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## Q. 5.1

DYNAMIC 6EHA VIOR OF THE LIQUID LEVEL IN A STORAGE TANK.

A 250 liter tank used for liquid storage is configured as in Figure 5.1 so that its mathematical model and its dynamic behavior are as discussed in Section 5.1.1.

The cross-sectional area of the tank is 0.25 m², and it may be assumed uniform. The outlet valve resistance, which we shall assume to be linear for simplicity, has a value c = 0.1 m²/min.

The entire system was initially at steady state with the inlet flowrate $F_{i}$ at 37 liters/min (0.037 m³/min). The following questions are now to be answered:

1. What is the initial value of the liquid level in the tank?
2. If the inlet flowrate was suddenly changed to 87 liters/min (0.087 m³ /min), obtain an expression for how the liquid level in the tank will vary with time.
3. When will the liquid level in the tank be at the 0.86 m mark? And to what final value will the liquid level ultimately settle?

## Verified Solution

1.   From Eq. (1.13), we observe that at steady state we have:

$A_{c} \frac{dh}{dt} = F_{i} – ch$        (1.13)

$F_{is} = ch_{s}$

If we now introduce the given values for the flowrate and the resistance c, we have the required value for the initial steady-state liquid level:

$h_{s} = 0.37 m$

2.   Recalling the transfer function for this process given in Eq. (5.7), and introducing the given numerical values, we find that the steady-state gain and time constant are given by:

$g(s) = \frac{1/c}{(A_{c} /c)s + 1}$        (5.7)

K = 1/0.1 = 10 (min /m²)

$\tau$ = 0.25/0.1 = 2.5 (min)

The magnitude of the step input in the flow rate is (0.087 – 0.037) = 0.05 m³ / min. Thus from Eq. (5.13) the expression representing the change in liquid level in response to this input change is:

$y(t) = AK (1 – e^{-t/ \tau})$        (5.13)

$y(t) = 0.5 (1 – e^{-0.4t})$

Note that this is in terms of y, the deviation of the liquid level from its initial steady-state value. The actual liquid level time behavior is represented by:

$h(t) = 0.37 + 0.5 (1 – e^{-0.4t})$

3.   Introducing h = 0.86 into the expression shown above, it is easily solved for t to give the time required to attain to the 0.86 m mark as 9.78 min. The final value to which the liquid level eventually settles is easily obtained as 0.37 + 0.5 = 0.87 m.