A two-position mode controller is used to control the water level in a tank by opening or closing a valve which in the open position allows water at the rate of 0.4 m3 / s to enter the tank. The tank has a cross-sectional area of 12 m2 and water leaves it at the constant rate of 0.2 m3 / s. The valve opens when the water level reaches 4.0 m and closes at 4.4 m. What will be the times taken for (a) the valve opening to closing, (b) the valve closing to opening

Answer:

The right solution is:

(a) 24 sec

(b) 24 sec

Explanation:

The given values are:

Water inlet's flow rate,

= 0.4 m³/s

Water outlet's flow rate,

= 0.2 m³/s

(a)

The volume of water filled will be:

= [tex]Area(height)[/tex]

= [tex]12(4.4-4)[/tex]

= [tex]12\times 0.4[/tex]

= [tex]4.8 \ m^3[/tex]

Opening to closing time taken will be:

= [tex]\frac{4.8}{0.2}[/tex]

= [tex]24 \ sec[/tex]

(b)

Inlet's flow rate,

= 0

Outlet's flow rate will be:

= 0.2 m³/s

Closing to opening time taken will be:

= [tex]\frac{4.8}{0.2}[/tex]

= [tex]24 \ sec[/tex]

a. The time taken for the valve opening to closing is 24 seconds.

b. The time taken for the valve closing to opening is 24 seconds.

a.

First we need to determine the volume of the water

= Area * height

= 12*(4.4-4)

= 12*0.4

= 4.8

Now the time taken should be

= 4.8/0.2

= 24 seconds.

b. The time taken should be

= 4.8/0.2

= 2.4 seconds

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