§ 176. Direct-Acting Hydraulic Lift (fig. 171).—This is the simplest of all kinds of hydraulic motor. A cage W is lifted directly by water pressure acting in a cylinder C, the length of which is a little greater than the lift. A ram or plunger R of the same length is attached to the cage. The water-pressure admitted by a cock to the cylinder forces up the ram, and when the supply valve is closed and the discharge valve opened, the ram descends. In this case the ram is 9 in. diameter, with a stroke of 49 ft. It consists of lengths of wrought-iron pipe screwed together perfectly water-tight, the lower end being closed by a cast-iron plug. The ram works in a cylinder 11 in. diameter of 9 ft. lengths of flanged cast-iron pipe. The ram passes water-tight through the cylinder cover, which is provided with double hat leathers to prevent leakage outwards or inwards. As the weight of the ram and cage is much more than sufficient to cause a descent of the cage, part of the weight is balanced. A chain attached to the cage passes over a pulley at the top of the lift, and carries at its free end a balance weight B, working in T iron guides. Water is admitted to the cylinder from a 4-in. supply pipe through a two-way slide, worked by a rack, spindle and endless rope. The lift works under 73 ft. of head, and lifts 1350 lb at 2 ft. per second. The efficiency is from 75 to 80%.
The principal prejudicial resistance to the motion of a ram of this kind is the friction of the cup leathers, which make the joint between the cylinder and ram. Some experiments by John Hick give for the friction of these leathers the following formula. Let F = the total friction in pounds; d = diameter of ram in ft.; p = water-pressure in pounds per sq. ft.; k a coefficient.
F = k p d
k = 0.00393 if the leathers are new or badly lubricated;
= 0.00262 if the leathers are in good condition and well lubricated.
Since the total pressure on the ram is P = 1⁄4πd2p, the fraction of the total pressure expended in overcoming the friction of the leathers is F/P = .005/d to .0033/d, d being in feet.
Let H be the height of the pressure column measured from the free surface of the supply reservoir to the bottom of the ram in its lowest position, Hb the height from the discharge reservoir to the same point, h the height of the ram above its lowest point at any moment, S the length of stroke, Ω the area of the ram, W the weight of cage, R the weight of ram, B the weight of balance weight, w the weight of balance chain per foot run, F the friction of the cup leather and slides. Then, neglecting fluid friction, if the ram is rising the accelerating force is
P1 = G (H − h) Ω − R − W + B − w (S − h) + wh − F,
and if the ram is descending
P2 = G (Hb − h) Ω + W + R − B + w (S − h) − wh − F.