Page:EB1911 - Volume 14.djvu/78

Rh (b) A second method is to obtain a rough value of d by assuming = . This value is

d&#8202;′ = 5√ (32Q2 / g2i) 5√ = 0.6319 5√ (Q2 / i) 5√.

Then a very approximate value of is

′ = (1 + 1/12d&#8202;′);

and a revised value of d, not sensibly differing from the exact value, is

d&#8202;″ = 5√ (32Q2 / g2i) 5√ ′ = 0.6319 5√ (Q2 / i) 5√ ′.

(c) Equation 7 may be put in the form

d = 5√ (32Q2 / g2i) 5√ (1 + 1/12d). (9)

Expanding the term in brackets,

5√ (1 + 1/12d) = 1 + 1/60d − 1/1800d&#8202;2

Neglecting the terms after the second,

and

§ 85. Arrangement of Water Mains for Towns’ Supply.—Town mains are usually supplied oy gravitation from a service reservoir, which in turn is supplied by gravitation from a storage reservoir or by pumping from a lower level. The service reservoir should contain three days’ supply or in important cases much more. Its elevation should be such that water is delivered at a pressure of at least about 100 ft. to the highest parts of the district. The greatest pressure in the mains is usually about 200 ft., the pressure for which ordinary pipes and fittings are designed. Hence if the district supplied has great variations of level it must be divided into zones of higher and lower pressure. Fig. 96 shows a district of two zones each with its service reservoir and a range of pressure in the lower district from 100 to 200 ft. The total supply required is in England about 25 gallons per head per day. But in many towns, and especially in America, the supply is considerably greater, but also in many cases a good deal of the supply is lost by leakage of the mains. The supply through the branch mains of a distributing system is calculated from the population supplied. But in determining the capacity of the mains the fluctuation of the demand must be allowed for. It is usual to take the maximum demand at twice the average demand. Hence if the average demand is 25 gallons per head per day, the mains should be calculated for 50 gallons per head per day.

§ 86. Determination of the Diameters of Different Parts of a Water Main.—When the plan of the arrangement of mains is determined upon, and the supply to each locality and the pressure required is ascertained, it remains to determine the diameters of the pipes. Let fig. 97 show an elevation of a main ABCD, R being the reservoir from which the supply is derived. Let NN be the datum line of the levelling operations, and Ha, Hb the heights of the main above the datum line, Hr being the height of the water surface in the reservoir from the same datum. Set up next heights AA1, BB1, representing the minimum pressure height necessary for the adequate supply of each locality. Then A1B1C1D1 is a line which should form a lower limit to the line of virtual slope. Then if heights ɧa, ɧb, ɧc are taken representing the actual losses of head in each length a, b, c of the main, A0B0C0 will be the line of virtual slope, and it will be obvious at what points such as D0 and E0, the pressure is deficient, and a different choice of diameter of main is required. For any point z in the length of the main, we have

Pressure height = Hr − Hz − (ɧa + ɧb + ɧz).

Where no other circumstance limits the loss of head to be assigned to a given length of main, a consideration of the safety of the main from fracture by hydraulic shock leads to a limitation of the velocity of flow. Generally the velocity in water mains lies between 1 and 4 ft. per second. Occasionally the velocity in pipes reaches 10 ft. per second, and in hydraulic machinery working under enormous pressures even 20 ft. per second. Usually the velocity diminishes along the main as the discharge diminishes, so as to reduce somewhat the total loss of head which is liable to render the pressure insufficient at the end of the main.

J. T. Fanning gives the following velocities as suitable in pipes for towns’ supply:—

§ 87. Branched Pipe connecting Reservoirs at Different Levels.—Let A, B, C (fig. 98) be three reservoirs connected by the arrangement of pipes shown,—1, d1, Q1, v1; 2, d2, Q2, v2; 3, d3, Q3, v3 being the length, diameter, discharge and velocity in the three portions of the main pipe. Suppose the dimensions and positions of the pipes known and the discharges required.

If a pressure column is introduced at X, the water will rise to a height XR, measuring the pressure at X, and aR, Rb, Rc will be the lines of virtual slope. If the free surface level at R is above b, the reservoir A supplies B and C, and if R is below b, A and B supply C. Consequently there are three cases:—

To determine which case has to be dealt with in the given conditions, suppose the pipe from X to B closed by a sluice. Then there is a simple main, and the height of free surface h′ at X can be determined. For this condition

ha − h′ = (v12 / 2g) (41 / d1) = 32Q′21 / g2d15;

h′ − hc = (v32 / 2g) (43 / d3) = 32Q′23 / g2d35;

where Q′ is the common discharge of the two portions of the pipe. Hence

(ha − h′) / (h′ − hc) = 1d35 / 3d15,

from which h′ is easily obtained. If then h′ is greater than hb, opening the sluice between X and B will allow flow towards B, and the case in hand is case I. If h′ is less than hb, opening the sluice will allow flow from B, and the case is case III. If h′ = hb, the case is case II., and is already completely solved.