Problem 1: A iquid storage tank has an area of 1 m2 and a normal m /hr. How does

Question

Problem 1: A iquid storage tank has an area of 1 m2 and a normal m /hr. How does depth of liquid storage scharge of 20 changesed to with time if the inlet flow rate to the tank is Changed to 25 m3/hr? The outflow Fout is given by Fout10 /hy as where h) is the liquid level in the tank. (a) What is the steady-state liquid level in the tank? (Mark 1) (b) Derive an unsteady-state model for the liquid level in the tank Clearly list all of your assumptions. (Mark 1.5) (c) Obtain the transfer function after linearization and show th effect of the change in input flow rate. Calculate the depth o liquid after 6 hours. (Mark 1.3) (d) Derive and solve an unsteady state model if The outflow Fout given by (Mark 1) Fout = 10 h(t) in out

Explanation / Answer

By Mass balance, In-Out+Generation-Consumption=Accumulation

a) Since, There is no reaction inside tank, therefore, Generation = Consumption = 0

Because, Steady state, Therefore, Accumulation = 0

Therefore, In = Out

In = 25 m3/hr

Out = 10*h^0.5

Therefore, 25 = 10*h^.5

Therefore, steady state height = 6.25 meter

b) For Unsteady state, accumulation = d(V)/dt

Volume of tank at time, t = Area*Height(h(t))

Therefore, 25-10*h^.5 = 1*dh/dt -(1)

At t=0, Vout = 10*h^.5 = 20m3/hr => h =4 m

Solving the Non linear ordinary linear differential eqn (1)

=>t = -h^.5/5-ln|2*h^.5-5|)/2+c

=>c=2/5

Therefore, t=-h^.5/5-ln|2*h^.5-5|)/2+2/5 -(2)

c) @t=2.66 hrs, h=6.25m, and In flow = Out flow and system will reach steady state. After that, at any time height will remain constant at 6.25 m only

=>t =6, h=6.25 m

d) F(out) = 10 *h

=>25-10h=1*dh/dt

Therefore,t =-Ln|2*h-5|)/10+0.1

in this case, at t = 0.26 hrs, In =Out and system will reach steady state.

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