Question

An alarm goes ON when a pressure sensor voltage rises above 4.00V. The pressure sensor outputs 20mV/kPa and has a time constant of 4.9s. How long after the pressure rises suddenly from 100kPa to 400kPa does the light go ON? Sketch the input and corresponding output/response of the pressure sensor

Answer 1

Input pressure at beginning : 100 kPa

so corresponding sensor output voltage = 20 mV/kPa * 100 kPa = 2 V

Now final input pressure : 400 kPa

so corresponding sensor output voltage = 20 mV/kPa * 400 kPa = 8 V

Now if we consider the step response of first order system, the standard response of the system is given by:

V(t) = V(0) + (1 - e^-t/T) V' ..... 1

where V(t) = Voltage at time t

V(0) = Initial voltage = 2 V

T= time constant of the system = 4.9 s

V'= Voltage difference between final and initial voltage = 8 - 2 = 6V

Now alarm goes ON when pressure sensor voltage rises above 4.00V

So, from equation 1 we have:

4 = 2 + 6(1 - e^-t/4.9)

(1 - e^-t/4.9) = 2/6 = 1/3

e^-t/4.9 = 2/3

-t/4.9 = -0.405

So, t = 1.9845 secs

So, alarm goes ON after t = 1.9845 secs counted after input pressure change.

Response of the pressure sensor: