Rational Psychrometric Formulae/Appendix No. 3

by Willis H. Carrier

APPENDIX No. 3

PROOF OF FORMULA [5]

$G={\frac {5284(t+459.64)D_{s}}{P-e}}$

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

[34]

${\frac {W_{1}}{W_{2}}}={\frac {p_{1}S_{1}}{p_{2}W_{2}}}$ and $P=p_{1}+p_{2}$

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

${\frac {W_{s}}{1}}={\frac {S_{w}p_{s}}{1(P-p_{s})}}$, or $W_{s}={\frac {S_{w}}{p_{s}}}{P-p_{s}}{\frac {W_{s}}{1}}={\frac {S_{w}p_{s}}{1(P-ps)}}$, or $W_{s}{\frac {S_{w}p_{s}}{P-p_{s}}}$

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. Also

[35]

$S={\frac {V_{a}}{V_{s}}}=V_{a}D_{s}$

where

$V_{s}$ = specific volume of saturated steam (volume of 1 lb.) at a given temperature, $t$, Pressure $e$

$V_{a}$, = specific volume of air at the same pressure and temperature

$D_{s}$ = density of saturated steam in lb. per cu. ft. at temperature $t$

But we have for air

${\frac {p^{o}}{T}}=53.35$

[36]

$Va=53.35\left({\frac {t+459.64}{P_{s}}}\right)$

Therefore

[37]

$S_{1}={\frac {53.35C_{s}(t+459.64)}{p_{s}}}D_{s}$

Hence, substituting in [34]

[38]

$W={\frac {53.35(t+459.64)}{(P-p)}}D$

$G=7000W$

$(P-e)$ in. mercury = $(144\times 0.4908)(P-p)$ lb. per sq. ft.

Hence

[39]

$G={\frac {5284(t+459.64)D_{s}}{(P-e)}}$