CIRCULATION TIMES AND THEORY OF INDICATOR-DILUTION METHODS 



J93 



time -» 



FIG. 7. Graphic representation of fact that /o" [i — H(t)]dt is mean transit time. /J [i — H{s)] ds 

 is stun of the two shaded areas, of which j\ s k{s) ds is the lower. For the hmit of the lower area to 

 remain finite as / — » co, the area of the upper rectangle must approach zero, which it does. [From 

 Meier & Zierler (ig).] 



cause, from assumption e, the distribution of traversal 

 times for indicator particles is the same as for fluid 

 particles, of those fluid particles which entered the 

 system at time zero, the fraction leaving per unit time 

 is h(t). Of the indicator introduced during the interval 

 between s and s -\- ds time units before /, that is, in 

 the vicinity of time / — s, the fraction eliminated per 

 unit time at time t is h(s). Therefore, of the indicator, 

 / ds, introduced between s and s + ds time units 

 before t, the amount leaving per unit time at time / 

 is h{s) •/ ds. 



Summing for all such time intervals before t, the 

 rate at which indicator leaves at / is Jo / h{s) ds. But 

 the rate at which indicator leaves the system is also 

 F C(t). Therefore 



C(0 



U' 



his) ds 



Combining equations 16 and 17, 



an 



- H(t) 

 F 



Hit) = C{t)/Cr, 



(17) 



(18) 



(■9) 



Incidentally, since \o I'is) ds = i, lim,^„ H{t) = i. 



Therefore, as / becomes large, C{t) approaches I/F. 



Substituting H(t) for C{t)/Cma^ in equation 15, 



V = F f 

 Jo 



[i - H(0] dt 



(20) 



We must now prove the identity of Jq [i — //(/)] 

 dt with I = \^ t hit) dt. This is seen by an integration 

 by parts: 





[i - His)] ds = t[i - Hit)] + / s his) ds 



f 



Jo 



f 



Jo 



[i - H(s)] ds - 



lim t [i 



Hit)] + 



r 



s his) ds 



Figure 7 illustrates the graphic meaning of the parts 

 of the integral. It is evident from figure 7 that the 

 integrals must both be finite or both infinite (which 

 would yield infinite volume) and that for the case of 

 finite integrals lim,-.^ t [i — //(<)] = o. Because the 

 volume must be finite. 





[i - His)] ds 



-r 



s his) ds = i 



(21) 



Since the derivation of equation 15, from simple 

 consideration of input-output relations, is independent 

 of equation 10, the identity shown in equation 21 is 

 an independent demonstration of the basic relation- 

 ship, \olume = flow multiplied by mean transit 

 time. 



From equations 6 and 17, 



Cit) 



I r^ F , I r' 



= - / - ds) ds = - \ 



^ Jo 1 ll 



cis) ds 



(22) 



