When Will the Alarm Sound for a First Order Transfer Function?

[koala]

Homework Statement

The dynamic behavior of a pressure sensor/transmitter can be expressed as a first-order 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?

Homework Equations

Pm'(s)/P'(s)=1/(30s+1)

The Attempt at a Solution

This is what I did however I don't think it's correct... :

If P'(s)=(50-35)/s=15/s
then Pm'(s)=[1/(30s+1)]*P'(s) --> Pm'(s)=15/[s(30s+1)] --> Inverse Laplace --> Pm'(t)=15(1-e^(-t/30))
Pm'(t)=Pm(t)-[P(t) steady state] --> Pm(t)=15[1-e^(-t/30)]+35
45=15*[1-e^(-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.

 

[rude man]

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

 

[koala]

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

 

[rude man]

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.

 

[DeeNos]

As a first-order transfer function, the dynamic behavior of the pressure sensor/transmitter can be described by the equation Pm'(s)/P'(s)=1/(30s+1), where Pm' and P' are both measured in psi and the time constant is in seconds. In this case, we are given that an alarm will sound if Pm exceeds 45 psi. To determine the time at which the alarm will sound, we can use the given information that the process is initially at steady state and then P suddenly changes from 35 to 50 psi at 1:10 PM.

To solve for the time at which the alarm will sound, we can first use the equation Pm'(s)/P'(s)=1/(30s+1) to express Pm' in terms of P'. By substituting P'=(50-35)/s=15/s, we get Pm'(s)=15/[s(30s+1)]. We can then use the inverse Laplace transform to get the expression Pm'(t)=15(1-e^(-t/30)). Since we are interested in the time at which Pm exceeds 45 psi, we can set Pm'(t)=45 and solve for t. This gives us t=30 seconds, meaning that the alarm will sound at 1:10:30 PM.

It is important to note that this solution assumes that the pressure change from 35 to 50 psi occurs instantaneously at 1:10 PM. In reality, there may be a slight delay in the pressure change, which would affect the time at which the alarm will sound. Additionally, other factors such as sensor accuracy and response time may also impact the exact timing of the alarm. Therefore, it is important to consider all relevant factors when analyzing the behavior of a first-order transfer function.