What is the Efflux Velocity from a Tank with Water and Oil Layers?

[Amith2006]

Sir,
A tank is filled with water (density 1000 kg/m^3) and oil(density 900 kg/m^3). The height of water layer 1 metre and that of oil layer is 4 metre. If g=9.8 m/s^2, then what is the velocity of efflux from an opening at the bottom of the tank?
I solved it the following way.
Let d1 be the density of water and d2 be the density of oil. Let h1 and h2 be the height of water and oil layer respectively.
Total pressure head = Pressure head due to 1 metre water + Pressure head due to 4 metre oil
=[ h1d1g]/[d1g] + [h2d2g]/[d1g]
= h1 + [h2d2]/d1
= 1 + [4x900]/1000
= 4.6 metres of water column
Pressure energy is converted into kinetic energy.
P = ½(d1)(v^2)
[4.6x1000xg]= ½(1000)(v^2)
v^2 = 2x4.6x9.8
v = 9.5 m/s^2
Is it right? Here the symbol ^ denotes power.

 

[Andrew Mason]

Sir,
A tank is filled with water (density 1000 kg/m^3) and oil(density 900 kg/m^3). The height of water layer 1 metre and that of oil layer is 4 metre. If g=9.8 m/s^2, then what is the velocity of efflux from an opening at the bottom of the tank?
I solved it the following way.
Let d1 be the density of water and d2 be the density of oil. Let h1 and h2 be the height of water and oil layer respectively.
Total pressure head = Pressure head due to 1 metre water + Pressure head due to 4 metre oil
=[ h1d1g]/[d1g] + [h2d2g]/[d1g]
= h1 + [h2d2]/d1
= 1 + [4x900]/1000
= 4.6 metres of water column
Pressure energy is converted into kinetic energy.
P = ½(d1)(v^2)
[4.6x1000xg]= ½(1000)(v^2)
v^2 = 2x4.6x9.8
v = 9.5 m/s^2
Is it right? Here the symbol ^ denotes power.

Your units are not right and you are overlooking atmospheric pressure. The net force acting on the water is difference between tank pressure and atmospheric pressure x area of the opening.

The potential energy density of the water at the opening is $P_{tank}$. The potential energy density of the water outside the opening is given by the ambient (atmospheric) pressure $P_a$. By conservation of energy (ie. Bernouilli's equation) pressure difference is converted into a Kinetic energy density of the water ($KE/V = KE\rho_w/m = \frac{1}{2}\rho_w v^2$

So:
$$P_{tank} - P_a = \frac{1}{2}\rho_w v^2$$

AM

 

[Amith2006]

Why is no one responding to my question?