A vessel has compartments A and B communicating by an orifice `250 cm^2`, its centre being 1m above bottom of vessel. The cross section of A and B are `10 m^2` and `20m^2`. At certain time, water stands 4m in A and 2m in B. How soon reafter water will attain common level.

SOLUTION:

Here,

Area of orifice,`a=150 **10^(-4) m^2`

Area,`A_1=10 m^2`

Area,`A_2=20 m^2`

Let, 

h be the difference between water level on either side of the profile.

y be the depth of water in A at any instant.

dy be the change in depth of A during the small time interval dT.

Now, decrease in volume of water in A is,

`=A_1*dy=-10dy`

Similarly increase in volume of water in B is,

`=A_2*dy_2=20dy_2....Show More

SOLUTION:

Here,

Area of orifice,`a=150 **10^(-4) m^2`

Area,`A_1=10 m^2`

Area,`A_2=20 m^2`

Let, 

h be the difference between water level on either side of the profile.

y be the depth of water in A at any instant.

dy be the change in depth of A during the small time interval dT.

Now, decrease in volume of water in A is,

`=A_1*dy=-10dy`

Similarly increase in volume of water in B is,

`=A_2*dy_2=20dy_2`

Since the decrease in volume of A corresponds to the increase in volume of B,

`dy_2=10/20*dy=1/2*dy`

Now the total change in level of water in small time dt is,

`dh=dy-(-dy_2)=3/2dy`

`dy=2/3dh`

Discharge through orifice in time dT is,

`=C_d*a*root ()(2gh) dT`.... (i)

Thus the decrease in volume of water in time dT is,

`=A_1*dy`

`=20**2/3dh`... (ii)

Thus from (i) and (ii),

`-10**2/3**dh=C_d*a*root ()(2gh) dT`

or,`\int_{0}^{T}dT=-20/(3**C_d**a**root ()(2g))*\int_{h}^{0} h^(-1/2) dh`

or,`T=20/(3**0.62**150**10^(-4)**root ()(2g))*2root ()(h)` ,where h=2

`T=457.7 Sec`