deliver against a 10' head h00.00 100' - 10" pipe 0 $.05/# 390.00 3 - a" bend 0 10¢/# 2M.60 3 - 30" pieces 4“ pipe @ 6¢/# 8.00 3 - u" gate valves 0 $20.00 60.00 3 — 5" gate valves @ $20.00 60.00 3 - 5" check valves 0 $35.00 105.00 3 — 30“ pieces 5“ pipe @ 6¢ 10.50 u a 5" bends 0 10¢/# ' 35.00 2 - 5" tees @ 6¢/# 2h.00 120' - 5" pipe @ 611/”é 225.00 2 — 5" gate valves 0 $20.00 “0.00 2 - mechanical sludge removers 70' x 10' x 8' @ $1000 ZSMLDO 150' ~ 8" pipe 0 5¢/# 300.00 2 — 25 G,P.M. sludge pump @ 3350.00 1500.00 -01.. 6 - 8" bends 0 10¢/# 0 120.00 200' 8" pipe o 5¢/# . 000.00 2 - 8" tees 0 10¢/# 60.00 2 e 8" crate valves 0 $50.00 100.00 200' - h“ pipe 0 6¢/# 17h.00 h - 90° a h“ bends @ 10¢/# . 32.00 5 - aerators, 18000 C.F. circulation @ $1000.00 5000.00 100' - 8" pipe 0 5¢/# ' ' 200.00 2 - 8” bends 0 10¢/# _ 00.00 1 _ 30' mechanical sludge remover 800.00 100' — 8“ pipe 0 5¢/# 200.00 300' e 8” pipe o 6¢/# 180.00 50' a 8" pipe 0 5¢/# 100.00 500' - 2" pipe 0 6¢/# 300.00 1000' - 8" pipe 0 5¢/# 2000.00 h - S” bends 0 10¢/# 80.00 1 - 8“ tee 0 10¢/# . 30.00 1 - 10 G.P.M. hot water pump 30.00 Cover, sludge digester 1500.00 1 Boiler (150,000 B.T.U. Cap) 600.00 Flame Trap, Pressure Relief Valves, etc. 200.00 1 automatic flow recorder 350.00 1 Chlorinator & connections 1200 00 Total $18,338.50 Estellpsstlsf P1ant= ‘ Service Building ~ 0 3,h00.00 Concrete including steel 5,650.80 Form work 2,662.00 -02- lExcavation Mechanical Equipment Engineer‘s fees (design, supervision, etc.) Total Qapitalized Cost: Page 20, Babbit & Dolan CC.=C$Q_*D ' r C a first cost = $3h,500 t 1,3h0.80' 18,338.50 $31.392.50 ..112112.25 $3”.531.75 o 3 operating cost a $S/million gal. a 0.503 x 360 x g a $1, 2.010/year. r a rate of interest per period a 3% C = 38509 D = depreciation = Fff:_FTC3 Cl QI)JU' 1 = D j ‘- 1£5%9.. 21300 1. 2 0.0. a 21300 . lhho . 30500 = $57,200 Annual Expense: A 2 Cr & 0 Cr p *Il-rr -1 Ap a 1400 a 1032 + 638 = $3110 annual expense. BIBLIOGRAPHY H. L. Frost - "Design of a Sewerage System for East Jordan, Michigan" Professor F. R. Theroux - Lecture Notes & Interview E. J, Postal - sanitation engineer, State Parks Division A. C. 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