§ 74. Case of a Uniform Pipe connecting two Reservoirs, when all the Resistances are taken into account.—Let h (fig. 82) be the difference of level of the reservoirs, and v the velocity, in a pipe of length L and diameter d. The whole work done per second is virtually the removal of Q cub. ft. of water from the surface of the upper reservoir to the surface of the lower reservoir, that is GQh foot-pounds. This is expended in three ways. (1) The head v2/2g, corresponding to an expenditure of GQv2/2g foot-pounds of work, is employed in giving energy of motion to the water. This is ultimately wasted in eddying motions in the lower reservoir. (2) A portion of head, which experience shows may be expressed in the form ζ0v2/2g, corresponding to an expenditure of GQζ0v2/2g foot-pounds of work, is employed in overcoming the resistance at the entrance to the pipe. (3) As already shown the head expended in overcoming the surface friction of the pipe is ζ(4L/d) (v2/2g) corresponding to GQζ (4L/d) (v2/2g) foot-pounds of work. Hence
GQh = GQv2/2g + GQζ0v2/2g + GQζ·4L·v2/d·2g;
| h = (1 + ζ0 + ζ·4L/d) v2/2g. v = 8.025 √ [hd / {(1 + ζ0)d + 4ζL} ]. |
(5)
If the pipe is bell-mouthed, ζ0 is about = .08. If the entrance to the pipe is cylindrical, ζ0 = 0.505. Hence 1 + ζ0 = 1.08 to 1.505. In general this is so small compared with ζ4L/d that, for practical calculations, it may be neglected; that is, the losses of head other than the loss in surface friction are left out of the reckoning. It is only in short pipes and at high velocities that it is necessary to take account of the first two terms in the bracket, as well as the third. For instance, in pipes for the supply of turbines, v is usually limited to 2 ft. per second, and the pipe is bellmouthed. Then 1.08v2/2g = 0.067 ft. In pipes for towns’ supply v may range from 2 to 41⁄2 ft. per second, and then 1.5v2/2g = 0.1 to 0.5 ft. In either case this amount of head is small compared with the whole virtual fall in the cases which most commonly occur.
When d and v or d and h are given, the equations above are solved quite simply. When v and h are given and d is required, it is better to proceed by approximation. Find an approximate value of d by assuming a probable value for ζ as mentioned below. Then from that value of d find a corrected value for ζ and repeat the calculation.
The equation above may be put in the form
h = (4ζ/d) [{ (1 + ζ0) d/4ζ} + L] v2/2g;
(6)
from which it is clear that the head expended at the mouthpiece is equivalent to that of a length