Page:Journal of the American Society of Mechanical Engineers, Volume 33.pdf/680

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]

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