Question

Flow in a tank is in steady state with inlet and output flow rate of 20 m3/min. The area of the tank 2 m2 and the resistance in the outlet flow pipe is 3 min/m2. If at t=0, the inlet flow rate started increasing at a constant rate of 2 m3/min, what will be the output flow rate after 4 min? please solve

Answer 1

Answer:

Step-by-step explanation:

                        .

Answer 2

Answer:

Consider the tank as shown in fig . The volumetric flow rate is Fi and the outlet volumetric flow rate is F0. In the outlet stream there is a resistance to flow , such as a pipe, valve, or weir. Let outlet flow rate Fo is related linearly to the hydroststic pressure of the liquid level h, through the resistance R.

Fo=hR=driving force for flowresistance to flow

The tank has the capacity to store mass, the total mass balance gives,

ρFi−ρFo=d(ρAh)dt

⟹Fi−Fo=Adhdt,

⟹Fi−hR=Adhdt,_________ (1)

At, steady state Fis−hsR=Adhsdt, ____ (2)

Subtracting (2) from (1) , we get

(Fi−Fis)−(h−hs)R=Ad(h−hs)dt

⟹ F′i−h′R=Adh′dt [ F′i=Fi−Fis ; h′=h−hs]

⟹F′i¯(s)−h′¯(s)R=Ash′¯(s)

⟹RF′i¯(s)=h′(s)[RAs+1]

⟹h′¯(s)F′i¯(s)=KPτPs+1 [ KP=R ; τ=RA ]

∙ The cross- sectional area of the tank , A , is a measure of its capacitance to store mass, thus larger the value of A, the larger the storage capacity of the tank.

∙ Since τP=AR , we can say that for the tank we have

( time constant ) = ( storage capacitance )x ( resistance to flow ).

Pure Capacitive system:

Consider the tank discussed above, with the following difference:

The effluent flow rate Fo is determined by a constant-displacement pump

In such case, total mass balance around the tank yields

Adhdt=Fi−Fo ; _______ (3)

At steady state, Adhsdt=Fis−Fo ______ (4)

Subtracting (4) from (3) , we get Adh′dt=F′i;

Which yield following transfer function G(s)=h′¯(s)F′i¯(s)=1sA

Q:1

A tank having a cross-sectional area is 2 ft2 is operating at steady state with an inlet flow rate of 2.0cfm. The flow head characteristics is shown in fig

(a) Find the transfer function H(s)/Q(s).

(B) If flow to the tank increases from 2.0 to 2.2 cfm according to a step change, calculate the level h two minutes after the change occurs.

Solution :

(a) Since, the above system resembles first-order system with a capacity for mass storage, the transfer function H(s)Q(s)=RRAs+1

From the above graph between qo vs h , its slope = 1R

1R=2.4−1.01−0.3=1.40.7=2; ⟹R=0.5

Thus H(s)Q(s)=RRAs+10.50.5×2s+1=0.5s+1 [ RA = 1]

(b) Step change of 0.2 cfm is applied , then Q(s)=0.2s

H(s)=0.5s+10.2s=0.1s(s+1)

⟹H(t)=0.1(1−e−t)

Thus H after 2 minutes =h(t) = 1+ 0.1(1−e−2)=1.086ft [ hs= 1 ft]