Page:Journal of the American Society of Mechanical Engineers, Volume 33.pdf/671

EFFECT OF CHANGE IN BAROMETRIC PRESSURE

68⁠Suppose that air in which the vapor pressure is $e_{0}$ is compressed from a barometric pressure $P_{0}$ to a barometric pressure $P$, then the partial pressure of both the air and the vapor are increased proportionally and $e={\frac {e_{0}P}{P_{0}}}$. The temperature corresponding to saturation at $e$ is the temperature of the dew point at pressure $P$.

69⁠The per cent of isothermal saturation becomes

[29]

$R=\left({\frac {e}{e_{2}}}\right)={\frac {e_{0}}{e_{2}}}{\frac {P}{P_{0}}}={\frac {R_{0}P}{P_{0}}}$

where $e_{2}$ is the saturated vapor pressure corresponding to the dry bulb temperature $t$.

TOTAL HEAT CURVE

70⁠This curve shows the sensible heat in the air above a base temperature of 0 deg. fahr., plus the latent heat contained in the water vapor at saturation, but not including the heat of the liquid. Since the wet-bulb temperature, or adiabatic lines contain all points having the same total heat (neglecting heat of liquid), the curve serves to determine the total heat in the air under any and all conditions represented by the chart. This is of great convenience in calculating refrigeration required to cool and de-humidify air. For example, suppose it is required to find the refrigeration necessary to cool 1 lb. of air containing 98 grains of moisture and having a dry bulb temperature of 95 deg., to a final temperature of 40 deg. saturated. We find from the chart that the wet-bulb temperature is 75 deg. The total heat corresponding to a saturation temperature of 75 deg. is 37.8 B.t.u., while the total heat at 40 deg. is 15.3 B.t.u. The difference, 22.5 B.t.u. is the refrigeration required per pound of air.

The author wishes to acknowledge his indebtedness to his assistants, Mr. Theodore A. Weager and Mr. Frank L. Busey, for the actual work of computation and the construction of the diagrams.