DESIGN OFA RAPID-MIX BASIN AND FLOCCULATION BASIN

DESIGN OFA RAPID-MIX BASINAND FLOCCULATION BASIN 1.0 Mdg 3785 m3/d of equalized secondary effluent from a municipal waste water ment facility is to receive tertiary treatment through a d

DESIGN OFA RAPID-MIX BASIN

AND FLOCCULATION BASIN

1.0 Mdg (3785 m3/d) of equalized secondary effluent from a municipal waste water ment facility is to receive tertiary treatment through a direct filtration process which in-cludes rapid mix with a polymer coagulant, flocculation and filtration Size the rapid mixand flocculation basins necessary for direct filtration and determine the horsepower of therequired rapid mixers and flocculators

treat-Calculation Procedure:

1 Determine the required volume of the rapid mix basin

A process flow diagram for direct filtration of a secondary effluent is presented in Fig 6.This form of tertiary wastewater treatment is used following secondary treatment when anessentially "virus-free" effluent is desired for wastewater reclamation and reuse

The rapid mix basin is a continuous mixing process in which the principle objective is

to maintain the contents of the tank in a completely mixed state Although there are merous ways to accomplish continuous mixing, mechanical mixing will be used here Inmechanical mixing, turbulence is induced through the input of energy by means of rotat-ing impellers such as turbines, paddles, and propellers

nu-The hydraulic retention time of typical rapid mix operations in wastewater treatmentrange from 5 to 20 seconds A value of 15 seconds will be used here The required volume

of the rapid mix basin is calculated as follows:

Volume (V) = (hydraulic retention time)(wastewater flow)

V= (I* s)(l x 10* gal/day) = 1?4 m 24 ft3 (0>68 m3)

86,400 s/d

2 Compute the power required for mixing

The power input per volume of liquid is generally used as a rough measure of mixing fectiveness, based on the reasoning that more input power creates greater turbulence, andgreater turbulence leads to better mixing The following equation is used to calculate therequired power for mixing:

where G = Mean velocity gradient (s l )

P = Power requirement (ft-lb/s) (kW)

fju = Dynamic viscosity (lb-s/ft2) (Pa-s)

V= Volume of mixing tank (ft3) (m3)

G is a measure of the mean velocity gradient in the fluid G values for rapid mixing

oper-ations in wastewater treatment range from 250 to 1,500 s"1 A value of 1,000 s"1 will beused here For water at 6O0F (15.50C), dynamic viscosity is 2.36 x IQ-5 lb-s/ft2 (1.13 x10~3 Pa-s) Therefore, the required power for mixing is computed as follows:

P = G 2 IJiV= (1,000 s~1)2(2.36 x 10-5 Ib-s/ft2)(24 ft3) = 566 ft-lb/s

= 1.03 horsepower (0.77 kW)

Use the next largest motor size available = 1.5 horsepower (1.12 kW) Therefore, a 1.5horsepower (1.12 kW) mixer should be used

3 Determine the required volume and power input for flocculation

The purpose of flocculation is to form aggregates, or floes, from finely divided matter.The larger floes allow a greater solids removal in the subsequent filtration process In thedirect filtration process, the wastewater is completely mixed with a polymer coagulant inthe rapid mix basin Following rapid mix, the flocculation tanks gently agitate the waste-water so that large "floes" form with the help of the polymer coagulant As in the rapidmix basins, mechanical flocculators will be utilized

For flocculation in a direct filtration process, the hydraulic retention time will rangefrom 2 to 10 minutes A retention time of 8 minutes will be used here Therefore, the re-quired volume of the flocculation basin is

F= (Sn1In)(I x 1O^ gal/day) = a l ? 4 3 f t 2 l

l,440mm/d

G values for flocculation in a direct filtration process range from 20 to 100 s"1 A value of

80 s"1 will be used here Therefore, the power required for flocculation is

P = G 2 VV= (80 s~1)2(2.36 x 10-5)(743 ft3)

= 112 ft-lb/s = 0.2 horsepower (0.15 kW)Use the next largest motor size available = 0.5 horsepower (0.37 kW) Therefore, a 0.5horsepower (0.37 kW) flocculator should be used

