Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 5

Q. 5.3

RECTANGULAR PULSE RESPONSE OF THE LIQUID LEVEL SYSTEM.

Let us return once again to the system of Example 5.1. The flowrate is again to be changed suddenly to 87 liters/min (0.087 m² /min), but this time we hold it at this value only for 2.5 min; returning it to its initial value of 37 liters/min (0.037 m³ /min).

What is the maximum value the liquid level will attain during the course of this experiment?

Step-by-Step

Verified Solution

The expression for the peak value attained by a first-order system in response to a rectangular pulse input of magnitude A and duration b is given by:

y_{max} = y(b) = AK(1  –  e^{-b/ \tau})

(Note that this expression is in terms of deviations from the initial steady-state value for the liquid level.)

Since in this example, A = 0.05, b = 2.5, and from Example 5.1 we recall that K = 10, \tau = 2.5, with an initial liquid level of 0.37 m, the peak value for the liquid level is obtained from:

h_{max} = 0.37  +  0.5 (1  –  e^{-1})

or

h_{max} = 0.686  m