Page:Journal of the American Society of Mechanical Engineers, Volume 33.pdf/679



84In this equation
 * $$G$$ = grains of moisture per lb. of pure air at saturation
 * $$t$$ = temperature of saturation in deg. fahr.
 * $$t + 459.64$$ = absolute temperature
 * $$D_s$$ = density inlb. per cu. ft. of saturated water vapor at temperature $$t$$
 * = reciprocal of specific volume of steam
 * $$P$$ = 29.92 = assumed standard of barometric pressure in in. of mercury.
 * $$e$$ = vapor pressure of saturated water vapor
 * $$5284$$ = constant of the equation

85According to the law of gaseous mixtures the total pressure is equal to the sum of the gaseous pressures of the component parts, and the weights of the two components are in proportion to the products of their respective pressures times their speciﬁc weights

For a mixture of 1 lb. of air and water vapor saturated at a given temperature, therefore

where


 * Weight of air = 1 lb.
 * Speciﬁc weight of air = 1
 * $$(P - p_s)$$ = total barometric pressure in lb. per sq. ft.
 * $$W_s$$ = weight of saturated water vapor in 1 lb. pure air
 * $$p_s$$ = pressure of saturated water at given temperature $$t$$, in lb.per sq. ft.
 * $$S_1$$ = speciﬁc weight of water vapor at temperature $$t$$, compared with air at same temperature and pressure; $$p_s$$ or $$e$$.
 * $$W$$ = weight of moisture vapor in mixture containing 1 lb. pure air.