Steam Power Plant Engineering (Paperback)

This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1910 edition. Excerpt: ...Hc absorbed by the air in passing through the cooling device, B.T.U. per hour, is Ht = Hc + Ht. (102) Neglecting radiation and other minor losses, the heat H, absorbed by the air must be equal to the heat given up by the circulating water, or H, = H. (103) Example: Determine the quantity of air passing through the cooling tower per hour and the circulating water lost by evaporation in a power plant operating under the following conditions: Engines indicate 500 H.P. and consume 20 lbs. steam per I.H.P. hour; temperature of the injection water, discharge water and outside air, 90,122 and 72 F., respectively; barometer 29.5; relative humidity of air entering and leaving tower 70 and 90 per cent respectively; vacuum at condenser 25 inches. Determine also the weight of water evaporated in per cent of that circulated and of the condensed steam. In the problem, These values are obtained from Steam Tables and from Air Tables (Table 58). Substitute these values in equations (96) to (103) thus: (96), p, = 29.5-0.79 X 0.7 = 28.95. (96a), p, = 29.5-2.74 X 0.9 = 27.03. (97), w = H X 0.0747 V0. = 0.0722 y0. (97a), w = 0.001224 X 0.7 V0 = 0.000857 V0. By assumption, t2 being 10 to 20 degrees lower than ( in average practice when the range le--is greater than 30 degrees. t Marks and Davis: the values in Table 58 are Regnault's. Og V.28.95 460 + 112 Vl 27.03 460 + 72 0 = 1.152; that is, each cu. ft. of dry air entering the cooling-tower is increased in volume to 1.152 cu. ft. as it leaves. (98a), w2 = 0.003978 X 0.9 X 1.152 Vt = 0.004125 V0. (98b), w3 = 0.004125 V0-0.000857 V0 = 0.003268 V0. The total heat to be abstracted from the steam (see equation (84), page 347) is H-500 X 20 (1120.1-122 + 32) = 10,300,000 B.T.U. per hour. (99), But W (122-90)...