Rational Psychrometric Formulae/Appendix No. 4

by Willis H. Carrier

APPENDIX No. 4

DERIVATION OF FORMULA [6] GIVING THE EQUATION OF THE ADIABATIC SATURATION LINE

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

86Assuming 1 lb. of pure air having the temperature $t$ containing $W$ lb. of moisture with the corresponding dew point hand vapor pressure $e_{1}$ having a resultant adiabatic saturation temperature of $t'$, assume also a moisture increment $dW$ under adiabatic conditions resulting in a temperature increment of $-dt$. This moisture increment $dW$ is evidently evaporated at a vapor pressure $e_{1}$ corresponding to temperature $t_{1}$ and superheated to temperature $t$. The temperature of the liquid is evidently constant at temperature $t'$, from principle C. The total heat of the vapor in the increment is $H_{1}dW+C_{ps}(1-t_{1})dW$, where $H_{1}$ is the total heat of steam corresponding to temperature $t_{1}$ and vapor pressure $e_{l}$, and $C_{ps}(t-t_{1})dW$ is the heat required to superheat from saturation temperature $t_{1}$ to dry-bulb temperature $t$. The heat of the liquid evaporated, however, is $q'dW$ corresponding to temperature of saturation $t'$.

87 The total heat interchange required to evaporate $dW$ under these conditions is therefore

[40]

$\Sigma =\{H_{1}-q'+[C_{ps}(t-t_{1})]\}dW$

The change in sensible heat of 1 lb. of air and $W$ lb. of water vapor due to the temperature increment $-dt$ is

[41]

$\Sigma =-(C_{pa}+WC_{ps})dt$

Since the change is adiabatic these values may be related by the equation

[42]

$\{H_{1}-q'+[C_{ps}(t-t_{1})]\}dW-(C_{pa}+WC_{ps}dt=0$

[43]

$\int _{W}^{W'}\{H_{1}-q'+[C_{ps}(t-t_{1})]\}dW=\int ^{t'}(C_{pa}+WC_{ps})dt$

in which $H_{1}$ and </math>t_1</math> are variables corresponding to the variable $W$ while $t$ is a variable related to $W$ by the different equation. A constant corresponding to $t'$ is $q'$ while $C_{ps}$, may be taken approximately as a mean between its values at $t_{1}$ and at $t'$ and $C_{ps}$ as a mean between its values at $t$ and at $t'$. The temperature of saturation is $t'$, and $W'$ is the corresponding moisture content at saturation.

88It is not necessary, however, to solve this equation in this form as this relationship may be simplified.

[44]

$\{H_{1}-q'+[C_{ps}(t-t_{1})]\}dW=\{H_{1}-q'+[C_{ps}(t'-t_{1}))+(C_{ps}(t-t')]\}dW$

It may be shown thermodynamically, assuming steam to be a perfect gas. that [45]

$H_{1}-q'+C_{ps}(t'-t_{1})]=H'-q'=r'$

89This may also be demonstrated approximately for the range of temperatures under discussion by computation from the values given in the steam tables of Marks and Davis, as in Table 6:

TABLE 6 COMPARISON OF ACTUAL VALUES OF $r'$ WITH VALUES OF $r'$ COMPUTED FROM THE TOTAL HEAT AT DIFFERENT TEMPERATURES $t_{1}$
$t'=80{\text{deg., }}r'=1046.7$
$t'$	$t_{1}$	$C_{ps}$	$H_{1}$	$q'$	$r'$ (Computed)
80	70	0.44365	1090.2	48.03	1046	.70
80	60	0.44356	1085.9	48.03	1046	.74
80	50	0.44347	2081.4	48.03	1046	.78
80	40	0.44347	1076.9	48.03	1046	.5
$t'=100{\text{deg., }}r'=1035.6$
100	90	1099.2	0.44401	67.97	1035	.67
100	80	0.44392	1094.8	67.97	1035	.7
100	70	0.44383	1090.8	67.97	1035	.64
100	60	0.44374	1085.9	67.97	1035	.67
100	50	0.44365	1081.4	67.97	1035	.81
100	40	0.44356	1076.9	67.97	1035	.54
$t'=120{\text{deg., }}r'=1024.4$
120	110	0.44419	1108.0	87.91	1024	.4
120	100	0.44410	1103.6	87.91	1024	.57
120	90	0.44401	1099.2	87.91	1024	.52
120	80	0.44392	1094.4	87.91	1024	.64
120	70	0.44383	1090.3	87.91	1024	.58
120	60	0.44374	1085.9	87.91	1024	.50
120	50	0.44365	1081.4	87.91	1024	.54
120	40	0.44356	1076.9	87.91	1024	.47
		$r'=H_{1}-q'+[C_{pa}(t'-t_{1})]$
	or	$H'=H_{1}+[C_{ps}(t'-t_{1})]$

Hence substituting in equation [43]

[46]

$r'\int _{W}^{W'}dW+C_{ps}\int _{W}^{W'}(t-t')dW=C_{pa}\int _{t'}^{t}dt+C_{ps}\int _{t'}^{t}Wdt$

[47]

$r'(W'-W)+C_{pa}\int _{W}^{W'}\left[\int _{t'}^{t}dt\right]dW=C_{pa}(t-t')+C_{ps}W\int _{t'}^{t}dt+C_{ps}\int _{W}^{W'}\left[\int _{t'}^{t}dt\right]dW$

[48]

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

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.