It is common practice to taper the energy input to flocculation basins so that floes tially formed will not be Broken as they leave the flocculation facilities In the above ex-ample, this may be accomplished by providing a second flocculation basin in series withthe first The power input to the second basin is calculated using a lower G value (such as

ini-50 s"1) and hence provides a gentler agitation

Related Calculations If the flows to the rapid mix and flocculation basin vary

significantly, or turn down capability is desired, a variable speed drive should be providedfor each mixer and flocculator The variable speed drive should be controlled via an out-put signal from a flow meter immediately upstream of each respective basin

It should be noted that the above analysis provides only approximate values for mixerand flocculator sizes Mixing is in general a "black art," and a mixing manufacturer isusually consulted regarding the best type and size of mixer or flocculator for a particularapplication

SIZING A POLYMER DILUTION/

FEED SYSTEM

1.0 Mgd (3,785 m3/d) of equalized secondary effluent from a municipal wastewater ment facility is to undergo coagulation and flocculation in a direct filtration process Thecoagulant used will be an emulsion polymer with 30 percent active ingredient Size thepolymer dilution/feed system including: the quantity of dilution water required, and theamount of neat (as supplied) polymer required

treat-Calculation Procedure:

1 Determine the daily polymer requirements

Depending on the quality of settled secondary effluent, organic polymer addition is oftenused to enhance the performance of tertiary effluent filters in a direct filtration process:

see Design of a Rapid Mix Basin and Flocculation Basin Because the chemistry of the

wastewater has a significant effect on the performance of a polymer, the selection of atype of polymer for use as a filter aid generally requires experimental testing Commontest procedures for polymers involve adding an initial polymer dosage to the wastewater(usually 1 part per million, ppm) of a given polymer and observing the effects Dependingupon the effects observed, the polymer dosage should be increased or decreased by 0.5ppm increments to obtain an operating range A polymer dosage of 2 ppm (2 parts poly-mer per 1 x 106 parts wastewater) will be used here

In general, the neat polymer is supplied with approximately 25 to 35 percent activepolymer, the rest being oil and water As stated above, a 30 percent active polymer will beused for this example The neat polymer is first diluted to an extremely low concentrationusing dilution water, which consists of either potable water or treated effluent from thewastewater facility The diluted polymer solution usually ranges from 0.005 to 0.5 per-cent solution The diluted solution is injected into either a rapid mix basin or directly into

a pipe A 0.5 percent solution will be used here

The gallons per day (gal/day) (L/d) of active polymer required is calculated using thefollowing:

Active polymer (gal/day) = (wastewater flow, Mgd)

x (active polymer dosage, ppm)Using the values outlined above:

Active polymer = (1.0 Mgd)(2 ppm) = 2 gal/day active polymer (pure polymer)

= 0.083 gal/hr (gal/h) (0.31 L/h)

2 Find the quantity of dilution water required

The quantity of dilution water required is calculated using the following:

^M , , „ , active polymer, gal/h

Dilution water (gal/h) =

————^—^——*—.—-% solution used (as a decimal)Therefore, using values obtain above:

Dilution water = — —— = 16.6 gal/h (62.8 L/h)

0.005

3 Find the quantity of neat polymer required

The quantity of neat polymer required is calculated as follows:

, , „ N active polymer, gal/h

% active polymer in emulsion as suppliedUsing values obtained above:

Neat polymer = 0'083 gal/h = 0.277 gal/h (1.05 L/h)

This quantity of neat polymer represents the amount of polymer used in its "as supplied"form Therefore, if polymer is supplied in a 55 gallon (208.2 L) drum, the time required touse one drum of polymer (assuming polymer is used 24 h/d, 7 d/wk) is:

55 galTime required to use one drum of polymer = ————— = 200 h = 8 days

assuming the efficiency (E) of each stage is the same.

Calculation Procedure:

1 Find the efficiency of the trickling filters

The modern trickling filter, shown in Fig 7, consists of a bed of highly permeable

medi-um to which microorganisms are attached and through which wastewater is percolated ortrickled The filter media usually consists of either rock or a variety of plastic packing ma-terials The depth of rock varies but usually ranges from 3 to 8 feet (0.91 to 244 m) Trick-ling filters are generally circular, and the wastewater is distributed over the top of the bed

by a rotary distributor

Filters are constructed with an underdrain system for collecting the treated wastewaterand any biological solids that have become detached from the media This underdrain sys-tem is important both as a collection unit and as a porous structure through which air cancirculate The collected liquid is passed to a settling tank where the solids are separatedfrom the treated wastewater In practice, a portion of the treated wastewater is recycled todilute the strength of the incoming wastewater and to maintain the biological slime layer

in a moist condition

The organic material present in the wastewater is degraded by a population of croorganisms attached to the filter media Organic material from the wastewater is ab-sorbed onto the biological slime layer As the slime layer increases in thickness, the mi-croorganisms near the media face lose their ability to cling to the media surface The

mi-FIGURE 7 Cutaway view of a trickling filter (Metcalf & Eddy, Wastewater

Engineer-ing: Treatment, Disposal, and Reuse, 3rd Ed., McGraw-Hill.)

