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Differentiating both sides; 



Q = a -^ (AP ) (5.9) 



^ dt c 



Substituting (Equations 5.8 and 5.9 into 5.6, one obtains: 



AP = P, - P^ + AP + a(-— + — -) :rr AP /r -.^n 



12 c 'b^ B^ dt c (5.10) 



where aP is the difference in pressure between the two sensors. If 

 the change, aP, occurs instantaneously, it can be shown that the 

 solution to Eq. 5.10 is 



AP = AP(1 - e"^D^) , (5.11) 



substitution into Eq. 5.10 yields: 



AP = AP^ + a(^ + ^) K^ AP e-V ^^^^^^ 



where Kp"^ is the "folding time" of the response. Solving, 



