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First Order Transfer Functionby koala
Tags: chemical engineering, dynamic behavior, pressure sensor, steady state, transfer function 
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#1
Sep2111, 04:23 PM

P: 3

1. The problem statement, all variables and given/known data
The dynamic behavior of a pressure sensor/transmitter can be expressed as a firstorder transfer function (in deviation variables) that relates the measured value Pm to the actual pressure, P: Pm'(s)/P'(s)=1/(30s+1). Both Pm' and P' have units of psi and the time constant has units of seconds. Suppose that an alarm will sound if Pm exceeds 45psi. If the process is initially at steady state, and then P suddenly changes from 35 to 50 psi at 1:10 PM, at what time will the alarm sound? 2. Relevant equations Pm'(s)/P'(s)=1/(30s+1) 3. The attempt at a solution This is what I did however I don't think it's correct... : If P'(s)=(5035)/s=15/s then Pm'(s)=[1/(30s+1)]*P'(s) > Pm'(s)=15/[s(30s+1)] > Inverse Laplace > Pm'(t)=15(1e^(t/30)) Pm'(t)=Pm(t)[P(t) steady state] > Pm(t)=15[1e^(t/30)]+35 45=15*[1e^(t/30)]+35 > t=33.3 s > So the time would be 1:10:33 PM. As I said I don't believe it's correct so any help would be greatly appreciated. 


#2
Sep2111, 04:49 PM

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PF Gold
P: 4,759

I see nothing wrong with what you did unless it's an arithmetic error.



#3
Sep2111, 05:03 PM

P: 3

Are you sure? I want to make sure my method is correct and that this is the correct way to do this problem.



#4
Sep2111, 06:38 PM

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PF Gold
P: 4,759

First Order Transfer Function
I am sure. I have been doing Laplace transforms and transfer functions for 40 years!
I would have done one thing differently: ignored the 35 psi initial pressure, calling it zero instead. Then the aiming pressure would be 10. Transfer functions by definition do not have initial conditions associated with them, so by concentrating on the transfer function itself you are less likely to slip up. This particular transfer function is the simplest one in existence (other than a constant) so take my word for it, it's better that way. 


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