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

Find T and R

T= ART = A ( hs )

qsT = 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

FOR AN IMPULSE OF 55 GAL = 7.352 ft3

Q(s) = 7.352

H (s) = (7. 352) (0.1768) = 1.299835s + 1 5s + 1

H (s) = 0.25996672s + 1/5

H (t) = 0.25996672 e-t/5

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 THE TANK WILL NOT OVERFLOW

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, 000Period = 100% Period= ?

55 gal = 7. 352 ft3 = (% period) (100) (10,000)100

% period = 0.0007352

Flow Rate Response

Amplitude = 10,000Period = 100% period = 0.0007352

c) For output flow

Q0(s) = 1 = Q0(s) = 7.352Ts + 1 5s + 1

Q0(s) = 1.4704s + 1/5

Q0(t) = 1.4704 e-t/5

Qo(actual) = 5.3466 ft3+ 1.4704e-t/5

Use simulink to verify.