Page:Journal of the American Society of Mechanical Engineers, Volume 33.pdf/682

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]

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