liquid then washes the slime off the media, and a new slime layer starts to grow The nomenon of losing the slime layer is called "sloughing" and is primarily a function of theorganic and hydraulic loading on the filter

phe-Two possible process flow schematics for a two-stage trickling filter system areshown in Fig 8

The NRC equations for trickling filter performance are empirical equations which areprimarily applicable to single and multistage rock systems with recirculation

The overall efficiency of the two-stage trickling filter is calculated using:

influent BOD5 - effluent BOD5Overall efficiency = — x 100

ClarifierEffluent

Recycle RecycleInfluent

Also, overall efficiency = El + E 2 (I -E 1 ), and E 1 =E 2

where E1 = The efficiency of the first filter stage, including recirculation and settling (%)

E 2 = The efficiency of the second filter stage, including recirculation and settling

(%)

Substituting E1 for E2 , setting up as a quadratic equation, and solving for E 1 :

E\ - 2E 1 + 0.917 = O = E1 = 0.712 or 71.2%

Therefore, the efficiency of each trickling filter stage is 71.2 percent

2 Analyze the first stage filter

For a single stage or first stage rock trickling filter, the NRC equation is

100

El= 1+0.0561 /E

V VF where W= BOD5 loading to the filter, Ib/d (kg/d)

V= Volume of the filter media, 103 ft3 (m3)

F= Recirculation factor

2a Compute the recirculation factor of the filter

Recirculation factor represents the average number of passes of the influent organic ter through the trickling filter The recirculation factor is calculated using

mat-F= 1 +*[1 + (fl/10)]2

where R = Recirculation ratio = QJQ

Q r = Recirculation flow

Q = Wastewater flow

Using values from above, the recirculation factor is

F= _ J L ± 2 _ _=2 0 8[1+(2/1O)]2

2b Compute the BOD 5 loading for the first stage filter

The BOD5 loading for the first stage filter is calculated using

W - (Influent BOD5, mg/L)(Wastewater flow, Mgd)(8.34 Ib/Mgal/mg/L)

Using values from above, the BOD5 loading for the first stage filter is

W= (240 mg/L)(1.0 Mgd)(8.34) - 2,002 Ib BOD5/d (908.9 kg/d)

2c Compute the volume and diameter of the first stage filter

Therefore, the volume of the first stage trickling filter is calculated as follows:

71.2 = 1QQ — V= 18.51 103 ft3 (523.8 m3)

1+0.0561 /2,2002 Ib/d

V F(2.08)Using the given depth of 7 feet (2.1 m), and a circular trickling filter, the area and diame-ter of the first stage filter are

Area=^^ 18.51 x l03 f t 3= 2 > 6 4 4 f t 2 diameter=5g-02 ft(177m)

depth 7 ft

3 Analyze the second stage filter

The BOD5 loading for the second stage trickling filter is calculated using

W = (I-E1)W

where W = BOD5 loading to the second stage filter

W=(I- 0.712)(2,002 Ib/d) = 577 Ib BOD5/d (261.9 kg/d)

The NRC equation for a second stage trickling filter is

Area = 64'33 * 1^ ft3 = 9,190 ft2 diameter - 108.17 ft (32.97 m)

7 ft

4 Compute the BOD 5 loading and hydraulic loading to each filter

The BOD5 (organic) loading to each filter is calculated by dividing the BOD5 loading bythe volume of the filter in 103 ft3 (m3):

First stage filter: BOD5 loading = Jffig^ = 108.2 -^

(1.74kg/m3-d)

Second stage filter: BOD5 loading = J™^ = 8.97 -^

(0.14kg/m-d)

BOD5 loading for a first stage filter in a two stage system typically ranges from 60 to

120 lb/103 ft3-d (0.96 to 1.93 kg/m3-d) The second stage filter loading typically rangesfrom 5 to 20 lb/103 ft3-d (0.08 to 0.32 kg/m3-d)

The hydraulic loading to each filter is calculated as follows:

