[zeta]´ = [alpha](1 + 1/12d´);
and a revised value of d, not sensibly differing from the exact value,
is
d´´ = [root 5](32Q²/g[pi]²i) [root 5][zeta]´
= 0.6319 [root 5](Q²/i) [root 5][zeta]´.
(c) Equation 7 may be put in the form
d = [root 5](32[alpha]Q²/g[pi]²i) [root 5](1 + 1/12d). (9)
Expanding the term in brackets,
[root 5](1 + 1/12d) = 1 + 1/60d - 1/1800d² ...
Neglecting the terms after the second,
d = [root 5](32[alpha]/g[pi]²) [root 5](Q²/i)·{1 + 1/60d}
= [root 5](32a/g[pi]²) [root 5](Q²/i) + 0.01667; (9a)
and
[root 5](32a/g[pi]²) = 0.219 for new pipes
= 0.252 for incrusted pipes.
[Illustration: FIG. 96.]
[Illustration: FIG. 97.]
§ 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.
[Illustration: FIG. 98.]
§ 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 H_a, H_b ... the heights of the main above
the datum line, H_r 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 [h]_a,
[h]_b, [h]_c ... are taken representing the actual losses of head in
each length l_a, l_b, l_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 = H_r - H_z - ([h]_a + [h]_b + ... [h]_z).
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 1½ and 4½ 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:--
Diameter in inches 4 8 12 18 24 30 36
Velocity in feet per sec. 2.5 3.0 3.5 4.5 5.3 6.2 7.0
§ 87. _Branched Pipe connecting Reservoirs at Different Levels._--Let
A, B, C (fig. 98) be three reservoirs connected by the arrangement of
pipes shown,--l1, d1, Q1, v1; l2, d2, Q2, v2; h3, 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:--
I. R above b; Q1 = Q2 + Q3.
II. R level with b; Q1 = Q3; Q2 = 0
III. R below b; Q1 + Q2 = Q3.
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
h_a - h´ = [zeta](v1²/2g)(4l1/d1)
= 32[zeta]Q´² l1/g[pi]²d1^5;
h´ - h_c = [zeta](v3²/2g)(4l3/d3)
= 32[zeta]Q´²l3/g[pi]²d3^5;
where Q´ is the common discharge of the two portions of the pipe.
Hence
(h_a - h´)/(h´ - h_c) = l1d3^5/l3d1^5,
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 h_b, opening the sluice
will allow flow from B, and the case is case III. If h´ = h_b, the
case is case II., and is already completely solved.
The true value of h must lie between h´ and h_b. Choose a new value of
h, and recalculate Q1, Q2, Q3. Then if
Q1 > Q2 + Q3 in case I.,
or
Q1 + Q2 > Q3 in case III.,
the value chosen for h is too small, and a new value must be chosen.