Page:Journal of the American Society of Mechanical Engineers, Volume 33.pdf/683

90The same result may be obtained by equating the total heat in the air in any state with its total heat when in the state of adiabatic saturation. The total heat in a mixture of 1 lb. of pure air and saturated water vapor at a temperature $t'$ calculated from a base temperature of 0 deg. fahr. and deducing the heat of the liquid, $q'$, which as we have shown is unaffected by the adiabatic change, is

[49]

$\Sigma =C_{pa}t'+r'W'$

91The total heat under any other adiabatic condition, where temperature is $t$ and moisture $W$, is

[50]

$\Sigma =C_{pa}t+[(H_{1}-q_{1})+C_{ps}(t'-t_{1})]W$

which is substantially equivalent to

[51]

$\Sigma =C_{pa}t+r'W+C_{ps}(t-t')W$

Therefore since the change is adiabatic we may equate [47] and [49].

[52]

$C_{pa}t+r'W+C_{ps}(t-t')W=C_{pa}t'+r'W'$

[53]

$C_{pa}(t-t')+C_{ps}(t-t')W=r'(W'-W)$

where

$(t-t')$ = the true wet-bulb depression

$(W'-W)$ = the moisture absorbed per lb. of pure air when it is adiabatically saturated from an initial dry-bulb temperature to and an initial moisture content $W$

$C_{pa}$ = mean specific heat of air at constant pressure between temperature $t$ and $t'$

$C_{ps}$ = specific heat of steam at constant pressure between $t$ and $t'$

$r'$ = latent heat of evaporation at wet-bulb temperature $t'$

This is identical with equation [20] obtained by the differential method.