(1 + R)(0Hydraulic loading = — —T-TT

(Area)(l ,440mm/d)First stage filter:

(1+2)(1 x 106 gal/day)Hydraulic loading = —: ——

y 6 (2,644 ft2)( 1,440 min/d)

= 0.79 gal/min-ft2 (0.54 L/s-m2)Second stage filter:

(1+2)(1 x 106 gal/day)Hydraulic loading = —: :

J (9,190 ft2)( 1 ,440 mm/d)

= 0.23 gal/min-ft2 (0.156 L/s-m2)Hydraulic loading for two stage trickling filter systems typically ranges from 0.16 to 0.64gal/min-ft2 (0.11 to 0.43 L/s-m2)

Related Calculations In practice, the diameter of the two filters should be

rounded to the nearest 5 ft (1.52 m) to accommodate standard rotary distributor nisms To reduce construction costs, the two trickling filters are often made the same size.When both filters in a two stage trickling filter system are the same size, the efficiencieswill be unequal and the analysis will be an iterative one

mecha-DESIGN OFA PLASTIC MEDIA

TRICKLING FILTER

A municipal wastewater with a flow of 1.0 Mgd (694 gal/min) (3,785 m3/d) and a BOD5

of 240 mg/L is to be treated in a single stage plastic media trickling filter without recycle.The effluent wastewater is to have a BOD5 of 20 mg/L Determine the diameter of the fil-ter, the hydraulic loading, the organic loading, the dosing rate, and the required rotationalspeed of the distributor arm Assume a filter depth of 25 feet (7.6 m) Also assume that atreatability constant (k20/2o) °f 0.075 (gal/min)0 5/ft2 was obtained in a 20 foot (6.1 m)high test filter at 2O0C (680F) The wastewater temperature is 3O0C (860F)

Calculation Procedure:

1 Adjust the treatability constant for wastewater

temperature and depth

Due to the predictable properties of plastic media, empirical relationships are available topredict performance of trickling filters packed with plastic media However, the treatabil-

ity constant must first be adjusted for both the temperature of the wastewater and thedepth of the actual filter.

Adjustment for temperature The treatability constant is first adjusted from the given

standard at 2O0C (680F) to the actual wastewater temperature of 3O0C (860F) using thefollowing equation:

^30/2O = £20/20 0r~2°

where ^30720 = Treatability constant at 3O0C (860F) and 20 foot (6.1 m) filter depth

Ar20720= Treatability constant at 2O0C (680F) and 20 foot (6.1 m) filter depth

6 = Temperature activity coefficient (assume 1.035)

T = Wastewater temperature

Using above values:

£30/20 = (0.075 (gal/min)0 VfF)(1.035)30-20 = 0.106 (gal/min)0 5/ft2

Adjustment for depth The treatability constant is then adjusted from the standard depth of

20 feet (6.1 m) to the actual filter depth of 25 feet (7.6 m) using the following equation:

£30/25 = £30/20(^1/^2)*

where &30725 = Treatability constant at 3O0C (860F) and 25 foot (7.6 m) filter depth

£30/20 = Treatability constant at 3O0C (860F) and 20 foot (6.1 m) filter depth

D1 = Depth of reference filter (20 feet) (6.1 m)

D 2 = Depth of actual filter (25 feet) (7.6 m)

jc = Empirical constant (0.3 for plastic medium filters)

Using above values:

£30/25 = (0.106 (gal/mm)0-5/ft2)(20/25)°-3

= 0.099 (gal/min)0 5/ft2 (0.099 (L/s)° 5/m2)

2 Size the plastic media trickling filter

The empirical formula used for sizing plastic media trickling filters is

^=exphftD(a,n^i

where S e = BOD5 of settled effluent from trickling filter (mg/L)

S 1 = BOD5 of influent wastewater to trickling filter (mg/L)

&20 = Treatability constant adjusted for wastewater temperature and filter depth =(£30/25)

n = Empirical constant (usually 0.5)

Rearranging and solving for the trickling filter area (A):

/ -In(S IS ) \ i/n

™ (kw2s)D )

Using values from above, the area and diameter of the trickling filter are

A = 694 gal/min ( ^^ft J^ = 6"'6 ff 29'9 ft (9>1 m)

3 Calculate the hydraulic and organic loading on the filter

The hydraulic loading (QIA ) is then calculated:

