Modern Gasworks Practice

340 MODERN GASWORKS PRACTICE (1) The h.p. required to overcome the additional friction on the guides, due to the change of direction of the chain at the upward bend. (2) The h.p. required to lift the conveyed material through the height h. (3) The load lifted = w.l. Speed of lifting (vertically) = S X - g p __ wl x Sh __ wSh ’ ~ 33,000 X 17 33,000 _____ _ Ch x __ 37 Ch 33,000 33,000 (4) Work per minute absorbed by friction. Fig. 249.— Diagram of Fokces kor H.P. Calculation. Let a = angle of inclmation (see Fig. 249) 6 — angle of contact in radians 7 —arc— __ h (neari ) for small angles = 0'5 for 30° radius l Let P, = total friction under new conditions F = coefficient of friction as before (0-25) Then, — P = added friction (at bend). Additional work due to friction per minute — (Pj — P) S p but = eve, where e = the base of Naperian logs = 2-718 • P1~F == e™ — 1 ' ’ P Pj-P = P(eF9-l) So that S (Pi-P) = SP (eI7’-l) Thus, the additional power to be overcome due to this friction can be found by multiplying the predetermined h.p. (S.P.) by— 1 (eF9-l). For an angle of 30 deg. this factor (2-718* —1) = *13, which indicates that the