# Rational Psychrometric Formulae/Appendix No. 3

Rational Psychrometric Formulae by Willis H. Carrier
Appendix No. 3

APPENDIX No. 3

PROOF OF FORMULA [5]

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

84In this equation

${\displaystyle G}$ = grains of moisture per lb. of pure air at saturation
${\displaystyle t}$ = temperature of saturation in deg. fahr.
${\displaystyle t+459.64}$ = absolute temperature
${\displaystyle D_{s}}$ = density inlb. per cu. ft. of saturated water vapor at temperature ${\displaystyle t}$
= reciprocal of specific volume of steam
${\displaystyle P}$ = 29.92 = assumed standard of barometric pressure in in. of mercury.
${\displaystyle e}$ = vapor pressure of saturated water vapor
${\displaystyle 5284}$ = 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]

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

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

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

where

Weight of air = 1 lb.
Speciﬁc weight of air = 1
${\displaystyle (P-p_{s})}$ = total barometric pressure in lb. per sq. ft.
${\displaystyle W_{s}}$ = weight of saturated water vapor in 1 lb. pure air
${\displaystyle p_{s}}$ = pressure of saturated water at given temperature ${\displaystyle t}$, in lb.per sq. ft.
${\displaystyle S_{1}}$ = speciﬁc weight of water vapor at temperature ${\displaystyle t}$, compared with air at same temperature and pressure; ${\displaystyle p_{s}}$ or ${\displaystyle e}$.
${\displaystyle W}$ = weight of moisture vapor in mixture containing 1 lb. pure air. Also
[35]

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

where

${\displaystyle V_{s}}$ = speciﬁc volume of saturated steam (volume of 1 lb.) at a given temperature, ${\displaystyle t}$, Pressure ${\displaystyle e}$
${\displaystyle V_{a}}$, = speciﬁc volume of air at the same pressure and temperature
${\displaystyle D_{s}}$ = density of saturated steam in lb. per cu. ft. at temperature ${\displaystyle t}$

But we have for air

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

[36]

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

Therefore

[37]

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

Hence, substituting in [34]

[38]

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

${\displaystyle G=7000W}$

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

Hence

[39]

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