Hydraulic loading = 694 gal/min/699.6 ft2 - 0.99 gal/min-ft2 (0.672 L/s-m2)For plastic media trickling filters, the hydraulic loading ranges from 0.2 to 1.20gal/min-ft2 (0.14 to 0.82 L/s-m2)

The organic loading to the trickling filter is calculated by dividing the BOD5 load tothe filter by the filter volume as follows:

(1.0 Mgd)(240 mg/L)(8.34 lb-L/mg-Mgal)Organic loading = (699.6 ft2)(25 ft)(103 fWlOOO ft3)

= 114 To^Fd (557kg/m2'd)For plastic media trickling filters, the organic loading ranges from 30 to 200 Ib/103

ft3-d (146.6 to 977.4 kg/m2-d)

4 Determine the required dosing rate for the filter

To optimize the treatment performance of a trickling filter, there should be a continualand uniform growth of biomass and sloughing of excess biomass To achieve uniformgrowth and sloughing, higher periodic dosing rates are required The required dosing rate

in inches per pass of distributor arm may be approximated using the following:

Dosing rate = (organic loading, lb/103 ft3-d)(0.12)

Using the organic loading calculated above, the dosing rate is:

Dosing rate = (114 lb/103 ft3-d)(0.12) = 13.7 in/pass (34.8 cm/pass)

Typical dosing rates for trickling filters are listed in Table 2 To achieve the typicaldosing rates, the speed of the rotary distributor can be controlled by (1) reversing the lo-cation of some of the existing orifices to the front of the distributor arm, (2) adding re-versed deflectors to the existing orifice discharges, and (3) by operating the rotary distrib-utor with a variable speed drive

5 Determine the required rotational speed of the distributor

The rotational speed of the distributor is a function of the instantaneous dosing rate andmay be determined using the following:

= L6(6r)

" (WPK)

TABLE 2 Typical Dosing Rates for Trickling Filters

Organic loading Ib BOD 5 /! O 3 Dosing rate,

ft 3 -d (kg/m 2 -d) in/pass (cm/pass)

<25 (122.2) 3 (7.6) 50(244.3) 6(15.2) 75(366.5) 9(22.9)

100 (488.7) 12 (30.5) 150(733.0) 18(45.7)

200 (977.4) 24 (60.9)

(Wastewater Engineering: Treatment, Disposal, and Reuse, Metcalf

& Eddy, 3rd Ed.)

where n = Rotational speed of distributor (rpm)

Q T = Total applied hydraulic loading rate (gal/min-ft2) (L/s-m2) = Q + QR

Q = Influent wastewater hydraulic loading rate (gal/min-ft2) (L/s-m2)

Q R = Recycle flow hydraulic loading rate (gal/min-ft2) (L/s-m2) Note: recycle is sumed to be zero in this example

as-A = Number of arms in rotary distributor assembly

DR = Dosing rate (in/pass of distributor arm)

Assuming two distributor arms (two or four arms are standard), and using values fromabove, the required rotational speed is:

1.6(0.99 gal/min-ft2

n = ,^,^n- / T" = °-058 1(2)(13.7 in/pass) P111This equates to one revolution every 17.2 minutes

SIZING OFA ROTARY-LOBE SLUDGE PUMP

A municipal wastewater treatment facility produces approximately 6,230 gal/day (23.6

m3/d) of aerobically digested sludge at 5 percent solids The digested sludge is pumped to

a sludge dewatering facility which is operated 4 h/d, 7 d/wk; 1000 ft (304.8 m) of 3 in(76.3 mm) equivalent length pipe exists between the aerobic digester and the dewateringfacility The equivalent length of pipe includes all valves, fittings, discharge pipe, andsuction pipe lengths in the sludge piping system The static head on the sludge pump is 10feet (3.0 m) Size a rotary-lobe pump transfer sludge from the digester to the dewateringfacility, including pump discharge and head condition, rpm and motor horsepower (kW)

Calculation Procedure:

1 Find the flow rate required for the sludge pump

A schematic of the sludge handling system is illustrated in Fig 9 Rotary-lobe pumps arepositive-displacement pumps in which two rotating synchronous lobes push the fluidthrough the pump Although these types of pumps are advertised as self-priming, they are

FIGURE 9 Sludge handling system.

generally located so as to have a suction head as shown in Fig 9 A schematic of a rotarylobe pump is shown in Fig 10

The required flow rate for the sludge pump using the 4 h/d, 7 d/wk operation schemeis:

6,230 gal/dayFlow rate (gal/min) = (4 ^60 ^ - 26 ga./min (1.64 L/s)

2 Compute the headless in the piping system

The head loss through the piping system is calculated using the Hazen-Williams formula:

/ 100 \i.85/ Q185 \i L \

where H = Dynamic head loss for clean water, ft (m)

C = Hazen-Williams constant (use 100)

Q = Flow rate in pipe, gal/min (L/s)

D = Pipe diameter, in (mm)

L = Equivalent length of pipe, ft (m)

Note: There is an SI version of the Hazen-Williams formula; use it for SI calculations

DISCHARGE

SUCTION FIGURE 10 Rotary-lobe pump schematic.

