(1) The document describes a liquid level process in a tank where the rate of mass flow in equals the rate of mass flow out plus the rate of accumulation. (2) It finds the time constant T and recovery ratio R for the tank to be 5 minutes and 0.1768 ft3/min respectively. (3) For an impulse of 55 gallons, it plots the height response and flow rate response using Simulink with the height reaching a maximum of 1.205 feet, so the tank will not overflow.

3. FOR LIQUID-LEVEL PROCESSES,
T = AR
[ Rate of mass flow in] – [Rate of mass flow out] = [Rate of
accumulation of mass in tank]
p q (t) – p q0 (t) = d (p Ah)
dt

4. FindT and R
T= AR
T = A ( hs )
qs
T = Vs = 200 gal = 5 min
qs 40 gal/min
R = Vs = 200 gal = 26.733 ft3
A Qs (∏(3 ft2) (40 gal/min) (∏ (3 ft)2) (5.346 ft3/min)
= 0.1768 ft /ft3/min

5. FOR AN IMPULSE OF 55 GAL = 7.352 ft3
Q(s) = 7.352
H (s) = (7. 352) (0.1768) = 1.29983
5s + 1 5s + 1
H (s) = 0.25996672
s + 1/5
H (t) = 0.25996672 e-t/5

6. For question (a)
a) Will the tank overflow?
Answer: No
Find initial height.
H = 200 gal = 26. 733 ft3
A ∏ ( 3 ft2 )
H = 0.9455 ft
at t = 0 maximum change in height happens
h = 0.9455 + H (t)
h = 0.9455 + 0.25946672 e-t/5
h = 1.205 ft THETANKWILL NOT OVERFLOW

7. b ) Plot the height as f (t) starting at t = 0
Equation is f(t) = 0 .9445 + 0.25996672 e-t/5
Use simulink to show
Height Response
Assume Amplitude or % Period and Period
Amplitude = 10, 000
Period = 100
% Period= ?

8. 55 gal = 7. 352 ft3 = (% period) (100) (10,000)
100
% period = 0.0007352
Flow Rate Response
Amplitude = 10,000
Period = 100
% period = 0.0007352

9. c) For output flow
Q0(s) = 1 = Q0(s) = 7.352
Ts + 1 5s + 1
Q0(s) = 1.4704
s + 1/5
Q0(t) = 1.4704 e-t/5
Qo(actual) = 5.3466 ft3+ 1.4704e-t/5
Use simulink to verify.