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

Respuesta :

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.

Calculation of the time taken:

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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