Aerobic

Digester Rotary LobeSludge Pump SludgeHolding

Tank

Sludge DewateringFacility (Centrifuge) To Off SiteDisposal3" Pipe

10 feet (3m)

Using values given above, the dynamic head loss in the piping system at the designflow rate of 26 gal/min (1.64 L/s) is:

/ 100 \i.85/ (26 gal/min)1-85 \ / 1000 ft \

H ~ (°-2083)(^) ( oV** X-IOT) = 4'12 A (L26 m)

This represents head loss in the sludge piping system for clean water only To

deter-mine head loss when pumping sludge at 5 percent solids, a multiplication factor (k) is used The value of k is obtained from empirical curves (see Fig 11) for a given solids

content and pipeline velocity The head loss when pumping sludge is obtained by plying the head loss of water by the multiplication factor

multi-The velocity of 26 gal/min (1.64 L/s) in a 3 in (76.3 mm) diameter pipe is

Velocity, m/s

Velocity, ft/s

FIGURE 11 Head-loss multiplication factor for

dif-ferent pipe velocities vs sludge concentration

(Met-calf & Eddy, Wastewatwer Engineering: Treatment,

Disposal and Reuse, 3rd Ed., McGraw-Hill.)

26 gal/minVel°dty = (7.48 gal/tfX(^4X3 in/12 in/ft)')(60 s/min) = L2 *" (0'37 "^Using Fig 11 with a velocity of 1.2 ft/s (0.37 m/s) and a solids content of 5 percent, the

multiplication factor (k) is 12 Therefore, the dynamic head loss for this system when

pumping 5 percent solids is:

Dynamic Head Loss5o/0 = (Dynamic Head Losswater)(&)

Dynamic Head Loss5o/0 = (4.12')(12) = 49.44 feet (15.1 m)

Use 50 feet of head loss (15.2 m)The total head loss is the sum of the static head and the dynamic head Ioss5o/0 Therefore,the total head loss (Total Dynamic Head or TDH) for the system is

10'+ 50'= 60 feet (18.3m)This translates to a discharge pressure on the pump of

TDH = (60')/(2.31 ft/psi) = 26 IMn2 (179.1 kPa)Therefore, the design condition for the rotary lobe pump is 26 gal/min (1.64 L/s) at 26lb/in2 (179 IkPa)

3 Choose the correct pump for the application

At this point, a rotary lobe pump manufacturers catalog is required in order to choose thecorrect pump curve for this application This is accomplished by choosing a pump per-formance curve that meets the above design condition An example of a manufacturerscurve that satisfies the design condition is shown in Fig 12

Plotting a horizontal line from 26 gal/min (1.64 L/s) on the left to the 26 lb/in2 (179.1kPa) pressure line and reading down gives a pump speed of approximately 175 rpm Thismeans that for this pump to deliver 26 gal/min (1.64 L/s) against a pressure of 26 lb/in2(179.1 kPa), it must operate at 175 rpm

The motor horsepower required for the rotary lobe pump is calculated using the lowing empirical formula (taken from the catalog of Alfa Laval Pumps, Inc of Kenosha,WI):

fol-W r (NKSf) (?)(#/)!

HP= ^- [(0.043X^)(P)+ ^- + f^]

Hp = Motor horsepower

Af= Pump speed (rpm)

q = Pump displacement, gal/100 revolutions (L/100 revs)

P = Differential pressure or TDH, psi, (kPa)

Sf= Factor related to pump size which is calculated using: Sf= (#)(0.757) + 3 Nf= Factor related to the viscosity of the pumped liquid which is calculated using Nf= 2.2 ^/Viscosity (cp)

The pump speed was found in Fig 12 to be 175 rpm; the pump displacement is takenfrom the pump curve and is 25 gal/100 revs (94.6 L/100 revs); the differential pressure or