Let it be required to determine the nozzle layout for one acre of sprinkling filters with 5 feet available head on the nozzles.
The selection of the type of nozzle and the size of opening is a matter of judgment and experience. Nozzles with large openings are less liable to clog and fewer nozzles are needed than where small nozzles are used, but the distribution of sewage is not so even as with the use of small nozzles. In this example Taylor circular spray nozzles will be selected. Fig. 170 shows that a Taylor circular spray nozzle will discharge 22.3 g.p.m. under a head of 5 feet, and that the economical nozzle spacing will be 15.3 feet. The least number of nozzles at this spacing required for a bed of one acre in area is found as follows: In Fig. 172, let n equal the number of nozzles in a horizontal row, counting half-spray nozzles as ½, and let m equal the number of rows counting rows of half-spray nozzles as half rows.[[162]] Then the number of nozzles, N, equals mn, and 15.3m × 13.2n equals 43,560 or mn equals 215.
Fig. 172.—Typical Sprinkler Nozzle Layout.
The next step should be the design of the dosing tank and siphon. It is possible to design a tank which will give equal distribution over equal areas of filter surface. It has been found, however, that the expense of this refinement is unwarranted as there are a number of outside factors which tend to overcome the theoretical design. The effect of wind, unequal spacing, and irregularities in the elevation of the nozzles have a tendency to offset refinements in the design of a dosing tank. It is therefore the general practice to slope the sides of the tank at an angle of about 45 degrees as previously stated. The dosing tank is generally designed to have a capacity which will give a complete cycle of operation once in 15 minutes. In the ordinary design the factors given are the rate of inflow and the given time of filling. In the following example the time of filling will be taken as 10 minutes, the time of emptying as 5 minutes, and the rate of flow as 1,000,000 gallons per day. The capacity of the tank will therefore be 1,000,000
24 x 6 = 7,000 gallons. The diameter of the siphon to be selected can be computed as follows:
| Let | Q | = the capacity of the tank in cubic feet; |
| q1 | = the rate of discharge of the siphon in cubic feet per second; | |
| q2 | = the rate of inflow to the tank in cubic feet per second; | |
| q | = the rate of emptying the tank in cubic feet per second = (q1 − q2); | |
| A | = the cross-sectional area of the free surface of the water in the tank at any instant, in square feet; | |
| a | = the cross-sectional area of the siphon in square feet; | |
| b | = the small dimension of the base of the tank in feet; | |
| h | = the head of water, in feet, on the discharge siphon; | |
| h1 | = the initial head of water, in feet, on the siphon; | |
| h2 | = the final head of water in feet, on the siphon; | |
| t | = the time, in seconds, required to empty the tank, |
| then | dQ = -Adh = q1dt − q2dt, |
| and | dt = dQ q = − Adh q1 − q2, |
| but | q1 = 0.4 A √(2gh),[[163]] |
| therefore |  |
| but | A = 4h2 + 4bh + b2, |
| therefore |  |
The integration of this expression is tedious. Its solution for siphons between 6 inches and 12 inches operating under heads commencing from 3 feet to 6 feet, with a time of emptying of 5 minutes and time of filling of 10 minutes is given in Fig. 173. In the example given the rate of inflow is 1.55 sec. feet and the head is 5 feet. Then from Fig. 173 the size of the siphon to be used is 12 inches. Where a siphon of the size required to empty the tank in the time fixed is not available, combinations of available sizes can sometimes be used.
Fig. 173.—Diagram for the Determination of the Capacities of Dosing Tanks for Trickling Filters.
Time of emptying, 5 minutes. Time of filling, 10 minutes. Shape of tank is a right pyramid or a truncated right pyramid with all four sides making an angle of 45 degrees with the vertical. All horizontal cross-sections are squares.